TITLE: Calderon-Zygmund decomposition on manifolds? QUESTION [8 upvotes]: The classical Calderon-Zygmund decomposition says that if $f\geq 0$ is $L^1$ on a cubes $B$, with average value $\alpha$, then there is a sequence of disjoint cubes $B_j$, such that the average of $f$ on each $B_j$ is in between $\alpha$ and $2^n \alpha$, and $f\leq \alpha$ a.e. away from $\bigcup_j B_j$. I am wondering if there is a similar result on (compact) manifolds, for example with a volume doubling condition. If one uses a Vitali type argument, it seems there is no control on the radius (they could be too small), so no upper bound in the average value of $f$ on (small) balls obtained by Vitali - in contrast to the $2^n\alpha$ upper bound of Calderon-Zygmund. REPLY [9 votes]: Calderon-Zygmund theory generalises without much difficulty to doubling metric measure spaces (or more generally to "spaces of homogeneous type"). See for instance Chapter 1 of Stein, Elias M., Harmonic analysis: Real-variable methods, orthogonality, and oscillatory integrals. With the assistance of Timothy S. Murphy, Princeton Mathematical Series. 43. Princeton, NJ: Princeton University Press. xiii, 695 pp. (1993). ZBL0821.42001. The generalisation of the Calderon-Zygmund decomposition in this setting is given in Section 1.4. In more recent years, most of Calderon-Zygmund theory has also been extended to the non-doubling case, though one has to make some natural modifications to the statements in order to avoid trivial counterexamples. One reference is Chapter 2 of Tolsa, Xavier, Analytic capacity, the Cauchy transform, and non-homogeneous Calderón-Zygmund theory, Progress in Mathematics 307. Cham: Birkhäuser/Springer (ISBN 978-3-319-00595-9/hbk; 978-3-319-00596-6/ebook). xiii, 396 p. (2014). ZBL1290.42002. The Calderon-Zygmund decomposition is given as Lemma 2.14 of that book, though the statement may look slightly different from the classical one. One common trick in this subject is to replace balls by roughly equivalent "dyadic cubes" that have good geometric properties, such as nesting. See for instance Hytönen, Tuomas; Kairema, Anna, What is a cube?, Ann. Acad. Sci. Fenn., Math. 38, No. 2, 405-412 (2013). ZBL1288.30066. for a brief introduction to this topic.<|endoftext|> TITLE: Does "agreement on cardinalities" imply second-order elementary substructurehood? QUESTION [7 upvotes]: Say that a logic $\mathcal{L}$ satisfies the weak test property iff for all $\mathfrak{A}\subseteq\mathfrak{B}$ we have $(1)\implies(2)$ below: For each $\mathcal{L}$-formula $\varphi$ with parameters from $\mathfrak{A}$ we have $$\vert\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}\vert=\vert\varphi^\mathfrak{B}\vert.$$ (In this case write "$\mathfrak{A}\trianglelefteq_{\mathcal{L}}^{\mathsf{Card}}\mathfrak{B}$.") $\mathfrak{A}\preccurlyeq_\mathcal{L}\mathfrak{B}$. This is a massive weakening of the Tarski-Vaught test, which says that we get elementarity merely from $\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}$ being nonempty whenever $\varphi^\mathfrak{B}$ is nonempty. By contrast, $\mathfrak{A}\trianglelefteq_\mathcal{L}^\mathsf{Card}\mathfrak{B}$ is a highly restrictive hypothesis (and so the corresponding implication is weaker): as long as $\mathcal{L}$ is "reaosnable" it immediately implies, for example, that $\vert\mathfrak{A}\vert=\vert\mathfrak{B}\vert$ via the formula $x=x$. My question is: Does second-order logic have the weak test property? Producing interesting instances of $\trianglelefteq_{\mathsf{SOL}}^\mathsf{Card}$, even before trying to also prevent $\preccurlyeq_{\mathsf{SOL}}$, seems very difficult; on the other hand, I see absolutely no reason why $\mathsf{SOL}$ should have the weak test property. In fact there is a whole spectrum of variants of the test property which seem interesting to me. For each class $X$ of cardinals and pair of structures $\mathfrak{A}\subseteq\mathfrak{B}$, say $\mathfrak{A}\trianglelefteq_\mathcal{L}^X\mathfrak{B}$ iff for each $\mathcal{L}$-formula $\varphi$ with parameters from $\mathfrak{A}$ and each $\kappa\in X$ we have $\vert\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}\vert<\kappa\iff \vert\varphi^\mathfrak{B}\vert<\kappa$; then the weak test property at $X$ is the implication $\trianglelefteq_\mathcal{L}^X\implies \preccurlyeq_\mathcal{L}$. The Tarski-Vaught test itself corresponds to $X=\{1\}$, while the weak test property corresponds to $X=\mathsf{Card}$. If the main question above happens to have a positive answer - which would surprise me quite a bit! - I would be further interested in which $X$s are "sufficient" to ensure $\preccurlyeq_\mathcal{L}$. REPLY [2 votes]: No, second order logic does not have the weak test property: let $\mathfrak{B}=(\mathbb{R},{<})$ (that is, the real numbers with the only predicate being the usual "less than" order) and let $\mathfrak{A}=(\mathbb{R}\backslash\{0\},{<})$. Then $\mathfrak{B}\models$"I am a complete linear order" (completeness as in "for every $<$-downward closed set $X$ such that $X\neq\mathbb{R}$, there is a least upper bound for $X$, and likewise symmetrically"), whereas $\mathfrak{A}$ does not satisfy this, so $\mathfrak{A}\not\equiv_{\mathrm{SOL}}\mathfrak{B}$, and hence $\mathfrak{A}\not\preccurlyeq_{\mathrm{SOL}}\mathfrak{B}$. But property 1 does hold for $(\mathfrak{A},\mathfrak{B})$. For for simplicity let's first consider the case that the arity of $\varphi$ is 1. Let $x_10$ is small enough then for each $x\in(-\varepsilon,\varepsilon)$ we can produce an automorphism $\pi:\mathfrak{B}\to\mathfrak{B}$ which fixes $x_1,\ldots,x_n$ but with $\pi(0)=x$.) For arity $k>0$ it is similar, with $k$-dimensional rectangles.<|endoftext|> TITLE: Graphs from the point of view of Riemann surfaces QUESTION [6 upvotes]: I was listening to the lecture "Graphs from the point of view of Riemann surfaces" by Prof. Alexander Mednykh. I am looking for references for the basics of this topic. Any kind of suggestions is highly appreciated. Thank you in advance. REPLY [10 votes]: It seems that the question is about the following lecture notes: http://math.nsc.ru/conference/g2/g2r2/files/pdf/Lecture-8.pdf At the start of these notes, five papers are mentioned. Based on the topic, the authors and the listed years, I believe that these refer to the following papers: Roland Bacher, Pierre de la Harpe, and Tatiana Nagnibeda, The lattice of integral flows and the lattice of integral cuts on a finite graph. Bulletin de la Société Mathématique de France, 125(2):167–198, 1997. https://doi.org/10.24033/bsmf.2303 Hajime Urakawa, A discrete analogue of the harmonic morphism and Green kernel comparison theorems. Glasgow Mathematical Journal, 42(3):319–334, 2000. https://doi.org/10.1017/S0017089500030019 Matthew Baker and Serguei Norine, Harmonic morphisms and hyperelliptic graphs. International Mathematics Research Notices, 2009(15):2914–2955, 2009. https://doi.org/10.1093/imrn/rnp037 Lucia Caporaso, Algebraic and combinatorial Brill–Noether theory. In: Valery Alexeev, Angela Gibney, Elham Izadi, János Kollár, and Eduard Looijenga (editors), Compact moduli spaces and vector bundles, Contemporary Mathematics, pages 69–85. American Mathematical Society, 2012. https://bookstore.ams.org/conm-564 Scott Corry, Maximal harmonic group actions on finite graphs. Discrete Mathematics, 338(5):784–792, 2015. https://doi.org/10.1016/j.disc.2014.12.016 (The publication years of the 4th and 5th paper do not quite match the publication years listed in the lecture notes. However, the lecture notes do match the year that the corresponding preprints were first announced on arXiv.) In addition to the papers 1–3 listed above, I consider the following papers to be essential reading on this topic: Matthew Baker and Serguei Norine, Riemann–Roch and Abel–Jacobi theory on a finite graph, Advances in Mathematics, 215(2):766–788, 2007. https://doi.org/10.1016/j.aim.2007.04.012 Matthew Baker, Specialization of linear systems from curves to graphs, Algebra & Number Theory, 2(6):613–653, 2008. https://doi.org/10.2140/ant.2008.2.613 Filip Cools, Jan Draisma, Sam Payne, and Elina Robeva, A tropical proof of the Brill–Noether theorem, Advances in Mathematics, 230(2):759–776, 2012. https://doi.org/10.1016/j.aim.2012.02.019 You may also want to take a look at the 2016 survey by Baker and Jensen: Matthew Baker and David Jensen, Degeneration of linear series from the tropical point of view and applications. In: Matthew Baker, Sam Payne (editors), Nonarchimedean and Tropical Geometry, Simons Symposia, pages 365–433, Springer, 2016. https://doi.org/10.1007/978-3-319-30945-3_11 Furthermore, I am aware of the existence of two textbooks on the more combinatorial side of things, but I'm not sure that these will contain what you are looking for: Scott Corry and David Perkinson, Divisors and Sandpiles: An Introduction to Chip-Firing, American Mathematical Society, 2018. Caroline J. Klivans, The Mathematics of Chip-Firing, Mathematical Association of America, 2018. Finally, depending on your level, you might also enjoy the following "light reading": Matt Baker, Riemann–Roch for graphs and applications, blog post, 2013. https://mattbaker.blog/2013/10/18/riemann-roch-for-graphs-and-applications/ Kevin Hartnett, Tinkertoy models produce new geometric insights, Quanta Magazine, 2018. https://www.quantamagazine.org/tinkertoy-models-produce-new-geometric-insights-20180905/ Jan Draisma and Alejandro Vargas, On the gonality of metric graphs, Notices of the American Mathematical Society, 68(5):687–695, 2021. https://doi.org/10.1090/noti2277 David Jensen, Chip firing and algebraic curves, Notices of the American Mathematical Society, 68(11):1875–1881, 2021. https://doi.org/10.1090/noti2378 To get an overview of recent developments in this field, I would suggest to start with the excellent expository article by Jensen (number 15 on the list), followed by the survey by Baker and Jensen (number 9 on the list). After that, either follow the references in those papers that you find interesting, or come back to list and take a look at some of the classics (1–3 and 6–8).<|endoftext|> TITLE: Implicit function theorem with continuous dependence on parameter QUESTION [10 upvotes]: Let $X,Y$ be Hilbert spaces and $P$ a topological space$^1$ and $p_0\in P$. Let $f:X\times P\to Y$ be a continuous map such that for any parameter $p\in P$, $f_p:= f|_{X\times \{p\}}:X\to Y$ is smooth. Suppose also that $f_p\to f_{p_0}$ in $C^3_{loc}(X,Y)$ for $p\to p_0$ (i.e. once we fix an arbitrary compact $K\subset X$, we have uniform convergence in $C^3$-norm over it) Suppose that $Df_{p_0}(x_0):X\to Y$ is an isomorphism, we would like to prove that under these assumptions there exists a continuous implicit function, i.e. Show that exists $U_{x_0}\times U_{p_0}\subset X\times P$ product neighbourhood of $(x_0,p_0)$ and continuous function $x:U_{p_0}\to U_{x_0}$ such that the zero locus of $f|_{U_{x_0}\times U_{p_0}}$ is the graph of $x$, $\{(x(p),p)\ | \ p\in U_{p_0}\}$. If false please provide an example Notice that the existence of the neighbourhood is already a difficult part. Indeed $C^1_{loc}$ convergence implies that there is $U_{p_0}$ such that for any $p\in U_{p_0}$, $Df_{p}(x_0)$ is isomorphism so we can apply the inverse function theorem but the size of the "inversion"-neighbourhood can shrink to zero as $p\to p_0$. In order to prevent this, I thought to use the $C^3_{loc}$ convergence, because we can estimate the radius of the neighbourhood using the second derivative in a neighbourhood of $x_0$, however the convergence is only on compact sets therefore I do not know how to use it to produce a bound for $D^2f_p(x)$ for $\|x-x_0\| TITLE: Tohoku and cohomology of toposes QUESTION [8 upvotes]: In McLarty's The Rising Sea: Grothendieck on simplicity and generality I found the following quote: The same, Grothendieck knew, would work for cases yet unimagined. He notes that Tohoku [Grothendieck 1957] already gave foundations for the cohomology of any topos [Grothendieck 1985–1987, p. P41n.]. That context was hardly foreseen as he wrote Tohoku in 1955. This is one more proof that it was the right idea of cohomology. In which sense gave Tohoku a foundation for the cohomology of any topos? In particular, which theorem in Tohoku proves or constructs the cohomology of toposes? [Grothendieck 1985–1987, p. P41n.] is Récoltes et Semailles. (I can spot the passage in which Grothendieck refers to Tohoku, but this doesn't answer my question.) I could swear I heard the claim before that Tohoku is the only place in the literature which shows that toposes have cohomology (of course without mentioning the word "topos"), although I can't recall at the moment where I heard that. REPLY [9 votes]: As requested: By Theorem 1.10.1 in Tohoku, an Grothendieck abelian category has enough injectives. Sheaves of abelian groups on a Grothendieck topos form a Grothendieck abelian category. By Theorem 2.2.2 in Tohoku, one may then take derived functors of global sections. As mentioned in the comments, though, sheaf cohomology on toposes is developed quite a bit elsewhere (e.g. in SGA 4).<|endoftext|> TITLE: Generating function of product of binomial coefficients QUESTION [5 upvotes]: Let $m, n\in \mathbb{N}$ and $|x| < 1$. I look for hints to derive an analytic formula for $$f_{m,n}(x) = \sum_{k \in \mathbb{N}} {n + k \choose k} {m + k \choose k} x^{k}. $$ REPLY [4 votes]: Hjalmar Rosengren gave a nice formula for $f_{m,n}(x)$ based on the theory of hypergeometric series. Here I provide a direct elementary proof of the same based on generating series. Let us introduce the differential operator $$D_r:=\frac{1}{r!}\frac{\partial^r}{\partial x^r}.$$ Then $$\sum_{k=0}^\infty\binom{n+k}{k}x^k=D^n\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^{n+1}},$$ hence by the general Leibniz rule \begin{align*}\sum_{k=0}^\infty\binom{m+k}{k}\binom{n+k}{k}x^k &=D^m\left(\frac{x^m}{(1-x)^{n+1}}\right)\\ &=\sum_{k=0}^m D^{m-k}(x^m)D^k\left(\frac{1}{(1-x)^{n+1}}\right)\\ &=\sum_{k=0}^m\binom{m}{k}\binom{n+k}{k}\frac{x^k}{(1-x)^{k+n+1}}\\ &=\frac{1}{(1-x)^{m+n+1}}\sum_{k=0}^m\binom{m}{k}\binom{n+k}{k}x^k(1-x)^{m-k}\\ &=\frac{1}{(1-x)^{m+n+1}}\sum_{k=0}^m D^n(x^{n+k})\binom{m}{k}(1-x)^{m-k}. \end{align*} Here we can interpret $D^n(x^{n+k})$ as the coefficient of $t^n$ in the polynomial $(x+t)^{n+k}\in\mathbb{Z}[x][t]$, hence by the binomial theorem, the $k$-sum on the right-hand side equals $$[t^n]\sum_{k=0}^m (x+t)^{n+k}\binom{m}{k}(1-x)^{m-k}=[t^n](x+t)^n(1+t)^m.$$ Expanding $(x+t)^n$ and $(1+t)^m$ by the binomial theorem, and then collecting the coefficient of $t^n$ in the product of the resulting two sums, we get that $$\sum_{k=0}^\infty\binom{m+k}{k}\binom{n+k}{k}x^k=\frac{1}{(1-x)^{m+n+1}}\sum_{k=0}^{\min(m,n)}\binom{m}{k}\binom{n}{k}x^k.$$<|endoftext|> TITLE: Functions of $\mathbb{R}^d$ preserving convexity of sets QUESTION [17 upvotes]: Consider a function $f : \mathbb{R}^d \to \mathbb{R}^d$, with $d\geq 2$, such that: $f$ is injective, For any convex set $A$ of $\mathbb{R}^d$, $f(A)$ is also convex. What can we say about $f$ ? In particular, is $f$ necessarily affine ? I tend to think yes, but I can't prove it. REPLY [22 votes]: See Meyer, Walter; Kay, David C., A convexity structure admits but one real linearization of dimension greater than one, J. Lond. Math. Soc., II. Ser. 7, 124-130 (1973). ZBL0271.52002. Theorem 4. If $V$ and $W$ are real vector spaces with $\dim V>1$, and $f:V\to W$ is a one-to-one mapping which preserves convexity, then $f$ is either linear (if $f(0)=0$) or is the translate of a linear map.<|endoftext|> TITLE: First cohomology of tangent sheaf of rational curve QUESTION [5 upvotes]: Let $C$ be a reduced, connected, projective and purely one-dimensional scheme of finite type over a field $k$. Suppose that $C$ is rational, i.e. that the normalisation of $C$ is a disjoint union of copies of $\mathbb{P}^1_k$. Let $T_C = \mathcal{H}om(\Omega^1_C,\mathcal{O}_C)$ be the tangent sheaf of $C$. Question: Is it true that $H^1(C,T_C)$ vanishes? You may additionally assume that $C$ has only planar singularities, but I'm not sure this is needed. The question is true if $C$ is smooth. For context, a positive answer would imply that $C$ has no locally trivial deformations. REPLY [10 votes]: Let $C$ be the union of 5 lines in general position in $\mathbb{P}^2$ (hence with 10 pairwise intersection points $P_{ij}$, $1 \le i < j \le 5$) and let $F$ be the equation of $C$. We have the standard exact sequence $$ 0 \to \mathcal{O}_C(-5) \stackrel{dF}\to \Omega_{\mathbb{P}^2}\vert_C \to \Omega_C \to 0. $$ Taking its dual we obtain an exact sequence $$ 0 \to T_C \to T_{\mathbb{P}^2}\vert_C \stackrel{dF}\to \mathcal{O}_C(5) \to \bigoplus_{1 \le i < j \le 5} \mathcal{O}_{P_{ij}} \to 0. $$ Now an easy computation of Euler characteristics gives $$ \chi(T_{\mathbb{P}^2}\vert_C) = 5, \quad \chi(\mathcal{O}_C(5)) = 20, \quad \chi(\mathcal{O}_{P_{ij}}) = 1, $$ hence $\chi(T_C) = 5 - 20 + 10 = -5$, and since $C$ is 1-dimensional, it follows that $H^1(C,T_C)$ is non-zero.<|endoftext|> TITLE: An integral indexed by two partitions that mysteriously vanishes QUESTION [9 upvotes]: Let $\alpha,\beta\vdash n$ and define the polynomial $$f_{\alpha,\beta}(x)=\sum_{\lambda \vdash n}\chi_\lambda(n)\chi_\lambda(\alpha)\chi_\lambda(\beta)x^{\ell(\lambda)-1},$$ where $\chi_\lambda$ are irreducible characters of the permutation group and $\ell$ is the length of a partition. I am interested in the integral $$I_{\alpha,\beta}=\int_0^\infty \frac{f_{\alpha,\beta}(x)}{(1+x)^{n+1}}dx,$$ and I have observed with surprise that $I_{\alpha,\beta}=0$ if $\ell(\alpha)+\ell(\beta)>n+1$. How to prove this? I know that $\chi_\lambda(n)=0$ unless $\lambda$ is a hook, i.e. of the form $(n-k,1^k)$, and that there is a formula for $\chi_\lambda$ when $\lambda$ is a hook. But I was not able to use that effectively (notice that $f_{\alpha,\beta}$ is never zero, only the integral is zero). Another nice observation is that $f_{(1^n),(1^n)}(1)=0$ if $n$ is even and equals $(-1)^k{2k\choose k}$ if $n=2k-1$. Why? Oh, I see. The integral is not relevant because it can be done to give $$I_{\alpha\beta}=\frac{1}{n!}\sum_{\lambda \vdash n}\chi_\lambda(n)\chi_\lambda(\alpha)\chi_\lambda(\beta)(n-\ell(\lambda)-1)!(\ell(\lambda)-1)!$$or $$I_{\alpha\beta}=\frac{1}{n}\sum_{\lambda \vdash n}\frac{1}{\chi_\lambda(1^n)}\chi_\lambda(n)\chi_\lambda(\alpha)\chi_\lambda(\beta),$$ and this is proportional to the number of factorizations of the full cycle with one factor in the conjugacy class $\alpha$ and the other in class $\beta$. REPLY [4 votes]: Your nice observation is fairly easily explained: as you say $\chi_\lambda(n) = 0$ unless $\lambda$ is a hook partition of the form $(n-r,1^r)$ for some $r$. In this case, by the Murnaghan–Nakayama rule, we have $\chi_{(n-r,1^r)}(n) = (-1)^r$. Since there are $\binom{n-1}{r}$ standard Young tableaux of shape $(n-r,1^r)$, we have $\chi_{(n-r,1^r)}(1^n) = \binom{n-1}{r}$. Therefore $$f_{(1^n),(1^n)}(1) = \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r}^2. $$ The right-hand side counts pairs $(X,Y)$ of subsets of $\{1,\ldots, n-1\}$ such that $|X|+|Y| = n-1$, weighted by $(-1)^{|X|}$. Let $X \circ Y = (X \backslash Y) \cup (Y \backslash X)$ be the symmetric difference of $X$ and $Y$. Moving the maximum element in $X \circ Y$ to the opposite set (so from $X$ to $Y$ if it is in $X$, and from $Y$ to $X$ if it is in $Y$) defines a sign-reversing involution on those pairs $(X,Y)$ with $X \not= Y$. The contribution from such pairs to the sum is therefore zero. We can have $X = Y$ only if $n-1 = 2k$ is even and $|X|=|Y| = k$; in this case the number of such pairs is $\binom{n-1}{k} = \binom{2k}{k}$, with sign $(-1)^k$. (I think you omitted the sign in your question.)<|endoftext|> TITLE: Can $\mathsf{Ord}$ be weakly compact from a second-order perspective? QUESTION [6 upvotes]: Working in $\mathsf{ZFC}$ + "There is a weakly compact cardinal" and letting $\kappa$ be the least weakly compact cardinal, say that a logic $\mathcal{L}$ is loraxian iff every $\mathcal{L}$-definable-over-$V_\kappa$ subtree of $2^{<\kappa}$ has an $\mathcal{L}$-definable-over-$V_\kappa$ branch. Enayat and Hamkins showed (among other things) in $\mathsf{ZFC}$ that first-order logic is not loraxian, but I don't see how to use their techniques to address other logics - the key point being that everything leading up to case $2$ of Theorem $2.6$ is highly $\mathsf{FOL}$-specific. As is my wont, I'm specifically curious about the situation with respect to second-order logic: Is second-order logic consistently loraxian? All I've been able to see is the following two basic observations: In contrast with case $2$, case $1$ of E/H's Theorem $2.6$ is fairly coarse: roughly speaking, by considering a tree whose nodes code choice functions for the $V_\alpha$s with $\alpha<\kappa$, we have that $\mathcal{L}$ is loraxian only if there is some weakly compact $\kappa$ such that $V_\kappa$ has an $\mathcal{L}$-definable-over-$V_\kappa$ well-ordering. Meanwhile, there is a silly red herring here. My initial guess was that $\mathsf{SOL}$ would obviously be loraxian since we can define the nodes which belong to some path. However, since $\kappa$ is weakly compact in reality this is actually no power whatsoever. What posed a problem for $\mathsf{FOL}$ isn't an inability to detect when a node should extend to a path, but rather (very roughly) the issue of piecing together a single path in a consistent way - and I don't see that this goes away for $\mathsf{SOL}$. More generally I'm interested in anything on loraxian logics, but $\mathsf{SOL}$ seems like a natural starting point. REPLY [4 votes]: If $\kappa$ is weakly compact and there is a wellorder of $V_{\kappa+1}$ definable over $V_{\kappa+1}$ without parameters, then second order logic is Loraxian for $V_\kappa$: the least branch through a definable tree is definable. In $L$ (or in fact any of the known canonical inner models), there is such a wellorder. You didn't ask, but I think it is consistent for second order logic to fail to be Loraxian. After a preparatory forcing, one can add a Cohen subset $G$ of $\kappa$ without destroying its weak compactness. The forcing can be factored as adding a Suslin tree $T$ and then adding $G$ as a branch through $T$. I am not an expert here, but I think one can probably force over $V[G]$ to make $T$ definable over $V_{\kappa+1}$, arranging that $G$ remains undefinable and $\kappa$ remains weakly compact. Probably one can arrange that any subset of $V_\kappa$ definable over $V_{\kappa+1}^{V[G,H]}$ is in fact be definable over $V_{\kappa+1}^{V[T]}$ from $T$. This is an (admittedly sketchy) local version of the arguments from Cheng-Friedman-Hamkins's "Large cardinals need not be large in HOD."<|endoftext|> TITLE: Minimal injective extension is rigid QUESTION [6 upvotes]: Let $V$ be an operator system. Definition 1: A pair $(W, \kappa)$ is called extension of $V$ if $W$ is an operator system and $\kappa: V \to W$ is a unital complete isometry. Definition 2: An extension $(W,\kappa)$ of $V$ is called injective extension if $W$ is an injective operator system. Definition 3: An extension $(W, \kappa)$ of $V$ is called rigid if for every unital completely positive map $\varphi: W \to W$ with $\varphi \kappa = \kappa$, we necessarily have $\varphi = \iota_W$. Definition 4: An injective extension $(W,\kappa)$ of $V$ is called minimal if $\kappa(V)\subseteq W_1 \subseteq W$ with $W_1$ injective implies that $W_1 = W.$ Question: Is a minimal injective extension necessarily rigid? Is there a quick way to see this? I have a rather long argument along the lines of stuff in Paulsen where we use minimal seminorms, but maybe there is an easier way. If it helps, I know that for injective extensions, rigidity = essentiality. REPLY [3 votes]: Hamana's proof (Theorem 3.5 in Injective Envelopes of Operator Systems, PubL RIMS, Kyoto Univ. 15 (1979), 773-785) is fairly direct. Consider the partial ordering on the space $\Xi = \{ \phi \in UCP(W, W) \mid \phi \kappa = \kappa \}$ given by $\phi \prec \psi$ when $\| \phi(x) \| \leq \| \psi(x) \|$ for all $x \in W$. First, note that every element in $\Xi$ dominates a minimal element in $\Xi$ since if $\{ \phi_i \}_i$ is a decreasing net in $\Xi$, then taking any concrete realization $W \subset \mathcal B(\mathcal H)$ we have $E \phi \prec \phi_i$ for all $i$, where $E: \mathcal B(\mathcal H) \to W$ is any ucp idempotent and $\phi$ is any limit point of $\{ \phi_i \}_i$ in $UCP(W, \mathcal B(\mathcal H))$, which is compact in the topology of pointwise ultraweak convergence. Second, note that any minimal element $\phi \in \Xi$ is an idempotent. Indeed, for any $N \geq 1$ and $x \in W$ we have $\| \frac{1}{N} \sum_{n = 1}^N \phi^n( x ) \| \leq \| \phi(x) \|$ and so by minimality it follows that this is equality. Considering $x = y - \phi(y)$ we have $\| \phi(y) - \phi^2(y) \| = \| \frac{1}{N} \sum_{n = 1}^N \phi^n(y - \phi(y) ) \| \leq \frac{2}{N} \| y \| \to 0$. Thus, if every idempotent in $\Xi$ is the identity, it follows that the identity is the only map in $\Xi$.<|endoftext|> TITLE: For every ring R, is there a block-diagonal canonical form for a square matrix over R? QUESTION [8 upvotes]: This question asks whether there exists an analogue of the Jordan decomposition for an arbitrary ring $R$. This analogue is not necessarily the Jordan-Chevalley decomposition, which is unnecessarily strong. This follows from this question, but you don't need to read it. Given a ring $R$, let $J(R)$ be the monoid whose elements are all square matrices over $R$, and where the monoid operation is $\oplus$ denoting direct sum of matrices. Let $A { \sim_\text{S}} B$ mean that there exists an invertible matrix $P$ such that $PAP^{-1} = B$. Is the monoid $J(R)/{ \sim_\text{S}}$ always a free abelian monoid? For perfect fields, the answer is yes by the Jordan-Chevalley decomposition. What about for every ring? Is there a counterexample? [edit] I've referenced this question in this paper: Towards a Singular Value Decomposition and spectral theory for all rings. REPLY [11 votes]: Your question is equivalent to whether the category $\mathcal{E}$ of pairs $(V,f)$ consisting of a finitely generated free (right) $R$-module and an endomorphism $f$ of $V$ is a Krull-Schmidt category, i.e., an additive category where every object decomposes as a direct sum of finitely many indecomposable objects and the decomposition is unique up to isomorphism and reordering. (The category $\cal E$ is also equivalent to the category of right $R[t]$-modules which are f.g. free $R$-modules, and it is rarely Krull-Schmidt, but I won't use this point of view.) Here is one possible counterexample, phrased using the category $\mathcal{E}$ of pairs $(V,f)$ above: Take $R$ to be a Dedekind domain admitting a non-free rank-$1$ projective module $L$ such that $L\oplus L\cong R\oplus R$ (equivalently, $L$ represents an element of order $2$ in the Picard group of $R$). Let $f_L : L^2\to L^2$ be defined by $f_L(x,y)=(0,x)$, and define $f_R:R^2\to R^2$ similarly. The isomorphism $L\oplus L\cong R\oplus R$ gives rise to an isomorphism $$(L^2,f_L)\oplus (L^2,f_L) \cong (R^2,f_R)\oplus (R^2,f_R)$$ in $\cal E$. One readily checks that ${\rm End}_{\cal E}(L^2,f_L)\cong R[\epsilon|\epsilon^2=0]$ and ${\rm End}_{\cal E}(R^2,f_R)\cong R[\epsilon|\epsilon^2=0]$, so the endomorphism rings of $(L^2,f_L)$ and $(R^2,f_R)$ contain no nontrivial idempotents. This means that these objects are indecomposable in $\cal E$. On the other hand, $(L^2,f_L)\ncong (R^2,f_R)$ because $\ker f_L\cong L\ncong R\cong \ker f_R$. Consequently, the monoid $(\cal E/\cong, \oplus)$ (which is isomorphic to $(J(R)/\sim, \oplus)$ in your question) is not a free abelian monoid (because $2x=2y$ implies $x=y$ in a free abelian monoid). One the other hand, a sufficient (but not necessary) condition for the category $\cal E$ to be Krull-Schmidt is that the endomorphism ring of every object $(V,f)$, i.e., the centralizer of $f$ in ${\rm End}_R(V)\cong {\rm M}_n(R)$, is a semiperfect ring. For example, if $R$ is commutative and aritinian as in Benjamin Steinberg's comment, then ${\rm End}_{\cal E}(V,f)$ will be an $R$-subalgebra of ${\rm End}_R(V)$, hence artinian, and in particular semiperfect. The semiperfectness ${\rm End}_{\cal E}(V,f)$ for all f.g. free $V$ is actually true even if $R$ is non-commutative one-sided artinian, and even if $R$ is just semiprimary. This appears implicitly in a paper of mine (page 20 & Thm. 8.3(iii) & Remark 2.9). When $R$ is commutative noetherian and local, one can use a theorem of Azumaya (Theorem 22) and a little work to show that this property is equivalent to $R$ being henselian.<|endoftext|> TITLE: Convergence speed of the tail of distribution using Tauberian remainder theorem QUESTION [6 upvotes]: This question may be related to this one. Now I try to make some statistical estimator using Laplace transform, but I face the following serious problem. Let $f$ be some one-sided probability distribution defined on $[0,\infty)$, and $\hat{f}$ be its Laplace transform. Now assume that we only have information of $\hat{f}(s)$ near $s=0$ (for example, $\hat{f}(s) = \hat{g}(s) / (2-\hat{g}(s))$ with known probability distribution $g(x)$), and using this, we want to find the convergence speed of the tail: \begin{align*} \int_x^{\infty} f(x)\, dx = O(?). \end{align*} According to Tauberian remainder theory in J. Korevaar's book (Examples 2.3., page 348), he said If $|\hat{f}(s) - \hat{f}(0)| \le Cs^{\alpha}$ with some $\alpha > 0$, then the tail rate is $O(1/\log x)$; If $|\hat{f}(s) - \hat{f}(0)| \le Ce^{-\alpha/s}$ with some $\alpha > 0$, then the tail rate is $O(1/\sqrt{x})$. In fact, this is a disaster for statisticians because any higher-order Taylor approximation of $\hat{f}(s)$ near $s=0$ cannot guess whether the tail is light or super super heavy ($1/\log x$). So the question is the following. Question. What other condition on $\hat{f}$ near $0$ is required to guarantee the tail has at least a power-tail $O(x^{-\beta})$? If the power tail cannot be guaranteed by any of information about $\hat{f}$ near $s=0$, then what condition is required for $g$? Any help would be appreciated. (suggesting books, papers, or anything!) Thanks for the reading, REPLY [3 votes]: Let $f$ be a pdf on $[0,\infty)$. Let $\hat f$ be the Laplace transform of $f$, so that \begin{equation} \hat f(s)=\int_0^\infty f(x)e^{-sx}\,dx \end{equation} for real $s\ge0$. Suppose that \begin{equation} |\hat f(s)-\hat f(0)|\le Cs^a \end{equation} for some real $a,C>0$ and all real $s\ge0$. Then for all real $s>0$ \begin{equation} Cs^a\ge|\hat f(s)-\hat f(0)|=\int_0^\infty f(u)(1-e^{-su})\,du \ge(1-e^{-1})\int_{1/s}^\infty f(u)\,du, \end{equation} whence for all real $x>0$ \begin{equation} \int_x^\infty f(u)\,du\le C_1/x^a, \end{equation} where $C_1:=C/(1-e^{-1})\in(0,\infty)$, as desired. (Your difficulty was due to the fact that you tried to use general Tauberian theorems, valid without the nonnegativity condition. On the other hand, all pdf's are of course nonnegative. Once this nonnegativity condition is taken into account, everything becomes much simpler and better.)<|endoftext|> TITLE: Algebraic topology and homotopy theory with simplicial sets instead of topological spaces QUESTION [10 upvotes]: To quote Kerodon: In fact, it is possible to develop the theory of algebraic topology in entirely combinatorial terms, using simplicial sets as surrogates for topological spaces. A similar quote can be found in the mathscinet review for Kan's On c. s. s. complexes: In recent years it has become evident that for most purposes in homotopy theory it is more convenient to use semi-simplicial complexes instead of topological spaces. For instance, I know that to special simplicial sets called Kan complexes one can assign higher homotopy groups and prove and analogue of Whitehead's theorem. This certainly demonstrates that one can do some homotopy theory with simplicial sets / Kan complexes. If I open an introductory book on algebraic topology or homotopy theory (such as Hatcher's), do all the main theorems admit analogues in the world of simplicial sets or Kan complexes (replacing topological spaces)? I'd be totally happy if you could give me, say, four theorems in algebraic topology / homotopy theory that can be phrased for simplicial sets, together with the original reference. I'd also be interested in whether these theorems are more algebraic topology or more homotopy theory (I don't really know the difference). REPLY [7 votes]: It certainly used to be the case that if you went to an algebraic topology research talk, when the speaker said, "Let $X$ be a space," then there was about a 50% chance that they really meant "Let $X$ be a simplicial set": simplicial sets are that intertwined with homotopy theory. More precisely, Quillen's model category framework allows you to start with a category $C$, say simplicial sets or topological spaces, add some extra structure, and then define what is called the "associated homotopy category," $\text{Ho}\,C$. Quillen showed further that the homotopy category for simplicial sets is equivalent to the homotopy category for topological spaces, and therefore if you want to study homotopy theory, you can use either topological spaces (with CW complexes as a distinguished subcategory) or simplicial sets (with Kan complexes as a distinguished subcategory) interchangeably. Even better, you can switch back and forth depending on which setting is more convenient for proving the result you happen to be interested in at the time.<|endoftext|> TITLE: Embedding torsors of elliptic curves into projective space QUESTION [5 upvotes]: Suppose I have a genus 1 curve $C$ over a field $k$. If $C$ has a point, then we can embed it into the projective plane by a Weierstrass equation. Now let us suppose that $C$ does not have a point (so that it is a non trivial torsor for it's Picard group). Can I still embed $C$ into the projective plane? I guess not but there is apparently a theorem of Lang-Tate that we can always find an effective divisor of some degree over $k$ (what is a reference?) so we can embed it into some high dimensional projective space. Can we always embed C into a Severi Brauer variety of dimension $2$? REPLY [11 votes]: Suppose that $C \subset X$ is a smooth projective curve of genus $1$ embedded in a Brauer-Severi surface over a field $k$. We have $C^2 = 9$ since this holds after passing to the algebraic closure, where it is embedded as a curve of degree $3$ in the projective plane. So we deduce that $C$ admits a divisor of degree $9$. It thus just suffices to write down a curve of genus $1$ without a divisor of degree $9$. The example of Piotr Achinger works here. The given curve has a divisor of degree $4$ and cannot have a divisor of degree $9$, since otherwise it would have a divisor of degree $1$ as $\gcd(4,9)=1$.<|endoftext|> TITLE: Higher regularity of solutions of non-linear elliptic PDE QUESTION [6 upvotes]: Let $\Omega\subset \mathbb{R}^n$ be a bounded domain with infinitely smooth boundary. Let $u\in C^2(\bar \Omega)$ be a solution of the Dirichlet problem for the non-linear equation \begin{eqnarray} F(x,u,\nabla u,\nabla^2 u)=0 \mbox{ in } \Omega,\\ u=\phi \mbox{ in } \partial \Omega. \end{eqnarray} Let the equation be elliptic with respect to $u$. Let us assume that $F,\phi$ are infinitely smooth. QUESTION. Is it true that the assumption $u\in C^{2,\alpha}(\bar \Omega)$ for some $0<\alpha<1$ implies that $u\in C^\infty(\bar \Omega)$? A reference would be very helpful. Special cases are also of interest. Remark. So far I was able to find in literature two special cases of this statement. (1) In the above generality in follows that the solution $u\in C^\infty(\Omega)$ (i.e. smooth in the interior of $\Omega$, not including the boundary). This is Lemma 17.16 in the Gilbarg-Trudinger book. (2) The question has positive answer (including the boundary) for $F$ of the form $$F(x,u,\nabla u,\nabla^2 u)=G(\nabla^2u)-f(x).$$ This is Prop. 5.1.10 in Qing Han's book. REPLY [5 votes]: It is true and well-known (assuming $F$ is e.g. uniformly elliptic). The idea is sketched in ch. 9 of the book by Caffarelli-Cabre, but I am not sure of a precise reference at this level of generality. Interior smoothness follows from interior Schauder estimates (applied to difference quotients of $u$ and its derivatives, successively). To extend to the boundary, one uses boundary Schauder estimates. One can reduce to the case of zero boundary data and flat boundary (that is, $\Omega = B_1^+$ and $u = 0$ on $\{x_n = 0\}$) after subtracting $\phi$ and performing a diffeomorphism, which don't change the class of equations under consideration. The difference quotient method and boundary Schauder estimates show that $u_i$ are $C^{2,\alpha}$ up to the flat part of the boundary for $i < n$. By the uniform ellipticity of the equation, $u_{nn}$ can be written as a smooth function of $u_{ij}$ for $(i,\,j) \neq (n,\,n)$, $\nabla u$, $u$, and $x$. All of these quantities are $C^{1,\,\alpha}$ up to the flat part of the boundary, hence $u \in C^{3,\,\alpha}$ up to the boundary. Higher regularity follows after differentiating the equation (the coefficients of the differentiated equation are now $C^{1,\alpha}$) and applying a similar procedure. (One in fact only needs $u \in C^2\left(\overline{\Omega}\right)$; the first step uses instead the Calderon-Zygmund $W^{2,\,p}$ estimate for the equation solved by the difference quotients to get $C^{2,\alpha}$ regularity via embeddings (take $p > n$), and then proceeds as before).<|endoftext|> TITLE: Estimating the growth of the Taylor coefficients given the growth of the function at the boundary QUESTION [11 upvotes]: Let $f(z)=\sum a_nz^n$ be a Taylor series that converges for $|z|<1$ and satisfies $$ |f(z)|\le \frac{1}{(1-|z|)^{k}} $$ for some fixed $k>0$. Question: What can I deduce about the growth of the Taylor coefficients $a_n$? Partial result: By judiciously selecting the location of the contour in the formula $a_n=\oint z^{-n}f(z)\tfrac{dz}{2\pi i z}$, namely, by performing the integration over the contour $|z|=\tfrac{n}{n+k}$ [which is the minimum of $|z|^{-n}(1-|z|)^{-k}$], I can get the "trivial bound" $|a_n|< c\cdot n^k$. But I suspect that this is not sharp. In particular, the growth of the Taylor coefficients of $(1-z)^{-k}$ is only $n^{k-1}$. Not $n^k$. More precise formulation of the question: What is the optimal $k'>0$ such that $$ |f(z)|\le \frac{1}{(1-|z|)^{k}}\quad\Rightarrow\quad |a_n|< c\cdot n^{k'} $$ for all $f(z)=\sum a_nz^n$. From the above arguments, I know that $k-1\le k'\le k$. REPLY [12 votes]: The optimal exponent is $k$. Such examples are given by sparse power series. This is actually trivial in the case $k=0$ (which was not included in the OP). Then we can simply take $f(z)=\sum j^{-2} z^{N(j)}$, say. This is obviously bounded, and the coefficients $a_n$ will not satisfy $|a_n|\lesssim n^{-\epsilon}$ for any $\epsilon>0$ if $N(j)$ increases fast enough. For positive $k$, we can similarly consider something like $$ f(z)=\sum n^{-2} \left(n^n\right)^k z^{n^n} . $$ Using calculus to find the maximum, we see that $$ x^k(1-\delta)^x \le C\delta^{-k} . $$ Thus $f$ satisfies the desired bound, but the coefficients do not satisfy $|a_n|\lesssim n^{k-\epsilon}$ for any $\epsilon>0$.<|endoftext|> TITLE: Is it OK to send a manuscript to good mathematicians in the field of your paper even if they don't know you before QUESTION [9 upvotes]: I am wondering if it is OK to send a finished manuscript to some mathematicians in your field for initial references or feedbacks in addition to just posting it in Arxiv, even if they do not know you completely, before submitting to a journal? REPLY [13 votes]: Informing specialists about a new manuscript posted on the arXiv is a common practice. Once it is posted there is no need to send a file. But on my opinion, asking explicitly a person you do not know to read and comment is too intrusive and impolite. You can write something like this: "I would like to bring to your attention the new preprint... I would appreciate any comments". Or "I will be grateful for any comments".<|endoftext|> TITLE: Is choice over definable sets equivalent to AC over axioms of ZF-Reg.? QUESTION [10 upvotes]: If we add the following axiom schema to ZF-Reg., would the resulting theory prove $\sf AC$? Definable sets Choice: if $\phi$ is a formula in which only the symbol $``y"$ occurs free, then: $$\forall X (X=\{y \mid \phi\} \to \\\exists f (f:X \setminus \{\emptyset\} \to \bigcup X \land \forall x (f(x) \in x)))$$ If not, then which form of choice this is equivalent to, over axioms of ZF-Reg.? The same question but over axioms of ZF was answered to the positive, and since ZF-Reg. is consistent with a principle stating that every class whose union is the universe won't be well-orderable, then this would kill the least fashioned argument used above. It was noted that this matter is complicated, and that the answer might be to the negative, hence this question! REPLY [11 votes]: Yes, indeed this kind of choice in general doesn't imply $\mathsf{AC}$ over $\mathsf{ZF}-\mathsf{Reg}$. I will reason in $\mathsf{ZFC}$ and construct an interpretation of $\mathsf{ZF}-\mathsf{Reg}$, where $\mathsf{AC}$ fails, but choice for definable sets holds. The idea is to define a modified permutation model $M$ of $\mathsf{ZFU}+\lnot\mathsf{AC}$, where internally the collection of all urelements is a proper class and from external perspective any two urelements could be swapped by an automorphism. Definable choice holds in $M$, since due to the presence of automorphisms and the fact that urelements form a proper class, the only definable sets are sets whose transitive closure contains no urelements. Finally we convert $M$ to the desired model $M'$ of $(\mathsf{ZF}-\mathsf{Reg})$ by transforming each urelement $a$ into the set $\{a\}$. Now let me outline the construction of $M$ in more details. We fix a countable set of urelements $U$ together with a function $r\colon U\to \mathbb{Q}$ such that $|r^{-1}(q)|=\aleph_0$, for all $q\in \mathbb{Q}$. We consider a group $G$ of permutations of $U$ consisting of all permutations $\pi\colon U\to U$ such that $r\circ \pi=r$. Let filter $\mathcal{F}$ on the lattice of subgroups of $G$ be generated by all the subgroups $\mathsf{Fix}_S=\{\pi\in G\mid \forall a\in S(\pi(a)=a)\}$, where $S$ is a finite subset of $U$. We naturally define the universe $V_U$ of sets with urelements from $U$. Naturally we define the support function $\mathsf{supp}\colon V_U\to \mathcal{P}(U)$ and the action of $G$ on $V_U$. We say that $x\in V_U$ is symmetric if $\{\pi \in G\mid \pi x= x\}\in \mathcal{F}$. And we say that $x\in V_U$ is hereditarily symmetric if it is symmetric and all elements of the transitive closure $\mathsf{TC}(x)$ are symmetric. We say that $x\in V_U$ is bounded if the set of rationals $r[\mathsf{supp}(x\cup\mathsf{TC}(x))]$ is bounded from above. The universe $M$ is a subuniverse of $V_U$ consisting of all $x\in V_U$ that are hereditarily symmetric and bounded. Since $U$ itself isn't bounded, the urelements form a proper class in $M$. The proof that $M$ satisfies $\mathsf{ZFU}$ is essentially the standard proof that permutation models satisfy $\mathsf{ZFU}$. The proof that $M$ doesn't satisfy $\mathsf{AC}$ is also essentially standard: there are no well ordering of the set $r^{-1}(0)$, since otherwise we would be able to split $r^{-1}(0)$ into two disjoint infinite subsets and they wouldn't be symmetric. For any two $a,b\in A$ we easily find a permutaion $\sigma$ of $A$ such that $\sigma(a)=b$, $\sigma(b)=a$, $\forall c,d\in A(r(c)=r(d)\iff r(\sigma(c))=r(\sigma(d)))$, and for any $S\subseteq A$ the set $r[S]$ is bounded from above iff $r[\sigma[S]]$ is bounded from above. Obviously, such a permutation naturally extends to an automorphism of $M$ swapping $a$ with $b$. Note that in fact I choose $M$ in such a manner that it satisfies even collection (and not merely replacement). Namely, consider the models $M_q\subseteq M$, for $q\in\mathbb{Q}$ that consist of all $x\in M$ such that $r[\mathsf{supp}(x)]$ is bounded from above by some $q' TITLE: Equivalences of $n$-categories QUESTION [7 upvotes]: This question is an extension of my previous question last year (see [2020]) in which I asked about the (consensus of a) definition of a weak $n$-category. Here are some background: while strict $n$-categories are easily defined, they are not sufficient for $n>2$. Therefore weak $n$-categories need to be defined. What a definition of a weak $n$-category should satisfy was proposed in [BD1995]. However, many proposals have since been given (see [Lei2001] or [2020]). And as David White pointed out in [2020], we had not reached to a consensus yet. This question focuses on a smaller part of the problem. Question: In order to prove that different models of $n$-categories to be equivalent, there must be a well-defined notion of a $(n+1)$-category to start with. So how is it possible to really prove the equivalence? I guess this relates to a philosophical problem that in order to justify (anything) one needs to justify the setting in which we justify. Before entering the realm of formalized arguments, we can postulate some desired results (as in [BD1995], or called "specification" in compsci's term), but nothing can stop people from building different settings (or called "implementation" in compsci's term). How could such problem be resolute without brute-force translating results from different foundations? [2020]: Definition of a $n$-category [BD1995]: Higher-dimensional Algebra and Topological Quantum Field Theory-[John C. Baez and James Dolan]-[arXiv:q-alg--9503002] [Lei2001]: A Survey of Definitions of n-Category-[Tom Leinster]-[Theory and Applications of Categories, Vol. 10, 2002, No. 1, pp 1-70] REPLY [7 votes]: As Marc Hoyois indicates in the comments, historically this was a major obstruction, past tense "was". My feeling is that these days, there is a nice perspective that whatever weak $n$-categories are, they are the objects of some $(\infty,1)$-category $n\mathrm{Cat}$. (Of course, weak $n$-categories are the objects of more than an $(\infty,1)$-category. But the extra $n$ dimensions in $n\mathrm{Cat}$ should be recoverable from looking at exponential objects.) It was a substantial feat to develop a theory of $(\infty,1)$-categories, but it has been more or less done. Moreover, surely $n\mathrm{Cat}$ will be not just some weird $(\infty,1)$-category, but in fact a presentable $(\infty,1)$-category, and these can be "presented" by model categories. So the question is "just" one of finding the correct Quillen equivalence class of model categories, where "correct" means that it should match your intuition about weak $n$-categories. Note that I am not saying that there is complete consensus about which presentable $(\infty,1)$-category deserves the name $n\mathrm{Cat}$. I certainly have opinions on the matter (namely: set $0\mathrm{Cat}:= \mathrm{Set}$, and define inductively $n\mathrm{Cat}$ to be the $(\infty,1)$-category of $(\infty,1)$-categories enriched in $(n{-}1)\mathrm{Cat}$), but I don't have the sociological data to conclude that my opinions are shared by the majority, let alone that there exists a consensus. I should also emphasize that, although I do believe there to be a single correct answer to the question of which presentable $(\infty,1)$-category should be called "$n\mathrm{Cat}$", this answer doesn't satisfy, or at least doesn't obviously satisfy, all natural desiderata. Most notably, it is not very algebraic. I mean, it isn't very non-algebraic — it is about as algebraic as is the notion of "Kan simplicial set" — but it isn't very algebraic either. However, I counter that the search for a highly algebraic theory of $n$-categories is most likely a fool's errand. Any such theory will in particular include a highly algebraic description of homotopy $n$-types. Postnikov and Whitehead provide a "lowly algebraic" description of homotopy $n$-types, and I am pessimistic about there being anything better.<|endoftext|> TITLE: Axiom of choice in linear algebra QUESTION [10 upvotes]: It is well known that the Axiom of Choice is needed to prove that every vector space has a basis (in fact this statement is equivalent to the AC). But what about the apparently weaker statement that every subspace of a vector space has a complementary subspace, or, equivalently, every short exact sequence of vector spaces splits? (This may actually not be a weaker statement after all, since it would apply also to a vector space whose cardinality is not an aleph.) REPLY [6 votes]: Your question is essentially answered in a math.SE post by Asaf Karagila, but I think it is worth spelling out a subtle point. The axiom of multiple choice (or MC for short) says: For every set $X$ of nonempty sets, there exists a function $f$ on $X$ such that for every $x\in X$, $f(x)$ is a finite nonempty subset of $x$. In Lemma 2 of Some theorems on vector spaces and the axiom of choice (Fund. Math. 54 (1964), 95–107), M. N. Bleicher showed that MC is implied by the following statement: For some field $F$, every subspace of a vector space over $F$ has a complementary subspace. In fact, Bleicher's results combined with a result of M. C. Armbrust (An algebraic equivalent of a multiple choice axiom, Fund. Math. 74 (1972), 145–146) imply that the above statement is equivalent to MC. The subtle point is that the above equivalence can be proved not just in ZF, but in some other set theories such as ZFU (a variant of ZF with "urelements" or "atoms"). So this equivalence is "robust" in some sense. On the other hand, ZF proves the equivalence of MC with the axiom of choice (for a non-paywalled proof, see Theorem 2.18 in Kevin Barnum's note, The axiom of choice and its implications), but this equivalence is "delicate" because if you switch to a slightly different set theory, MC might be strictly weaker than AC. So the answer to your question is that your "apparently weaker" statement is equivalent to AC if you work over ZF. But over some other closely related set theories, I'm not sure the answer is known. Some further information may be found in Paul Howard's paper, Bases, spanning sets, and the axiom of choice (Math. Logic Q. 53 (2007), 247–254), and Marianne Morillon's paper, Linear extenders and the axiom of choice (Comm. Math. Univ. Carol. 58 (2017), 419–434).<|endoftext|> TITLE: Geometrically rational variety over a finite field QUESTION [6 upvotes]: Let $k=\mathbb{F}_q$ be a finite field, and let $X$ be a smooth projective variety over $k$. Suppose that $X_{\overline{k}}$ is birational to $\mathbb{P}^n_{\overline{k}}$, do we know (1)If $X$ is necessarily birational to $\mathbb{P}^n_k$? (2)If $X$ necessarily has a $k$-point? REPLY [6 votes]: (1) No: There exist minimal cubic surfaces over finite fields (see for example https://arxiv.org/abs/1611.02475). Such surfaces are non-rational over the ground field. (2) Yes: This is a special case of a more general result of Esnault: https://arxiv.org/abs/math/0207022 This proves the congruence $\#X(\mathbb{F}_q) \equiv 1 \bmod q$, which clearly implies the existence of a rational point.<|endoftext|> TITLE: Is it ever unnecessary to mathematically formalize a concept? QUESTION [12 upvotes]: From my understanding, mathematics sometimes gives rise to new physical/tangible laws and the converse is also true. In particular, physical phenomena give rise to new mathematics. In all of the cases that I have seen, the mathematics is usually formalized. That is, definitions, lemmas, theorems and their proofs are developed in either of the two cases mentioned above. Are there ever cases where formally defining physical phenomena in mathematical language is unnecessary? For example, deep learning has recently been formalized. Is this formalism necessary for developing new techniques in the field? I can see how it would make sense if the formal theory was developed first. REPLY [3 votes]: The post asks: "Are there ever cases where formally defining physical phenomena in mathematical language is unnecessary?" The key clarification is: necessary for what? Mathematical rigor may require formal definitions -- but this may be the only goal where the formal definitions are necessary. A successful application does not require formal definitions. Motivating good mathematics does not require formal definitions. Even doing good mathematics does not require formal definitions. First example. Consider the top-voted question in mathematical physics here on MathOverflow ("what is an integrable system?". The top-voted answer includes the telling phrase "I'm being notably vague here." So for doing research mathematics by the standards of MathOverflow, formalization is apparently not necessary. Second example. The field of deep learning existed before the formalization that you describe, and some people continue in the field without using that formal approach. So the field does not seem like an example of necessary formalization. Maybe ten years from now we will say that most good work in deep learning uses the formalization; in that case we might say it was necessary. More likely, we will say that someone proposed a formalization, and some people used it, but most contributors to the field did not use it. Third example. Would Newton's definitions, as in Ian Bruce's translation of the Principia here, qualify as formal mathematical definitions? I: The quantity of matter is a measure of the same arising jointly from the density and magnitude. II: The quantity of motion is a measure of the same arising from the velocity and quantity of matter jointly. III: The innate force of matter is the resisting force, by which each individual body, however great it is in itself, persists in its state either of rest or of moving uniformly straight forwards. IV: The impressed force is the action exercised on the body, to changing the state either of rest or of motion uniform in direction. V: It is the centripetal force, by which bodies are drawn, impelled, or tend in some manner from all sides towards some point, as towards a centre. I'd say that I, IV and V are or could be formal definitions in the usual sense (mass := density * volume, momentum := velocity * mass, centripetal force := force towards a fixed point), but that II and III are not. So formal definitions were apparently not a requirement for Newton’s work, the canonical example of analyzing physical phenomena in mathematical language.<|endoftext|> TITLE: Representation stability for systems of braid group representations QUESTION [5 upvotes]: In general, if I understand correctly, the representation theory of the braid groups is quite complicated, and there's no classification of the irreducibles. However, the braid groups form a sort of system of groups just as the symmetric groups do, and so one can ask about representation stability for coherent systems of representations of braid groups. Editing in response to Andy's comments below: it may be that "representation stability" doesn't mean anything because we can't decompose braid group representations in the same way we can decompose symmetric group representations. So a more basic question would be: is there any sensible way to talk about systems of braid group representations stabilizing that doesn't involve decomposition into irreducibles? This is a very basic question, mostly just a reference request. Note that I am not asking about representation stability for the braid groups themselves. REPLY [7 votes]: A good way to handle systems of braid group representations is to consider the category of functors $\mathcal{C}\to R\textrm{-Mod}$, where $\mathcal{C}$ is a category with the braid groups as automorphisms. The braid groupoid $\beta$ (ie the groupoid with natural numbers as objects and braid groups as automorphisms) is then a subcategory of such $\mathcal{C}$. Note that $\beta$ itself is not quite satisfactory for such $\mathcal{C}$ since a functor $\beta\to R\textrm{-Mod}$ encodes a family of representations where the representations of $B_{n}$ is independent of the one of $B_{n+1}$. In other words, we would like $\mathcal{C}$ to encode compatibilities between the representations. There already exist good candidates for such category: the partial braid category, see Section 2.3 of Palmer https://arxiv.org/pdf/1308.4397.pdf; the Quillen’s bracket construction applied to $\beta$, denoted by $\mathcal{U}\beta$, and applied by Randal-Williams and Wahl https://arxiv.org/abs/1409.3541 [RWW]. The latter has the significant advantage that a large class of classical families of representations of the braid groups define functors $\mathcal{U}\beta\to R\textrm{-Mod}$: the Burau representations; see Example 4.3 of [RWW]. the Tong-Yang-Ma and Lawrence-Krammer-Bigelow representations; see Section 1.2 of https://arxiv.org/pdf/1702.08279.pdf [S1]. the whole family of the Lawrence-Bigelow representations; see Section 5.2.1.1 of https://arxiv.org/pdf/1910.13423.pdf [PS]. Also, there are notions of polynomiality which allows us to characterise and prove more properties on these systems of representations: the notion of (strong) polynomiality, a.k.a finite degree coefficient systems: the Burau representation is of degree $1$ (see Example 4.15 of [RWW]), the Tong-Yang-Ma representation is of degree $1$ and Lawrence-Krammer-Bigelow representations is of degree $2$ (see Propositions 3.25 and 3.33 of [S1]). This is the appropriate notion to prove twisted homological stability result see [RWW]. the notion of weak polynomial functors, which has originally been introduced for symmetric monoidal categories (for instance $FI$) by Djament and Vespa https://arxiv.org/abs/1308.4106, and generalised to categories of the same type as $\mathcal{U}\beta$ (namely pre-braided monoidal categories) in https://arxiv.org/pdf/1709.04278.pdf (see Section 4.2). An advantage of this notion is that it reflects more accurately than the strong polynomiality the behaviour of functors in the stable range. For instance, Church, Miller, Nagpal and Reinhold https://arxiv.org/pdf/1706.03845.pdf compute the weak polynomial degree (named "stable degree" in this paper) of some FI-modules. Moreover, denoting by $Pol_{d}(\mathcal{U}\beta)$ the category of weak polynomial functor of degree less or equal to $d$, we can define the quotient category $$Pol_{d+1}(\mathcal{U}\beta)/Pol_{d}(\mathcal{U}\beta).$$ These quotient categories provide a new tool to handle families of representations with a sensible way to classify them. In particular, it doesn’t involve decomposition into irreducibles. See also Palmer https://arxiv.org/pdf/1712.06310.pdf for a comparison of the various instances of the notions of twisted coefficient system and polynomial functor. Hence weak polynomiality might be viewed as a refinement of representation stability phenomena and a sensible to talk about system of braid group representations.<|endoftext|> TITLE: Bijective proof of recurrence for rooted unlabeled trees QUESTION [11 upvotes]: Would've been a better question for Christmas than Thanksgiving, but alas... Let $t_n$ denote the number of rooted, unlabeled trees on $n$ vertices (OEIS A000081). These are the isomorphism classes of rooted trees under root-preserving isomorphisms. Let $T(z) = \sum_{n\geq 1} t_n z^n$ be the corresponding generating function. In 1937, using his enumeration under symmetry theorem, Pólya showed that $$ T(z) = z \prod_{i=1}^{\infty}e^{\frac{T(z^i)}{i}}.$$ By differentiating this identity one obtains the recurrence $$ (n-1)\cdot t_n = \sum_{i=1}^{n-1}t_{n-i}\sum_{m \mid i}mt_{m}$$ for $n> 1$. This is such a nice recurrence that I wonder: Question: Is there a bijective proof of this recurrence for $t_n$? REPLY [5 votes]: Late edit: having now read through the OP comments, I can see that my proof is essentially a carbon copy of @darij.grinberg's approach (although my derivation was independent). I'm okay to delete this answer once/if darij chooses to post theirs. Pick a canonical ordering of unlabelled rooted trees, say, with lexicographical comparison of tuples $(n, T_1, \ldots, T_k)$, where $n$ is the number of vertices, $T_1, \ldots, T_k$ is the non-descending sequence of children subtrees. Throughout $T, T_1, T_2$ are unlabelled rooted trees in canonical form (that is, subtrees of any vertex are ordered as above). Let $A_n$ be the set of pairs $(T, v)$ with $|T| = n$, $v$ is a non-root vertex of $T$. Also, let $B_n$ be the set of tuples $(T_1, T_2, k, u)$ such that $|T_1| + k|T_2| = n$, $u$ is a vertex of $T_2$. Observe that $|A_n|$ and $|B_n|$ are LHS and RHS of the recurrence in OP, more readily seen by rewriting $(n - 1)t_n = \sum_{m = 1}^{n - 1}mt_m \sum_{0 < km < n} t_{n - km}$. The bijection between $A_n$ and $B_n$ is as follows: we associate $(T_1, T_2, k, u)$ to $(T, v)$ by: $T_2$ = the subtree of $T$ containing $v$, $u$ = the respective vertex of $T_2$, $k$ = the number of children subtrees of $T$ isomorphic to $T_2$ not later than the copy containing $v$ in the ordered sequence of children subtrees, $T_1$ = the result of removing the $k$ children subtrees from $T$. we associate $(T, v)$ to $(T_1, T_2, k, u)$ by inserting $k$ copies of $T_2$ as children subtrees of the root of $T_1$ (and naming the result $T$), and picking the respective vertex $u$ in the $k$-th copy as $v$.<|endoftext|> TITLE: Can the nth projective space be covered by n charts? QUESTION [38 upvotes]: That is, is there an open cover of $\mathbb{R}P^n$ by $n$ sets homeomorphic to $\mathbb{R}^n$? I came up with this question a few years ago and I´ve thought about it from time to time, but I haven´t been able to solve it. I suspect the answer is negative but I´m not very sure. Also, is there an area of topology which studies questions like this one? REPLY [16 votes]: It seems worth giving the cup-length argument, as it's relatively short and sweet. Suppose $\mathbb{R}P^n=U_1\cup\cdots\cup U_n$, with each $U_i\approx\mathbb{R}^n$, and let $c\in H^1(\mathbb{R}P^n;\mathbb{Z}/2)$ be the generator. For each $i$ the inclusion-induced map $H^1(\mathbb{R}P^n;\mathbb{Z}/2)\to H^1(U_i;\mathbb{Z}/2)$ is trivial, so by the long exact cohomology sequence of the pair $(\mathbb{R}P^n,U_i)$ there exists a relative cohomology class $c_i\in H^1(\mathbb{R}P^n,U_i;\mathbb{Z}/2)$ whose image in absolute cohomology is $c$. But then by the naturality of relative cup products, $c^n$ is the image of $$ c_1\cdots c_n\in H^n(\mathbb{R}P^n,U_1\cup\cdots\cup U_n;\mathbb{Z}/2)=0, $$ and therefore $c^n=0$, a contradiction. As Aleksander Milivojevic points out in the comments, the relevant area of topology is the study of Lusternik--Schnirelmann category and related invariants.<|endoftext|> TITLE: Properness for uncountable models QUESTION [6 upvotes]: There are certain generalizations of the notion of a proper forcing to uncountable cardinals in the context of forcing iterations. For example, the ones introduced by Eisworth, and by Roslanowski and Shelah. They usually considered such a notion under some cardinal arithmetic assumptions and in the context of iterated forcings. I am simply looking for a reference for the following definition: Definition: Suppose $\mathcal S$ is a set of uncountable elementary submodels in some large $H_\theta$, and $\mathbb P$ is a forcing notion belonging to every model in $\mathcal S$. As in the definition of a proper forcing, let us say that $\mathbb P$ is $\mathcal S$-proper if, for every $M\in\mathcal S$ and every $p\in\mathbb P\cap M$, there is an $(M,\mathbb P)$-generic condition $q\leq p$. I considered the above definition and the following theorem as folklore, but I was asked by a referee to provide a reference for them. Theorem: Suppose $\kappa$ is a regular cardinal, and $\mathcal S\subseteq\mathcal P_\kappa(H_\theta)$ is stationary. If $\mathbb P$ is $\mathcal S$-proper, then $\mathbb P$ preserves $\kappa$. I think calling it $\kappa$-properness can be confusing for several reasons, including the fact that the set $\mathcal S$ is just stationary. Question 1: Does the above notion have a name in the literature? If not, is my terminology convenient? What about "$\mathbb P$ is proper for $\mathcal S$"? Question 2: What is the most appropriate work to cite for the above theorem? REPLY [4 votes]: The concept first appeared in Shelah's paper Independence results The theorem you have stated should be folklore, but you may see Tapani Hyttinen and Mika Rautila, The canary tree revisited for a proof. Of course in the above cited papers, the definitions are not given for some stationary set, but that is clear how to modify them for this case. Finally, a very good reference to look at is the following paper by Roslanowski: Shelah's search for properness for iterations with uncountable supports<|endoftext|> TITLE: $2$-norm of idempotent matrix QUESTION [8 upvotes]: Suppose $n > 1$ is an integer. Let $P \in \mathbb C^{n \times n}$ be a matrix such that $P^2=P$ and $1\leqslant {\rm rank}(P) TITLE: Analogue of Grauert's upper semi-continuity for Bott–Chern cohomology QUESTION [6 upvotes]: In Coherent analytic sheaves, one has the following theorem due to Grauert: Let $f: X \rightarrow Y$ be a holomorphic family of compact complex manifolds with connected complex manifolds $X, Y$ and $V$ a holomorphic vector bundle on $X$. Then for any integers $q, d \geq 0$, the set $$ \left\{y \in Y: h^{q}\left(X_{y},\left.V\right|_{X_{y}}\right) \geq d\right\} $$ is an analytic subset of $Y$. Hence, we can take $V=\Omega^p$ to get the hodge number. I wonder if the above assertion still holds for $(p,q)$-Bott–Chern cohomology? In other words, for the same conditions, is the set $$ \left\{y \in Y: h^{p,q}_\text{BC}\left(X_{y}\right) \geq d\right\} $$a complex analytic subset of $Y$? Notice that one cannot apply Grauert's theorem to Bott–Chern cohomology, since now we just have the following isomorphism (one can refer to Demailly's book Basic results on sheaves and analytic sets for more details): $$ H_\text{BC}^{p, q}(X) \cong \mathbb{H}^{p+q-1}\left(X, \mathscr{L}_{X}^{\bullet}\right). $$ REPLY [5 votes]: The semi-continuity is true for elliptic complexes: if $(C, d_t)$ is a continuous family of elliptic complexes, parametrized by $t\in \mathbb R$, the cohomology of $(C, d_t)$ is semicontinuous in $t$. However, Bott–Chern cohomology are cohomology of an elliptic complex: M. Schweitzer, Autour de la cohomologie de Bott–Chern, arXiv:0709.3528. This takes care of semicontinuity in the real analytic setting, but I suppose you need your jumping loci to be complex analytic. To see that the jumping loci are complex analytic, you could use an exact sequence $$ H^{*}(\Lambda^{p, q-1}(M), \bar\partial)\oplus \overline{H^{*}(\Lambda^{q, p-1}(M), \bar\partial)} \rightarrow H_{BC}^{p,q}(M) \rightarrow H^{p+q}(M). $$ I don't have a citation for this, except my own paper with Ornea, Morse–Novikov cohomology of locally conformally Kähler manifolds, Theorem 4.7, where we proved it for BC-cohomology with coefficients in a local system; but I suppose it's rather well known. I don't expect that the theorem that you want (that all jumping loci are complex analytic) is published anywhere, but this exact sequence is a good starting point.<|endoftext|> TITLE: Is this infinite product entire? QUESTION [5 upvotes]: Let $(z_i)$ be a square-summable sequence which is even summable but not absolute summable, i.e. $\sum_{i=1}^{\infty} \vert z_i \vert = \infty$,$\sum_{i=1}^{\infty} \vert z_i \vert^2 < \infty$ and $\sum_{i=1}^{\infty} z_i$ exists. I would like to ask if the following function $$f(\mu):=\prod_{i=1}^{\infty}(1+\mu^2 \vert z_i \vert^2 - 2 \mu \Re(z_i))$$ is necessarily entire? I must say that I do not even know if this product exists away from the real axis. On the real axis it is clear that it exists by using that $(1+x) \le e^x$ such that $\vert f(\mu) \vert \le e^{\mu^2 \sum_i \vert z_i \vert^2 -2\mu \Re \sum_i z_i }.$ Assuming it was entire, does there exist a similar growth bound on $\vert f(\mu) \vert$ as the one I obtained on the real axis? REPLY [11 votes]: This function is (on the real line, at least) the product of $$ \exp( \mu^2 \sum_{i=1}^\infty |z_i|^2 - 2 \mu \Re(\sum_{i=1}^\infty z_i)) \quad (1)$$ and the Hadamard type product $$ \prod_{i=1}^\infty E_1( 2 \mu\Re z_i - \mu^2 |z_i|^2) \quad(2)$$ where $E_1$ is the first elementary factor $$ E_1(z) := (1-z) \exp(z).$$ The expression (1) is clearly entire in $\mu$; the product (2) is locally uniformly convergent (from the standard bound $E_1(z) = 1+O(|z|^2)$ when $|z| \leq 1$) and so is also entire. A refinement of this analysis (using for instance the upper bound $|E_1(z)| \leq \exp(O(|z|^2))$ for all $z$) also gives growth bounds comparable to the ones you already located in the real case. An alternate factorisation is $$ \exp( - 2 \mu \Re \sum_{i=1}^\infty z_i ) \prod_{i=1}^\infty E_1(\mu z_i) E_1(\mu \overline{z_i}).$$ Thus this function is an order two entire function with zeroes precisely at $1/z_i, 1/\overline{z_i}$ (counting multiplicity), which specifies the function uniquely up to quadratic exponential factors (such as (1)) by the Hadamard factorisation theorem.<|endoftext|> TITLE: Is every Zariski closed subgroup a stabilizer? QUESTION [10 upvotes]: Let $ G $ be a linear algebraic group. Is it true that a subgroup $ H $ of $ G $ is Zariski closed if and only if there exists a representation $ \pi: G \to \mathrm{GL}(V) $ and a vector $ v \in V $ such that the stabilizer $ G_v:=\{g \in G: \pi(g)v=v \} $ is equal to $ H $? I think one implication is clear since $ G_v $ is certainly a subgroup and the equation $ \pi(g)v=v $ is polynomial in the matrix entries. Thus any stabilizer must be Zariski closed. I am not sure of the reverse implication. Is it really true that every Zariski closed subgroup of $ G $ arises as the stabilizer of some vector in some representation? REPLY [23 votes]: Chevalley's theorem (see Theorem 4.19 in Milne) is very close to this. It says Let $G$ be a linear algebraic group and let $H$ be a Zariski closed subgroup. Then there is a representation $V$ of $G$ and a one dimensional subspace $L$ of $V$ such that $H$ is the stabilizer of $\mathbb{P}(L)$ in the action of $G$ on $\mathbb{P}(V)$. So this is close to what you asked for, but uses projective rather than linear representations. To see that you can't get exactly what you asked for, work over a field of characteristic zero, take $G = \text{SL}_2$ and let $H$ be the Borel subgroup $B:=\left[ \begin{smallmatrix} \ast & \ast \\ 0 & \ast \end{smallmatrix} \right]$. The classification of $\text{SL}_2$-representations is well known, and we see that $V^B = V^{\text{SL}_2}$ for any $\text{SL}_2$-representation $V$.<|endoftext|> TITLE: Is there a noncommutative Gaussian? QUESTION [20 upvotes]: In classical probability theory, the (multivariate) Gaussian is in some sense the "nicest quadratic" random variable, i.e. with second moment a specified positive-definite matrix. I do not know how to make this precise, but non-precisely what I mean is that 1. Gaussian shows up everywhere, and 2. it is universal/canonical/... in some sense, e.g. as in the central limit theorem. My question is whether for many noncommutative probability spaces (an algebra $A$ over $\mathbf{C}$ and a map $E:A\to\mathbf{C}$, with conditons), there also exists a "nicest quadratic" random variable $X\in A$, satisfying analogous properties to the Gaussian. REPLY [5 votes]: If one does not insist on having a notion of "independence" in the background, but takes, as asked in the question, the Wick/Isserlis formula (which expresses general moments in terms of second moments via a nice combinatorial formula) as guiding principle, then there are also other nice non-commutative versions of multivariate Gaussian distributions. A prominent one is given by the q-Gaussian distribution, which interpolates between the classical Gaussian and the free Gaussian (i.e., semicircular) distribution.<|endoftext|> TITLE: The radical of $kG$-modules QUESTION [8 upvotes]: $\DeclareMathOperator\Rad{Rad}$Let $k$ be a finite field of $p$ elements. Let $G$ be an elementary abelian p-group and $V$ a $kG$-module corresponding to the representation $\alpha:G\rightarrow \mathrm{GL}_{n}(k)$. Denote the radical of $V$ by $\Rad(V)$ which is defined to be the intersection of all maximal submodules of $V$. Note that $\Rad^{i}(V)=\Rad( \Rad^{i-1}(V) )$. Now, let $s TITLE: Osculating circle QUESTION [5 upvotes]: (This question may be too elementary for this site — I'm fine if it needs to be moved to math.stackexchange.) If I approximate a nice planar curve by a straight line, the tangent, then the second derivative tells me which side of the line the curve lies on locally. If instead I approximate the curve by a circle — the osculating circle — what determines the answer to the analogous question? Which side of the osculating circle does the curve lie on locally? Update: I'm accepting one of the answers below, but it's surprising and with my poor geometric intuition not particularly easy. This situation is the exact opposite of the tangent line analogy I made in the question. In that case the graph of a function lies locally on one side or the other of the tangent line at generic points, and only crosses at inflection points. But for the osculating circle it is the reverse: the curve crosses the circle at generic points, and only lies (locally) inside or outside the circle at maxima or minima of the curvature. The can be deduced from the Tait–Kneser theorem, as the answer below claims. A nice explication by Ghys, Tabachnikov and Timorin can be found at Osculating curves: around the Tait–Kneser Theorem. The theorem itself says wherever the curvature is monotonic (e.g. when parametrized by arclength), the osculating circles are pairwise disjoint and nested. It takes some geometric visualization to realize this means the circles can not lie on one side of the curve as the point of contact moves. This can be a challenge to see on actual graphics, because the curve and circle agree up to second order terms. A Mathematica demonstration may be found here which allows one to select in particular the example of an ellipse which was mentioned in the comments. REPLY [2 votes]: In retrospect I realize one does not need the full strength of Tait-Kneser. The curve and the circle agree up to quadratic order. So the error, the difference between the two, looks like a constant times $s^3+O(s^4)$ when parametrized by arclength $s$ measured from the point of contact. The error has to change sign when $s=0$.<|endoftext|> TITLE: Does Lefschetz pencil always exist in char $p$? QUESTION [7 upvotes]: Let $X\subset \mathbb{P}^n_k$ be a smooth projective variety, a point $p\in \mathbb{P}^{n,\vee}_k$ gives rise to a hyperplane $H_p\subset \mathbb{P}^n$, hence an intersection $X_p:=H_p\cap X$. We say a line $L\subset\mathbb{P}_k^{n,\vee}$ is a Lefschetz pencil if (1) There exists $0,\infty \in L(k)$ such that $H_0, H_\infty, X$ intersect transversally. (2) For $p\in L-\{s_1,...,s_r\}$, the section $X_p$ is smooth, $H_{s_i}\cap X$ has exactly one quadratic ordinary singularity. When $k$ is an algebraically closed field of $\mathrm{char}(k)=p$, do we know if Lefschetz pencils always exist? If not, can we relax the assumption by allowing $L$ be a smooth curve in $\mathbb{P}^{n,\vee}_k$? (From Theorem 3 here, Lefschetz pencils always exist over any field after $d$-uple embedding of $X$) REPLY [8 votes]: I think you are asking whether a Lefschetz pencil exists without re-embedding. Then the answer is no. In cor. 3.5.0 of expose XVII of SGA 7, Katz gives a necessary and sufficient condition for a Lefschetz pencil to exist. In the case of a hypersurface $X=V(F)$ in $\mathbb{P}^n$ and $char k=p\not=2$, the condition amounts to the Gauss map $\phi:X\to (\mathbb{P}^n)^\vee$ defined by $\phi(x_0,\ldots, x_n) = (\partial F/\partial x_i)$ being seperable. But he shows this condition fails for $F= \sum x_i^{p^n+1}$.<|endoftext|> TITLE: Does there always exist a categorical extension of $ZFC_2$ with no set models? QUESTION [5 upvotes]: $ZFC_2$, i.e. second-order Zermelo-Fraenkel set theory with Choice, has only one proper class model upto isomorphism, namely $V$. But it may or may not also have set models. If $V$ has no inaccessible cardinals, then $ZFC_2$ has no set models, making $V$ its only class model upto isomorphism. But if $V$ has inaccessible cardinals then $ZFC_2$ can have some set models of the form $V_\kappa$ for some inaccessible cardinal $\kappa$. My question, no matter what the truth is about the nature, existence, and number of inaccessible cardinals in $V$, does there always exist an extension of $ZFC_2$ which has no set models, making $V$ its only class model upto isomorphism? Or are there some conditions under which no consistent extension of $ZFC_2$ has no set models? REPLY [6 votes]: Since "consistent" is a weird notion in the context of second-order set-theories and moreover we can't even directly talk about a second-order theory being true of $V$ within $V$, I think it's usefully demystifying to rephrase the question in a "set-ish" way as follows: Is it consistent with $\mathsf{ZFC}$ that there is some inaccessible cardinal $\kappa$ such that for every second-order theory $T$ in the language of set theory, if $V_\kappa\models T$ then $V_\alpha\models T$ for some $\alpha<\kappa$? Note that whether or not $V_\gamma\models S$ for $\gamma\le\kappa$ and $S$ a second-order set theory is detected by the first-order diagram of $V_{\kappa+1}$, so the above question does make sense. As Elliot Glazer observes, there is in this case a simple counting argument we can employ: if there are more than continuum-many inaccessibles, then some pair of inaccessibles $\alpha<\kappa$ have $V_\alpha\equiv_{\mathsf{SOL}}V_\kappa$. Of course this is somewhat unsatisfying. What we have is a set-theoretic assumption which guarantees the existence of some appropriate $\kappa$, but we don't have a concrete property which would identify such a $\kappa$. So at this point it's natural to ask: Is there a natural set-theoretic property - e.g. an already-studied large cardinal property - which guarantees that any $\kappa$ with that property has $V_\alpha\equiv_{\mathsf{SOL}}V_\kappa$ for some $\alpha<\kappa$? Note that, trivially, such a property would have to be non-second-order-definable. And this pushes us into the realm of very strong properties indeed (see e.g. the discussion here).<|endoftext|> TITLE: Are there good mutually interpretable axioms for synthetic Euclidean and hyperbolic geometry? QUESTION [14 upvotes]: There are familiar analytic equiconsistency proofs for Euclidean and hyperbolic geometry.  Those proofs are so robustly geometric that it seems like they must have synthetic analogues. Looking into the literature, though, I wonder if I am too optimistic about this.  The most common rigorous axiomatizations for synthetic Euclidean and hyperbolic geometry, so far as I can tell, are Hilbert's from his Foundations of Geometry, with two changes.  They omit the non-elementary axiom of continuity, and they add something to assure that circles will actually have points of intersections with lines and other circles that they cross.  The only difference, in these axiomatizations, between Euclidean and hyperbolic geometry is in the axiom of parallels. But this synthetic hyperbolic geometry is incapable of some constructions that we take pretty much for granted both in Euclidean geometry and also (more to my point) in analytic hyperbolic geometry.   Notably $n$-section of lines, see trisection-of-a-hyperbolic-line-segment.  Maybe the analytic equiconsistency proofs really do not transfer well to synthetic geometry. I know Greenberg's discussion of axiomatic issues in Greenberg. But I do not have all his references at hand.  Judging from the ones I do have (including Hartshorne) it seems likely that  his discussion of equiconsistency for the two geometries (pp.213-214) refers to analytic presentations of geometry. Can I find mutually interpretable axioms for synthetic Euclidean and hyperbolic geometry? Edit: Erik Walsberg's comment about Tarski's axioms answers my title question, even though not in the way I had in mind when I wrote the text.  My text was ambiguous about "synthetic" methods, in just the way that Tarski What is elementary geometry? means when he says   "In colloquial language the term elementary geometry is used loosely [...with] no well determined meaning."    Hilbert, Tarski, and Greenberg all show that the important logical distinction characterizing elementary methods is not between using or not using coordinates in some field.  It is between using or not using higher order notions like point-set continuity and limits.  First order algebraic considerations on fields (notably pythagorean fields) are already implicit in Euclid and  central to successful first-order elementary geometry.   The work that led me to this question is about interpretation in first order logic, and not about compass (or horocompass) and straightedge or other such construction methods, and really not about avoiding coordinates.  So my title was true to my actual concern.  Some of my text concerning synthetic methods was less relevant (though those questions too intrigue me).  Walsberg's comment is really an answer both to the question as titled and to my actual concern, though he correctly saw I had another issue about methods also in mind. REPLY [4 votes]: One should not forget Arthur Cayley's claim "And thus projective geometry is all geometry". The treatise of Oswald Veblen and John Wesley Young, "projective geometry", follows that path. The first volume is dedicated to synthetic projective geometry. The second one builds both euclidean and non-euclidean geometry in that context, the first one by choosing a line (to be the line at infinity) and an involution on that line (to be the orthogonal relation on the set of directions), the second one by choosing a conic (to be the ideal points of the absolute circle at infinity). Then Veblen studies the group of transformations leaving invariant these datas, by pure synthetic methods. This is Klein original idea. Both geometries are consistent because they are derived from projective geometry, and Veblen provides a set of axioms for both. Nowadays, projective geometry is often built from affine geometry (the set of lines through a point), and that closes the loop if really you want to go from euclidean to non-euclidean geometry. But really, projective geometry should not be set aside, at least that's what the ancients say. From the introduction of the second volume of Veblen and Young, The ideal of such books should be not merely to prove every theorem rigorously but to prove it in such a fashion as to show in which spaces it is true and to which geometries it belongs. The treatise is 990+ pages long and certainly achieves its goal. From the synthetic viewpoint, modern treatises such as Hartshorne and Greenberg certainly pale in comparison to the works of geometers from the beginning of the XX century.<|endoftext|> TITLE: Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order? QUESTION [31 upvotes]: Consider the hierarchy of relative geometric constructibility by straightedge and compass. Namely, given a geometric figure $B$, a set of points in the plane, we define that geometric figure $A$ is constructible from $B$, written as $$A\leq B,$$ if from points in $B$ using straightedge and compass we may construct every point in $A$. This is a partial preorder on geometric figures, considered as sets of points in the Euclidean plane. The order gives rise to a corresponding strict order notion $A TITLE: The character table of the symmetric group modulo m QUESTION [11 upvotes]: Let $S_n$ be the symmetric group and $M_n$ the character table of $S_n$ as a matrix (in some order) for $n \geq 2$. Question: Is it true that the rank of $M_n$ as a matrix modulo $m$ for $m \geq 2$ is equal to the number of partitions of $n$ by numbers that are not divisible by $m$? Here the matrix modulo $m$ is obtained by replacing each number $l$ by its canonical representative mod $m$ (for example -1 mod 3 =2) and then calculate the rank of the obtained matrix as a matrix with integer entries. This seems to be true for $m=2,3,4$ by some computer experiments. For $m=2$ the sequence of ranks (for $n \geq 2$) starts with 2,3,4,5,6,8,10,12,15,18,22 , for $m=3$ it starts with 2,4,5,7,9,13,16,22 and for $m=4$ it starts with 2,3,4,6,9,12,16,22,29. For example for $n=5$ and $m=3$, the matrix $M_5$ looks as follows: \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & 1 & -1 \\ 2 & 0 & 2 & -1 & 0 \\ 3 & -1 & -1 & 0 & 1 \\ 3 & 1 & -1 & 0 & -1 \end{bmatrix} modulo 3 the matrix is given by \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 2 \\ 2 & 0 & 2 & 2 & 0 \\ 0 & 2 & 2 & 0 & 1 \\ 0 & 1 & 2 & 0 & 2 \end{bmatrix} and this matrix has rank 4. The paritions of $n=5$ are given by [ [ 1, 1, 1, 1 ], [ 2, 1, 1 ], [ 2, 2 ], [ 3, 1 ], [ 4 ] ] and thus there are 4 partitions of 5 whose parts are not divisble by $m=3$. REPLY [21 votes]: When $m$ is prime there is a simpler proof. The Smith normal form of the character table of $S_n$ is computed at Problem 14 here (solution here). From this it follows that the rank of the character table mod $m$ equals the number of partitions of $n$ for which every part has multiplicity less than $m$. By a simple generating function argument (generalizing Euler's well-known proof for the case $m=2$) this is also the number of partitions of $n$ for which no part is divisible by $m$.<|endoftext|> TITLE: Conceptual definition of the extension of a connection to 1-forms QUESTION [25 upvotes]: I have a question that arose while reading Milnor's "Characteristic Classes". I will use the notation from that book. Let $M$ be a smooth manifold and let $\zeta$ be a complex vector bundle on $M$. Milnor defines a connection on $M$ to be a map $\nabla\colon C^{\infty}(\zeta) \rightarrow C^{\infty}(\tau_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying the Leibniz identity, where $\tau_{\mathbb{C}}$ is the complexified tangent bundle of $M$. In Lemma 4 of Appendix C, he proves that such a connection can be extended to a map $\hat{\nabla}\colon C^{\infty}(\tau_{\mathbb{C}}^{\ast} \otimes \zeta) \rightarrow C^{\infty}(\wedge^2 \tau_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying an appropriate Leibniz rule. However, his proof is just a definition in local coordinates, with the details left to the reader. I verified these details, though they were a bit of a pain. However, I feel like there must be a more conceptual definition of $\hat{\nabla}$ that makes no reference to local coordinates. Does anyone know one? REPLY [3 votes]: Although the answer is already given, it may be helpful to point out how the coordinate definition contains the multilinear definition, in a single step of work. From the definition $$(d^\nabla s)_{ij}^\alpha=\frac{\partial s_j^\alpha}{\partial x^i}-\frac{\partial s_i^\alpha}{\partial x^j}+\Gamma_{i\beta}^\alpha s_j^\beta-\Gamma_{j\beta}^\alpha s_i^\beta$$ you can contract with $X^iY^j$ to get (using product rule from calculus twice) $$(d^\nabla s)(X,Y)^\alpha=\Big(\frac{\partial (s(Y))^\alpha}{\partial x^i}-s_j^\alpha\frac{\partial Y^j}{\partial x^i}\Big)X^i-\Big(\frac{\partial (s(X))^\alpha}{\partial x^j}-s_i^\alpha\frac{\partial X^i}{\partial x^j}\Big)Y^j+\Gamma_{i\beta}^\alpha s(Y)^\beta X^i-\Gamma_{j\beta}^\alpha s(X)^\beta Y^j.$$ The first and fifth terms form the definition of $\nabla_X(s(Y))$, the third and sixth terms form the definition of $\nabla_Y(s(X))$, and the second and fourth terms define $-s([X,Y]).$ So you have $$(d^\nabla s)(X,Y)=\nabla_X(s(Y))-\nabla_Y(s(X))-s([X,Y]).$$<|endoftext|> TITLE: Independent families on $\omega$ with an additional splitting property QUESTION [9 upvotes]: Let $\mathscr{F}$ be an independent family, that is, a collection of subsets of $\omega:=\{0,1,\ldots\}$ such that, for all integers $n\ge 1$ and all distinct $X_1,\ldots,X_n \in \mathscr{F}$, then $$ \forall (e_1,\ldots,e_n) \in \{0,1\}^n,\quad X_1^{e_1} \cap \cdots \cap X_n^{e_n} \,\,\text{ is infinite} $$ where $X^0:=X$ and $X^1:=\omega\setminus X$ for all $X\subseteq \omega$. It is well known that independent families of cardinality $\mathfrak{c}$ exist, see e.g. here for eight different proofs of this fact (Theorem 3.4). Independent families with additional properties (e.g., selective independent families) have been studied, e.g., here and here. Here we ask about a variant of these properties: Question. Let $\Omega$ be the collection of all $X\subseteq \omega$ which are neither finite nor cofinite. Does there exist an independent family $\mathscr{F}$ of cardinality $\mathfrak{c}$ and a bijection $f: \Omega \to \mathscr{F}$ such that $$ \forall X\in \Omega, \quad X^0 \cap f(X) \,\,\text{ and }\,\, X^1 \cap f(X)\,\,\text{ are both infinite}\,? $$ It is clear that if $\mathscr{F}$ is an arbitrary independent family of cardinality $\mathfrak{c}$ and $f: \Omega \to \mathscr{F}$ an arbitrary bjection, then the answer is negative: indeed, if $f(X)\subseteq X$ for some $X\in \Omega$, then $X^1 \cap f(X)$ would be empty. REPLY [4 votes]: Edit: My previous answer had a mistake. Here is an updated answer which still contains some hopefully useful information: Assuming $\operatorname{cov}(\mathcal{M}) = \mathfrak{d}$, there is a positive answer. Recall that $\mathfrak{d}= \kappa$ is least such that there is a sequence $\langle K_\alpha : \alpha < \kappa\rangle$ of compact subsets of $2^\omega$ such that $\bigcup_{\alpha < \kappa} K_\alpha = \Omega$. Consider the partial order $\mathbb{P}$ consisting of monotone functions $h \colon 2^{\leq n} \to 2^{<\omega}$, for some $n \in \omega$, ordered by extension. If $G$ is $\mathbb{P}$-generic over $M$ and $H := \bigcup_{h \in G} h$, then $f \colon 2^\omega \to 2^\omega$ such that $f(x) = \bigcup_{n \in \omega} H(x\restriction n)$ is a continuous bijection between $2^\omega$ and $f''2^\omega$. We will recursively construct continuous functions $f_\alpha \colon K_\alpha \to \Omega$ for $\alpha < \kappa$. Namely, given $\langle f_\beta: \beta < \alpha \rangle$ for $\alpha < \kappa$, let $M_\alpha$ be an elementary submodel of some large $H(\theta)$ of size $\vert \alpha \vert < \kappa$ containing all $f_\beta$ for $\beta < \alpha$ and $K_\alpha$. Force with $\mathbb{P}$ over $M_\alpha$ and get a generic continuous function $f \colon 2^\omega \to 2^\omega$. This is possible since $\mathbb{P}$ is a countable poset and $\alpha < \operatorname{cov}(\mathcal{M})$. Let $f_\alpha := f \restriction K_\alpha$. Finally, let $F \colon \Omega \to \Omega$ be such that $F(x) = f_\alpha(x)$ for $\alpha$ least such that $x \in K_\alpha$. To see that $F''\Omega$ is independent we note that: Claim Let $c$ be Cohen over $M$ and $B \in M$ an analytic independent family. Then $B \cup \{ c \}$ is independent. This is Lemma 5.8 here and is essentially due to an argument of Arnie Miller that can be found in here. Since $f_\alpha(x_0), \dots, f_\alpha(x_k)$ are mutually Cohen generic over $M_\alpha$, any Boolean combination is Cohen as well and can be added to any finite union of the compact sets $f_\beta ''K_\beta \in M$, for $\beta < \alpha$, to form an independent set. Thus also the union of all $f_\beta '' K_\beta$, $\beta \leq \alpha$ is independent. Finally, to see that $x \in K_\alpha$ hits and avoids $f_\alpha(x)$ infinitely, make the following genericity argument: Let $h \in \mathbb{P}$ be arbitrary, say the domain of $h$ is $2^{\leq n}$ and the range is contained in $2^{\leq m}$. By compactness, there is $k > \max(m,n)$ such that for any $x \in K_\alpha$, $x(i) = 1$ for some $i \in (\max(m,n),k]$. We can simply extend $h$ to $h'$ of domain $2^{\leq k}$ such that $h'(s)$ hits $s$ above $\max(m,n)$ for every $s \in 2^{\leq k}$. Similarly for avoiding instead of hitting. Since $K_\alpha \in M_\alpha$, this genericity argument can be made over $M_\alpha$.<|endoftext|> TITLE: Reference request: Generic k3 surface has Picard number 1 QUESTION [5 upvotes]: I keep running into the statement that "the generic k3 surface has Picard rank 1". For instance the answer of this question (end) and this paper (following Example 1.1) or this paper (proof of Theorem 3.5) all state this fact without reference. I simply cannot track down a reference for this (I tried e.g. Huybrechts) -- perhaps because it is very simple. I would greatly appreciate a reference, or if very simple an argument. REPLY [11 votes]: Welcome new contributor. I am just writing my comment as an answer, and expanding on the observation of Prof. Arapura. For a smooth, projective scheme $X$ over a field $k$, the space of first order deformations of $X$ as a $k$-scheme is naturally isomorphic to the $k$-vector space $H^1(X,T_X)$. For a locally free sheaf $E$ on $X$ of rank $r>0$, the "Atiyah extension" is an element in $\text{Ext}^1_{\mathcal{O}_X}(E,E\otimes_{\mathcal{O}_X}\Omega^1_{X/k})$ whose "characteristic polynomial" has coefficients in $\text{Ext}^r_{\mathcal{O}_X}(\mathcal{O}_X,\Omega^r_{X/k}) = H^r(X,\Omega^r_{X/k})$ that are equal to the images of the degree $r$ Chern classes of $E$ under the cycle class map from $\text{CH}^r(X)$ to the "de Rham cohomology groups" $H^r(X,\Omega^r_{X/k})$. In particular, the "pairing" of the Atiyah extension with an element of $H^1(X,T_X)=\text{Ext}^1_{\mathcal{O}_X}(\Omega^1_{X/k},\mathcal{O}_X)$ gives an element in $\text{Ext}^2_{\mathcal{O}_X}(E,E)$. As proved in references on deformation theory (e.g., Illusie's book on the cotangent comples or Grothendieck's earlier LNM on the good 2-term truncation of the cotangent complex), this element is the obstruction to lifting $E$ to a locally free sheaf on the corresponding first-order deformation of $X$. Altogether, this defines a $k$-linear map, $$ H^1(X,T_X) \to \text{Hom}_k(\text{Ext}^1_{\mathcal{O}_X}(E,E\otimes_{\mathcal{O}_X}\Omega^1_{X/k}), \text{Ext}^2_{\mathcal{O}_X}(E,E)). $$ Now consider the case when $E$ is an invertible sheaf. The $k$-linear map above reduces to the form, $$ H^1(X,T_X) \to \text{Hom}_k(H^1(X,\Omega^1_{X/k}),H^2(X,\mathcal{O}_X)). $$ For a polarized K3 surface $X$ over $\mathbb{C}$, using the trivialization of $\Omega^2_{X/\mathbb{C}}$, this map is "equivalent to" the usual cup-product pairing, $$ H^1(X,\Omega^1_{X/\mathbb{C}}) \to \text{Hom}_{\mathbb{C}}(H^1(X,\Omega^1_{X/\mathbb{C}}),H^2(X,\Omega^2_{X/\mathbb{C}}). $$ The cup-product pairing is a perfect pairing. Thus, this $\mathbb{C}$-linear map is an isomorphism of $\mathbb{C}$-vector spaces. Finally, since $H^1(X,\mathcal{O}_X)$ vanishes for a K3 surface, the cycle class map from the Picard group to $H^1(X,\Omega^1_{X/\mathbb{C}})$ is injective (in characteristic $p$, the kernel of this cycle class map obviously contains the $p$-power image of the Picard group, so this is one place characteristic $0$ is crucial). Thus the induced $\mathbb{C}$-linear map, $$ H^1(X,T_X) \to \text{Hom}_{\mathbb{Z}}(\text{Pic}(X),H^2(X,\mathcal{O}_X)), $$ is surjective. In particular, if the Picard rank is at least $1$, if we fix a saturated rank $1$ sublattice, there are first order deformations such that the corresponding "obstruction map" on $\text{Pic}(X)$ has kernel precisely equal to this saturated rank $1$ sublattice. Since also $H^2(X,T_X)$ vanishes, these first order deformations extend. Therefore, there are deformations of the K3 surface over a $1$-dimensional base such that the Picard lattice of a very general member of the family equals the saturated rank $1$ sublattice, i.e., the Picard rank is $1$. In applications, typically we choose the saturated rank $1$ sublattice to be generated by an ample divisor class, so that the deformation is a family of projective K3 surfaces (rather than just Kaehler K3 surfaces that are not projective). Some part of this is included in the proof of Claim 3.5 of my note about Artin's axioms. Artin's axioms, composition and moduli spaces http://www.math.stonybrook.edu/~jstarr/papers/moduli4.pdf<|endoftext|> TITLE: What is the distribution of $2M_1-B_1$ where $M_t$ is the maximum process of the the Brownian motion $B_t$ QUESTION [5 upvotes]: Let $B_t$ be a standard Brownian motion and let $M_t:=\sup _{s\le t}B_s$ be the maximum process. What is the distribution of $2M_1-B_1$? is it elementary? REPLY [4 votes]: Yes, the pdf of this distribution is \begin{equation} u\mapsto 2u^2 f(u)\,1(u>0) \tag{1} \end{equation} where $f$ is the standard normal pdf. Indeed, by Proposition 2, \begin{equation} G(m,b):=P(M_1>m,B_1\le b)=P(B_1>2m-b)=1-F(2m-b) \end{equation} for real $m,b$ such that $m>b_+:=\max(0,b)$, where $F$ and $f$ are the standard normal cdf and pdf, respectively. So, for the joint pdf $g$ of $(M_1,B_1)$ we have \begin{equation} g(m,b)=-\frac{\partial^2 G(m,b)}{\partial m\,\partial b} =\frac{\partial^2 F(2m-b)}{\partial m\,\partial b} =2(2m-b)f(2m-b) \end{equation} if $m>b_+$, with $g(m,b)=0$ otherwise. So, for \begin{equation} U:=2M_1-B_1 \end{equation} and all real $u>0$ we have \begin{equation} \begin{aligned} &P(U\le u) \\ &=\iint_{\mathbb R^2}dm\,db\,g(m,b)\,1(m>b_+,\,2m-b TITLE: What tools should I use for this problem? QUESTION [16 upvotes]: Suppose we have $d$ cylindrical metal bars, with radius $l$, attached orthogonal to a support in random places: Now we have to attach bars with radius $k$ EVENLY SPACED, with distance $p$ between their centers, in the same support and without any bar being in top of other: Determine the possible values for $p$. ($k$ is not necessarily greater than $l$.) (The way that I think the) Problem: Given a increasing finite sequence $\{a_n\}_{n=1}^d$ of $d$ real numbers and positive real numbers $l$ and $k$, find $p$ and $q$ such that the sequence $\{b_n\}_{n=0}^\infty=\{pn+q\}$ has the property: $$(a_i-l,a_i+l)\cap(b_j-k,b_j+k)=\emptyset\quad \forall i,j.$$ That is: we have open intervals around each $a_i$ with size $2l$ and we must find $p$ and $q$ such that the intervals with size $2k$ around numbers of the form $\{pn+q\}$ has no intersection with the first ones. Question: How should I approach the problem to find ALL nontrivial solutions? (Note that $p>a_d-a_1$ will trivially solve the problem.) Edit: I was not clear about looking for all nontrivial solutions. (Edit) What I expect as an answer: A path to the solution to the problem, that is: an expression for $p$ and $q$ depending on the given data. Something like: "You formulate in the manner A, use results B and calculate C. I leave for you the calculations and minor details." Edit: As user Timothy Chow pointed out, to give a bound on the number $d$ of bars can drastically make a difference on the possibility for such expression. So consider that $d \le 15$. About the problem: A student of my class on differential equations asked me that problem and said that would be very helpful to have a 'very nice formula' for $p$ and $q$. He is doing university in engineering and already works in a big company here (Brazil). At first it appeared to me to be a 'very simple-trivial-easy' problem because of the finiteness of the first sequence, but the fact that the $a_n$'s are not necessarily evenly spaced make all too hard. About the post itself: I had no idea of how to ask for help with that problem, so I asked in meta.mathoverflow about it and user @fedja very kindly answered me how to do it. Any help with a better title and appropriate tags will be very welcome. My attempt to solve the problem: The problem looks like an optimization problem, so I tried to define a function $F(p,q)$ such the solutions correspond to the minimum of this function. One option is: $$F(p,q)=\int_{\mathbb{R}}f(x)g(x)\,dx,$$ where $$f(x) = \begin{cases} 1, & \text{if $x \in (a_i-l,a_i+l)$ } \\ 0, & \text{if $x \notin (a_i-l,a_i+l)$ } \end{cases}$$ and $$g(x) = \begin{cases} 1, & \text{if $x \in (b_i-k,b_i+k)$ } \\ 0, & \text{if $x \notin (b_i-k,b_i+k)$ } \end{cases}.$$ Certainty $F(p,q)\ge0$ $\forall p,q$. Also $F(p,q)=0$ is a solution to the problem. As $f(x)\neq0$ only for a finite number of finite intervals this becomes $$F(p,q)=\sum_{n=0}^d\int_{a_n-l}^{a_n+l}g(x)\,dx.$$ That $g(x)$ can be written as a Heaviside function applied to a cosine function: $$g(x)=H\left(\cos\left(\frac{2\pi}{p}x-\frac{2\pi}{p}q\right)-\cos\left(\frac{2\pi k}{p}\right)\right).$$ So we have $$F(p,q)=\sum_{n=0}^d\int_{a_n-l}^{a_n+l}H\left(\cos\left(\frac{2\pi}{p}x-\frac{2\pi}{p}q\right)-\cos\left(\frac{2\pi k}{p}\right)\right) \, dx.$$ For this integral I tried to use analytic approximations to the step function like the ones in the Wikipedia article, but none of the expressions seems to lead to an integral that can be expressed in closed form. I still have a feeling that this problem could be solved in closed form. Since I had my master degree (about 10 years ago) I had never again contact to some areas of math like abstract algebra (fields, rings, etc). In all that time I essentially taught Linear Algebra and Differential Equation courses and just started my PhD in Physics, so all my experience is way too limited. I feel like someone with experience in Algebra would say "Oh, that is very simple application of Fermat's little theorem and euclidean division...". Or may be some one with experience in Measure theory "It's a easy calculation with the Lebesgue-Sobolev formula...". Any help will be very welcome! REPLY [3 votes]: If $m=k+\ell$, then the question reduces to when the intervals $(a_i-m,a_i+m)\mod p$ do not cover the full interval $[0,p]$. The answer is, of course, a union of finitely many intervals and the interesting question is how many they can be and how to compute them fairly quickly. Note that the interesting range of $p$ is $[2m,2md]$ ($p<2m$ gives just too dense set to avoid a single initial bar and $p>2md$ is long enough to admit $q$ by simple length comparison). It may be convenient to denote $\eta=1/p$ and to ask for which eta the intervals $(\eta a_j-\eta m, \eta a_j+\eta m)\mod 1$ cover or do not cover $[0,1]$. The first (unfortunate) observation is that the number of constituting intervals in that set can be arbitrarily large even for $d=2$. Indeed, if you think of $a_1=0$ and $a_2=a\gg 1$ with $m=\frac 14$, say, then we have one stationary slowly expanding interval $(-0.25\eta,0.25\eta)$ and the other interval that goes around at the huge speed $a$ and also slowly expands when $\eta$ goes from $1$ to $2$ (the interesting range). That crazy rotation gives us about $a$ switches from covering to not covering, so no "linear in $d$" (or even depending on $d$ alone) computation time is feasible. The number of intervals can certainly be as large as $\frac{a_d-a_1}{m}$. Fortunately, we can easily describe all possble endpoints of those intervals. They are all of the kind $\frac{a_j-a_i\pm 2m}{n}$ where $i\le j$ and $n$ is a positive integer. To fit into the interval $[2m,2md]$ we must have $n\le\frac{a_j-a_i}m+2$, so we have about $\frac{a_d-a_1}m$ choices of $n$ and about $d^2$ choices of $i,j$. On each interval between these endpoints the answer to the covering question remains the same, so we can just pick a particular value of $p$ for each interval and check what happens with that $p$. This is not hard but still takes $d\log d$ time (sorting the intervals modulo $p$ in the increasing order and going left to right over the resulting partition in the search of an empty spot). Thus, with this naive approach, the running time becomes $$ \frac{a_d-a_1}{m}d^3\log d\,. $$ As I showed, the first factor cannot be avoided, so the only question is whether we can reduce the second (depending of $d$) factor to something smaller.<|endoftext|> TITLE: Another generalization of parity of Catalan numbers QUESTION [6 upvotes]: Recently, a question by T. Amdeberhan gathered up many enjoyable proofs that a Catalan number $C_n$ is odd if and only if $n=2^r-1$. Noam D. Elkies' answer considered $F=\sum_{n=0}^\infty C_n x^{n+1}$. Using the facts that this generating function satisfies $F=x+F^2$, and that $(a+b)^{2^m} \equiv a^{2^m}+b^{2^m} \pmod 2$, he showed that $$F \equiv x+x^2+x^4+x^8+x^{16} + ... \pmod 2,$$ proving the assertion. Now, please consider the generating functions $F_j$ defined to satisfy $F_{j} = x+(F_{j})^j$ with $j$ an integer greater than $1$. Define $C^{(j)}_n$ by $F_{j} = \sum_{n=0}^\infty C^{(j)}_n x^{n+1}$, with $C^{(j)}_0 = 1$. My question is: for which $j$ does the following statement $S$ hold? $$C^{(j)}_n \equiv 1 \pmod j \iff n = j^r - 1 \tag{S}$$ Here are my thoughts. First, Noam D. Elkies' answer for $j=2$ generalizes immediately for the case of $j=p$, where $p$ is prime. This is because $(a+b)^{p^m} \equiv a^{p^m} + b^{p^m} \pmod p$, which follows by induction using the key step of $(a+b)^p \equiv a^p + b^p \pmod p$. This last assertion follows from $\binom{p}{l} \equiv 0 \pmod p$ for $00$, then $k\mid i$ and $d_{tk}=1$ for all $t\leq \frac{i}k$. Hence, $m=(r-1)k$ and thus $n=\frac{j^r-1}{j-1}$. Vice versa, if $n=\frac{j^r-1}{j-1}$ for some $r$, then we can use generalization of Lucas theorem given by Theorem 1 in Granville (1997) to conclude that $\binom{jn+1}{n}\equiv 1\pmod{j}$. Namely, it can be seen that in our case we have either $M_i=0$ or $R_i=0$ for all $i$, and thus in the r.h.s. of the congruence given by this theorem all terms equal to 1. So, it remains to address the case when $j$ is not a prime power.<|endoftext|> TITLE: Maximally continuous extension of continuous functions from $\mathbb Q$ to $\mathbb R$ QUESTION [10 upvotes]: Let $f: \mathbb Q \to \mathbb R$ be a continuous function. An extension of $f$ is a function $\tilde f: \mathbb R \to \mathbb R$ such that $\tilde f = f$ on $\mathbb Q$. We say an extension $\tilde f$ of $f$ is maximally continuous if for any other extension $g$ of $f$, we have that if $g$ is continuous at $x \in \mathbb R$, then so is $\tilde f$. Question: For any continuous function $f: \mathbb Q \to \mathbb R$, does there exist a maximally continuous extension of $f$? Remark: One can always obtain an extension that is continuous at every point in $\mathbb Q$, see for example, the answer in this post by Fedor Petrov. REPLY [8 votes]: There even exists a largest set $X$ to which $f$ can be continuously extended. The trick is the following result (which I state here in more generality, to point out which topological assumptions one needs): Theorem. Let $X,Y$ be topological spaces, where $Y$ is $T_3$, and let $D \subseteq X$ be dense. Let $f: D \to Y$ be continuous and assume that the following property is satisfied for each point $x \in X \setminus D$ (equivalently, each point $x \in X$): $(*)$ There exists $y_x \in Y$ such that, for each net $(x_j)$ in $D$ that converges to $x$, the net $(f(x_j))$ converges to $y_j$. Then $f$ has a (obviously, unique) continuous extension to $X$. (I gave a proof of this in a topology course a year ago, but unfortunately I don't know a reference - maybe somebody else can help out with a reference?) Application to your situation. For $D = \mathbb{Q}$, let $X$ be the set of all $x \in \mathbb{R}$ which satisfy property $(*)$. Then $f$ can be continuously extended to $X$, and this extension is clearly the largest one.<|endoftext|> TITLE: Example of usual Laplacian does not respect bidegree for general hermitian manifolds QUESTION [5 upvotes]: We know that the Kähler identity $\Delta=2\Delta_{\partial}=2\Delta_{\bar{\partial}}$ on a Kähler manifold $(X,g)$ implies that the usual Laplacian $\Delta:=dd^*+d^*d$ respects the bidegree, i.e. for any $(p,q)$-form $\alpha$ on $(X,g)$, the form $\Delta(\alpha)$ is still a $(p,q)$-form. But for general hermitian manifolds, $\Delta$ does not respect bidegree. Are there any famous concrete examples for this? Another question is: if $\Delta$ respects bidegrees, can we show that $(X,g)$ is Kähler? Or is there any non-Kähler hermitian manifold, whose Laplacian respects bidegrees? REPLY [4 votes]: If the Hermitian metric is not Kähler, the Laplacian won't respect bi-degree. In fact, a more general result is true: If an almost Hermitian metric is not Kähler, then its Laplacian will not respect bi-degree. Here's a simple explicit (i.e., 'concrete') example: Let $X = \mathbb{C}^2$ with standard holomorphic coordinates $z_1 = x_1 + i\,x_2$ and $z_2=x_3 + i\,x_4$. Now let $f$ be a smooth function on $X$ and consider the positive definite $(1,1)$-form $$ \omega = \mathrm{e}^{2f}\mathrm{d}x_1\wedge \mathrm{d}x_2 + \mathrm{d}x_3\wedge \mathrm{d}x_4 $$ with associated Hermitian metric $g = \mathrm{e}^{2f}({\mathrm{d}x_1}^2 + {\mathrm{d}x_2}^2) + {\mathrm{d}x_3}^2 + {\mathrm{d}x_4}^2$. It's a straightforward calculation to show that $\Delta\omega$ has type $(1,1)$ if and only if $$ \frac{\partial^2f}{\partial x_2\partial x_3} +\frac{\partial^2f}{\partial x_1\partial x_4} = \frac{\partial^2f}{\partial x_1\partial x_3} -\frac{\partial^2f}{\partial x_2\partial x_4} = 0. $$ (These are the conditions that the (2,0) and (0,2) pieces of $\Delta\omega$ vanish.) The point is that, because $\omega = \ast\omega$, the formula for the Laplacian simplifies to $$ \Delta\omega = -d{*}d{*}\omega-*d{*}d\omega = -(1+*)(d{*}d\omega), $$ so $\Delta\omega$ is minus twice the projection of $d{*}d\omega$ into its self-dual part, which, since it is self-dual, is the sum of a multiple of $\omega$ itself with the real part of a $(2,0)$-form. The above two equations represent the real and imaginary parts of the coefficient of the $(2,0)$ part. To answer the OP's final question, I'll state the following result, which is stronger than needed: Proposition: Let $(M,J,g)$ be an almost Hermitian manifold and suppose that its $g$-Laplacian $\Delta$ carries $(0,1)$-forms to $(0,1)$-forms. Then $J$ is $g$-parallel, i.e., $(M,J,g)$ is Kähler. Proof: (sketch) Let $\alpha\in\Omega^1(M)$ be any $1$-form on $M$. Define $J^*\alpha$ by the property $J^*(\alpha)(v) = \alpha(Jv)$. Since, by hypothesis, $\Delta$ carries $(0,1)$-forms to $(0,1)$-forms, it follows that $\Delta$ carries $(1,0)$-forms to $(1,0)$-forms, and in particular, it follows that $\Delta(J^*\alpha) = J^*(\Delta\alpha)$. Now, it's a straightforward fact that if $L:TM\to TM$ is any smooth vector bundle map (it doesn't have to be a complex structure or compatible with $g$), then $\Delta(L^*\alpha) = L^*(\Delta\alpha)$ holds for all $1$-forms $\alpha$ if and only if $L$ is parallel with respect to the metric $g$, i.e., its covariant derivative with respect to $g$ (when $L$ is thought of as a section of $TM\otimes T^*M$), vanishes. [The point is that the difference $\Delta(L^*\alpha) - L^*(\Delta\alpha)$ is first order in $\alpha$, and it's enough to check what the condition is on the covariant derivatives of $L$ at one point $p$ when one considers the 1-forms $\alpha$ that vanish at $p$. Then it's easy.] Once one knows this, then the desired result immediately follows, since an almost Hermitian structure $(M,J,g)$ is Kähler if and only if $J$ is parallel with respect to $g$. $\square$ Note: I had forgot about this argument when I answered this question last December. In fact, I showed this argument to Jeremy Daniel and Xiaonan Ma back in 2013, who incorporated it into their arXiv paper Characteristic Laplacian in sub-Riemannian geometry as Remark 2.3 and Lemma 2.4<|endoftext|> TITLE: Expected number of compositions needed to get constant function QUESTION [12 upvotes]: This is somewhat inspired by Factoring a function from a finite set to itself. Fix natural number $n$ and let $[n] := \{1,2,\ldots,n\}$. Set $g_0 \colon [n]\to [n]$ to be the identity, and for $i \geq 1$ define $g_i := f_i \circ g_{i-1}$ where the $f_i\colon [n] \to [n]$ are chosen (independently and) uniformly at random among all functions from $[n]$ to $[n]$. What is the expected value of the smallest $t$ for which $g_t$ is a constant function? (More generally, what is the distribution of this random variable $t$?) EDIT: As Peter Taylor explained, it is easy to view this also as a Markov chain on $[n]$ where at time $t$ our state $a_t$ is the size of the image of $g_t$. And as I mentioned in the comments then the trajectory of this Markov chain $(a_1-1,a_2-1,\ldots)$ gives a random partition with part sizes $\leq n-1$; the expected time to a constant function is the expected length of this partition. There is also a natural $q$-analog of this problem, where instead of random functions $[n]\to [n]$ we look at random linear functions $\mathbb{F}_q^n\to \mathbb{F}_q^n$. This gives a Markov chain on $\{0,1,\ldots,n\}$ where our state $a_t$ is the dimension of the image of $g_t$. Of course now the transition probabilities of the Markov chain involve the parameter $q$ (and should recover the previous case with $q=1$). REPLY [5 votes]: This question was completely settled by J.A. Fill here: https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.8.641 REPLY [4 votes]: I expect the answer is $(2+o(1))n$. As Peter Taylor says: We have a Markov process on the set $[n]:=\{ 1,2, \ldots, n \}$ where the transition probability from $i \to j$ is the probability that a randomly selected function from $[i] \to [n]$ will have image of size $j$. Fix some $m \geq 2$ and consider running this Markov process starting at $m$. I will show that the expected time to reach $1$ is $(2-2/m+o(1)) n$. For $k$ fixed, the probability of the transition $k \to k$ is $1-\binom{k}{2} \tfrac{1}{n} + O(1/n^2)$, the probability of a transition $k \to k-1$ is $\binom{k}{2} \tfrac{1}{n} + O(1/n^2)$ and the probability of a transition $k \to \ell$ for $\ell \leq k-2$ is $O(1/n^2)$. Consider the simplified process where the probability of transitioning $k \to k$ is $1-\binom{k}{2} \tfrac{1}{n}$, the probability of $k \to k-1$ is $\binom{k}{2} \tfrac{1}{n}$ and the probability of $k \to \ell$ for $\ell \leq k-2$ is $0$. The expected time for this process to go from $k$ to $k-1$ is $\tfrac{2n}{k(k-1)}$ so the expected time to go from $m$ to $1$ is $2n \sum_{k=2}^m \tfrac{1}{k(k-1)} = 2n (1-1/m)$. We can think of the original process as applying the simplified process, and then changing our mind with probability $O(1/n^2)$ at each step. But since we only take $O(n)$ steps, the probability that we ever change our mind is $O(1/n)$, so we have the same expected time for the original process as for the simplified process. (I am skipping over some details, but I'm pretty sure I can fill them in.) Now, it is tempting to send $m \to \infty$ and conclude that the expected time from $n$ to $1$ is $(2+o(1))n$. I think you should be able to rigorously prove a lower bound by this route without working too hard. I want to provide nonrigourous arguments that $2$ actualy is the right constant. Fix $\beta$ in $[1, \infty)$ and suppose the Markov process is at position $n/\beta$. The probability that a fixed element in $[n]$ is in the image of a random map $[n/\beta] \to [n]$ is $1-(1-1/n)^{n/\beta} \approx 1-e^{-\beta^{-1}}$. So, roughly, the Markov process goes from $\alpha n$ to $(1-e^{-\beta^{-1}})n=n/(1-e^{-\beta^{-1}})^{-1}$. The iteration $\beta \mapsto (1-e^{-\beta^{-1}})^{-1}$ approaches $\infty$. So, if we fix some $R>0$, I expect the Markov process to get below $n/R$ in a finite number of steps. Now, how long should I expect the transition from $n/R$ to $n/S$ to be, if $R < S$ are fixed and large? We have $$(1-e^{-\beta^{-1}})^{-1} = \beta + 1/2 + O(\beta^{-1}) \ \text{as}\ \beta \to \infty.$$ This suggests that the time to go from $\beta = R$ to $\beta=S$ is roughly $2(S-R)$. Sending $S \to n$ suggests the answer to the original question should be $(2+o(1))n$. (Here I am not at all sure I can fill in the details.) I haven't actually given a proof, but I've analyzed both the part of the Markov chain where $k = O(1)$, and the part where $k \sim n$, and found consistent results.<|endoftext|> TITLE: The source of the Integral QUESTION [7 upvotes]: Wolfram alpha calculates the integral $$\int\limits_0^\infty \frac{x^2\ln{x}}{e^x-1}dx=2\zeta^\prime(3)+3\zeta(3)-2\gamma\zeta(3).$$ However, I need to cite the source of this identity (the table of integrals, or the article where this integral was calculated). Could you indicate any? REPLY [18 votes]: One way to get the claimed value of the given integral, $J$ in notation below, is by starting from the standard relation $$ \begin{aligned} \zeta(s) &= \frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx\ ,\qquad \text{ so }\\ \zeta'(s) &= \frac\partial{\partial s} \left(\ \frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx \ \right) \\ &= \frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}\; \ln x}{e^x-1}\; dx - \underbrace{\frac{\Gamma'(s)}{\Gamma(s)}}_{=\psi(s)}\zeta(s)\ ,\qquad\text{leading to } \\ \zeta'(3)&=\frac 1{\Gamma(3)}\underbrace{\int_0^\infty \frac {x^2\; \ln x}{e^x-1}\; dx}_{\text{our integral }J} - \psi(3)\zeta(3)\ . \\[2mm] &\qquad\text{Extracting $J$ from above,}\\[2mm] J &=\Gamma(3)\; \zeta'(3) \ +\ \Gamma(3)\; \psi(3)\; \zeta(3) \\ &=2\zeta'(3)+(3-2\gamma)\zeta(3)\ , \end{aligned} $$ where we have used the relations $\displaystyle\psi(x+1)=\frac 1x+\psi(x)$ and $\psi(1)=-\gamma$, e.g. from here. Explicitly, this gives the value for $\displaystyle\psi(3)=\frac 12+\psi(2)=\frac 12+1+\psi(1)=\frac 12(3-2\gamma)$. $\square$ The above shows how to get similar relations when there is an other power of $x$ instead of $x^2$ in the numerator of then integrand, and gives an interpretation of the involved coefficients.<|endoftext|> TITLE: How many non-isomorphic abelian subgroups of the permutation group $S_n$? QUESTION [11 upvotes]: I am interested in how many (pairwise non-isomorphic) subgroups of the permutation group $S_n$ are abelian. ($n \in \mathbb{N}$ arbitrary and possibly big) Are you aware of any references which treat this problem? REPLY [2 votes]: This is a (cw) post to illustrate Claim 2 from the answer by Sean Eberhard. Below is the plot of the list $(\frac1n,\frac{\log f(n)}{\pi\sqrt{2/3}(n/\log n)^{1/2}})$ for $n$ from $1000$ to $10000$. I used the list of values from OEIS A023893. As you see after $n$ about 3000 the sequence sort of changes behavior; at about $n=5000$ (to be precise, at $5226$) it achieves a (local ?) maximum and then starts to drop with increasing speed. It looks so strange that I suspected some error in the computation of values. But then I recomputed it myself in Mathematica using power series expansion and obtained precise coincidence. Update I now extended computations to $n\leqslant100000$ and it seems that the tendency continues, so that at least in the range from $10000$ to $100000$ $\log f(n)$ grows qualitatively slower than $\sqrt{\frac n{\log n}}$... Update 2 With the constant $2\pi/\sqrt3$ instead of $\pi\sqrt{2/3}$ it is<|endoftext|> TITLE: On the fundamental solution for elliptic PDE QUESTION [5 upvotes]: In the well known paper by Littman-Weinberger-Stampacchia "Regular points for elliptic equations with discontinuous coefficients", the authors were able to prove the validity the following statement: given a bounded, measurable and uniformly elliptic matrix $A$ (roughly speaking $\lambda\,\mathrm{Id}\leq A \leq L \,\mathrm{Id}$, for some $\lambda \leq L$) and $\Gamma$ the fundamental solution in $\mathbb{R}^n$, that is $\Gamma\geq 0$ and $$\label{ciap} -\mathrm{div}(A\nabla \Gamma)=\delta_0 \quad\mbox{in }\mathbb{R}^n $$ then $$ \frac{C_1}{|x|^{n-2}}\leq \Gamma(x)\leq \frac{C_2}{|x|^{n-2}}\quad\mbox{in }\mathbb{R}^n, $$ for some $C_1,C_2$ depending only on $n,\lambda$ and $L$. I was wondering if it is true that, if $A$ is smooth, then $$ |\nabla \Gamma|(x)\leq \frac{C_3}{|x|^{n-1}}\quad\mbox{in }\mathbb{R}^n, $$ for some $C_3$ depending only on $n,\lambda$ and $L$. Indeed, in the book "Harmonic Analysis Techniques for Second Order Elliptic Boundary Value Problems" by Kenig, he states in Theorem 1.2.8 that a similar estimate holds for the Green function of operator with smooth coefficient $A$ in smooth domain $\Omega$ for some constant $C_3$ depending only on $n, \lambda $ and $L$ (he does not mention the possible dependence from the diameter of $\Omega$ but the proof is omitted): if $-\mathrm{div}(A\nabla G)=\delta_0$ in $B_1$ and $G=0$ on $\partial B_1$ then $$ |\nabla G|(x)\leq \frac{C_3}{|x|^{n-1}}\quad\mbox{for }x \in B_1. $$ Otherwise, is there a reference where one could find the proof of this last result? REPLY [5 votes]: In the paper The Green Function for uniformly elliptic equations Manuscripta Mathematica Gruter & Widman proved - among other things - pointwise estimates about the gradient of the Green function under the Dini continuity assumption on the leading coefficients. This is in section 3 Theorem 3.3<|endoftext|> TITLE: Is the injective envelope functorial? QUESTION [6 upvotes]: Let $A$ and $B$ be unital $C^*$-algebras, so we can view these as operator systems, and it makes sense to consider their injective envelopes $I(A)$ and $I(B)$. These injective envelopes become $C^*$-algebras for the Choi-Effros product. Given a unital $*$-morphism $f: A \to B$, is it true that there exists a unique unital $*$-morphism $\overline{f}: I(A) \to I(B)$ that extends $f$? Once the above question is answered positively (if the answer is positive), the following will probably be easy: Is this construction functorial? I.e. is $I(-)$ a functor from the category of unital $C^*$-algebras to the category of unital $C^*$-algebras (with morphisms unital $*$-homomorphisms? A reference is more than enough for me to be satisfied with an answer. REPLY [7 votes]: As mentioned by Chris, injective envelopes are brutal when seen as C$^*$-algebras. Let $A=\text{UHF}(2^\infty)$ and $B$ the hyperfinite II$_1$ factor. Take $f$ to be the inclusion map. We have $I(B)=B$, while $I(A)$ is a wild AW$^*$ factor of type III. If $g:I(A)\to B$ is a $*$-homomorphism and $\tau$ is the trace on $B$, then $\gamma=\tau\circ g$ is a trace on $I(A)$. In a type III AW$^*$-factor any projection $p$ can be halved, so there exist $p_1,p_2$ with $p=p_1+p_2$ and $p\sim p_1\sim p_2$, which gives us the usual $$ \gamma(p)=\gamma(p_1)+\gamma(p_2)=2\gamma(p), $$ and so $\gamma(p)=0$ for any projection $p$. Thus $\gamma=0$. As $\tau$ is faithful, $g=0$. In summary, $f$ is a $*$-monomorphism that admits no extension to a $*$-homomorphism, and in fact the only $*$-homomorphism $I(A)\to I(B)$ is the zero homomorphism.<|endoftext|> TITLE: Conjectures inspired by AI QUESTION [23 upvotes]: Today in Nature a paper described how AI guided mathematicians to make highly non-trivial conjectures, which they managed to prove, one in Knot Theory involving a new invariant, the other in Representation Theory. The proof of the former result is in this paper, while the latter result is in this paper. The github repository is here. What are other areas where AI-inspired conjectures have a great chance of being discovered? I would especially be interested by topics related to Dynamical Systems. REPLY [4 votes]: The PO seems to be especially interested in dynamical systems. ` `Here is an article where a deep learning Long Short-Term Memory (LSTM) network (see this reference for the architecture) can be used to reconstruct an underlying dynamical system from a set of data points without prior assumptions. It should be pointed out that most models generated by deep learners are not easy to understand. However, recent works goes in the direction of explainability across all Machine Learning, and there are tools even for deep networks (there is a substantil literature in this space, and also tools. For instance, if someone likes Pytorch, as I do, for his/her experiments, one may leverage SHAP or other tools, see this basic article here . So, summing up: a researcher could use a recurrent deep network to ' interpolate" data points and generate a non-linear dynamics, then leverage some explainer to extract a simplified and more human digestible model. As one can run all of the above in a distributed fashion, this sequence would provide a great help both in creating models, and testing out hypothesis.<|endoftext|> TITLE: example of "really" non-existent transferred model structure QUESTION [8 upvotes]: I am looking for an example where a transferred model structure fails to exist, even if one is willing to work with semi-model category. But let me be more precise: Let's say I have a combinatorial model category $C$, a locally presentable category $D$ and an adjunction : $$ L: C \rightleftarrows D : U$$ A classical (at least - mentioned on the nLab) necessary and sufficient condition (in this case) for the existence of a transferred model structure on $D$ is that one has the following two: (A) For every object $X \in D$ such that $U(X)$ is fibrant, there exists a "path object" $X \overset{a}{\to} P \overset{p}{\to} X \times X$ such that $U(a)$ is a weak equivalence and $U(p)$ is a fibration. (B) There exists a "fibrant replacement" functor and natural transformation $X \overset{a_x}{\to} FX$ on $D$, such that $U(FX)$ is fibrant and $U(a_x)$ is a weak equivalence. I know examples where condition (B) fails, but I can't find an example where (A) fails. Do you know one ? Some details and motivations: In practice, it appears that condition (A) is often almost free and condition (B) is the hard one. For example, if $C$ is a simplicial model category, $D$ is simplicially enriched (with cotensor) and the adjunction is a simplicial adjunction, you can take $P$ to be the cotensor $P = X^{\Delta[1]}$. The same applies with other enrichement. Now, it also appears that condition (A) is sufficient to build a "transferred model structure" on D, at least if one is willing to work with right semi-model category and slightly generalizing what one means by transferred model structure. So failure of condition (B) isn't really a deal breaker, but just an additional hassle. This being said, I can't find a single example where condition (A) fails. REPLY [4 votes]: The usual example in operad theory is when $C$ is a combinatorial, monoidal model category and $D$ is the category of commutative monoids in $C$. Unless $C$ satisfies a strong condition (that in my thesis, I called the commutative monoid axiom) guaranteeing symmetric powers are homotopically well-behaved, $D$ won't even have a semi-model structure. For example, if $C = Ch(\mathbb{F}_p)$ is chain complexes over a field $k$ of characteristic p, then it is easy to show that $D$ can't have a transferred semi-model structure. You know that, if it did, then the generating trivial cofibrations would be of the form $Sym(J)$ where $J$ is the set of generating trivial cofibrations in $C$, and $Sym$ is the free commutative monoid function (L in your notation). Recall that maps in $J$ look like $0\to D(n)$ where $D(n)$ is the chain complex with one copy of $k$ in degrees $n$ and $n-1$, and identity boundary map. Let's be quite explicit. Take $p=2$. Then $Sym(0)=k \to Sym(D(n))$ is not a weak equivalence, because if $y\in D(n)$ is non-zero then $y^2 \in Sym(D(n))$ is a cycle of degree $2n$ which is not a boundary. This is Example 3.7 in Model Categories and Simplicial Methods.<|endoftext|> TITLE: Is there an efficient generalized algorithm to find at least one binary word with the maximum rotational imbalance and the full $\{0, 1\}$-balance? QUESTION [7 upvotes]: Assuming that $x$ is a sequence of $l$ bits (i.e. a binary word of length $l$) and $0 \le m < l$, let $R(x, m)$ denote the result of the left bitwise rotation (i.e. the left circular shift) of $x$ by $m$ bits. For example, if $x = 0100110001110000$, then $l = 16$ and $$\begin{array}{l} R(x,0) = x = {\rm{0100110001110000}},\\ R(x,1) = {\rm{1001100011100000}},\\ R(x,2) = {\rm{0011000111000001}},\\ \ldots \\ R(x,15) = {\rm{0010011000111000}}. \end{array}$$ Let $A \oplus B$ denote the result of the bitwise “exclusive or” operation for two sequences of $l$ bits. For example, $$0100110001110000 \oplus 1010010001000010 = 1110100000110010.$$ Let $H(x)$ denote the number of non-zero bits in $x$ (i.e. the Hamming weight of $x$). Assuming that $x$ is an $l$-bit word, let $f(x)$ denote the minimal element (the smallest number) in the tuple $$\begin{array}{l} (H(x \oplus R(x, 1)),\\ H(x \oplus R(x, 2)),\\ \ldots, \\ H(x \oplus R(x, l - 2)),\\ H(x \oplus R(x, l - 1))). \end{array}$$ For example, $$\begin{array}{l} f(10011110100010010011) = 10,\\ f(11111111111111111111111111111111) = 0,\\ f(10000000000000000000000000000000) = 2,\\ f(10011100001111100000011111110000) = 8,\\ f(01000110111111100011100100100101) = 16. \end{array}$$ Question: assuming that for any even natural number $n \ge 2$ there exists a $2n$-bit word $x$ such that $f(x) = n$ and $H(x) = n$, what can be a time- and space-efficient algorithm which, given an arbitrary (even) $n \ge 2$, allows to find at least one such $x$? For example, if $n = 10$, it is easy to check all $20$-bit words and find (in the lexicographic order) $$\begin{array}{l} x_0 = 00000101011011001111,\\ x_1 = 00000101011110011011,\\ x_2 = 00000101101110011101,\\ x_3 = 00000101110011101101,\\ \ldots,\\ x_{718} = 11111010100001100100,\\ x_{719} = 11111010100100110000. \end{array}$$ But as the value of $n$ grows, it becomes infeasible to check the huge amount of elements, even if one skips any element $x$ such that $H(x) \neq n$. For example, how can one find a $256$-bit word $x$ such that $f(x) = H(x) = 128$? Let $S_n (n = 2, 4, 6, \ldots)$ denote cardinality of the set of solutions for the corresponding $n$. Note that each solution $x$ is an element of the family of solutions, and this family contains exactly $2n$ elements, namely, $x$ and $(2n-1)$ its rotations. So we can pick a single (e.g. lexicographically first) element of a family and call it a canonical solution (all other elements of the corresponding family can be generated from this solution in a trivial way). Then let $C_n (n = 2, 4, 6, \ldots)$ denote cardinality of the set of canonical solutions for the corresponding $n$. If my computations are correct, we have $$\begin{array}{l} S_2 = 4 = 2^2,\\ C_2 = 1 = 2^0,\\ S_4 = 48 = 2^5 + 2^4,\\ C_4 = 6 = 2^2 + 2^1,\\ S_6 = 144 = 2^7 + 2^4,\\ C_6 = 12 = 2^3 + 2^2,\\ S_8 = 768 = 2^9 + 2^8,\\ C_8 = 48 = 2^5 + 2^4,\\ S_{10} = 720 = 2^9 + 2^7 + 2^6 + 2^4,\\ C_{10} = 36 = 2^5 + 2^2,\\ S_{12} = 5376 = 2^{12} + 2^{10} + 2^8,\\ C_{12} = 224 = 2^7 + 2^6 + 2^5,\\ S_{14} = 3360 = 2^{11} + 2^{10} + 2^8 + 2^5,\\ C_{14} = 120 = 2^6 + 2^5 + 2^4 + 2^3,\\ S_{16} = 4096 = 2^{12},\\ C_{16} = 128 = 2^7. \end{array}$$ Interestingly, six of the first eight elements of the tuple $(C_2, C_4, \ldots)$ are highly composite numbers: $1, 6, 12, 48, 36, 120$. I cannot explain the fact that $S_{16}$ and $C_{16}$ are powers of two, but it does not look like a random coincidence. REPLY [4 votes]: The function $f(x)$ is closely related to the notion of autocorrelation, which for a binary sequence $x$ of length $|x|=N$ and shift $w$ can be expressed as $$\textbf{AC}_x(w) := N - 2H(x\oplus R(x,w)).$$ The values of $\textbf{AC}_x(w)$ for various non-trivial shifts (i.e. $1\leq w\leq N-1$) are called out-of-phase autocorrelation values. So, $$f(x) = \min_{1\leq w\leq N-1} H(x\oplus R(x,w)) = \frac{N}2 - \frac12\max_{1\leq w\leq N-1} \textbf{AC}_x(w).$$ For the known constructions for optimal binary sequence w.r.t. autocorrelation, I refer to the paper Cai and Ding (2009). In particular, for $N=q-1\equiv 0\pmod{4}$, where $q$ is a power of prime, the Sidelnikov–Lempel–Cohn–Eastman construction produces a sequence $x$ with $H(x)=\frac{N}2$ and out-of-phase autocorrelation values $\{0,-4\}$, thus giving $f(x)=\frac{N}2$ as requested in the question. The value $N=256$ well fits into this construction since $q=257$ is prime. Here is one such 256-bit word: 0011101101010000101000110101100110101001111011001111111100011111010110011111100000100000011011001011011001110001000001000011110100001011010001010001001001100011110011100101011101011010100110000010100100101010110011101001111110011000001101111101000000111011<|endoftext|> TITLE: How quasirandom are the nonabelian finite simple groups? QUESTION [10 upvotes]: A group is $d$-quasirandom if every nontrivial complex representation has dimension at least $d$. Gowers introduced quasirandomness in this paper and proved that every nonabelian finite simple group of order $n$ is $\sqrt{\log n}/2$-quasirandom. Question: What is the correct (asymptotic) lower bound for the quasirandomness of nonabelian finite simple groups? Gowers indicates that a theorem of Jordan (Theorem 14.12 here) implies a slightly better bound. Indeed, this theorem gives that a nonabelian simple group of order $n$ has a nontrivial representation of dimension $d$ only if $n\leq (d!)12^{d(\pi(d+1)+1)}$, which implies that these groups are $\Omega(\sqrt{\log n\log\log n})$-quasirandom. Of course, the correct bound should be obtainable using the classification of finite simple groups. The sporadic groups are irrelevant to the asymptotics, and the alternating groups are $\Omega(\log n/\log\log n)$-quasirandom. It remains to study the groups of Lie type. What I want can be read off the first column of the generic character tables of these groups, but to my surprise, these aren't known yet (see LeechLattice's answer here). Interestingly, the answer here describes how to find an upper bound on the smallest nontrivial representation in these cases, whereas I want a lower bound. REPLY [12 votes]: Suppose $G$ is a finite simple group of order $n$ with a nontrivial representation of degree $d$. Then $G$ is isomorphic to a subgroup of $U(d)$. By Collins's sharp version of Jordan's theorem (https://www.degruyter.com/document/doi/10.1515/JGT.2007.032/html), $G$ has an abelian normal subgroup of index at most $(d+1)!$, which must be trivial since $G$ is simple, so $|G| \leq (d+1)!$. Rearranging, $d \gtrsim \log n / \log\log n$. Collins's work builds on work of Weisfeiler that I think was unfinished by the time of his disappearance. Edit: Specializing to simple $G$ actually reduces Collins's paper to a reference to a reference to the paper of Seitz and Zalesskii (https://www.sciencedirect.com/science/article/pii/S0021869383711324?via%3Dihub) mentioned by David Craven in the comments, so that's really the heart of the matter. We thereby get the slightly stronger bound $n \leq (d+1)!/2$ (for sufficiently large $d$). Apart from alternating groups I think you can read out a much stronger bound like $n \leq \exp O((\log d)^2)$, or $d \geq \exp \Omega( (\log n)^{1/2})$ (I am guessing the next worst case is $\mathrm{SL}_n(2)$).<|endoftext|> TITLE: Representing $x^3-2$ as a sum of two squares QUESTION [15 upvotes]: Prove that there exist infinitely many integers $x$ such that integer $P(x)=x^3-2$ is a sum of two squares of integers. Ideally, I am looking for a proof method that also applies for other $P(x)$, such as, for example, $P(x)=x^3+x+1$. For $x=4t+3$, $P=(4t+3)^3-2$ is $1$ modulo $4$. By a well-believed (but difficult) Bunyakovsky conjecture, $P$ is a prime infinitely often, and every prime that is $1$ modulo $4$ is a sum of two squares. To find an unconditional proof, it suffices to find polynomials $A(t)$, $B(t)$ and $C(t)$ with rational coefficients such that $A(t)^2 + B(t)^2 = C(t)^3-2$. Are there any good heuristics to find such polynomials? Is there any computer algebra system that helps to guess a solution of a polynomial equation over $Q[t]$? One way to guess $C(t)$ is to form a set $S$ of all integers up to (say) $10^5$ that are sums of two squares, and look for polynomials $C(t)$ such that $C(t)^3-2$ belong to $S$ for all small $t$. I then checked all polynomials of degree up to $4$ and coefficients up to $12$, and found, for example, a polynomial $D(t)=3 + 8 t + 12 t^2 + 8 t^3 + 4 t^4$ such that $D(t)^3-1$ is always a sum of two squares. But no $C(t)$ found in this range, and increasing the degree and/or coefficients makes the enumeration infeasible. Are there methods to find a polynomial with all values in the given set, that are better than enumeration? REPLY [15 votes]: The answer is similar to one provided here. The approach is elementary and proves stronger fact that it can be expressed as sum of two coprime squares. We consider the product $(n+2)(n^3-2)$ which is equal to $n^{4}-2n+2n^3-4$. Now we observe that $$(n+2)(n^3-2)=(n^2+n-7)^2+(13n^2+12n-53).$$ Proceeding as shown in the link we can easily get that $13n^2+12n-53$ is a perfect square for infinitely many $n\equiv 3\pmod{4}$ with first solution as $13(3^2)+12(3)-53=10^2$. If one chooses such a $n$ then clearly $n^3-2$ doesn't have any prime divisor of form $4k+3$ as $\gcd(n^2+n-7,13n^2+12n-53)\mid 25\cdot 59$ and $59$ doesn't divide $n^3-2$ if one looks at the general solution of the equation. Since, $n^3-2$ doesn't have any prime divisor of form $4k+3$ and $n\equiv 3\pmod{4}$ we can infer from the folklore result that it can be expressed as sum of two coprime squares.<|endoftext|> TITLE: The localization of the span category QUESTION [5 upvotes]: Suppose one has a model category $C$ with its class of weak equivalences $W$. It is possible to form a separate homotopical category $(C_{\operatorname{span}},W_{\operatorname{span}})$ which has objects given by the objects of $C$ and morphisms from $x$ to $z$ given by spans $x \overset{\simeq} \twoheadleftarrow y \twoheadrightarrow z$ where the first arrow is an acyclic fibration and the second arrow is a fibration. The class of weak equivalences $W_{\operatorname{span}}$ is given by the spans consisting of two acyclic fibrations. The composition of two spans is defined by pullback. In case it is not obvious, the purpose of the fibration requirement is to ensure that the composite of two such spans is still a span with the backwards arrow an equivalence. It is well known that we may take a Hammock localization of a relative category in order to obtain a simplicially enriched category. The process involves considering commutative diagrams of zig-zags. Is it the case that the Hammock localization of $(C,W)$ is equivalent to the Hammock localization of $(C_{\operatorname{span}},W_{\operatorname{span}})$? REPLY [2 votes]: Here's half an answer. See Prop 5.2 of Dwyer and Kan's Function Complexes in Homotopical Algebra for a proof that the hammock localization of $C$ is the same as the hammock localization of the non-full subcategory of fibrations in $C$. Then the remaining question is whether introducing the backwards weak equivalences affects the hammock localization. I think in order to see this you will have to spell out exactly what bicategory of spans you're using (what its universal property is), but I think the fact that this doesn't affect the hammock localization should follow from considering the universal properties involved.<|endoftext|> TITLE: Isometric imbedding of a 2-disk into Euclidean 3-space QUESTION [6 upvotes]: Let us call a cap the intersection of the boundary of 3-dimensional convex compact set $K$ in $\mathbb{R}^3$ with a half-space bounded by a plane $H$ such that the orthogonal projection to $H$ of this intersection is contained in $K\cap H$. QUESTION. Given a metric on a closed 2-dimensional disk which has non-negative curvature in the sense of Alexandrov. Can the disk be isometrically imbedded into $\mathbb{R}^3$ as a cap? A reference would be helpful. REPLY [7 votes]: Take doubling of the disc, we obtain a metric on the sphere. By Perelman's theorem it had nonnegative curvature in the sense of Alexandrov. Therefore, by Alexandrov's theorem, it is isometric to a convex surface in the Euclidean space. This convex surface is unique up to congruence (Pogorelov's theorem). Therefore, the involution of our sphere extends to a reflection of the space. In particular, the boundary of the disc lies in one plane. Postscript. You may also proceed as suggested by Joseph O'Rourke --- approximate the disc by polyhedral space, apply Alexandrov's theorem for polyhedral surfaces (which has uniqueness for free) and pass to the limit.<|endoftext|> TITLE: A problem concerning a divergent function on $[0, 1]$ QUESTION [6 upvotes]: This problem was posted on another forum and was given at the 1992 Miklós Schweitzer Competition. This competition is known for its very difficult problems and this one seems no exception. I also can't find a solution anywhere online. Here is the problem: Let $E \subset [0,1]$ be Lebesgue measurable with measure $\lvert E\rvert < 1/2$. Define $$h (s) = \int _ {E^c} \frac{dt}{{(s-t)}^2}$$ where $E^c = [0, 1]\setminus E$. Prove that there exists $t \in E^c$ such that $$\int_E \frac {ds} {h (s) {(s-t)} ^ 2} \leq c {\lvert E \rvert} ^ 2$$ with some constant $c$ that is independent of $E$. The tricky part of the problem is the quadratic convergence to zero with the measure of $E$. It is easy to see that the left hand side of the inequality is linearly bound with $\lvert E\rvert$. Any ideas (or links to sites with solutions to Miklós Schweitzer problems?) REPLY [8 votes]: I asked a colleague in Hungary, and he found the solution here (on page 170): http://real-j.mtak.hu/9393/1/MTA_MatematikaiLapok_1992.pdf It is in Hungarian, but with some effort and Google translator, finally I understood. EDIT This is the translation of the argument above. I keep the same notation but write $E^c$ for $[0,1] \setminus E$ insetad of $\overline E$. a) Let $$F=\{t \in E^c: |I \cap E| \leq K|E||I|\}$$ for every interval $t \in I$. In other words, $F$ consists of all points for which the maximal function of $\chi_E$ is less than $K|E|$. Using the maximal inequality, one selects $K$ (independent of $E$) such that $|F| \approx 1$, hence $|E_1|:=|E^c \cap F| \geq 1/2$. b) Given $s \in E$ let $I_s$ be an interval containing $s$ such that $|I_s \cap E|=(1/2)|I_s|$. Such a $I_s$ exists in all Lebesgue points of $E$, again using that $|E| <1/2$ and a continuity argument. c) If $s \in E$ and $t \in E_1$, then applying a) to $J=I_s \cup [s,t]$ (or the other way around) we get $(1/2)|I_s|=|I_s \cap E| \leq |J\cap E| \leq K|E|(|t-s|+|I_s|)$. Then $|t-s| \geq \frac{|I_s|}{4K|E|}$ if $K|E| \leq 1/4$. d) Suppose that $K|E| \leq 1/4$. Then $$ \int_{E_1} \frac{dt}{(s-t)^2} \leq 2 \int_{\frac{|I_s|}{4K|E|}}^\infty \frac{du}{u^2}=\frac{8K|E|}{|I_s|}$$ but also, using b), $$h(s)=\int_{E^c} \frac{dt}{(s-t)^2}\geq \int_{I_s \cap \geq E^c} \frac{dt} {(s-t)^2} \geq \frac{1}{|I_s|^2} \frac 12 |I_s|=\frac{1}{2|I_s|}.$$ Summing up, $$ \int_{E_1} \frac{dt}{(s-t)^2} \leq 16 h(s) K|E|.$$ e) If $K|E| \geq 1/4$ then $$ \int_{E_1} \frac{dt}{(s-t)^2} \leq \int_{E^c} \frac{dt}{(s-t)^2} =h(s) \leq 4h(s) K|E| \leq 16 h(s) K|E|.$$ f) Finally, $$ \int_{E_1}dt\int_{E}\frac{ds }{h(s)(s-t)^2}=\int_{E}\frac{ds}{h(s)}\int_{E_1}\frac{ dt}{(s-t)^2} \leq 16K|E|^2 $$ and there is a point $t \in E_1$ for which the statement holds with $c=32 K$ since $|E_1| \geq 1/2$. PS This does not deserve badges for maths...maybe for translating from hungarian!<|endoftext|> TITLE: Hartshorne's proof of Halphen's theorem QUESTION [5 upvotes]: Apologies if this is not quite at the level of MathOverflow, but it has already been asked at MSE and gone unresolved for several years despite a bounty. Hartshorne states the theorem as follows: Proposition IV.6.1. A curve $X$ of genus $g\geq 2$ has a nonspecial very ample divisor $D$ of degree $d$ if and only if $d\geq g+3$. The necessity is shown, and then sufficiency. The idea is to show that the set $S$ of divisors $D \in X^d$ such that there exists $D' \sim D$ and points $P,Q \in X$ with $E = D'-P-Q$ an effective special divisor has dimension $\leq g+2$. Because $d\geq g+3$ this means there is some $D\notin S$ that is nonspecial and very ample of degree $d$. Hartshorne shows that the set of divisors of the form $E+P+Q$ in $X^d$ that are nonspecial with $E$ a special effective divisor has dimension $\leq g+1$. The part that confuses me comes afterwards. Namely, as $E$ is special the Riemann-Roch tells us that $\dim |E| \geq d-1-g$, and similarly that $\dim |E+Q+P| = d-g$. Because the difference between these two dimensions is at most 1, this somehow implies that the set of divisors $S$ as above has dimension $\leq g+2$. I don't understand this implication. We have divisors of the form $E+Q+P$, each of which gives a linear system of dimension $d-g$, and all of which form a set of divisors of dimension $\leq g+1$. How does the difference in dimension of $E$ and $E+P+Q$ tell us the dimension of $S$? REPLY [2 votes]: I think this Hartshorne's explanation is very rough. By using Picard variety, this step can be proved more clearly as follows. Write $\mathrm{Pic}^d(X)$ for the scheme which parametrized all line bundles of degree $d$ on $X$, $\mathrm{Div}^d(X)$ for the scheme which parametrized all effective divisors of degree $d$ on $X$, and $D_{\mathrm{univ}}\subset X\times \mathrm{Div}^d(X)$ for the universal effective divisor of degree $d$. Then, by the universality of $\mathrm{Pic}^d(X)$, the line bundle $\mathcal{O}_{X\times \mathrm{Div}^d(X)}(D_{\mathrm{univ}})$ induces a morphism $$ \varphi_d:\mathrm{Div}^d(X) \to \mathrm{Pic}^d(X). $$ This morphism can be written as $\varphi_d(D) = \mathcal{O}_X(D)$. Hence each fiber of $\varphi_d$ is linearly equivalent class. In particular, for any $L\in \mathrm{Pic}^d(X)$, it holds that $\dim(\varphi_d^{-1}(L)) = \dim H^0(X,L) - 1$. Write $\mathrm{SpDiv}^d\subset \mathrm{Div}^d(X)$ for the closed subscheme determied by all special effective divisors. Then, by Riemann-Roch, for any $D\in \mathrm{SpDiv}^d$, it holds that $$ \dim(\varphi_d^{-1}(\mathcal{O}_X(D))) = 1+d-g+l(K-D)-1 \geq 1+d-g. $$ Since $\dim(\mathrm{SpDiv}^d) = g-1$, it holds that $$ \dim(\varphi_d(\mathrm{SpDiv}^d)) \leq (g-1)-(1+d-g) = 2g-2-d. $$ Now, let us consider the following two morphisms: \begin{align*} f:X\times X\times \mathrm{Div}^{d-2}(X)\to \mathrm{Div}^d(X), (P,Q,D)\mapsto P+Q+D, \\ g:X\times X \times \mathrm{Pic}^{d-2}(X) \to \mathrm{Pic}^d(X), (P,Q,L)\mapsto L\otimes \mathcal{O}(P+Q). \end{align*} Then we obtain the following commutative diagram: $$ \require{AMScd} \begin{CD} X\times X\times \mathrm{SpDiv}^{d-2} @>{\subset}>> X\times X\times \mathrm{Div}^{d-2}(X) @>{f}>> \mathrm{Div}^d(X) \\ @. @V{\psi}VV @VV{\varphi_d}V \\ @. X\times X\times \mathrm{Pic}^{d-2}(X) @>{g}>> \mathrm{Pic}^d(X), \end{CD} $$ where $\psi := \mathrm{id}_X\times \mathrm{id}_X \times \varphi_{d-2}$. Write $T:= g(\psi(X\times X\times \mathrm{SpDiv}^{d-2}))\subset \mathrm{Pic}^d(X)$. Then $$\dim(T) \leq 2g-2-(d-2)+2 = 2g-d+2.$$ Since $\dim(\mathrm{Div}^d)) = d$, and $\dim(\mathrm{Pic}^d) = g$, the dimension of general fibers of $\varphi_d$ are $d-g$. Hence $$\dim(\varphi_d^{-1}(T)) \leq (2g-d+2)+(d-g) = g+2,$$ (where we note that if $T \subset \varphi_d(\mathrm{SpDiv}^d)$, then the dimension of general fibers of $\varphi_d^{-1}(T) \to T$ are $>d-g$, however, in this case, since $\varphi_d^{-1}(T)\subset \mathrm{SpDiv}^d$, it holds that $\dim(\varphi_d^{-1}(T)) \leq \dim(\mathrm{SpDiv}^d) = g-1 < g+2$). Moreover, by our construction, the scheme $\varphi_d^{-1}(T)$ parametrizes all effective divisors $D\subset X$ which are linearly equivalent to $E+P+Q$, the sum of a special effective divisor $E\subset X$ and points $P,Q\in X$. This is desired conclusion.<|endoftext|> TITLE: Is every matrix involution over a UFD diagonalisable? QUESTION [5 upvotes]: Let $A$ be a UFD, that is also a $k$-algebra, where $k$ is a field of characteristic $\not=2$ (for instance polynomials over $k$). Is every involution in $\mathrm{GL}_n(A)$ diagonalisable? This is of course true over the field of fractions of $A$. In this question it is explained that $$M=\left(\begin{array}{rr} 3&-1\\-1 & 3\end{array}\right)$$ is diagonalisable over $\mathbb{Q}$ but not over $\mathbb{Z}$. However, it is not an involution. REPLY [8 votes]: Here is an answer based on my comment, and Geoff Robinson's earlier comment. Let $A$ be a domain with $2\in A^\times$, and let $M$ be an $n\times n$ matrix with $M^2=I$. It is convenient to consider $M$ as an $A$-endomorphism of $A^n$ and $I$ as the identity endomorphism of $A^n$. Put $S_+ =\{x\in A^n\,:\,Mx=x\}$ and $S_-=\{x\in A^n\,:\,Mx=-x\}$. As noted in Geoff Robinson's comment, since $2\in A^\times$, we have $A^n=S_+\oplus S_-$. (This follows readily from the fact that $x=\frac{1}{2}(x+Mx)+\frac{1}{2}(x-Mx)$ for all $x\in A^n$ and $M^2=I$.) In particular, both $S_+$ and $S_-$ are f.g. projective $A$-modules. Suppose that every f.g. projective $A$-module is free. Then $S_+$ and $S_-$ are free. Let $E_+$ and $E_-$ be bases for these spaces. Then $E=E_+\cup E_-$ is a basis of $A^n$ consisting of eigenvalues of $M$, which means that $M$ is a diagonalizable. On the other hand, if there exists a f.g. projective $A$-module $P$ which is not free, then we can construct a non-diagonalizable involution $M:A^n\to A^n$ as follows. There exists a f.g. $A$-module $Q$ such that $A^n\cong P\oplus Q$ for some $n\in\mathbb{N}$. Now take $M$ to be the matrix corresponding to $\mathrm{id}_P\oplus (-\mathrm{id}_Q)$. For this $M$, we have $S_+=P$ and $S_-=Q$. If $M$ were diagonalizable, then $A^n$ would have a basis $E$ consisting of vectors $x\in A^n$ with $Mx\in \{\pm x\}$ (here we need $A$ to be a domain). Consequently, $E = (E\cap S_+)\cup (E\cap S_-)$, which means that $E_+:=E\cap S_+$ is a basis of $S_+=P$. This contradicts our choice of $P$, so $M$ is not diagonalizable. There are examples of UFDs admitting f.g. projective modules which are not free --- see this question, for instance. Quoting from this answer, one can take $A=\mathbb{C}[a,b,c,x,y,z]/(ax+by+cz-1)$, the module $P$ to be the kernel of $(\alpha,\beta,\gamma)\mapsto a\alpha+b\beta+c\gamma:A^3\to A$, and $Q=(x,y,z)A$. One has $P\oplus Q=A^3$, so it should be possible to write the resulting $M\in \mathrm{M}_3(A)$ explicitly.<|endoftext|> TITLE: What do the components of $\operatorname{Hom}(\pi_1(S),\operatorname{SL}_n(\mathbb{R}))$ look like? QUESTION [15 upvotes]: Let $S$ be a closed orientable surface of genus at least $2$. I'm interested in the connected components of $\operatorname{Hom}(\pi_1(S),\operatorname{SL}_n(\mathbb{R}))$ for $n$ at least $3$. I know that for $n$ odd there is only $3$ connected components. My first question is: For $n$ even, do we still have $3$ connected components? Suppose $n$ is odd. There is the connected component containing the trivial representation, the Hitchin component and another one. My second question is: What do we know about the connected component which doen't contains the trivial or Hitchin representations? Can we construct an explicit representation in it? Can any representation in it be deformed to have image in a compact group? Thanks REPLY [5 votes]: $\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\R{\mathbb{R}}$Just to set the notation, as in the OP's question, let $S$ be a closed orientable surface of genus $g$ at least 2, and $G$ is a Lie group. The question concerns connected components of $\Hom(\pi_1(S),G)$ with respect to the subspace topology where we identify $\Hom(\pi_1(S),G)$ with a subset of $G^{2g}$ by evaluation. The question in particular concerns the case $G=\SL(n,\mathbb{R})$, however this case is very much related to three other cases, so I will summarize all four cases. We assume $n\geq 2$ everywhere below. The first paper to mention is: Goldman, W., Topological components of spaces of representations. Invent. Math. 93 (1988), no. 3, 557–607. In that paper, among other foundational results, it is shown that for $G=\SL(2,\mathbb{R})$ there are $2^{2g+1}+2g-3$ components, and for $G=\PSL(2,\mathbb{R})$ there are $4g-3$ components. Then next foundational paper to mention (as already noted by others) is: Hitchin, N., Lie groups and Teichmüller space. Topology 31 (1992), no. 3, 449–473. In that paper, for $G=\PSL(n,\mathbb{R})$ and $n\geq 3$, there are 3 components if $n$ is odd, and 6 if $n$ is even. Remark: Since $\PSL(n,\mathbb{R})=\SL(n,\mathbb{R})$ when $n$ is odd, this also gives 3 components for $G=\SL(n,\mathbb{R})$ for all odd $n$. Next, let's mention the paper in Eugene Xia: Xia, E., Components of $\Hom(\pi_1,\PGL(2,\mathbb{R}))$. Topology 36 (1997), no. 2, 481–499. In that paper, it is shown there are $2^{2g+1}+4g-5$ components for $G=\PGL(2,\mathbb{R})$. Generalizing the above result, we have: Oliveira, A., Representations of surface groups in the projective general linear group. Internat. J. Math. 22 (2011), no. 2, 223–279. In this paper, it is shown there are $2^{2g+1}+2$ components for $G=\PGL(n,\mathbb{R})$ when $n\geq 4$ and even (which completes the picture for this group given prior results listed above). Lastly, we mention (out of order): Bradlow, S.; García-Prada, O.; Gothen, P., Representations of surface groups in the general linear group. Proceedings of the XII Fall Workshop on Geometry and Physics, 83–94, Publ. R. Soc. Mat. Esp., 7, R. Soc. Mat. Esp., Madrid, 2004. In this paper, it is shown there are $3(2^{2g})$ components when $G=\GL(n,\mathbb{R})$ and $n\geq 3$. Finally to answer your question. As explained to me by a much more knowledgeable friend (André Oliveira), implicit in the latter two papers is that for $G=\SL(n,\mathbb{R})$ there are $2^{2g+1}+2$ components when $n\geq 4$ and even. Here what I understood (any errors are my own): There is only one topological type, determined by the second Stiefel-Whitney class (which takes 2 values). For one value the moduli space is connected. This component contains only representations homotopic to those which factor through $\SO(n)$. For the other value, there is one component like the previous one, but also $2(2^{2g})$ components coming from a choice of a square root of the canonical line bundle for each of the two Teichmüller components in the $\PSL(n,\R)$ case. Remarks: For the $\GL(2,\R)$ case see: Baraglia, D.; Schaposnik, L., Monodromy of rank 2 twisted Hitchin systems and real character varieties. Trans. Amer. Math. Soc. 370 (2018), no. 8, 5491–5534. For $g=1$ and any real reductive $G$, or $g\geq 2$ and $G$ compact or complex reductive, the number of components is equal to the order of $\pi_1([G,G])$. For the $g=1$ case, see Proposition 5.1 (and the references in its proof) in Sikora, A., Character varieties of abelian groups. Math. Z. 277 (2014), no. 1-2, 241–256. For the $g\geq 2$ and $G$ complex reductive (with references for the compact case), see the Appendix of Lawton, S.; Ramras, D., Covering spaces of character varieties. With an appendix by N. Ho and C. Liu. New York J. Math. 21 (2015), 383–416. For non-orientable $S$ and $G=\PSL(2,\R)$ or $\PGL(2,\R)$ see Palesi, F., Connected components of spaces of representations of non-orientable surfaces. Comm. Anal. Geom. 18 (2010), no. 1, 195–217. and also Palesi, F., Connected components of PGL(2,R)-representation spaces of non-orientable surfaces. Geometry, topology and dynamics of character varieties, 281–295, Lect. Notes Ser. Inst. Math. Sci. Natl. Univ. Singap., 23, World Sci. Publ., Hackensack, NJ, 2012.<|endoftext|> TITLE: Criteria for operators to have infinitely many eigenvalues QUESTION [6 upvotes]: Normal compact linear operators on Hilbert spaces have infinitely many (counting multiplicities) eigenvalues by the spectral theorem. For non-normal operators this no longer has to be true. There exist even examples of compact operators without eigenvalues such as weighted shifts and the Volterra operator $Tf(t) = \int_0^t f(s) \ ds$ on $L^2(0,1),$ which when applied to a polynomial basis can also be interpreted as a shift. So a perhaps bold guess would be that we want to avoid some generalized infinite Jordan blocks. I would therefore like to understand: Do there exist other criteria not assuming normality that imply the existence of infinitely many eigenvalues for compact operators on Hilbert spaces? REPLY [2 votes]: The only criterion I know is based on a Theorem in the second book of Dunford and Schwartz, see Theorem X1.6.29 and following. If the resolvent of an Hilbert-Schimdt operator satisfies some decay estimates on some rays dividing the complex plane, then the span of the generalized eigenfuntions is dense. The theorem generalizes to operators having trace-class powers and to closed operators with trace-class powers of the resolvent.<|endoftext|> TITLE: Compressible Ebin-Marsden? QUESTION [8 upvotes]: In Ebin and Marsden's paper Groups of Diffeomorphisms and the Motion of an Incompressible Fluid, there is a footnote on the first page indicating that non-homogeneous cases and the case of compressible fluid mechanics will be treated in a forthcoming paper. Having taken a search through MathSciNet I did not see anything obvious as the mentioned followup. Does anyone have a reference as to which paper that is referring to? REPLY [6 votes]: The compressible case uses semidirect products of groups (group of diffeomorphisms times functions). To my knowledge, the first paper that discusses this in detail is Marsden, Ratiu, Weinstein: Semidirect Products and Reduction in Mechanics (PDF). There the authors state By 1980 it was known that the equations for compressible flow on R3 were formally Lie-Poisson equations for the semidirect product... So even back then it was already folklore.<|endoftext|> TITLE: $\sum a_n = 0$ but $\sum \frac{a_n}{n} = \infty$ QUESTION [6 upvotes]: I'm hoping to find an explicit construction for a sequence such that $\sum a_n = 0$ but $\sum \frac{a_n}{n} = \infty$, or a proof that one cannot exist. So far, I have a good idea of how we can increase the value of the sum up to infinity. For example, we could construct a sequence like this (a_n shown in red): It stays constant for a while, and then that constant part is canceled by a large negative term. Because the flat part becomes smaller and smaller, the partial sums (shown in green) will eventually go to zero. We now have that $\sum \frac{a_n}{n} > 0$, since the negative terms occur later, they will be smaller than the positive terms that occurred earlier. However, this method can only get something finite, more is needed to get the sum all the way to infinity. Another thought is that we could rearrange the sum, but flipping the order in which the negative terms appear. For instance, here I reverse the order of every group of 5 negative terms, so that larger negative terms occur later (flipped shown in black) I'm fairly certain this is still insufficent to get out an infinite value (mainly since the effect will be neglible on the tail, so the partial sums of $\sum \frac{a_n}{n}$ still converge). What's interesting, is that this rearrangement hasn't effected the value of the original sum, we still have $\sum a_n = 0$. Now, I know that by the Reimann rearrangement theorem, there is some rearrangement $\sigma (n)$ such that $\sum a_n = 0$ but $\sum \frac{a_{\sigma(n)}}{\sigma(n)} = \infty$ if $\frac{a_n}{n}$ is conditionally convergent. But, what I'm truly curious about, is if there is some $a_n$ such that $\sum a_{n} = 0$ and $\sum \frac{a_{n}}{n} = \infty$ when we are forced to evaulate the sum without rearranging (so, we sum the n=1 term, then the n=2 term, then the n=3 term, etc.). I'm thinking this is possible, perhaps by reversing a larger and larger amount of consecutive terms, up to infinity at the tail, and being careful about making sure the original sum still converges, though I don't have a good idea of one would actually do this correctly. REPLY [6 votes]: This is just a more detailed version of Christian Remling's comment. Define, for each positive integer $n$, the partial sum $$\displaystyle A(n) = \sum_{k \leq n} a_k.$$ By the definition of limits of series, it follows that $0 = \lim_{n \rightarrow \infty} A(n)$. By abuse of notation, we also denote by $A(x)$ to be $\sum_{k \leq x} a_k$ even when $x$ is not a positive integer. Then partial summation gives $$\displaystyle \left \lvert \sum_{n \leq x} \frac{a_n}{n} \right \rvert = \left \lvert \frac{1}{x} A(x) + \int_1^x \frac{1}{t^2} A(t) dt \right \rvert \leq \frac{|A(x)|}{x} + |A(x)| \left(1 - \frac{1}{x} \right) = |A(x)|,$$ which tends to $0$ as $x \rightarrow \infty$. Actually, we just need that the right hand side is bounded. This suffices for the proof.<|endoftext|> TITLE: Algebraic atlas on smooth manifolds QUESTION [10 upvotes]: A real/complex rational atlas on a smooth closed manifold $M$ is an atlas with charts homeomorphic to Euclidean open sets in $\Bbb{R}^n$/$\Bbb{C}^n$ covering $M$ and real/complex rational transition maps. A real/complex rational structure is a maximal collection of compatible mentioned real/complex atlases. Two rational structures are called isomorphic if there is a topological automorphism $\sigma$ of $M$ such that each coordinate representation of $\sigma$ under a rational chart pair (from each structure) is rational. The existence of a rational structure does not imply being algebraic because every complex torus admits one (the transition maps are translations, thus polynomial), so I want to ask about how rigid these structures are. Q$1$: Are there topological invariants classifying whether a smooth manifold admits a real/complex rational structure? Or on the contrary every smooth manifold admits one? Q$2$: Obviously a real/complex rational structure belongs to a smooth/holomorphic structure. Consider the converse problem: what is the moduli space of real/complex rational structures compatible with the smooth/holomorphic structure on $M$? Specifically, is it true that the number of complex rational structures belonging to the holomorphic structure of $M$ is at most $1$? Counterexamples are welcome. Q$3$: A related question exists on this website, and the first answer gives a sheaf-sense definition of rational structures. However its conclusion -- the algebraic structure of $\Bbb{C}^n$ can be pullbacked onto $M$ -- seems to contradict my complex torus example. I hope someone can help. REPLY [15 votes]: The answers to your question in the case $n=1$ are well-known. In higher dimensions, the answers are less complete, but something is known. For example, in the real case when $n=1$, there is only one smooth, connected compact $1$-manifold, the circle, and, for each natural number $k\ge1$, there is a rational structure $\mathcal{R}_k$, which is the rational structure induced on the $k$-fold connected cover of $\mathbb{RP}^1\simeq S^1$ (which is smoothly diffeomorphic to $S^1$, of course). Two rational structures $\mathcal{R}_i$ and $\mathcal{R}_j$ on the circle are isomorphic if and only if $i=j$, and every rational structure on the circle is isomorphic to some $\mathcal{R}_i$. Meanwhile, in the complex case when $n=1$, specifying a rational structure on a compact Riemann surface of genus $g$ (i.e., a topological surface of genus $g$ with a fixed underlying holomorphic structure) is easily seen to be equivalent to specifying a projective structure on the surface. For $g>1$, it is known that the moduli of projective structures on a given compact Riemann surface is equivalent to the moduli of quadratic holomorphic differentials, a complex vector space of dimension $3g{-}3$. For example, see R. C. Gunning's On uniformization of complex manifolds: The role of connections. In particular, these provide counterexamples sought by the OP. In higher dimensions the situation is more complicated because the pseudogroup of rational maps with a rational inverse, i.e., the so-called birational pseudogroup, in either $\mathbb{R}^n$ or $\mathbb{C}^n$ is not well-understood when $n>1$. However, when $n=2$, the question of when a given compact complex surface has a rational structure is well-understood since we have a classification of compact complex surfaces, by the work of Kodaira. Not all such surfaces have a rational structure; for example, a K3 surface does not.<|endoftext|> TITLE: Proving algebraicity of compact Riemann surfaces without Chow's theorem QUESTION [15 upvotes]: I am trying to write a report for a complex analysis class where I prove Riemann-Roch and apply it to prove algebraicity of compact Riemann surfaces. While writing this, I found that Riemann-Roch quite nicely yields a very ample line bundle and hence a holomorphic embedding into projective space. However, it was not so obvious to me why the image of a holomorphic embedding $f: X \longrightarrow \mathbb{CP}^N$ (for $X$ a compact Riemann surface) is actually an algebraic variety. The resources I have consulted tend to resolve this either by not mentioning algebraicity (as Forster does) or appealing to Chow's theorem (as Griffiths and Harris does). Also, Miranda does a strange thing where an "algebraic curve" is not the zero set of polynomials, but a Riemann surface whose meromorphic functions separate points and tangents. I find the algebraicity of compact Riemann surfaces to be a very striking and beautiful fact, but Chow's theorem is quite advanced for what I am intending this report to be, and I don't believe I can include a proof of it. Furthermore, Chow's theorem applies very generally to any analytic subvariety of $\mathbb{CP}^N$. As I am only interested in the case of $\dim X = 1$, I was wondering if a more elementary argument for algebraicity is out there. In attempting to find such an argument, I tried to compute the Zariski closure of $im(f)$ and use that as a candidate, but I was not able to show equality. Specifically, I wrote $f$ as $x \mapsto [f_0(x) : \dots : f_N(x)]$. Given that the transcendence degree of the field of meromorphic functions $\mathcal M(X)$ over $\mathbb C$ is $1$, we can ensure that the map $\mathbb C[x_0, \dots, x_N] \longrightarrow \mathcal{M}(X)$ sending $x_i \mapsto f_i$ has a nontrival kernel $\mathfrak p$ of height $n$. As such, letting $Z(\mathfrak p)$ be the variety in $\mathbb{CP}^N$ defined by $\mathfrak p$, we can show that $im(f) \subseteq Z(\mathfrak p)$. By dimension considerations, I would expect to have equality $im(f) = Z(\mathfrak p)$, but it's not at all obvious to me why the reverse inclusion must hold. $Z(\mathfrak p)$ will be the Zariski closure of $im(f)$, so what I'm asking is of course tantamount to showing that $im(f)$ is Zariski closed. I was also recommended to try using Riemann-Roch to get a finite covering $X \longrightarrow \mathbb{CP}^1$ and use this to prove algebraicity. I suspect that I am to analyze the associated map on sheaves of holomorphic functions $f^*: \mathcal O_{\mathbb{CP}^1} \longrightarrow \mathcal O_X$ and exploit finiteness as well as the fact that $\mathcal O_{\mathbb{CP}^1}$ is the analytification of an algebraic sheaf. But this too is not clear to me. Is there any known elementary argument I can use? REPLY [9 votes]: Given an embedding $X\to \Bbb P^n$, it's possible to construct polynomial equations for the image. Theorem (Narasimhan 9.6.2): Let $X$ be a compact Riemann surface and $f$, a nonconstant meromorphic function on $X$. Then, there exists an integer $p>0$ with the following property: For any $F$ meromorphic on $X$, there exist rational functions $a_1,\dots,a_p$ (in one variable) such that $$(F(x))^p+a_1(f(x))(F(x))^{p-1}+\cdots+a_p(f(x))=0 \quad \text{on } X.$$ The idea of the proof is that the map $f:X\to\Bbb P^1$ is a covering map with fibers of constant size $p$ over some set $U\subset \Bbb P^1$ with finite complement in $\Bbb P^1$. Let $a_k(z)$ be the $k^{th}$ elementary symmetric polynomial in the values of $F$ along the $p$ points in $f^{-1}(z)$. Then the $a_k(z)$ are meromorphic functions on $\Bbb P^1$ and hence rational, and the relation above holds everywhere by continuity. Letting $f$ and $F$ vary among the meromorphic functions you used to embed $X$ in $\Bbb P^n$, after clearing denominators you get polynomial equations for the image. Alternatively, a little extra work with the above result shows that the field of meromorphic functions on $X$ is a finitely generated field of transcendence degree one over $\Bbb C$, which is exactly an algebraic function field. By the standard theory of such objects, it's equivalent to a smooth projective curve over $\Bbb C$ where the (closed) points correspond to (nontrivial) discrete valuations trivial on $\Bbb C$, and this correspondence between points and valuations also holds on the Riemann surface side. Narasimhan, Raghavan; Nievergelt, Yves, Complex analysis in one variable., Boston, MA: Birkhäuser. xiv, 381 p. (2001). ZBL1009.30001.<|endoftext|> TITLE: Explicit estimates for $N(T,\chi)$ (not $N(T,\chi)+N(T,\overline{\chi})$) QUESTION [7 upvotes]: Let $N(T,\chi)$ denote the number of zeros of $L(s,\chi)$ with imaginary part between $0$ and $T$, with any zero with imaginary part equal to $T$ or to $0$ (not that the latter kind really exists) counting as half a zero. Here I am following the convention in Montgomery-Vaughan, rather than that in part of the literature, where $N(T,\chi)$ means what I would call $N(T,\chi) + N(T,\overline{\chi})$. The explicit literature generally (McCurley, Trudgian, Bennett-Martin-O'Bryant-Rechnitzer...) generally bounds $N(T,\chi) + N(T,\overline{\chi})$. The question is: what kind of explicit bounds we can extract from their proofs for $N(T,\chi)$? The first step is easy: we can express $N(T,\chi)$ as $\text{main term} + S(T,\chi)-S(0,\chi)$, as in Montgomery-Vaughan, Thm. 14.5, where $S(T,\chi) = \frac{1}{\pi} \arg L(1/2+iT,\chi)$. One would then decompose $$S(T,\chi)-S(0,\chi) = \frac{1}{\pi} \left(\arg L(\sigma+i T,\chi)|_{\sigma=\sigma_0}^{1/2} + \arg L(\sigma_0+i t,\chi)|_{t=0}^T + \arg L(\sigma,\chi)|_{\sigma=1/2}^{\sigma_0}\right)$$ for some $\sigma_0>1$ of our choice. The literature gives the bound $2 \log \zeta(\sigma_0)$ on $\left|\arg L(\sigma_0+it)|_{t=-T}^T\right|$. The reason is a mystery to me -- it is obvious that $2 \sum_p \arcsin p^{-\sigma}$ is a tighter upper bound on $\left|\arg L(\sigma_0+it)|_{t=-T}^T\right|$ (and it is easy to compute). I do not know how to do better than $2 \sum_p \arcsin p^{-\sigma}$ as an upper bound on $\left|\arg L(\sigma_0+it)|_{t=0}^T\right|$, and suspect one cannot, in general, as $t$ and $\chi$ could conspire. The bulk of the explicit literature deals with bounding $\arg L(\sigma+i T,\chi)|_{\sigma=\sigma_0}^{1/2}$. Is there a better bound on $\arg L(\sigma,\chi)|_{\sigma=1/2}^{\sigma_0}$ than what one would get just by setting $T=0$? REPLY [2 votes]: For the first question about $2\log\zeta(\sigma_0)|$, I think the reasoning is this: For $\sigma>2$ we have $$|L(s,\chi)-1|\le \sum_{n=2}^\infty\frac{1}{n^\sigma}=\zeta(\sigma)-1<1.$$ Hence $\log L(s,\chi)$ can be defined by the power series $$\log L(s,\chi)=-\sum_{k=1}^\infty \frac{1}{k}(1-L(s,\chi))^k$$ In particular $|\Im \log L(2+it,\chi)|<\pi/2$ and coincide with $\arg(L(2+it,\chi))$. Also this is equal to $$\log L(2+it)=\sum_{n=2}^\infty \frac{\Lambda(n)}{\log n}\frac{\chi(n)}{n^{2+it}}$$ It follows that $$|\arg L(2+it)|\le |\log L(2+it)|\le \sum_{n=2}^\infty \frac{\Lambda(n)}{\log n}\frac{1}{n^{2}}=\log\zeta(2).$$ Now it follows that $$|\arg L(2+it,\chi)-\arg L(2,\chi)|\le 2\log\zeta(2).$$<|endoftext|> TITLE: Example of a non-$\infty$-category whose homotopy category is a groupoid QUESTION [5 upvotes]: What is an example of a simplicial set $S$ such that its homotopy category $hS$ is a groupoid, but such that $S$ is not an $\infty$-category? I know that if $S$ is an $\infty$-category, then $S$ is a Kan complex if and only if $hS$ is a groupoid. My question is about what happens if $S$ is not an $\infty$-category. REPLY [9 votes]: Since the homotopy category depends only on the 2-skeleton of the simplicial set, the easiest thing to do is to take the 2-skeleton of an ∞-groupoid. For example, let $E$ be the nerve of the contractible groupoid on two objects. This has as a set of $n$-simplices the set $\{a,b\}^{[n]}$ of functions (of sets) from $[n]=\{0<... TITLE: Terminology for a set that does not surject onto $\omega$ (in ZF) QUESTION [9 upvotes]: Short question: Is there a standard term for a set $F$ such that there does not exist a surjection $F \twoheadrightarrow \omega$ (in the context of ZF)? More detailed version: Consider the following four notions of “finiteness” in ZF, the third of which is the one I am asking about and will be arbitrarily named “P-finite” here: “$F$ is finite” means any of the following equivalent statements: there exists $n\in\omega$ and a bijection $n \xrightarrow{\sim} F$, there exists a bijection $E \xrightarrow{\sim} F$ with $E\subseteq\omega$ and no bijection $\omega \xrightarrow{\sim} F$, every nonempty subset of $\mathscr{P}(F)$ has a maximal element. “$F$ is T-finite” means: every chain in $\mathscr{P}(F)$ has a maximal element. “$F$ is P-finite” [nonstandard terminology which I'd like a standard term form] means any of the following equivalent statements: $\mathscr{P}(F)$ is Noetherian under inclusion (i.e., any increasing sequence $A_0 \subseteq A_1 \subseteq A_2 \subseteq \cdots$ of subsets of $F$ is stationary), $\mathscr{P}(F)$ is Artinian under inclusion (i.e., any decreasing sequence $A_0 \supseteq A_1 \supseteq A_2 \supseteq \cdots$ of subsets of $F$ is stationary), there does not exist a surjection $F \twoheadrightarrow \omega$. “$F$ is D-finite” (i.e., Dedekind-finite) means any of the following equivalent statements: there is no bijection of $F$ with a proper subset of it, there is no injection $\omega \hookrightarrow F$. (I gave several equivalent conditions to emphasize the parallel between these four notions.) We have finite $\Rightarrow$ T-finite $\Rightarrow$ P-finite $\Rightarrow$ D-finite, and none of the implications I just wrote is reversible. (To construct a permutation model with a P-finite set that is not T-finite, start with a set of atoms in bijection with $\mathbb{R}$ and use the group of permutations given by continuous increasing bijections $\mathbb{R} \xrightarrow{\sim} \mathbb{R}$ and the normal subgroup given by pointwise stabilizers of finite sets.) Surely these four notions, and the implications and nonimplications I just mentioned must appear somewhere in the literature, as well as possibly others. My question is, what is the standard name for “P-finiteness”, and where are its properties, including what I just wrote, discussed in greater detail? REPLY [4 votes]: I suggest the terminology "power Dedekind finite" by Andreas Blass in his paper Power-Dedekind Finiteness, and I use this terminology throughout all my papers. By Kuratowski's celebrated theorem, a set $F$ does not map onto $\omega$ if and only if the power set of $F$ is Dedekind finite. So this terminology does make sense. I do not like the terminology "weakly Dedekind finite", since it is in fact stronger than Dedekind finiteness. If we use "strongly Dedekind finite", then it contains a adverb "strongly" which is different from the adverb "weakly" used in its dual notion "weakly Dedekind infinite".<|endoftext|> TITLE: Ricci curvature : beyond heat-like flows QUESTION [15 upvotes]: Let me give you some context first: just a few days ago I found some intriguing references to Ricci flows in the setting of directed graphs. There are at least two versions of Ricci curvature in the discrete realm (one being the Ollivier-Ricci curvature, the other the Forman-Ricci, see here for reference (*)), and as it turns out, they are both useful in graph analytics. To be a tad more specific, one application leads to a new method for determining communities (the so-called Ricci communities, for the interested ones there is even a github Python implementation which can easily be used for hands-on explorations ), whereas another quite useful one is used to get rid of "bottlenecks" in graph messaging ( thereby solving some critical issue in Graph deep learning see picture below). https://towardsdatascience.com/over-squashing-bottlenecks-and-graph-ricci-curvature-c238b7169e16 Now, if I understand them correctly, the associated Ricci flow, just like in the differentiable realm, acts as a kind of "curvature heat-like operator", a diffusion which tends to smoothen out the curvature across the underlying geometrical object. Perhaps naively, it occurred to me this: why confining ourselves to diffusion? (note: I am aware of the centrality of the Ricci flow in the proof of the Poincare conjecture) Could one replace the Ricci flow with some kind of PDE (or a difference equation in the finite setting) for the curvature change modeled on completely different PDEs? For instance, what about a kind of wave equation? Now the questions (and I apologize if this is too naive, I am coming from the data science world, my knowledge of Riemannian geometry does not go beyond standard grad courses): Have such curvature flow involving non-heat-like PDEs been investigated in the world of Riemannian geometry? I would think the answer is in the affirmative, but I just do not happen to know it. Are there any references for generalized curvature flows in discrete metric spaces and particularly in weighted directed graphs? Any help is most welcome. (*) actually in the referenced article there are three discrete Ricci curvatures, but I haven't wrapped my mind around the third one yet. REPLY [7 votes]: A small number of authors have considered hyperbolic versions of the standard flows, see e.g. "Wave character of metrics and hyperbolic geometric flow" by De-Xing Kong and Kefeng Liu and related articles. I am not familiar with the literature on discrete curvature flows. Some papers that look interesting are "Super Ricci flows for weighted graphs" by Matthias Erbar & Eva Kopfer and "Simplicial Ricci flow" by Warner Miller, Jonathan McDonald, Paul Alsing, David Gu & Shing-Tung Yau. I would suspect that undirected graphs are more natural here than directed graphs, just based on analogy to Riemannian metrics.<|endoftext|> TITLE: Moebius function of finite abelian groups QUESTION [8 upvotes]: I am wondering if there is any literature on general formula of the Moebius function of subgroup lattices of any finite abelian group $G$? What I know is When $G$ is cyclic, the Moebius function is simply the classical number theoretic one. When $G=(\mathbb Z/p\mathbb Z)^r$, the formula involves the number of $k$-dimensional linear subspace of $G$. But is there some formula for any finite abelian group? REPLY [19 votes]: Since every interval of the subgroup lattice $\mathcal{L}(G)$ of a finite abelian group $G$ is isomorphic to the subgroup lattice of some finite abelian group, we can restrict ourselves to $\mu(\hat{0},\hat{1})$, where $\hat{0}$ is the bottom element (the trivial subgroup of $G$) and $\hat{1}$ is the top element (the group $G$ itself) of $\mathcal{L}(G)$. If $G$ and $H$ have relatively prime orders, then $\mathcal{L}(G\times H)= \mathcal{L}(G) \times \mathcal{L}(H)$. Thus (EC1, second ed., Proposition 3.8.2) $\mu_{G\times H}(\hat{0},\hat{1}) = \mu_G(\hat{0},\hat{1})\mu_H(\hat{0},\hat{1})$, so we can assume that $G$ has prime power order $p^n$. Then $\mathcal{L}(G)$ is atomic (i.e., $\hat{1}$ is a join of atoms, or $G$ is generated by the subgroups of order $p$) if and only if $G$ is elementary abelian (a product of groups of order $p$). Since for any finite lattice $L$, $\mu(\hat{0},\hat{1})\neq 0$ implies that $L$ is atomic (EC1, Corollary 3.9.5), we have $\mu(\hat{0},\hat{1})=0$ unless $G$ is elementary abelian. Finally, if $G$ is elementary abelian of order $p^n$, then it is well-known (e.g., EC1, equation (3.34)) that $\mu(\hat{0},\hat{1})=(-1)^n p^{{n\choose 2}}$.<|endoftext|> TITLE: Is the union of a compact and the relatively compact components of its complementary in a manifold compact? QUESTION [7 upvotes]: I was thinking of a way to prove this and I realised that for my approach the lemma from the title would be useful, and it´s an interesting question on its own. Obviously it is true if the manifold is compact or $\mathbb{R}^n$, but I don´t see it in general. This is the precise question: Let $M$ be a connected manifold and $X$ a compact inside it, then is the union of $X$ and all the relatively compact components of $M\setminus X$ compact? P.S. I already asked this question in MSE a few days ago, but after asking a few people I thought it could be more appropiate for MO. REPLY [3 votes]: Actually, there is an elementary proof of this. I will imitate the one given in O.Forster Lectures on Riemann Surfaces. We assume $M$ is connected. Let $Y$ be equal to the union of $X$ with all the relatively compact components of $M \setminus X$ Let $U$ be a relatively compact, open subset containing $X$ and let $C_j$, $j \in J$ be the connected components of $M \smallsetminus X$. Let $bU$ be the boundary of $U$, which is compact and disjoint from $X$. Claim 1. Every $C_j$ meets $U$: If $C_j \subset M \setminus U$, its closure in $M$ is also contained in $M \setminus U$, but $C_j$ is a connected component of $M \setminus U$ so $C_j = \overline{C_j}$ which conttradicts connectedness of $M$. Claim 2. Only finitely many $C_j$ intersects $bU$: This is because $bU$ is compact and the $C_j$ are open and disjoint, anc cover $bU$. Claim 3. $Y$ is closed: Let $J_0$ consist of the indices corresponding to relatively compact components. Then $M \setminus Y = \bigcup_{j \not \in J_0} C_j$, which is open. By Claim 2, we can find $j_1, \ldots , j_k \in J_0$ be such that $C_{j_i}$ intersects $bU$. Then , by Claim 1 again, any other $C_j$ is contained in $U$. Therefore, $$ Y \subset U \cup C_{j_1} \cup \ldots \cup C_{j_k}$$ RHS is relatively compact by the choice of $U$ and the $j_i$, and LHS is closed by Claim 3, so LHS is compact too.<|endoftext|> TITLE: Uniqueness of the solution to some SDE QUESTION [5 upvotes]: Consider the stochastic differential equation as follows: $$X_t=X_0+t+\int_0^t\frac{dW_s}{1+m(s)},\quad \forall t\ge 0,~~~~~~~~~~~~~~~(\ast)$$ where $X_0>0$ is square integrable and $m(t)=\mathbb P[\inf_{0\le s\le t}X_s>0]$ for all $t\ge 0$. Can we prove that $(\ast)$ admits at most one solution $(X,m)$ (assuming its existence)? Any answer, comments or references are highly appreciated. REPLY [4 votes]: $\newcommand{\F}{\mathcal{F}}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$According to a comment by the OP, $X_0$ and $(W_t)$ are independent. So, without loss of generality (wlog), $X_0$ is a real number $x_0>0$. Let $\F$ denote the set of all nonincreasing functions from $[0,\infty)$ to $[0,1]$. Define the (nonlinear) operator $F$ from $\F$ to $\F$ as follows: for each $f\in\F$ and each real $t\ge0$, \begin{equation*} F(f)(t):=P(\inf_{s\in[0,t]}X^f_s>0), \tag{1} \end{equation*} where \begin{equation*} X^f_t:=x_0+t+\int_0^t\frac{dW_s}{1+f(s)}. \tag{2} \end{equation*} We have to show that the equation \begin{equation*} F(f)=f \tag{3} \end{equation*} has at most one solution $f$ in $\F$. Before doing this, let us prove, for completeness, the existence of a solution of (3). Here we just detail the argument in the comment by user GJC20 on this previous answer. Let $f_0:=0$ and $f_n:=F(f_{n-1})$ for all natural $n$. Then $f_1=F(f_0)\ge0=f_0$ and hence, by that previous answer, $f_n\ge f_{n-1}$ for all natural $n$. So, the uniformly bounded sequence $(f_n)$ converges (as $n\to\infty$) pointwise to some $\bar f\in\F$, which implies $f_{n+1}=F(f_n)\to F(\bar f)$ pointwise. So, $F(\bar f)=\bar f$, that is, $\bar f$ is a solution of (3). To prove that (3) has at most one solution in $\F$, suppose that functions $f_1$ and $f_2$ in $\F$ are solutions of (3). Let \begin{equation*} t_0:=\sup\{t\in[0,\infty)\colon f_1(s)=f_2(s)\ \forall s\in[0,t)\}. \tag{4} \end{equation*} Then wlog $t_0<\infty$. Take any $h\in(0,1)$. To obtain a contradiction, suppose that \begin{equation*} \ep:=\|f_1-f_2\|>0, \tag{5} \end{equation*} where $\|g\|:=\sup\{|g(t)|\colon t\in[0,t_0+h]\}$. The crucial point is to use the same kind of time change as in the mentioned previous answer: The process \begin{equation*} \text{$(X^{f_i}_t)$ equals the process $(x_0+t+W_{\tau_i(t)})$ in distribution,} \tag{*} \end{equation*} where \begin{equation*} \tau_i(t):=\int_0^t\frac{ds}{(1+f_i(s))^2}; \tag{6} \end{equation*} here and in what follows, $i\in\{1,2\}$. Note that $\frac1{(1+z)^2}$ is $2$-Lipschitz in $z\ge0$. Therefore and in view of (6), (4), and (5), \begin{equation*} |\tau_1(t)-\tau_2(t)|\le\int_{t_0}^{\max(t_0,t)} ds\,\Big|\frac1{(1+f_1(s))^2}-\frac1{(1+f_2(s))^2}\Big| \le 2h\ep \tag{7} \end{equation*} for all $t\in[0,t_0+h]$. Since $\frac14\le\frac1{(1+f_i)^2}\le1$, the functions $\tau_i$ are Lipschitz-continuous and strictly increasing on $[0,\infty)$ from $\tau_i(0)=0$ to $\tau_i(\infty-)=\infty$, and the inverse functions $\tau_i^{-1}$ are defined, strictly increasing, and $4$-Lipschitz on $[0,\infty)$. It follows that for all $u\in[0,\tau_2(t_0+h)]$ \begin{equation*} |\tau_1^{-1}(u)-\tau_1^{-1}(u)|\le4\sup_{t\in[0,t_0+h]}|\tau_1(t)-\tau_2(t)|\le8h\ep, \tag{8} \end{equation*} in view of (7). By (*), for all real $t\ge0$, \begin{equation*} F(f_1)(t)-F(f_2)(t)=D_1(t)+D_2(t), \tag{9} \end{equation*} where \begin{equation*} \begin{aligned} D_1(t)&:=P(\inf_{u\in[0,\tau_1(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &-P(\inf_{u\in[0,\tau_2(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \end{aligned} \end{equation*} and \begin{equation*} \begin{aligned} D_2(t)&:=P(\inf_{u\in[0,\tau_2(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &-P(\inf_{u\in[0,\tau_2(t)]}(x_0+\tau_2^{-1}(u)+W_u)>0). \end{aligned} \end{equation*} Using e.g. (as an overkill) Lemma 8, p. 407 (or its Russian original, Lemma 8, p. 423) and (8), we get \begin{equation*} \|D_2\|\le C\Big(1+\frac1{\sqrt{t_0+h}}\Big)h\ep\le2C\sqrt h\,\ep, \tag{10} \end{equation*} where $C$ is some universal positive real constant. Let $\tau_{\min}(t):=\min(\tau_1(t),\tau_2(t))$ and $\tau_{\max}(t):=\max(\tau_1(t),\tau_2(t))$. Take any $t\in[0,t_0+h]$ and then let $u_1:=\tau_{\min}(t)$, $\de:=\tau_{\max}(t)-\tau_{\min}(t)$, $x_1:=x_0+\tau_1^{-1}(u_1)$, and $G:=1-\Phi$, where $\Phi$ is the standard normal cdf. Let $G\big(\frac{x_1}{\sqrt{u_1}}\big):=0$ if $u_1=0$. Then \begin{equation*} \begin{aligned} &|D_1(t)| \\ &=P(\inf_{u\in[0,\tau_{\min}(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &-P(\inf_{u\in[0,\tau_{\max}(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &=P(\inf_{u\in[0,\tau_{\min}(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0, \\ &\inf_{u\in(\tau_{\min}(t),\tau_{\max}(t)]}(x_0+\tau_1^{-1}(u)+W_u)\le0) \\ &\le P(\inf_{u\in[0,u_1]}W_u>-x_1,\inf_{u\in(u_1,u_1+\de]}W_u\le-x_1) \\ &=P(\inf_{u\in[0,u_1+\de]}W_u\le-x_1) -P(\inf_{u\in[0,u_1]}W_u\le-x_1) \\ &=2G\Big(\frac{x_1}{\sqrt{u_1+\de}}\Big)-2G\Big(\frac{x_1}{\sqrt{u_1}}\Big) \\ &\le\frac\de{x_1^2}\le\frac\de{x_0^2}\le\frac{2h\ep}{x_0^2}. \end{aligned} \tag{11} \end{equation*} The first inequality in (11) follows because the function $\tau_1^{-1}$ is increasing and $x_0+\tau_1^{-1}(u_1)=x_1$. The fourth, last equality in (11) follows by the reflection principle. The second inequality there follows because $\frac{\partial}{\partial u}G\big(\frac x{\sqrt u}\big)\le\frac1{2x^2}$ all real $x>0$ and $u>0$. The last inequality in (11) follows by (7), since $\de=\tau_{\max}(t)-\tau_{\min}(t)$. Collecting (5), (9), (11), and (10), for any real $h\in(0,1)$ such that $\frac{2h}{x_0^2}+2C\sqrt h<1$, we have \begin{equation*} \ep=\|f_1-f_2\|=\|F(f_1)-F(f_2)\|\le\|D_1\|+\|D_2\|\le\Big(\frac{2h}{x_0^2}+2C\sqrt h\Big)\ep<\ep, \end{equation*} which is the mentioned desired contradiction with (5). So, $\ep=0$, which means that $f_1(s)=f_2(s)\ \forall s\in[0,t_0+h)$, which contradicts (4). This final contradiction shows that $t_0=\infty$ in (4), and we are done.<|endoftext|> TITLE: Questions on 'Improved bounds for the sunflower lemma' QUESTION [5 upvotes]: I have been reading 'Improved bounds for the sunflower lemma' by Alweiss, Lovett, Wu and Zhang (Ann. of Math., Vol. 194(3), 2021), and have some gaps in my understanding of the paper. They are as follows: In Lemma: 2.8 (pg. 802), the authors bound the number of bad pairs $(W,S_{i})$ by a product of four bounds obtained earlier- (i) the bound on the number of sets of the form $W \cup S_{i}$, (ii) the number of sets of the form $A = S_{i} \cap S_{j}$ (once $S_{j}$ is chosen and fixed in a certain manner), (iii) the number of sets in $\mathcal{F}^{'}$ that contain the set $A$, and (iv) the number of sets of the form $S_{i} \cap W$. Each of the bounds in i)-iv) are clear to me, but I am unable to understand how the product is indeed a bound for the number of bad pairs, both heuristically and in terms of constructing an injection from the set of bad pairs into the set encoded as above. In Lemma 2.10 (pg. 803), a new weighting $\sigma^{'}$ is defined on $\mathcal{F}^{'}$. Am I right in thinking that $\forall \, S \, \in F^{'}: \sigma^{'}(S) = 1$? In this case, the spreadness criteria seems to go through. If not, how is $\sigma^{'}$ defined explicitly? On pg. 804, the authors construct a recursively-defined, strictly-decreasing sequence of integers, $(w_{i})_{i =1}^{r}$, with $w_{i+1} = \lfloor (1 - \varepsilon)w_{i} \rfloor + 1$ for some small, fixed $\varepsilon$, uniformly bounded below by $w^{*}.$ It can be shown that $w^{*} \geq \frac{1}{\varepsilon}$. The authors claim that $\exists K > 0: r \leq \frac{K \, \log(w)}{\varepsilon}.$ I am unable to prove this remark, and would prefer, if possible, a constructive proof. My Attempt: Here are some observations, which don't seem to lead to a proof of what is required. We have that $w_{i} \geq (1 - \varepsilon)^{i}w.$ Moreover, we have that $w_{i} - w_{i+1} \geq \varepsilon w_{i} - 1,$ and hence, we also have that $$w_{i} - w_{i + 1} \geq \varepsilon w_{r} - 1 \geq \varepsilon w^{*} - 1.$$ Thus, the distance between each pair of successive terms is at least $\varepsilon w^{*} - 1$, and so, assuming that this is positive, the number of integers in the sequence, i.e., $r$ is bounded above by $$\frac{w - w^{*} + 1}{\varepsilon w^{*} - 1}.$$ We also observe that $$1 \leq w_{i} - w_{i + 1} \leq \varepsilon w_{i} \leq \varepsilon w_{0} = \varepsilon w,$$ and hence, a lower bound for $r$ is $$\frac{w - w^{*} + 1}{\varepsilon w}.$$ On pg. 808, the proof of Theorem: 4.2 concludes by stating that $\mathcal{F}$ cannot be $C \, \log(w)$ spread for large enough $C$. This apparently follows from the improved version of Theorem 2.5 from Rao's paper (also proved independently by Tao). Below is my argument- I am not certain that it is correct. Any help on any or all of these questions is appreciated. Edit: I would especially appreciate any insights regarding the first question, even if you don't have the time to answer the others. REPLY [6 votes]: Let me try to answer your questions. We can recover the pair $(W, S_i)$ from the four quantities you mentioned. So the number of their combinations upper bounds the number of pairs $(W, S_{i})$, which is why we multiply the number of options for each one. Indeed, we switch from a set system with not necessarily uniform weights to a multi set system with uniform weights to simplify the proof. You are asking for a constructive proof. Assume that $w_{i} \geq \frac{2}{\varepsilon}$. Then $w_{i+1} \leq (1 - \varepsilon) w_{i} + 1 \leq (1-\frac{\varepsilon}{2}) w_{i}$. So you reach $w_{i} < \frac{2}{\varepsilon}$ after at most $O\left(\frac{\log w}{\varepsilon}\right)$ steps. I don't understand your question. If you replace our Theorem 2.5 with Rao's result it removes all the $\log\log$ terms we have, which is the point we are making.<|endoftext|> TITLE: Identity involving a quadratic term inside the Pochhammer symbol QUESTION [9 upvotes]: This identity came up in my research: $$ \sum_{m=1}^n m^2 \frac{(\frac{xy}n + m-1)_{2m-1} (n+m-1)_{2m-1}}{(x+m)_{2m+1} (y+m)_{2m+1}} = \frac{n^2}{(x^2-n^2) (y^2 - n^2)}. $$ Here $n$ is a fixed positive integer and $x,y$ are variables. So for each $n$ this is an identity of rational functions in $x,y$. I denote by $(x)_n$ the falling factorial (Pochhammer symbol) $$ (x)_n = x(x-1)\cdots (x-n+1). $$ It may be worth pointing out that all the falling factorials that appear are of this form: $$ (x+m-1)_{2m-1} = x \prod_{i=1}^{m-1} (x^2-i^2) $$ This identity looked very foreign to me because of the quadratic argument $\frac{xy}n$ inside the falling factorial. I managed to prove it eventually, but I wonder if there is a simple/standard approach or if this is known and/or related to something interesting. EDIT: If you prefer binomial identities, by taking the partial fraction expansions this reduces to $$ \sum_{m=\max(a,b)}^n \binom{2m}{m-a} \binom{2m}{m-b} \binom{\frac{ab}{n}+m-1}{2m-1} \binom{n+m-1}{2m-1} = 0 $$ for positive integers $a,b,n$ satisfying $a TITLE: Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$? QUESTION [5 upvotes]: Problem: Given three positive integers $0 < n_1 < n_2 < n_3$. Is there always a real number $x$ such that $$\cos n_1 x + \cos n_2 x + \cos n_3 x < -2?$$ REPLY [17 votes]: The answer is negative. If $n_3=n_1+n_2$, then $$\cos n_1 x + \cos n_2 x + \cos n_3 x\geq -2.$$ Indeed, the left-hand side equals $$(1+\cos n_1 x)(1+\cos n_2 x)-1-\sin n_1 x\sin n_2 x,$$ where $1+\cos n_j x\geq 0$, and $\sin n_1 x\sin n_2 x\leq 1$. Furthermore, if $n_2=2 n_1$ and $n_3=3 n_1$, then $$\min_{x\in\mathbb{R}}\,(\cos n_1 x + \cos n_2 x + \cos n_3 x)=-\frac{17+7\sqrt{7}}{27}\approx-1.31557.$$ More generally, the minimum of the Dirichlet kernel is known to great precision, see e.g. here.<|endoftext|> TITLE: What is an example of two Banach spaces $X,Y$ such that $X$ embeds isometrically but not linearly into $Y$? QUESTION [11 upvotes]: By a result of Godefroy and Kalton if $X,Y$ are separable Banach spaces and $X$ embeds isometrically into $Y$, then $X$ embeds with a linear isometry into $Y$. Is this result known to fail for nonseparable spaces? That is, is there a known example of two (necessarily nonseparable) Banach spaces $X,Y$ such that $X$ embeds isometrically into $Y$, but such that there is no linear isometric embedding of $X$ into $Y$? This question was previously asked on MSE but received no answer there. REPLY [13 votes]: Yes, if $H$ is a nonseparable Hilbert space then it embeds isometrically into the Arens-Eells space ${\rm AE}(H)$, but not linearly isometrically, or even linearly homeomorphically. See Theorem 5.21 of my book Lipschitz Algebras (second edition). As I explain in the notes to that chapter, a more general version of this statement was claimed in a paper of Godefroy and Kalton, but their proof is erroneous and, as far as I can tell, not fixable. However, some of the ideas of my argument are based on theirs.<|endoftext|> TITLE: Hodge dual of de Rham cohomology and singular cohomology QUESTION [8 upvotes]: We know that the de Rham cohomology is isomorphic to the singular cohomology, does the Hodge dual of differential forms induce a dual operation on de Rham cohomology, hence also on singular cohomology? Namely, Is the Hodge dual of a closed form also a closed form? Is the Hodge dual of an exact form also an exact form? Is there a Hodge dual of de Rham cohomology and singular cohomology? REPLY [8 votes]: The Hodge * operator action on cohomology is generally speaking metric-dependent, hence * is not well-defined without fixing the metric. There are some caveats. On complex curves, for example, the Hodge * operator is complex rotation, which depends only on complex structure. This gives an example of a manifold for which the action of * is metric-dependent: indeed, the action of complex rotation on $H^1$ determines the biholomorphism class of the complex curve (Torelli). On compact 2n-dimensional manifolds, the *-operator in the middle cohomology is determined by the conformal structure.<|endoftext|> TITLE: In which dimensions is a strongly causal Lorentzian manifold determined conformally by its causal structure? QUESTION [7 upvotes]: Let $M$ be a strongly causal Lorentzian manifold. If $M$ has dimension 4, a theorem of Hawking, King, and McCarthy (see Thm 5) says that $M$ is determined up to conformal isomorphism by its class of null geodesic curves (where the parameterization of the null geodesic is forgotten) and thence by the causal relation $J^+$ on $M$. Question: Does this theorem also hold in dimensions other than 4? My hunch would be that it continues to hold in higher dimensions, but in lower dimensions I'm not so sure. At any rate, I don't see how to adapt Hawking, King, and McCarthy's argument (which involves choosing coordinates using a certain configuration of null geodesics) to dimension 2. EDIT: As came up in the comments, I should clarify that a priori I don't want to assume anything about the smooth structure on $M$. In fact, I would prefer not to assume anything about the topology either -- just take $M$ as a set of points, equipped with the relation $J^+$. From this data, when can one recover the topological, smooth, and conformal structures on $M$? REPLY [3 votes]: In 2 dimensions the question posted in this comment has a negative answer. Consider the standard Minkowski space with double null coordinates $(u,v)$ in which the metric is $ds^2 = - du~dv$. Any strictly increasing bijection $\phi: \mathbb{R}\to \mathbb{R}$ induces a mapping $(u,v) \mapsto (\phi(u),v)$ that is an causal isomorphism. Choosing $\phi$ to be continuous and not differentiable you get an example of an $\mathscr{M}$-homeomorphism that preserves null geodesics and is not a diffeomorphism (in the terminology of Hawking-King-McCarthy; compare to Theorem 5).<|endoftext|> TITLE: Can the forcing-absolute fragment of SOL have a strong Lowenheim-Skolem property? QUESTION [5 upvotes]: Previously asked and bountied at MSE: Let $\mathsf{SOL_{abs}}$ be the "forcing-absolute" fragment of second-order logic - that is, the set of second-order formulas $\varphi$ such that for every (set) forcing $\mathbb{P}$ and every (set-sized) structure $\mathfrak{A}$ (in the ground model $V$) we have $$\mathfrak{A}\models\varphi\quad\iff\quad\Vdash_\mathbb{P}(\mathfrak{A}\models\varphi).$$ Note that this is in fact definable since the second-order theory of a structure in $V_\kappa$ is calculated, uniformly, in $V_{\kappa+1}$; by contrast, the set of forcing-absolute second-order properties of the universe is not so nice (to put it mildly). Under a "mild" large cardinal assumption, $\mathsf{SOL_{abs}}$ has a weak downward Lowenheim-Skolem property for countable languages, namely every satisfiable countable $\mathsf{SOL_{abs}}$-theory $T$ has a countable model. This is just a consequence of the appropriate amount of large-cardinal-given absoluteness applied to the $\Sigma^1_{\omega+1}$ sentence "$T$ has a countable model," with $L(\mathbb{R})$-absoluteness being more than enough. However, I don't see how to get the full downward Lowenheim-Skolem property for countable languages: Does $\mathsf{ZFC}$ + [large cardinals] prove that every structure in a countable language has a countable $\mathsf{SOL_{abs}}$-elementary substructure? (To clarify since there is a stronger notion of second-order elementary equivalence in use as well, I mean the "weak" version of elementarity here: $\mathfrak{A}\preccurlyeq_{\mathsf{SOL_{abs}}}\mathfrak{B}$ iff for every $\mathsf{SOL_{abs}}$-formula $\varphi(x_1,...,x_n)$ with only object variables we have $\varphi^\mathfrak{A}=\varphi^\mathfrak{B}\cap\mathfrak{A}^n$.) The problem is that for a given structure $\mathfrak{A}$, the property "is a countable substructure of $\mathfrak{A}$" might be too wild for our large cardinal assumption to get useful purchase. REPLY [8 votes]: Yes, and a proper class of Woodin cardinals suffices. For $n<\omega$ and $X$ a set of ordinals, $M_n(X)$ denotes the minimal iterable proper class model $M$ of ZFC with $n$ Woodin cardinals above $\mathrm{rank}(X)$ and $X\in M$. Because we have a proper class of Woodin cardinals, $M_n(X)$ exists for every set $X$ of ordinals. If $x$ is a real, then for $n$ even, $M_n(x)$ is boldface-$\Sigma^1_{n+2}$-correct, and for $n$ odd, is boldface-$\Sigma^1_{n+1}$-correct. All forcing below is set forcing. Claim 1: Given a structure $\mathfrak{B}$ and a set $B$ of ordinals coding $\mathfrak{B}$, and given an element $x\in\mathfrak{B}$ and a second order formula $\varphi$, if "$\mathfrak{B}\models\varphi(x)$" is forcing absolute then for all sufficiently large $n<\omega$, $M_n(B)\models$"It is forced by $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$ that $\mathfrak{B}\models\varphi(x)$". (Here '"$\mathfrak{B}\models\varphi(x)$" is forcing absolute' just means that the truth value of "$\mathfrak{B}\models\varphi(x)$" is independent of which generic extension of $V$ we are working in, but here $\mathfrak{B}$ and $x$ are fixed.) (Note I'm not saying that the converse of the claim holds.) Proof: Let $G$ be $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$-generic over $V$, hence also over $M_n(B)$. Then $V[G]\models$"$\mathfrak{B}\models\varphi(x)$". Since $\mathfrak{B}$ is countable in $V[G]$, this is just a projective statement there, about some reals in $M_n(B)[G]$. But one can show that (with our large cardinal assumption) $M_n(y)$ is not changed by forcing, so $(M_n(B))^V=(M_n(B))^{V[G]}$, and $V[G]\models$"$M_n(B)[G]=M_n(B,G)$", so $M_n(B)[G]$ is at least boldface-$\Sigma^1_{n+1}$-correct in $V[G]$. So for sufficiently large $n$, $M_n(B)[G]\models$"$\mathfrak{B}\models\varphi(x)$", as desired. Now working in $V$, let $\pi:H\to V_\gamma$ be elementary, with $\gamma$ a sufficiently large ordinal, and $H$ transitive and countable, and $\mathrm{rg}(\pi)$ containing $B,\mathfrak{B}$. The proper class mice $M_n(B)$ are determined by set-sized mice $M_n^\#(B)$, and these are definable from $B$ over $V_\gamma$, so are also in $\mathrm{rg}(\pi)$. Let $\pi(\bar{B},\bar{M}_n^\#(\bar{B}))=(B,M_n^\#(B))$. Then because $\pi$ is elementary, in fact $\bar{M}_n^\#(\bar{B})=M_n^\#(\bar{B})$, i.e. the collapse of $M_n^\#(B)$ is the true $M_n^\#(\bar{B})$. Claim 2: $\pi\upharpoonright\bar{\mathfrak{B}}:\bar{\mathfrak{B}}\to\mathfrak{B}$ is elementary with respect to forcing absolute second order formulas. Proof: Suppose $\bar{\mathfrak{B}}\models\varphi(\bar{x})$ where $\varphi$ is forcing absolute, but $\mathfrak{B}\models\neg\varphi(x)$ where $x=\pi(\bar{x})$. Clearly $\neg\varphi$ is also forcing absolute, so in particular, "$\mathfrak{B}\models\neg\varphi(x)$" is forcing absolute, so by Claim 1, for sufficiently large $n$, $M_n(B)$ models 'it is forced by $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$ that $\mathfrak{B}\models\neg\varphi(x)$'. So $M_n(\bar{B})$ models the same about $\bar{\mathfrak{B}}$ and $\bar{x}$ (as the theory gets encoded into the $\#$ versions). We have an $(M_n(\bar{B}),\mathrm{Coll}(\omega,\mathrm{rank}(\bar{B})))$-generic $g\in V$ (as $M_n^\#(\bar{B})$ is countable). So for large $n<\omega$, $M_n(\bar{B})[g]$ models "$\bar{\mathfrak{B}}\models\neg\varphi(\bar{x})$". But $M_n(\bar{B})[g]=M_n(\bar{B},g)$, so this model is boldface-$\Sigma^1_{n+1}$ correct, so with $n$ large enough, this gives that $\bar{\mathfrak{B}}\models\neg\varphi(\bar{x})$, a contradiction.<|endoftext|> TITLE: Examples of locally compact groups that do not admit enough finite dimensional representations QUESTION [14 upvotes]: I apologize in advance if this is well-known, but I can't seem to find the answer in the literature. Let me be precise about my question. I am looking for concrete examples of locally compact Hausdorff groups $G$ such that that there exists $a, b \in G$ with $a \ne b$, but for any continuous representation $\pi : G \to \operatorname{GL}_n(k)$ with $k$ a valuated field and $n$ a positive integer, we have $\pi(a) = \pi(b)$. I am also interested in a weaker form: the case where $k$ is the field $\mathbb{C}$ of complex numbers. If it turns out that there are no such groups, please indicate a reference or sketch a proof. Also a relevant statement which my gut tells me should be true, but I don't know enough Lie theory yet to give a complete proof: for every real Lie group $G$, there are enough finite dimensional, continuous complex representations of $G$ to distinguish points of $G$. I was thinking maybe this can be proved by exploiting the close relationship between representations of a Lie group and representations of its Lie algebra. It would be nice if someone could give a sketch if this can indeed be achieved or describe why this "hand-waving" actually does not work. Edit: Sorry for the poor choice of terminology. To truly get the "weaker form" as YCor pointed out, I should replace "valuated field" by "a field $k$ with an absolute value" $|\cdot|: k \to \mathbb{R}_{\ge 0}$, such that (1) $|ab| = |a| |b|$; (2) $|a| = 0$ if and only if $a=0$; (3) $|1 + a| \le 2$ for all $a$ with $|a| \le 1$ (or equivalently by a well-known argument, the triangle inequality). To exclude the discrete topology induced by this absolute value, we exclude the trivial absolute value that $|a| = 1$ for all $a \in k^\times$. REPLY [12 votes]: There is an example which satisfies something much stronger: there exist nontrivial groups $G$ such that any homomorphism (not even necessarily continuous) $\pi:G\to GL_n(k)$ for any field $k$ (and, in fact, any commutative integral domain) is trivial, so in particular for any $a,b\in G$ we have $\pi(a)=\pi(b)$. Since any group can be given a locally compact Hausdorff topology, namely the discrete topology, these will in particular answer your question. As for examples of such groups $G$, we can take any finitely generated which has no finite quotients, e.g. the Higman group mentioned in the comment by Terry Tao. This result is apparently due to Mal'cev, but I don't have a reference at hand so here is a sketch of an argument. It is enough to show that if you have a finitely generated subgroup $\Gamma$ of a group $GL_n(R)$ for some integral domain $R$ (e.g. the image of $G$ under a representation), then $\Gamma$ is residually finite (hence trivial, if $\Gamma$ is a quotient of $G$ which has no finite quotients). The main idea of the proof is to replace $R$ by some ring which is finitely generated over $\mathbb Z$, meaning it is a quotient of a polynomial algebra over $\mathbb Z$. Then $R$ is a Jacobson ring, so in particular the intersection of all maximal ideals of $R$ is trivial, and moreover for all maximal ideals $m$ of $R$ we have $R/m$ finite. Now if $\gamma\in\Gamma$ is any nontrivial element, then there is a maximal ideal $m$ not containing all coefficients of $\gamma$, so $\gamma$ has nontrivial image in the finite group $GL_n(R/m)$.<|endoftext|> TITLE: (update) Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$? QUESTION [20 upvotes]: Problem: Given three positive integers $0 < n_1 < n_2 < n_3$ such that $$n_1 + n_2 \ne n_3, \quad n_2 \ne 2n_1, \quad n_3 \ne 2n_1, \quad n_3 \ne 2n_2,$$ is there always a real number $x$ such that $$\cos n_1 x + \cos n_2 x + \cos n_3 x < -2?$$ This is a follow-up of the question in MO: Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$? I checked that (brute force), the statement is true for all positive integers $1\le n_1 < n_2 < n_3 \le 1000$ with $n_1 + n_2 \ne n_3, \, n_2 \ne 2n_1, \, n_3 \ne 2n_1, \, n_3 \ne 2n_2$. For $n_3 > 1000$, I randomly generate some $n_1, n_2, n_3$ without finding a counterexample. I also found a related problem in MO: The maximum of a real trigonometric polynomial. Perhaps the problem here is easier to deal with. My idea is to consider the case $\cos n_1 x = -1$ or $\cos n_2 x = -1$ or $\cos n_3 x = -1$. For example, $\cos n_3 x = -1$ leads to $x = \frac{(2k_3 + 1)\pi}{n_3}$ where $0\le k_3 \le n_3; k_3\in \mathbb{Z}$; We consider $f(k_3) = \cos \frac{(2k_3 + 1)n_1\pi}{n_3} + \cos\frac{(2k_3 + 1)n_2\pi}{n_3} - 1$. However, I failed to go proceed. I also tried non-negative trigonometric polynomials and found this article: "Extremal Positive Trigonometric Polynomials", https://www.dcce.ibilce.unesp.br/~dimitrov/papers/main.pdf Theorem 4 gives the necessary and sufficient condition for $f(\theta) = a_0/2 + \sum_{k=1}^n a_k \cos k \theta$ to be non-negative on $\mathbb{R}$. REPLY [25 votes]: In principle this problem can be resolved numerically in finite time, by exploiting the dichotomy between structure (small linear relations between the frequencies $n_1,n_2,n_3$) and randomness (equidistribution), though I do not know if the approach below can actually be implemented in a feasible amount of time. (One could in fact use quantitative equidistribution theorems on tori for this, such as Proposition 1.1.17 of this book of mine, but I will take a more explicit approach here as it will likely give better numerical constants.) We first claim that if $\cos(n_1 x) + \cos(n_2 x) + \cos(n_3 x) \geq -2$ for all $x$ then there must be a non-trivial linear relation $a_1 n_1 + a_2 n_2 + a_3 n_3 = 0$ between the $n_1,n_2,n_3$ with integers $a_1,a_2,a_3$ with $|a_1| + |a_2| + |a_3| \leq C$ for some effectively computable absolute constant $C$. Indeed, use the Weierstrass approximation theorem to find a polynomial $P: {\bf R} \to {\bf R}$ such that $$ \int_0^1 \int_0^1 \int_0^1 P( \cos(2\pi \theta_1) + \cos(2\pi \theta_2) + \cos(2\pi \theta_3) )\ d\theta_1 d\theta_2 d\theta_3 > \sup_{-2 \leq x \leq 3} P(x)$$ (this can be done by choosing $P$ to be large and positive on $[-3,-2]$ and small on $[-2,3]$). Then we have $$ \int_0^1 \int_0^1 \int_0^1 P( \cos(2\pi \theta_1) + \cos(2\pi \theta_2) + \cos(2\pi \theta_3) )\ d\theta_1 d\theta_2 d\theta_3 \neq \int_0^1 P( \cos(2\pi n_1 \theta) + \cos(2\pi n_2 \theta) + \cos(2\pi n_3 \theta) )\ d\theta.$$ But expanding out the polynomial and extracting the Fourier coefficients we see that the LHS and RHS in fact agree unless there is a non-trivial collision $a_1 n_1 + a_2 n_2 + a_3 n_3 = 0$ with $|a_1| + |a_2| + |a_3| \leq \mathrm{deg} P$. In order to make this algorithm run in as feasible a time as possible it is desirable to get $\mathrm{deg} P$ as small as one can; presumably this can be done numerically since for each fixed choice of degree, one has to solve a linear program in the coefficients of $P$. Next, once one has restricted to the case $a_1 n_1 + a_2 n_2 + a_3 n_3=0$ for some fixed $a_1,a_2,a_3$, one can perform a similar argument to see that either one has $\cos(2\pi \theta_1) + \cos(2\pi \theta_2) + \cos(2\pi \theta_3) \geq -2$ on the entire hyperplane $\{ (\theta_1,\theta_2,\theta_3): a_1 \theta_1 + a_2 \theta_2 + a_3 \theta_3 =0\}$ (in which the answer to your question is negative), or else there must be a second constraint $b_1 n_1 + b_2 n_2 + b_3 n_3 = 0$ with $(b_1,b_2,b_3)$ independent of $(a_1,a_2,a_3)$ and with $|b_1|+|b_2|+|b_3|$ also bounded. After reducing $n_1,n_2,n_3$ to lowest terms, this leaves one with an explicit finite list of candidate triples $(n_1,n_2,n_3)$ for which the question can be decided by case-by-case check. (To reduce the number of cases, one can normalise $0 \leq a_1 \leq a_2 \leq a_3$ (with $n_1,n_2,n_3$ now arbitrary integers) rather than $0 < n_1 < n_2 < n_3$.)<|endoftext|> TITLE: Oriented bordism in higher dimensions (e.g. $12 \leq d \leq 28$) QUESTION [8 upvotes]: The classification of oriented compact smooth manifolds up to oriented cobordism is one of the landmarks of 20th century topology. The techniques used there form the part of the foundations of differential topology and stable homotopy theory. It is a popular knowledge to find the oriented bordism groups $\Omega_d^{SO}$ in dimensions lower or equal to 8, http://www.map.mpim-bonn.mpg.de/Oriented_bordism, which we have $$\Omega_0^{SO}=\mathbb{Z}$$ $$\Omega_1^{SO}=0$$ $$\Omega_2^{SO}=0$$ $$\Omega_3^{SO}=0$$ $$\Omega_4^{SO}=\mathbb{Z}$$ $$\Omega_5^{SO}=\mathbb{Z}/2$$ $$\Omega_6^{SO}=0$$ $$\Omega_7^{SO}=0$$ $$\Omega_8^{SO}=\mathbb{Z} \oplus \mathbb{Z}$$ $$\Omega_9^{SO}=\mathbb{Z}/2 \oplus \mathbb{Z}/2$$ $$\Omega_{10}^{SO}=\mathbb{Z}/2$$ $$\Omega_{11}^{SO}=\mathbb{Z}/2$$ do we happen to know other dimensions in the literature? For any $d \leq 28$? Also are the manifold generators already known for those $d \leq 28$? Are manifold generators systematically constructible? References (precisely which pages) are surely welcome! Thank you in advance. REPLY [15 votes]: Most of the main results needed for this calculation can be found in Wall's paper "Determination of the oriented cobordism ring", but this note by Gwynne might be helpful to express this in more modern language. Here is the exective summary: All of the homotopy groups are a direct sum $\Bbb Z^r \oplus \Bbb (Z/2)^s$. Bordism classes of oriented manifolds are completely determined by their Pontrjagin and Stiefel-Whitney numbers. The mod-2 cohomology of $MSO$ is the same as the mod-2 cohomology of $BSO$, a polynomial ring on $w_2, w_3, \dots$ whose Poincare series is $$ \prod_{i \geq 2} \tfrac{1}{1-t^i}. $$ Rationally, the ring is a polynomial algebra $\Bbb Q[x_4, x_8, x_{12}, \dots]$ on generators in degrees that are a power of $4$. This tells us the rank $r$ of each group. The Poincare series for the free part of $\Omega^{SO}_*$ is thus $$ p_{free}(t) = \prod_{j \geq 1} \tfrac{1}{1-t^{4i}}. $$ $2$-locally, the bordism spectrum $MSO$ is a wedge of suspensions of Eilenberg--Mac Lane spectra $H\Bbb Z/2$ and $H\Bbb Z$. This allows us to write $$ H^*(MSO) \cong \bigoplus_\text{free summands}H^*(H\Bbb Z) \oplus \bigoplus_\text{torsion summands} H^*(H\Bbb Z/2). $$ Turning this into a Poincare series expression using the Poincare series for the cohomology of Eilenberg--Mac Lane spectra, we can solve for the Poincare series of the torsion part in $\Omega^{SO}_*$. $$ p_{tors}(t) = \left[(1-t) \prod_{k \geq 2, k \neq 2^i-1} \left(\tfrac{1}{1-t^k}\right)\right] - \left[\frac{1}{1+t}\prod_{k \geq 1}\left(\tfrac{1}{1-t^{4k}}\right)\right] $$ I asked Mathematica for a calculation of these groups out to degree 28. Assuming I didn't make a typo, here they are. $$ \begin{array}{c|l} n & \Omega^{SO}_n \\ \hline 0 & \Bbb Z\\ 1 & 0\\ 2 & 0\\ 3 & 0\\ 4 & \Bbb Z\\ 5 & \Bbb Z/2\\ 6 & 0\\ 7 & 0\\ 8 & \Bbb Z^2\\ 9 & (\Bbb Z/2)^2\\ 10 & \Bbb Z/2\\ 11 & \Bbb Z/2\\ 12 & \Bbb Z^3\\ 13 & (\Bbb Z/2)^4\\ 14 & (\Bbb Z/2)^2\\ 15 & (\Bbb Z/2)^3\\ 16 & \Bbb Z^5 \oplus \Bbb Z/2\\ 17 & (\Bbb Z/2)^8\\ 18 & (\Bbb Z/2)^5\\ 19 & (\Bbb Z/2)^7\\ 20 & \Bbb Z^7 \oplus (\Bbb Z/2)^{20}\\ 21 & (\Bbb Z/2)^{15}\\ 22 & (\Bbb Z/2)^{11}\\ 23 & (\Bbb Z/2)^{15}\\ 24 & \Bbb Z^{11} \oplus (\Bbb Z/2)^{10}\\ 25 & (\Bbb Z/2)^{28}\\ 26 & (\Bbb Z/2)^{22}\\ 27 & (\Bbb Z/2)^{31}\\ 28 & \Bbb Z^{15} \oplus (\Bbb Z/2)^{23}\\ \end{array} $$ (The OEIS doesn't seem like anybody interested in bordism theory has invested the effort into adding this type of information.)<|endoftext|> TITLE: Homology of singular chain complex modulo subdivision QUESTION [5 upvotes]: Let $S_p(X)$ be the $p$-th singular chain group and $\mathcal S(X)$ be the singular chain complex of a topological space $X$. There is a barycentric subdivision operator (which is also a chain map) $\operatorname{sd}: S_p(X) \to S_p(X)$ as is defined in standard topology textbooks such as Munkres. There is a chain homotopy from $\operatorname{sd}$ to $\operatorname{id}_{\mathcal S(X)}$. Let $D_p$ be the subgroup of $S_p(X)$ generated by $T - \operatorname{sd}(T)$ such that $T \in S_p(X)$. Then $\mathcal D:= (D_p)$ is a chain subcomplex of $\mathcal S(X)$. I wonder whether the homology of the quotient chain complex $\mathcal S(X)/\mathcal D$ is isomorphic to the homology of $\mathcal S(X)$. In fact, this is how I used to understand the proof of the excision theorem of singular homology "intuitively". However, now I begin to doubt this intuition. There are two possible ways I can think of to prove the statement above. One is to show $\mathcal D$ is acyclic and use zig-zag lemma, the other is to construct a chain homotopy between $\mathcal S(X)/\mathcal D$ and $\mathcal S(X)$. This question has been posted in MSE (link) for two days, but there is no answer. REPLY [7 votes]: I believe $\mathcal D$ is not acyclic, unless $X$ is a discrete space, and therefore $S(X)/\mathcal D$ is not quasi-isomorphic to $S(X)$. In fact, assuming that I did not make a mistake in the claim below, we have the following description of $\mathcal D$: it is zero in degree zero, and is isomorphic to $S(X)$ in positive degrees. It follows that $H_0(\mathcal D)=0$, $H_1(\mathcal D)$ is the huge group $S_1(X)/\partial S_2(X)$, and for $i>1$ $H_i(\mathcal D)\cong H_i(X)$. However, the inclusion $\mathcal D\hookrightarrow S(X)$ induces zero in homology. It is obvious that $D_0$ is the zero group. For higher degrees, we have the following claim Claim: for all $n>0$, the operator $T\mapsto T-sd(T)$ defines an injective homomorphism $S_n(X)\to S_n(X)$. It follows that $D_n\cong S_n(X)$ for all $n>0$. Sketch of proof (hope it is correct, I did not check every detail): Define a partial ordering on the set of singular $n$-simplices of $X$. Given two singular simplices $\sigma, \tau\colon \Delta^n \to X$, we say that $\sigma\le \tau$ if $\sigma=\tau$, or there exists a contracting linear map $h\colon \Delta^n\to \Delta^n$ such that $\sigma=\tau \circ h$. By contracting we mean that there exists a $0\le r<1$ such that $\mbox{dist}(h(x),h(y))\le r\cdot \mbox{dist}(x, y)$. It is not hard to check that this is a partial order. Moreover $\sigma=\sigma\circ h$ for some contracting map only if $\sigma$ is constant. Now let $0\ne T\in S_n(X)$. We can write $T=\Sigma_{i=1}^m n_i \sigma_i$, where $\sigma_1, \ldots, \sigma_m$ are pairwise distinct singular simplices, and $n_i$ are non-zero integers. If all $\sigma_i$ are constant, then $T-sd(T)=T\ne 0$. Otherwise, among the $\sigma_i$s there exists a non-constant simplex $\sigma_{i_0}$ that is maximal with respect to the order that we defined. The singular chain $T-sd(T)$ is a linear combination of $\sigma_i$s and simplices that are smaller than $\sigma_i$. It is clear that there is no summand that can cancel $\sigma_{i_0}$, so $T-sd(T)\ne 0$.<|endoftext|> TITLE: Infinite family of different prime knots with trivial Alexander polynomial QUESTION [5 upvotes]: I am looking for infinite families of prime knots that have all Alexander polynomial equals to 1. I wrote "families" (and not "family") since perhaps there are different constructions out there. Moreover, is there such a family for which all knots are of genus one (just as the Whitehead doubles)? REPLY [2 votes]: Another family of examples is given by the "generalised Kinoshita-Terasaka" knots, here is a picture from Lickorish' "An introduction to Knot Theory". Here $d$ is assumed to be even. Of course, this is not a family of genus 1 knots.<|endoftext|> TITLE: Proper morphisms that are closed immersion on a fiber QUESTION [5 upvotes]: I am interested in a morphism of $S$-schemes $f : X \to Y$ such that $X$ and $Y$ are proper over $S$ and there is some $s_0 \in S$ such that $f : X_{s_0} \to Y_{s_0}$ is a closed immersion. Is it true that there is an open neighborhood $U \subset S$ of $s_0$ so that $X_U \to Y_U$ is a closed immersion? Because $f$ is automatically proper, I hoped to use the characterization of closed immersions as proper monomorphisms. Using an argument on formal neighborhoods of the fibers, I think I can show that $f : X \to Y$ is unramified at each point of the fiber and hence should be unramified on some neighborhood of the fiber. Maybe I can also show that $f$ is radicial in a similar fashion? I am okay with assuming that $X, Y$ are noetherian and flat over $S$ if this helps. REPLY [9 votes]: This is proved in EGA III, tome 1, Proposition 4.6.7(i).<|endoftext|> TITLE: Recurrence of ergodic processes QUESTION [5 upvotes]: Let $(X_1,X_2,\ldots)$ be a stationary ergodic process with each $X_n$ a real random variable taking values in $[-1,+1]$. Suppose that $\mathbb{E}[X_n]=0$. Let $S_n = \sum_{k=1}^n X_k$. Is the process $(S_1,S_2,\ldots)$ necessarily recurrent, in the sense that there exists some $M$ such that almost surely $|S_n| \leq M$ infinitely often? REPLY [4 votes]: Yes. I will use some different notation, but the idea is the same. Let $(\Omega,\mu)$ be a probability space, and let $\sigma \colon \Omega \to \Omega$ be an ergodic measure preserving transformation. Let $f$ be a measurable function taking values in $[-1,1]$ such that $\int f\,d\mu=0$. Write $S_nf(\omega)=f(\omega)+\ldots+f(\sigma^{n-1}\omega)$. I claim that for $\mu$-a.e. $\omega$, $|S_nf(\omega)|\le c$ infinitely often for any $c>\frac 12$. Here is a proof by contradiction. Suppose that $c>\frac 12$ and there is a set of positive measure $A$ such that $|S_nf(\omega)|\le c$ only finitely many times for $\omega\in A$. Then since $c-(-c)>1$, then for $\omega\in A$, either $S_nf(\omega)>c$ for all sufficiently large $n$ or $S_nf(\omega)<-c$ for all sufficiently large $n$. Negating $f$ if necessary, we may pick an $N>0$, and a subset $B$ of $A$ of positive measure such that $S_nf(\omega)>c$ for all $n\ge N$. Now we consider the induced (first return) dynamical system on $B$, which is also ergodic. The $k$th return of $\omega$ to $B$ satisfies $t_k(\omega)/k\to 1/\mu(B)$. But by definition of $B$, we see (by splitting up the summation) that if $\omega\in B$, $$ S_{t_{kN}}f(\omega)= \sum_{j=0}^{k-1} S_{t_{(j+1)N}(\omega)-t_{jN}(\omega)} f(\sigma^{t_{jN}(\omega)})>ck. $$ This implies $S_{t_{kN}(\omega)}f(\omega)/t_{kN}(\omega)= [S_{t_{kN}(\omega)}f(\omega)/k] / [t_{kN}(\omega)/k ]$ has a positive limit superior. That contradicts the Birkhoff ergodic theorem.<|endoftext|> TITLE: Find the order of a class of finite matrices over finite fields QUESTION [9 upvotes]: Consider matrices $M$ of size $L\times L$ over a finite field $\mathbb{Z}_p$, for simplicity focus on $p$ prime. The size $L$ is even. We want to find the order of a specific class of matrices, namely we want to find the smallest non-zero integer $n$ such that $$M^n=1,$$ where 1 is here the identity matrix of size $L\times L$. The matrix $M$ has a specific block structure, which originates from a specific linear cellular automaton. We have $$M=AB,$$ where $$A= \begin{pmatrix} 1 & 1 & & && & \\ 1 & -1 && & & & \\ & & 1 & 1 && &\\ & & 1 & -1 && & \\ &&&& \ddots & & \\ & & & &&1 & 1\\ & & & &&1 & -1\\ \end{pmatrix}$$ and $$ B= \begin{pmatrix} -1 && & & && 1&\\ & 1 & 1 && && \\ & 1 & -1 && && \\ &&& \ddots & & & \\ & & &&1 & 1&\\ & & &&1 & -1&\\ 1 && & & && 1\\ \end{pmatrix}$$ You can see $B$ as $CAC^{-1}$ where $C$ is a cyclic shift over $L$ variables. Separately $A$ and $B$ have simple properties, but their alternating product becomes complicated. I did some numerical testing, choosing first $\mathbb{Z}_3$. The interesting thing is (which also motivates me to look deeper into this) that the order $n$ seems to have a complicated behaviour as a function of $L$ and it is not clear to me what is the precise source of this. Sometimes $n$ is very big, seemingly exponentially growing with $L$. For example $n(L=46)=354292=2^2\cdot 23\cdot 3851$. Or $n(L=58)=9565940=2^2\cdot 5\cdot 29\cdot 16493$. But if $L$ is divisible by 6 then we get much lower numbers, $n(60)=120$ for example. I understand that it is natural to see the prime $p$ somehow reflected in the function $n(L)$, but what I don't understand is how the big primes mentioned above can enter the game. Update: Further numerical experiments showed that the smallest orbit happens when $L=2p^m$, in which case $n=2L$, we have simple linear growth. It gets complicated and quickly growing for most other $L$. The origin of the problem is the following cellular automaton. Consider $L$ variables $s_j\in \mathbb{Z}_p$, with $j=1,2,\dots,L$. We define a dynamics on the model, such that we perform two operations cyclically. In each operation we group two neighbouring variables into a pair, and we alternate the ways how we make the pairing. For each pairing we perform the update $$(s_j,s_{j+1})\to (s_j+s_{j+1},s_j-s_{j+1})$$ What changes is that at every second update $j$ is even or odd. If you represent this linear operation with matrices, you get $A$ and $B$ and we want to iterate the product $M=AB$. REPLY [2 votes]: Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity in a suitable extension of ${\mathbb F}_p$. The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$ (which turns out to be a discrete Fourier analysis). Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s. The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the exceptional case that this lcm is odd (unlikely) and the order of the solutions of $t^2-6t+1=0$ divide $\ell$. In other words, defining the vectors $v_\omega=(1,\omega,\ldots,\omega^{\ell-1})$, we see that the subspaces defined by $y,z\in{\rm Span}(v_\omega)$ are stable under both $A$ and $B$, and we are able to compute the spectrum of $AB$ by studying its restriction to these $2$-dimensional spaces. Notice that the action on the spaces associated with $v_{\omega}$ and $v_{\omega^{-1}}$ are conjugate to each other.<|endoftext|> TITLE: Infinite dimensional Lie algebras with trivial homology QUESTION [5 upvotes]: The basic question is: Does vanishing of homology with trivial coefficients imply triviality of an infinite-dimensional Lie algebra? My question is motivated by acylic groups in group theory. In particular, there is a an acylic group $\textrm{Aut}_f(\mathbb{R})$ of autohomeomorphisms of $\mathbb{R}$ with finite support. In the same vein is it true that Lie algebra of finite vector fields on $\mathbb{R}$ is acyclic? I know that the cohomology of all vector fields is described by Goncharova's theorem, but I don't know the answer for finite vector fields (although I have seen that the answer is mentioned as known somewhere). This is an extension of a post on MSE. REPLY [8 votes]: There exist acyclic infinite-dimensional Lie algebras, i.e., for which the trivial homology vanishes in all nonzero degree (recall that $H_0$ of every Lie algebra is always 1-dimensional over the ground field $K$). Lemma. Let $\mathfrak{g}$ be the increasing union of Lie subalgebras $\mathfrak{g}_n$. Suppose that for fixed $k$, we have $H_k(\mathfrak{g}_n)=0$ for all $n$ large enough (say $n\ge N$). Then $H_k(\mathfrak{g})=0$. Proof: this is essentially formal "diagram chasing". If $b\in Z_k(\mathfrak{g})\subset\bigwedge^k\mathfrak{g}$ (spaces of $k$-cycles and $k$-chains), there exists $n$, which we can choose $\ge N$, such that $b\in \bigwedge^k\mathfrak{g}_n$. Then $d_k(b)=0$, so $b\in Z_k(\mathfrak{g}_n)$. Since $H_k(\mathfrak{g}_n)=0$, there exists $c\in\Lambda^{k-1}\mathfrak{g}_n$ such that $d_{k-1}(c)=b$. Hence $b\in B_k(\mathfrak{g})$ (space of $k$-boundaries). This proves the lemma. Corollary. Under the same assumptions, if for some sequence $(N_k)$, we have $H_k(\mathfrak{g}_n)=0$ for all $n\ge N_k$ and all $k\ge 1$, then $H_k(\mathfrak{g})=0$ for all $k$. $\Box$ So it is enough to have such an "increasing sequence of Lie algebras" $(\mathfrak{g}_n)$ at disposal. We can't choose them finite-dimensional (since $H_1$ and $H_3$ can't be both zero then, at least in char. zero). However there are examples in the literature: For instance choose $\mathfrak{g}_n$ as the Lie algebra of formal power series vector fields $$\sum_{i=1}^nf_i\frac{\partial}{\partial x_i},\quad f_i\in K[\![x_1,\dots,x_n]\!].$$ Gelfand and Fuks proved in [1] (see [2], Corollary 1) that $H_k(\mathfrak{g}_n)=0$ for all $1\le k\le n$. References: [1] I. M. Gel′fand and D. B. Fuks, Cohomologies of the Lie algebra of formal vector fields, Izv. Akad. Nauk SSSR Ser. Mat. 34 (1970), 322–337 (Russian). MR 0266195 [2] Victor Guillemin and Steven Shnider, Some stable results on the cohomology of the classical infinite-dimensional Lie algebras, Trans. Amer. Math. Soc. 179 (1973), 275-280. Link at AMS site, unrestricted access<|endoftext|> TITLE: Asymptotics of truncated logarithm on a cricle QUESTION [7 upvotes]: Consider $f_n(x) = \min_{|z|=x} \Re \sum_{j=1}^{n} \frac{z^j}{j}$, a real function of positive variable $x>0$. I am interested in lower bounds on $f_n(x)$. Specifically, I ask: what lower bounds can be given on $f_n(x)$ in the regime where $x=1+O(1/n)$ and $n$ tends to $\infty$? Trivially, $f_n(x) \ge - \sum_{j=1}^{n} \frac{x^j}{j}$ which has order of magnitude $-\log n$ (times a constant) in the aforementioned regime. However, equality cannot be achieved here, since $z^j$ cannot equal to $-x^j$ for both $j=1$ and $j=2$. I don't even know if $f_n(x)$ is eventually negative. REPLY [2 votes]: So, we have a function $u_n(z) = \Re \sum_{j = 1}^n \frac{z^j}{j}$. As a real part of an analytic function, it is a harmonic function. We are interested in its behaviour on the circle $|z| = x = 1 + \frac{c}{n}$, where $c > 0$ is some constant and we want to estimate the minimal value of $u_n$ on it. Let me start with the upper bound from my comment: $u_n$ is a harmonic function, thus it satisfies the minimum principle in the form $$\min_{|z| \le x} u_n(x) = \min_{|z| = x} u_n(x)$$ (if you only heard of the maximum principle, then it is it applied to $-u_n$). Thus, we have $f_n(x) \le u_n(-1)$. On the other hand $u_n(-1)\to -\log(2)$, so $\limsup f_n(x) \le -\log(2)$ (in reality, $u_n(-1) \le -\log(2) + \frac{1}{n}$, so actually $f_n(x) \le -\log(2) + \frac{1}{n}$). Now for the lower bound. According to the Lemma 2 from the link provided by Lucia, we have $f_n(1) \ge -\log(2) - \frac{2}{n}$. For $z$ with $|z| = x$ let us denote $w = \frac{z}{|z|}$. We have $$u_n(z) \ge u_n(w) - |u_n(z) - u_n(w)| \ge -\log(2) - \frac{2}{n} - |u_n(z)-u_n(w)|.$$ So, if we can give a uniform upper bound on $|u_n(z)-u_n(w)|$ then we get a uniform lower bound on $f_n(x)$. We have $$u_n(z) - u_n(w) = \sum_{j = 1}^n \frac{z^j-w^j}{j} = \sum_{j=1}^n w^j \frac{x^j-1}{j}.$$ Since $x = 1 + \frac{c}{n}$ for $j\le n$ we have $1\le x^j\le 1 + \frac{Cj}{n}$ for some $C$ depending on $c$ (something like $C = e^c$ should work). Plugging this in and taking the absolute values (remember that $|w| = 1$) we get $$|u_n(z) - u_n(w)| \le \sum_{j = 1}^n \frac{C}{n} = C,$$ which proves the desired bound. Note though that the final constant $D$ in $f_n(x) \ge -D$ depends on $c$ even if we are concerned with only the big values of $n$. I don't know if it can be made approaching $\log(2)$ (it definitely can't be smaller by the minimum principle argument from above).<|endoftext|> TITLE: Martingales converging in probability but not a.s QUESTION [10 upvotes]: It is known that a random series $$ \sum_{n\geq 1} X_n $$ whose terms $X_n$ are independent converges a.s. if and only if it converges in probability. Is it true that a martingale $(Y_n)$ converges a.s. if and only if it converges in probability? If not, are there any counter-examples? Thanks. REPLY [11 votes]: $\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\F}{\mathcal{F}}\newcommand{\Om}{\Omega}\newcommand{\Z}{\mathbb{Z}}$A counterexample can be obtained as follows. Let $T_1,T_2,\dots$ be independent (say) geometrically distributed random variables with fast growing means, say with $ET_i=2^i$. Let $Y_n=0$ for $0\le n\le T_1$. For time moments $n$ after that, let the values of $Y_n$ coincide with the positions of a simple random walk $W^{(1)}_\cdot$ starting from $0$ at time $T_1$ -- but only till the time, say $\nu_1$, of the first return of $W^{(1)}_\cdot$ to $0$. This is the first "step". After this, let $Y_\cdot$ stay at $0$ for time $T_2$. After that, let the values of $Y_n$ coincide with the positions of a simple random walk $W^{(2)}_\cdot$ starting in state $0$ at time $\nu_1+T_2$ -- but only till the time, say $\nu_2$, of the first return of $W^{(2)}_\cdot$ to $0$, where $W^{(2)}_\cdot$ is independent of $W^{(1)}_\cdot$. This is the second "step". Continue doing such "steps" indefinitely (assuming that the walks are independent of the $T_j$'s). Thus, we get a martingale $(Y_n)$, with respect to a certain filtration of $\sigma$-algebras. The walks occur increasingly rarely; they start at very uncertain times (with standard deviations $\asymp2^i$ and corresponding very flat distributions); and the walks last comparatively short times. Therefore, for any given large time moment $n$, $P(Y_n\ne0)$ is small. So, $Y_n\to0$ in probability. However, because the walks occur infinitely many times, clearly $Y_n\not\to0$ almost surely. Here are formal details. Construction: Let $T_1,T_2,\dots$ be independent geometrically distributed random variables (r.v.'s) (defined on some probability space $(\Om,\F,P)$) with means $ET_i=1/p_i$, where $0S_k,W^{(k)}_n=0\}, \tag{3} \end{equation*} where \begin{equation*} W^{(k)}_n:=\sum_{i\in\N}R_i\,1(S_kS_k,W^{(k)}_n=0\}<\infty$ for all $k\in\N$ almost surely (a.s.), because the simple random walk is recurrent. So, $S_k<\infty$ for all $k\in\N$ a.s. Now, for all $n\in\N$ let \begin{equation*} Y_n:=\sum_{k\in\N}1(S_kF_{k-1}$ a.s. for all $k\in\N$. To complete the construction, for each $n\in\N$ let \begin{equation*} \F_n:=\si(R_1,\dots,R_n,\{1(T_k\le j)\colon k\in\N,j\in\N,j\le n\}), \end{equation*} the $\si$-algebra generated by $R_1,\dots,R_n,\{1(T_k\le j)\colon k\in\N,j\in\N,j\le n\}$. Clearly, $(\F_n)_{n\in\N}$ is a filtration. Showing that $((Y_n,\F_n))_{n\in\N}$ is a martingale: Abusing notation as is commonly done, for a r.v. $X$ let us write $X\in\F_n$ to mean that $X$ is $\F_n$-measurable. Then obviously $1(F_0\le n)=1(0\le n)\in\F_0:=\{\Om,\emptyset\}\subseteq\F_1$ for all $n\in\N$. Using now the relations \begin{equation*} 1(S_k\le n)=\sum_{j=1}^{n-1}1(T_k=j)1(F_{k-1}\le n-j), \end{equation*} $1(T_k=j)=1(T_k\le j)-1(T_k\le j-1)$, and \begin{equation*} \{F_k\not\le n\}= \bigcap_{m=1}^n\Big(\{S_k\not\le m-1\}\cup\Big\{\sum_{i=1}^m R_i\,1(S_k\le i-1)\ne0\Big\}\Big) \end{equation*} for natural $k,j$ (which follow by (3) and (3a)), we conclude by induction on $k$ that \begin{equation*} \{S_k\le n\}\in\F_n,\quad \{F_k\le n\}\in\F_n \tag{5} \end{equation*} for all natural $k,n$. Hence, by (4) and (3a), the sequence $(Y_n)_{n\in\N}$ is adapted to the filtration $(\F_n)_{n\in\N}$. Moreover, by (4), for all $n\in\N$ we have \begin{equation*} Y_{n+1}-Y_n=\sum_{k\in\N}R_{n+1}\,1(S_k\le n,F_k\not\le n); \end{equation*} also, $R_{n+1}$ is independent of $\F_n$. So, in view of (5), $((Y_n,\F_n))_{n\in\N}$ is a martingale. Showing that $Y_n\to0$ in probability (as $n\to\infty$): By (4), \begin{equation*} 1(Y_n\ne0\}=\sum_{k\in\N}1(S_k TITLE: Approximation for a series involving the derivative of a Jacobi theta function QUESTION [5 upvotes]: I’ve considered the diffusion equation $$\frac{\partial f(x,t)}{\partial t}=\frac12 \frac{\partial^2 f(x,t)}{\partial x^2}$$ with the conditions $f(x,0)=\delta(x)$ and $f(-1,t)=f(1,t)=0\ \forall t>0$ and I’ve found the solution $$f(x,t)=\sum_{n=0}^\infty \cos\left[ \left( n+\frac12 \right)\pi x \right] e^{-\frac12 (n+1/2)^2\pi^2 t}$$then I’ve considered the function$$\Lambda(t)=-\frac d{dt}\int_{-1}^1f(x,t)dx=\sum_{n=0}^\infty (-1)^n \left(n+\frac12\right)\pi\ e^{-(n+1/2)^2\pi^2 t/2}=\\=\frac\pi2\left(e^{-\frac{\pi^2} 8t}-3\,e^{-\frac{9\pi^2} 8t}+5\,e^{-\frac{25\pi^2} 8t}-\ldots\right)=\frac \pi 4 \vartheta_1'(0,e^{-\pi^2 t/2})$$where $\vartheta_1'(u,q)=\dfrac{\partial}{\partial u}\vartheta_1(u,q)$ and $$\vartheta_1(u,q)=2\,q^{1/4}\sum_{n=0}^\infty (-1)^n q^{n(n+1)}\sin\left[\left(2n+1\right)u \right]$$is a Jacobi theta function. It turns out that $$\lim_{t\to0^+}\frac{\Lambda(t)}{g(t)}=1$$ where $g(t)=\sqrt{\dfrac2{\pi t^3}}e^{-\frac1{2t}}$ (so Mathematica says and the approximation is very good as the values of $\Lambda$ and $g$ differ for less than $1\%$ for $00$. Then use Poisson's summation formula $$\sum_{-\infty}^\infty f(n)=\sum_{-\infty}^\infty \hat{f}(2\pi n),$$ where $$\hat{f}(s)=\int_{-\infty}^\infty f(x)e^{-isx}dx$$ is the Fourier transform. This Fourier transform can be explicitly computed: $$\hat{f}(s)=\frac{1}{\sqrt{2\pi}}t^{-3/2}e^{is/2}\left((s/\pi+1)e^{-(s/\pi+1)^2/(2t)}-(s/\pi-1)e^{-(s/\pi-1)^2/(2t)}\right).$$ The trick is that in the Fourier transform your parameter $t$ will stand in the denominator of the exponent, so the series $\sum\hat{f}(n)$ will be asymptotic to the sum of $3$ terms (with $n=0,\pm1$), z when $t\to 0$, $$\sum_{-\infty}^\infty\hat{f}(2\pi n)\sim \hat{f}(0)+\hat{f(2\pi)}+\hat{f}(-2\pi)\sim 2\hat{f}(0)=2\sqrt{\frac{2}{\pi t^3}}e^{-1/(2t)},\quad t\to 0+,$$ and since $\Lambda(t)$ is $1/2$ of this sum, we obtain the answer that wrote. (In the computation of Fourier transform, I started with the $e^{-y^2/(2t)}$, and then used the transformation rules of Fourier transform: multiplied my function on $y$, then on $\sin y$, and then scaled by $\pi$ and added $1/2$ to the argument). Remark. This approximation is related to the famous computation of the age of Earth by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the spherical Earth, while the asymptotics corresponds to the flat Earth approximation. It is unclear from his papers on the subject, whether Kelvin knew the exact solution and Poisson's formula.<|endoftext|> TITLE: Classification of 3-dimensional manifolds with boundary QUESTION [13 upvotes]: It is well-known that every closed, connected and orientable 3-manifold $\mathcal{M}$ can uniquely be decomposed as $$\mathcal{M}=P_{1}\#\dots\# P_{n}$$ where $P_{i}$ are prime manifolds, i.e. manifolds which can not be written as a non-trivial connected sum of two 3-manifolds. My main question is the following: Is there are similar result for 3-dimensional compact, connected and orientable manifolds with connected boundary? I know that in this case there are two types of connected sums, namely: the internal connected sum, which is defined as the connected sum of manifolds without boundary by choosing balls, which are purely in the interior. This also includes the hybrid case, i.e. the connected sum of a manifold with boundary with a manifold without boundary and the boundary connected sum, where one choses balls on the boundaries and glues them together. In any case, can any 3-manifold with boundary (with the properties written above) be decomposed in some kind of connected sum? As a second, very related question: If I fix the boundary to be some compact, oriented surface of arbitrary genus, is there some kind of classification/decomposition result classifying all manifolds having a boundary homeomorphic to our chosen boundary surface? (This question was previously posted to MathStackExchange) REPLY [9 votes]: With regard to the first question: There are a couple of versions of unique decompositions for 3-manifolds with boundary, with respect to boundary connected sum. See Gross, Jonathan L. The decomposition of 3-manifolds with several boundary components. Trans. Amer. Math. Soc. 147 (1970), 561–572, and Swarup, G. Ananda Some properties of 3-manifolds with boundary. Quart. J. Math. Oxford Ser. (2) 21 (1970), 1–23.<|endoftext|> TITLE: An abelian group associated to divisors of an integer $N$ QUESTION [7 upvotes]: A divisor $d$ of a natural integer $N$ defines a permutation of $\{0,\ldots,N-1\}$ by considering $$x\longmapsto \pi_{d\vert N}(x)=\left\lfloor \frac{x}{d}\right\rfloor+\frac{N}{d} \left( x\pmod d\right)$$ for $x \pmod d$ in $\{0,\ldots,d-1\}$. One can show that the group $A(N)$ generated by all permutations $\pi_{d\vert N}$ associated to divisors of $N$ is abelian. (The group $A(N)$ is in fact generated by $\pi_{p\vert N}$ for $p$ running through all prime divisors of $N$.) Example: $A(p^k)$ is cyclic of order $k$ if $p$ is prime. It seems that $A(N)$ is fairly small: I have no example of $N\geq 3$ such that $A(N)$ has more than $N-2$ elements. (It is however not cyclic in general.) What does the group $A(N)$ measure? REPLY [3 votes]: All the $\tau$'s fix $0$ so we can look at how they behave on the set $\{1,2,\dots,N-1\}$. We have $$\tau_{d|N}(kd+r)=k+\frac{rN}{d}=\left[\frac{N}{d}(kd+r)\right]_{\pmod{N-1}}$$ where $[a]_{\pmod{N-1}}$ is the unique integer in $\{1,2,\dots, N-1\}$ congruent to $a\pmod{N-1}$. So $\tau_{d|N}$ is the same as multiplication by $\frac{N}{d}$ modulo $N-1$. Therefore we see that the group $A(N)$ is the subgroup of $(\mathbb Z/(N-1)\mathbb Z)^{\times}$ generated by the divisors of $N$, therefore $$|A(N)|\le \varphi(N-1)\le N-2$$ as predicted by the OP.<|endoftext|> TITLE: When must a set of sections which is Zariski dense in the generic fiber also be dense in some special fiber? QUESTION [6 upvotes]: Let $f : X\rightarrow S$ be a flat finite type morphism of schemes with $S$ integral and Noetherian. Let $\eta\in S$ be the generic point. Let $\{\sigma_i\}$ be a collection of sections of $f$ (possibly infinite), which are Zariski dense in $X_\eta$. I'm interested in additional conditions on $f,\{\sigma_i\}$ under which one or both of the following properties are satisfied: P1: There exists a point $s\in S - \{\eta\}$ such that $\{\sigma_i\}$ is dense in the fiber $X_s$. P2: There exists a nonempty open $U\subset S$ such that P1 holds for every $s\in U$. Clearly a necessary condition is that $S - \{\eta\}$ needs to be ``large'' (e.g., P1 will often fail if $S$ is the spec of a discrete valuation ring). From now on lets assume $S - \{\eta\}$ is Zariski dense in $S$. My intuition is that under mild additional assumptions there should be some kind of semicontinuity result for the dimensions of the Zariski closures inside fibers. In particular, the set of $s\in S$ such that $\{\sigma_i\}$ is not Zariski dense in $X_s$ should be closed in $S$. However I'm not aware of any results in this direction. Here are some specific questions: (1) Do P1, P2 hold under the above assumptions? (2) What if we also assume $f$ has geometrically irreducible fibers? (3) What if $X$ is a semisimple affine algebraic $S$-group scheme? REPLY [7 votes]: Building on Yosemite Stan's example, for any scheme over any number ring, even if you take all the sections, it will still not be Zariski dense in any special fiber, because all sections go through the rational points of the special fiber, which are not dense as the base field is finite. Of course, there are many examples of schemes over number rings with Zariski dense global sections. Similar examples work for $\mathbb A^1$ over any curve of dimension $1$ over a countable field. There are countably many closed points, order them, and choose the $n$th section to agree with at least one of the previous $n-1$ sections at all of the first $n-1$ closed points. Then in the $n$'th point there will be at most $n$ distinct sections, so they won't be Zariski dense. Over an uncountable field, the same construction proves that any countable set of points can be the set where the sections aren't dense, so there will not be an open set where the sections are dense. However, one can prove there is at least one closed point where the sections are dense. First, pick a countable subset of the sections that is Zariski dense at the generic point. Then, for every $d$, by Zariski density we can find some finite set of sections that don't satisfy any nontrivial degree $d$ equation over the generic point, and then the set where those sections do satisfy some nontrivial degree $d$ equation is contained in a proper closed subset. Over un uncountable field, the complement of the union of countably many proper closed subsets will be nonempty, and any point in that nonempty closed set does the trick.<|endoftext|> TITLE: Distinguishing poly-free groups QUESTION [7 upvotes]: I have two discrete groups $G_1$ and $G_2$ sitting in the following exact sequences: $1\to H_1\to G_1\to K_1\to 1$ and $1\to H_2\to G_2\to K_2\to 1$. $H_1$, $K_1$, $H_2$ and $K_2$ are all non-abelian free groups of ranks $k+n$, $k$, $k+l+n$ and $k+l$ respectively. Also $k,l>1$ and $n\geq 1$. Somehow I feel that $G_1$ and $G_2$ are not isomorphic! May be there is some easy way to see if it is true or not. Edit: The Hochschild-Serre spectral sequence gives the following. $0\to H_1(H_1, {\Bbb Z})_{K_1}\to H_1(G_1, {\Bbb Z})\to H_1(K_1, {\Bbb Z})\to 0$ $0\to H_1(H_2, {\Bbb Z})_{K_2}\to H_1(G_2, {\Bbb Z})\to H_1(K_2, {\Bbb Z})\to 0$ with action of $G_i$, $i=1,2$, is trivial on $\Bbb Z$. The action of $K_i$ on $H_1(H_i, {\Bbb Z})$, $i=1,2$, giving the co-invariant $ H_1(H_i, {\Bbb Z})_{K_i}$ is mysterious! Edit: Let $S_1$ and $S_2$ be two $2$-manifolds with non-abelian free fundamental groups of ranks $k$ and $k+l$ respectively. Consider the configuration space $C(S_i)$ of $2$-tuple of ordered different points in $S_i$, $i=1,2$. Then taking the projection to one coordinate gives a fibration $C(S_i)\to S_i$ with fiber $S_i$ minus a point (Fadell-Neuwirth fibration theorem). Hence we get two exact sequences as above with $n=1$: $G_i=\pi_1(C(S_i))$, $K_i=\pi_1(S_i)$ and $H_i=\pi_1(S_i- \{\mbox{point}\})$ for $i=1,2$. Furthermore, consider the configuration spaces of $m$-tuple of ordered different points of $S_i$. Then the claim is that the fundamental groups are not isomorphic. These groups are poly-free and hence the title of this thread. REPLY [5 votes]: Euler characteristic is multiplicative in the setting of your exact sequences $1\to H_i\to G_i\to K_i\to 1$, i.e. $\chi(G_i)=\chi(H_i)\chi(K_i)$. (You can see this directly by building a model for each $G_i$ as a graph of graphs, or by more sophisticated arguments.) In your case, this gives $\chi(G_1)=\chi(H_1)\chi(K_1)=(k-1)(k+n-1)$ while $\chi(G_2)=\chi(H_2)\chi(K_2)=(k+l-1)(k+n+l-1)$. In particular, if $l>0$ we can see that $\chi(G_2)>\chi(G_1)$, which distinguishes the two groups.<|endoftext|> TITLE: What are the strongest arguments for a genuine quantum computing advantage? QUESTION [34 upvotes]: Despite having become a fairly mature field with enormous sums of money dumped into research and development, there does not as yet exist a formal proof that quantum computation actually provides an advantage. Having learned a bit about quantum circuits, quantum error correcting codes, and all the related basics of quantum information theory in university, the only arguments that were presented to me for quantum supremacy were: The existence of Shor's algorithm, The seeming infeasibility of simulating quantum systems with classical computers, Neither of which are I find particularly compelling, especially when compared to the evidence for $P \neq NP$ for example. My question is, what are the current "best" arguments for quantum advantage, assuming an ideal quantum computer (either by direct evidence for an advantage, or against the implications there being no advantage)? Given the lack of a hard proof, I am being deliberately vague in my use of the word "best". I would hope that given the billions of dollars dumped into the field, that there would be extremely strong conjectures and circumstantial evidence of quantum advantage being real/its negation being false. Edit: By "ideal quantum computer", I mean a quantum computer in the abstract, similar to how we might consider an abstract turing machine or other classical computer model as opposed to any specific physical architecture with all the related engineering concerns. REPLY [4 votes]: In addition to Shor's algorithm and to Grover algorithm let me mention two important pieces of information that abstract quantum computer have for some specific goals superior computational powers. The ability of quantum computers, and even quantum circuits of bounded depth to perform sampling tasks that goes beyond the polynomial hierarchy! (Evidence that ${\bf QuantumSampling \not \subset PH}$.) The evidence that the class of decision problems that quantum computers can solve goes beyond the polynomial hierarchy! (Evidence that ${\bf BQP \not \subset PH}$) The first item is a long story related to early works of Terhal and DiVincenzo (2004), Aaronson and Arkhipov (2013) and Bremner, Jozsa, and Shepherd (2011). Related also to Boson Sampling. The second item (with a weaker type of evidence) is related to a 2018 work by Raz and Tal. ${\bf QuantumSampling}$ (which is one aspect of the power of quantum computers that reflects the power of quantum computers for sampling) may be in a sense stronger than ${\bf BQP}$ (The power of quantum computers for decision problems) since it is a plausible conjecture that ${\bf QuantumSampling \not \subset P^{BQP}}$. (And even ${\bf QuantumSampling} \not \subset {\bf PH^{BQP}}$.)<|endoftext|> TITLE: Decomposition of manifolds with toroidal boundary QUESTION [7 upvotes]: Let $\mathcal{M}$ be a compact, connected, oriented 3-manifolds with non-empty connected boundary $\partial\mathcal{M}$. Then, following this article, it is stated that $\mathcal{M}$ can be written as $$\mathcal{M}=P_{1}\#_{\partial}\dots\#_{\partial}P_{n}$$ where $\#_{\partial}$ denotes the boundary connected sum and where $P_{i}$ are $\partial$-prime manifolds, i.e. 3-manifolds with non-empty connected boundary, which are not homeomorphic to a 3-ball and for which a decomposition like $P=Q_{1} \#_{\partial}Q_{1}$ implies that either $Q_{1}$ or $Q_{2}$ is a closed 3-ball. So this is basically a generalization of the famous prime decomposition ("Kneser-Milnor theorem") to the case of manifolds with boundary. Now, lets say I only consider manifolds $\mathcal{M}$ with the property that $\partial\mathcal{M}$ is homeomorphic to the 2-torus $T^{2}=S^{1}\times S^{1}$. Then, if I understand the theorem above correctly, $\mathcal{M}$ has to be a prime-manifold: Suppose that $\mathcal{M}$ can be decomposed as $\mathcal{M}=P_{1}\#_{\partial}P_{2}$ for two prime manifolds. However, the boundary connected sum has the property that $\partial\mathcal{M}=(\partial P_{1})\# (\partial P_{2})$, which is not possible in our case, since $\partial\mathcal{M}=T^{2}$ and the 2-torus cannot be obtained as the connected sum of two other manifolds. What I am wondering is the following: Is every compact, connected, oriented 3-manifold $\mathcal{M}$ with non-empty connected boundary $\partial\mathcal{M}\cong_{\mathrm{homeo.}} T^{2}$ of the form $$\mathcal{M}\cong_{\mathrm{homeo.}}\overline{T^{2}}\#\mathcal{N},$$ where $\overline{T^{2}}=S^{1}\times D^{2}$ denotes the solid torus and where $\mathcal{N}$ is a closed, orientable and connected 3-manifold. The connected sum here is the internal one. (This is a follow-up question to this MathOverflow post. ) REPLY [5 votes]: The situation is similar to the one of closed manifolds. One defines "boundary-prime" manifolds as those that cannot be decomposed nontrivially in a boundary-connected sum. Note that if $M$ is connected, has nonempty boundary and is not prime, then $M$ is never boundary-prime. Namely, take a 2-sphere $S\subset M$ separating $M$ into two components none of which is a ball. Connect $S$ to $\partial M$ by a 1-handle $D^2\times [0,1]$ ($D^2\times \{0\}\subset S$, $D^2\times \{1\}\subset \partial M$). Now, remove from $S$ the open disk equal to $int(D^2)\times \{0\}$ and add to $S$ the annulus $\partial D^2\times [0,1]$. The resulting surface $S'$ is a 2-dimensional disk with $\partial S'\subset \partial M$. This disk will cut $M$ in two components, none of which is a ball. This works no matter what $\partial M$ is, in particular, if $\partial M=T^2$. Thus, in what follows (until the concluding paragraph), I will assume that $M$ is prime. Lemma. If $\partial M$ is a torus, then $M$ is necessarily boundary-prime. Proof. Let $D\subset M$ be a properly embedded disk splitting $M$ in two components, none of which is a 3-ball. (Splitting means that you remove from $M$ an open tubular neighborhood of $D$.) Since $D$ separates $M$, $\partial D$ separates $T^2=\partial M$, hence, bounds a disk $D'\subset T^2$. Taking the union $D\cup D'$ we obtain a non-properly embedded 2-sphere in $M$. Pushing the disk $D'$ slightly into $M$, we obtain a 2-sphere $S\subset M$ disjoint from the boundary. Since $D$ was splitting $M$ into two submanifolds none of which is a 3-ball, the same holds for $S$. Hence, $M$ is not prime, contradicting the standing assumption. qed We continue the discussion of prime manifolds $M$ with toral boundary. Such a manifold still can have compressible boundary. However, if this is the case, a boundary-compressing $D$ disk in $M$ is necessarily nonseparating. (Since every separating loop in the torus $T^2$ bounds a disk in $T^2$.) Cutting $M$ open along $D$ results in a manifold $M'$ with spherical boundary. If $M'$ is homeomorphic to the 3-ball, then $M$ itself is a solid torus, $\hat{T}=D^2\times S^1$. Otherwise, attaching $B^3$ along $\partial M'$ results in a closed 3-manifold $N$ which is not $S^3$. Then $M= N\# \hat{T}$. But there is one more possibility, namely, $\partial M$ is incompressible. There are many manifolds like that, for instance, the exterior of any nontrivial knot in $S^3$. Hence, we obtain the trichotomy for 3-manifolds $M$ which need not be $\partial$-prime. To conclude: Suppose that $M$ is a connected (not necessarily prime) 3-manifold and $\partial M$ is homeomorphic to $T^2$. Then one of the following mutually exclusive properties holds: $M$ is not prime, equivalently, is not $\partial$-prime. $M=\hat{T}$. $M$ is prime and $\partial M$ is incompressible.<|endoftext|> TITLE: Inserting points into a polygon QUESTION [5 upvotes]: I want to insert $n$ points into arbitrary polygon $P$ described by ordered list of its vertexes $v_1, v_2, ..., v_m$. Each inserted point must distanced from the others on distance at least $d$. In case when $n$ it to large to do it, we should be able to detect such situation. Here is example solution for $n=8$ (green circle diameter is equal to some $d$). I try to find something about this problem on internet but no luck. So also I try myself to figure out how to do it - my idea was to split polygon into convex parts and put points on each part but there is problem to set proper distance between points in two parts (and putting points into convex part is also not trivial for me). Any idea? REPLY [2 votes]: I don't think there is an easy greedy algorithm for this, circle packing is a hard problem and most results are about simple instances e.g. circle packing in a circle, square, equilateral triangle and it is one of the problems in the book "Unsolved problems in geometry there is an upperbound on $n$ of $\lfloor{\frac{A + Pd}{\pi d^2}}\rfloor ​+ 1$ where $A, P$ are the area and perimetar of the polygon, which can be proved easily for the case of convex polygons, but that upperbound is not always possible I suggest to try to use randomized algorithms, I also do think that genetic algorithms can get great results for this problem<|endoftext|> TITLE: Can one detect a cyclic and separating vector for a concrete $C^*$-algebra using a dense subalgebra? QUESTION [9 upvotes]: Let $A$ be a $C^*$ algebra of operators acting on some Hilbert space $H$, and $A_0$ is a norm dense $*$-subalgebra of $A$. Suppose there exists some unit vector $\xi \in H$, such that (i) $A_0 \xi$ is dense in $H$; (ii) $a\xi = 0$ if and only if $a=0$ for all $a \in A_0$, i.e. $\xi$ is separating for $A_0$. Question: is $\xi$ also separating for $A$, i.e. is it true that $a \xi = 0$ implies $a=0$ for all $a \in A$? It is clear that if the vector state $\omega_\xi$ of $\xi$ is a trace on $A$, then the answer is affirmative by (i). More generally, if $\omega_\xi$ is a KMS state (so that we have sufficient control of $\omega_\xi$ from being away from a trace using suitable automorphisms of $A$), then the answer is still affirmative. But in general, I don't know the answer, and suspect that there is a counter-example. REPLY [2 votes]: Here is another example, following a somewhat naive intuition that there could be a suitably chosen involutive algebra generated by operators with sufficiently large range, thus having a high chance of admitting a cyclic and separating vector, yet its norm closure contains sufficiently many rank-one projections, thus such a vector could not be separating for the closure. Let $H=\ell^2(\mathbb{Z})$ and $(\delta_k)$ the standard orthonormal basis. Denote the corresponding matrix unit by $e_{i,j}$, i.e. $e_{i,j}(\sum_k \xi_k \delta_k) = \xi_j \delta_i$ and pose $e_k = e_{k,k}$. Consider two self-adjoint operators $S, T \in B(H)$ defined as follows: $S(\delta_0) = \delta_0$, and $S(\delta_k) = \frac{1}{2k}\delta_k$ for $k\ne 0$; or equivalently, $S = e_{0} + \sum_{k \ne 0} \frac{1}{2k}e_k$ with resepct to the norm topology. $T(\delta_0) = 0$, and $T(\delta_k) = - \delta_{-k}$; or equivalently, $T = -\sum_{k \ne 0} e_{k,-k}$ with respect to the strong topology. We calculate $$ ST = - \sum_{k \ne 0} \frac{1}{2k} e_{k,-k} = \sum_{k \ne 0} \frac{1}{2k} e_{-k,k} = -TS. $$ Also note that $T^2 = 1 - e_0$, while $Te_0 = e_0 T = 0$ and $S^k e_0 = e_0 S^k = e_0$ for all $k \ge 0$, with the usual convention that $S^0=1$. Let $\mathcal{A}_0$ be the unital (involutive since $S$, $T$ are self-adjoint) subalgebra generated by $S$ and $T$, and let $\mathcal{A}$ be the norm closure of $\mathcal{A}_0$. By the above calculation, we see that $\mathcal{A}_0$ is linearly spanned by $e_0$, $S^i$ with $i \ge 0$ and $S^i T$ with $i \ge 0$. Let $\xi = \delta_0 + \sum_{k > 0}\frac{1}{k}(\delta_k + (-1)^k \delta_{-k}) = (\xi_k) \in H$. We claim that $\xi$ is separating for $\mathcal{A}_0$. Indeed, let $x = a e_0 + \sum_{i=0}^N (b_i S^i + c_i S^i T) \in \mathcal{A}_0$, with $a$, $b_i$, $c_i$ being coefficients. To prove the claim, we only need to show that $x \xi = 0$ implies $x = 0$. We calculate $$ x \xi = \left(a+ \sum_{i=0}^N b_i\right) \delta_0 + \sum_{k \ne 0}\sum_{i=0}^N (2k)^{-i} (b_i \xi_k - c_i \xi_{-k}) \delta_k. $$ Suppose $x \xi = 0$. For $k > 0$, we compare the coefficients of $\delta_k$ in $x \xi$. If $k$ is odd, we have $\xi_k = - \xi_{-k} \ne 0$ and $\sum_{i=0}^N (2k)^{-i}(b_i + c_i)=0$. Take any $N+1$ different odd $k$ and note that the corresponding Vandermonde determinant does not vanish, we get $b_i + c_i = 0$ for all $0 \le i \le N$. Similarly, if $k$ is even, then $\xi_k = \xi_{-k} \ne 0$, and $\sum_{i=0}^N (2k)^{-i}(b_i - c_i)=0$. This time, taking $N+1$ different even $k$ shows that $b_i - c_i = 0$ for all $0 \le i \le N$. Hence $b_i = c_i = 0$ for all $0 \le i \le N$. Comparing the coefficient of $\delta_0$ now gives $a=0$ too. Thus $x=0$, and $\xi$ is indeed separating for $\mathcal{A}_0$. Since $\mathcal{A}_0$ is norm dense in $\mathcal{A}$, cyclicity of $\xi$ for $\mathcal{A}_0$ is equivalent to that for $\mathcal{A}$, which we now establish. By the definition of $S$, we have $\| S - e_0 \| = \frac{1}{2}$ and $e_0 (S - e_0) = (S - e_0) e_0 = 0$. Hence for all positive integer $n$, we have $e_0 + (S - e_0)^n = S^n \in \mathcal{A}_0$, which converges to $e_0$ in norm. Thus $e_0 \in \mathcal{A}$. Similarly, $(2(S - e_0))^{2n+1} \in \mathcal{A}_0$ converges in norm to $(e_1 - e_{-1}) \in \mathcal{A}$, and $(2(S - e_0))^{2n} \in \mathcal{A}_0$ converges in norm to $(e_1 + e_{-1}) \in \mathcal{A}$. Thus both $e_1$ and $e_{-1}$ are in $\mathcal{A}$. Continue in this way (i.e. raising $4(S - e_0 - \sum_{0 < |k| \le 1} \frac{1}{2k}e_k)$ to odd powers and even powers then taking limits), we get $e_2, e_{-2} \in \mathcal{A}$. Just repeat the procedure, by induction, we see that every $e_k \in \mathcal{A}$. Since $\xi_k \ne 0$, we have $\delta_k = \xi_k^{-1}e_k \xi \in \mathcal{A}\xi$ for all $k \in \mathbb{Z}$, so $\xi$ is cyclic for $\mathcal{A}$, hence for $\mathcal{A}_0$. Finally, by definition, $-Te_{-1} \delta_{-1} = \delta_1$ so $0 \ne e_1 - Te_{-1} = e_1 + e_{1,-1} \in \mathcal{A}$, and $$ (e_1 + e_{1,-1})\xi = (\xi_1 + \xi_{-1})\delta_1 = 0 $$ showing $\xi$ is not separating for $\mathcal{A}$. Normalizing $\xi$ to a unit vector answers the question.<|endoftext|> TITLE: Automorphism group of compact almost complex manifold QUESTION [7 upvotes]: Does the automorphism group of a compact almost complex manifold carry a (canonical) Lie group structure? Part 3 of Theorem 4.1 in *"The automorphism group of a homogeneous almost complex manifold" by J. Wolf (link at AMS site) says that in a specific group-invariant setting the automorphism group is compact if and only if the almost complex structure is not integrable. Is this a general phenomenon? REPLY [8 votes]: See Kobayashi, Transformation Groups, Theorem 4.1 page 16, where the theorem is proved that the group of automorphisms of a smooth compact almost complex manifold is a finite dimensional Lie group acting smoothly.<|endoftext|> TITLE: Birational geometry over finite fields QUESTION [5 upvotes]: I apologize in advance since probably my questions are very naive. I would like to understand some central notions in birational geometry, that are clear to me over the complex numbers, for varieties over finite fields. Let $X\subset\mathbb{P}^N$ be an $n$-dimensional variety over $\mathbb{F}_p$, and let $\mathbb{P}^n_{\mathbb{F}_p}$ be the $n$-dimensional projective space over $\mathbb{F}_p$. Then $\mathbb{P}^n_{\mathbb{F}_p}$ has $p^{n+1}-1$ points. Assume that there is a birational map $\mathbb{P}^n_{\mathbb{F}_p}\dashrightarrow X$. Does this imply that $X$ has at most $p^{n+1}-1$ points or even exactly $p^{n+1}-1$ points? What if the map $\mathbb{P}^n_{\mathbb{F}_p}\dashrightarrow X$ is dominant but not necessarily birational? Now, take a variety $X_{\mathbb{Q}}$ defined over $\mathbb{Q}$ and let $X_{\mathbb{F}_p}$ be its reduction modulo $p$. Assume that there is a dominant rational map $\mathbb{P}^n_{\mathbb{Q}}\dashrightarrow X_{\mathbb{Q}}$. Does this imply that there is also a dominant rational map $\mathbb{P}^n_{\mathbb{F}_p}\dashrightarrow X_{\mathbb{F}_p}$? REPLY [4 votes]: As explained in the comment of Karl Schwede, you cannot bound the number of points of $X(\mathbb{F}_p)$ if you only assume that $X$ is rational over $\mathbb{F}_p$ (i.e. that there exists a birational map $X\dashrightarrow \mathbb{P}^n_{\mathbb{F}_p}$ for some $n$, necessarily equal to $\mathrm{dim}(X)$). In any dimension $n\ge 2$, you simply blow-up a $\mathbb{F}_p$-point in $\mathbb{P}^n$ and obtain more $\mathbb{F}_p$-points: you replace one point by $\lvert \mathbb{P}^{n-1}(\mathbb{F}_p)\rvert$ points, so add at least $p$ points. You can repeat the process as many times as you want, so there is no upper bound for the number of $\mathbb{F}_p$-points. For the lower bound, as you can find in this answer: https://mathoverflow.net/q/409410 the number of points of a smooth geometrically rational variety is equal to $1$ modulo $p$ and thus you get a lower bound to be $1$. For smooth rational ones, one might ask if you get at least the same number of $\mathbb{F}_p$ points as $\mathbb{P}^n$. This is not true by taking the simple following example: in dimension $2$, take an irreducible polynomial $P\in \mathbb{F}_p[X]$, having two distinct roots $\xi_1,\xi_2$ in $\mathbb{F}_{p^2}$ and blow-up the points $[0:1:\xi_1]$, $[0:1:\xi_2]$ of $\mathbb{P}^2$. This is defined over $\mathbb{F}_p$ and does not add any $\mathbb{F}_p$-point. You then contract the strict transform of the line $x=0$ and get a smooth quadric in $\mathbb{P}^3$, isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ over $\mathbb{F}_{p^2}$ but not over $\mathbb{F}_{p}$. It will have less $\mathbb{F}_p$-points as $\mathbb{P}^2$. For the second question, it is not true as Ben C explained. There are rational varieties over $\mathbb{Q}$ (or $\mathbb{Z}$) whose reduction modulo $p$ are not anymore rational or even unirational, like $xy+pz^2$ given by Ben C.<|endoftext|> TITLE: Structure of coefficients of polynomials giving a specified Galois group QUESTION [12 upvotes]: For any $a = [a_0; \dots; a_n]\in \mathbb{P}^n(\mathbb{Q})$, the corresponding Galois group $G_a$ of $f(X) = a_n X^n + \cdots + a_1 X + a_0\in \mathbb{Q}[X]$ is a subgroup of $S_n$. (I'm interested primarily in the case where $f$ is irreducible here, but we can take $G_a = \text{Aut}(k/\mathbb{Q})$ for the splitting field $k$ of $f$ regardless.) For a given group $G \subset S_n$, what exactly can be said about the size of the set of points $a$ with $G_a = G$? The group $G = S_n$, for example, occurs exactly when the resolvent of $f$ is irreducible, and this happens for $a$ in a Zariski-dense set by Hilbert's irreducibility theorem. Generalizing from that, is there a specific sense in which $A(X) = \{a\in \mathbb{P}^n(\mathbb{Q}):\, G_a\subset X\}$ has $A(H)\subset A(G)$ "small" in $A(G)$ for $H < G$? That is, is there a result invoking some notion like the dimension of a variety (although we're necessarily working over $\mathbb{Q}$ here, or at least some other ground field that isn't algebraically closed), a thin set in the sense of Serre, etc. that makes this vague idea of "smallness" more precise? REPLY [16 votes]: Every Galois group is Zariski dense, so that's not a very exciting measurement. In fact, for any degree n field, minimal polynomials of generators of that field are Zariski dense. This follows from the fact that the map taking an element to its minimal polynomial is finite-to-one, thus sends Zariski dense sets to Zariski dense sets. (Or, more generally, for any rank n étale algebra, characteristic polynomials of nondegenerate elements are Zariski dense.) Typically the more precise question one asks is the number of such polynomials of height up to $H$, or really the asymptotics as a function of $H$. Taking Weil height, this is equivalent to asking the number of polynomials $\sum_{i=0}^n a_i X^i \in \mathbb Z[x]$ with a given Galois group, where the $a_i$ are relatively prime and $|a_i| TITLE: Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors? QUESTION [7 upvotes]: Let $ G $ be the real points of a linear algebraic group and $ G' $ a Zariski closed subgroup. Then is $ G/G' $ a Cartesian product $$ (K/K') \times F $$ where $ F $ is contractible? Here $ K,K' $ are maximal compacts of $ G,G' $. Some relevant information: Let $ G,G' $ be Lie groups with finitely many connected components (for example the real points of linear algebraic groups). They can be expressed as cartesian products $$ G= K \times E\, , \, G'=K' \times E' $$ where $ K,K' $ are maximal compacts and $ E,E' $ are contractible. According to [Mostow, G. D., Covariant fiberings of Klein spaces II, Amer. J. Math. 84 (1962), 466–474] $ G/G' $ is of the form $$ G/G' \cong K \times_{K'} F $$ where $ F \times E' \cong E $ and the $ \times_{K'} $ means taking the Cartesian product but then identifying $$ (k_0,f_0) \sim (k_0k,kf_0k^{-1}) $$ for all $ k \in K' $. We can essentially summarize this by saying that $ G/G' $ is a $ K' $ homogeneous vector bundle over $ K/K' $. Follow-up: Great answer. Exactly what I wanted. $ SO_3(\mathbb{C})/SO_2(\mathbb{C}) $ is the normal bundle over the 2 sphere (so 4 dimensional) and is nontrivial. Upon further reflection I found a less beautiful but more minimal counterexample. $ SE_2(\mathbb{R}) $ is a linear algebraic group. There is a Zariski closed subgroup given by taking $ x_{1,1}x_{2,2}=1 $ and $ x_{1,3}=0 $ and the quotient is the Moebius strip. REPLY [7 votes]: The answer is no. At least when $G$ and $H$ are semisimple, the quotient $G/H$ is diffeomorphic to the normal bundle of $K_G/K_H$ inside $G/H$ (where $K_G$ and $K_H$ denote respectively maximal compact subgroups of $G/H$), but this normal bundle might not be trivial. For a concrete example take $G= SO(n+1,\mathbb C)$ and $H= SO(n,\mathbb C)$ (seen as real algebraic groups). Then $G/H$ is the smooth complex affine quadric of dimension $n$ and $K_G/K_H = SO(n+1,\mathbb R)/SO(\mathbb R)$ is the real sphere of dimension $n$ inside it. In particular, it is a real form of $G/H$, so its normal bundle is isomorphic to the tangent bundle of the sphere, which is trivial if and only if $n= 1,3$ or $7$ !<|endoftext|> TITLE: Computing the Petersson norm of newforms of weight 2 from the symmetric square $L$-function QUESTION [9 upvotes]: Let $f \in S_2(\Gamma_0(N))$ be a newform with trivial character. I want to compute the Petersson norm $\lVert f\rVert^2$ of $f$, not normalized by $1/[\operatorname{SL}_2(\mathbf{Z}):\Gamma_0(N)]$, as in Gross–Zagier. From Numerical evaluation of the Petersson product of elliptic modular forms, I came across the formula $$\lVert f\rVert^2 = \frac{(k-1)!}{2^{2k - 1}\pi^{k + 1}}L(\operatorname{Sym}^2(f),2)$$ with $k = 2$. I implemented this in Magma. If $N$ is not square-free, I guess the correct Euler factor for the symmetric square at $p^2 \mid N$ by testing if the functional equation for the symmetric square is satisfied with $1 \pm x$ or $1 \pm px$. However, comparing with the result of PARI/gp (which is normalized by $1/[\operatorname{SL}_2(\mathbf{Z}):\Gamma_0(N)]$, so removing that normalization), it seems that I have to multiply my result by $N$ if $N$ is square-free, the reason for which I don't understand (maybe it's a convention of the implementation of the symmetric square $L$-function in Magma?). It is even worse for $N$ not square-free, e.g. $f \in S_2(\Gamma_0(125))^+$ or $S_2(\Gamma_0(147))^{w_3,w_{49}}$, where the normalization factors seem to be $125$ and $147 \cdot 7/8$, respectively. PARI's code is hard to read, and Petersson scalar products are not implemented for $N \neq 1$ in Sage. Can someone please shed light on this? REPLY [8 votes]: Let $f$ be a newform of weight $k$, level $q$, and nebentypus $\chi$, where $\chi$ is a primitive Dirichlet character modulo $q_1 \mid q$, and let $f(z) = \sum_{n = 1}^{\infty} \lambda_f(n) n^{\frac{k - 1}{2}} e(nz)$ be the Fourier expansion of $f$, where the Hecke eigenvalues $\lambda_f(n)$ are normalised such that $|\lambda_f(n)| \leq d(n)$. Let $d\mu(z) = \frac{dx \, dy}{y^2}$ and let $E(z,s) = \sum_{\gamma \in \Gamma_{\infty} \backslash \Gamma_0(q)} \Im(\gamma z)^s$ denote the real analytic Eisenstein series for $\Gamma_0(q)$. Then by unfolding, $$\int_{\Gamma_0(q) \backslash \mathbb{H}} |y^{k/2} f(z)|^2 E(z,s) \, d\mu(z) = \sum_{n = 1}^{\infty} |\lambda_f(n)|^2 n^{k - 1} \int_{0}^{\infty} y^{s + k - 1} e^{-4\pi ny} \, \frac{dy}{y}.$$ We make the change of variables $y \mapsto y/(4\pi n)$. Since $\Gamma(s) = \int_{0}^{\infty} y^s e^{-s} \, \frac{dy}{y}$, we arrive at the identity $$\int_{\Gamma_0(q) \backslash \mathbb{H}} |y^{k/2} f(z)|^2 E(z,s) \, d\mu(z) = (4\pi)^{1 - s - k} \Gamma(s + k - 1) \sum_{n = 1}^{\infty} \frac{|\lambda_f(n)|^2}{n^s}.$$ We take the residue of both sides at $s = 1$. Since the residue of $E(z,s)$ at $s = 1$ is $$\frac{1}{\operatorname{vol}(\Gamma_0(q) \backslash \mathbb{H})} = \frac{1}{[\mathrm{SL}_2(\mathbb{Z}) : \Gamma_0(q)] \operatorname{vol}(\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H})} = \frac{3}{\pi q \prod_{p \mid q} (1 + p^{-1})},$$ we deduce that $$\int_{\Gamma_0(q) \backslash \mathbb{H}} |y^{k/2} f(z)|^2 \, d\mu(z) = \frac{\pi q \prod_{p \mid q} (1 + p^{-1}) \Gamma(k)}{3 (4\pi)^k} \operatorname*{Res}_{s = 1} \sum_{n = 1}^{\infty} \frac{|\lambda_f(n)|^2}{n^s}.$$ Next, we use the fact that $$\sum_{n = 1}^{\infty} \frac{|\lambda_f(n)|^2}{n^s} = \frac{\zeta^q(s) L^q(s,\operatorname{ad} f)}{\zeta^q(2s)} \prod_{p \mid q} \sum_{r = 0}^{\infty} \frac{|\lambda_f(p^r)|^2}{p^{rs}},$$ where I write $L^q(s,\pi)$ to denote the $L$-function with the Euler factors dividing $q$ omitted, and $L(s,\operatorname{ad} f)$ denotes the adjoint $L$-function; if the nebentypus of $f$ is trivial, this is the same as the symmetric square $L$-function $L(s,\operatorname{sym}^2 f)$. Since $\operatorname*{Res}_{s = 1} \zeta^q(s) = \prod_{p \mid q} (1 - p^{-1})$, whereas $\zeta^q(2) = \frac{\pi^2}{6} \prod_{p \mid q} (1 - p^{-2})$, we arrive at the identity $$\int_{\Gamma_0(q) \backslash \mathbb{H}} |y^{k/2} f(z)|^2 \, d\mu(z) = \frac{2 q\Gamma(k)}{\pi (4\pi)^k} L^q(1,\operatorname{ad} f) \prod_{p \mid q} \sum_{r = 0}^{\infty} \frac{|\lambda_f(p^r)|^2}{p^r}.$$ To simplify this any further, we need to use the local theory of automorphic representations of $\mathrm{GL}_2(\mathbb{A}_{\mathbb{Q}})$ at each prime $p \mid q$. If $p \parallel q$ but $p \nmid q_1$, then the local component $\pi_p$ at $p$ of the automorphic representation $\pi_f$ associated to $f$ is a twist-minimal special representation (i.e. a special representation associated to an unramified character of $\mathbb{Q}_p^{\times}$), in which case $|\lambda_f(p^r)| = p^{-r/2}$, so that $\sum_{r = 0}^{\infty} |\lambda_f(p^r)|^2 p^{-r} = (1 - p^{-2})^{-1}$. In this case, we have that $L_p(1,\operatorname{ad} f) = (1 - p^{-2})^{-1}$ as well. If $p \mid q$ but $p \nmid \frac{q}{q_1}$, so that $p^k \parallel q$ and $p^k \parallel q_1$, then $\pi_p$ is a twist-minimal principal series representation (i.e. $\pi_p = \omega_1 \boxplus \omega_2$ with one of $\omega_1,\omega_2$ unramified), in which case $|\lambda_f(p^r)| = 1$, so that $\sum_{r = 0}^{\infty} |\lambda_f(p^r)|^2 p^{-r} = (1 - p^{-1})^{-1}$. In this case, we have that $L_p(1,\operatorname{ad} f) = (1 - p^{-2})^{-1}$ as well. In all remaining cases (so that $\pi_p$ is supercuspidal or a non-twist-minimal principal series representation or a non-twist-minimal special representation), $\lambda_f(p^r) = 0$ for all $r \geq 1$, so that $\sum_{r = 0}^{\infty} |\lambda_f(p^r)|^2 p^{-r} = 1$. However, we do not necessarily have that $L_p(1,\operatorname{ad} f) = 1$ as well, and in fact this only occurs when $\pi_p$ is supercuspidal and not twist-invariant by the unramified quadratic character of $\mathbb{Q}_p^{\times}$. Here we can determine the local factors at $s = 1$ of $L(s,\operatorname{ad} f) = L(s,f \otimes \widetilde{f})/\zeta(s)$ via work of Gelbart and Jacquet (namely Corollary (1.3) and Proposition (1.4)). REPLY [2 votes]: My guess is that the formula you are trying to use is only valid for $N=1$, and thus needs correction in general. Maybe Shimura's paper can help sort this out. https://doi.org/10.1002/cpa.3160290618 In (2.1), which Shimura writes for $\Gamma_1(N)$ but that doesn't matter when the character is trivial, his definition is $$\langle f,f\rangle={3/\pi\over [SL_2(Z):\Gamma_0(N)]}\int_\Phi |f|^2 dx dy.$$ Then in (2.5) you have $$\langle f,f\rangle={\Gamma(2)\over (4\pi)^2}\cdot\mathop{\rm res}\limits_{s=2} D(s,f,f)$$ where by the last display of Section 1 he defines $$D(s,f,f)=\sum_{n=1}^\infty {a_n^2\over n^s}.$$ Now a comparison of Euler products gives that the local factors of $D(s,f,f)\zeta(2s-2)$ and $L(s,Sym^2 f)\zeta(s-1)$ match, at least away from $p$ that divide $N$ (this discrepancy is the issue that David Loeffler raises). This Euler product comparison is mentioned in another paper of Shimura, see (0.4) of https://doi.org/10.1112/plms/s3-31.1.79 Anyway, this gives the answer up to the bad factors, namely $$\int_\Phi |f|^2 dx dy={[SL_2(Z):\Gamma_0(N)]\over 3/\pi}\langle f,f\rangle$$ $$=[SL_2(Z):\Gamma_0(N)]{\pi\over 3}{1\over (4\pi)^2}\mathop{\rm res}\limits_{s=2} D(s,f,f)$$ $$=[SL_2(Z):\Gamma_0(N)]{\pi\over 48\pi^2}{1\over\zeta(2)}L(2,Sym^2 f)\prod_{p|N} C_p$$ $$=N\prod_{p|N}(1+1/p)\cdot{1\over 8\pi^3}L(2,Sym^2 f)\prod_{p|N}C_p$$ Note that this matches your asserted formula when $N=1$ and $k=2$. In the more general case, considering a bad prime $p|N$, the Euler factor from $\zeta(s-1)/\zeta(2s-2)$ evaluated at $s=2$ exactly cancels out factor of $(1+1/p)$ in the index formula. Meanwhile, the Euler factor of $L(s,Sym^2f)$ when $p$ exactly divides $N$ is $(1-1/p^s)^{-1}$, as is the Euler factor of $D(s,f,f)$ in this case (since $a_p^2=1$). Finally, when $p^2|N$, the Euler factor of $D(s,f,f)$ is trivial since $a_p^2=0$, while that of $L(s,Sym^2f)$ can be known either by theory or trial-and-error computation. For the theoretical side, one can presuably work with the $p$-minimal twist of $f$ where this minimality allows twists with nontrivial Nebentypus - see 2.1 of Coates and Schmidt, particularly (2.12). https://doi.org/10.1515/crll.1987.375-376.104 I think one aspect is that if $v_p(N)$ is odd then the Euler factor of $L(s,Sym^2f)$ is trivial; while if $v_p(N)$ is even and $f$ is itself $p$-minimal then the factor is $(1+p/p^s)^{-1}$; and otherwise the Euler factor comes from that of the $p$-minimal twist (though perhaps not completely transparently, again with this $(1+p/p^s)^{-1}$ possibly appearing).<|endoftext|> TITLE: Are all free ultrafilters 'the same' in some sense? QUESTION [29 upvotes]: Consider the set of ultrafilters $\beta(\mathbb N)$ on $\mathbb N$. Any function $f\colon\mathbb N\to\mathbb N$ extends to a function $\beta f\colon \beta \mathbb N \to \beta\mathbb N$. We say that two ultrafilters $\mathcal U$ and $\mathcal V$ are isomorphic if there is some bijection $f$ with $f(\mathcal U) = f(\mathcal V)$. Since there are only $2^{\aleph_0}$ many bijections of $\mathbb N$, but $2^{2^{\aleph_0}}$ many ultrafilters on $\mathbb N$, we know that there are many isomorphism classes of free ultrafilters. On the other hand, in any proof that I have seen using ultrafilters, it does not seem to matter which ultrafilter is chosen. This leads me to the following Question: is there some way in which all free ultrafilters are the 'same'? I have thought of some possibilities what it could mean for ultrafilters to be the 'same'. We can see any ultrafilter $\mathcal U$ as an ordered set, using the partial order $\subseteq$. I can imagine that if $\mathcal U$ and $\mathcal V$ are free ultrafilters, they are isomorphic as partial orderings. This seems pretty weak though. Another possibility would be to consider the action of $\operatorname{Homeo}(\beta\mathbb N)$ on $\beta\mathbb N$. Does it act transitively? It might be interesting to consider the Rudin–Keisler ordering $\leq_{\text{RK}}$ on $\beta\mathbb N$. It is defined by $\mathcal U\leq_{\text{RK}} \mathcal V$ iff there is a function $f\colon\mathbb N\to\mathbb N$ with $\beta f(\mathcal V) = \mathcal U$. It is known that there exist free ultrafilters that are not minimal for the Rudin–Keisler ordering, while it is independent of ZFC whether there exists free ultrafilters that are not minimal. Presumably, a minimal ultrafilter is not the 'same' as a not-minimal ultrafilter. However, even then it might be consistent with ZFC that all free ultrafilters are the 'same'. REPLY [5 votes]: It is consistent that all ultrafilters are the same in the following sense: for any free ultrafilters $\mathcal U,\mathcal V$ on $\omega$ there exists a finite-to-one map $f:\omega\to\omega$ such that $f(\mathcal U)=f(\mathcal V)$. The latter statement is called NCF, the Near Coherence of Filters. In some situation it is very helpful principle. It has been thoroughly studied by Andreas Blass, see his papers: I, II, III.<|endoftext|> TITLE: Are $E_k$ monoids higher categories? QUESTION [5 upvotes]: The May Recognition Theorem establishes an equivalence between the $\infty$-categories The $\infty$-category of grouplike $E_n$ monoids The $\infty$-category of pointed $(n-1)$-connected spaces There is also an equivalence between the $\infty$-categories The $\infty$-category of $E_1$ monoids The $\infty$-category of pointed $(\infty,1)$-categories whose core is a connected space Is there analogous generalization to higher dimension? E.g. something like an equivalence between The $\infty$-category of $E_n$ monoids The $\infty$-category of pointed $(\infty,n)$-categories that are sufficiently trivial in dimensions below $n$ ? REPLY [6 votes]: This is closely related to the Baez–Dolan stabilization hypothesis. There are numerous proofs of this statement. One line of reasoning is to establish a general 1-category statement first: given a symmetric monoidal presentable (∞,1)-category $C$, the (∞,1)-category of $C$-enriched categories with one object is equivalent to the (∞,1)-category of ∞-monoids in $C$, i.e., algebras over the operad $\def\E{{\rm E}} \E_1$ in $C$. Iterating this result $n$ times, we obtain that the (∞,1)-category of categories enriched in the (∞,1)-category of categories enriched in … (repeat $n$ times) … in $C$, with a single $k$-morphism for all $0≤k TITLE: Minimal surface enclosing two congruent balls QUESTION [8 upvotes]: Let $B_1$ and $B_2$ be two unit-radius balls in $\mathbb{R}^3$ whose centers are separated by a distance $d \ge 2$. Q. For sufficiently small $d$, is the minimal area surface enclosing $B_1$ and $B_2$ formed by spherical caps joined to a catenoid of revolution? If $d$ is large enough, the minimal surface is just two disjoint spheres. If $d$ is small, then it seems natural that there is a circular ring on the surface of each ball at which a catenoid is tangent, as illustrated in a 2D profile below. Is this known to be the minimal surface? If so, is there a calculation for the position of the rings?        See also the related question: Minimal surface enclosing balls. REPLY [2 votes]: This is (mostly) just an answer to your final question. To piece together the surface of the ball with the catenoid, we have to satisfy the following relations, where $\pm x$ denote the horizontal coordinates of the dotted lines in your figure, counted from the midpoint between the balls, and $c$ is the integration constant in the catenoid solution $r(z)=c\cosh (z/c)$: $$ \sqrt{1-(d/2 -x)^2 } = c\cosh (x/c) $$ $$ \frac{(d/2 -x)}{\sqrt{1-(d/2 -x)^2 } } = \sinh (x/c) $$ These two conditions fix $c$ and $x$; plotting $x$ as a function of $d$ (I'm allowing the balls to penetrate one another and am thus plotting starting at $d=0$), There ceases to be a solution at about $d=2.399357285$. I have checked that this corresponds to the usual point where the ratio of $x$ to the ring radius goes outside the bounds within which a catenoid solution exists. However, already before reaching that maximal value of $d$, the area of the catenoid solution begins to exceed the area of the disjoint sphere solution. This happens at $d=2.319947$. Nonetheless, we therefore see that there is an interval in $d$ above $d=2$ where the catenoid solution has smaller area than the disjoint sphere solution.<|endoftext|> TITLE: Knots: locally flat, PL and smooth QUESTION [6 upvotes]: In the classical dimension (knots in $S^3$), it is considered standard (I think?) that the following sets are in bijective correspondence: locally flat knots up to ambient isotopy; PL-knots up to PL ambient isotopy; smooth knots up to smooth ambient isotopy. This should be "by work of Moise" but I am having trouble finding references. The textbooks of Burde-Zieschang (Proposition 1.10) and Kawauchi (Appendix A) seem to indicate that the set of PL-knots up to ambient isotopy coincides with the set of PL-knots up to PL-ambient isotopy. Some authors also work with tame knots which are defined as knots that are ambient isotopic to a PL-knot. So, it seems that the previous two sets also agree with the set of tame knots up to either ambient or PL ambient isotopy. Does someone know where I can find the statements that involve the locally flat and smooth cases? The closest I could get was 1.11.6 and 1.11.7 of Cromwell's book that seems to sketch a proof that the set of PL-knots up to isotopy coincides with the set of smooth knots up to ambient isotopy (the category mixing is unfortunate). Most textbooks appear to stick to the PL knots. Related but distinct MO posts include: Reference for a fact (?) on homeomorphic knot complements as well as Reference request: A knot is tame if and only if it has a tubular neighbourhood REPLY [3 votes]: Tame knots are PL was proved by Bing and Moise in 1954.<|endoftext|> TITLE: Question about Suzuki's theory of exceptional characters QUESTION [5 upvotes]: $\DeclareMathOperator\Irr{Irr}$As elegant as Suzuki's theory is, the set up requires that the number of conjugacy classes of $p$-elements in a cyclic T.I. (as an example) Sylow $p$-subgroup $P$ of $G$, $t$, is at least 2, in order to produce at least two exceptional characters of $N_G(P)$ of equal degrees which in turn generates a virtual character which vanishes off $p$-elements. My question is, what happens when $t = 1$? Is there an analogous theory to treat this case since the end result is that $\Irr(G)$ is still largely controlled by $N_G(P)$? (For instance McKay's conjecture still holds.) To illustrate, the theory of exceptional characters would work perfectly for $G = A_5$ for $p = 5$ since $N_G(P) \cong D_{10}$ have two exceptional characters of degree 2. However when $G = S_5$, it seems that the theory won't work since in this case $N_G(P) \cong C_5 \rtimes C_4$ has only 1 non-linear character, of degree 4. But $\Irr(G)$ and $\Irr(N_G(P))$ are strongly connected in this case. Is there a similar theory that addresses cases like $G = S_5$ with $p = 5$? I'd be grateful to know any literature on such theory if it exists. And I suppose when $t=1$ it's not always possible to find a normal subgroup $N$ of $G$ for which the theory of exceptional characters would work, as in the $A_5 \unlhd S_5$ case. Is that right? (I just learned this website is for research-level questions, so I moved the question from Math Stack Exchange.) REPLY [3 votes]: The theory of blocks with cyclic defect group addresses this question in slightly more generality. The beginning of this theory was Richard Brauer's Annals paper in the early 1940s which determined the structure of principal blocks of defect $1$, and showed, in particular that there is a bijection between irreducible characters in the principal $p$-block of $G$ and the principal $p$-block of $N_{G}(P)$, where $P$ is a Sylow $p$-subgroup of $G$ (of order $p$). The full character theory of $p$ blocks with cyclic defect groups was completed circa 1967 by E.C. Dade, following earlier advances by J.A. Green and J.G. Thompson. However, the full character-theoretic conclusions require methods from modular representation theory (at present), and (as far as I am aware), can't be deduced purely by ordinary (i.e., complex) character-theoretic methods. Expositions of the cyclic defect theory can be found in many modern texts on representation theory (perhaps texts by M. Isaacs and G. Navarro (and maybe also one by D. Goldschmidt) use mainly complex character-theoretic methods in their treatment of the cyclic defect theory).<|endoftext|> TITLE: Do surface groups embed into PSL_2 over a real quadratic integer ring? QUESTION [7 upvotes]: $\DeclareMathOperator\PSL{PSL}$ Let $ \mathbb{Z} $ be the ring of integers and $ \mathbb{R} $ the field of real numbers. Let $ \Sigma_g $ be a surface of genus $ g \geq 2 $. Let $ \pi_1(\Sigma_g) $ be the fundamental group of the surface. There are many way to embed $ \pi_1(\Sigma_g) $ into $\PSL_2(\mathbb{R}) $. There are, however, no ways to embed $ \pi_1(\Sigma_g) $ into $ \PSL_2(\mathbb{Z}) $. Given some $ g \geq 2 $, is there a good way (an algorithm) to find a real algebraic integer $ \alpha $ such that $ \pi_1(\Sigma_g) $ embeds in $ \PSL_2(R) $? Here $ R $ is the ring $$ R:=\mathbb{Z}[\alpha] $$ (Preferably $ \alpha $ is just a (real) quadratic extension. That is, $ \alpha $ is the root of some polynomial $ x^2+bx+c $ where $ b^2-4c \geq 0 $ and $ b,c \in \mathbb{Z} $.) History of the question: The original question claimed that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ and asked for embeddings into $ \PSL_2(\mathbb{F}) $ where $ \mathbb{F} $ is a finite degree field extension of $ \mathbb{Q} $. The claim that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ is false. In fact there are many such embeddings. The first edit of the question fixed this and asked instead for an embedding into $ \PSL_2(R) $ for $ R $ a finite rank extension of $ \mathbb{Z} $ by algebraic integers. The current version of the question is the second edit. REPLY [4 votes]: Here is, at Moishe Kohan's request, an optimal answer to the original question (which asked for a representation over any number field; as i noted in the comments there exists plenty of surface group representations with rational coefficients, as follows from a theorem of Takeuchi). To prove the existence of single a $\mathrm{PSL}_2(\mathbb Q)$-representation of a surface group is quite simpler than the full proof of Takeuchi's theorem (which is not super hard itself in the torsion-free case). Of course it suffices to prove it in genus 2. To do so observe first that it is possible to find two matrices $A_1, B_1 \in \mathrm{SL}_2(\mathbb Q)$ such that they generate a discrete subgroup of $\mathrm{SL}_2(\mathbb R)$ and $$ A_1B_1A_1^{-1}B_1^{-1} = \left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right) $$ for some $a \in \mathbb Q \setminus \{0, \pm 1\}$ and the quotient $\langle A_1, B_1 \rangle \backslash \mathbb H^2$ is a one-holed torus (the set of matrices $A, B$ satisfying the last condition is an open set, so rational matrices are dense there, and we can always conjugate to diagonalise the commutator over $\mathbb Q$). Geometrically, in the half-plane model, this means that a fundamental domain for the action of $\langle A_1, B_1 \rangle$ on the convex hull of its limit set is a polygon with one side on the geodesic $c$ the geodesic from 0 to $\infty$ and the adjacent sides are orthogonal to $c$; the boundary component of the quotient torus (the convex core of $\langle A_1, B_1 \rangle \backslash \mathbb H^2$, let us call it $T$) is the image of $c$. Now let $\sigma$ be the reflexion in $c$, and $A_2, B_2$ be the conjugates of $A_1, B_1$ by $\sigma$. Then $\langle A_2, B_2 \rangle \subset \mathrm{PSL}_2(\mathbb Q)$, and the quotient of $\mathbb H^2$ by $\Gamma = \langle A_1, B_1, A_2, B_2 \rangle$ is the double of $T$, that is a genus 2 surface. Hence $\Gamma$ is a genus 2 surface group inside $\mathrm{PSL}_2(\mathbb Q)$ which is discrete in $\mathrm{PSL}_2(\mathbb R)$. Takeuchi's argument for the full theorem is similar to this, except using Calabi--Weil rigidity instead of the geometric argument, and being more precise about commutators. It is also possible to prove it using Fenchel--Nielsen coordinates.<|endoftext|> TITLE: Irreducibility of linear sections of the commuting variety QUESTION [8 upvotes]: The irreducibility of the commuting variety $\{(A,B) \in \mathcal{M}_{n}(\mathbb{C})^2, \ AB = BA \}$ is well-known (see for instance On Dominance and Varieties of Commuting Matrices by Gerstenhaber). I am interested in the irreducibility of some special linear sections of the commuting varieties. Namely, let $W_1$, $W_2$ be two $k$-dimensional subspaces of $\mathbb{C}^n$ ($n$ and $k$ are fixed). Is the variety: $$ \{(A,B)\in \mathcal{M}_{n}(\mathbb{C})^2, \ AB = BA, \ W_1 \subset \operatorname{Ker}(A), \ W_2 \subset \operatorname{Ker}(B) \}$$ known to be irreducible? Or perhaps are there examples where it is not? REPLY [8 votes]: It is already reducible in the toy case $n=2, k=1$ where $W_1, W_2 \subseteq \mathbb{C}^2$ are two distinct lines. Without loss of generality, have them be spanned by the standard basis vectors respectively, so that $$A = \begin{bmatrix}0 & a_{12} \\ 0 & a_{22} \end{bmatrix}, \quad B = \begin{bmatrix}b_{11} & 0 \\ b_{21} & 0 \end{bmatrix}.$$ The equation $AB = BA$ now reads $$\begin{bmatrix}a_{12}b_{21} & 0 \\ a_{22}b_{21} & 0 \end{bmatrix} = \begin{bmatrix}0 & b_{11}a_{12} \\ 0 & b_{21}a_{12} \end{bmatrix}.$$ (As you can see, $AB=BA=0$.) But we have $$(a_{22}b_{21}, a_{12}b_{21}, a_{12}b_{11}) = (a_{22},a_{12}) \cap (a_{12},b_{21}) \cap (b_{11},b_{21})$$ of ideals in four variables, so there are three irreducible components of dimension $2$ here. In contrast, in the case $W_1 = W_2$, the equation $AB=BA$ instead has the form of the single equation $ab'=a'b$, which is not only irreducible, but of a different dimension and degree. So I imagine it's complicated in general, depending on $n,k$ and $\dim(W_1 \cap W_2)$. I wouldn't be surprised if there's a story here similar to Springer fibers, which vary in dimension and reducibility, but in a very nice way. edit: Investigations in Macaulay2 for the case $n=4, k=2$ give the following: If $W_1 + W_2 = \mathbb{C}^4$, then the variety has six irreducible components (two of dimension 9, four of dimension 8). If $\dim(W_1 \cap W_2) = 1$, then the variety has eight irreducible components (three of dimension 9 and five of dimension 8). If $W_1 = W_2$, then the variety has two irreducible components, both of dimension 10, but with different degrees of generators.<|endoftext|> TITLE: A geometric law of large numbers QUESTION [9 upvotes]: Question: Choose independent uniformly distributed random variables $X$ and $Y$ on a closed Riemannian manifold $(M,g).$ The geodesic midpoint of $X$ and $Y$ is well-defined a.e., and is distributed according to some probability measure $\Pi_1$ on $M.$ We can then repeat the process and choose two independent random variables distributed according to $\Pi_1,$ and the geodesic midpoint of these will again be defined a.e. and will be distributed according to a new probability measure $\Pi_2$ on $M.$ If we iterate this process, does it converge and can we characterize the limit? Hypothesized solution for convex surfaces: For convex hypersurfaces $M$ in $\mathbb R^d$ I hypothesize the sequence of measures converges to a volume form associated with a round spherical metric on $M.$ Further detail for motivation Law of large numbers in Euclidean space Let $X$ and $Y$ be independent identically distributed random variables on the real line with probability distribution $\Pi_1.$ The midpoint $(X+Y)/2$ of the line joining $X$ to $Y$ is a new random variable with distribution $\Pi_2$ (which can be computed from $\Pi_1$ by convolution and scaling). Now pick two independent random variables distributed according to $\Pi_2$ and iterate the procedure. The law of large numbers says precisely that the sequence of distributions $\Pi_1,\Pi_2,\dots$ so defined will converge to a point mass at the mean of $\Pi_1.$ Indeed, $\Pi_N$ is just the distribution of $(X_1+\cdots + X_{2^N})/2^N,$ where $X_1,\dots, X_{2^N}$ are IID with distribution $\Pi_1.$ We may even characterize the rate of convergence through the central limit theorem, which can be proved using Fourier analysis for instance. Set-up on a general manifold The same arguments are fine in Euclidean spaces of all dimensions. But this procedure may in fact be implemented in any Riemannian manifold $(M,g),$ using the metric to define the midpoint between two points. Let us keep it simple and assume that $M$ is a closed manifold, and let $\Pi$ be the uniform probability measure on $M$ (the measure obtained by rescaling the volume element coming from $g$ to have unit mass). If $X$ and $Y$ are independent uniformly distributed random variables on $M,$ then we may define $Z$ to be the midpoint of the shortest geodesic segment between $X$ and $Y.$ The point $Z$ is well-defined a.e. (likely some application of Sard's theorem to the exponential function will prove this), and determines a new probability distribution $\Pi_1$ on $M$ that is absolutely continuous with respect to the uniform measure $\Pi.$ We may now iterate this process, choosing two independent random variables distributed according to $\Pi_1$ and letting $\Pi_2$ be the distribution of their geodesic midpoint (which is defined a.e. by absolute continuity of $\Pi_1$ with respect to $\Pi$), and so on. The result is a sequence $\Pi_1,\Pi_2,\Pi_3,\dots$ of probability measures on $M,$ each absolutely continuous with respect to volume. The question is whether this sequence converges, to what, and if possible how fast. Easy examples Any curve On a closed $1$-dimensional manifold, there is no intrinsic geometry and the evolution plainly results in the uniform measure for each iteration. In other words, $\Pi = \Pi_1 = \Pi_2 = \cdots,$ where $\Pi$ is the uniform measure on $M.$ The reason is that the operations preserve the translational symmetry under which all closed curves are invariant. Constant positive curvature: spheres of any dimension In case of a sphere $S^d,$ the procedure also plainly results in the uniform measure for each iteration. This must be true because the operation of picking the midpoint is invariant under the sphere's rotational symmetries and so the resulting measure must be as well. Constant zero curvature: flat torus of any dimension On the flat torus, I expect the story to be the same based on symmetry but have not reasoned through it carefully. Non-trivial example 1: sphere with a spike Let $M$ be a $2$-dimensional sphere with a large thin spike protruding from it. Assume for simplicity that $M$ has unit volume, and let $\epsilon$ be the total area of the spike. Using the notation from the problem statement, we can see that the first measure $\Pi_1$ assigns smaller area to the spike, of the order $\epsilon^2$: speaking approximately, the midpoint of $X$ and $Y$ will live on the spike only when $X$ and $Y$ both live on the spike. This happens with probability $\epsilon^2$ given $X$ and $Y$ are independent. Non-trivial example 2: dumbbell with a long neck Let $M^2$ be the combination of two $2$-dimensionsal spheres that are very far apart connected by a very thin tubular neck. Assume $M$ has unit volume and the neck has total area $\epsilon.$ We will see that the area of the neck grows under our procedure to have area approximately $1/2+\epsilon - \epsilon^2.$ Let $X,Y,$ and $\Pi_1$ be as in the problem statement. Then, roughly speaking, the geodesic midpoint of $X$ and $Y$ is contained on the neck whenever $X$ and $Y$ are not contained on the same sphere. Each sphere has volume roughly $(1-\epsilon)/2$ and the chance that $X$ and $Y$ are contained together in one of them is roughly $(1-\epsilon)^2/2.$ Therefore the complementary probability is $1-(1-\epsilon)^2/2 = 1/2 +\epsilon - \epsilon^2,$ which is roughly the chance that the midpoint of $X$ and $Y$ is found on the neck. Some possible extensions Non-compact manifolds The same procedure makes sense on any Riemannian manifold equipped with a probability measure that is absolutely continuous with respect to the volume induced by the metric. (Perhaps all that is needed is a metric measure space.) Therefore we may ask the same questions on a non-compact manifold provided we first choose an absolutely continuous probability measure, as we do in Euclidean space A geometric evolution based on Minkowski's theorem for convex bodies The Minkowski problem in geometry assures us that a convex body in $\mathbb R^d$ is uniquely determined by its Gaussian curvature (or by the push-forward of volume to the sphere under the Gauss map). Now begin our iteration with a convex manifold $M$ in $\mathbb R^d.$ Choose $X$ and $Y$ on the sphere $S^d$ to be independent and identically distributed according to the pushforward of $M$'s volume measure under the Gauss map. We may now carry out our midpoint procedure on the sphere resulting in a new measure $\Pi_1$ on the sphere that is absolutely continuous with respect to volume. The first question is whether this new measure meets the conditions of Minkowski's theorem and determines a convex body (its centroid must be at the origin and it must not be concentrated on a great sub-sphere). Assuming it does, we obtain a new convex body with each iteration of the procedure, and we may ask two questions Does a convex body converge to the sphere under this procedure? Is there a 'natural' continuous geometric evolution equation which 'extends' this discrete evolution? REPLY [3 votes]: Updated more quantitatively: This repeated averaging can also lead to a distribution concentrated at a single point, even if $M$ is topologically a 2-sphere. Consider the unit sphere (or the earth); let $N$ be the northern hemisphere, and let $C$ be the convex hull of the equator. So $N$ is a positively curved surface and $C$ is a flat surface going through the center of the sphere. Let $M=N\cup C$, or some smoothed version. Since $\text{area}(N)=2 \pi$ and $\text{area}(C)=\pi$, the uniform distribution on $M$ assigns a weight of $\frac23$ to $N$ and $\frac13$ to $C$. But repeated averaging would lead to almost all the weight going to the center of $C$. The key insight is that in this manifold, $N$ is not convex. For some pairs of points just north of the equator, the shortest path connecting them in $M$ will go through $C$. (In the earthly $M$, the shortest path from Aceh to Bogotá would not be the great-circle path through Europe, but the path going down to the equator and then close to the center of the earth.) Similarly, take a blue strip and a green strip, each of area $\epsilon$ with points roughly $\pi/6$ away from the equator in $N$ and $C$ respectively. What is the chance that a random blue-green pair of points will have a midpoint in $C$? Points with similar longitudes will have midpoints roughly evenly split between $N$ and $C$. But for points with different longitudes, the angular distance will be traversed more quickly in $C$ than in $N$ (since the path will stay closer to the axis of the earth), and the midpoints will be in $C$ more often than in $N$. We can analyze this more quantitatively using the area-preserving map $f:[-\frac12,1] \times [0, 2\pi] \to M$ given by $$f(u,v)=\left(\sqrt{1+\min(2u,-u^2)},\ v,\ \max(u,0)\right)$$ We can identify several regions where the midpoint of $f(u,v)$ and $f(u',v')$ is in $C$: In $11$% (or $1/9$) of the cases, both points are in $C$, so the midpoint is obviously in $C$. In ~$34$% of the cases, one point is in $N$, one point is in $C$, and the midpoint is in $C$. (This came from a numerical test: given $n\in N$ and $c\in C$, I checked many points $e$ on the equator, to see whether $d_N(n,e) \arcsin(u)+\arcsin(u')+2\sin(\frac{v-v'}{2})$$ so the midpoint is in $C$. In total at least $47$% of the midpoints are in $C$ after one averaging, even though $C$ started with only $33$% of the weight. Repeating this will lead to a distribution with almost all the weight in $C$, and eventually with almost all the weight near the center of $C$, the center of the sphere. For another example, consider a cone with its base. This has a single edge, and is flat elsewhere, so the calculations are (difficult with the cut locus but) as simple as possible. Repeated averaging on this surface will also concentrate a uniform distribution on the base and then in the center of the base.<|endoftext|> TITLE: Do we know any examples of complex surfaces where we have explicit knowledge of the Chern–Weil functions? QUESTION [8 upvotes]: Let $X$ be a compact complex surface (smooth). Let $\gamma_1, \gamma_2$ denote the Chern–Weil functions. That is, if $\omega$ is a Kähler form on $X$ with volume form $\omega^2$, then $\gamma_1, \gamma_2$ are functions such that $$c_1(X) = \int_X \gamma_1 \omega^2, \quad c_2(X)=\int_X\gamma_2\omega^2.$$ In other words, the Chern–Weil functions are the functions for which the Chern classes are given by their $L^1$-norms. We have a good list of examples where we know constraints of the form $$a c_1^2(X) \leq b c_2(X) \leq k c_1^2(X)$$ for $a,b,k\in \mathbb{R}$. I'm asking for something stronger; namely: Do we know concrete examples of surfaces for which $\gamma_1^2$ and $\gamma_2$ are known? If you give me, for instance, a Barlow surface, or Beauville surface of some type, or some other surface, can we get information on the relations between $\gamma_1^2$ and $\gamma_2$? I do not mean "well, you can't have this relation, because integrating it would violate the Chern class inequality" — I'm not asking for non-relations, I'm asking for relations on the Chern–Weil functions. I'd be very surprised and pleased if such knowledge exists, but I'm doubtful. REPLY [6 votes]: If you know the volume form, then you are asking for explicit formulas for the Chern-Weil representatives of $c_1$ and $c_2$. These would come from explicit formulas for the curvature. The calculation of the curvature for a hypersurface of ${\mathbb C}P^3$, with respect to the (restriction of) the Fubini-Study metric was carried out by Al Vitter in On the curvature of complex hypersurfaces, Indiana Univ. Math. J. 23 (1973/74), 813–826."<|endoftext|> TITLE: Representation ring of the symmetric group $S_n$ in the limit as $n \to \infty$ QUESTION [6 upvotes]: Let $S_n$ denote the symmetric group on $n$-letters and let $\mathrm{Rep}(S_n)$ denote its representation ring. For every $n$ restriction along the inclusion $S_{n-1} \to S_n$ induces a ring homomorphism $\mathrm{Res}^{S_n}_{S_{n-1}} :\mathrm{Rep}(S_n) \to \mathrm{Rep}(S_{n-1 })$. Considering these together we get a diagram of the following form $$\dots \to \mathrm{Rep}(S_n) \to \mathrm{Rep}(S_{n-1 }) \to \dots \to \mathrm{Rep}(S_2) \to \mathrm{Rep}(S_1) \simeq \mathbb{Z}$$ Question: Is there a generators and relations description of the limit $\underset{n }{\varprojlim} \,\mathrm{Rep}(S_n)$ as ring? Edit: Will Sawin has pointed out in the comments that the limit, must have uncountably many generators in any presentation. Let me slightly modify the question in the hopes of making it more answerable. Lets consider instead of $\mathrm{Rep}(S_n)$ its completion with respect to the augmentation ideal $\widehat{\mathrm{Rep}}(S_n):=\varprojlim_m \mathrm{Rep}(S_n)/I^m$ where $I:= \ker(\dim : \mathrm{Rep}(S_n) \to \mathbb{Z})$. We still have a tower as above and we can ask the following variation. Question: Is there an explicit countable presentation of the limit $\underset{n }{\varprojlim} \,\widehat{\mathrm{Rep}}(S_n)$ in some suitable category of complete topological rings? REPLY [7 votes]: It follows from Hooks generate the representation ring of the symmetric group by Ivan Marin that $\operatorname{Rep}(S_n)$ is generated, and thus the inverse limit is topologically generated, by the elements $$x_k = \sum_{i=0}^k (-1)^i \binom{n+i -2}{i}\wedge^{k-i} (\operatorname{std} ) $$ So the ring is some kind of completion of $\mathbb Z[x_1,x_2,\dots ]$. Moreover, the augmentation ideal of this ring is profinite, since the augmentation ideal of $\operatorname{Rep}(S_n)$ modulo any power of the augmentation ideal of $\operatorname{Rep}(S_n)$ is finite. So the simplest guess is that it is obtained from $\mathbb Z[x_1, x_2,\dots]$ by the inverse limit over all ideals which are finite-index subgroups of the "augmentation ideal" $(x_1,x_2,\dots)$. However, this is not correct, because the sign representation is an element of the inverse limit which squares to $1$ but is nontrivial modulo $I^2$ for all $n$. However, the completion of $\mathbb Z[x_1,x_2,\dots]$ along all ideals which are finite-index subgroups of $(x_1,x_2,\dots)$ does not contain a nontrivial element which squares to $1$. A first step to understand this completion would be to understand it modulo $I^2$. In other words, what is the inverse limit over $n$ of $I_n /I_n^2$, where $I_n$ is the augmentation ideal in the representation ring of $\operatorname{Rep}(S_n)$? This is some kind of countably-topologically-generated profinite abelian group, but which one? As an alternate, more well-behaved ring, I propose to restrict attention to representations whose norms have at most polynomial growth in $n$, where the norm of a class $\alpha$ is $$\min\left\{ \dim V + \dim W \mid V, W \textrm{ representations of }S_n, [V]-[W]=\alpha \right\} .$$ I expect that this restriction of the inverse limit is a polynomial ring in the $x_k$ plus one more generator, the sign representation, that squares to one, but I haven't checked this.<|endoftext|> TITLE: Solution to sixth order equation QUESTION [6 upvotes]: I'm dealing with the expression $x = \frac{1}{3}y(y+1)(2y+1)^2(2y^2+2y+1)$. What is this approximately, if one is explicitly writing y in terms of x? There's no general formula for sixth powers unfortunately. Also can one given an approximation of this so that the difference between the true y and the approximation go to zero? (Not just the ratio). REPLY [14 votes]: For $x,y>0$ there is a unique solution $y(x)$ to $x = \frac{1}{3}y(y+1)(2y+1)^2(2y^2+2y+1)$ given by $$y=\tfrac{1}{2} 3^{-1/3} \sqrt{\frac{\left(\sqrt{11664 x^2-3}+108 x\right)^{2/3}+3^{1/3}}{\bigl(\sqrt{11664 x^2-3}+108 x\bigr)^{1/3}}}- \tfrac{1}{2}.$$ Here is a plot of $y$ versus $x$.<|endoftext|> TITLE: Given the skeleton of an inscribed polytope. If I move the vertices so that no edge increases in length, can the circumradius still get larger? QUESTION [10 upvotes]: Let $P\subset \Bbb R^n$ be an inscribed convex polytope, that is, all its vertices are on a common sphere of radius $r$. Let $G$ be the edge-graph of $P$. For convenience, assume $V(G)=\{1,\dotsc,s\}$. Let $\ell_{ij}$ denote the length of the edge of $P$ corresponding to $ij\in E(G)$. Question. Let $p_1,\dotsc,p_s\in\Bbb R^{m}$ be points so that the points are on a common sphere $S$, $\lVert p_i-p_j\rVert\le\ell_{ij}$ for all $ij\in E(G)$, the center of $S$ lies in the convex hull $\operatorname{conv}\{p_1,\dotsc,p_s\}$. Is it then true that the radius of $S$ is at most $r$? If no, does this change if $n=m$? In other words, are the skeleta of inscribed polytopes "as expanded as possible" for the given edge-lengths? Note that the condition on the convex hull is necessary. Without this we could choose an arbitrarily large sphere $S$, and place all the $p_1,\dotsc,p_s$ in an arbitrarily small patch of $S$, so that $S$ is their circumsphere. The case $n=m=2$ I will demonstrate my general ideal on the case $n=m=2$, which I hope to somehow generalize to all cases with $n=m$. I do not yet have an idea for $m>n$. Let $P\subset\Bbb R^2$ be an inscribed polygon with circumradius $r$ and vertices $v_1,...,v_s$ in circular order. Let $\alpha_i$ be the angle between $v_i$ and $v_{i+1}$ (indices mod $s$) as seen from the circumcenter. Then $\alpha_1+\dots+\alpha_s=2\pi$. Suppose now that we have such a set of points $p_1,...,p_s$ with circumradius $r'>r$. Let $\beta_i$ be the angle between $p_i$ and $p_{i+1}$ as seen from the circumcenter. Since $\|p_i-p_{i+1}\|\le \|v_i-v_{i+1}\|$ but also $\|p_i\|>\|v_i\|$ it is easy to see that $\beta_i<\alpha_i$. In particular, $\beta_1+\cdots+\beta_s<2\pi$, and the closed polyline with vertices $p_1,...,p_s$ must have zero winding number around the circumcenter. But since the convex hull of the $p_i$ contains the origin, there are three points $p_{i_1},p_{i_2},p_{i_3}$ with $i_13$, somehow using the face structure provided by the polytope. One might then be able to construct a similar argument using space angles (aka fractions of the sphere, as described by Matt F. in the comments). REPLY [3 votes]: The answer to the question is unfortunately no, in the case of $n=m=3$. There is a simple example to illustrate this. Let $P$ be the cube with vertices $(\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2})$. Very obviously, every edge is of length $1$, and a quick calculation shows this is inscribed upon a sphere of radius $\frac{\sqrt{3}}{2}$. We now describe the locations of 8 vertices, one for each sign-string of length 3. Vertex $V_{sign}$ with $s=\#$positive entries of $sign$ is at the point $(sin(\frac{s\pi}{3}),cos(\frac{s\pi}{3}),0)$. It is easy to check that the lengths of the edges is exactly preserved. However, these vertices lie on a sphere of radius $1 > \frac{\sqrt{3}}{2}$. There are three things wrong with this example, two of which are immediately fixable. First, $\vec{0}$ is not interior. A small reduction of the radius can fix this. Secondly, the convex hull of these points is not full dimensional. Again a small reduction of the radius combined with some small perturbation in the third coordinate fixes this. Finally, the images of the faces of $P$ form a complex which does not contain the origin. This is both what allows the example to exist, and makes it a not satisfying example. Perhaps an additional condition which could contribute to a winding number argument could be: the cone generated by all edges incident to a vertex contains the origin. This could replace that the origin is in the convex hull of the points.<|endoftext|> TITLE: Super mixed Hodge structures? QUESTION [7 upvotes]: It's common in subjects that have some version of the "yoga of weights" that you have a functor called "Tate twist" and that the most natural version of it seems like it should be the square of something which obviously doesn't exist. For example, in the category of mixed Hodge structures, there's the natural Tate Hodge structure $\mathbb{Q}(k)$ (which is, for example, the Hodge structure on the top degree of a smooth $k$-dimensional projective variety). You'd like there to be a Hodge structure $\mathbb{Q}(\frac{1}{2})$ such that $\mathbb{Q}(\frac{1}{2})^{\otimes 2k}=\mathbb{Q}(k)$, but there's no way that such a thing could exist in the standard definition of mixed Hodge structures, because Hodge structures of pure odd weight must be even dimensional. A way of fixing this is suggested by Beilinson, Ginzburg and Soergel - Koszul Duality Patterns in Representation Theory on page 514: in essence, you consider two copies of the category of mixed Hodge structures with one of them formally Tate twisted by $(\frac{1}{2})$. I was writing a paper that used this idea, and thought of an explanation of this I like quite a bit better, and now am wondering if I'm really the first to have thought of it. The idea of this: you think of the two vector spaces in MHS as the even and odd parts of a super vector space (i.e. a $\mathbb{Z}/2$-graded vector space). The definition of a Hodge structure on an even vector space is the usual one; the definition of a Hodge structure on a purely odd vector space is the usual one, but now the Hodge filtration is indexed by elements of $\mathbb{Z}+\frac{1}{2}$. So, for example, $\mathbb{Q}(\frac{1}{2})$ is pure of weight 1, with $F_{-1/2}=\mathbb{C}$ and $F_{1/2}=0$ (with the purity condition being exactly that these guys are transverse). Has anyone encountered this approach before? Once I saw it, it seemed like obviously the right thing to do. REPLY [6 votes]: I am not sure this is relevant to you, and you may know this already, but in the theory of monodromic ("exponential") mixed Hodge structures, there is a very natural (even) square root of the Tate motive. See Kontsevich-Soibelman's Cohomological Hall algebra, exponential Hodge structures and motivic Donaldson–Thomas invariants, Section 3.4: Square root of Tate motive. Most papers on DT theory will use this, see e.g. Davison-Meinhardt, Cohomological Donaldson–Thomas theory of a quiver with potential and quantum enveloping algebras, Section 2.1 for a summary.<|endoftext|> TITLE: K3 surfaces with small Picard number and symmetry QUESTION [5 upvotes]: I am looking for examples of K3 surfaces that have a low Picard rank and at least one holomorphic involution. Here, low is no mathematically precise concept. I want to do computations with Monad bundles and for that lower is better for me. For Picard rank bigger than 5 those computations become too difficult for me. For example, the very general branched double cover of $\mathbb{CP}^2$ branched over a sextic has $Pic \simeq \mathbb{Z}$ and the map that swaps the two sheets of the cover is a holomorphic involution. The post Picard groups of quartic K3 surfaces contains more examples of K3 surfaces with Picard ranks 1, 2, and 3, but I didn't find any holomorphic involutions of the K3 surfaces mentioned there. REPLY [6 votes]: In Section 9 of the paper I. Shimada: An algorithm to compute automorphism groups of (K3) surfaces and an application to singular (K3) surfaces, Int. Math. Res. Not. 2015, No. 22, 11961-12014 (2015) ZBL1333.14034 there are many examples of complex elliptic K3 surfaces $X$ with Picard rank 3 and having (infinite) automorphism group containing involutions (in fact, $\operatorname{Aut}(X)$ contains a copy of $\mathbb{Z}/2 \ast \mathbb{Z}/2$).<|endoftext|> TITLE: How many square roots can a non-identity element in a group have? QUESTION [53 upvotes]: Let $G$ be a finite group. Let $r_2\colon G \to \mathbb{N}$ be the square-root counting function, assigning to each $g\in G$ the number of $x\in G$ with $x^2=g$. Perhaps surprisingly, $r_2$ does not necessarily attain its maximum at the identity for general groups, see Square roots of elements in a finite group and representation theory. I'm interested in whether $r_2(g)$ can attain a value above $0.999|G|$ for some non-identity element $g\in G$. Update: Thanks to everybody who participated in the discussion. The lemma proved here influenced greatly the statement of Theorem 4.2 in https://arxiv.org/pdf/2204.09666.pdf . Proposition 3.12 in this same paper is essentially the answer posted by GH from MO. REPLY [21 votes]: Here is a streamlined and simplified version of the posts by Saúl Rodríguez Martín and Emil Jeřábek. Theorem. Assume that $G$ is a finite group, and $r_2(g)>(3/4)|G|$ holds for some $g\in G$. Then $G$ is an elementary abelian $2$-group, and $g$ is the identity element. Proof. Fix any element $y\in G$, and consider the sets $$S=\{x\in G: x^2=g\},\qquad T=\{x\in S:xy\in S\}.$$ By the union bound, $$|G\setminus T|\,\leq\, 2|G\setminus S|<|G|/2,$$ hence $|T|>|G|/2$. For any $x\in T$, we have $(xy)^2=x^2$, which implies that $$xyx^{-1}=(xy)x^{-1}=(xy)^{-1}x=y^{-1}.$$ So $\{x\in G:xyx^{-1}=y^{-1}\}$ contains more than half of the elements of $G$, whence it contains all elements of $G$. In particular, $y=y^{-1}$, which shows that $G$ is an elementary abelian $2$-group. Moreover, $g$ is the identity element, since the identity element is the only square in $G$.<|endoftext|> TITLE: Etale fundamental group of the circle QUESTION [8 upvotes]: What is the étale fundamental group of the circle $X({\bf R})$, where $$ X(k) = \{(x,y) \in k^2 \mid x^2+y^2 = 1\}? $$ I know that there is a sequence $$ 1 \rightarrow \pi_1^{et}(X({\bf C})) \rightarrow \pi_1^{et}(X({\bf R})) \rightarrow Gal({\bf C}/{\bf R}) \rightarrow 1 $$ with $Gal({\bf C}/{\bf R})= \{z\mapsto z, z\mapsto \bar{z}\}$. The first group is the profinite completion of $\bf Z$ since $X({\bf C})$ is the projective complex line with two points removed, but I don't know if the sequence splits or not. More generally, what can be said about that sequence for general smooth varieties over ${\bf R}$? REPLY [10 votes]: As Donu explained, the sequence splits by choosing an $\mathbf R$-point of $X$. So the only question remaining is what the $\operatorname{Gal}(\mathbf C/\mathbf R)$-action on $\pi_1^{\text{ét}}(X_{\mathbf C})$ is. I claim that the action is trivial, because the two points at infinity $V(x^2+y^2)$ are not defined over $\mathbf R$. (By contrast, for $\mathbf G_{m,\mathbf R}$ we get the nontrivial action where a generator $\sigma \in \operatorname{Gal}(\mathbf C/\mathbf R)$ acts by $-1$, using the same argument as below.) Indeed, consider the tower $Y_n \to X_{\mathbf C} \to X$ where $Y_n \to X_{\mathbf C}$ is the unique cover of degree $n$. The composite $Y_n \to X$ is Galois as $\pi_1^{\text{ét}}(X_{\mathbf C}) \trianglelefteq \pi_1^{\text{ét}}(X)$ is normal and $n\mathbf Z \subseteq \mathbf Z$ is characteristic (in other words, any conjugate $Y'_n$ would contain $X_{\mathbf C}$ as $X_{\mathbf C} \to X$ is normal, hence $Y'_n \cong Y_n$ as $X_{\mathbf C}$-covers since $Y_n$ is the unique degree $n$ cover). We need to compute $\operatorname{Gal}(Y_n/X)$. This depends on some choices of isomorphisms: we choose $X_{\mathbf C} \stackrel\sim\to \mathbf G_{m,\mathbf C}$ via $(x,y) \mapsto x+yi$, and we write $t = x+yi$. Under this identification, the generator $\sigma$ of $\operatorname{Gal}(X_{\mathbf C}/X) = \operatorname{Gal}(\mathbf C/\mathbf R)$ acts on $\mathbf G_{m,\mathbf C}$ via $x+yi \mapsto x-yi$, i.e. the $\mathbf C$-semilinear map $\sum_j c_jt^j \mapsto \sum_j \bar c_jt^{-j}$. Then $Y_n$ can be identified with $\operatorname{Spec} \mathbf C[t^{\pm1/n}]$. The Galois group of $Y_n$ over $Y_1$ is $\mu_n(\mathbf C)$, where $\zeta \in \mu_n(\mathbf C)$ acts as the $\mathbf C$-linear map $t \mapsto \zeta t$. The conjugation action of $\sigma \in \operatorname{Gal}(X_{\mathbf C}/X)$ on $\operatorname{Gal}(Y_n/X_{\mathbf C})$ is therefore trivial, as $$(\sigma \circ \zeta \circ \sigma^{-1})(t) = (\sigma \circ \zeta)(t^{-1}) = \sigma (\zeta^{-1}t^{-1}) = \overline{\zeta^{-1}}t = \zeta t.$$ Thus, we see that the conjugation action of $\operatorname{Gal}(\mathbf C/\mathbf R)$ on $\pi_1^{\text{ét}}(X_{\mathbf C})$ is trivial. $\square$<|endoftext|> TITLE: Finite *covering* groups that act freely on some sphere QUESTION [6 upvotes]: A remarkable result (reviewed here) -- going back, at least, to P. A. Smith, developed by Cartan & Eilenberg and Milnor, and culminating in the theorem of Madsen, Thomas & Wall -- characterizes those finite groups which admit a free continuous action on a topological sphere in at least one dimension. I am wondering about the following more general question: Which finite groups are covered by (i.e.: receive a surjective homomorphism from) a finite group that acts freely on some topological sphere? For example (here), there are dihedral groups of order $2 p$ (for a prime number $p$) which thus violate Milnor's "$2 p$-condition" and hence cannot have any free action on any sphere; but their double cover by the corresponding binary dihedral group does have a free action on the 3-sphere (namely by regarding the binary dihedral group as a subgroup of $\mathrm{SU}(2) \simeq S^3$). So I am wondering how generic this situation is. Maybe all finite groups can be covered by ones that act freely on some sphere? Looking at the above example, one sees that the double cover of $\mathrm{SO}(3)$ by $\mathrm{SU}(2)$ remains non-trivial on all non-cyclic subgroups, whereby the order of all non-cyclic subgroups is multiplied by 2 under the extension, so that Smith/Cartan&Eilenberg's $p^2$-condition and Milnor's $2 p$-condition are enforced. This seems to suggest a strategy for how to understand finite covering groups freely acting on some sphere in more generality. Maybe it's easy, but I am not sure yet. REPLY [10 votes]: For a group to act freely on a sphere (even on a homotopy sphere), it must not have subgroups of the form $\mathbb Z/p \times \mathbb Z/p$ (this is the $p^2$-condition you mention). It follows that its Sylow subgroups are all cyclic or generalized quaternion. Thus, for groups whose cover acts freely on a (homotopy) sphere, their Sylow subgroups must be cyclic, generalized quaternion, or dihedral (as a Sylow subgroup of a quotient is a quotient of a Sylow subgroup by Sylow's thorems). Groups satisfying this condition were called $\mathcal P'$-groups by Wall, and classified in his article On the structure of finite groups with periodic cohomology. This is only a necessary condition and not (necessarily) sufficient, but it seems very stringent. One could look through this list for examples where the 2p condition fails, to see if it fails also on every cover.<|endoftext|> TITLE: Subcountability QUESTION [5 upvotes]: In these slides of a talk Giovanni Curi shows that the generalized uniformity principle follows from Troesltra’s uniformity principle and from the subcountability of all sets, which are both claimed to be consistent with CZF. Subcountability’s consistency with CZF is not surprising in light of counterintuitive results like that subsets of finite sets aren’t necessarily finite, but it seems to have a different flavor. What are the intuitions or motivations for subcountability? What references prove that subcountability is consistent with CZF? REPLY [6 votes]: An intuition for ESC (every set is subcountable, i.e., a subquotient of the natural numbers) in a predicative framework is that everything is built up from below starting with natural numbers, so we may assume that every set can be represented as a set of codes (natural numbers) quotiented out by an equivalence relation (denoting equality of whatever the codes represent). For the consistency of CZF + ESC (indeed, CZF + REA + ESC), see Michael Rathjen's Choice principles in constructive and classical set theories, Thm. 8.3: http://www1.maths.leeds.ac.uk/~rathjen/acend.pdf<|endoftext|> TITLE: Second countable vs. $G_\delta$-diagonal QUESTION [11 upvotes]: Here A Question on a second countable $T_2$ space, Paul asked if every second countable Hausdorff space has a $G_\delta$-diagonal. In the comments Brian M. Scott answered that, at the time (2015), the answer to that question might not be known. Interestingly enough, here Does second countable and functionally Hausdorff imply submetrizable?, Taras Banakh showed two years later that, as a consequence of a more general result, every second countable functionally Hausdorff space is submetrizable (in particular, every such space has a $G_\delta$-diagonal). My question is whether the answer to Paul's original question is already known. I couldn't find anything definitive about it. REPLY [10 votes]: There exists a counterexample to this question of Paul. It suffices to find a second-countable Hausdorff space $X$ that has two properties: (1) the space $X\times X$ is Baire; (2) for any nonempty open sets $U,V\subseteq X$ we have $\overline U\cap\overline V\ne\emptyset$. Such a space $X$ cannot have $G_\delta$-diagonal. Indeed, assuming that the diagonal $\Delta_X$ is of type $G_\delta$ in $X\times X$, we can write $(X\times X)\setminus \Delta_X$ as the union $\bigcup_{n\in\omega}F_n$ of closed subsets $F_n$ of $X\times X$. Property (2) implies that $X$ has no isolated points and hence the diagonal $\Delta_X$ is nowhere dense in $X\times X$. Since the space $X\times X$ is Baire, there exists $n\in\omega$ such that $F_n$ has non-empty interior in $X\times X$. Then there are nonempty open sets $U,V$ in $X$ such that $U\times V\subseteq F_n$ and hence $\overline U\times\overline V\subseteq F_n$. Then $(\overline U\times \overline V)\cap\Delta_X=\emptyset$ and hence $\overline U\cap\overline V=\emptyset$, which contradicts the property (2). For the space $X$ one can take the projective space of the countable product of lines $\mathbb R^\omega$. So, $X$ is the quotient space of $\mathbb R^\omega\setminus\{0\}^\omega$ by the equivalence relation $x\sim y$ iff $\mathbb Rx=\mathbb Ry$. Observe that the quotient map $q:\mathbb R^\omega\setminus\{0\}^\omega\to X$ is open. This fact can be used to show that $X$ is Hausdorff and its square $X\times X$ is Baire. Moreover, for any nonempty open set $U\subseteq X$ the closure $\overline{U}$ contains the image $q[(\{0\}^n\times \mathbb R^{\omega\setminus n})\setminus\{0\}^\omega]$ for some $n\in\omega$, which implies that the condition (2) is satisfied.<|endoftext|> TITLE: Solving $\Delta \text{tr}(h) - \mathrm{div}(\mathrm{div}(h)) + \text{tr}(h) = f$ on $S^2$ QUESTION [6 upvotes]: $\DeclareMathOperator\ddiv{div}\DeclareMathOperator\tr{tr}\newcommand{\conf}{\mathrm{conf}}$Consider this PDE on a symmetric tensor $h$ on $S^2$: $$\Delta \text{tr}(h) - \ddiv(\ddiv(h)) + \tr(h) = f$$ where $f \in L^2(S^2)$ and $\Delta$, $\ddiv$ and $\tr$ are with respect to the round metric on $S^2$. I wish to show that there exists at least one solution to this. If we assume for simplicity that $h = \frac{1}{2} \tr(h) g_{S^2}$, then the PDE becomes $$\Delta \tr(h) + 2\tr(h) = 2f$$ which doesn't have a solution for every $f$ since $-2$ is an eigenvalue of $\Delta$ (I am assuming that $\Delta + \lambda$ is not surjective if $-\lambda$ is an eigenvalue; is this correct?). I am not sure how to approach this. One approach is decomposing $h$ into its trace part and a conformal Lie derivative of a vector field $X$: $h = \frac{1}{2} \tr(h) g_{S^2} + \mathcal{L}_{\conf}X$. Then the PDE becomes: $$\frac{1}{2}\Delta \tr(h) - \ddiv(\Delta_{\conf}X) + \text{tr}(h) = f$$ where $\Delta_{\conf}$ is the conformal laplacian on vector fields. I am not able to continue. Any help is appreciated. REPLY [4 votes]: Your second-order differential operator appears when one takes the variation of the scalar curvature of the sphere by a symmetric tensor $h\mapsto \frac{d}{dt}|_{t=0}\mathrm{Sc}_{g+th}$. In constant sectional curvature, such operators were studied for instance by Calabi (60'), and then in constant scalar curvature by Ebin (69'). Edit: I have taken another look at Ebin's paper and figured out a similar, yet simpler argument than my original one (you can still find it below). I think this argument provides an answer to your question in any geometry: let $(M,g)$ be an arbitrary closed Riemmanian manfiold. Define a linear second order operator $\gamma:S^{2}(T^{*}M)\rightarrow C^{\infty}(M)$ by $\gamma h=-\Delta_{g}tr_{g}h+\mathrm{div}\mathrm{div}h-(h,Ric_{g})_{g}$. Consider its $L^{2}$-dual $\gamma^{*}:C^{\infty}(M)\rightarrow S^{2}(T^{*}M)$. It can be checked to assume the form $\gamma^{*}f=\mathrm{Hess}_{g}f-fRic_{g}-\Delta_{g}fg$. The symbol of $\gamma$ is $|\xi|^{2}tr+i_{\xi}i_{\xi}$ while the symbol of $\gamma^{*}$ is $\xi\otimes\xi+|\xi|^{2}I$. Thus the symbol of $\gamma\gamma^{*}$ is a constant multiplied by $|\xi|^{4}$, so $\gamma\gamma^{*}$ is an elliptic fourth order operator (a "bilaplacian"). Therefore, if you want to solve $\gamma h=f$ then replace $h=\gamma^{*}g$ and solve $\gamma\gamma^{*} g=f$. This has a solution if and only if $f$ satisfies a compatibility condition: it must be $L^{2}$-orthogonal to the finite dimensional kernel of $\gamma\gamma^{*}$, which in this case is equal to the kernel of $\gamma^{*}$. In your original case of the sphere, $Ric_{g}=g$ and so $(h,Ric_{g})_{g}=tr_{g}h$. My original answer is below: Here is what I suggest to solve your question in the special case of $S^{2}$. If you have a Riemmanian manifold with constant sectional curvature, $(M,g)$, consider the second order differential operator $H_{g}=\frac{1}{2}{(d^{\nabla^{g}}d^{\nabla^{g}}_{V}+d^{\nabla^{g}}_{V}d^{\nabla^{g}}})-\frac{1}{2}\kappa g\wedge$. $\kappa$ here is the sectional curvature of the space, so in your case it can be taken to be $\kappa=1$. The wedge product here is known as the Kulkarni-Numizu product of symmetric tensor fields (there is a more generlized version of this wedge product, operating on "double forms", also adressed by Kulkarni, but this notion will do). $d^{\nabla^{g}}$ and $d^{\nabla^{g}}_{V}$ here are the exterior covariant derviatvie of the first and second index of a symmetric tensor, respectively, where we think of symmetric tensors as examples of vector-valued differential forms $\Omega^{1}(M ; T^{*}M)$. The image of $H_{g}$ lies in $\Omega^{2}(M;\Lambda^{2}T^{*}M)$, and in the case where $M$ is two-dimensional every element in this space is fully determined by its "scalar curvature", namely if $\sigma\in\Omega^{2}(M;\Lambda^{2}T^{*}M)$ then $\sigma=\frac{1}{4}(tr_{g}tr_{g}\sigma) g\wedge g$. In the case where $\kappa=1$, a direct calcultion shows that taking the trace twice from $H_{g}$ yields the operator $tr_{g}tr_{g}H_{g}=-\Delta_{g}tr_{g}+\mathrm{div}\mathrm{div}-tr_{g}$, so solving your equation is equivelent of solving $H_{g}h=\frac{1}{4} f g\wedge g$, where $g$ is the metric of the sphere. Replacing $h=H^{*}_{g}\psi$ for $\psi\in\Omega^2(M;\Lambda^{2}T^{*}M)$ where $H_{g}^{*}$ is the formal $L^{2}$ dual of $H_{g}$ yields the equation $H_{g}H^{*}_{g}\psi=\frac{1}{4} f g\wedge g$. Note how since $H_{g}$ operates on symemtric tensors, the image of $H^{*}_{g}$ is a symmetric tensor in $\Omega^{1}(M;T^{*}M)$. Another calculation then shows that the principle symbol of $H_{g}H^{*}_{g}$ in the case where $M$ is two dimensional is $|\xi|^{4}$. Thus this is an elliptic fourth order differential opertor (a "bilaplcian"), and so the equation $H_{g}H^{*}_{g}\psi=\frac{1}{4} f g\wedge g$ is solvable for $\psi$ if and only if $ \frac{1}{4} f g\wedge g$ is orthogonal to the kernel of $H_{g}H^{*}_{g}$. By duality, this kernel is equal to $\mathrm{ker} H^{*}_{g}$. I am not sure if this kernel is trivial when $M$ is simply connected. If not, then this orthogonality yields a compatability condition which $f$ must satisfy in order for the equation to have a solution.<|endoftext|> TITLE: How the solve the equation $\frac{(a+b\ln(x))^2}{x}=c$ QUESTION [5 upvotes]: I need to solve the equation $$\frac{(a+b\ln(x))^2}{x}=c$$ where $a$, $b$, and $c$ are given. It is known that $a$ and $b$ are fixed and satisfy some condition such that the left hand side is decreasing. So $x$ is uniquely determined by $c$ when $c$ is chosen in certain range. A related problem is $$\frac{(a+b\ln(x))}{x}=c$$ for which the solution is $$x=-\frac{bW(-\frac{ce^{-a/b}}{b})}{c}$$ where $W(z)$ is the product log function. Any hint about how to solve the first equation? REPLY [9 votes]: Step-by-step solution with Lambert W. The goal is to get something of the form $\color{red}{ue^u = v}$ then re-write it as $\color{blue}{u=W(v)}$. $$ \frac{(a+b\ln(x))^2}{x}=c \\ \frac{(a+b\ln(x))}{\sqrt{x}}=\pm\sqrt{c} \\ (a+b\ln(x))e^{-\ln(x)/2}=\pm\sqrt{c} \\ (a+b\ln(x))\exp\left(-\frac{a}{2b}-\frac{\ln(x)}{2}\right) =\pm\sqrt{c}\exp\left(\frac{-a}{2b}\right) \\ (a+b\ln(x))\exp\left(-\frac{a+b\ln(x)}{2b}\right) =\pm\sqrt{c}\exp\left(\frac{-a}{2b}\right) \\ \color{red}{-\frac{a+b\ln(x)}{2b}\exp\left(-\frac{a+b\ln(x)}{2b}\right) =\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)} \\ \color{blue}{-\frac{a+b\ln(x)}{2b} = W\left(\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)\right)} \\ \ln(x) = \frac{-a-2bW\left(\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)\right)}{b} \\ x = \exp\left(-\frac{a}{b}-2 W\left(\mp\frac{\sqrt{c}}{2b}\exp\left(\frac{-a}{2b}\right)\right)\right) $$ The other example mentioned... $$ \frac{a+b\ln(t)}{t}=c \\ (a+b\ln(t))e^{-\ln(t)} = c \\ (a+b\ln(t))\exp\left(-\frac{a}{b}-\ln(t)\right) = c \exp\left(-\frac{a}{b}\right) \\ (a+b\ln(t))\exp\left(-\frac{a+b\ln(t)}{b}\right) = c \exp\left(-\frac{a}{b}\right) \\ \color{red}{-\frac{a+b\ln(t)}{b}\exp\left(-\frac{a+b\ln(t)}{b}\right) = -\frac{c}{b} \exp\left(-\frac{a}{b}\right)} \\ \color{blue}{-\frac{a+b\ln(t)}{b} = W\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)} \\ \ln(t) = -\frac{a+bW\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)}{b} \\ t = \exp\left(-\frac{a+bW\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)}{b}\right) $$ We may question the other solution given in the OP. In fact, this solution is equal to that solution: Claim $$ \exp\left(-\frac{a+bW\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right)}{b}\right) = -\frac{b}{c}W\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right) \tag1$$ Why? $$ \text{Let}\quad Q = W\left(-\frac{c}{b} \exp\left(-\frac{a}{b}\right)\right). \\ \text{Then}\quad Qe^Q = -\frac{c}{b}\exp\left(-\frac{a}{b}\right) \\ -\frac{b}{c}Q = \exp\left(-\frac{a}{b}\right)e^{-Q} \\ -\frac{b}{c}Q = \exp\left(-\frac{a+bQ}{b}\right) \\ \text{which is $(1)$.} $$ Challenge: Simplity the first solution in the same way: $$ x = \left[\frac{2b}{\sqrt{c}}W\left(\mp\frac{\sqrt{c}}{2b}\exp\left(-\frac{a}{2b}\right)\right)\right]^2 $$<|endoftext|> TITLE: Do power sums determine the variables? QUESTION [15 upvotes]: In my analysis research, I came across the following problem. Given $n$ positive real numbers $x_1,\dots,x_n$, consider the $n$-many power sums $$ p_3 = x_1^3 + x_2^3 + \dots + x_n^3 , $$ $$ p_5 = x_1^5 + x_2^5 + \dots + x_n^5 , $$ $$ \vdots $$ $$ p_{2n+1} = x_1^{2n+1} + x_2^{2n+1} + \dots + x_n^{2n+1} . $$ Do the values of the power sums $p_3,p_5,\dots,p_{2n+1}$ uniquely determine $x_1,\dots,x_n$ (up to reordering)? I was wondering if this problem exists in the literature? I know the answer to this problem is "yes" in the case of the first $n$ power sums $p_1,p_2,\dots,p_{n}$ by Newton's identities: the power sums $p_1,p_2,\dots,p_{n}$ determine the elementary symmetric polynomials $e_1,\dots,e_n$ via explicit formulas, from which we can construct a degree-$n$ polynomial with roots $x_1,\dots,x_n$ and appeal to the fundamental theorem of algebra. I believe the answer is yes, and I've checked some special cases using a computer. Using analysis techniques I can easily get a local uniqueness statement. Indeed, the function $f:(x_1,\dots,x_n) \mapsto (p_3,p_5,\dots,p_{2n+1})$ has a Jacobian matrix $Df$ that looks like a Vandermonde matrix, and this makes it easy to compute its determinant (and minors). In particular, on the simplex $\{ 0 < x_1 < x_2 < \dots < x_n \} \subset\mathbb{R}^n$ the Jacobian matrix $Df$ has nonzero determinant, and so the inverse function theorem tells us that $f$ is locally injective. Moreover, this argument applies to all minors of $Df$, and so $Df$ is a strictly totally positive matrix. Such matrices turn out to be diagonalizable with distinct positive eigenvalues, but I haven't been able to conclude that $f$ is globally injective. REPLY [10 votes]: Let $\gamma (x) = (x^3, x^5, \dots, x^{2n+1})$. By rearranging both sides (using that all the polynomials have odd degree and therefore are antisymmetric), the problem can be restated as: Let $0\le a,b$, $a+b \le 2n$ integers, and $x_1, \dots x_a,y_1, \dots y_b \in \mathbb{R}_{\ge 0}$. Then $$ \sum_{i=1}^a \gamma(x_i) = \sum_{i=1}^b \gamma(y_i) $$ has no nontrival solutions. From what I can see from the MathSciNet review, that is precisely the content of [1]. I haven't been able to find that paper, but I found it as a reference in [2], which also contains the idea of the proof (in a slightly more general setting) in the second half of Section 3. The key idea is using that the Jacobian matrix is a strictly totally positive matrix, indeed! [1] J. Steinig, On some rules of Laguerre's, and systems of equal sums of like powers. Rend. Mat. (6) 4 (1971), 629–644 (1972). [2] S. W. Drury and B. P. Marshall, Fourier restriction theorems for degenerate curves. Mathematical Proceedings of the Cambridge Philosophical Society, 101(3), 541-553.<|endoftext|> TITLE: Finding vector fields on $S^2$ with equal divergence QUESTION [5 upvotes]: Let $\mathfrak{X}_{CK}^{\perp}$ be the space of vector fields on $S^2$ that are $L^2$-orthogonal to conformal Killing vector fields. Let $\mathfrak{X}_{CK}$ be the 6-dimensional space of conformal Killing vector fields on $S^2$. Can we find a vector field $Y \in \mathfrak{X}_{CK}^{\perp}$ and a vector field $W \in \mathfrak{X}_{CK}$ that is not Killing such that $$\mathrm{div}(Y) = \mathrm{div}(W)$$ REPLY [5 votes]: I think that this is not possible: Per my comment on Divergence of conformal Killing vector fields on $S^2$ and the spherical harmonics you want to solve $$ \textrm{div} (Y) = -2a\cdot x $$ for $Y$ orthogonal to conformal-KVF's and $a \in \mathbb{R}^3$ fixed (nonzero). Suppose you can do this. Then, we find that $$ Y = a^T + W $$ for $W$ divergence free. From this question https://math.stackexchange.com/questions/1898371/divergence-free-vector-field-on-a-2-sphere you can see that $W = J \nabla f$ for some function $f \in C^\infty(S^2)$ (for $J$ the complex structure on $S^2$). Now, by assumption, $Y$ is orthogonal to $a^T$: \begin{align*} 0 & = \int_{S^2} Y\cdot a^T \\ & = \int_{S^2}|a^T|^2 + (J \nabla f) \cdot a^T \\ & = \int_{S^2}|a^T|^2 - \nabla f \cdot J a^T \\ & = \int_{S^2}|a^T|^2 + f \textrm{div}(J a^T). \end{align*} However, note that $Ja^T$ is a KVF and thus divergence free. This implies that $\int_{S^2}|a^T|^2 = 0$. This is a contradiction.<|endoftext|> TITLE: RELU representation of $\max(x,y,z)$ QUESTION [6 upvotes]: Here is a question that occurred to me while learning about neural networks. For $t\in\mathbb{R}$ put $t_+=\max(0,t)$, so $t_+=t$ if $t\geq 0$ and $t_+=0$ if $t\leq 0$. (This is RELU=rectified linear unit in neural network language.) Note that $t=t_+-(-t)_+$ and $|t|=t_++(-t)_+$. For the maximum of two variables we have \begin{align*} \max(x,y) &= x + (y-x)_+ \\ &= \left(x+y+(x-y)_++(y-x)_+\right) /2 \\ &= \left((x+y)_+-(-x-y)_++(x-y)_++(y-x)_+\right)/2 \end{align*} Is there a similar expression for $\max(x,y,z)$ as a linear combination of terms $\phi(x,y,z)_+$ with $\phi$ linear? I suspect not, but I have not found a proof. More generally, is there a nice characterisation of the functions of several variables that do admit such a representation? REPLY [2 votes]: Here is a simpler proof, inspired by YCor's approach. Fix a finite-dimensional vector space $V$ and let $F$ be the class of functions $V\to\mathbb{R}$ in question. First, by using the rule $t_+=(t+|t|)/2$ and combining terms appropriately, we see that any $f\in F$ can be represented as $$ f(x) = L_0(x) + |L_1(x)| + \dotsb + |L_n(x)| - |L_{n+1}(x)| - \dotsb - |L_{n+m}(x)|, $$ where $L_0,\dotsc,L_{n+m}$ are linear and $L_1,\dotsc,L_{n+m}$ are pairwise linearly independent. As $f(x)-L_0(x)$ is even, it follows that $f$ is differentiable at $x$ iff it is differentiable at $-x$. The $\max$ function is differentiable at $(1,0,0)$ but not at $(-1,0,0)$, so it does not lie in $F$. For a sharper statement, we can proceed as follows. Let $P$ be the set of piecewise-linear maps $f\colon V\to\mathbb{R}$. For $f\in P$ and $x,v\in V$ put $$ Q(f,x)(v) = \lim_{t\to 0^+} t^{-1}\left(f(x+tv)+f(x-tv)-2f(x))\right). $$ Then $f\in F$ iff there are distinct hyperplanes $W_1,\dotsc,W_p TITLE: Ideas for introducing Galois theory to advanced high school students QUESTION [48 upvotes]: Briefly, I was wondering if someone can suggest an angle for introducing the gist of Galois groups of polynomials to (advanced) high school students who are already familiar with polynomials (factorisation via Horner, polynomial division, discriminant and Vieta's formulas for quadratic equations). I am struggling to find a coherent approach that makes use of their current understanding, e.g., whether to give an intro on group theory, or to describe permutation operations on roots. This is all intended for a short course introduction and not an extended program. Any advice would be quite helpful. REPLY [2 votes]: Please take a look at Math Girls 5: Galois Theory, which I translated and was published just last month. If you aren't familiar with the series, these books, originally published in Japan, are novels in that they present fictional characters (high school students who explore various topics in math for fun) and have a thin veneer of plot, but most of the content is mathematics. MG5 isn't a comprehensive presentation of Galois Theory, but rather focuses on Galois' original paper regarding solutions of equations. The bulk of that comes in the final chapter, with preceding chapters introducing permutations, symmetric polynomials, and other requisite information. There's a lengthy fan review of the book on Amazon.com giving a much more detailed description of the content, if you're interested. You can also use Amazon's "Look Inside" feature to browse the table of contents. Because these books are written with the aim of being accessible to motivated high school students, perhaps they will give you some ideas regarding how to present this challenging content?<|endoftext|> TITLE: Sequence of $k^2$ and $2k^2$ ordered in ascending order QUESTION [6 upvotes]: Let $\eta(n)$ be A006337, an "eta-sequence" defined as follows: $$\eta(n)=\left\lfloor(n+1)\sqrt{2}\right\rfloor-\left\lfloor n\sqrt{2}\right\rfloor$$ Sequence begins $$1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1$$ Let $a(n)$ be A091524, $a(m)$ is the multiplier of $\sqrt{2}$ in the constant $\alpha(m) = a(m)\sqrt{2} - b(m)$, where $\alpha(m)$ is the value of the constant determined by the binary bits in the recurrence associated with the Graham-Pollak sequence. Sequence begins $$1, 1, 2, 2, 3, 4, 3, 5, 4, 6, 7, 5, 8, 6, 9, 7, 10, 11, 8, 12, 9, 13$$ Then we have an integer sequence given by $$b(n)=(a(n))^2\eta(n)$$ Sequence begins $$1, 2, 4, 8, 9, 16, 18, 25, 32, 36, 49, 50, 64, 72, 81, 98, 100, 121, 128, 144, 162, 169$$ I conjecture that $b(n)$ is a sequence of $k^2$ and $2k^2$ ordered in ascending order. Is there a way to prove it? REPLY [9 votes]: Denote by $f(n)$ the sequence of squares and double squares in ascending order. We have to prove that $f(n)=b(n)=(a(n))^2\eta(n)$. Consider two cases. $f(n)=k^2$. Then the number of squares and double squares not exceeding $k^2$ equals $n$, that is, $n=k+\lfloor k/\sqrt{2}\rfloor$. Therefore $n2n-k$, and $\lfloor n\sqrt{2}\rfloor\geqslant 2n-k$. On the other hand, $n+1>k(1+1/\sqrt{2})$ that analogously yields $(n+1)\sqrt{2}>2(n+1)-k$ and $\lfloor (n+1)\sqrt{2} \rfloor\leqslant 2n-k+1$. Since also $\lfloor (n+1)\sqrt{2} \rfloor\geqslant \lfloor n\sqrt{2} \rfloor+1\geqslant 2n-k+1$, this implies that $\eta(n)=1$. According to OEIS we have $a(n)=a(\lfloor k(1+1/\sqrt{2}\rfloor)=k$, thus $(a(n))^2\eta(n)=k^2$ as needed. $f(n)=2k^2$. Then the number of squares and double squares not exceeding $2k^2$ equals $n$, that is, $n=k+\lfloor k\sqrt{2}\rfloor$. So, $n(n+1)+k$. This yields $\eta(n)=2$. Again by OEIS we get $a(n)=a(\lfloor k(1+\sqrt{2}\rfloor)=k$ and $(a(n))^2\eta(n)=2k^2$.<|endoftext|> TITLE: Does there exist a closed manifold with vanishing reduced rational cohomology but nonvanishing odd torsion cohomology? QUESTION [17 upvotes]: Question: Let $p$ be an odd prime. Does there exist a closed manifold $M$ with $\widetilde H^\ast(M; \mathbb Q) = 0$ but $\widetilde H^\ast(M; \mathbb F_p) \neq 0$? When $p = 2$, an example is given by $\mathbb R \mathbb P^2$. After discussion, this question turns out to be equivalent to this other question. That is, over at that question Saal Hardali explained that if $M$ is a closed manifold, then the chromatic type of $M$ at a prime $p$ is either 0 or 1. Both possibilities are realized at $p=2$; the question is whether chromatic type 1 is realized at odd primes. Chromatic type 0 just means having nonvanishing rational (co)homology. So the question is whether there exists a closed manifold $M$ which is rationally contractible but whose $p$-localization is nontrivial for an odd $p$. This turned out to be mistaken, thanks to Ben Wieland for pointing this out at the other question. Side Question: When $p=2$, what are some other examples of $M$ with $\widetilde H^\ast(M;\mathbb Q) = 0$ but $\widetilde H^\ast(M;\mathbb F_2) \neq 0$ besides $M = \mathbb R \mathbb P^{2n}$ and products thereof? REPLY [19 votes]: As mme noted in the comments, such examples cannot exist in odd dimensions, for Euler characteristic reasons. They can't exist in dimension 2 either, by classification. I claim that in all other dimensions $2n > 2$ we have (plenty of) examples. Let $N$ be a rational homology $2n$-sphere, that is a $2n$-manifold with $H_*(N; \mathbb{Q}) = H_*(S^{2n}; \mathbb{Q})$. For every prime $p$ there exists a rational homology $2n$-sphere with $\dim_{\mathbb{F}_p} H_*(N;\mathbb{F}_p) > 2$. For instance, you can take a spun lens space (any spun rational homology $2n-1$-sphere would do). Now, the integral homology of $M = \mathbb{RP}^{2n} \# N$ splits as a direct sum of that of the two summands in all dimensions strictly between 0 and 2n, and it vanishes in dimension 2n (because $M$ is non-orientable) and it is $\mathbb{Z}$ in dimension 0 (because $M$ is connected). (This is Exercise 6 in Section 3.3 of Hatcher's Algebraic topology.) That is, $M$ is a rational homology ball, and its homology has as much $p$-torsion as that of $N$. For the side question, if we choose $N$ to have no 2-torsion in its homology (e.g. spinning an odd lens space should do the trick), this gives plenty of examples.<|endoftext|> TITLE: Is the set of powerful numbers piecewise syndetic? QUESTION [8 upvotes]: Recall that a subset $A \subset \mathbb Z_+$ of positive integers syndetic if there exists a $d>0$ such that every positive integer has distance at most $d$ to an element of $A$. It is called piecewise syndetic if it is the intersection of a syndetic set with a subset of $[0,\infty)$ containing arbitrarily long intervals. Let us call $n \in \mathbb Z_+$ powerful if for every prime $p$, the multiplicity $\nu_p(n) \neq 1$. (Or, equivalently: $n$ can be expressed as a product of a square and a cube.) Here is the question: is the set of powerful numbers piecewise syndetic? REPLY [14 votes]: The answer is no. A set $S$ to be piecewise syndetic iff there is an integer $d$ such that there exist intervals $I$ of arbitrary length such that distances between elements of $S\cap I$ are bounded by $d$. In particular, $|S\cap I|\geq\frac{1}{d}|I|$. I will show no such $d$ exists. For any prime $p$, the fraction of those which are either not divisible by $p$ or divisible by $p^2$ is equal to $1-\frac{p-1}{p^2}$. Further, these conditions are independent - formally, if we consider take $P=p_1\dots p_k$, the product of first $k$ primes, then from the Chinese remainder theorem, in any interval $I$ of length $P^2$ the fraction of integers $n$ in $I$ such that $v_{p_i}(n)\neq 1$ for all $i$ is equal to $$C:=\prod_{i=1}^k\left(1-\frac{p-1}{p^2}\right).$$ This gives an upper bound of $CP^2$ for the number of powerful numbers an interval of length $P^2$. Now, as $k$ tends to infinity, then $C$ tends to zero (this essentially follows from the fact sum of reciprocals of primes diverges), in particular for large enough $k$ it becomes smaller than $\frac{1}{d}$, so no interval $I$ of length $P^2$ contains more than $\frac{1}{d}|I|$ powerful numbers.<|endoftext|> TITLE: Do we have the Oka coherence theorem for finite group actions? QUESTION [6 upvotes]: We first consider the sheaf of holomorphic functions $\mathcal{O}(\mathbb{C}^n)$ on $\mathbb{C}^n$. By Oka coherence theorem, $\mathcal{O}(\mathbb{C}^n)$ is coherent over itself. Now we consider a finite group $G$ acting on $\mathbb{C}^n$ and let $\pi: \mathcal{C}^n\to \mathbb{C}^n/G$ be the projection. We define a sheaf $\bar{\mathcal{O}}$ on $\mathbb{C}^n/G$ as follows: for any open subset $U\subset \mathbb{C}^n/G$, we define $$ \bar{\mathcal{O}}(U):=\{f\in \mathcal{O}(\pi^{-1}(U))|f \text{ is }G-\text{invariant.}\} $$ My question is: is this sheaf $\bar{\mathcal{O}}$ coherent over itself too? REPLY [8 votes]: This would be true. You need two facts: Grauert's theorem that coherent sheaves are preserved by proper direct images. This implies $\pi_*\mathcal{O}_{\mathbb{C}^n}$ is coherent. Sub modules of coherent sheaves are coherent. Therefore $$\tilde{\mathcal{O}} = \pi_*\mathcal{O}_{\mathbb{C}^n}^G\subset \pi_*\mathcal{O}_{\mathbb{C}^n}$$ is coherent<|endoftext|> TITLE: From Delzant polytope to lattice polytope QUESTION [5 upvotes]: By definition, an $n$-dimensional Delzant polytope $P$ is not necessarily a lattice polytope. But is there a natural way (or operations) to turn $P$ into a lattice polytope using the fact that the edge vectors incident to any vertex of $P$ form an integral basis of $\mathbb{Z}^n$? And what do these operations mean in symplectic geometry/topology? The background of this question is: when people explain the equivalence between symplectic toric manifold and smooth projective toric variety, they often assume the moment polytope to be both lattice and Delzant. So I feel there must be a reason for this assumption. REPLY [5 votes]: There is a way to turn a Delzant polytope into a lattice polytope, but there is no natural or canonical way. Note that the Delzant condition implies that the normal vector to each facet (codimension 1 face) is integral. By letting the defining equation for the hyperplane containing each facet rational (by slightly moving hyperplanes), you can make the coordinates of all vertices rational without changing the combinatorial structure. Now the polytope is integral after scaling. As you see, this is far from being canonical. The above operation changes the cohomology class represented by the symplectic form. So you can think of it as choosing a different symplectic form on the same smooth manifold. If you view a symplectic toric manifold as a symplectic reduction, you are taking a different regular value of the moment map. Some people regard symplectic toric manifolds as smooth projective varieties, but strictly speaking, that is wrong. Any symplectic toric manifold is diffeomorphic to a smooth projective variety, but it may not be symplecally embedded into a projective space equipped with the Fubini-Study form $\omega_{FS}$. In fact, by Kodaira embedding theorem, a symplectic toric manifold $(M, \omega)$ can be (symplectically) embedded into $\left(\mathbb{P}^N, \omega_{FS}\right)$ for some $N$ if and only if $[\omega] \in H^2(M, \mathbb{Z})$. Even if we allow scalar multiples of $\omega_{FS}$, the class $[\omega]$ can only be a multiple of an integral one. For an easy example, consider $(S^2 \times S^2, \sigma \oplus \lambda\sigma)$ where $\sigma$ is a volume form on $S^2$ and $\lambda>0$ is irrational. It can never be embedded into a projective space preserving the symplectic structure. You can take rational $\lambda$ to find an embedding, but then you are dealing with a different symplectic manifold. REPLY [3 votes]: If you have a lattice polytope then you get a projective toric variety. If your polytope is not a lattice polytope then you can still construct a symplectic toric manifold (e.g. by performing symplectic cuts) but it won't be the underlying symplectic manifold of a projective variety. To see this, note that the cohomology class of the symplectic form on a projective toric variety is integral because its ultimately pulls back from the Fubini-Study form on projective space. If your polytope is not lattice then one of the edges has non-integer affine length, so the symplectic area of the corresponding sphere is not an integer, so there cohomology class of the symplectic form is not integral. Let $a_1,\ldots,a_N$ be the affine lengths of the edges of your polytope. If the ratios $a_i/a_j$ are all rational then you can rescale to make them integers and this rescaling of your polytope will be a lattice polytope. Otherwise, I guess (though I didn't check) you should be able to change the constants in the inequalities that define your polytope by an arbitrarily small amount to get these ratios to be rational, and hence find a small deformation of your symplectic toric manifold which is (a rescaling of) a projective toric variety.<|endoftext|> TITLE: Growth of the word norm for elementary matrices in $\rm SL_3 (\mathbb{Z})$ QUESTION [5 upvotes]: This is a reference request, since the answer is probably well known, but I could not find it. Given a finitely generated group $\Gamma$ with a generating set $S$, define the word norm $l = l_S : \Gamma \rightarrow \mathbb{N}$ to be $$l (g) = \min \lbrace k : \exists s_1,...,s_k \in S, g = s_1 ... s_k \rbrace ,$$ i.e., $l(g)$ is the distance between $e$ and $g$ in the Cayley graph of $\Gamma$ (w.r.t the generating set $S$). My question: Let $\Gamma = \rm SL_3 (\mathbb{Z})$ (with some finite generating set). For $1 \leq i,j, \leq 3, i \neq j$ and $m \in \mathbb{Z}$, denote $e_{i,j} (m)$ to be the elementary matrix with $1$'s along the main diagonal, $m$ in the $(i,j)$-entry and $0$ in all other entries. What can one say about the growth rate of $l (e_{i,j} (m))$? My naive attempt for an answer gives me $l (e_{i,j} (m)) = O (\log^3 (m))$: For convenience, we fix the generating set $S = \lbrace e_{i,j} (\pm 1), e_{i,j} (\pm 2) : 1 \leq i,j, \leq 3, i \neq j \rbrace$. Using commutator (Steinberg) relations $$ e_{i,j} (2^{2^{r+1}}) = [e_{i,k} (2^{2^{r}}), e_{k,j} ( 2^{2^{r}})]$$ it is not hard to show by induction on $r$ that $l (e_{i,j} (2^{2^r})) \leq 4^{r} $. Again by the commutator relations, it follows that for every $2^r \leq d < 2^{r+1}$ it holds that $l (e_{i,j} (2^{d})) \leq 4^{r+1} \leq 4 d^2$. Thus it follows that for every $2^d \leq m < 2^{d+1}$, $$l (e_{i,j} (m)) \leq \sum_{t =0}^d l (e_{i,j} (2^t)) \leq 4 \sum_{t =0}^d t^2 = 4 \frac{d (d+1) (2d +1)}{6} = O (\log^3 (m)).$$ As noted above - this computation is quite naive. Are there better known results? REPLY [8 votes]: The answer is that this is $\simeq\log(m)$. Where $f\sim g$ means that $f\preceq g\preceq f$ and $f\preceq g$ means that eventually $f\le cg$ for some $c>0$. This is a particular case of a result of Lubotzky, Mozes and Raghunathan. But it's easy to prove directly. First since the matrix norm of $e_{ij}(1)^m$ grows linearly and the matrix norm is submultiplicative, one immediately sees that $|e_{ij}(1)^m|\succeq \log(m)$. Now we use that $m\ge 3$, so we can suppose $(i,j)=(1,3)$. Then consider the subgroup $\Gamma$ of matrices that are identity, except the entries 13,23 that are arbitrary, and the block 11 12 21 22 which is an arbitrary integral power of $\begin{pmatrix}2&1\\1& 1\end{pmatrix}$. Then $\Gamma$ is a cocompact lattice in the 3-dimensional group SOL. It is immediate that in the group SOL, if $v$ is in the normal abelian subgroup (written additively), then the word length of $v$ (with respect to any compact generating subset) is $\simeq \log(\|v\|)$. In particular, for fixed nonzero $v$ the word length of $mv$ (=$v^m$, switching back to multiplicative notation) is $\simeq\log(m)$. Since cocompact lattices are undistorted, we deduce that the same holds in $\Gamma$. Hence $|e_{13}(1)^m|\preceq \log(m)$ in $\Gamma$, and hence in the larger group $\mathrm{SL}_n(\mathbf{Z})$.<|endoftext|> TITLE: Motivation for Laver's use of large cardinals to show finite combinatorial properties of Laver tables QUESTION [10 upvotes]: Laver showed in 1995 that the period of the first row of certain Laver tables is unbounded, assuming that a rank-into-rank cardinal exists. The most accessible proof of his result that I was able to find is in chapter 12 of Patrick Dehornoy's Braids and Self-Distributivity (Springer 2000). The proof is quite technical. Laver defines an algebra on the models of set-theory - technically, on certain elementary embeddings of huge sets. He defines two operations on such elementary embeddings, quotients out certain large infinities to get finite results, and investigates their properties. My question is: why should elementary embeddings of ZFC have anything to do with the periodicity of these finite tables? More generally, is there some guiding intuition I can use to make sense of Laver's very complex construction? I find it difficult to understand how Laver could have plowed through this weird and intricate technical construction without having some reason to think it would lead to some kind of specific finitary result. REPLY [4 votes]: Richard Laver was a set theorist throughout his entire career, so he was originally motivated to investigate very large cardinals and the resulting finite algebras from a set theoretic perspective, but today it is probably best to think of the Laver tables $A_{n}$ as algebraic objects within a much larger class of algebraic objects. With this algebraic perspective, large cardinals are a source of examples of such algebraic objects, and large cardinals also provide the consistency strength to prove more theorems. Red herrings As one looks more closely at the Laver tables and similar structures, one observes that some ideas are essential while others are non-essential or just special cases of a more general idea including the following notions. The Laver tables, algebras generated by 1 elementary embedding-The Laver tables are simply the one-generator finite algebraic structures within much broader classes of algebraic structures such as the nilpotent left-distributive algebras. The Laver tables are analogous to the finite cyclic groups while the broader class of algebraic structures is analogous to the class of all groups. You get a more accurate analogy by relating the Laver tables to the cyclic $p$-groups while the broader class of algebraic structures is analogous to the class of all $p$-groups and related objects such as pro-$p$-groups (in this analogy, $\mathcal{E}_{\lambda}$ corresponds to a dense $G_{\delta}$-subset of a pro-$p$-group). And yes, these generalizations of Laver tables have the same kind of intricate structure and periodicity that appears with the Laver tables. Set theory-While the Laver tables originally arose from set theory, they along with similar algebraic structures can easily be studied without any reference to large cardinals. Furthermore, some of these structures similar to Laver tables cannot arise from large cardinals. Composition-The composition of elementary embeddings and a more general composition like operation in the Laver tables should be thought of as operations that are constructed from the self-distributive application operation. Rank-into-rank embeddings-One can construct self-distributive algebras such as the Laver tables from $n$-huge cardinals rather than rank-into-rank cardinals. A linearly ordered set of critical points-The notion of a critical point is essential for Laver tables, but it can be generalized quite a bit. The direct product of two Laver tables $A_{5}\times A_{6}$ will have a partially ordered but not linearly ordered set of critical points. In algebra, one would typically like to study classes of algebraic structures that are closed under taking products (or at least finite product), quotients, and substructures. One should therefore consider algebraic structures with a partially ordered set of critical points. Starting from nilpotence The nilpotent left-distributive algebras are algebraic structures that resemble the algebras of elementary embeddings and Laver tables. The notion of a nilpotent left-distributive algebras is very easy to define, but it precisely captures the notion of a critical point and composition operation as well as other notions related to Laver table for finite algebras (infinite algebras will require a different axiomatization though, but that is a completely open research direction). I admit that very little is known about nilpotent left-distributive algebras (these structures will become more understandable in the future when people write papers on these structures), but they so far seem to be one of the most important classes of algebraic structures that contains the Laver tables but does not contains objects like non-trivial quandles which are quite different from the Laver tables. If $(X,*)$ is a left-distributive algebra, then define terms $x^{[n]},x_{[n]}$ for $n\geq 1$ recursively by letting $x^{[1]}=x_{[1]}=x$ and $x^{[n+1]}=x*x^{[n]},x_{[n+1]}=x_{[n]}*x$. Suppose that $(X,*)$ is a left-distributive algebra. We say that an element $x\in X$ is a left-identity if $x*y=y$ for each $y\in Y$. We say that a subset $L\subseteq X$ is a left-ideal if $x*y\in L$ whenever $y\in L$. Let $\mathrm{Li}(X)$ be the collection of all left-identities in $(X,*)$. Then we say that $(X,*)$ is a nilpotent left-distributive algebra if $\mathrm{Li}(X)$ is a left-ideal, and for each $x\in X$, there exists an $n$ where $x^{[n]}\in\mathrm{Li}(X)$. The algebra $\mathcal{E}_{\lambda}/\equiv^{\gamma}$ is nilpotent because of the following prominent result: Theorem (Kunen inconsistency): Suppose that $j:V_{\alpha}\rightarrow V_{\alpha}$ is a non-trivial elementary embedding. Let $\lambda=\lim_{n\in\omega}j^{n}(\mathrm{crit}(j))$. Then $\alpha=\lambda$ or $\alpha=\lambda+1$. Let us now obtain critical points and a composition operation from nilpotent left-distributive algebras. If $(X,*)$ is a left-distributive algebra, then for each $n\in\omega$, define an operation $*_{n}$ recursively by letting $x*_{0}y=y,x*_{n+1}y=x*_{n}(x*y)=x*(x*_{n}y)$. Each $*_{n}$ is self-distributive. If $(X,*)$ is a nilpotent-self-distributive algebra, then the sequence $(x*_{n}y)_{n}$ is eventually constant for each $x,y\in X$, so let $x*_{\infty}y=\lim_{n\rightarrow\infty}x*_{n}y$. Define a relation $\preceq$ on $(X,*)$ by setting $x\preceq y$ if and only if $x*_{\infty}y\in\mathrm{Li}(X)$. Then $\preceq$ is a pre-ordering. Let $\simeq$ be the equivalence relation defined by letting $x\simeq y$ iff $x\preceq y\preceq x$. Let $\mathrm{crit}[X]=X/\simeq$, and let $\mathrm{crit}(x)$ denote the equivalence class containing $x$ whenever $x\in X$. Then $\mathrm{crit}[X]$ is partially ordered by letting $\mathrm{crit}(x)\leq\mathrm{crit}(y)$ if and only if $x\preceq y$. One can define an implication operation on $\mathrm{crit}[X]$ by letting $\mathrm{crit}(x)\rightarrow\mathrm{crit}(y)=\mathrm{crit}(x*_{\infty}y).$ Suppose that $X$ is a self-distributive algebra where $\mathrm{Li}(X)$ is a non-empty left-ideal. Suppose that $L$ is a poset. Then we say that a mapping $\phi:X\rightarrow L$ is a critical point operator if it satisfies the following conditions: i. $\phi(x)\leq\phi(y)\leftrightarrow\phi(x)\leq\phi(x*y)$. ii. $\phi(y)\leq\phi(x*y)$. iii. $\phi(y)\leq\phi(z)\rightarrow\phi(x*y)\leq\phi(x*z)$. iv. $\phi(x)=1$ if and only if $x\in\mathrm{Li}(X)$. v. $\phi(x)=\phi(x*x)$ if and only if $x\in\mathrm{Li}(X).$ vi. For all $x\in X$, there is a $c\in X$ with $c*c\in\mathrm{Li}(X)$ and $\phi(x)=\phi(c)$. If $X$ is a nilpotent LD-system, then the mapping $\mathrm{crit}:X\rightarrow\mathrm{crit}[X]$ is a critical point operator. Proposition: Suppose that $X$ is a finite left distributive algebra, and $\phi:X\rightarrow L$ is a critical point operator. Then $(X,*)$ is nilpotent, and $\phi(x)\leq\phi(y)$ if and only if $\mathrm{crit}(x)\leq\mathrm{crit}(y)$ The above proposition shows that the finite nilpotent left-distributive algebras are just the finite left-distributive algebras with a sensible (though partially ordered) notion of critical points. One can obtain nilpotent self-distributive algebras from rank-into-rank embeddings without even invoking Kunen's inconsistency result. If $X$ is a nilpotent self-distributive algebra, and $\alpha\in\mathrm{crit}[X]$, then there is some $c\in X$ with $c*c\in\mathrm{Li}(X)$ and $\mathrm{crit}(c)=\alpha.$ Define a congruence $\equiv^{\alpha}$ on $X$ by letting $x\equiv^{\alpha}y$ if and only if $c*x=c*y$. The nilpotent left-distributive algebras can also be endowed with a composition operation in some sense. An LD-monoid is an algebraic structure $(X,*,\circ,1)$ where $(X,*)$ is an LD-system, $(X,\circ,1)$ is a monoid, and $x*1=1,1*x=x$, $x\circ y=(x*y)\circ x$, $x*(y\circ z)=(x*y)\circ(x*z)$, $(x\circ y)*z=x*(y*z)$. A nilpotent LD-monoid is an LD-monoid $(X,*,\circ,1)$ where $(X,*)$ is nilpotent and $\text{Li}(X)=\{1\}$. If $(X,*,\circ,1)$ is a nilpotent LD-monoid, then $(\mathrm{crit}[X],\rightarrow,\wedge,1)$ is a Heyting semilattice where $\mathrm{crit}(x)\wedge\mathrm{crit}(y)=\mathrm{crit}(x\circ y)$. Suppose that $(X,*)$ is a nilpotent left-distributive algebra where $\text{Li}(X)$ is a non-empty left-ideal. Let $\simeq$ be the smallest equivalence relation on $\bigcup_{n\in\omega}X^{n}$ that satisfies the following conditions whenever $m\geq 0,n\geq 0$, and $x_{1},\dots,x_{m},y_{1},\dots,y_{n}\in X$: i. $(x_{1},\dots,x_{m},a,y_{1},\dots,y_{n})\simeq(x_{1},\dots,x_{m},y_{1},\dots,y_{n})$ whenever $a\in\mathrm{Li}(X)$. ii. $(x_{1},\dots,x_{m},a,b,y_{1},\dots,y_{n})\simeq(x_{1},\dots,x_{m},a*b,a,y_{1},\dots,y_{n})$ whenever $a,b\in X$ Let $\text{LDM}(X)=\bigcup_{n\in\omega}X^{n}/\simeq$. We shall let $[x_{1},\dots,x_{m}]$ denote the equivalence class containing $(x_{1},\dots,x_{m})$. Then $\text{LDM}(X)$ is a LD-monoid with operations defined by $[x_{1},\dots,x_{m}]\circ[y_{1},\dots,y_{n}]=[x_{1},\dots,x_{m},y_{1},\dots,y_{n}]$ and $[]*[y_{1},\dots,y_{n}]=[y_{1},\dots,y_{n}]$ and $[x_{1},\dots,x_{m+1}]*[y_{1},\dots,y_{n}]= [x_{1},\dots,x_{m}]*[x_{m+1}*y_{1},\dots,x_{m+1}*y_{n}]$. If $(X,*)$ is a nilpotent LD-system, then $\text{LDM}(X)$ is a nilpotent LD-monoid, and the mapping $e:X\rightarrow\text{LDM}(X)$ defined by $e(x)=[x]$ is a homomorphism. This is therefore a construction that allows one to obtain a composition operation from the self-distributive operation. Proposition: A finite left-distributive algebra $(X,*)$ where $\text{Li}(X)$ is a left-ideal is nilpotent if and only if $\text{LDM}(X)$ is finite. The nilpotent left-distributive algebras generated by a single element in a sense are just the Laver tables. Proposition: Suppose that $X$ is a finite monogenic left-distributive algebra such that $\mathrm{Li}(X)\neq\emptyset$. Then $X\simeq A_{n}$. Proposition: The Laver tables $A_{n}$ are up-to-isomorphism, the only finite monogenic nilpotent left-distributive algebras. Theorem: Suppose that $X$ is a monogenic nilpotent left-distributive algebra. Then either $X$ is isomorphic to a Laver table $n$ or $X$ is infinite and for all $n$, we have $X/\equiv^{\alpha_{n}}\simeq A_{n}$ where $\alpha_{n}=\mathrm{crit}(x_{[2^{n}]})$. Theorem: Suppose that there exists a rank-into-rank cardinal. Then the Laver tables $A_{n}$ are the only nilpotent monogenic left-distributive algebras. Laver-like algebras For the order of operations, the implied parentheses are grouped on the left so that $a*b*c*d=((a*b)*c)*d$. A left-distributive algebra $(X,*)$ is Laver-like precisely when $\mathrm{Li}(X)$ is a left-ideal and whenever $x_{n}\in X$ for each $n\in\omega$, there exists some $N$ where $x_{0}*\dots*x_{N}\in\mathrm{Li}(X)$. Every Laver-like algebra is nilpotent, and if $(X,*)$ is Laver-like, then $\mathrm{crit}[X]$ is well-ordered. Every quotient algebra $\mathcal{E}_{\lambda}/\equiv^{\gamma}$ of rank-into-rank embeddings is Laver-like (this is the Laver-Steel theorem). Theorem: Every Laver-like generated by a single element is isomorphic to a Laver table $A_{n}$. Proposition: If $(X,*)$ is a reduced Laver-like algebra, then mapping $e_{X}:X\rightarrow\text{LDM}(X)$ is an isomorphism (so there exists some composition operation $\circ$ such that $(X,*,\circ,1)$ is an LD-monoid). The Laver-like algebras therefore more closely resemble the algebras of elementary embeddings, but there are still reduced Laver-like algebras that cannot arise from elementary embeddings. Construction of new nilpotent left-distributive algebras from old ones The construction of the Laver tables is actually a special case of a technique for constructing new nilpotent self-distributive algebras from old ones. While there are plenty of ways to construct new nilpotent left-distributive algebras from old ones, it is most fruitful to construct new nilpotent left-distributive algebras that have the same number of generators as the old algebras but which have a higher poset of critical points (more specifically if $Y$ is the new left-distributive algebra, and $X$ is the old one, then $\textrm{crit}(Y)$ is isomorphic to $\textrm{crit}(X)\cup\{\mu\}$ where $\mu$ is a new element defined by letting $\mu>\gamma$ for each $\gamma\in\textrm{crit}(X)$). The best way to achieve this is to begin with a finite left-distributive algebra $(X,*)$ that is generated by a system $(x_{a})_{a\in A}$ and obtain a new finite left-distributive algebra $(Y,*)$ that is generated by $(y_{a})_{a\in A}$ and where $\mathrm{crit}[Y]\setminus\{1\}$ has a maximum element $\gamma$ and there there is a (necessarily surjective) homomorphism $\phi:Y\rightarrow X$ with $\ker(\phi)=(\equiv^{\gamma})$ and where $\phi(y_{a})=x_{a}$ for each $a\in A$. Observe that in this case, $X$ is isomorphic to a subalgebra $\{c*y\mid y\in Y\}$ of $Y$ where $\mathrm{crit}(c)=\gamma$. A way to build the algebra $Y$ from $X$ algorithmically is to start off with $X$ and then repeatedly construct new algebras $Z$ that contain $X$ as a subalgebra but where $Z$ is generated by $X\cup\{r\}$ for some element $r\in Z$ and where $Z\setminus\{r\}$ is a subalgebra of $Z$ that can be written as a sub-direct product of previously constructed algebras. When one applies this technique of extending algebras to larger algebras one element at a time, one goes from $A_{n}$ to $A_{n+1}$ by traversing through all the intermediate algebras $\{x\in A_{n+1}\mid x\geq r\}$ where $1\leq r\leq 2^{n}+1$.<|endoftext|> TITLE: How do you generate math figures for academic papers? QUESTION [13 upvotes]: Good day! I am looking for any tool that would allow me to generate a figure similar to the figures embedded in the paper by King et al. (2020) titled "Trigonometry: a brief conversation." King, C., Evelyn, T., Ye, F., & Carvajal, B. (2020). Trigonometry: A Brief Conversation. Open Educational Resources. https://academicworks.cuny.edu/qb_oers/167/ REPLY [15 votes]: To complement Kostya's answer, here is a way to turn hand-drawings into something looking vaguely professional using inkscape, very quickly (and for free). Draw something on paper. Take a photo in good light conditions, and drag the photo file into inkscape. Select the photo in inkscape and press Shift + Alt + B at the same time (this starts the bitmap function). Tweak the parameters until you're happy with them (click "update" to preview), then click "OK". Delete the underlying image, an svg (vector graphics) picture will have been generated on top of it. Go to "document properties" to tweak the page size. Press Ctrl + Shift + E to open the export function, and export it. You get an image (png or jpg) which you can include using various packages into a LaTeX file. This gives the middle image (the above process took ~10 seconds): Because it's an svg file, you can use the node tool (select the line drawing, then press N) to tweak the image a bit, change its colour, etc., e.g. giving the above picture on the right. Here is another example, based on a random maths drawing I found online. The last one took about a minute to (badly) edit. Tip: do not delete the inkscape file, so that you can re-edit the image later if you like. REPLY [2 votes]: I find Inkscape a bit too heavy and use a much smaller vector drawing program for Windows called Mayura Draw in combination with equally antiquated psfrag package.<|endoftext|> TITLE: Tzitzeica surface QUESTION [5 upvotes]: A Tzitzeica surface has the property that the ratio of the surface’s Gaussian curvature and the fourth power of the distance from the origin to the tangent plane at any arbitrary point of the surface is constant. My question is: are there Tzitzeica surfaces with constant negative Gaussian curvature? REPLY [4 votes]: The answer is 'no'. Suppose that $M\subset\mathbb{E}^3$ is a smooth connected surface. If the ratio of the Gauss curvature $K$ and $p^4$ is constant (where $p(x)$ is the distance from $T_xM$ to the origin is constant) and $K$ is constant and nonzero, then $p$ is also constant. However, if $p$, which is known as the support function of the surface, is constant and the Gauss curvature $K$ is nonzero, then the surface is a portion of a sphere centered at the origin, and hence $K$ is a positive constant.<|endoftext|> TITLE: Rate of decrease of the Fourier transform of standard mollifiers QUESTION [6 upvotes]: What is the the rate of decrease of $|\widehat{f_p}(t)|$ (as $t\to\infty$), where $p\in(0,\infty)$, $$\widehat{f_p}(t):=\int_{\mathbb R} e^{itx}f_p(x)\,dx,$$ and $$f_p(x):=e^{-1/(1-x^2/p)^p}1(|x|<\sqrt p),$$ so that $f_p\in C^\infty$, $f_p'(0)=0$, and $f_p''(0)=-2/e$ for all $p$. Here are the graphs $\{(x,f_p(x))\colon|x|<\sqrt p\}$ for $p=1$ (red), $p=2$ (green), and $p=3$ (blue): The decrease seems to be very fast, almost to $0$ in a finite time $t>0$. Below are parts of the graphs of the functions $\ln\ln\dfrac1{|\widehat{f_p}|}$ for $p=1$ (red), $p=2$ (green), and $p=3$ (blue): This seems to almost contradict Hardy's uncertainty principle -- which implies, as noted in Dmitry Krachun's comment, that $\widehat{f_p}(t)$ cannot decay faster than $e^{-ct^2}$ for any $c>0$. REPLY [3 votes]: So, morally, the only real way to bound the decay of $\hat{f}$ is to obtain some bounds on the derivatives of $f$. Let me do a little dilation and consider $f(x) = \exp(-1/(1-x)^p - 1/(1+x)^p)$, it does a well-known thing on the Fourier side and doesn't affect general decay properties. And fortunately for us, bounds for the derivatives of this function were written out by Arie Israel in this beautiful paper (I think it's a third time I've given this reference to someone, see also this MO question for related matters). Specifically, Lemma 2 of his paper says that $$|f^{(k)}(x)| \le (16p)^k k^{(1+1/p)k}.$$ (he considers the case $p\in \mathbb{N}$ but his methods should work for any $p > 0$). We have $$|\hat{f}(\xi)| \le 2\frac{||f^{(k)}||_{L^\infty}}{|\xi|^k}.$$ Factor of $2$ here is because $\text{supp} f = [-1, 1]$ has length $2$ and there also might be some powers of $2\pi$ thrown here and there depending on the normalization of the Fourier transform. Neither of those will affect our argument of course. Combining this with the above estimate we get $$|\hat{f}(\xi)| \le 2 \left(\frac{16p k^{(1+1/p)}}{|\xi|}\right)^k.$$ We want to choose $k\in \mathbb{N}$ so that the number in the brackets is some constant, say it's at most $\frac{1}{2}$ and at least $\frac{1}{4}$ (this can always be done for $|\xi|$ big enough). This gives us $$|\hat{f}(\xi)| \le 2^{1-k}.$$ On the other hand we have $$k \ge \left(\frac{|\xi|}{64p}\right)^{p/(p+1)},$$ which gives us the desired decay in terms of $|\xi|$ of the form $$|\hat{f}(\xi)|\le A\exp(-B|\xi|^{p/(p+1)})$$ for some $A, B>0$. Note that $\hat{f}$ can't have exponential decay because otherwise $f$ would be analytic in a strip and thus not compactly supported; one can actuallly deduce even more from the general results, see Denjoy-Carleman Theorem and Beurling-Malliavin Theorem in my answer to the above MO question. But here I think it is probable that the above decay is the correct one up to the values of $A$ and $B$: indeed, if $\hat{f}$ decays very fast then we can obtain better pointwise bounds on the derivatives of $f$ and the Israel's bounds are probably sharp, although I'm not sure how to prove it.<|endoftext|> TITLE: In CZF (w/ Subset Collection removed) the Powerset axiom Implies Subset Collection QUESTION [7 upvotes]: The Subset Collection axiom: $$ \forall a \forall b \exists c \forall u [\forall x \in a \exists y \in b (\psi(x,y,u)) \longrightarrow \exists d \in c (\forall x \in a \exists y \in d (\psi(x,y,u)) \land \forall y \in d \exists x \in a (\psi(x,y,u)))] $$ is often digested by considering the equivalent (in CZF - Subset Collection) axiom of Fullness. Let $mv(B^A)$ be the class of all sets $R \subseteq A \times B$ satisfying $\forall a \in A \exists b \in B (\langle a,b \rangle \in R)$. A set $C$ is full in $mv(B^A)$ if $C \subseteq mv(B^A)$ and $$ \forall R \in mv(B^A) \exists S \in C (S \subseteq R). $$ The Fullness axiom states: For all sets $A$ and $B$ there exists a set $C$ such that $C$ is full in $mv(B^A)$. For completness the Powerset axiom states: $\forall x \exists y \forall z(z \subseteq x \longleftrightarrow z \in y$). Now to show that the Powerset axiom implies Subset Collection (in CZF - Subset Collection) it suffices to show that the Powerset axiom implies Fullness. This proof is claimed to be obvious in many papers. I understand this may be so but I am still having some trouble with it. The difficulty I am having is deciding which set to apply the Powerset axiom to. For $mv(B^A)$ is only assumed to be a class (although the Powerset axiom is equivalent to it being a set this result typically comes later so I don't think that is neccesary.) I must admit I am new to proving these meta-set-theoretic results so I am aware the answer is likely right under my nose. I would be delighted for someone to offer some wisdom. REPLY [3 votes]: (2022-04-06 EST: I once changed the terminology "Image" in my answer to "subimage." Thus please remember that "image" in the previous comments must be "subimages.") A previous answer by aws pointed out how to prove Subset Collection from Powerset: (…) If you want to think of it as a special case of power set, then what it is saying is that given sets $a$ and $b$ it asserts the existence of a collection of subsets of $b$ that contains something like the image of every multi-valued function from $a$ to $b$. For example, if we had the powerset axiom, we could take $c$ to be the powerset of $b$. Let me provide more details on aws's explanation. For sets $a,b$ and a multi-valued function $R:a\rightrightarrows b$ (which can be a proper class), call a set $c$ an subimage of $R$ if it satisfies For every $x\in a$, we can find $y\in b$ such that $(x,y)\in R$, and If $y\in c$, then there is $x\in a$ such that $(x,y)\in R$. Their formal statements are: if $R$ is defined by a formula $\phi(x,y)$ then $\forall x\in a\exists y\in c\ \phi(x,y)$, and $\forall y\in c\exists x\in a\ \phi(x,y)$. (If $R$ is a set, then $\phi(x,y)$ must be $(x,y)\in R$.) You can see that if $R$ is functional (that is, if $R:a\to b$ is a function) then $c$ is exactly the image of $R$. Unlike functions, however, a subimage of a multi-valued function is not in general unique: Example. Let $a,b=3=\{0,1,2\}$ and consider $R=\{(0,0),(0,1),(1,1),(2,2)\}$. You can see that $R$ is a multi-valued function from $3$ to $3$, but both of $\{1,2\}$ and $\{0,1,2\}$ are subimages of $R$. As I explained in your previous question, the axiom of subset collection states the following: For any class family of relations $R_u\subseteq a\times b$ parametrized by $u$, we can find a set $c$ such that if $R_u\colon a\rightrightarrows b$, we can find $d\in c$ such that $d$ is a subimage of $R_u$. You can check that if $\psi(x,y,u)$ defines $R_u$ in the sense that $(x,y)\in R_u$ if and only if $\psi(x,y,u)$, then the above statement is just an informal rephrase of Subset Collection. Then the Subset Collection is an immediate corollary of Powerset: observe that every subimage of $R_u$ is a subset of $b$, so $c=\mathcal{P}(b)$ witnesses Subset Collection. Added 2022-04-06 EST: I found that the following one would be a simpler but equivalent statement for Subset Collection while pertaining to the notion of subimage: For any $a$ and $b$, we can find a set $c$ full in subimages of multi-valued functions from $a$ to $b$, in the sense that if $r\colon a\rightrightarrows b$, then we can find a subimage $d\in c$ of $r$. For the one direction, Subset Collection applied to $u$ implies the above statement. On the other direction, let $R_u\colon a\rightrightarrows b$ be a class multi-valued function with a parameter $u$. Then $\mathcal{A}(R_u)\colon a\rightrightarrows a\times b$. (See below for the definition of $\mathcal{A}$.) By Strong Collection, we have $r$ such that $r$ is a subimage of $\mathcal{A}(R_u)$, so $r\subseteq R_u$ and $r\colon a\rightrightarrows b$ (it follows from the lemma I stated below.) Now let $c$ is full in subimages of multi-valued functions from $a$ to $b$, and if $d\in c$ is a subimage of $r$. Then it is straightforward to see that $d$ is a subimage of $R$. You may also ask how to connect Fullness and Subset Collection in the way I referred. I found that the following notion is prevalent in proofs about Subset Collection and Fullness: Definition. Let $R:a\rightrightarrows b$ be a multi-valued function. Define $\mathcal{A}(R):a\rightrightarrows a\times b$ by $$\mathcal{A}(R)=\{(a,(a,b)) \mid (a,b)\in R\}.$$ Then we can prove the following: Lemma. Let $R:a\rightrightarrows b$ be a multi-valued function, then the following holds: $\mathcal{A}(R):a\rightrightarrows s \iff R\cap s: a\rightrightarrows b$, and $\mathcal{A}(R):a\leftleftarrows s \iff s\subseteq R$. (The proof of my lemma is not too hard, but tedious. See my previous blog post or Lemma 2.8 of my preprint.) We can see that $R$ is a subimage of $\mathcal{A}(R)$. In fact, $R$ is a maximal subimage (or, just the image) of $\mathcal{A}(R)$, in the sense that if $s$ is a subimage of $\mathcal{A}(R)$ then $s\subseteq R$. (Observe that $s$ is a subimage of $\mathcal{A}(R)$ if and only if $\mathcal{A}(R):a\rightrightarrows s$ and $\mathcal{A}(R):a\leftleftarrows s$ by definition.) Then we can view the implication of Fullness from Subset Collection in the following way: consider the following (definable) collection of multi-valued functions: $$\mathcal{R} = \{\mathcal{A}(R) \mid R\in\operatorname{mv}(\sideset{^a}{}b)\}.$$ By Subset Collection, there is a collection $C$ of subimages of $\mathcal{R}$. For each $r:a\rightrightarrows b$, let $s\in C$ be a subimage of $\mathcal{A}(r)$, that is, we have $\mathcal{A}(r):a\rightrightarrows s$ and $\mathcal{A}(r):a\leftleftarrows s$. By the lemma, we have $s\subseteq r$. Hence $C$ is a full subset of $\operatorname{mv}(\sideset{^a}{}b)$.<|endoftext|> TITLE: Does there exist an automorphism of a finitely generated group with periodic points of unbounded period? QUESTION [11 upvotes]: Does there exist a finitely generated group $G$ and an automorphism $\Phi\colon G \to G$ such that there are $\Phi$-periodic elements with unbounded period? If $G$ is merely countable and not finitely generated there are easy examples: consider a free or free abelian group of countably infinite rank, let $x_1,x_2,\ldots$ be a standard generating set, and consider the automorphism that cyclically permutes the first $2$ generators, the next $3$ generators, the next $4$ generators and so on. REPLY [11 votes]: This can be done with an inner automorphism of the Baumslag-Solitar group $BS(2, 3) = \langle a, t | (a^2)^t = a^3 \rangle$, namely conjugation by $a$. The smallest positive integer $n$ such that $a^{t^{-k}}$ (with $k > 0$) commutes with $a^n$ is $n = 2^k$ (this follows from Britton's lemma). REPLY [11 votes]: Yes, and even with an inner automorphism. Namely, start from your example of a countable group $G$ with elements $x_n$ ($n$ ranging over some arbitrary subset) and automorphism $f$ such that $x_n$ lies in an $n$-cycle for $f$. Let $G'$ be the semidirect product $G\rtimes\langle f\rangle$. Finally, embed $G'$ into a finitely generated group $H$ (every countable group embeds into a f.g. group, an old result of Higman-Neumann-Neumann). Then in $H$, for the inner automorphism defined by $f$, the element $x_n$ lies in an $n$-cycle. [NB: as $G$ can be chosen to be abelian and hence $G'$ metabelian, $H$ can be chosen to be 4-step solvable, by the Neumann-Neumann proof of the previous embedding theorem.] Note: there are no examples among linear groups (= subgroup of $\mathrm{GL}_d$ over a field) with an inner automorphism. Indeed, for the inner automorphism $i_g$ defined by an element $g$, the centralizer of $g^{n!}$ is an increasing sequence of subgroups. But in a matrix group the centralizers in the algebra of matrices are subspaces. So there is no strictly infinite sequence of centralizers. Hence the finite $i_g$-cycles have bounded length. Here's another example, due to B.H. Neumann 1937, with the additional feature of being residually finite. Namely, consider a set $X$ consisting of the infinite disjoint union of sets $X_n$ ($n$ ranging over an infinite set $I$ of integers $\ge 5$), each $X_n$ being in bijection with $\{1,\dots,n\}$. Let $c$ be the element acting as the $n$ cycle $n\mapsto 1\mapsto 2\mapsto\dots$ on each $X_n$, and $t$ the element acting as the transposition $1\mapsto 2\mapsto 1$ on each $X_n$. Let $G_I$ be the subgroup of permutations of $X$ generated by $\{t,c\}$. Then $G_I$ is residually finite (since it acts faithfully with finite orbits). Also $G_I$ contains, for each $n\in I$, as normal subgroup, the alternating group $\mathrm{Alt}_n$ of $X_n$ (acting trivially elsewhere). Let $s_n$ be the 3-cycle $1\mapsto 2\mapsto 3\mapsto 1$ acting on $X_n$, identity elsewhere. Then we see that $s_n\in G_I$, and the inner automorphism defined by $c$ admits $s_n$ as element of period exactly $n$.<|endoftext|> TITLE: Current status on Richardson's model (growth model) QUESTION [6 upvotes]: A very simple stochastic growth model on a lattice is the Richardson's model (Actually originally defined by Murray Eden in the 60s). Each point of the lattice can be either occupied or vacant, once they are occupied they remain so forever, and vacant points become occupied at a rate equal to the fraction of the occupied neighbours (so a point can become occupied only if at least one of its neighbours is occupied). Eventually all points will be occupied, but the limiting shape of the aggregation of occupied points roughly looks like a circle. There is a link (1) that discuss a bit more about this model and gives a few papers. However none of these papers are recent. Question: I would be curious to know if there are still some interesting open problems on this model, and more interestingly, if despite its simplicity this model can accurately describes any real biological phenomena. (I can easily imagine that this model can be seen as a special case of some epidemic model on a lattice, but this is not really what this post aims at) Here is a picture (black vacant, white occupied) that I took from Eden's original paper link: https://services.math.duke.edu/~rtd/survey/survc1.html original article : Eden, Murray. "A two-dimensional growth process." Proceedings of the fourth Berkeley symposium on mathematical statistics and probability. Vol. 4. University of California Press Berkeley, 1961. REPLY [6 votes]: Minor remark: I have always known this model as the Eden model, it seems that most probabilists currently refer to it as such. To answer your question, yes there are many interesting open questions, mainly regarding the fluctuations of the interface around its limiting shape. The Eden model is the prototypical example of a model conjectured to belong to the KPZ universality class. This immediately yields a wealth of conjectures. For example, after a time of order $t$, the deviations from a limiting shape of radius $t$ (the exact shape itself isn't known I believe) should be of order $t^{1/3}$. The law of this fluctuation in any fixed direction is further conjectured to converge to the GUE Tracy-Widom distribution as $t \to \infty$. If, instead of starting with one single occupied site, one starts with an occupied half-space, then the law of the interface between occupied and empty sites, suitably recentred and rescaled, is conjectured to converge to the KPZ fixed point. All of these conjectures are expected to be extremely hard since the Eden model has no known integrable structure, so that the exact calculations that allowed to show similar results for some other models aren't available. You'll find more details in the various review articles by people like Quastel, Spohn, Corwin, etc.<|endoftext|> TITLE: Tensor product of a von Neumann algebra and $L_\infty $ QUESTION [10 upvotes]: Let $R$ be the hyperfinite $II_1$-factor. We know that $R$ is isomorphic to $R\otimes R$. So, $L_\infty(0,1) \otimes R$ is a von Neumann subalgebra of $R$. I am not sure whether it is sure for any type $II_1$ von Neumann algebra $M$, i.e., is $L_\infty(0,1) \otimes M$ a von Neumann subalgebra of $M$? REPLY [11 votes]: If $\mathbb F_I$ denotes the free group on $I$ generators with $\lvert I \rvert > 1$, then $L^\infty(0, 1) \overline \otimes L \mathbb F_I$ is not isomorphic to a von Neumann subalgebra of $L \mathbb F_I$. For $\lvert I \rvert > \aleph_0$ this is Corollary 6.4 in [S. Popa: Orthogonal pairs of subalgebras in finite von Neumann algebras, J. Op. Th. 9(1983), 253-268]. The general case $\lvert I \rvert > 1$ is Theorem 1 in [N. Ozawa: Solid von Neumann algebras, Acta Math. 192 (2004), 111-117].<|endoftext|> TITLE: Over which (graded) rings are all modules decomposable into indecomposables? QUESTION [9 upvotes]: A module is decomposable if it is the direct sum of two modules. The process of splitting summands off of a decomposable module does not need to terminate, so infinitely generated modules do not typically split into sums of indecomposable modules. But they do over certain rings, and I wonder which kinds of rings. Clearly, fields are okay, but even rings as simple as $\mathbb Z$ are not: an infinite product of $\mathbb Z$ is not free. On the other hand, if we look at nonnegatively graded, connected $k$-algebras and their categories of nonnegatively graded modules, there seem to be more examples. I think some Zorn yoga shows that any nonnegatively graded module over $k[t]$, with $t$ in positive degree, splits as a sum of cyclic modules. Are there more examples? What about, let's say, graded modules over the Steenrod algebra? REPLY [4 votes]: Regarding the question about the Steenrod algebra, it is not true that every non-negatively graded module for the $\text{mod }2$ Steenrod algebra is a direct sum of indecomposable modules. I haven't checked the $\text{mod }p$ Steenrod algebra for odd $p$, but I would be astonished if something similar didn't work. First, note that not every (ungraded) $\mathbb{F}_2[x]$-module is a direct sum of indecomposables. This follows from more general results, since $\mathbb{F}_2[x]$ is not Artinian, or (more directly) at least one of the proofs of the corresponding fact about a countably infinite product of copies of $\mathbb{Z}$ also shows that a countably infinite product of copies of $\mathbb{F}_2[x]$ is not a direct sum of indecomposable $\mathbb{F}_2[x]$-modules. Next, I will describe a full exact embedding of the category of $\mathbb{F}_2[x]$-modules into the category of non-negatively graded modules for the Steenrod algebra. So applying this to the example above will give a non-negatively graded module for the Steenrod algebra that is not a direct sum of indecomposables. Let $V$ be a $\mathbb{F}_2[x]$-module. I will contruct a graded module $\bigoplus_{n\geq0}V_n$ for the Steenrod algebra with $$V_n=\begin{cases}V&(0\leq n\leq3)\\ 0&\text{(otherwise).} \end{cases}$$ The only Steenrod squares $\text{Sq}^i$ that can act nontrivially are for $i\leq3$, so the only Adem relations that will need to be checked are $\text{Sq}^1\text{Sq}^1=0$ and $\text{Sq}^1\text{Sq}^2=\text{Sq}^3$. When $i TITLE: Homogeneous representations of compact manifolds QUESTION [5 upvotes]: There is a classification of effective transitive groups actions on the sphere by compact connected Lie groups, compare Besse "Einstein manifolds" 7.13 Examples. Are there similar results for $\mathbb{C}P^n$ and $\mathbb{H}P^n$? REPLY [3 votes]: It seems that Table 3, p. 185 of Gorbatsevich, V. V.; Onishchik, A. L. Lie transformation groups. Lie groups and Lie algebras, I, 95–235, Encyclopaedia Math. Sci., 20, Springer, Berlin, 1993. contains the answer to your question. It contains Onishchik's classification of transitive actions of compact Lie groups on manifolds of rank 1. According to this table, only $SU(n+1)$ and, for $n$ odd, $Sp(\frac12(n+1))$ act transitively on $\mathbb CP^n$. On $\mathbb HP^n$ only $Sp(n+1)$ is acting transitively.<|endoftext|> TITLE: Chalkboard eraser QUESTION [5 upvotes]: I just started my first year of university and because I'm visually impared I have trouble seeing what's written on the chalkboard. I've partially solved this problem by purchasing chalk from hagoromo and asking my professors to use it. (Leaves a wider and nicer line and erases more cleanly). But the chalkboard still gets dirty from chalk which reduces the contrast between the board and the chalk, which in turn makes it harder for me to see. My university supplies what look to be whiteboard erasers for the blackboard which I suspect isn't a good idea. I'm looking for suggestions for premium chalkboard eraser brands or DIY options. Thanks EDIT: My university washes all the boards regardless of department every day at the end of the day. (At least in my campus). REPLY [12 votes]: The Hagoromo chalk, now produced in South Korea, is well accompanied by a professional Korean microfibre eraser (600,000 fibers per square inch), as explained by professor Bayer on his Chalk page. For more on chalk, there is this older MO post.<|endoftext|> TITLE: Which convex bodies roll straight? QUESTION [9 upvotes]: Let $K$ be a convex body in $\mathbb{R}^3$. Suppose $K$ is held at some position and orientation on an inclined plane, and released. Let there be sufficient friction so that it rolls without slippage. My question is: Q. If $K$ rolls along a straight line, i.e., if the point of contact along the inclined plane is a single straight line, what can we conclude about the shape of $K$? In other words, which $K$, when properly oriented, roll straight?               (Figure from Which convex bodies roll along closed geodesics?) It seems that if $K$ is a smooth surface of revolution about an axis $X$, and $K$ has reflective symmetry about a plane orthogonal to $X$ (as in the above illustration), then $K$ rolls straight. But perhaps a wider class of bodies also roll straight. Perhaps reflective symmetry is not necessary; perhaps equal moments of inertia about $X$ in the two halves suffice? Or would any asymmetry cause a wobble in the footprint? I would be interested in learning of any class of shapes that roll straight, especially non-symmetric shapes. REPLY [2 votes]: Assuming that we are considering idealized physical objects, then a criterion for convex bodies to roll on straight lines would be the existence of closed planar geodesics of which the containing plane is orthogonal to an inertial axis and that also contains the center of gravity. Instable equilibria do not pose a problem; in the real world a sufficiently high rotational velocity will provide the necessary stability. Another physical effect is that the curve on which the body rolls may depend on speed whenever the center of gravity is not contained in the plane in which the closed "contact geodesic" is contained: take a cone that is rotating around its axis of symmetry; its rolling motion will become more linear with increasing rotational velocity. What also must be guaranteed for linear rolling motion is that the convex body must be oriented in a way that renders the inertial axis parallel to the (tangent of the) plane's height-lines at the initial point of contact.<|endoftext|> TITLE: Three questions about three functions similar to $\sin,\cos$ QUESTION [8 upvotes]: In The Basel problem revisited? a question about the function, similar to sinc, $f(x)$ was asked: $$f(x) = \prod_{n=1}^\infty \left ( 1+ \frac{x^3}{n^3} \right ) = \prod_{n=1}^\infty \left ( 1+ \frac{x}{n} \right ) \left ( 1+ \frac{x}{n}\omega \right )\left ( 1+ \frac{x}{n} \omega^2 \right )$$ which was shown to be equal to: $$f(x) = \sum_{k=0}^\infty x^{3k}\zeta_k(3)$$ where we have defined: $$\zeta_k(3) = \sum_{1\le n_1 TITLE: On the smallest open Diophantine equations: beyond Hilbert's 10 problem QUESTION [19 upvotes]: In 2018, Zidane asked What is the smallest unsolved diophantine equation? The suggested way to measure size of the equation is substitute 2 instead of all variables, absolute values instead of all coefficients, and evaluate. In the Mathoverlow question Can you solve the listed smallest open Diophantine equations? I list the current smallest equations for which it is open whether there exists any integer solution at all (Hilbert's 10 problem). However, there are some famous equations, like $x^3+y^3+z^3=3$ of size $H=2^3+2^3+2^3+3=27$, for which the Hilbert's 10 problem is trivial (in this example, $x=y=z=1$ is a solution), but the equation can hardly be classified as solved, because we do not even know whether the solution set is finite. Here, I consider more general problem: for a given polynomial Diophantine equation, determine whether the solution set is finite, and if so, list all the solutions. This is a much better approximation of our intuition what does it mean to solve an equation, but still avoids a subtle issue what counts as an acceptable description of the solution set if it is infinite (see What does it mean to solve an equation? for some discussion of this). Selected solved equations. The smallest equation that required a new idea turned out to be $y^2+z^2=x^3-2$ with $H=18$, see Representing $x^3-2$ as a sum of two squares for the proof that it has infinitely many integer solutions. Equations $ y(z^2-y)=x^3+2 $ and and $ xyz=x^3+y^2-2 $ with $H=22$ has been listed as open and then solved by Tomita, see the answer below. Smallest open equations. The current smallest open equations are the equations $$ y(z^2-y)=x^3-2 $$ and $$ xyz=x^3+y^2+2 $$ with $H=22$. These are the only remaining open equations with $H \leq 22$. One may also study equations of special types. For example, the current smallest open symmetric equation (that is, invariant under cyclic shift of the variables) is $$ x^2y+y^2z+z^2x=1 $$ with $H=25$. The current smallest open equations in two variables are $$ y^3+y=x^4+x $$ and $$ y^3-y=x^4-x $$ with $H=28$, while the current smallest open 3-monomial equation is $$ x^3y^2=z^3+2 $$ with $H=42$. For the listed equations, the Hilbert 10th problem is trivial, because there are some obvious small solutions. The question, for each of the listed equations, is whether the solution set is finite or infinite, and if finite, list the solutions. The plan is to list new smallest open equations once these ones are solved. The solved equations will be moved to the "solved" section. REPLY [9 votes]: This is a partial solution. The equation $$y(z^2-y)=x^3+2\tag{1}$$ has infinitely many integer solutions. Since $y = 1/2z^2 \pm 1/2\sqrt{z^4-4x^3-8}$, the expression $z^4-4x^3-8$ must be a perfect square. On the other hand, substitute $x=-3n^2-2n-2$ and $z=3n+1$ to $z^4-4x^3-8$, then we get $$z^4-4x^3-8 = (25+8n+12n^2)(1+2n+3n^2)^2,$$ where $n$ is arbitrary integer. Hence we must find integer solutions of $$v^2 = 25+8n+12n^2\tag{2}.$$ We know equation $(2)$ has infinitely many integer solutions (Gauss's theorem, Mordell's book p.57). Recursive solutions are given as follows. \begin{align*}(v_0,n_0)&=(\pm 5,0),\\ (v_{k+1},n_{k+1}) &= (7v_k + 24n_k + 8,2v_n + 7n_k + 2).\end{align*} k x y z [12],[-458, 10510, 37] [12],[-458, -9141, 37] [172],[-89098, 26729099, 517] [172],[-89098, -26461810, 517] [2400],[-17284802, 71887487358, 7201] [2400],[-17284802, -71835632957, 7201] [33432],[-3353162738, 194174947774195, 100297] [33432],[-3353162738, -194164888285986, 100297] [465652],[-650496286618, 524648022526642094, 1396957] [465652],[-650496286618, -524646071037782245, 1396957] [-8],[-178, 2654, 23] [-8],[-178, -2125, 23] [-108],[-34778, 6538075, 323] [-108],[-34778, -6433746, 323] [-1500],[-6747002, 17535455598, 4499] [-1500],[-6747002, -17515214597, 4499] [-20888],[-1308883858, 47355417946259, 62663] [-20888],[-1308883858, -47351491294690, 62663] [-290928],[-253916721698, 127949397813283390, 872783] [-290928],[-253916721698, -127948636063118301, 872783]<|endoftext|> TITLE: $p-1$ elements in $\mathbb{Z}_p\times\mathbb{Z}_p$ with a sum $(0,0)$ QUESTION [7 upvotes]: Given prime $p\ge 11$, $S$ is a subset of $\mathbb{Z}_p\times\mathbb{Z}_p$ with $3p-3$ elements. Prove: $S$ has a subset $T$ with $p-1$ elements, such that$\sum_{x\in T}x\equiv (0,0)\pmod{p}$. REPLY [5 votes]: In this post a sum over sets means the additive-combinatorial sum, i.e. $\sum A_i=\{a_1+...+a_i : a_1 \in A_1, ..., a_i \in A_i\}$. Lemma. Let $(a_1,a_2,... ,a_{2p−2})$ be a sequence of $2p−2$ elements of $\mathbb Z_p$, where $p$ is a prime. Then either there exists a subsequence $A$ of length $p-1$, in which the sum of the elements equals zero, or $p$ elements take the same value in the sequence. Proof: We may assume $a_1 \leq a_2 \leq ... \leq a_{2p-2}$. If there are $p$ consecutive values, then we are in the second case, otherwise $a_1 \neq a_{p}$, $a_2 \neq a_{p+1}$, $...$, $a_{p-1} \neq a_{2p-2}$, and we can define the sets $B_i=\{a_i,a_{i+p-1}\}$. By the Cauchy-Davenport theorem, $|\sum B_i|-1 \geq \sum (|B_i|-1)$ if $\sum B_i$ is not the whole $\mathbb Z_p$, but this is impossible because $\sum (|B_i|-1) = p-1$. So $0\in ∑B_i$ and it's possible to pick one element in each $B_i$ to get a zero sum. Now we prove the main theorem. Let $X$ be a sequence where the number of occurences of $x$ is one less than the number of occurences of $x$ in $S$ as the first index, i.e. $|\{i:X_i=x\}|=|\{y: (x,y)\in S\}|-1$. The sequence has length $2p-3$ if every element appears in $S$ as the first index, and at least $2p-2$ otherwise. If $X$ has length $2p-3$, we may assume that the elements we choose have first indices $1,2,3...(p-1)$, and we try to find appropriate second indices. Let $Y_i$ be the set of second indices of the elements of $S$ having first index $i$ ($i \neq 0$). We may assume $\sum |Y_i| \geq 2p-2$, for otherwise we can choose all the elements $(0,x)$ ($x \neq 0$). Thus $\sum (|Y_i|-1) \geq p-1$. By the Cauchy-Davenport theorem, $|\sum Y_i|-1 \geq \sum (|Y_i|-1)$ if $\sum Y_i$ is not the whole $\mathbb Z_p$, but this is impossible because $\sum (|Y_i|-1) \geq p-1$. So $0\in ∑Y_i$ and there is an appropriate choice of second indices. If $X$ has length at least $2p-2$, we may find a $(p-1)$-subsequence $A$ in $X$ that sum to zero, or there are $p$ elements with the same first index (this case is trivial). The sequence $A$ is our choice of first indices, and the point is again trying to find appropriate second indices. Let $Y_i$ be the set of k-sums of second indices of the elements of $S$ having first index $i$ ($i \in A$), where $k$ is the number of occurences of $i$ in $A$. As $k$ is smaller than the number of elements in $S$ having first index $i$ (by the definition of $X$), the set $Y_i$ has size at least (number of occurences of $i$ in $A$)+1. Now $|\sum Y_i|-1 \geq \sum (|Y_i|-1)$ if $\sum Y_i$ is not the whole $\mathbb{Z} _p$, but this is impossible because $\sum (|Y_i|-1) \geq |A| = p-1$. So $0\in ∑Y_i$ and there is an appropriate choice of second indices.<|endoftext|> TITLE: Compactly supported sections of coherent sheaves and the dualizing complex QUESTION [8 upvotes]: Suppose $U$ is a (possibly singular) scheme and $X$ is a compactification (potentially unnecessary at least in characteristic $0$). Let $\pi:X\to *$ be the map to the point (though one can consider more general maps as well). There is a classically known pro-algebraic "compactly supported global sections" functor defined by Deligne in the appendix to Hartshorne's "Residues and Duality". Namely, one has a functor $$\pi_!: \mathrm{Coh}_U\to D^b \mathrm{ProFinVect},$$ given informally by taking a sheaf $F$ to the fiber of the map $$\Gamma(U, F)\to \Gamma(\mathring{\delta}_X, F),$$ where $\mathring{\delta}_X: = \widehat{X\setminus U}\cap U$ is the punctured formal boundary. (More explicitly, one defines a functor on coherent sheaves on $X$ and applies it to any continuation of $F$, or equivalently, to $j_*F$ as an ind-object. It's easy to see that this can be done fully inside the $\infty$-category of complexes of pro-coherent sheaves on $X$.) This functor is described in the condensed language in Lecture 11 of Scholze and Clausen's Lectures, though here my understanding is limited (and the lecture works under a smoothness condition, which is surely unnecessary in defining $\pi_!$). Since $\pi_!$ is valued in complexes of pro-vector spaces, one can define the dual contravariant functor $(\pi_!)^*:D^b Coh(U)\to D^b Vect$ (as a functor of $\infty$-categories). As $U$ varies, this forms a presheaf of "distributions on $F$" (which I think is a sheaf in general) which, for $X$ smooth, agrees with the Serre dualizing sheaf. Write $S_X$ for the sheaf of distributions on the constant sheaf, i.e. (the sheafification of) $$S_X:U\mapsto (\pi_!^U)^*(\mathcal{O}_X).$$ I have three questions about this construction. What is the name of the resulting complex for a general proper (and arbitrarily singular) $X$? I want to call it the dualizing complex, but have only seen that defined under some homological singularity restrictions on $X$ like the Gorenstein property. Is there a way to interpret this construction in the condensed language (e.g., is this what would be called $\pi^!\mathcal{O}$ in the Clausen-Scholze lectures? Is it clear that their construction works in the non-smooth context?) The construction in the Deligne appendix only defines $\pi_! F$ as a complex of pro-vector spaces, not pro-finite vector spaces. It is very easy to make it pro-finite-dimensional (and thus with an ind-finite dimensional dual) by rewriting his construction $\infty$-categorically, but I have never seen this done. Is there a reference to this? REPLY [4 votes]: Isn't the dualizing complex defined in general in the proper case by taking applying the right adjoint of $\pi_\ast$ to $k$? That's what I'll take as the definition anyways. The Gorenstein property just means that the dualizing complex is invertible. It is true that this can be computed by the formula you write down, and one way to see this is as in our lecture notes. (We do define compactly supported cohomology without the smoothness assumption.) So yes, this can be interpreted in our setting. I know only few references on those compactly supported cohomology groups, and I'm not aware of any that work $\infty$-categorically. And as you guessed, you actually don't have to sheafify this construction, it's already a sheaf.<|endoftext|> TITLE: Books/websites which have motivating stories of mathematicians overcoming hardships in life QUESTION [55 upvotes]: Edit 1: I have received a lot of great answers. I am not accepting any answer because I think there might be in future that some user want to contribute any new answer, as in my opinion some users might find the answers useful to them to keep their head high in time of despair. Thank you very much to all! I am a person living in a 3rd world country who has done a master's in mathematics and is preparing for grad school in math. Unfortunately, I fell into depression due to horrible harassment by 2 professors against which no action could be taken due to nepotism and corruption in my country, family issues and had to take a break. I don't get much support from my parents or friends as research in mathematics doesn't yield much money and there are a lot of other high paying jobs. I don't care much about their opinions. I live in a very very capitalistic country and people here respect only money. I like to study mathematics as I am very much interested in it. I am taking therapy and medications. I have realized that I need some motivating instances to help keep me going and so that I can motivate myself when I am low due to my depression. It is my humble request to you to suggest books, websites, and/or blogs of real life instances of mathematicians overcoming challenges and hardships in life, both mathematical and non-mathematical. I have been reading Men of Mathematics by E.T. Bell but it is mostly about mathematicians of older centuries, not the 20th and 21st centuries, although it's still very good read. REPLY [2 votes]: Niels Abel surely deserves a mention here. Mario Livio gives a detailed account of his and Galois' lives, but unfortunately not of their mathematical results, in The equation that couldn't be solved. Abel came from a very poor background and was always short of money. The Norwegian government gave him money to travel Europe to meet Gauss and other prominent mathematicians of the day, but demanded that he repay it on his return to Norway. He died from tuberculosis at the age of 26, days before Crelle obtained a position for him as a professor in Berlin. (Galois was a hot-head who would inevitably have got himself killed one way or another.)<|endoftext|> TITLE: Generalisation of Cauchy's mean value theorem QUESTION [14 upvotes]: I apologise in advance if this is an elementary question more fitted for Math Stack Exchange. The reason why I have decided to post here is that the question I am used to seeing on that site are not of the open-ended format of the one I am asking. It is now the second time I have been studying Calculus (first self-taught, now in school) and we are going over the proof of Cauchy's mean value theorem (the precursor to l'Hopital's rule). I do understand the proof, and the intuitive explanation about parametrised curves in a plane, but I still think the statement of the theorem looks relatively obscure. Why are we considering a ratio and not something else? This led me to try and generalise, and this is where we get to my question. I first tried finding a function $h: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that for any functions $f,g: D \subset \mathbb{R} \rightarrow \mathbb{R}$ satisfying Cauchy's mean value theorem's hypotheses, for any interval $[a,b] \subset D$, there exists $x \in [a,b]$ such that $h(f'(x), g'(x))=h(f(b-a), g(b-a))$. Beyond making a few tries and finding a few counterexamples, I realised this wasn't really in the spirit of a mean value theorem: we are trying to make an analogy, if we may use this term, between $f'(x)$ and $f(b)-f(a)$, while in both Lagrange and Cauchy's mean value theorems the analogy is made between $f'(x)$ and $\frac{f(b)-f(a)}{b-a}$. So I started looking for $h$ such that there exists $x$ such that $$ h(f'(x), g'(x))= h\left(\frac{f(b)-f(a)}{b-a}, \frac{g(b)-g(a)}{b-a}\right). $$ But this didn't really lead me anywhere. The question I'm asking is precisely this: can we say anything more about functions $h(x,y) \neq \frac{x}{y}$ satisfying these statements? Suppose we simplify even further, and consider, for example, only the functions $h_{\alpha,\beta}(x,y) = x^{\alpha}y^{\beta}$. Can we maybe prove that only those with $\alpha = k, \beta = -k$ for some $k$ work? (in addition, clearly, to those with $\alpha\beta = 0$) Is this even interesting to investigate? Thanks in advance for helping me. My knowledge doesn't really go far beyond Calculus and Linear Algebra (say, Spivak and Axler's books) but I will try to understand your replies. REPLY [19 votes]: When I reviewed this question a few days ago, I thought there was something sounding familiar in it but I did not remembered what it was: now I have remembered. This problem was fully solved by Alessandro Faedo in paper [1]: in his ZBMath review, Peter Bullen says The author determines all the functions $F(X,Y)$ such that, for every $f(x)$ and $g(x)$ continuous in $[a, b]$ with $f^\prime(x)$ and $g^\prime(x)$ defined for $a0$. Then the function $$ F(Y)=\psi(Y)-Y\psi^\prime(Y_0) $$ has a local minimum in $Y=Y_0$ since $$ F^\prime(Y_0)=0\; \wedge \; F^{"}(Y_0)>0 $$ Now choose a function $g\in C^1([a,b])$ satisfying the following properties: its values at the endpoints of $[a,b]$ satisfy the following relation $$ \frac{g(b)-g(a)}{b-a} =Y_0, $$ $|Y_0 -g^\prime(x)|<\delta$ for all $x\in[a,b]$ and a sufficiently small $\delta>0$, $Y_0\neq g(x)$ for all $x$ belonging to a subset of $[a,b]$ of positive (Lebesgue) measure. For example we can define an indexed family of such functions as $$ g(x) =Y_0 x +\varepsilon \sin\frac{2\pi x}{b-a}\quad 0<\varepsilon<\delta $$ Finally define $f^\prime(x)=\psi(g^\prime(x))$: this implies that $$\DeclareMathOperator{\Dm}{\operatorname{d}\!} f(x)= f(a) + \int\limits_a^x \psi(g^\prime(x))\Dm x \iff \frac{f(b)-f(a)}{b-a} = \frac{1}{b-a}\int\limits_a^b \psi(g^\prime(x))\Dm x. $$ Then, for any $g$ satisfying the above properties, we have $$ \begin{split} \frac{1}{b-a}\int\limits_a^b F(g^\prime(x))\Dm x &= \frac{1}{b-a}\int\limits_a^b \psi(g^\prime(x))\Dm x - \frac{\psi^\prime(Y_0)}{b-a}\int\limits_a^b g^\prime(x)\Dm x\\ & = \frac{1}{b-a}\int\limits_a^b \psi(g^\prime(x))\Dm x - \psi^\prime(Y_0)\frac{g(a)-g(b)}{b-a}\\ &> \frac{1}{b-a}\int\limits_a^b F(Y_0)\Dm x = F(Y_0) =\psi(Y_0) -Y_0 \psi^\prime(Y_0) \\ &= \psi\left(\frac{g(a)-g(b)}{b-a}\right) - \psi^\prime(Y_0) \frac{g(a)-g(b)}{b-a} \end{split} $$ This implies that $$ \begin{eqnarray} \frac{1}{b-a}\int\limits_a^b \psi(g^\prime(x))\Dm x & > & \psi\left(\frac{g(a)-g(b)}{b-a}\right)\\ &\Updownarrow &\\ \frac{f(b)-f(a)}{b-a} & > &\psi\left(\frac{g(a)-g(b)}{b-a}\right)\label{2}\tag{2} \end{eqnarray} $$ while $f^\prime(x) -\psi(g^\prime(x)) =0 $ for each $x\in [a,b]$. This finally implies, contradicting the hypothesis, that $\Phi (X,Y) =X - \psi(Y)$ does not satisfy \eqref{1} thus it is not a Cauchy function: therefore it must be $\psi^{"}(x)=0$ for all $x\in[a,b]$. $\blacksquare$ I said this is the core result of the paper since the main theorem, proved in the following section of the paper, follows from an application of the ideas developed in this lemma. §6. The main theorem (pp. 496-497). The characterization of Cauchy functions given by Faedo is expressed by the following Theorem. Let $\Phi(X,Y)\in C^2(\Bbb R^2\setminus E)$ where $E$ is defined as above. Then a necessary condition for $\Phi(X,Y)$ to be a Cauchy function is that each level curve defined by the equation $$ \Phi(X,Y)=c,\quad c=\text{const.}\label{3}\tag{3} $$ is a piecewise linear curve $\Gamma_c$ whose vertex are points $(X,Y)$ for which, simultaneously, $$ \frac{\partial\Phi}{\partial X}=0\;\wedge\;\frac{\partial\Phi}{\partial Y}=0. $$ Proof. Let $(X_0,Y_0)$ be a point in $\Bbb R^2\setminus E$ for which the partial derivatives $$ \dfrac{\partial\Phi}{\partial X}\bigg|_{(X_0,Y_0)}\text{ and }\dfrac{\partial\Phi}{\partial Y}\bigg|_{(X_0,Y_0)} $$ are not simultaneously equal to zero. For example and without restriction to generality, let $$ \left.\frac{\partial\Phi}{\partial X}\right|_{(X_0,Y_0)}\neq 0. $$ Since $\Phi(x_0,Y_0)=c$, equation \eqref{3} defines a function $X=\psi(Y)$ with $X_0=\psi(Y_0)$ and $$ \Phi(\psi(Y),Y)-c =0 $$ at least in a neighborhood of $Y_0$, thus can find a $\delta >0$ such that if $|Y-y_0| <\delta$ then also $|X_0-\psi(Y)|<\delta$. Moreover, since $\frac{\partial\Phi}{\partial X}\neq 0$, the function $X\mapsto\Phi(X,Y_0)$, considered in a neighborhood of $X=X_0$, takes the value $c$ only in $X=X_0$. From here on it is possible to proceed as in the proof of the lemma in §5: assume that $\psi^\prime(Y_0)>0$ and put $$ \overline X = \frac{f(b)-f(a)}{b-a},\; \overline Y= \frac{g(b)-g(a)}{b-a}. $$ We have that $$ \frac{g(b)-g(a)}{b-a} = Y_0 $$ and moreover $$ \frac{f(b)-f(a)}{b-a} = \frac{1}{b-a}\int\limits_a^b \psi(g^\prime(x))\Dm x. $$ with $|X_0-\psi(g^\prime(x))|<\delta$, and thus $$ \Bigg|\frac{f(b)-f(a)}{b-a} - X_0\Bigg|=\Bigg| \frac{1}{b-a}\int\limits_a^b [\psi(g^\prime(x))-X_0]\Dm x\Bigg|<\delta. $$ This implies $|\overline{X}-X_0|<\delta$ and $|\overline{Y}-Y_0|<\delta$ and due to the arbitrariness of $\delta$ we have $$ \Phi(\overline{X},\overline{Y}) \neq \Phi(X_0,Y_0) = c. $$ Equation \eqref{2} of §5 thus proves that $$ \Phi\left(\frac{f(b)-f(a)}{b-a}, \frac{g(b)-g(a)}{b-a}\right) \neq \Phi(f^\prime (x), g^\prime(x))=c\quad\forall x\in[a,b], $$ thus it must be $\psi^{"}(Y_0)$ for otherwise $\Phi(X,Y)$ is not a Cauchy function. Thus $\psi(Y)$ must be a linear function, and this implies that $$ \psi^\prime(Y) = \text{const.} = - \frac{{\partial \Phi}/{\partial Y}}{{\partial \Phi}/{\partial X}} $$ thus ${{\partial \Phi}/{\partial X}}\neq 0$ implies ${{\partial \Phi}/{\partial Y}}\neq 0$, therefore $X=\psi(Y)$ is defined for every $X$ except those for which ${\partial \Phi}/{\partial X} = {\partial \Phi}/{\partial Y} = 0$. $\blacksquare$ §7 and §8. Properties of the piecewise linear level curve $\Gamma_c$ and rational Cauchy functions. In §7 Faedo proves that on the points $(X_0,Y_0)$ where the gradient of the Cauchy function $\Phi$ does not vanish, two level curves $\Gamma_c$ and $\Gamma_{c_0}$ with $c\neq c_0$ do not intersect nor self-intersect. In the last paragraph, the Author shows that the only rational Cauchy functions are those belonging to the classes $\Phi_1$ defined in §2 and $\Phi_2$ defined in §3. Reference [1] Sandro Faedo, "Sul teorema di Cauchy degli incrementi finiti" [On Cauchy's theorem about finite increments] (Italian), Rendiconti di Matematica, VI Serie 10 (1977), 489-499 (1978), MR0480904, Zbl 0384.26002. REPLY [10 votes]: One may investigate such generalizations, but let's make a step backwards to your first question: Why are we considering a ratio? The reason is that those ratios are what we mainly care about, as we want to compare finite differences and differentials. And, as Dieudonné remarks in Foundations of Modern Analysuis, the point is not the existence of the point $x$, a result which is only true for scalar functions, and of which we usually can't say anything more than $x$ lies somewhere between $a$ and $b$. Rather, it is the inequality that follows from it, and which holds true even for vector valued curves: the "true" Mean Value Theorem is the inequality $$ \|f(b)-f(a)\|\le(b-a)\sup_{a\sup_{a TITLE: Reference on the Chern-Simons theory and WZW models for mathematicians QUESTION [11 upvotes]: I would like to ask if there are any beginner friendly references for learning CS theory and WZW models. It seems that most mathematical texts on the subjects begin with convenient definitions that are easy to start off with. But it seems to me that there are a lot of interesting topics that are very hard to find any reference on. When there is a reference, it is usually written by physicists for physicists and very hard to understand. Some of the things that I am confused with are: Quantization of WZW models. Why the space of states should be $$\bigoplus_{\lambda\ \text{integrable}} H_\lambda \otimes H_\lambda^*$$ seems only to be explained by Gawędzki in his lectures Wess-Zumino-Witten Conformal Field Theory, in the book Constructive Quantum Field theory, edited by Velo and Wightmann. Gawędzki constructs a line bundle on the loop group $LG$ to take care of the ambiguity in the action, and then identifies $e^{-iS}$ as a section of that line bundle. He then goes on to use Feynman's path integral. However, already, this seems to involve some gaps to me. State-Field Correspondence, Covariance of primary fields in WZW models. One can consider a 'pair of pants' to give a incomplete argument, but how could one show that the correspondence is actually one-to-one? The covariance rules that the primary fields should satisfy are also just given as definitions of primary fields in most mathematical texts. What are 'unitary' representations for Lie algebras? It seems that this depends on the type of Lie algebra that is being dealt with. How to show CS/WZW correspondence. There does not seem to be a mathematical reference for this as well. I would be very grateful if anyone could give me a list of references that could help me out on these, and maybe more. REPLY [3 votes]: A general reference for Chern-Simons theory written for mathematicians are the two papers by D. S. Freed: D. S. Freed: Classical Chern-Simons Theory, Part 1. Advances in Mathematics, 113(2):237–303, 1995. Preprint: arXiv:hep-th/9206021. D. S. Freed: Classical Chern-Simons theory, Part 2. Houston Journal of Mathematics, 28(2):293–310, 2002. see here. The first part covers connected and simply connected gauge groups and the second part covers arbitrary compact Lie groups. REPLY [2 votes]: I'm more familiar with the physics oriented literature, so Fuchs' book Affine Lie Algebras and Quantum Groups - An Introduction, with Applications in Conformal Field Theory might suffer from the same problems you refer to. Perhaps worth a look, even thought it does not cover Chern-Simons theory.<|endoftext|> TITLE: Is the comultiplication of a compact quantum group always injective? QUESTION [7 upvotes]: Let $(A, \Delta)$ be a compact quantum group in the sense of Woronowicz. Is it true that the comultiplication $\Delta : A \to A \otimes A$ always injective? This is true for both the universal (because one has a counit) and the reduced (because the Haar state is faithful) version, but the general case seems more delicate. In particular, is it true for the dual of a discrete group $\Gamma$ realized as a compact $C^*$-algebraic quantum group using a $C^*$-norm that lies strictly between the universal and the reduced $C^*$-norms? What about the case where $\Gamma$ is the free group on two generators? REPLY [6 votes]: No, the comultiplication need not be injective. When $\Gamma$ is a countable group and $\pi : \Gamma \to \mathcal{U}(H)$ is a faithful unitary representation with the property that $\pi \otimes \pi$ is weakly contained in $\pi$, we write $A = C^*_\pi(\Gamma)$ and there is a unique unital $*$-homomorphism $\Delta : A \to A \otimes A$ satisfying $\Delta(\pi(g)) = \pi(g) \otimes \pi(g)$ for all $g \in \Gamma$. Then, $(A,\Delta)$ is a compact quantum group in the sense of Woronowicz. Now $\Delta$ is faithful if and only if $\pi$ is weakly contained in $\pi \otimes \pi$. That need not be the case, as the following example with $\Gamma = \mathbb{F}_2$ shows. For every $0 < \rho < 1$, we consider the function $\varphi_\rho : \mathbb{F}_2 \to \mathbb{R}$ given by $\varphi_\rho(g) = \rho^{|g|}$, where we use the word length $|g|$. By [Haa, Lemma 1.2], $\varphi_\rho$ is a positive definite function and we define $\pi_\rho$ as the associated cyclic representation. Denote by $\lambda$ the regular representation. By [Haa, Theorem 3.1], we have that $\pi_\rho$ is weakly contained in $\lambda$ if and only if $\rho \leq 1/\sqrt{3}$. We now claim that with $\rho = 2/3$, the representation $\pi = \pi_\rho \oplus \lambda$ provides a counterexample for the faithfulness of $\Delta$. By construction, $\pi \otimes \pi$ is weakly equivalent with $(\pi_\rho \otimes \pi_\rho) \oplus \lambda$. Since $\rho^2 < 1/\sqrt{3}$, it follows from [Haa, Theorem 3.1] that $(\pi \otimes \pi) \sim \lambda \prec \pi$. But since $\rho > 1/\sqrt{3}$, we have $\pi_\rho \not\prec \lambda$, so that $\pi \not\prec \pi \otimes \pi$. [Haa] U. Haagerup, An example of a nonnuclear C$^*$-algebra, which has the metric approximation property. Invent. Math. 50 (1978/79), 279-293.<|endoftext|> TITLE: Branching rule of $S_n$ and Springer theory QUESTION [12 upvotes]: Let $u\in\mathrm{GL}_n$ be a unipotent element, let $\mathcal{B}_u$ be the variety of Borel subgroups containing $u$, and let $d=\dim \mathcal{B}_u$. Then Springer theory tells us that $H^{2d}(\mathcal{B}_u,\overline{\mathbb{Q}}_{\ell})$ is an irreducible representation of $S_n$ in a natural way, and that all irreducible representations of $S_n$ come this way. Now, combining with the branching rule of symmetric groups, viewing as representations of $S_{n-1}$ one has that \begin{equation} H^{2d}(\mathcal{B}_u,\overline{\mathbb{Q}}_{\ell})\cong\bigoplus_{u'}H^{2d'}(\mathcal{B}_{u'},\overline{\mathbb{Q}}_{\ell}), \end{equation} where $u'$ runs over some unipotent elements of $\mathrm{GL}_{n-1}$ whoese corresponding Young diagrams are obtained from that of $u$ by removing a corner box. Question. Is there a purely geometric proof of the above displayed formula? REPLY [4 votes]: This is a nice question. I have never seen this before. Let us write $$\mathcal B_u = \{ V_0 \subset V_1 \subset \cdots \subset V_{n-1} \subset V_n = \mathbb C^n : u V_i \subset V_i \}. $$ Let $ \lambda $ be the Jordan type of $ u $. Then we can partition $ B_u $ into locally closed subsets depending on the Jordan type of $ u \rvert_{V_{n-1}} $; such a Jordan type $ \mu$ must necessarily be obtained by removing one box from $ \lambda$. So we have $$ \mathcal B_u = \bigsqcup_\mu B_u^\mu \quad B_u^\mu = \{ V_\bullet : u\rvert_{V_{n-1}} \text{ has Jordan type $ \mu$ } \}.$$ Let $ G(n-1, n)_u $ denote the set of all $n-1$-dimensional $u$-invariant subspaces of $ \mathbb C^n $ and partition $ G(n-1,n)_u $ into locally closed subsets $ G(n-1,n)_u^\mu $ according to the Jordan type of the restriction of $ u $ to the subspace. We have $ \mathcal B_u \rightarrow G(n-1,n)_u $ and $ \mathcal B^\mu_u $ is the preimage of $ G(n-1,n)^\mu_u$. Note that the fibre of $ \mathcal B_u \rightarrow G(n-1,n)_u $ is $ \mathcal B_{u\rvert_{ V_{n-1}}}$. Now, here are two facts that I don't see right away, but which must be true: Each piece $ \mathcal B_u^\mu $ has the same dimension. $ G(n-1,n)^\mu_u $ is irreducible and simply-connected (or at least that the bundle $\mathcal B^\mu_u \rightarrow G(n-1,n)\mu_u$ is topologically trivial on components). Edit: I think these facts should be contained in the classic paper by Spaltenstein https://www.sciencedirect.com/science/article/pii/S138572587680008X Assuming these facts, we get $$H(\mathcal B_u) = \bigoplus_\mu H(\mathcal B_u^\mu) = \bigoplus_\mu H(\mathcal B_{u(\mu)})$$ where again the direct sum ranges over all partitions made by deleting one box from $\lambda$ and where $ u(\mu)$ denotes a unipotent element of Jordan type $ \mu$, and $ H(X) $ denotes top (co)homology. This gives the desired decomposition.<|endoftext|> TITLE: The group structure on $[X,S^n]$ induced by the framed bordism QUESTION [7 upvotes]: I'm concerned about the group structure on $[X,S^n]$, i.e. the set of homotopy classes of continuous maps from $X$ to $S^n$. On the one hand, $[X,Y]$ has a group structure that is natural with respect to $X$ if and only if $Y$ is an H-space. The naturality is in the sense that $f:X'\to X$ induces a homomorphism $f_{\star}:[X,Y]\to[X',Y]$. It's known that $S^n$ is an H-space only for $n=0,1,3,7$. Thus, for a general $X$, there's no natural group structure on $[X,S^n]$ when $n$ takes other values. On the other hand, when $X$ is a closed smooth $d$-manifold, the Pontryagin-Thom construction establishes a bijection between $[X,S^n]$ and $\Omega_{d-n}^{fr}(X)$, i.e. the (unstable) $(d-n)$th framed bordism set of $X$. When $d<2n-1$, two transverse $(d-n)$-submanifolds in $X$ have no intersection and the disjoint union induces a group structure on $\Omega_{d-n}^{fr}(X)$. Then we can make $[X,S^n]$ into a group via the Pontryagin-Thom bijection. Hence $[X,S^n]$ is a group when $X$ is a closed smooth manifold whose dimension $<2n-1$. I would like to know how "natural" or how "unnatural" this group structure is: (1) if $X'$ is a closed smooth manifold whose dimension $<2n-1$, does a smooth map $f:X'\to X$ induce a group homomorphism? (2) When $n=0,1,3,7$, does this group structure coincide with the H-space induced group structure? Edit 1 The question has been edited to correct a mistake pointed out by Gregory. Edit 2 As mentioned by Tyrone, $[X,S^n]$ is a cohomotopy group when $X$ is a complex of dimension $<2n-1$, with the multiplication induced by the folding map. It reduces to the H-space induced group if $n=0,1,3,7$. My Q2 essentially asks whether cohomotopy groups coincide with framed bordism groups (when the latter applies). REPLY [9 votes]: This is an answer to question (2). Let $n=0,1,3,7$ and $i=1,2$ and $d\leq 2n-2$. Let $e=(1,0,\ldots,0)\in S^n$ Let $f_i:X\rightarrow S^n$ be two maps representing framed submanifolds $(M_i,\nu_i)$. Let $T_i$ be tubular neighborhoods of $M_i$. By the assumptions on the dimensions, the maps can be chosen in such a way that $T_1\cap T_2=\emptyset$. The value $-e\in S^n$ is regular for both $f_1,f_2$ $f_i^{-1}(\{-e\})=M_i$ and the framing induced by the differential is $\nu_i$. $f_i(X\setminus T_i)=e$. Let $g:X\rightarrow S^n$ be the product map $g(x)=f_1(x)f_2(x)$. By the choices made above $-e$ is a regular value and $g^{-1}(\{-e\})=M_1\cup M_2$. The induced framing on $M_1\cup M_2$ is $\nu_1,\nu_2$ on the respective components. This shows that the group structures coincide.<|endoftext|> TITLE: For which planar topological spaces $Z$ does there exist a hyperbolic group $\Gamma$ with $\partial \Gamma \cong Z$? QUESTION [8 upvotes]: Recall a topological space is called planar if it can be embedded in $S^2$. I'm interested in understanding hyperbolic groups with planar boundaries. In [1], it is shown that if a one-ended hyperbolic group has 1-dimensional planar boundary, then this boundary is either a circle or a Sierpinski carpet. It is also well known that there exists hyperbolic groups with boundary homeomorphic to $S^2$, and moreover the only 0-dimensional such space is the Cantor set. My question is: What other planar spaces arise as boundaries of hyperbolic groups? Does a nice "list" of these spaces exist, or is this a hopeless problem? What if we restrict ourselves to just one-ended groups? Any references would be appreciated. Thanks! References [1] Kapovich, Michael, and Bruce Kleiner. "Hyperbolic groups with low-dimensional boundary." Annales scientifiques de l'Ecole normale supérieure. Vol. 33. No. 5. 2000. REPLY [8 votes]: There are many further examples with local cut points, which can be obtained by amalgams over $\mathbb{Z}$, as @YCor suggests in his comment. Perhaps the easiest example is obtained by gluing three one-holed tori along their boundary. The resulting group $\Gamma$ is hyperbolic, and its boundary cannot be a Sierpinski carpet or $S^2$, since it has a local cut point, coming from either of the limit points of the amalgamating copy of $\mathbb{Z}$). On the other hand, that local cut point has valence 3, so the boundary also cannot be a circle. An even richer family of examples can be constructed via the following construction. Take any simple closed curve $\gamma$ on the boundary of a handlebody $U$, and take the double $\Gamma=\pi_1(U)*_{\langle\gamma\rangle} \pi_1(U)$ . This will be a 3-manifold group, and hence its boundary will be planar. By the Convergence Group Theorem of Tukia--Casson--Jungreis--Gabai, the boundary of $\Gamma$ is only a circle if $\Gamma$ is a surface group, which in turn only occurs if $\gamma$ was a ``surface element" of $\pi_1(U)$. And the boundary won't be a Sierpinski carpet, because it has a local cut point. It's complicated to describe the $\partial\Gamma$ for this construction in general, but you can make a start by looking at the papers of Cashen--Macura and Cashen, who explain how to construct the decomposition space $D(\gamma)$. The cut-point structure of $D(\gamma)$ gives information about the cut-point structure of $\partial\Gamma$. As far as I know, the problem of characterising all planar compacta that arise as boundaries hasn't been worked out, but I suspect it is tractable, and would make a very nice thesis problem, say. One would also want to make use of Bowditch's theorem, which explains how local cut points in the boundary correspond to cyclic splittings, and the Strong Accessibiltiy Theorem of Louder--Touikan, which asserts (under mild hypotheses) that hierarchies of cyclic splittings terminate.<|endoftext|> TITLE: "Classification" of (orientable) 3-manifolds with genus-g-surface as their boundary QUESTION [5 upvotes]: This is in some sense a generalization of the question I asked some time ago. I am very sorry if this question is too basic for MathOverflow, but I just started learning about some more detailed things of the topology of 3-manifolds quite recently for some project and hence, I am still missing some concepts. My general question is the following: I am interested in the class of all compact, orientable and connected 3-dimensional topological manifolds $\mathcal{M}$ with boundary $\partial\mathcal{M}\cong_{\mathrm{homeo.}}\Sigma_{g}$ where $g\in\mathbb{N}_{0}$ and $\Sigma_{g}$ denotes the orientable genus-g surface. Is there a way to "divide" this class into different types? Obviously, I am not asking for a classification in the strict sense, but just a way to distinguish such manifolds by their construction. As an example, in Moishe Kohan's great answer to the question I linked above, he explained that all 3-manifolds (with the properties as above) with $\partial\mathcal{M}=T^{2}$, where $T^{2}=S^{1}\times S^{1}$ denotes the 2-torus, is of one (and only one) of the following types: $\mathcal{M}$ is homeomorphic to the solid torus $\overline{T}^{2}=D^{2}\times S^{1}$. $\mathcal{M}$ is homeomorphic to $\overline{T}^{2}\#\mathcal{N}$, where $\mathcal{N}$ is some closed, connected and orientable 3-manifold, which is not $S^{3}$ and where $\#$ denotes the (internal, oriented) connected sum. $\mathcal{M}$ has incompressible boundary. Is a similar trichotomy also true for the more general case with $\partial\mathcal{M}\cong_{\mathrm{homeo.}}\Sigma_{g}$? REPLY [5 votes]: You can find the discussion of the characteristic compression body in Section 3.3 of Bonahon’s survey. See Theorem 3.7 for the irreducible case, which together with the uniqueness of connect sum in Theorem 3.1 I think gives the sort of generalization of the trichotomy you are seeking. Either the manifold has incompressible boundary, or it has compressible boundary and it is a handlebody, or it has compressible boundary and is obtained from a compression body (with genus g compresible boundary) attached to some manifolds with incompressible boundary. I think this gives a reference for John Pardon’s comment.<|endoftext|> TITLE: Are free $E_\infty$-algebras in spaces closed under pullbacks? (as a full subcategory of all $E_\infty$-algebras in spaces) QUESTION [12 upvotes]: Let's call an $E_\infty$-algebra $A$ in spaces free if there is a space $A_0$ and an equivalence of $E_\infty$-algebras: $ \coprod_{n \ge 0} (A_0)^n_{h\Sigma_n} \simeq A. $ Consider a diagram of $E_\infty$-algebras in spaces $$ A \longrightarrow B \longleftarrow C $$ and assume that each of them is free in the above sense. However, the maps $A \to B \leftarrow C$ are not required to be free, i.e. they don't have to send $A_0$ to $B_0$ or $C_0$ to $B_0$. In this setting, is the homotopy pullback $A \times_B C$ always free? I've convinced myself in a very roundabout way that this should be true, but would be very grateful for a proper proof or even better a reference. I think a hands-on approach using the formula for the free $E_\infty$-algebra might work, but the combinatorics will get quite tricky. As an example, consider the case where $A, B, C$ are all free on a point and the maps $A \to B \leftarrow C$ both send the generator to twice the generator. Then I think the resulting pullback $A \times_B C$ is free on a space with infinitely many components. Let me give some more details on what happens in this example. I'll think of all the involved 1-types as groupoids, so $A, B, C$ are all represented by the groupoid of finite sets and bijections $\mathrm{FB}$, which is symmetric monoidal under disjoint union. In my example the two maps correspond to the functor $F: \mathrm{FB} \to \mathrm{FB}$ that sends $a \mapsto a \times \{0,1\}$. Now the homotopy pullback can be computed in terms of groupoids. The resulting groupoid $\mathcal{G}$ has as objects triples $(a, b, \varphi)$ where $a, b \in \mathrm{FB}$ are finite sets and $\varphi: a \times \{0,1\} \cong b \times \{0,1\}$ is a bijection. Morphisms are tuples of bijections compatible with $\varphi$. Define the set of connected components of such an object $(a, b, \varphi)$ as the pushout $\pi_0(a,b,\varphi) = a \amalg_{a \times \{0,1\}} b$ where the right-hand map is $\varphi$ composed with projection to $b$. This defines a symmetric monoidal functor $\pi_0: \mathcal{G} \to \mathrm{FB}$. Let's say an object is connected if $\pi_0(a,b,\varphi)$ is a point. Then I claim that $\mathcal{G}$ is freely generated under disjoint union by connected objects. This is probably easiest to see by thinking of the objects as directed graphs with red and blue edges, where the vertices are $a \times \{0,1\}$, the blue edges are $(a, 0) \to (a, 1)$, the red edges are $\varphi^{-1}(b,0) \to \varphi^{-1}(b,1)$. Only graphs where every vertex is incident to exactly one blue and one red edge are allowed. Now this category is a free symmetric monoidal groupoid on the connected graphs, of which there are countably infinitely many. REPLY [4 votes]: This is not an answer, but I think it's really interesting to look at an example where this statement fails for commutative monoids and the analogous statement for $E_\infty$-algebras works out. This is too long for a comment, so here it goes: By reverse engeneering the theorem about Krull monoids that @TimCampion mentions in his comment, we can find the following minimal example. Consider the pullback of free commutative monoids: $$ \mathbb{N}^2 \xrightarrow{\;\;+\;\;} \mathbb{N} \xleftarrow{\;\;\cdot2\;\;} \mathbb{N} $$ The pullback is $\{(x,y) \in \mathbb{N}^2 | x + y \equiv 0 \mod 2\}$. This can't be a free commutative monoid: any generating set must contain $\{(2,0), (1,1), (0,2)\}$ (and this is a generating set), so it would have to be free on at least three generators. On the other hand we know that it's group-completion is $\mathbb{Z}^2$, so if it was free it would have to have two generators. Why does this not give a counter-example to the $E_\infty$-algebra question? Let's model the free $E_\infty$-algebra on a point by $\mathrm{FB}$, the groupoid of finite sets and bijections, and accordingly the free $E_\infty$-algebra on two points by $\mathrm{FB}^2$, the groupoid of pairs of finite sets. Then the analogous diagram to the above would be: $$ \mathrm{FB}^2 \xrightarrow{\;\;\amalg\;\;} \mathrm{FB} \xleftarrow{\;\;\times \{1,2\}\;\;} \mathrm{FB} $$ An object in the homotopy pullback groupoid can be described as a triple $(X, Y, Z)$ of finite sets together with a bijection $\varphi:X \amalg Y \cong Z \times \{1,2\}$. We can forget about $\varphi$ and $Z$ and instead remember the directed pairing induced by pairing $\varphi^{-1}(z,1)$ with $\varphi^{-1}(z,2)$. So the pullback is the groupoid of tuples $(X, Y)$ together with a directed pairing on $X \amalg Y$. This is symmetric monoidal under disjoint union and as a symmetric monoidal groupoid it's generated by four objects: $(\{a,b\}, \emptyset)$, $(\{a\}, \{b\})$, $(\{b\}, \{a\})$ and $(\emptyset, \{a,b\})$ each with the directed pairing $a \to b$. So in this $E_\infty$-algebra setting the pullback is actually free on four generators.<|endoftext|> TITLE: Generating function for Schur polynomials QUESTION [11 upvotes]: Consider the generating function $$ G_n(x_1,x_2,\ldots,x_n, t_1,t_2,\ldots,t_n) =\sum_{\lambda}s_{\lambda}(x_1,x_2,\ldots, x_n) t_1^{\lambda_1}t_2^{\lambda_2} \cdots t_n^{\lambda_n}, $$ where the sum is over all partitions $\lambda=(\lambda_1, \lambda_2,\ldots, \lambda_n)$ and $s_\lambda$ is a Schur polynomial. For small $n$ such generating function is easy to find, for example for $n=2$ by direct calculation we have $$ G_2(x_1,x_2, t_1,t_2)=\sum_{\lambda}s_{\lambda}(x_1,x_2) t_1^{\lambda_1}t_2^{\lambda_2}=\frac{1}{(1-x_1 t_1)(1-x_2 t_1)(1-x_1 x_2 t_1 t_2)}. $$ If we put $t_1=t_2=1$ then we come to well-known Littlewood identity $$ \sum_{\lambda}s_{\lambda}(x_1,x_2)=\frac{1}{(1-x_1)(1-x_2)(1-x_1 x_2)}. $$ Question. Is there any close expression for the generating function $G_n$ for arbitrary $n?$ REPLY [14 votes]: This is done in my paper The character generator of SU(n). I believe there was an essentially the same previous MO question, but I am unable to find it.<|endoftext|> TITLE: What shapes can be gears? QUESTION [39 upvotes]: I am interested in gears. Sadly most of the writing on this is very practical and does not get into abstract theory. I have been trying to formalize these ideas to be able to ask what shapes can function as gears and at what ratios. I have had two attempts to formalize this question. For my first formalization we can think of the functions $f:[0,2\pi) \to (0,\infty)$ as defining the gears outlined by the polar expression $r=f(\theta)$. My question becomes for what values $a\in [0,\infty)$ is there a function $g:[0,2\pi) \to (0,\infty)$ that generates a gear that meshes with our first gear and rotates $a$ times for every time the first gear rotates. I included $0$ as a possible value of $a$ to represent linear gears. This formalization falls short because it only looks at particular shapes. In this formalization it is also difficult to define what it means for two gears to mesh. For my second and preferred formalization we think of a gear as an ordered pair $G = (X,c)$ where $X$ is a compact subset of $\mathbb{R}^2$ and $c \in \mathbb{R}^2$. This is meant to represent the gear made of the points in $X$ that rotates around the point $c$. The set $X$ is not necessarily connected and $c$ is not necessarily in $X$ because of objects like lantern gears. There are a lot of ways to formalize a rotation. I chose to use elements of $S^1$ and apply them to the gear object. Now given the gears $G_1$ and $G_2$ their meshing space is $\{(a,b) \in S^1 \times S^1\mathrel\vert aG_1\cap bG_2 = \emptyset\}$. It is interesting to note that $S^1 \times S^1$ is the torus. So the question from before becomes whether or not there are continuous functions from $S^1$ to this subset of the torus, and if so how many and members and of what homotopy classes. Edit: There are a lot of questions that could be asked about the meshing space. One could be dimensionality. If the space is made of one-dimensional curves that means the gears line up perfectly. If the space is for example the whole torus that means the gears do not touch each other at all. Also upon further playing around, I think it makes more sense to have $X$ be open and bounded. If $X$ is closed then by the intersection condition the gears could not touch each other. This also removes situations with single disconnected points. I was wondering if anyone had an answer to my question, under either formalization, or had a smarter way to formalize this question. REPLY [6 votes]: if you look for non-circular gear you at least find the wiki article that also addresses the mathematics; there are also links to publication on the subject.<|endoftext|> TITLE: Does this group construction preserve finite presentability? QUESTION [13 upvotes]: Suppose $G$ is a group. Consider the set $G^G$ of all functions $G \to G$, which forms a group under elementwise multiplication. Now, for all $g \in G$ let’s define $c_g \in G^G$ as the constant function $c_g(x) \equiv g$, and $id \in G^G$, as the identity map $id(x) = x$. Now, consider the subgroup $E(G) = \langle \{c_g | g \in G\} \cup \{id\} \rangle$. $E(G)$ preserves several “finiteness” properties of $G$: If $G$ is finite then $E(G)$ is also finite. Proof: $|E(G)| \leq |G^G| = |G|^{|G|}$ If $G$ is finitely generated then $E(G)$ is also finitely generated. Proof: If $A$ is a generating set of $G$, then $\{c_g | g \in A\} \cup \{id\}$ is a generating set of $E(G)$. If $G$ is residually finite then $E(G)$ is also residually finite. Proof: Consider the following class of maps $\pi_g: E(G) \to G, f \mapsto f(g)$ for all $g \in G$. All $\pi_g$ are homomorphisms and each non-trivial element of $E(G)$, maps to a non-zero element of $G$ under some of $\pi_g$. The rest follows from residual finiteness of $G$. However, there is also a fourth “finiteness” property I am interested in but do not know how to deal with: If $G$ is finitely presented, does that mean that $E(G)$ is also finitely presented? I suspect, it should be, but have no idea how to prove it. This question on MSE REPLY [13 votes]: It seems the answer is no, $E(G)$ can fail to be finitely presented even if $G$ is finitely presented. I claim that a counterexample is given by $G=F_2\times F_2$. First, as @SeanEberhard explains in the comments, $E(F_2\times F_2)$ is isomorphic to the subgroup $H$ of $F_3\times F_3$ generated by $\{(x,1),(y,1),(1,x),(1,y),(z,z)\}$, where $\{x,y,z\}$ is a generating set for $F_3$. Now I claim that $H$ is not finitely presented. First I claim that $H$ contains the commutator subgroup of $F_3\times F_3$. Any simple commutator in $F_3\times F_3$ is an ordered pair of simple commutators in $F_3$, so it suffices to show that any $(g,1)$ for $g$ a simple commutator in $F_3$ lies in $H$ (and by a parallel argument also $(1,g)$). This is easy to see because given any word in $(x,1),(y,1),(z,1)$ (and their inverses) representing an element of $[F_3,F_3]\times\{1\}$, we can replace each instance of $(z,1)$ with $(z,z)$ (and $(z^{-1},1)$ with $(z^{-1},z^{-1})$) and get a word representing the same element of $[F_3,F_3]\times\{1\}$ but now lying in $H$. Now that we know $H$ contains the commutator subgroup, we can use the Bieri-Neumann-Strebel-Renz invariant $\Sigma^2(F_3\times F_3)$ to conclude that $H$ is not finitely presented. Indeed, $H$ lies in the kernel of the homomorphism $\chi\colon F_3\times F_3 \to \mathbb{Z}$ sending $(z,1)$ to $1$, $(1,z)$ to $-1$, and all of $(x,1),(y,1),(1,x),(1,y)$ to $0$ (I guess it actually equals this kernel), and the computation of $\Sigma^2(F_3\times F_3)$ (see, e.g., Bux-Gonzalez) reveals that $[\chi]$ is not in $\Sigma^2(F_3\times F_3)$, since the "dead" edge from $(x,1)$ to $(1,x)$ in the defining flag complex has empty "living link". (Actually, I think $\Sigma^2(F_3\times F_3)$ is empty, so probably we didn't even need to isolate a particular $\chi$, and in fact every infinite index subgroup of $F_3\times F_3$ containing the commutator subgroup fails to be finitely presented, but I'll just leave this analysis as-is.) Edit: By request and for the sake of being self-contained, here is the proof that $E(F_2\times F_2)\cong H$, compiled from @SeanEberhard's comments. Here $\{x,y,z\}$ is a free basis of $F_3$ and $H$ is the subgroup of $F_3\times F_3$ generated by $\{(x,1),(y,1),(1,x),(1,y),(z,z)\}$. Let $\{a,b\}$ be a free basis of $F_2$. We have an epimorphism $\phi\colon F_3\to E(F_2)$ given by sending $x$ to $c_a$, $y$ to $c_b$, and $z$ to $id_{F_2}$, and we claim it is injective (hence an isomorphism). Indeed, given any non-empty reduced word in $c_a$, $c_b$, and $id_{F_2}$ (and their inverses) representing an element $f$ of $E(F_2)$, since $F_2$ is mixed identity-free, we can evaluate $f$ at some element of $F_2$ to get a non-trivial element of $F_2$, which means $f$ is a non-trivial element of $E(F_2)$. Hence $F_3 \cong E(F_2)$. This also shows $F_3\times F_3 \cong E(F_2)\times E(F_2)$. Let us identify $H$ with its isomorphic image in $E(F_2)\times E(F_2)$, generated by $\{(c_a,c_1),(c_b,c_1),(c_1,c_a),(c_1,c_b),(id_{F_2},id_{F_2})\}$. Now consider the monomorphism $\psi \colon F_2^{F_2} \times F_2^{F_2} \to (F_2\times F_2)^{F_2\times F_2}$ given by sending $(f,g)$ to the function $f\times g$ defined by $(f\times g)(w,v):=(f(w),g(v))$. The restriction of $\psi$ to the subgroup $H$ is an isomorphism onto its image, which is generated by $c_{(a,1)}$, $c_{(b,1)}$, $c_{(1,a)}$, $c_{(1,b)}$, and $id_{F_2\times F_2}$, hence is $E(F_2\times F_2)$ as desired.<|endoftext|> TITLE: Is the strict topology on the multiplier algebra of a $C^*$-algebra always finer than the ultrastrong-$*$ topology? QUESTION [5 upvotes]: Let me be precise about what I mean in the title. Let $A$ be a $C^*$-algebra, which we identify with its image of its universal representation $(\pi, H)$, so the second dual of $A$ is canonically identified with the von Neumann algebra $A''$ generated by $\pi(A) = A$. Denote $\widetilde{A}$ the minimal unitalization of $A$ in $A''$, and $M(A)$ the maximal one, i.e. $M(A) = \{x \in A'' \mid xA \cup Ax \subset A\}$, which is of course a copy of the multiplier algebra of $A$. It is known (see e.g. Pedersen's book, Theorem 3.12.9) that $M(A)_{\mathrm{sa}} = (\widetilde{A}_{\mathrm{sa}})_m \cap (\widetilde{A}_{\mathrm{sa}})^m$, where $(\widetilde{A}_{\mathrm{sa}})_m$ denotes the set of strong limits in $A''$ of bounded decreasing nets in $\widetilde{A}_{\mathrm{sa}}$, and $(\widetilde{A}_{\mathrm{sa}})^m$ the set of strong limits of bounded increasing nets in $\widetilde{A}_{\mathrm{sa}}$. Of course $M(A)_{\mathrm{sa}} \subset A''_{\mathrm{sa}}$ and the inclusion is strict in general. There are a variety of topologies on $M(A)$. Besides the ones that inherited from $B(H)$ as a subspace, one also often considers the strict topology on $M(A)$, which is the locally convex topology defined by the family of semi-norms $\|(\cdot)a\|$ and $\|a(\cdot)\|$ with $a$ running through $A$. It is also known that $M(A)$ is in fact the strict completion of $A$ as a locally convex space. Now von Neumann's double commutation theorem says that the ultra-strong-$*$ closure of $A$ in $B(H)$ is $A''$. On $M(A)$, denote the ultra-strong-$*$ topology that is inherited from $H$ by $\sigma^*$, and the strict topology by $\beta$. If $\sigma^* = \beta$, then $M(A)$ is ultra-strongly-$*$ complete since it is strictly complete, in particular $M(A)$ is ultra-strongly-$*$ closed in $B(H)$, forcing $M(A) = A''$, but we know this is not always the case, although we do have $M(A) \subset A''$ by the above discussion, i.e. the $\beta$-closure of $A$ is always contained in the ultra-strong-$*$ closure of $A$. Since finer topology gives smaller closure, my question is, is it true that we always have $\beta$ finer than $\sigma^*$? Here's a special case to get started. Note that if $A = K(H)$, then $A''=B(H)$. In this case, it is fairly easy to see that the strict topology on $M(A)$ is finer than the strong-$*$ topology on $M(A)$. Indeed, take any rank-one projection $p_\xi$ onto $\mathbb{C}\xi$ with $\xi$ being an arbitrary unit vector of $H$, one has, for any $x \in B(H) = M(K(H)) = K(H)''$, that $\|xp_\xi\| = \|xp_\xi x^*\|^{1/2} = \|x\xi\|$ and $\|x^*p_\xi\| = \|x^* p_\xi x\|^{1/2} = \|x^*\xi\|$. As $p_\xi \in K(H)$, and $\|(\cdot)\xi\|$ together with $\|(\cdot)^*\xi\|$ form a generating family of semi-norms of the strong-$*$ topology, we've shown that indeed the strict topology on $M(K(H)) = B(H)$ is finer than the strong-$*$ one. It is known that the strict topology, the strong-$*$ topology, the ultra-strong-$*$ topology and the Arens-Mackey topology on $B(H)$ all agree on bounded parts. So it is reasonable to compare the strict topology to other topologies that are finer than the strong-$*$ one but still agree with it on bounded parts. Two such topologies are already mentioned, the ultra-strong-$*$ topology and the Arens-Mackey topology. We know the dual of $B(H)$ equipped with the strict topology is exactly the predual $B(H)_*$ of $B(H)$, so the Arens-Mackey topology is finer than the strict one. The question remains about the comparison between the ultra-strong-$*$ topology and the strict topology, even in this special case where we do have $M(A)=A''$. REPLY [6 votes]: The answer is "yes". Use the standard technique: if necessary, replace $H$ by $H\otimes\ell_2$ to ensure that for each $\omega\in B(H)_*$ (the predual of $B(H)$, the trace-class operators on $H$) there is $\xi\in H$ so that $\omega(y) = (y(\xi)|\xi)$ for each $y\in A''$. As $A$ acts non-degenerately on $H$ (because $A''$ must be unital) and as replacing $H$ by $H\otimes\ell_2$ does not change this, we may apply the Cohen–Hewitt factorization theorem. This allows us to find $a\in A, \xi'\in H$ with $a(\xi') = \xi$. Now let $(x_i)$ be a net in $M(A)$ which converges strictly (that is, for the $\beta$ topology) to $x\in M(A)\subseteq A''$. This means that $x_i a \rightarrow xa$ and $a^*x_i\rightarrow a^*x$ in norm, so also $x_i^*a\rightarrow x^*a$, in norm. In particular, $$ \lim_i \omega((x_i-x)^*(x_i-x)) = \lim_i \|(x_i-x)\xi\|^2 = \lim_i \|(x_i-x)a(\xi')\|^2 = 0, $$ and similarly $\omega((x_i-x)(x_i-x)^*) \rightarrow 0$. As $\omega$ was arbitrary, this shows that $x_i\rightarrow x$ in the $\sigma$-strong$^*$-topology, that is, for the $\sigma^*$ topology. In fact, we can avoid the "standard technique". Restriction of functionals defines a quotient map $B(H)_* \rightarrow M_*$ from the trace-class operators to the (unique) predual of our von Neumann algebra $M$. Then the ultra-strong$^*$-topology is given by the seminorms $$ M\ni x \mapsto ( \omega(x^*x) + \omega(xx^*) )^{1/2} $$ as $\omega$ varies through the positive part of $M_*$. In this way, we see that there is no dependence on $H$. (I guess there is an extra argument needed here: that a positive member of $M_*$ lifts to a positive trace-class operator). In our case, $M=A''\cong A^{**}$ the bidual, as the representation of $A$ is universal. So $M_* \cong A^*$ the dual of $A$. For $\mu\in A^*$ and $a\in A$ define $a\cdot\mu\in A^*$ to be the functional $A\ni b\mapsto \mu(ba)$. There are various ways to show that the linear span of $\{ a\cdot\mu : a\in A, \mu\in A^* \}$ is norm-dense in $A^*$. (For example, polar decomposition of functionals and the GNS construction.) We can hence directly apply Cohen--Hewitt to $A^*$ to show that for each $\mu\in A^*$ there is $a\in A, \mu'\in A^*$ with $\mu = a\cdot\mu'$. Similarly we can regard $A^*$ as a right $A$-module, and then the same argument shows that we can find $a'\in A,\mu''\in A^*$ with $\mu' = \mu''\cdot a'$, so that $\mu = a\cdot\mu''\cdot a'$ (this being a bimodule, so the order of left/right actions doesn't matter). Now the same argument works: if $(x_i)$ in $M(A)\subseteq A^{**}$ converges strictly to $x\in M(A)$ then $\|(x_i-x)a\| \rightarrow 0$ and $\|a'(x_i-x)^*\| = \|(x_i-x)a'^*\| \rightarrow 0$, and so $$ \mu((x_i-x)^*(x_i-x)) = \mu''(a'(x_i-x)^*(x_i-x)a) \rightarrow 0. $$ Thus we have converted the argument to not depend on $H$.<|endoftext|> TITLE: Extracting constant terms: is there a direct way? QUESTION [8 upvotes]: $\DeclareMathOperator\CT{CT}$ Let $\CT_t(f(t))$ denote the constant term of the Laurent polynomial of $f(t)$. Define the two functions $F(x_1,\dots,x_n)$ and $G(y)$ by $$F:=\prod_{i=1}^nx_i^{-1}(1-x_i)^{-2}\prod_{1\leq i2$ then consider the $(k-2)$nd factor of $\prod_{1\leq i TITLE: Open covering of $S^n$ by sets not containing antipodal points QUESTION [7 upvotes]: Given an $n$-dimensional sphere $S^n$ and an open cover such that none of the open sets contain antipodal points, does there exist a point on $S^n$ that belongs to at least $n+1$ open sets from the open cover? This is true for $S^1$, and it seems to be true for $S^2$. I am not sure if the Lusternik–Schnirelmann theorem would be useful here. I do not know too much about this field, so perhaps this problem is trivial. Any help would be appreciated! REPLY [5 votes]: This is not true for $\mathbb{S}^3$. Start with the central projection of 4-cube to $\mathbb{S}^3$. Consider a covering with one set $U$ that contains all vertices and edges + one open set that is very close to the remaining 2- and 3-cells. The multiplicity of this covering can be made to be 3. All open sets except $U$ do not contain opposite points. After a small perturbation, opposite points disappear in $U$. Since the perturbation is small, the property remains valid for the remaining open set.<|endoftext|> TITLE: Existence of a Gelfand triple involving the Arens–Eells space (aka Lipschitz free space) QUESTION [5 upvotes]: $\DeclareMathOperator\Lip{Lip}\DeclareMathOperator\AE{AE}$Background Gelfand triples. Let $\mathcal B$ be a Banach space, $\mathcal B^*$ its dual space, and $\mathcal H$ a Hilbert space. The triple $(\mathcal B,\mathcal H, \mathcal B^*)$ is called a Gelfand triple if the following embeddings are continuous $$ \mathcal B \hookrightarrow \mathcal H \hookrightarrow \mathcal B^*. $$ An example that I am familiar with is the triple $(BV(\Omega), L^2(\Omega), BV^*(\Omega))$, where $\Omega \subset \mathbb R^2$ is bounded and $BV(\Omega)$ is the space of functions of bounded variation. Arens-Eells space. Let $X$ be a compact pointed metric space with base point $e$. An elementary molecule is defined as follows (Nik Weaver, Lipschitz Algebras, 2nd ed.) $$ m_{pq} := \delta_p - \delta_q, $$ where $\delta_p, \delta_q$ are delta-functions placed at $p,q$. The Arens-Eells space $\AE(X)$ (also known as the Lipschitz-free space) is the completion of the linear span of elementary molecules with respect to the Arens-Eells norm $$ \|{m}\|_{\AE} := \inf \left\{\sum_{i=1}^n |{a_i}| d(p_i,q_i) \colon m = \sum_{i=1}^n a_i m_{p_iq_i} \right\}, $$ where $d(p,q)$ is the distance between $p,q \in X$. The dual of the Arens-Eells space is the $\Lip_0(X)$ space of all Lipschitz functions on $X$ vanishing at $e$, equipped with the following norm $$ \|f\|_{\Lip_0} := \Lip(f), $$ where $\Lip(f)$ denotes the Lipschitz constant. Question Is there a Hilbert space $\mathcal H$ such that $(\AE(X), \mathcal H, \Lip_0(X))$ form a Gelfand triple? It would suffice for me to think of $X$ as the unit ball in $\mathbb R^n$ with base point $0$, equipped with the euclidean metric. REPLY [6 votes]: Your question is equivalent to asking if there is an injective bounded linear operator from $AE(X)$ into a Hilbert space when $X$ is a compact metric space. The answer is "yes" because $AE(X)$ is separable. It is elementary to construct a nuclear injective linear operator from an arbitrary separable Banach space into a Hilbert space.<|endoftext|> TITLE: Internal language proof of Lawvere's fixed point theorem for cartesian closed categories QUESTION [5 upvotes]: This proof of Lawvere's fixed point theorem suggests (since it uses $\lambda$ notation) that it is written in the internal language of cartesian closed categories (which is the $\lambda$-calculus, as explained e.g. in Part I of Lambek and Scott's book Introduction to higher order categorical logic). However, one needs quantifiers in order to formulate the notion of point-surjective and the existence of a fixed point $s\colon 1\to B$. So strictly speaking, the proof can't take place in the internal language, since the $\lambda$-calculus doesn't have quantifiers and assumptions. (Higher-order logic admits quantifiers, but that is only available in elementary toposes and not in cartesian closed categories in general.) Question: Is there some way of making the proof formally work in some "internal language" of cartesian closed categories? Or is the $\lambda$ notation used in the proof just an informal explanation of the proof rather than an indication for the use of the internal language? REPLY [6 votes]: You're right that the statement of the theorem, and the entirety of the proof, don't fit inside the internal logic of a CCC. However, once given $f:B\to B$, the definition of $q$ and the proof that it is a fixed point of $f$ can take place inside that internal language.<|endoftext|> TITLE: Why aren‘t op and co switched? QUESTION [8 upvotes]: When reading through Loregian and Riehl - Categorical notions of fibration, on p. 3 there is a remark that confuses me about notation. Given a $2$-category $\mathcal C$ one usually defines $\mathcal C^\text{op}$ to be the category $\mathcal C$ but with inverted $1$-cells and $\mathcal C^\text{co}$ to be the category $\mathcal C$ but with inverted $2$-cells. Many constructions like for example limits and colimits seem to be named according to the co-notation but are mathematically op-dual. Similarly in the example above. Hence the question is, whether there are any interesting constructions that are named “correctly” and also whether I am even correct about this feeling about things being switched up here. REPLY [16 votes]: The problem is that for a long time there were only 1-categories and hence only one kind of duality, and sometimes people called it "op" and sometimes "co". Colimits and cofibrations were therefore the only possible notion of dual for limits and fibrations, and take place in the opposite category, which was also the only notion of dual for a category, and denoted "op" as being short for "opposite". Sometimes both "op" and "co" terminologies are in use for the same thing, e.g. "oplax" and "colax". One should not attempt to remedy this by departing from the standard definitions of $C^{\rm op}$ and $C^{\rm co}$ for 2-categories. Among other reasons, they align with standard notions of duality for enriched categories: for a general $V$ there is only one notion of opposite $V$-category, denoted $C^{\rm op}$, and when a 2-category is regarded as a $\rm Cat$-enriched category this produces the 1-cell dual. I think the most satisfying perspective on this (though not fully satisfying) is that the 2-cell dual of a 2-category, $C^{\rm co}$, corresponds to classical dual constructions (often denoted by "co") on its objects, regarded as generalized categories. For instance, while a comonad on a category $A$ is equivalently a monad on the category $A^{\rm op}$, a comonad in a 2-category $C$ (on an object $A\in C$) is equivalently a monad in $C^{\rm co}$. (A monad in $C^{\rm op}$ is just the same as a monad in $C$.)<|endoftext|> TITLE: Rational contraction and Proj of section ring QUESTION [5 upvotes]: I am reading the paper "Mori Dream Spaces and GIT" by Hu and Keel. https://arxiv.org/abs/math/0004017 I cannot understand the proof of Lemma 1.6 in it. Let $X$ be a normal projective variety. Assume that $D$ is a divisor on $X$ such that $R(X,D) = \bigoplus_{m \in \mathbb{Z}_{\ge0} } H^0(X,\mathcal{O}_X(mD))$ is finitely generated. Lemma 1.6 states that the natural rational map $\varphi_D \colon X \dashrightarrow Y := Proj(R(X,D))$ is a rational contraction, i.e, there exits a resolution $\Phi \colon \tilde{X} \to Y$ of $\varphi_D$ such that $\Phi_*(\mathcal{O}_{\tilde{X}} (E)) \simeq \mathcal{O}_Y$ for any effective exceptional divisor $E$. Moreover, $D$ can be written as $ D = \varphi_D^*(A) + F $ for an ample divisor $A$ on $Y$ and a $\varphi_D$-fixed divisor $F$. Outline of the proof of the paper is as follows: (1) The finite generation of $R(X,D)$ implies that $D$ can be written as $D = M+F$ for a movable divisor $M$ and an effective divisor $F$ such that $Sym_k(H^0(X,M)) \to H^0(X,kM)$ is surjective and $H^0(X,kM) \to H^0(X,kM +rF)$ is an isomorphism for any $k>0$ and $r>0$ after replacing $D$ by its multiple. (2) After taking appropriate resolution, we may assume that $\varphi_D$ is a regular and $M = \varphi_D^* A $. Now we can see $F$ is $\varphi_D$-fixed by the isoromorphism $H^0(X,kM) \simeq H^0(X,kM +rF)$. Here is my question : [1] How can wee see that $\varphi_D$ is a rational contraction? Is it a well-known result? [2] Why does the isomorphism $H^0(X,kM) \simeq H^0(X,kM +rF)$ implies that $F$ is $\varphi_D$-fixed? Can anyone who understands these results answer my questions? REPLY [4 votes]: Let's assume that $|kD|=|kM|+kF$ where $|M|$ is base point free and moreover $Sym ^kH^0(M)\to H^0(kM)$ is surjective for any $k>0$. This can be achieved replacing $X$ by an appropriate resolution and $k$ by a multiple. Let $f:X\to Y$ be the morphism induced by $|M|$ so that $f^*A=M$. For [1], we wish to show that $f_* \mathcal O _X=\mathcal O _Y$. The Stein factorization $X\to Z\to Y$ is defined by ${\rm Spec} f_* \mathcal O _X$. We have $H^0(X,kM)=H^0(f_*(kM))=H^0(f_* \mathcal O _X\otimes \mathcal O _Y(kA))$ and so $|kM|$ in fact defines the morphism $X\to Y$. Since $Sym ^kH^0(M)\to H^0(kM)$ is surjective, then $|M|$ defines the same morphism. For [2], Since $H^0(X,kM)=H^0(X,kM+rF)$, then $H^0(\mathcal O _Y(kA))\cong H^0(\mathcal O _Y(kA)\otimes f_* \mathcal O _X(rF))$. For $k\gg r$, the sheaf $\mathcal O _Y(kA)\otimes f_* \mathcal O _X(rF)$ is globally generated and so $f_* \mathcal O _X(rF)\cong \mathcal O _Y$. (See also Lemma 3.2 https://arxiv.org/pdf/1104.4981.pdf). Hope this helps.<|endoftext|> TITLE: Wild spheres in higher dimensions QUESTION [8 upvotes]: I was wondering whether anyone had any intuition (or references) as to particular reasons why there are no wild spheres of codimension 1 (where by wild we mean not flat) in higher dimensions $n>3$ that are wild at one point (i.e. locally flat everywhere but a point). Of course there are the famous examples by Fox and Artin in $n=3$ that have a finite number of wild points and the papers by Cantarelli and Edwards that show that wild simple closed curves in dimensions greater than $3$ contain a Cantor set of wild points, but I am interested if anyone has a reasonable idea as to why $\dim=3$ is so special. Also I do not know much about wild spheres in codimension other than 1 so I would be interested to find out whether an analogous statement for spheres of arbitrary codimension holds (i.e. if $\Sigma \subset S^n$ is a sphere of codimension $n-k$ that is wild in one point then does it hold that for $n>3$ $\Sigma$ is flat?) The case of a finite number of wild points is also interesting. REPLY [8 votes]: This is an attempt to answer the poster's questions. The initial question is about a result obtained by Cantrell in his doctoral dissertation. Here I quote the version presented in Daverman-Venema's book. The heart of the proof is "Mazur swindle" which Mazur used to prove the generalized Schonflies theorem. The exposition of the proof in Daverman-Venema's book is well written. Just as the poster observed the distinction between the cases $n=3$ and $n>3$ is interesting. Contrast with $n=3$ the theorem above is not true. There are well-known examples such as the classical Fox-Artin examples Ann. of Math. (2) 49 (1948), 979–990. MR0027512. An alternative resource is Daverman-Venema's book. Since flatness and wildness are local properties, one may tempt to readily extend Theorem 2.9.3 to imply that codimension one manifolds can have no isolated wild point. However, the extension is less trivial. Because Cantrell's argument depends strongly on the flatness of $\Sigma -p$. Using engulfing techniques ($n\geq 5$), Cernavskii initially obtained this extension, and later Kirby and Cernavskii both obtained an alternative approach, which works in all dimensions. Theorem. If $B_1$ and $B_2$ are flat $(n-1)$-cells in $\mathbb{R}^n$ such that $B_1\cap B_2 = \partial B_1 \cap \partial B_2$ is an $(n-2)$-cell that is flat in both boundaries, then $B_1 \cup B_2$ is flat. For Cernavskii's proof, see On singular points of topological imbeddings of manifolds and the union of locally flat cells. (Russian) Dokl. Akad. Nauk SSSR 167 1966 528–530. MR0198447 For Kirby's proof, see The union of flat (n−1)-balls is flat in $R^n$. Bull. Amer. Math. Soc. 74 (1968), 614–617. MR0225328 For more general version of the theorem, see Lemma 7.6.8 and Corollary 7.6.10 on P. 380 in Daverman-Venema's book. The following two corollaries due to Cernavskii and Kirby, which (almost) immediately follows, are the extensions of Theorem 2.9.3. Corollary. If $\Sigma$ is an $(n-1)$-sphere in $\mathbb{R}^n$ $(n\geq 4)$ and $W^\ast$ is the set of points of which $\Sigma$ is wildly embedded, then $W^\ast$ contains no isolated point. Corollary. If $\Sigma$ is an $(n-1)$-manifold in the interior of an $n$-manifold $(n\geq 4)$ and $W^\ast$ is the set of points at which $\Sigma$ fails to be locally flat, then $W^\ast$ is empty or uncountable. The reason I said "almost" is because to prove the first corollary above, Kirby utilized a result of Lacher (Locally flat strings and half-strings. Proc. Amer. Math. Soc. 18 (1967), 299–304. MR0212805), which is essentially Cantrell's theorem. As Danny Ruberman pointed out, one may consult Kirby's another paper On the set of non-locally flat points of a submanifold of codimension one. Ann. of Math. (2) 88 (1968), 281–290, where he sharpened the description of the wild set. Theorem. Suppose $\Sigma$ is an $(n-1)$-manifold in an $n$-manifold $N$ $(n\geq 4)$ such that $\Sigma$ is locally flat modulo a Cantor set $X \subset \Sigma$, where $X$ is tame both as a subset of $\Sigma$ and as a subset of $N$ (a.k.a. twice-flat). Then $\Sigma$ is locally flat. Again, one may check Daverman-Venema's book (Cor. 7.9.5, P. 404). Now, why is 3-dimensional case so special? One may consider a special case presented in Daverman-Venema's book (P. 92). The proof relies on the fact (also in Daverman-Venema's book) which may address the poster's question is For broader interests of this type of the question, as I made in the comments, one may consider engulfing techniques. Heuristically one can expand an open subset of a manifold to engulf a subpolyhedron $P$, provided certain dimension, connectivity and finiteness conditions are satisfied. Among those the codimension $P$ is at least 3. The full power of engulfing techniques require the limitation $n\geq 5$ because then there is room to control the top dimensional part of the space. Look ahead to Chapter 3 in Daverman-Venema's book we may find a lot of nice results obtained using engulfing and see why the dimension restriction matters. For the poster's second question, the answer would be negative. The counterexample can be found in Daverman-Venema's book. For codimension-two embeddings, an appropriate notion for the wildness is called "knotting". One can define that both globally and locally. The entire Chapter 6 of Daverman-Venema's book is dedicated to unknotting and flatterning.<|endoftext|> TITLE: Lines on quadric surfaces QUESTION [5 upvotes]: Consider a smooth quadric surface $Q\subset\mathbb{P}^3$ over a field $k$. Are there natural hypotheses one can put on $k$ in order to ensure the existence of a line defined over $k$ on $Q$? REPLY [5 votes]: Yes, it suffices that $k$ have no nontrivial quadratic extensions. Since the surface is geometrically $\mathbb P^1 \times \mathbb P^1$, the space of lines is geometrically a union of two copies of $\mathbb P^1$. Arithmetically, the components are defined over a quadratic extension of $k$. If there are no nontrivial quadratic extensions then they are defined over $k$. Each component is then a form of $\mathbb P^1$. Since the anticanonical bundle has degree $2$, a section of the anticanonical bundle has zero locus a scheme of length two. Because $k$ has no nontrivial extensions of degree $2$, the zero locus consists of either one or two $k$-points, so the space parameterizing lines has a $k$-point and thus there is a rational line. At least in characteristic not $2$, this condition is necessary, as if $k$ has a quadratic extension then it contains a nonsquare $a$ and the form $x^2 -y^2 + z^2 - a w^2$. Any line on this surface must, over $k(\sqrt{a})$, be a line on $(x+y)(x-y) + (z- \sqrt{a}w)( z+\sqrt{a}w)$, thus of the form $\alpha (x+y) + \beta (z-\sqrt{a}w) = \alpha (x-y) - \beta (z+\sqrt{a}w)=0$ or of the form $\alpha (x+y) + \beta (z+\sqrt{a}w) = \alpha (x-y) - \beta (z-\sqrt{a} w) =0$, but the Galois group exchanges these two types of lines so this is impossible.<|endoftext|> TITLE: If power of an ideal is locally free then it is locally free QUESTION [6 upvotes]: Let $X$ be a noetherian scheme and $\mathcal{I} \subset \mathcal{O}_X$ a coherent sheaf of ideals. Suppose that $\mathcal{I}^d$ is locally-free for some power $d$. Then the blowing up $\mathrm{Bl}_{\mathcal{I}^d} \to X$ is an isomorphism. However, $\mathrm{Bl}_{\mathcal{I}} \cong \mathrm{Bl}_{\mathcal{I}^d}$ as $X$-schemes so the map $\pi : \mathrm{Bl}_{\mathcal{I}} \to X$ is also an isomorphism. By definition of blowing up this implies that the inverse image of $\mathcal{I}$ which is equal to $\mathcal{I}$ is is locally-free. My question is about a direct algebraic proof of this fact translated into commutative algebra: Let $A$ be a noetherian local ring and $I \subset A$ be an ideal such that $I^d = (f)$ for some non zero-divisor $f$. Then $I = (g)$ for some non zero-divisor $g$. I can see how to prove this when $A$ is a UFD but I don't see how a direct proof would go of the general case. Is there an easy argument or is it necessary to argue with the Rees Algebra? REPLY [7 votes]: Here is a proof if $A$ is an integral domain. Let $x_1,\dots, x_n$ generate $I$. Then $\prod_{i=1}^n x_i^{e_i}$ generate $I^d =f$ for vectors $e_i$ of nonnegative integers satisfying $\sum_i e_i=d$. These generators are all multiples of $f$ and can't all lie in the maximal ideal times $f$ so, since $A$ is local, one of them must be a unit times $f$. Fix such a generator $\prod_{i=1}^n x_i^{e_i}$. Without loss of generality, $e_1>0$. Then for all $j$ from $1$ to $n$, $(x_j/x_1)\prod_{i=1}^n x_i^{e_i}$ is also in $I^d$, thus is a multiple of $f$, hence is a multiple of $\prod_{i=1}^n x_i^{e_i}$. Dividing by $x_1^{e_1-1} \prod_{i=2}^n x_i^{e_i}$, we see that $x_j$ is a multiple of $x_1$. Becuase this works for all $j$, $I=(x_1)$.<|endoftext|> TITLE: Proving that polynomials belonging to a certain family are reducible QUESTION [10 upvotes]: In an article, I've found the following result. Unfortunately, it was derived from a general, somewhat complicated theory, that would be cumbersome for this result alone. Assume that $\mathbb F_p$ is the field with $p$ elements, and that we choose an integer $a$ with $1< a < 4p$, such that $a$ be coprime to $p(p-1)$ (for example, $a = 2p - 1$ is suitable). Let $t$ be transcendental over $\mathbb F_p$, and set $K = \mathbb F_p(t)$. Consider the polynomial $$P(X) = X^{4p} +t^a$$ in $K[X]$. Then for every $f\in \mathbb F_p[t]$, $P(f)$ is reducible in $\mathbb F_p[t]$. I would like a direct proof of this result. Any idea? REPLY [9 votes]: It follows from this result of Swan: If $t$ divides $f(t)$ there is nothing to do. So assume that this is not he case. Then with $F(t)=P(f(t))=f(t)^{4p}+t^a$, $F'(t)=at^{a-1}$ and $F(t)$ is separable. From this one easily gets that the discriminant of $F(t)$ is a square in $\mathbb F_p$ (see the alternative formula for the discriminant at the beginning of the article of Swan). By Swan's result, the number of irreducible factors of $F(t)$ is congruent to $\deg F(t)=4p\deg f$ mod $2$, hence it is even and therefore $>1$. Note added later: In fact the special case needed here is already due to Dickson, as pointed out by Swan in his paper. The argument is easy: If $F(t)$ is separable of degree $n$ and irreducible, then its Galois group contains an $n$-cycle. If $n$ is even, then this $n$-cycle is an odd permutation, so the discriminant of $F(t)$ is not a square. (Assuming that we are working over a finite field of odd characteristic.)<|endoftext|> TITLE: Quotient of solid torus by swapping coordinates on boundary QUESTION [14 upvotes]: Let $T$ be the solid 2-torus and let $\sim$ be the equivalence relation on $T$ generated by the relation $\{(\alpha,\beta) \sim (\beta,\alpha) \mid \alpha, \beta \in S^1\}$ on the boundary $\partial T=S^1\times S^1$. What kind of space is $T/{\sim}$? I believe that it is $S^3$, but I can't prove it. I posted this question already here: https://math.stackexchange.com/questions/4339019/identifying-boundary-of-solid-torus-with-itself-by-swapping-coordinates Unfortunately, I didn't get a satisfying answer. Not sure if it's because it is harder than I thought or actually nobody cares. Maybe both :) REPLY [9 votes]: As yet another proof: let $M$ be the quotient of $\partial T$. Let $U$ be a small closed regular neighbourhood of $M$. Thus $U$ is a solid torus. Let $V$ be the closure of the complement of $U$. So $V$ is $T$, minus an open regular neighbourhood of its boundary. So $V$ is another solid torus. Let $S$ be the intersection of $U$ and $V$. So $S$ is a (genus one) Heegaard surface for the three-manifold. It is an “exercise” to show that the boundaries of the meridian disk in $U$ and $V$, respectively, meet once. Thus $S$ is the standard Heegaard splitting of the three-sphere.<|endoftext|> TITLE: Is $C_n$ infinitely log-convex? QUESTION [5 upvotes]: A sequence $a_n$ is called log-convex if $\mathcal{L}(a_n):=a_{n+1}a_{n-1}-a_n^2\geq0$ for all $n$; it is infinitely log-convex provided that all the iterates $\mathcal{L}^k(a_n)$ are still log-convex, $k\geq1$. Here $\mathcal{L}^2(a_n)=\mathcal{L}(\mathcal{L}(a_n))$, etc. Consider in particular the well-known sequence of Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$. It has been established that both $C_n$ and some of its "quantum" versions are log-convex. I would like to ask: QUESTION. Is it true that $C_n$ is infintely log-convex? It appears to be so. REPLY [11 votes]: I think so. If $a_n=\int_X f^n(x)d\mu(x)$ for a certain positive function $f$ on a measure space $(X,\mu)$, then $$a_{n-1}a_{n+1}-a_n^2=\int_{X\times X} f^{n-1}(x)f^{n+1}(y)d\mu(x)d\mu(y)-\int_{X\times X} f^{n}(x)f^{n}(y)d\mu(x)d\mu(y)\\=\frac12\int_{X\times X} f^{n}(x)f^{n}(y)\left(\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}-2\right)d\mu(x)d\mu(y)$$ has also a similar representation. But $$ C_n=4^n\frac{(2n-1)!!}{(2n)!!}\cdot \frac1{n+1}=\frac2{\pi}\int_0^{\pi/2} (2\sin x)^{2n}dx\int_0^{1} t^n dt. $$<|endoftext|> TITLE: Closed formula for $(-1)$-Baxter sequences QUESTION [5 upvotes]: The number of the so-called Baxter permutations of length $n$ is computed by $$a_n=\frac1{\binom{n+1}1\binom{n+1}2}\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}.$$ There has also been a $q$-analogue of this sequence $$a_n(q)=\frac1{\binom{n+1}1_q\binom{n+1}2_q}\sum_{k=0}^{n-1}q^{3\binom{k+1}2} \binom{n+1}k_q\binom{n+1}{k+1}_q\binom{n+1}{k+2}_q.$$ Hence $a_n(1)=a_n$. Now, if we focus on the special values at $q=-1$ then it appears that the resulting sequence does seem to exhibit small prime factors, suggesting that one might be able to find a closed formula. Notice that such is not true when $q\neq-1$. Here are the first few values of $a_n(-1)$ for $n\geq1$: $1, 0, 0, -2, -4, 0, 0, 18, 42, 0, 0, -240, -600, 0, 0, 3850, 10010, 0, 0, -68796,\dots$. So, I like to ask: QUESTION. Is there a nice formula for $a_n(-1)$? Of course, one may restrict to the non-zero values. REPLY [5 votes]: Absolute values $1,4,42,600,10010,183456,\ldots$ of the non-zero terms not covered by Brian Hopkins answer seem to coincide with the sequence $$n\longmapsto {3n-2\choose n-1}{2n-1\choose n-1}\frac{1}{2n-1}$$ which is absent of the OEIS. This corresponds to Fedor's case $n\rightarrow 4n+1$: $$\frac1{2n+1}\sum_{k=0}^{2n}(-1)^k\binom{2n+1}k\binom{2n+1}{k+1}\binom{2n}k =\frac{(-1)^n}{2n+1}\binom{3n+1}n\binom{2n+1}n.$$<|endoftext|> TITLE: What is the closure of the Eilenberg-MacLane spectra under limits? under colimits? QUESTION [6 upvotes]: Every bounded spectrum is in the closure of the Eilenberg MacLane spectra under finite co/limits. Thus every bounded below (resp. above) spectrum is in the closure of the EM spectra under limits (resp. colimits). Let $\mathcal L$ denote the closure of the EM spectra under limits, and let $\mathcal C$ denote the closure of the EM spectra under colimits. Note that $\mathcal L$ is also closed under (infinite) products, and $\mathcal C$ is also closed under (infinite) coproducts. Question 1: What is a good description of $\mathcal L$? Question 2: What is a good description of $\mathcal C$? Question 3: What is an example of a spectrum in $\mathcal L$ which is not a(n infinite) product of bounded-below spectra? Question 4: What is an example of a spectrum in $\mathcal C$ which is not a(n infinite) sum of bounded-above spectra? REPLY [5 votes]: This is an answer to Questions 3 and 4. Consider the map $$ H\mathbb Z/2 \xrightarrow{\ldots, Sq^{2^n},\ldots} \prod_{n=1}^\infty \Sigma^{2^n} H\mathbb Z/2.$$ This is a map to a product, where the $n$-th component is the map $Sq^{2^n}\colon H\mathbb Z/2 \to \Sigma^{2^n} H\mathbb Z/2.$ This is a map between products of bounded-below spectra which is not a product of maps between bounded-below spectra. As you suggested in a comment, I believe that the cofiber of this map is an element of $\mathcal C$ that is not a direct sum of bounded above spectra. It is enough to prove that the fiber of this map is not a direct sum of bounded above spectra. Let $F$ be the fiber. Moreover, for all $m\ge 0$ let $F_m$ be the fiber of the map $$ H\mathbb Z/2 \xrightarrow{\ldots, Sq^{2^n},\ldots} \prod_{n=1}^m \Sigma^{2^n} H\mathbb Z/2.$$ There is a tower of spectra $$\cdots F_m \to F_{m-1}\to \cdots $$ and $F$ is the homotopy limit of this tower. We want to prove that $F$ is not equivalent to a sum of bounded above spectra. I claim that it is enough to prove that none of the maps $F_m\to F_{m-1}$ has a homotopy section. To prove the claim, notice that $F_m$ is the $2^m-1$-th Postnikov section of $F$. Suppose $F$ splits as a sum (or equivalently a product) of bounded above spectra. Then each Postnikov section of $F$ admits a corresponding splitting of Postnikov sections of factors. Among the putative factors of $F$ there is one factor, let's call it $X$, that satisfies $\pi_0(X)\cong \mathbb Z/2$. All other factors have $\pi_0=0$. $X$ is bounded above. Let $m$ be the smallest positive integer such that $\pi_{2^m-1}(X)=0$. We have a splitting $F\simeq X\times Y$ for some spectrum $Y$. It induces a splitting of Postnikov sections $$P_{2^m-1} F\simeq P_{2^m-1}X\times P_{2^m-1}Y.$$ But by our construction, $P_{2^m-1} F=F_m$, $P_{2^m-1}X\simeq F_{m-1}$, and $P_{2^m-1}Y\simeq \Sigma^{2^m-1} H\mathbb Z/2$. It follows that the map $F_m\to F_{m-1}$ has a section. It remains to prove that none of the maps $F_m\to F_{m-1}$ has a homotopy section. The spectrum $F_m$ is the total fiber of the following square $$ \begin{array}{ccc} H\mathbb Z/2 & \xrightarrow{\ldots, Sq^{2^n}, \ldots} & \prod_{n=1}^{m-1} \Sigma^{2^n} H\mathbb Z/2\\ \quad \quad \downarrow Sq^{2^m} & & \downarrow \\ \Sigma^{2^m} H\mathbb Z/2 & \to & * \end{array} $$ Note that the fiber of the top map is $F_{m-1}$. There is a fibration sequence $$F_m \to F_{m-1} \to \Sigma^{2^m} H\mathbb Z/2$$ where the second map is a composition $$F_{m-1}\to H\mathbb Z/2\xrightarrow{Sq^{2^m}} \Sigma^{2^m}H\mathbb Z/2.$$ If the map $F_m\to F_{m-1}$ has a section, then this composition is null. But this is not possible, because the kernel of the homomorphism of graded groups $$\mathcal A\cong [H\mathbb Z/2, H\mathbb Z/2]_*\to [F_{m-1}, H\mathbb Z/2]_*$$ is the left ideal generated by $Sq^2, Sq^4, \ldots, Sq^{2^{m-1}}$, and it does not contain $Sq^{2^m}$. A context where such a spectrum occurs more or less naturally is the Adams spectal sequence, which in some incarnation probably can be called the Bousfield-Kan spectral sequence. Consider the cobar construction that gives rise to the ASS: $$H\mathbb Z/2 \Rightarrow H\mathbb Z/2\wedge H\mathbb Z/2 \Rightarrow \cdots H\mathbb Z/2^{\wedge k+1} \cdots$$ (I did not typeset the arrows properly, but it is supposed to be a cosimplicial object in spectra). The second stage of the Tot tower can be identified with the homotopy fiber of a map $$H\mathbb Z/2 \to H\mathbb Z/2 \wedge \overline{H\mathbb Z/2}$$ where $\overline{H\mathbb Z/2}$ is the cofiber of the map $S\to H\mathbb Z/2$. The spectrum $H\mathbb Z/2 \wedge \overline{H\mathbb Z/2}$ splits as a product of Eilenberg - Mac Lane spectra, and I believe the fiber of the map above is an example of a spectrum in $\mathcal C$ that is not a sum of bounded above spectra. Maybe A.S. had in mind something like this in their comment, but this is just a guess. Dually, you can construct a spectrum that is in $\mathcal L$ but is not a product of bounded below spectra by taking the fiber of the map $$\bigoplus_{n=0}^\infty \Sigma^{-2^n} H\mathbb Z/2 \xrightarrow{\ldots, Sq^{2^n}, \ldots} H\mathbb Z/2.$$<|endoftext|> TITLE: Different occurences of the word 'period' in algebraic geometry QUESTION [9 upvotes]: I have come across the word 'period' in several contexts and I wonder if these notions are related. (1) The period map and domain: Let $ \pi : X \rightarrow B $ be a proper holomorphic submersion of complex manifolds. If a fiber $ X_0 $ of $ \pi $ is (compact) Kahler, then the cohomology groups $ H^k(X_0, \mathbb{Z}) $ (modulo torsion) are Hodge structures of weight $ k $. Now if we consider a small enough neighborhood $ U $ of $ 0 \in B $, then all fibers of U are Kahler, the Hodge numbers $ h^{p,q} $ are constant, and there is a fixed isomorphism (#) $ H^k(X_b, \mathbb{C}) \cong H^k(X_0, \mathbb{C}) $ from Ehresmann's theorem. So the numbers $ a_p = \text{dim} F^pH^k(X_b, \mathbb{C}) $ are also constant. The period map then sends $ b \in U $ to the (image under (#) of the) flag $ \{F^pH^k(X_b, \mathbb{C})\}_p $ in the flag variety $ G(a_k , \ldots , a_1, H^k(X_0, \mathbb{C})) $. The period domain is the subset of the flag variety whose flags satisfy $ F^pH^k(X_0, \mathbb{C}) \oplus \overline{F^{k+1-p}H^k(X_0, \mathbb{C})} = H^k(X_0, \mathbb{C}) $. (2) Period: A real number is a period if it is an integral $ \int_{p \ge 0} q(x_1, \ldots, x_n) dx $ where $ p $ is a polynomial with rational coefficients and $ q $ is a rational function with rational coefficients. We can allow $p,q $ to be algebraic functions, this is equivalent. So abelian integrals $ \int_0^r \frac{dx}{\sqrt {4x^3 - ax - b}} $ are periods ($ r,a,b \in \mathbb{Q} $). (3) Period of a cohomology group: I came across this notion after reading Daniil Rudenko's question on this website. This one I do not understand at all. So my question is: How are these notions related? Surely (2) and (3) are related as you're supposedly computing an integral in (3) as well. And I suspect (1) and (3) are related as well as you're dealing with some cohomology group. I apologize if the question is a bit too general but I'd like to understand this. REPLY [9 votes]: The second and the third are pretty much equivalent. Indeed, "the period" in XIX century sense is essentially the same as the discrepancy between the branches of a multi-valued function, obtained as an integral of an algebraic function. If you take the Riemann surface associated with this algebraic function (that is, a branched covering of C where it is well-defined), the period becomes an integral of the holomorphic differential $fdz$ associated with this function over a closed loop, that is, an integral of a holomorphic 1-form over a closed cycle. The modern definition of "periods" is, more or less, "the pairing between holomorphic differential forms and integral homology". However, this notion can (and often is) extended to Hodge structures, and this is related to the first notion you mention. Define the Teichmuller space of a complex manifold as the space of all complex structures (here the complex structures are understood as endomorphisms of $TM$ satisfying $I^2=-Id$ and the integrability condition) up to isotopies: $Teich= \frac{Comp}{Diff_0}$. Never mind that this space might be horribly non-Hausdorff; the period map is naturally defined on $Teich$, and (if you are lucky) it would help to make sense even of the non-Hausdorff pathologies. The period map is the map taking $I\in Teich$ to the Hodge structure on $H^*(M,I)$, which is understood as a point in the appropriate flag space. This map is holomorphic, and, if you are lucky, defines a biholomorphism between $Teich$ and the space of Hodge structures ("period space"). This holds, unfortunately, only for complex tori, but weaker versions of this statement are true for K3, hyperkahler manifolds, complex curves and in some other cases. These results are called "global Torelli theorems". The local Torelli theorems are statements about the local structure of the period map, saying (in most cases) that is it locally an immersion; this is true, for instance, for Calabi-Yau manifolds. The modern definition of periods is not entirely equivalent to the traditional, because the Hodge structure contains more data than just holomorphic forms; however, if you know the Hodge structure on cohomology, you know the pairing between the holomorphic forms and the integer homology, hence it is an extension of the XIX century notion. If you restrict yourself to the first cohomology, the Hodge structure is a flag $H^{1,0}(M)\subset H^1(M)$; in this case the XIX century notion of "periods" coincides with the Hodge-theoretic notion.<|endoftext|> TITLE: What are the "hot" topics in mathematical QFT at the time? QUESTION [18 upvotes]: I am currently finishing my Master's studies in mathematical physics. One topic which always interested me a lot were modern mathematical approaches to Quantum Field Theory (QFT) as well as the mathematical structures arising from studying these theories. Hence, I would like to go into this direction for my Phd. For this, I would like to know, what are some "hot" topics in this area, in the sense of topics in which a lot of active research and effort is put into. Of course "mathematical QFT" is a little bit vague, but I am really open for everything, like algebraic QFT, axiomatic or constructive QFT, etc. Also, I was looking through a lot of websites of universities in order to localize places where things like this is are discussed. However, I was not very sucessful yet, since it seems that mathematical QFT is not so common as many other topics in physics or mathematics. So, if anyone knows some places where one could try to apply in this direction (preferably in (middle) europe), I would be happy if he/she can share his knowledge with me in the comments or in answers. (Some words to my background: In my university, there is no research group working in the mathematics of QFT. I did my thesis in some quantum gravity related stuff. On the physical side, the topic is related to holographic dualities in 3d quantum gravity approaches such as spin foam models. However, my main tasks were more on the mathematical side, especially in geometric and low-dimensional topology. I absolutely loved my project and I definitely can also imagine working more in that direction, but still I really would like to try out working on some project in the area of mathematical QFT, since this was always a topic I wanted to dive into and I guess, a Phd is the perfect way to do so. Hence, I unfortunately haven't research experience in things like algebraich QFT yet, however, since I did my Masters in mathematical physics, I did of course a lot of courses in the relevant directions of both theoretical physics and pure mathematics.) Thanks a lot! REPLY [8 votes]: Luckily for you, there is an online seminar where you can get a panorama of what is currently going on in the area of mathematical QFT. It is the webinar "Analysis, Quantum Fields, and Probability". To watch live upcoming seminars you will have to join their mailing list, so they can send you the relevant Zoom links. You can also watch the previous talks on their Youtube channel. QFT is a broad subject with many aspects that give rise to specific problems for mathematicians to work on. If you are interested in the renormalization group aspect, you might want to look at my previous answers: Reading list recommendation for a hep-ph student to start studying QFT at a more mathematically rigorous level? Formal mathematical definition of renormalization group flow<|endoftext|> TITLE: Definition of "classifying topos" QUESTION [9 upvotes]: Is there a geometric theory $T$ and a Grothendieck topos $\mathcal E$ such that (2) holds but (1) doesn't: $\mathcal E$ 2-represents the 2-functor $$\mathbf{GrothTop}\to\mathbf{Cat}$$ which sends a Grothendieck topos $\mathcal E$ to the category of models of $T$ in $\mathcal E$. $\mathcal E$ represents the 1-functor $$\mathrm h\mathbf{GrothTop}\to\mathrm h\mathbf{Cat}\to \mathbf{Set}$$ which can be obtained by truncating the above functor to the 1-categorical level and composing with the functor sending a category $\mathcal C$ to the set $\mathrm{Ob}(\mathcal C)/\mathord{\cong}$ of all objects of $\mathcal C$ up to isomorphism. Conversely, (1) implies (2), right? REPLY [8 votes]: This is a bit surprising to me, but the two statements turn out to be equivalent. A topos $\mathcal{E}$ is completely determined by the functor $$\mathbf{Geom}(-,\mathcal{E}) : \mathbf{GrothTop}^\mathrm{op} \to \mathbf{Cat}$$ (this is some kind of 2-Yoneda Lemma). But we can also look at the 1-category $\mathrm{h}\mathbf{GrothTop}$ in which we identify geometric morphisms that are isomorphic. Then the (1-categorical) Yoneda Lemma says that $\mathcal{E}$ is completely determined by the functor $$\mathbf{Geom}(-,\mathcal{E})/\!\cong\,\, : \mathrm{h}\mathbf{GrothTop}^\mathrm{op} \to \mathbf{Class}$$ to the category of classes (we have to work with classes rather than sets because there can be a proper class of geometric morphisms up to isomorphism between two toposes). In particular, suppose that $\mathbf{Geom}(-,\mathcal{E})/\!\cong\,\,$ agrees with the functor that sends each Grothendieck topos $\mathcal{F}$ to the collection of $T$-models in $\mathcal{F}$ up to isomorphism, for some geometric theory $T$. Then $\mathcal{E}$ is the classifying topos of $T$. To prove this, we need to use the fact that the classifying topos of a geometric theory always exists (thanks to @Mike Shulman for pointing this out).<|endoftext|> TITLE: Direct and inverse image terminology QUESTION [7 upvotes]: Let $f\colon X\to Y$ be a continuous map. Then $f$ induces a geometric morphism $f^\ast\dashv f_\ast\colon \mathrm{Sh}(X)\leftrightarrows\mathrm{Sh}(Y)$, whose left adjoint is called inverse image and whose right adjoint is called direct image. Why is the left adjoint called inverse image and why is the right adjoint called direct image? Especially the following makes it confusing: The direct image $f_\ast$ is defined by $$F\in\mathrm{Sh}(X)\mapsto (U\in\mathcal O(Y)\mapsto F(f^{-1}(U))).$$ Usually one calls $f^{-1}(U)$ the inverse image of $U$ under $f$. REPLY [10 votes]: There is a precise, almost literal, sense in which $f^* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$ generalises the inverse image as defined in elementary set theory. Observe that open subspaces $V \subseteq Y$ correspond to subterminal objects in $\textbf{Sh} (Y)$: the sheaf of sections of the inclusion $V \hookrightarrow Y$ is a subterminal object, and every subterminal object is isomorphic to one of this form. Since $f^*$ preserves finite limits, it preserves subterminal objects, and in fact it sends the sheaf corresponding to $V$ to the sheaf corresponding to $f^{-1} V$. So we may think of $f^* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$ as being an extension of the set-theoretic inverse image operation, and this justifies the name "inverse image functor". Once you have an "inverse image functor", there is a powerful temptation to call its partner the "direct image functor". My personal opinion is that the name "direct image functor" is unsuitable for the right adjoint of the inverse image functor. For one thing, $f_* : \textbf{Sh} (X) \to \textbf{Sh} (Y)$ is not a generalisation of the set-theoretic direct image operation, as I will now explain. Let $U$ be an open subspace of $X$. What do you suppose $f_*$ applied to the subterminal object corresponding to $U$ yields? It is a subterminal object, of course, but it is not (the subterminal object corresponding to) the direct image $$\exists_f U = \{ y \in Y : \exists x \in U . y = f (x) \} = \{ y \in Y : \exists x \in X . y = f (x) \land x \in U \}$$ which is not even guaranteed to be open if we do not assume $f : X \to Y$ is an open map. Instead, it corresponds to $$\forall_f U = \{ y \in Y : f^{-1} \{ y \} \subseteq U \} = \{ y \in Y : \forall x \in X . y = f (x) \Rightarrow x \in U \}$$ which is always open: indeed, $\forall_f U$ is the union of all open $V \subseteq Y$ such that $f^{-1} V \subseteq U$. Unfortunately, the reality is that $f^* : \textbf{Sh} (Y) \to \textbf{Sh} (X)$ does not always have a left adjoint that generalises $\exists_f$, so while it is $\exists_f$ that bears the name "direct image" in elementary set theory, for want of a better name it is $f_*$ that gets the name in topos theory.<|endoftext|> TITLE: Which arithmetical sentences have no counterexamples in the sense of Kreisel? QUESTION [5 upvotes]: It is a well-known fact that given a first-order sentence $\psi$ in prenex normal form $\forall x_1 \exists y_1 \forall x_2 \exists y_2 \dots \forall x_n \exists y_n \theta(x_1,\dots,x_n,y_1,\dots,y_n)$ (i.e., $\theta$ is quantifier free or at least $\Delta^0_0$ in the appropriate sense), the truth of $\psi$ in a given structure $M$ can be understood in terms of a game in which two players ($\forall$ and $\exists$) alternatively choose elements of $M$ corresponding to the bound variables of $\psi$, with $\forall$ choosing for the variables $x_i$ and $\exists$ choosing for the variables $y_i$. $\exists$ wins if and only if at the end the resulting variable assignment makes $\theta$ true in $M$. The characterization is of course that $M \models \psi$ if and only if $\exists$ has a winning strategy for this game. Such a strategy can be understood as a series of Skolem functions $f_1(x_1),f_2(x_1,x_2),\dots,f_n(x_1,x_2,\dots,x_n)$ on $M$ with the property that (after expanding $M$ with the $f_i$'s), $M \models \forall x_1\dots x_n \theta(x_1,\dots,x_n,f_1(x_1),\dots,f_n(x_1,\dots,x_n))$. In the specific case of arithmetic (i.e., the structure $(\mathbb{N},0,+,\cdot)$), these functions are closely related to the idea of realizability. If $\psi$ is provable in Heyting arithmetic, for instance, then one can extract computable functions $f_1,f_2,\dots,f_n$ giving a winning strategy for $\exists$. Kreisel showed that an analogous weaker statement is true of Peano arithmetic. A proof of $\psi$ in Peano arithmetic gives $\exists$ a computable procedure for beating any particular strategy for $\forall$, provided that $\exists$ is given $\forall$'s strategy in advance. More formally, given a proof of $\psi$ in Peano arithmetic, we can computably extract Turing indices $e_1,\dots,e_n$ such that for any functions $g_1 : \mathbb{N}^0 \to \mathbb{N}$, $g_2 : \mathbb{N}^1 \to \mathbb{N}$, $\dots$, $g_n : \mathbb{N}^{n-1} \to \mathbb{N}$, we have that $ \mathbb{N} \models \theta(a_0,\dots,a_n,b_0,\dots,b_n), $ where $$a_i = g_i(b_1,\dots,b_{i-1})$$ and $$b_i = \Phi_{e_i}^{g_1\dots g_n}(a_1,\dots,a_{i})$$ for each $i \in [1,n]$. ($\Phi_e^X(n)$ is of course the $e$th Turing functional applied to $n$ given $X$ as an oracle. Assume that we're using some fixed computable tupling function.) This fact is often called Kreisel's 'no-counterexample interpretation' of Peano arithmetic. Let's say that $\bar{e} = (e_1,\dots,e_n)$ is a weak classical realizer of $\psi$ if the above condition holds. (I don't know if this concept has an established name.) For the sake of this question, we'll write $\bar{e} \Vdash \psi$ to mean that $\bar{e}$ is a weak classical realizer of $\psi$. (Note that $(\exists \bar{e})\bar{e}\Vdash \psi$ is a lightface $\Pi^1_1$ condition on $\psi$.) I believe that the commonly studied notions of classical realizers for arithmetic (e.g., realizability in the sense of Krivine) are a priori stronger than this in that while they allow a certain amount of probing of $\forall$'s stragey, they do not necessarily allow arbitrary computation from it. We can now cosmetically restate Kreisel's result as saying that if $\mathsf{PA} \vdash \psi$, then $(\exists \bar{e})\bar{e} \Vdash \psi$. It's also not too hard to show that if $(\exists \bar{e})\bar{e} \Vdash \psi$, then $\mathbb{N} \models \psi$. (No amount of prep work can beat an unbeatable strategy.) I'm curious about the extent to which the conditions $\mathsf{PA} \vdash \psi$, $(\exists \bar{e})\bar{e} \Vdash \psi$, and $\mathbb{N} \models \psi$ differ. One immediate observation is that if $\psi$ is a true $\Pi^0_1$-sentence, then $(\exists \bar{e})\bar{e} \Vdash \psi$ (somewhat trivially), which means that there are many examples of $\mathsf{PA}$-unprovable sentences with weak classical realizers (a Gödel sentence for instance). This leads to a somewhat more nuanced question. Question 1. Is there a sentence $\psi$ such that $(\exists \bar{e})\bar{e} \Vdash \psi$, yet $\mathsf{PA} \cup \{\chi: \mathbb{N}\models\chi,~\chi\text{ a }\Pi^0_1\text{ sentence}\} \not \vdash \psi$? In the other direction, there is basically no way that $\mathbb{N} \models \psi$ implies $(\exists \bar{e})\bar{e} \Vdash \psi$, but finding an explicit example of this seems difficult. The theory of $\mathbb{N}$ is lightface $\Delta^1_1$, which is simpler than the apparent complexity of the set of sentences with weak classical realizers (lightface $\Pi^1_1$). Question 2. Is there a sentence $\psi$ such that $\mathbb{N} \models \psi$, yet $\neg (\exists \bar{e}) \bar{e} \Vdash \psi$? REPLY [3 votes]: In fact all true arithmetical sentences have weak classical realizers. Namely, a true arithmetical sentence $\psi$ $$\forall x_1\exists y_1 \ldots \forall x_n\exists y_n \theta(x_1,\ldots,x_n,y_1,\ldots,y_n)$$ is weakly realized by $(e_1,\ldots,e_n)$, where $\Phi_{e_i}^{\vec{g}}(y_1,\ldots,y_{i-1})$ performs an unbounded search for a tuple $(y_i,\ldots,y_n)$ such that $\theta(g_0(),\ldots,g_n(y_1,\ldots,y_n),y_1,\ldots,y_n)$ is true and then outputs $y_i$ from the first found tuple. This works since for any $g_1\colon\mathbb{N}^0\to\mathbb{N},\ldots,g_n\colon \mathbb{N}^{n-1}\to\mathbb{N}$ the sentence $\exists y_1,\ldots, y_n \theta(g_0(),\ldots,g_n(y_1,\ldots,y_n),y_1,\ldots,y_n)$ is true.<|endoftext|> TITLE: A rather curious identity on sums over triple binomial terms QUESTION [15 upvotes]: While exploring the Baxter sequences from my earlier MO post, I obtained a rather curious identity (not listed on OEIS either). I usually try to employ the Wilf-Zeilberger (WZ) algorithm to justify such claims. To my surprise, WZ offers two different recurrences for each side of this identity. So, I would like to ask: QUESTION. Is there a conceptual or combinatorial reason for the below equality? $$\frac1n\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2} =\frac2{n+2}\sum_{k=0}^{n-1}\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}.$$ Remark 1. Of course, one gets an alternative formulation for the Baxer sequences themselves: $$\sum_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}}{\binom{n+1}1\binom{n+1}2} =2\sum_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}}{\binom{n+1}1\binom{n+2}2}.$$ Remark 2. Yet, here is a restatement to help with combinatorial argument: $$\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2} =2\sum_{k=0}^{n-1}\binom{n+1}k\binom{n}k\binom{n+1}{k+2}.$$ REPLY [9 votes]: The combinatorial identity $$ \sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n+1}{k+1} \binom{n+1}{k+2} = 2 \sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n}{k} \binom{n+1}{k+2} $$ has a simple combinatorial proof. The left-hand side counts all triples $A,B,C \subseteq [n+1]$ whose sizes are $k,k+1,k+2$ for some $k$. These come in two types: $n+1 \in B$. The triple $A,B\setminus\{n+1\},C$ has sizes $k,k,k+2$, and the middle set is a subset of $[n]$. $n+1 \notin B$. The triple $\overline{C},\overline{B}\setminus\{n+1\},\overline{A}$ has sizes $\bar{k},\bar{k},\bar{k}+2$, where $\bar{k}=n-1-k$, and the middle set is a subset of $[n]$. The triples of each type are counted by the sum on the right-hand side.<|endoftext|> TITLE: A functional equation for a family of functions indexed by the symmetric group $S_3$ QUESTION [5 upvotes]: $\newcommand{\C}{\mathbb C}$A question asked recently was as follows: For the symmetric group $G:=S_3$, is it possible to construct functions $t_g\colon\C\to\C$ that satisfy the convolution identity \begin{equation} t_g(x+y)=\sum_{h\in G} t_{gh^{-1}}(x)t_h(y) \tag{*}\label{*} \end{equation} for all $g\in G$ and all complex $x$, $y$? This question was then deleted by the OP. Of course, identity \eqref{*} trivially holds if $t_g(x)=0$ for all $g$ and $x$. Otherwise, I think the question may be of interest to other users and therefore will provide an answer to it on this page. REPLY [4 votes]: I think there is another "degenerate" way to deal with the question as posed, for a general finite group $G$ (later note: which may be developed into more interesting solutions, as we have done in later edits). The key is the connection with idempotents (not necessarily central) of the group algebra $\mathbb{C}G.$ When $G$ is Abelian, the group algebra $\mathbb{C}G$ is clearly commutative, so all idempotents are central. Hence the case of non-central idempotents only arises when $G$ is non-Abelian). Notice that if such functions $t_{g}(x)$ exist for $ g \in G$ and $x \in \mathbb{C}$, then we must have $t_{g}(0) = t_{g}(0+0) = \sum_{h \in G} t_{gh^{-1}}(0)t_{h}(0)$. This means that the element $T$ of $\mathbb{C}G$, defined by $T = \sum_{g \in G} t_{g}(0)g$, satisfies $T^{2} = T$, so is either idempotent or zero. Going in the other direction (ignoring the trivial case $t_{g}(x) = 0$ for all $g,x$), suppose that we are given an idempotent $E = \sum _{g \in G} \lambda_{g}g$ of $\mathbb{C}G.$ Then we may define $t_{g}(x) = e^{x}t_{g}(0) = e^{x}\lambda_{g}$ for all $g \in G$ and all $x \in \mathbb{C}$. Then for any $g \in G$ and $x,y \in \mathbb{C}$, we have \begin{eqnarray*} \sum_{h \in G} t_{gh^{-1}}(x)t_{h}(y) & = & e^{x+y} \sum_{h \in G} \lambda_{gh^{-1}} \lambda_{h} = e^{x+y} \lambda_{g} \\ & = & e^{x+y}t_{g}(0) = t_{g}(x+y), \end{eqnarray*} using the fact that $E^{2} = E.$ One obvious case is when $E = \frac{1}{|G|} \left( \sum_{g \in G} g \right)$, which gives rise to the case that $t_{g}(x) = \frac{e^{x}}{|G|}$ for all $g \in G, x \in \mathbb{C}$ . Another easy case is when $E = 1_{G}$, which gives rise to the case $t_{g}(x) = e^{x}$ if $g = 1_{G}, 0$ otherwise (for $g \in G, x \in \mathbb{C}$). A more interesting example when $G = S_{3}$ and $E$ is the non-central idempotent $ \frac{1+(12)}{2} - \frac{1}{6} \left(\sum_{ \sigma \in S_{3}} \sigma \right)$, in which case ( for any $x \in \mathbb{C}$), we have $e^{-x}t_{g}(x) = \frac{1}{3}$ for $g = 1_{G}$ or $g = (12)$, and $e^{-x}t_{g}(x) = \frac{-1}{6}$ otherwise. This method gives existence of some solutions to the convolution equations, but there are other types of solution, as you already showed. In the solutions above we can also replace $e^{x}$ by $e^{\alpha x}$ for a fixed constant $\alpha$ (independent of $x$). Later edit: Let us try to develop these arguments further and obtain more information about general solutions to the convolution equations. Suppose that we solutions $\{t_{g}(x) : g \in $G$ \}$ for the convolution equations. For each $x \in \mathbb{C}$, let $E(x) = \sum_{g \in G} t_{g}(x) g$ in the group algebra $\mathbb{C}G$. We have seen that $E(0)$ is an idempotent of $\mathbb{C}G$. The convolution equations imply that $E(x) = E(0 + x) = E(0)E(x)$ and $E(x) = E(x+0) = E(x)E(0)$ for each $x \in \mathbb{C}$. Hence $E(x)$ always lies in the subalgebra $E(0)\mathbb{C}G E(0)$ of the group algebra $\mathbb{C}G$, and this subalgebra is isomorphic to the endomorphism algebra of the right $\mathbb{C}G$-module $E(0)\mathbb{C}G$. Also, $E(0)$ is the identity element of this algebra. Also, the convolution equations imply that $E(x)E(y) = E(x+y) = E(y+x) = E(y)E(x)$ for all $x,y \in \mathbb{C}$. We also remark that $E(x)$ is a unit in the algebra $E(0)\mathbb{C}G E(0)$, and that its inverse is $E(-x)$. Let us consider the case that $E(0)$ is a primitive idempotent of $\mathbb{C}G$. In that case, Schur's Lemma tells us that $E(0)\mathbb{C}GE(0)$ consists of scalar multiples of $E(0)$. In that case, for each $x \in \mathbb{C}$, we have $E(x) = \mu_{x}E(0)$ for some complex number $\mu_{x}$, and by the remarks above, we have $\mu_{x}\mu_{y} = \mu_{x+y}$ for all $x,y \in \mathbb{C}$. This forces $\mu_{x} = e^{\alpha x}$ for each $x \in \mathbb{C}$, for some constant $\alpha.$ Hence the only solutions for which $E(0)$ is a primitive idempotent of $\mathbb{C}G$ are the "degenerate" ones described earlier. This indicates how to deal with the case that $E(0)$ is a general idempotent of $\mathbb{C}G$. We may decompose $E(0)$ as a sum of mutually orthogonal primitive idempotents of $\mathbb{C}G$, say $E(0) = \sum_{j= 1}^{t} E_{j}(0)$, where $E_{i}(0) E_{j}(0) = \delta_{ij} E_{i}(0)$ for each $i,j.$ In the case that no two of the $E_{i}(0)$ are conjugate via a unit of $\mathbb{C}G$, the situation is reasonably clear. For then we see that $E(0) \mathbb{C}G E(0)$ is a commutative algebra which is the direct sum of the $1$-dimensional algebras $E_{j}(0)\mathbb{C}G E_{j}(0)$. Then we may see that there are complex constants $a_{j} : 1 \leq j \leq t$ such that $ E(x) = \sum_{j=1}^{t} e^{a_{j}x} E_{j}(0).$ Even later edit: In comments, it is asked what is necessary if we insist the determinant ${\rm det}[t_{gh^{-1}}(x)]$ is equal to $1$ for all $x$. We illustrate with one example. When $G = \langle u \rangle$ is cyclic of order $2$, there are just two mutually orthogonal idempotents in the group algebra $\mathbb{C}G$. These are $ E_{1} = \frac{1_{G} + u}{2} $ and $E_{2} = \frac{1_{G} - u}{2}. $ Then the above methods give a general solution to the convolution equations of the form $ \sum_{g \in G}t_{g}(x)g = e^{c_{1}x} E_{1} + e^{c_{2}x}E_{2}.$ This gives the determinant ${\rm det}[(t_{gh^{-1}}(x))])$ to be $e^{c_{1}x+c_{2}x}$. Hence this determinant is $1$ for all $x$, if and only if $c_{2} = -c_{1}.$ Incidentally, to partially address one question of the OP in comments, this example for the cyclic group of order two may be ``lifted" to a solution of the convolution equations for $G = S_{n}$ for any $n$. I don't give all details, but we may choose constants $c_{1}$ and $c_{2}$ and set $t_{g}(x) = \frac{1}{n!}\left( e^{c_{1}x} + e^{c_{2}x} \right)$ whenever $g$ is an even permutation and $t_{g}(x) = \frac{1}{n!} \left( e^{c_{1}x} - e^{c_{2}x} \right)$ if $g$ is an odd permutation. However, we then find that the matrix $[t_{gh^{-1}}(x)]$ is singular for all $x$ when $n > 2.$ Returning to the general solutions, in the case that two or more of the $E_{j}(0)$ are conjugate via a unit of the group algebra $\mathbb{C}G$, the analysis is more subtle. Latest edit (also addressing some issues in comments): Let us recall that for any finite group $G$, the group algebra $\mathbb{C}G$ is isomorphic to a direct sum of (mutually annihilating) full matrix algebras $ \bigoplus_{\chi} M_{\chi(1)}(\mathbb{C})$, where $\chi$ runs over the complex irreducible characters of $G$. This gives a decomposition of $1_{G}$ as a sum of mutually orthogonal primitive idempotents that is $1_{G} = \sum_{\chi} \sum_{j = 1}^{\chi(1)} E_{j}(\chi)$. Here, $E_{j}(\chi)$ is represented by an idempotent matrix of trace $1$ in any irreducible representation of $G$ which affords $\chi$, and is represented by $0$ in any irreducible representation of $G$ affording an irreducible character $\mu \neq \chi.$ Also, for each $\chi$, we have $\sum_{j=1}^{\chi(1)} E_{j}(\chi)$ is a central idempotent of $\mathbb{C}G$, and acts as the identity in any irreducible representation of $G$ affording character $\chi$. This decomposition of $1_{G}$ into a sum of mutually orthogonal primitive idempotents is not quite unique in general ( but it is when $G$ is Abelian). In general, it is unique up to conjugation by a unit of $\mathbb{C}G$ (and reordering the idempotents). For non-commutative $G$, we now outline how to construct some solutions to the convolution equations which have ${\rm det}([t_{gh^{-1}}(x)]) = 1$ for all $x$, but the list is not exhaustive in general. We take a set of pairwise distinct (this is not really necessary, but is useful for purposes of exposition) non-zero complex constants $\{ c_{j}(\chi) : \chi \in {\rm Irr}(G), 1 \leq j \leq \chi(1) \}$ with the property that $\sum_{\chi,j} \chi(1) c_{j}(\chi) = 0.$ We further insist that $\{ e^{c_{j}(\chi)} : \chi \in {\rm Irr}(G), 1 \leq j \leq \chi(1) \}$ still consists of $\sum_{\chi} \chi(1)$ distinct elements. This is always possible to achieve for non-trivial $G$. For example, we may take all but one of the $c_{j}(\chi)$ to be real, positive and pairwise distinct ( with distinct exponentials, all real and greater than one). The zero sum condition then forces the choice of the remaining $c_{j}(\chi)$, and it clearly must be negative with negative exponential. It is convenient to introduce exponentials of elements of the group algebra $\mathbb{C}G.$ To do this, we take an element $A$ of $\mathbb{C}G$, and take its component $A_{\chi}$ in the matrix algebra summand $M_{\chi}(\mathbb{C})$. Then we define $\exp(A)$ to be the unique element of $\mathbb{C}G$ which has component $\exp(A_{\chi})$ ( the usual matrix exponential) in $M_{\chi}(\mathbb{C}).$ As usual, if $A$ and $B$ are commuting elements of $\mathbb{C}G$, then we have $\exp(A+B) = \exp(A)\exp(B).$ Now for any complex number $x$, we define the element $E(x)$ to be $\exp{\left(\sum_{\chi} \sum_{j = 1}^{\chi(1)}c_{j}(\chi)x E_{j}(\chi) \right)}.$ Then we see that whenever $\sigma$ is an irreducible representation of $G$ affording character $\mu$, $\sigma(E(x))$ is a diagonalizable matrix with distinct eigenvalues $e^{c_{j}(\chi)x} : 1 \leq j \leq \chi(1)$ (strictly speaking, we need to avoid the case that $x$ has the form $\frac{2\pi i}{c_{r}-c_{s} }$ for any $r \neq s$). It follows (with the noted exceptions) that in the regular representation of $G$, $E(x)$ has the eigenvalue $e^{c_{j}(\chi)x}$ with multiplicity $\chi(1)$ for $1 \leq j \leq \chi(1)$, for each irreducible $\chi$. Since we assumed that $\sum_{\chi} \chi(1) \left( \sum_{j=1}^{\chi(1)} c_{j}(\chi) \right) = 0$, we see that the determinant of the image of $E(x)$ in the regular representation is $\prod_{\chi}\left( \prod_{j = 1}^{\chi(1)} e^{c_{j}(\chi)x} \right)^{\chi(1)} = 1$ for all $x$. But writing $E(x) = \sum_{g \in G} t_{g}(x) g$ in the group algebra $\mathbb{C}G$, we see that $E(x)$ is represented by the matrix $[t_{gh^{-1}}(x)]$ in the regular representation. Before treating the explicit case $G = S_{3}$, we remark that in general it is not straightforward to determine explicit formulae for the primitive idempotents of the group algebra $\mathbb{C}G$ when $G$ is a non-Abelian group. There is an explicit general procedure due to A. Young when $G$ is the symmetric group $S_{n}.$ However, for the case $G = S_{3}$, we may proceed directly. We know from the theory described above that there should be four mutually orthogonal primitive idempotents of the group algebra $\mathbb{C}G$ when $G$ is the symmetric group $S_{3}.$ Two of these are easy, but we also need to decompose the central idempotent corresponding to the degree $2$ irreducible character. This idempotent is $ X = \frac{3(1)- [(1) +(123) +(132)]}{3}.$ We have written the idempotent in this fashion to make it clear that it commutes with the idempotent $Y = \frac{(1) + (12)}{2}.$ Then we see easily that $XY = YX = \frac{(1) + (12)}{2} - \frac{ \sum_{g \in G} g}{6}.$ This is a primitive idempotent of $\mathbb{C}G$, and may serve as our idempotent $E_{1}(\chi)$, where $\chi$ is the degree $2$ irreducible character. The primitive idempotent $E_{2}(\chi)$ is then easily seen to be $\frac{(1)-(12)}{2} - \frac{ \sum_{g \in G} {\rm sign}(g) g}{6}.$ Hence we may set $E_{1}( \lambda_{0}) = \frac{\sum_{g \in G} g }{6}$, $E_{1}( \lambda_{1}) = \frac{\sum_{g \in G} {\rm sign}(g)g }{6}$, where $\lambda_{0}$ is the trivial character, $\lambda_{1}$ is the sign character. Then we may choose any four non-zero complex constants $c_{0},c_{1},c_{2},c_{3}$ with $c_{0} + c_{1} + 2(c_{2}+c_{3}) = 0$ and set $E(x) = e^{c_{0}x} E_{1}(\lambda_{0}) + e^{c_{1}x}E_{1}(\lambda_{1}) + e^{c_{2}x}E_{1}(\chi) + e^{c_{3}x}E_{2}(\chi)$. This leads to $t_{1}(x) = \frac{e^{c_{0}x}}{6} + \frac{e^{c_{1}x}}{6} + \frac{e^{c_{2}x}}{3} + \frac{e^{c_{3}x}}{3}$, $t_{(12)}(x) = \frac{e^{c_{0}x}}{6} - \frac{e^{c_{1}x}}{6} + \frac{e^{c_{2}x}}{3} - \frac{e^{c_{3}x}}{3}$, etc. , producing a solution to the convolution equations with ${\rm det}([t_{gh^{-1}}(x)]) = 1$ for all $x$.<|endoftext|> TITLE: Where can I find the following S. Shelah's paper? QUESTION [7 upvotes]: I've been trying to find the following article: "S. Shelah, Remarks on cardinal invariants in topology, General topology Appl. 7(3) (1977), 251-259". I tried to go directly to the journal page, but it turns out that Issue 3 isn't registered there (see: General Topology and its Applications). Is there any other place where I can find this paper? REPLY [9 votes]: please get the pdf here. note that the text itself starts on page 5.<|endoftext|> TITLE: What motivated Thom to relate the cobordism groups with some homotopy groups? QUESTION [8 upvotes]: I would like to know what motivated or led Thom to think that the (un)oriented cobordism groups would correspond with the homotopy groups of some structure (Thom spectum), or with the coefficient groups of a cohomology theory. REPLY [11 votes]: I did not know Thom so I can't speak to all his personal motivations. But I can speak as someone that has read much of his work carefully and I think I have some insights into this. The primary motivation for his theorem boils down to thinking carefully about the implicit function theorem, and asking when a subset of Euclidean space is the pre-image of a regular value of a smooth function. These are the problems most students grapple with when learning about smooth manifolds for the first time, and Thom's considerations flowed out of these kinds of naive considerations. Thom just went a little further than most. The first step is realizing a submanifold of $S^n$ is the pre-image of a regular value of a smooth function $$f : S^n \to S^m$$ if and only if it has no boundary, is compact and has a trivial tubular neighbourhood. And this is the key insight into the Pontriagin construction. The next step is asking about general manifolds, and this is where you take the step up to maps to Thom spaces. Specifically the structure of the normal bundle is key in the Pontriagin construction. So if you have a non-trivial normal bundle you need some feature of your map to take that into account. Grassman manifolds are the key object related to maps out of vector bundles. But once you see that Grassman manifolds are the key analogue to the Pontriagin construction that gives you your map defined in a tubular neighbourhood of your submanifold. Coning off as in the Pontriagin construction gives you the Thom space. From this point of view you can see his theorem as a direct extrapolation from the trivial tubular neighbourhood case. I think for people that like to think categorically it could maybe feel unintuitive since the Thom space is a departure from manifolds. As has been mentioned, spectra came after the fact, as the idea was germinating around that time. In Thom's paper he largely phrased things in the language of stable homotopy groups.<|endoftext|> TITLE: Breakthroughs in mathematics in 2021 QUESTION [112 upvotes]: This is somehow a general (and naive) question, but as specialized mathematicians we usually miss important results outside our area of research. So, generally speaking, which have been important breakthroughs in 2021 in different mathematical disciplines? REPLY [7 votes]: Since other answers mention works published in 2021, I think one can add to the list the proof of triviality of the $\phi^4$ quantum field theory in four dimensions: Michael Aizenman, Hugo Duminil-Copin, "Marginal triviality of the scaling limits of critical 4D Ising and $\lambda\phi_4^4$ models", Ann. of Math. (2) 194(1): 163-235 (July 2021).<|endoftext|> TITLE: Subgroups of the multiplicative group of a finite field satisfying a certain additive property QUESTION [7 upvotes]: Let $G \subseteq \mathbb F_p^*$ be a subgroup. Then $G$ is called almost trivial if $G \cap (2-G)$ consists of the element 1. Then I am wondering how big $G$ can be in terms of $p$. If $G$ is a random set containing 1 then according to the birthday paradox one expects that $G$ and $2-G$ have nontrivial intersection as soon as $\#G \gg \sqrt p$. I am wondering if maybe one could prove that all almost trivial subgroups of $\mathbb F_p^*$ have size at most $\tilde O(\sqrt p)$. Or if a $\sqrt p$ up to log factors bound is out of reach, I was wondering if there is $1/2 < \epsilon < 1$ and some constant $c$ such that for all almost trivial groups we have $\#G < cp^\epsilon$. I did some numeric experimentation by finding the largest almost trivial subgroup of $\mathbb F_p^*$ for all p < 80000. And the largest ratio that I could find for $\frac {\#G}{\sqrt p}$ was $\frac {884}{\sqrt {41549}} = 4.3368\ldots$ because $\mathbb F_{41549}$ contains an almost trivial subgroup of order 884, so a square root (possibly up to log factors) bound seams reasonable. P.S. The notion of being almost trivial is based on the notion of almost rational in the paper Almost rational torsion points on semistable elliptic curves by Frank Calegari. REPLY [13 votes]: If we let $S$ be the set of characters of $\mathbb F_p^\times$ trivial on $G$ then $$\sum_{\chi \in S} \chi(g) = \begin{cases} \frac{p-1}{|G|} & g\in G \\ 0 & g\notin G \end{cases}$$ so $$\sum_{\chi_1,\chi_2\in S} \sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g) = 0$$ if $G$ is almost trivial. Now if $\chi_1 =\chi_2=1$ then $\sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g)=p-3$ and otherwise, by the bound for Jacobi sums, $\left| \sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g) \right| \leq \sqrt{p}+1 $. Thus, if $G$ is almost trivial then $p-3 \leq ( ((p-1)/|G|)^2-1) (\sqrt{p}+1)$, or $p \leq c^2 p^{5/2} / |G|^2$ for a constant $c$, meaning $|G| \leq c p^{3/4}$, answering the weaker form of your question.<|endoftext|> TITLE: An extension of the Galois theory of Grothendieck QUESTION [18 upvotes]: This question is about Joyal and Tierney's famous An extension of the Galois theory of Grothendieck. One of the main results states (see the MathSciNet review by Peter Johnstone): Joyal and Tierney's theorem. Each Grothendieck topos is equivalent to the topos of equivariant sheaves on a groupoid in the category of locales. Question 1. What does this statement have to do with Grothendieck's Galois theory? I have to admit, I don't know much about Grothendieck's Galois theory and which kind of theorems it includes. In my mind, I usually identify the term "Grothendieck's Galois theory" with the following theorem (see Chua - The étale fundamental group for a concise introduction and SGA 1 for a complete treatment): Grothendieck's theorem. Let $X$ be a scheme and $x\in X$. Then the category of finite étale covers of $X$ is equivalent to the category of finite continuous $\pi_1(X,x)$-sets, where $\pi_1(X,x)$ is the étale fundamental group of $X$ at the base point $x$. This is related to Galois theory because whenever $k$ is a field, then the étale fundamental group of $\operatorname{Spec}(k)$ is the absolute Galois group of $k$. A more precise version of Question 1 would be: does Joyal and Tierney's theorem imply Grothendieck's theorem? (But an answer showing that Joyal and Tierney's theorem implies a modified version of Grothendieck's theorem would satisfy me as well.) Question 2. In an answer by Zhen Lin to "What does a proof in an internal logic actually look like?" it is mentioned that the proof of Joyal and Tierney's theorem uses the internal language of toposes. In which way does it use the internal language? Which kind of theorems are proved internally, how are they interpreted externally, and why are the external interpretations of these theorems helpful in proving Joyal and Tierney's theorem? Skimming the text I don't see any logic at all. REPLY [16 votes]: The point of view where this title comes from is that Grothendieck's theorem can be seen as a characterization of toposes of the form $BG$ for $G$ a profinite group. It shows that some toposes can be represented as $BG$. I think before Joyal–Tierney's paper it was also known how to generalize from profinite group to general localic groups. Joyal–Tierney's theorem shows that if you replace "pro-finite group" by "localic groupoid" then you actually get all Grothendieck toposes this way. You can't directly recover Grothendieck's theorem from Joyal–Tierney's theorem in the sense that the theorem as stated above doesn't tell you for which toposes the localic groupoid can be chosen to be a profinite group. But if you are familiar with the method used in the paper and how the groupoid is obtained (which in my opinion are even more important than the theorem itself) then it is fairly easy to recover Grothendieck's theorem. For example, it immediately follows from Joyal–Tierney's paper that a topos is of the form $BG$ for $G$ a localic groups if and only if it admits a point $* \to \mathcal{T}$ which is an open surjection (which does feel similar to Grothendieck's theorem in terms of a fiber functor). Regarding the use of internal logic, it is definitely not essential, it just makes everything simpler (at least if you are ok with its use) but one could do without it. The main way they use internal logic is that in the first sections they prove some results about sup-lattice and frames locales, that are later applied not to sup-lattices and frames, but to sup-lattices and frames in a topos $\mathcal{T}$. I believe there are also a few places where they make a claim about a morphism of locales $f:X \to Y$ and then only prove it when $Y$ is the point (sorry I don't have the paper with me to give precise reference). Another place where one can consider they use a bit of internal logic — though this one might be only at the level of intuition — is when they show that every topos admits an open cover by a locale. They do this by considering the topos as a classifying topos of some theory $T$ and considering the propositional theory $T'$ of "enumerated $T$-models", that is, $T$-models that are explicitly given as a subquotient of the natural numbers. Though if I remember correctly, they present the argument in a way that doesn't directly involve any logic… (and in any case there are other proofs of this results that are purely in terms of sites, for example the one in MacLane and Moerdijk's book "Sheaves in geometry and logic"). REPLY [8 votes]: $\DeclareMathOperator\Sh{Sh}$This is not an answer, but is too long for a comment — it is about question 1 specifically. The point is that you'll want to consider the finite étale topos of $X$, namely you take as a site the finite étale maps $Y\to X$, and étale covers as covers; and take sheaves on this site : $\Sh(X^\text{f,ét})$. This is a Grothendieck topos, and by Joyal–Tierney's result, it is sheaves on a localic groupoid $\mathcal G_X$. Now the idea is that $\mathcal G_X$ is supposed to be $B\pi_1^\text{ét}(X,x)$ if $X$ is connected and $x$ is a point of $X$. Indeed, suppose you know Grothendieck's theorem. Then the site I described is the site $\pi_1^\text{ét}(X,x)\text-\mathrm{FinSet}$ with topology given by the effective epimorphism topology, i.e. surjections of underlying (finite) sets. Now in general, if $G$ is a profinite group, and $BG$ the associated localic groupoid, then sheaves on $BG$ are the same as sheaves on $G\text-\mathrm{FinSet}$ with the effective epimorphism topology, and they're the same as continous $G$-sets. Ok, now, can you deduce Grothendieck's theorem from Joyal and Tierney's? I'm not sure, but here's what you can try: Start by proving that if $X$ is connected, so must be $\mathcal G_X$ — this should already bring you closer to a profinite group. Find a categorical interpretation of finite $\pi_1^\text{ét}(X,x)$-sets among all sheaves on the site (I think here "coherent" should be the relevant word, but I'm not sure), and try to see, for a localic groupoid $\mathcal G$, what this categorical interpretation corresponds to in sheaves on $\mathcal G$. Find some properties that $\Sh(X^\text{f,ét})$ "trivially" has that force $\mathcal G_X$ to be a profinite groupoid rather than an arbitrary localic groupoid (by "trivially" here I mean: don't reprove Joyal–Tierney and Grothendieck). If you can do these 3 things in a suitable way, you should be able to deduce Grothendieck's theorem from Joyal–Tierney's.<|endoftext|> TITLE: Major applications of the internal language of toposes QUESTION [14 upvotes]: What are the major applications of the internal language of toposes? Here are a few applications I know: Mulvey's proof of the Serre–Swan theorem in which he interprets the intuitionistically valid result of Kaplansky that each finitely generated projective module over a local ring has a finite basis in sheaf toposes. See Mulvey's paper Intuitionistic Algebra and Representations of Rings. Blechschmidt's proof of Grothendieck's generic freeness lemma. See his PhD thesis. In Sheaves in Geometry and Logic, Mac Lane and Moerdijk use the internal language in the proof that the elementary theory of the category of sets and bounded Zermelo set theory with choice are equiconsistent. Are there other big theorems in mathematics whose proofs use the internal language of toposes (or other types of categories like cartesian closed categories)? Some people might object with my usage of "uses", since each proof using the internal language can mechanically be rewritten so that it doesn't use the internal language anymore (by unwinding the interpretation of the internal language). In this case I want to put the question as: are there other big theorems in mathematics whose proofs benefit (e.g., increase the chance readers will understand the proof) from presenting it using the internal language of toposes? Let's create a list! REPLY [14 votes]: I don't know if this counts as an application of the internal language or as an avoidance of it, but I think it is worth listing anyway. In the development of homological algebra and homotopy theory in a Grothendieck topos, there is a technique of reducing proofs to the classical case by either: assuming that there are enough points and taking stalks everywhere, or applying Barr's theorem and passing to a Grothendieck topos that (is boolean and) satisfies the (external) axiom of choice. (I mention homological algebra and homotopy theory here because, if I recall correctly, the former strategy was used – in the first instance – in the foundations of those topics, before the development of "direct" proofs.) The first strategy is completely rigorous and unobjectionable except perhaps for its lack of generality. Because it reduces questions of truth (of certain propositions preserved by geometric morphisms and reflected by surjective geometric morphisms) to literally what happens in $\textbf{Set}$, it is unnecessary to know how the original theorems are proved (in $\textbf{Set}$); it is enough to know that they are true. In a sense, this is an avoidance of reasoning in the internal language. Instances of this can be found in: Artin and Mazur [1969, Étale homotopy], Verdier [1972, SGA 4 Exposé V], Brown [1973, Abstract homotopy theory and generalized sheaf cohomology], ... On the other hand, the only way I see of making rigorous the second strategy is to interpret it as an implicit claim that, for the theorems in question, there are classical proofs that can soundly interpreted not just in $\textbf{Set}$ but also any Grothendieck topos that satisfies the axiom of choice. In other words, classical theorems cannot be treated as black boxes for this strategy, and this is an implicit claim that it is possible to reason in the internal language (of a Grothendieck topos that satisfies the axiom of choice) the same way as in $\textbf{Set}$. Instances of this can be found in: Joyal [1984, Letter to Grothendieck], Jardine [1996, Boolean localization, in practice], ... It is also possible to continue avoiding having to think about proofs, in the fashion of the first strategy, by being more careful about what structures one really needs preserved/reflected to transfer theorems. For example, van Osdol [1977, Simplicial homotopy in an exact category] extended Brown's theory from toposes of sheaves on topological spaces to exact categories (including all Grothendieck toposes), essentially by observing that because the definitions only involve finite limits and regular epimorphisms, it is enough to assume that there are enough exact functors to $\textbf{Set}$, but this is guaranteed by (another theorem of) Barr. In a sense, this is even more of an avoidance of reasoning in the internal language: the proper internal language of exact categories is much more restrictive than that of a topos, so a proof in the internal language would be much more constrained than a proof in classical mathematics. But I think this still demonstrates the benefit of formulating propositions using the internal language, even if one does not formulate proofs in it.<|endoftext|> TITLE: Asymptotic behavior of $\sum_{k=1}^{\infty} \sqrt{\max\{1 - k^2/x^2,0\}}$ as $x\to\infty$ QUESTION [7 upvotes]: Let $f:\mathbb{R}\to\mathbb{R}$ be the function $$f(x) = \sum_{k=1}^{\infty} \sqrt{\max\left\{1 -\frac{k^2}{x^2},0\right\}}.$$ Numerical experiments suggests that there exists $n\in\mathbb{N}$ and a $1$-periodic function $p:[n,\infty)\to\mathbb{R}$ such that $$f(x) = - \frac{1}{2} + \frac{\pi}{4}x + \frac{p(x)}{\sqrt{x}}$$ for every $x\geq n$. Question: Can this be true? Edit (26.12.2021) Below is the portion of the graph of $p$. REPLY [4 votes]: This can not be true: $p(m+1)=p(m)$ for an integer $m$ yields that $\pi$ is Algebraic.<|endoftext|> TITLE: What is the name of this relative semidirect product of groups? QUESTION [21 upvotes]: We have two well known definitions of the semidirect product $N \rtimes H$ of groups: (Internal semidirect product) We write $G = N \rtimes H$ if $N$ is a normal subgroup of $G$, $H$ is another subgroup of $G$, $N \cap H = \{1\}$, and $G = NH$. (External semidirect product) Given a homomorphism $\phi: H \to \mathrm{Aut}(N)$ from $H$ to the automorphism group of $N$, denoted $h \mapsto \phi_h$, we write $N \rtimes H = N \rtimes_\phi H$ to be the set of pairs $(n,h)$ with $n \in N, h \in H$ and group law $(n_1,h_1) (n_2,h_2) = (n_1 \phi_{h_1}(n_2), h_1 h_2)$. The two are equivalent in the sense that every external semidirect product is an internal semidirect product (identifying $N$ with $\{ (n,1): n \in N\}$ and $H$ with $\{(1,h): h \in H\}$) and conversely every internal semidirect product is canonically isomorphic to an external semidirect product. I recently had occasion to use the version of this concept when the two groups $N,H$ are allowed to intersect [EDIT: in a group normal in $H$], which then imposes two additional compatibility conditions on the homomorphism $\phi$, but I was unable to locate the standard name for the concept, which I will for the sake of this question call the "semidirect product $N \rtimes_K H$ of $N,H$ relative to $K$". It can be defined internally or externally: (Internal relative semidirect product) We write $G = N \rtimes_K H$ if $N$ is a normal subgroup of $G$, $H$ is another subgroup of $G$, $N \cap H = K$ is a normal subgroup of $H$, and $G = NH$. (External relative semidirect product). Given a commuting square $\require{AMScd}$ \begin{CD} K @>> \iota > N\\ @V \iota V V @VV V\\ H @>>\phi> \mathrm{Aut}(N) \end{CD} of homomorphisms using the conjugation action of $N$ on itself [EDIT: with $\iota$ denoting inclusion maps, and obeying the additional compatibility condition $$ \phi_h(k) = h k h^{-1}$$ for all $h \in H$ and $k \in K$, which among other things makes $K$ a normal subgroup of $H$] we write $N \rtimes_{K,\phi} H = N \rtimes_{K,\phi} H = N \rtimes_K H$ to be the quotient of $N \rtimes_\phi H$ by $\{ (k, k^{-1}): k \in K \}$ (which one can check to be a normal subgroup of $N \rtimes_\phi H$). One can verify that every external relative semidirect product is an internal relative semidirect product, and conversely that every internal relative semidirect product is isomorphic to an external relative semidirect product. This is such a basic operation that it must already be well known, so I ask Question 1: What is the term for this operation in the literature? Also, Question 2: Does this operation have a category-theoretic interpretation (e.g., as the universal object for some diagram)? EDIT: One can also think of $N \rtimes_{K,\phi} H$ as the amalgamated free product $N \ast_K H$ quotiented by the relations $h n h^{-1} = \phi_h(n)$ for $h \in H$, $n \in N$. So maybe "amalgamated semidirect product" could be a better name than "relative semidirect product". REPLY [7 votes]: This construction appears in the literature as Remark 1.4.5 in "Pseudo-reductive groups" by Conrad, Gabber and Prasad. There, it is called a 'non-commutative pushout'. It is a generalization of their 'standard construction' which produces most of the pseudo-reductive groups.<|endoftext|> TITLE: Is every set being cardinal definable consistent with ZF + negation of Choice? QUESTION [11 upvotes]: Recall the definition of cardinal definable, where every set being cardinal definable is proved consistent relative to ZF + V=HOD. To re-iterate it: $Define: X \text { is cardinal definable} \iff \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda_1,.., \lambda_n <^\rho \kappa \ \exists \phi : \\ X=\{ y \in V_{\rho(\kappa)} \mid \phi^{V_{\rho (\kappa)}} (y,\lambda_1,..,\lambda_n)\}$ Where: $\lambda_i <^\rho \kappa \iff \rho(\lambda_i) < \rho(\kappa)$, and $\rho$ is the rank function; and "cardinal" is defined after Scott's as an equivalence class under bijection of sets of the lowest possible rank. Now, is the principle stating that every set is cardinal definable consistent with $\sf ZF + \neg AC$? A related question replacing cardinal by ordinal in the above question would lead to $\sf V=HOD$, which is known to prove $\sf AC$ and so would be inconsistent with $\sf ZF + \neg AC$. REPLY [12 votes]: This is consistent. Kanovei constructed a model $M$ with an infinite Dedekind finite set of reals which is lightface projectively definable. By descending to $L(R),$ we can further assume it satisfies $V=L(R).$ Clearly choice fails in this model. Since it satisfies $V=L(R),$ every set is definable from an ordinal and a real. Of course, any ordinal is definable from an $\aleph,$ so we just need to check that every real is cardinal definable. Let $A$ be the definable Dedekind finite set of reals, and fix a canonical surjection $f: A \rightarrow \omega$ (as exists for any infinite set of reals). Let $\langle q_n \rangle$ be an enumeration of the rationals. Define $A_r=f^{-1}(\{n: q_n TITLE: What classes of groups can arise as "symmetry groups of terms"? QUESTION [9 upvotes]: Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$ Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way. Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here): Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$? Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder. Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups? I really have no relevant information for this question, but it seems like a natural one to ask. Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version. Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details): If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group. If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$. Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here. One of the few negative results I know is that the class of finite cyclic groups is not of the form $\mathbb{G}(\mathfrak{A})$ for any $\mathfrak{A}$. To see this, suppose $t(x_1,...,x_k)$ is a $\mathfrak{A}$-term with $G_\mathfrak{A}(t)\cong C_2$. Then the term $$s(x_{1,1},...,x_{k,k}):=t(t(x_{1,1},...,x_{1,k}), ..., t(x_{k,1},...,x_{k,k}))$$ has $G_\mathfrak{A}(s)$ having too many elements of order $2$ to be cyclic. In fact, tweaking this argument we get that $\mathbb{G}(\mathfrak{A})$ consists entirely of cyclic groups iff $\mathbb{G}(\mathfrak{A})$ consists only of trivial groups. However, this sort of idea doesn't seem to be useful for either Q1 or Q2 above. REPLY [13 votes]: Let me edit this response in order to clarify what I am showing. First, I will begin with an example: $t(x_1,x_2,x_3,x_4,x_5,x_6) = ((((x_1x_2)x_3^{-1})x_3)x_4)x_4^{-1}$ is a group term. If you want to associate this term to a particular group, let it be the free group $F_6$ on $\{x_1,\ldots,x_6\}$. The term $t$ depends on $x_1$ and $x_2$. The variables $x_3$ and $x_4$ occur, but $t$ does not depend on them. The variables $x_5$ and $x_6$ do not occur. The permutations of indices that are symmetries of the original term $t$ are those from the full symmetric group $\textrm{Sym}(\{3,4,5,6\})$. One may eliminate fictitious variables by defining a term $s(x_1,x_2,x_3,x_4) = t(x_1,x_2,x_3,x_4,x_4,x_4)$. The terms $s$ and $t$ are the same, all variables of $s$ occur in $s$, and $s$ does not depend on its last two variables. The symmetries of $s$ are $\textrm{Sym}(\{3,4\})$. The term $r(x_1,x_2) = t(x_1,x_2,x_2,x_2,x_2,x_2)$ depends on all of its variables, which are the same variables that $t$ depends on, and $r$ acts the same way as $t$ with respect to those variables. The symmetry group of $r$ is trivial. More generally, suppose that $t(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n)$ depends on $x_1$--$x_i$, does not depend on $x_{i+1}$--$x_j$ although these variables occur, and the variables $x_{j+1}$--$x_n$ do not occur. The term $r(x_1,\ldots,x_i):=t(x_1,\ldots,x_i,x_i,\ldots,x_i)$ will be a term that depends on all variables and will have symmetry group equal to some $G\leq S_i$. The term $s(x_1,\ldots,x_j):=t(x_1,\ldots,x_i,\ldots,x_j,x_j,\ldots,x_j)$ will be a term equal to $t$, where all variables occur, and will have the symmetry group $G\times S_{j-i}$. The original term $t(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n)$ will have the symmetry group $G\times S_{n-i}$. The symmetry groups of terms are determined, up to symmetric group factors, by the symmetry groups of just those terms that depend on all variables. Moreover, the symmetric group factors can be determined if you know which variables of the term occur and which variables the term depends on. There are various questions that could be asked, such as What are the possible symmetry groups of terms? (Answer below: any concrete subgroup of $S_n$ is the symmetry group of an $n$-ary term that depends on all variables.) What are the possible classes of permutation groups on finite sets which are the symmetry groups of those terms of an algebra (i) which depend on all variables? (ii) in which all variables occur? (iii) are arbitrary? The question asked is (ii), but I will answer (i) here. I am going to explain why, if $\mathcal K$ is any class of groups whose members are subgroups of finite symmetric groups and $\mathcal K$ contains $S_1$, then there is an algebra $A$ such that the symmetry groups $G(t)$ of terms $t$ of $A$ that depend on all variables are exactly the groups in $\mathcal K$. (More precisely, if $G\in {\mathcal K}$ is a subgroup of $S_n$, then we will realize $G$ and all of its conjugates in $S_n$ as symmetry groups of terms of $A$ which depend on all variables.) We need to include $S_1$ in $\mathcal K$ because $t(x)=x$ has symmetry group $S_1$. For a given $G\in\mathcal K$, where $G\leq S_n$, define an $n$-ary operation $f_G(x_1,\ldots,x_n)$ on the set $\{a, b\}\cup \mathbb N^+ = \{a, b, 1, 2, 3, \ldots\}$ as follows: $$ f_G(\bar{u}) = \begin{cases} a & \textrm{if $\bar{u} = (\sigma(1),\ldots,\sigma(n))$ for some $\sigma\in G$;}\\ b & \textrm{otherwise.} \end{cases} $$ Let $A$ be the algebra on $\{a,b\}\cup \mathbb N^+$ whose operations are all operations of the form $f_G$, $G\in\mathcal K$. I claim that the following are true. Each $f_G$ depends on all of its variables. The symmetry group of $f_G(x_1,\ldots,x_n)$ is $G\;(\leq S_n)$. any term operation of $A$ that depends on all of its variables is obtained from one of the $f_G$'s by permuting the variables. Hence, its symmetry group is conjugate to $G$ in $S_n$. If $f_G(x_1,\ldots,x_n)$ is defined as above, and we identify two variables: $f_G(\underline{x_2},\underline{x_2},x_3,x_4,\ldots,x_n)$, then this operation is constant and its variable-set is $\{x_2,\ldots,x_n\}$. Its symmetry group in Noah's sense is $S_{n-1}$. Similarly, the composite $f_G(f_G(x_1,\ldots,x_n),y_2,\ldots,y_n)$ is constant and its variable-set is $\{x_1,\ldots,x_n,y_2,\ldots,y_n\}$, so its symmetry group is $S_{2n-1}$. Using tricks like this one can obtain every finite symmetric group as a symmetry group of a term of $A$, as long as $A$ has at least one operation $f_G$ of arity greater than $1$. The isomorphism types of Noah-symmetry groups turn out to be exactly those in $\mathcal K\cup \{S_n\;|\;n=1, 2, \ldots\}$.<|endoftext|> TITLE: Density of matrix coefficients of unitary representations of a locally compact group QUESTION [5 upvotes]: Let $G$ be a locally compact group, $C_0(G)$ the $C^*$-algebra of continuous functions on $G$ that vanish at infinity, $C_b(G)$ the $C^*$-algebra of bounded continuous functions on $G$. We know that $C_b(G)$ is the multiplier algebra of $C_0(G)$, and we denote the strict topology on $C_b(G) = \mathcal{M}\bigl( C_0(G) \bigr)$ by $\beta$. Now for a strongly continuous unitary representation $(\pi, H)$ of $G$, functions of the form $\omega_{\pi,\eta,\xi} : g \in G \to (\pi(g)\eta \mid \xi) \in \mathbb{C}$ are in $C_b(G)$, and we call them matrix coefficients of the representation $\pi$. Since we can form direct sum and tensor product of two strongly continuous unitary representations, as well as the contragredient representation, we see that (the linear span of) matrix coefficients of all strongly continuous unitary representations of $G$ form a $*$-subalgebra $A_0(G)$ of $C_b(G)$. Question. Is $A_0(G)$ strictly dense in $C_b(G)$, i.e. with respect to the $\beta$-topology? Note that in the compact case, $C_b(G) = C_0(G) = C(G)$ and the $\beta$-topology is the same as the norm topology on the $C^*$-algebra $C(G)$ of continuous functions on $G$. In this case, The answer to the question is affirmative by Peter-Weyl. In the case where $G$ is discrete, then one can check easily that all finitely supported functions on $G$ are already matrix coefficients of the left (or right) regular representation, so the answer to the question is again affirmative. Based on these considerations, here are some sub-questions with some bias on their possible answers. Q1. Does the question have an affirmative answer for unimodular $G$? Q2. Can we construct some counter-example for non-unimodular $G$? Q3. Does the question have an affirmative answer if $G$ is a real Lie group? What if the Lie group $G$ is nilpotent, or solvable, or semisimple/reductive? REPLY [8 votes]: First, some remarks that may help with literature-searching.$\newcommand{\fsnorm}[1]{{\Vert#1\Vert}_{\rm B}}$ $\newcommand{\supnorm}[1]{{\Vert#1\Vert}_\infty}$ The algebra you have denoted by $A_0(G)$ is known as the Fourier--Stieltjes algebra of $G$, and is usually denoted by $B(G)$, so I will do that from now on. $B(G)$ has been much studied: it turns out to be complete in a natural submultiplicative norm $\fsnorm{\cdot}$, so it is a commutative Banach algebra of functions on $G$; in fact it is also a dual Banach space for this norm, and multiplication is separately weak-star continuous, so it is an example of a dual Banach algebra. This norm dominates the supremum norm, so $\fsnorm{\cdot}$-convergence implies $\supnorm{\cdot}$-convergence. If one takes the $\fsnorm{\cdot}$-closure inside $B(G)$ of $B(G)\cap C_c(G)$, the resulting algebra is an ideal in $B(G)$ called the Fourier algebra of $G$, usually denoted by $A(G)$. One can show that $A(G)$ is a $\supnorm{\cdot}$-dense subalgebra of $C_0(G)$ that is closed under conjugation and separates points of $G$. (Aside: in general $B(G)\cap C_0(G)$ is larger than $A(G)$.) Given some $f\in C_b(G)$, an $\varepsilon>0$, and a finite set $E\subset C_0(G)$, it suffices to find $h\in B(G)$ such that $\supnorm{fg-hg}\leq \varepsilon$ for all $g\in E$. Since $C_0(G)$ is $\beta$-dense in $C_b(G)$ we can find $k \in C_0(G)$ such that $\supnorm{ fg-kg } \leq \varepsilon/2$ for all $g\in E$. Since $B(G)\cap C_c(G)$ is $\supnorm{\cdot}$-dense in $C_0(G)$ we can find $h\in B(G)\cap C_c(G)$ such that $\supnorm{k-h} \max_{g\in E} \supnorm{g} \leq\varepsilon/2$. Then $$ \supnorm{ fg -hg} \leq \supnorm{fg-kg} + \supnorm{k-h}\supnorm{g} \leq \varepsilon \quad\hbox{for all $g\in E$,} $$ as required.<|endoftext|> TITLE: Flatness of the closure of a closed subgroup of the generic fiber of an algebraic group inside an integral model of the ambient group QUESTION [6 upvotes]: $\DeclareMathOperator\GL{GL}$Let $K$ be a number field and $R$ its ring of integers. Let $G$ be a connected reductive closed subgroup of $\GL_{n,K}$. On p55 of Brian Conrad's notes Reductive group schemes, he claims that the schematic closure $\overline{G}$ of $G$ inside $\GL_{n,R}$ is a flat closed $R$-subgroup of $\GL_{n,R}$. First, I assume by schematic closure, he means the reduced induced structure on the topological closure. I can see that this closure $\overline{G}$ is an $R$-subgroup of $\GL_{n,R}$, and that it satisfies $\overline{G}(R) = \{g\in \GL_n(R) : g\cap \GL_{n,K}\in G\}$, where in the brackets we view $g$ as a closed subscheme of $\GL_{n,R}$. Why is $\overline{G}$ necessarily flat? (Is this obvious?) Is this also true if $R$ is an arbitrary domain? REPLY [9 votes]: $\DeclareMathOperator\GL{GL}$A colleague asked me this question some time ago and here is the answer I sent him. Let $R$ be a domain with fraction field $K$. Let $A$ be the $R$-algebra underlying the group scheme $\GL_{n,R}$ over $R$. Suppose given a closed subgroup scheme $H$ of $\GL_{n,K}$. Let $B$ be the underlying $K$-algebra of $H$. We are then given by assumption a surjection $\phi:A_K\to B$ and an injection $\lambda:A\to A_K$ (given by $a \mapsto a\otimes_K 1$). Let $\mu = \phi\circ\lambda$ and consider the $R$-module $\mu(A)$, which is a sub-$R$-module of $B$. The following fact can be checked from the definitions: $\mu(A)$ has a natural $R$-algebra structure, compatible with the $R$-algebra structure of B via $\lambda$. In fact the R-algebra $\mu(A)$ is (by definition) the underlying $R$-algebra of the scheme-theoretic image of $H$ in $\GL_{n,R}$. Furthermore, we have $\mu(A)_K\simeq B$ and $\mu(A)$ is (clearly) torsion free. Now consider the Hopf algebra structure of B, which is given by a morphism of $K$-modules $c:B\to B\otimes_K B$. It is easy to see (diagram chasing) that if $x\in \mu(A)\subseteq B$, then $c(x)$ lies in the image of the natural map $\mu(A)\otimes_R\mu(A)\to B\otimes_K B$. So if the natural map $\mu(A)\otimes_R\mu(A)\to B\otimes_K B$ is an injection, we obtain a natural map $\mu(A)\to \mu(A)\otimes_R\mu(A)$ and this then (checking this is elementary) defines a Hopf algebra structure on $\mu(A)$. In particular, if the natural map $\mu(A)\otimes_R\mu(A)\to B\otimes_K B$ is an injection, then the scheme-theoretic closure of $H$ in $\GL_{n,R}$ is indeed a subgroup scheme, which is reduced if $H$ is reduced (which is what would happen if $K$ is of characteristic 0). So the basic condition, which must be satisfied, is that the natural map $\mu(A)\otimes_R\mu(A)\to B\otimes_K B$ is injective. This will be true iff $\mu(A)\otimes_R\mu(A)$ is torsion free. For example, if $\mu(A)$ happens to be flat over $R$ then $\mu(A)\otimes_R\mu(A)$ will also be flat and thus torsion free. This is what will happen if $R$ is a Dedekind domain (where torsion free is equivalent to flat) or if $R$ is a valuation ring (where again torsion free is equivalent to flat). So if we work over either of these two rings, then the scheme-theoretic closure of $H$ in $\GL_{n,R}$ is indeed a subgroup scheme and it is even flat over $R$.<|endoftext|> TITLE: Question on OEIS A000085 QUESTION [7 upvotes]: The OEIS sequence A000085 is defined by $$ a_n \!=\! (n-1)a_{n-2} + a_{n-1} \;\text{with }\; a_0\!=\!1, a_1\!=\!1.$$ If $n$ of the form $b^2-b+1, b \in \mathbb{N}, b > 2, \;\text{then: }\;$ $$ \left\lfloor \frac{a_n}{a_{n-1}} \right\rfloor > \left\lfloor \frac{a_{n-1}}{a_{n-2}} \right\rfloor$$ How to prove this? REPLY [3 votes]: Let $q_n$ be the quotient $$q_n=\frac{a_n}{a_{n-1}}\ .$$ It may be useful to see the numerical values for the first terms of the $q$-sequence: $$ \scriptsize \begin{array}{|r|l||r|l||r|l||r|l||r|l|} \hline n & q(n) & n & q(n) & n & q(n) & n & q(n) & n & q(n) \\\hline 1 & 1.0000000 & 41 & 6.8807053 & 81 & 9.4846377 & 121 & 11.4876422 & 161 & 13.1779753 \\\hline 2 & \color{blue}{ 2.0000000} & 42 & \color{blue}{ 6.9586914} & 82 & 9.5401259 & 122 & 11.5330579 & 162 & 13.2173548 \\\hline 3 & \color{red}{ 2.0000000} & 43 & \color{red}{ 7.0356175} & 83 & 9.5952744 & 123 & 11.5782873 & 163 & 13.2566128 \\\hline 4 & 2.5000000 & 44 & 7.1117592 & 84 & 9.6500913 & 124 & 11.6233328 & 164 & 13.2957503 \\\hline 5 & 2.6000000 & 45 & 7.1869362 & 85 & 9.7045808 & 125 & 11.6681966 & 165 & 13.3347684 \\\hline 6 & \color{blue}{ 2.9230769} & 46 & 7.2613608 & 86 & 9.7587503 & 126 & 11.7128808 & 166 & 13.3736682 \\\hline 7 & \color{red}{ 3.0526316} & 47 & 7.3349008 & 87 & 9.8126038 & 127 & 11.7573877 & 167 & 13.4124509 \\\hline 8 & 3.2931034 & 48 & 7.4077213 & 88 & 9.8661482 & 128 & 11.8017192 & 168 & 13.4511174 \\\hline 9 & 3.4293194 & 49 & 7.4797254 & 89 & 9.9193876 & 129 & 11.8458774 & 169 & 13.4896687 \\\hline 10 & 3.6244275 & 50 & 7.5510426 & 90 & \color{blue}{ 9.9723281} & 130 & 11.8898645 & 170 & 13.5281060 \\\hline 11 & 3.7590564 & 51 & 7.6216021 & 91 & \color{red}{10.0249738} & 131 & 11.9336822 & 171 & 13.5664302 \\\hline 12 & \color{blue}{ 3.9262662} & 52 & 7.6915065 & 92 & 10.0773304 & 132 & \color{blue}{11.9773327} & 172 & 13.6046423 \\\hline 13 & \color{red}{ 4.0563388} & 53 & 7.7607042 & 93 & 10.1294019 & 133 & \color{red}{12.0208177} & 173 & 13.6427433 \\\hline 14 & 4.2048605 & 54 & 7.8292772 & 94 & 10.1811936 & 134 & 12.0641392 & 174 & 13.6807341 \\\hline 15 & 4.3294803 & 55 & 7.8971884 & 95 & 10.2327092 & 135 & 12.1072989 & 175 & 13.7186157 \\\hline 16 & 4.4646191 & 56 & \color{blue}{ 7.9645039} & 96 & 10.2839538 & 136 & 12.1502987 & 176 & 13.7563891 \\\hline 17 & 4.5837324 & 57 & \color{red}{ 8.0311975} & 97 & 10.3349311 & 137 & 12.1931404 & 177 & 13.7940551 \\\hline 18 & 4.7087680 & 58 & 8.0973227 & 98 & 10.3856456 & 138 & 12.2358257 & 178 & 13.8316147 \\\hline 19 & 4.8226559 & 59 & 8.1628614 & 99 & 10.4361009 & 139 & 12.2783562 & 179 & 13.8690687 \\\hline 20 & \color{blue}{ 4.9397378} & 60 & 8.2278576 & 100 & 10.4863015 & 140 & 12.3207337 & 180 & 13.9064181 \\\hline 21 & \color{red}{ 5.0487979} & 61 & 8.2922992 & 101 & 10.5362507 & 141 & 12.3629597 & 181 & 13.9436637 \\\hline 22 & 5.1594060 & 62 & 8.3562227 & 102 & 10.5859526 & 142 & 12.4050360 & 182 & \color{blue}{13.9808064} \\\hline 23 & 5.2640568 & 63 & 8.4196203 & 103 & 10.6354106 & 143 & 12.4469640 & 183 & \color{red}{14.0178471} \\\hline 24 & 5.3692538 & 64 & 8.4825227 & 104 & 10.6846284 & 144 & 12.4887454 & 184 & 14.0547865 \\\hline 25 & 5.4698949 & 65 & 8.5449253 & 105 & 10.7336094 & 145 & 12.5303816 & 185 & 14.0916254 \\\hline 26 & 5.5704717 & 66 & 8.6068541 & 106 & 10.7823571 & 146 & 12.5718742 & 186 & 14.1283648 \\\hline 27 & 5.6674683 & 67 & 8.6683071 & 107 & 10.8308746 & 147 & 12.6132247 & 187 & 14.1650054 \\\hline 28 & 5.7640320 & 68 & 8.7293062 & 108 & 10.8791653 & 148 & 12.6544344 & 188 & 14.2015481 \\\hline 29 & 5.8577107 & 69 & 8.7898516 & 109 & 10.9272322 & 149 & 12.6955049 & 189 & 14.2379934 \\\hline 30 & \color{blue}{ 5.9507395} & 70 & 8.8499619 & 110 & \color{blue}{10.9750786} & 150 & 12.7364375 & 190 & 14.2743424 \\\hline 31 & \color{red}{ 6.0413902} & 71 & 8.9096386 & 111 & \color{red}{11.0227073} & 151 & 12.7772336 & 191 & 14.3105957 \\\hline 32 & 6.1312693 & 72 & \color{blue}{ 8.9688979} & 112 & 11.0701214 & 152 & 12.8178946 & 192 & 14.3467540 \\\hline 33 & 6.2191477 & 73 & \color{red}{ 9.0277422} & 113 & 11.1173236 & 153 & 12.8584218 & 193 & 14.3828182 \\\hline 34 & 6.3061933 & 74 & 9.0861857 & 114 & 11.1643169 & 154 & 12.8988164 & 194 & 14.4187889 \\\hline 35 & 6.3915252 & 75 & 9.1442315 & 115 & 11.2111039 & 155 & 12.9390799 & 195 & 14.4546668 \\\hline 36 & 6.4760013 & 76 & 9.2018921 & 116 & 11.2576875 & 156 & \color{blue}{12.9792134} & 196 & 14.4904528 \\\hline 37 & 6.5589859 & 77 & 9.2591710 & 117 & 11.3040700 & 157 & \color{red}{13.0192183} & 197 & 14.5261474 \\\hline 38 & 6.6411159 & 78 & 9.3160793 & 118 & 11.3502543 & 158 & 13.0590957 & 198 & 14.5617514 \\\hline 39 & 6.7219299 & 79 & 9.3726209 & 119 & 11.3962428 & 159 & 13.0988469 & 199 & 14.5972655 \\\hline 40 & 6.8019052 & 80 & 9.4288057 & 120 & 11.4420380 & 160 & 13.1384730 & 200 & 14.6326903 \\\hline \end{array} $$ The code producing the table is given at the end. So (at least empirically so far) the sequence is increasing, has the asymptotic $O(\sqrt n)$, and jumps over the integers as shown at the colored places: for $\color{blue}n$ in the "blue" list $\color{blue}{1\cdot 2}=2$, $\color{blue}{2\cdot 3}=6$, $\color{blue}{3\cdot 4}=12$, $\color{blue}{4\cdot 5}=20$ ... an entry which is $\le$ then respectively $2,3,4,5$ ... , and for $\color{red}n$ in the "red" list $\color{red}{1\cdot 2+1}=3$, $\color{red}{2\cdot 3+1}=7$, $\color{red}{3\cdot 4+1}=13$, $\color{red}{4\cdot 5+1}=21$ ... an entry which is $\ge$ then respectively $2,3,4,5$ ... The OP question addresses only the "blue to red jumps". It would be nice to have control on the asymptotic of the $q$-sequence that reflects this observed behavior, so let us show in the sequel an explicit simple double inequality that separates the $q$-terms in pairwise disjoint intervals: $$ \color{blue} { \tag{$\dagger$} \underbrace{\sqrt{n+\sqrt {n-1}}}_{=A(n)} \le q_n \le \underbrace{\sqrt{n+\sqrt{n-\frac 12}}}_{=B(n)} \qquad\text{ for }n\ge 15\ . } $$ For the values for $n$ less than $15$ see the table. From the relation $\dagger$ we have than immediately the claimed property. If $n=b^2-b+1\ge 15$ then $$ \begin{aligned} q_n &\ge A(n)=\sqrt{b^2-b+1+\sqrt{b^2-b}} > b\ ,\\ q_{n-1} &\le B(n-1)=\sqrt{b^2-b+\sqrt{b^2-b-\frac 12}} < b\ . \end{aligned} $$ Here are some words on the proof of $(\dagger)$. (The main issue was finding the two functions $A$, $B$ in $O\left(\sqrt n +\frac 12 -\frac 1{8\sqrt n}\right)$ that allow an inductive proof.) For $n=15$ the above inequality is satisfied, computer check. Then we try to use inductively the relation $$ q_{n+1}= \frac{a_{n+1}}{a_n}= \frac{a_n+na_{n-1}}{a_n}= 1+\frac n{a_n/a_{n-1}}= 1+\frac n{q_n}\ , $$ so from the bounds for $q_n$ we would obtain the corresponding bounds for $q_{n+1}$ if the inequalities marked below with $(?)$ are true: $$ A(n+1) \ \overset{(?)}\le\ 1+\frac n{B(n)} \ \le\ 1+\frac n{q_n} \ \le\ 1+\frac n{A(n)} \ \overset{(??)}\le \ B(n+1) \ . $$ For $(?)$ we equivalently rewrite step by step: $$ \begin{aligned} A(n+1) &\overset?\le 1 + \frac n{B(n)} \ , \\ A(n+1)B(n) &\overset?\le n + B(n) \ , \\ (n+1+\sqrt n)\left(n+\sqrt{n-\frac 12}\right) &\overset?\le n^2 + 2n\sqrt{n+\sqrt{n-\frac 12}} + n+\sqrt{n-\frac 12} \ , \\ n\sqrt{n-\frac 12} + n\sqrt n + \sqrt{n\left(n-\frac 12\right)} &\overset?\le 2n\sqrt{n+\sqrt{n-\frac 12}} \ , \\ \sqrt{n\left(n-\frac 12\right)} + n + \sqrt{n-\frac 12} &\overset?\le 2\sqrt{n\left(n+\sqrt{n-\frac 12}\right)} \ , \\ \frac 12(n-1) +2n\sqrt{n\left(n-\frac 12\right)} + 2\sqrt n\left(n-\frac 12\right) &\overset?\le 2n^2 + 2n\sqrt{n-\frac 12} \ . \end{aligned} $$ And the last inequality can be shown, estimate first $$ 2n^2 - 2n\sqrt{n\left(n-\frac 12\right)} = 2n\cdot\frac {n^2 -n\left(n-\frac 12\right)} {n+\sqrt{n\left(n-\frac 12\right)}} \ge 2n\cdot\frac{\frac n2}{n+n}=\frac n2\ . $$ For $(??)$ we equivalently rewrite step by step: $$ \begin{aligned} 1 + \frac n{A(n)} &\overset?\le B(n+1) \ , \\ n+A(n) &\overset?\le A(n)B(n+1) \ , \\ n^2 + 2n\sqrt{n+\sqrt{n-1}} + n+\sqrt{n-1} &\overset?\le (n+\sqrt{n-1})\left(n+1+\sqrt{n+\frac 12}\right) \ , \\ 2n\sqrt{n+\sqrt{n-1}} &\overset?\le n\sqrt{n+\frac 12} + n\sqrt{n-1} + \sqrt{(n-1)\left(n+\frac 12\right)} \ , \\ 4n^2(n+\sqrt{n-1}) &\overset?\le n^2\left(n+\frac 12\right) + n^2(n-1) + (n-1)\left(n+\frac 12\right) \\ &\qquad +2n^2\sqrt{(n-1)\left(n+\frac 12\right)} +2n\left(n+\frac 12\right)\sqrt{n-1} \\ &\qquad\qquad +2n(n-1)\sqrt{n+\frac 12} \ , \\ 2n^3 + 2n^2\sqrt{n-1} &\overset?\le 2n^2\sqrt{(n-1)\left(n+\frac 12\right)} + 2n^2\sqrt{n+\frac 12} + \frac{n^2}2 \\ &\qquad - 2n\sqrt{n+\frac 12} + n\sqrt{n-1} -\frac{n+1}2 \ , \\ 2n^2\cdot \frac {n^2 -(n-1)\left(n+\frac 12\right)} {n + \sqrt{(n-1)\left(n+\frac 12\right)}} -\frac{n^2}2 &\overset?\le 2n^2\cdot\frac{\left(n+\frac 12\right)-(n-1)}{\sqrt{n+\frac 12}+\sqrt{n-1}} \\ &\qquad +n\sqrt{n-1}-2n\sqrt{n+\frac 12}-\frac{n+1}2 \ , \\[3mm] &\qquad\text{ and for the last relation, we estimate the LHS} \\ 2n^2\cdot \frac {n^2 -(n-1)\left(n+\frac 12\right)} {n + \sqrt{(n-1)\left(n+\frac 12\right)}} -\frac{n^2}2 &= 2n^2\left(\frac{\frac{n+1}2}{n+\sqrt{(n-1)\left(n+\frac 12\right)}}-\frac 14\right) \\ &\le n^2\left(\frac{n+1}{n+\left(n-\frac 14\right)}-\frac 12\right) =n^2\cdot\frac{2+\frac 14}{2\left(2n-\frac 14\right)}\in O(n)\ , \\ &\qquad\text{ and the first term on the RHS} \\ 2n^2\cdot\frac{\frac 32}{\sqrt{n+\frac 12}+\sqrt{n-1}} &\ge 2n^2\cdot\frac{\frac 32}{2\sqrt{n-\frac 14}} =\frac32\cdot\frac{n^2}{\sqrt{n-\frac 14}}\sim \frac 32n\sqrt n \ , \end{aligned} $$ and from the factor $\frac 32=1+\frac 12$ we need (only) the $1$ to compensate the other terms in $O(n\sqrt n)$ for "big values" of $n$. (Computations can be done explicitly.) Sage code producing the table: a = [1, 1] B_blue = [b^2 - b for b in [2..99]] B_red = [b^2 - b + 1 for b in [2..99]] for n in [2..1000]: a.append( a[-1] + (n-1)*a[-2] ) def q(n): return RR(a[n]/a[n - 1]) def q_color(n): if n in B_red : return r'\color{red}{%10.7f}' % q(n) if n in B_blue: return r'\color{blue}{%10.7f}' % q(n) return '%10.7f' % q(n) for k in [1..40]: print(' & '.join(['{} & {}'.format(j*40+k, q_color(j*40+k)) for j in [0..4]]), r'\\\hline') Computer check of the relation $A(n) \le q_n\le B(n)$ for $n$ between $15$ and $1000$, here $A$ is low and $B$ is up: def low(n): return sqrt(n + sqrt(n - 1 )) def up (n): return sqrt(n + sqrt(n - 1/2)) prod([ int( bool(low(n) < a[n]/a[n-1]) and bool(a[n]/a[n-1] < up(n)) ) for n in [15..1000] ]) And we get the $1$. Computer check of the relation inductively shown above: $$A(n+1) \le 1+\frac n{B(n)} \le q_{n+1}\le 1 +\frac n{A(n)} \le B(n+1)$$ for $n$ between $15$ and $999$. ok = True for n in [15..999]: qq = RR( a[n+1] / a[n]) if ( bool( RR(low(n+1)) < 1 + n/RR(up(n)) ) and bool( 1 + n/RR(up(n) ) < qq ) and bool( qq < 1 + n/RR(low(n)) ) and bool( 1 + n/RR(low(n)) < up(n+1) ) ): continue ok = False break print(f'ok = {ok}') and we get ok = True. (The flag remained as set initially.)<|endoftext|> TITLE: When is the opposite of the category of algebras of a Lawvere theory extensive? QUESTION [34 upvotes]: When is the opposite of the category of algebras of a Lawvere theory an extensive category? Any necessary or sufficient conditions on the Lawvere theory will be interesting to me. Here's why I'm interested. An extensive category acts to some extent like a category of "spaces" (e.g. topological spaces, schemes, sets, etc.). We can see this in various ways. First, in any extensive category, an object over $X + Y$ is the same as an object over $X$ together with an object over $Y$. Second, if such a category has binary products, they distribute over binary coproducts: $$ X \times (Y + Z) \cong X \times Y + X \times Z.$$ Third, in any extensive category, any object with a morphism to the initial object $0$ must be initial itself. All these are basic things we expect of a category of "spaces". Furthermore, algebras of a Lawvere theory are a pretty good formalization of the general idea of "algebraic gadgets" (e.g. monoids, rings, $R$-algebras, etc.). So, my question is an attempt at getting at the vaguer question "when does the opposite of a category of algebraic gadgets act like a category of spaces?" One classic example is the Lawvere theory of commutative rings. Its category of algebras is CommRing, and the opposite of that is the category of affine schemes, which is extensive. Another classic example is the Lawvere theory of commutative $R$-algebras for some commutative ring $R$. The opposite of its category of algebras is the category of affine schemes over $R$, which is again extensive. But there are also more exotic examples, some of which show up e.g. in Durov's approach to algebraic geometry. For example the Lawvere theories for commutative rigs, distributive lattices or $C^\infty$-rings also have categories of algebras with extensive opposites. So, it would be nice to get a clearer sense of the full range of Lawvere theories whose category of algebras have extensive opposites. Are there any that do not involve a commutative rig structure? REPLY [21 votes]: I offer the following summary/interpretation of Broodryk's results. In short, a category of algebras is coextensive if and only if there is a well behaved interpolation operation in the algebraic theory. By "interpolation operation" what I mean is an operation $t$ of arity $2 + \kappa$, together with two $\kappa$-tuples of constants, $\vec{e}$ and $\vec{e}{}'$, satisfying the following equations: $$\begin{aligned} t (x, x', \vec{e}) & = x \\ t (x, x', \vec{e}{}') & = x' \end{aligned}$$ The meaning of "well behaved" is more complicated. Let $A$ be an algebra, let $F (0)$ be the initial algebra (= free algebra with empty generating set), and let $\iota_1 : A + F (0) \times F (0)$ and $\iota_2 : F (0) \times F (0) \to A + F (0) \times F (0)$ be the coproduct insertions. We define a map $\delta_A : A \times A \to A + F (0) \times F (0)$ as follows: $$\delta_A (x, x') = t (\iota_1 (x), \iota_1 (x'), \iota_2 (\vec{e}, \vec{e}{}'))$$ (Here I am abusing notation. What I mean by $\iota_2 (\vec{e}, \vec{e}{}')$ is the image of the pair $(\vec{e}, \vec{e}{}')$ under the obvious composite $F (0)^\kappa \times F (0)^\kappa \to (F (0) \times F (0))^\kappa \to (A + F (0) \times F (0))^\kappa$.) This is not, prima facie, a homomorphism of algebras. We can think of it as an external pairing operation on $A$. (This will be made precise later.) However, if the category of algebras is coextensive, it is possible to choose $t, \vec{e}, \vec{e}{}'$ so that $\delta_A$ is an homomorphism and furthermore satisfies the following equation: $$\delta_A (x, x) = \iota_1 (x)$$ Conversely, the existence of such $t, \vec{e}, \vec{e}{}'$ guarantees that the category of algebras is coextensive. Example. For the theory of commutative rigs, we can take $\kappa = 2$, $t (x, x', y, y') = x y + x' y'$, $\vec{e} = (1, 0)$, and $\vec{e}{}' = (0, 1)$. The point is that, in $A$, $$\begin{aligned} t (x, x', 1, 0) & = x \\ t (x, x', 0, 1) & = x' \end{aligned}$$ and, in $A \otimes_\mathbb{N} (\mathbb{N} \times \mathbb{N})$ (the change in operator precedence by switching from $+$ to $\otimes$ here is confusing, but I digress), $$\delta_A (x, x') = t (x \otimes 1, x' \otimes 1, 1 \otimes (1, 0), 1 \otimes (0, 1)) = x \otimes (1, 0) + x' \otimes (0, 1)$$ i.e. $\delta_A : A \times A \to A \otimes_\mathbb{N} (\mathbb{N} \times \mathbb{N})$ is the obvious natural isomorphism. Example. For the theory of commutative rings we can have $\kappa = 1$. Indeed, we can take $t (x, x', y) = x y + x' (1 - y)$, $\vec{e} = 1$, $\vec{e}{}' = 0$. Note that $t$ is literally linear interpolation! In more detail now: Let $\mathcal{A}$ be a category of algebras, by which I mean a category equipped with a (strictly) monadic functor $U : \mathcal{A} \to \textbf{Set}$. Let $F : \textbf{Set} \to \mathcal{A}$ be left adjoint to $U$ and let $T = U F$. Broodryk implicitly only considers finitary algebraic theories (i.e. $\mathcal{A}$ is locally finitely presentable and $U : \mathcal{A} \to \textbf{Set}$ preserves filtered colimits) but I think his main results also apply to infinitary algebraic theories. First, as a warm up: Lemma. If a category has an initial object that is also a strict terminal object (i.e. every morphism with domain a terminal object is an isomorphism), then every object in that category is initial/terminal/zero. ◼ A trivial observation, to be sure, but turning it around tells us something about the set $T (0)$ of constants in the algebraic theory: Proposition. If $\mathcal{A}$ has a strict terminal object, then either: $T (0)$ is empty, or $T (0)$ has at least two elements, or $\mathcal{A}$ is trivial. Proof. $T (0)$ is the underlying set of the initial object $F (0)$ in $\mathcal{A}$. If $T (0)$ has exactly one element, then $F (0)$ is both an initial object and a strict terminal object in $\mathcal{A}$, in which case $\mathcal{A}$ is trivial. ◼ Next: Proposition. If $\mathcal{A}$ has codisjoint binary products then $T (0)$ is not empty. Proof. Let $A$ and $B$ be objects in $\mathcal{A}$. The product $A \times B$ is codisjoint if the following is a pushout square in $\mathcal{A}$: $$\require{AMScd} \begin{CD} A \times B @>{\pi_2}>> B \\ @V{\pi_1}VV @VVV \\ A @>>> 1 \end{CD}$$ Consider the case $A = B = F (0)$. Assume $T (0)$ is empty. Then $U (F (0) \times F (0)) \cong T (0) \times T (0)$ is also empty, hence both projections $F (0) \times F (0) \to F (0)$ are isomorphisms. But the pushout of an isomorphism is an isomorphism, so that implies $F (0) \to 1$ is an isomorphism, which is a contradiction. So $T (0)$ is not empty. ◼ Corollary. If $\mathcal{A}$ is coextensive, then either: $T (0)$ has at least two elements, or $\mathcal{A}$ is trivial. Proof. A coextensive category has both a strict terminal object and codisjoint binary products. ◼ So far so good – after all, a rig has two distinguished elements. But it is not obvious how to extract binary operations from the hypothesis that $\mathcal{A}$ is coextensive. It seems the best we can do is to get a well behaved interpolation operation. Theorem. $\mathcal{A}$ is coextensive if and only if the following conditions hold: There exist a regular cardinal $\kappa$ ($\le \lambda$ if $F (0) \times F (0)$ admits a generating set of $\le \lambda$ elements), an element $t \in T (2 + \kappa)$, and $\vec{e} \in T (0)^\kappa$ and $\vec{e}{}' \in T (0)^\kappa$, such that the following equations hold in every $A$ in $\mathcal{A}$, $$\begin{aligned} t (x, x', \vec{e}) & = x \\ t (x, x', \vec{e}{}') & = x' \end{aligned}$$ where $x$ and $x'$ are arbitrary elements of $A$ and (by abuse of notation) we have identified $t$ with the operation $U (A)^{2 + \kappa} \to U (A)$ it corresponds to, and $\vec{e}$ and $\vec{e}{}'$ with their images in $U (A)^\kappa$ under the unique morphism ${!} : F (0) \to A$. For every $A$ in $\mathcal{A}$, there is a morphism $\delta_A : A \times A \to A + F (0) \times F (0)$ in $\mathcal{A}$ (whose underlying map $U (\delta_A)$ is) given by $$\delta_A (x, x') = t (\iota_1 (x), \iota_1 (x'), \iota_2 (\vec{e}, \vec{e}{}'))$$ where $t, \vec{e}, \vec{e}{}'$ are as above and I have abused notation as in the introduction. Moreover, $\delta_A (x, x) = \iota_1 (x)$.　◻ I will sketch just the construction of $t$. Consider the following diagram in $\mathcal{A}$: $$\begin{CD} F (0) @<{\pi_1}<< F (0) \times F (0) @>{\pi_2}>> F(0) \\ @V{!_A}VV @V{\iota_2}VV @VV{!_A}V \\ A @<<{[\textrm{id}_A, \pi_1]}< A + F (0) \times F (0) @>>{[\textrm{id}_A, \pi_2]}> A \\ @| @V{\eta_A}VV @| \\ A @<<{\pi_1}< A \times A @>>{\pi_2}> A \end{CD}$$ The two squares in the top half are pushout squares (always). The morphism $\eta_A : A + F (0) \times F (0) \to A \times A$ is defined by the universal property of $A \times A$, and it is an isomorphism if $\mathcal{A}$ is coextensive. Consider the case where $A = F (2)$. Let $x$ and $x'$ be the two distinguished generators of $F (2)$. Then $(x, x')$ is an element of $U (F (2) \times F (2))$, and $U (\eta_X) : U (F (2) + F (0) \times F (0)) \to U (F (2) \times F (2))$ is a bijection, so it has a preimage $\bar{t}$. The hypothesis on $\kappa$ ensures there is an effective epimorphism $F (2 + \kappa) \to F (2) + F (0) \times F (0)$ and we can arrange for the first two distinguished generators of $F (2 + \kappa)$ to be mapped to the image under $\iota_1 : F (2) \to F (2) + F (0) \times F (0)$ of the two distinguished generators of $F (2)$ and the remaining $\kappa$ generators to be mapped into the image of $\iota_2 : F (0) \times F (0) \to F (2) + F (0) \times F (0)$. Let $(\vec{e}, \vec{e}{}')$ be a preimage in $T (0)^\kappa \times T (0)^\kappa$ of the images of the $\kappa$ generators, considered as an element of $U (F (2) + F (0) \times F (0))^\kappa$. Then, $[\textrm{id}_A, \pi_1] (\bar{t}) = t (x, x', \vec{e})$ by homomorphicity, and $[\textrm{id}_A, \pi_1] (\bar{t}) = \pi_1 (\eta_X (\bar{t})) = x$ because the bottom left square commutes, so $t (x, x', \vec{e}) = x$ as required. Similarly, $t (x, x', \vec{e}{}') = x'$. Since we have proved the claim for the two distinguished generators of $F (2)$, the claim follows for all pairs of elements in all algebras. The morphism $\delta_A : A \times A \to A + F (0) \times F (0)$ is simply the inverse of $\eta_A : A + F (0) \times F (0) \to A \times A$. This is the sense in which $\delta_A : A \times A \to A + F (0) \times F (0)$ is a pairing operation. Remark. We have $\vec{e} = \vec{e}{}'$ if and only if $\mathcal{A}$ is trivial. So this is another way of seeing that $T (0)$ has at least two elements if $\mathcal{A}$ is coextensive and not trivial. It seems difficult to come up with a novel example of an algebraic theory satisfying all three of the conditions in the theorem. The first condition – the existence of an interpolation operation – is easy enough to arrange. (We can just freely add such $t, \vec{e}, \vec{e}{}'$ to any existing algebraic theory and get a new one!) The hard part seems to be in the second and third conditions. Broodryk remarks that the first condition alone is enough to guarantee many of the properties of coextensivity.<|endoftext|> TITLE: Question concerning the coefficients of block idempotents QUESTION [6 upvotes]: Let $G$ be a finite group. Let $p$ be a prime number such that $p \mid |G|$. Let Irr$(G)$ denote the set of ordinary irreducible characters of $G$. For $\chi\in$ Irr$(G)$ define $e_{\chi} := \frac{\chi(1)}{|G|}\sum_{g\in G} {\chi(g^{-1})\cdot g}$. Let $B$ denote a $p$-block of $G$. Let Irr$(B)$ be the set of those ordinary irreducible characters of $G$ which lie in $B$. Define $e_B:=\sum_{\chi\in \text{Irr}(B)} {e_{\chi}}$. This can be rewritten as $$e_B = \sum_{g\in G} {\bigg(\underbrace{\frac{1}{|G|}\sum_{\chi\in \text{Irr}(B)} {\chi(1)\chi(g^{-1})}}_{=:a_g}\bigg)} \cdot g.$$ Question: Is it always true that $a_g$ lies in $\mathbb{Q}_p[\zeta]$ for $\zeta = \exp(\frac{2\pi i}{p^n-1})$ for some positive integer $n$? Example: Set $G:=A_5$ and $p:=5$. Then there are exactly two $p$-blocks of $G$. In the coefficients of some of the $e_{\chi}$ there is the number $z:=\exp(\frac{2\pi i}{5})$ involved. The number $z$ cannot be expressed via the $\zeta$ above and $e_{\chi}$ does not lie in $\mathbb{Q}_p[\zeta]$ for $\zeta$ as above. But the two block idempotents $e_{B_0}$ and $e_{B_1}$ do not involve $z$ anymore. In other examples, the ${e_{\chi}}'s$ add up nicely to ${e_{B_j}}'s$, or such $z$'s don't appear at all. If this is always the case, I would be fond of a reference. Thank you very much for the help. REPLY [9 votes]: Yes, this is true. By block orthogonality relations due to R. Brauer, it is true that $a_{g}=0$ whenever the element $g$ has order divisible by $p$. But when $g$ has order prime to $p$, it is clear that $a_{g}$ lies in $\mathbb{Q}[\omega]$ for some $t$-th root of unity $\omega$, where $t$ is not divisible by $p$. Furthermore, the same block orthogonality relations imply that when the order of $g$ is prime to $p$, then the class function $\theta = \sum_{\chi \in B} \chi(g^{-1} )\chi$ vanishes of all non-identity elements of $P$, a Sylow $p$-subgroup of $G$. Taking the inner product of $\theta$ (restricted to $P$) with the trivial character of $P$ then shows that $[G:P] a_{g}$ is an algebraic integer. These two facts give what you want. The necessary block orthogonality relations can be found in almost any text which deals with blocks and characters, eg those by G.Navarro or by M. Isaacs.<|endoftext|> TITLE: Relationship between non-standard computation and TM(oracle)? QUESTION [6 upvotes]: We know that there are non-standard models of arithmetic, and in such models there are non-standard proofs of standardly unprovable sentences. Now, we can imagine a version of representability relative to some non-standard model of some arithmetical theory such that the said function would be what I would call a "non-standard computation". At the same time we can imagine the said computation otherwise correct, as in it would be the same as adding some real/oracle to a TM. Is there a nice way to connect these concepts? Is it a trivial connection? Can it give us insights about the base theory X or the sentence S such that X + "sentence S" $\vdash$ TM(oracle) halts? REPLY [6 votes]: Let address the version of the question posed by Gro-Tsen in his comments. Sets (a) and (c) are the same, and these are exactly the standard system of the model $M$, as mentioned in the answer of Patrick Lutz. The standard system is equivalently defined as the set of traces on the standard numbers of any sequence that is definable from parameters in $M$, whether or not it is computable in $M$, since every (possibly nonstandard) finite initial segment of a definable sequence is an element of $M$. Any sequence as in (a) or (c) will be the standard part of a definable sequence in $M$ and thus in the standard system, and conversely, any set in the standard system is the standard part of a pseudo finite sequence in $M$, which can be hard-coded into a (nonstandard) program $e$ in $M$, fulfilling (a) or (c). Set (b) consists of the sets of natural numbers that are the trace of a set that is computably enumerable in $M$ by a standard program $e$, such that this program gives output on all standard input. This is equivalent to the sets of natural numbers such that both it and its complement are the standard parts of sets that $M$ thinks are computably enumerable using a standard program. In a model of true arithmetic, this would be exactly the computably decidable sets, since the operation of a standard program $e$ would be the same in $\mathbb{N}$ as in $M$ in this case. Thus, this set can be strictly smaller than the sets in (a) and (c). Meanwhile, the universal algorithm (see also Arithmetic potentialism and the universal algorithm (arXiv:1801.04599) shows that any given set of natural numbers can be placed into set (b) for a suitable choice of $M$, since the universal algorithm $e$, which is a standard program, can compute any set in a suitable model $M$ of PA, and by arranging that the program enumerate the set in increasing order, one will also be able to compute the complement with a standard program. Nevertheless, set (b) is never equal to the sets in (a) and (c), because the model $M$ can unify the sets of (b) into a single subset of $\mathbb{N}\times\mathbb{N}$, using program $e$ on slice $e$. This will be a set in the standard system, from which all sets in (b) are computable, but the standard system must have sets not computable from any given set. Set (d) consists of the standard parts of the sets that $M$ thinks are computably decidable by a standard program. In a model of true arithmetic, these are exactly the computably decidable sets, but by using the universal algorithm as above, we can place any given set into set (d). The difference between set (d) and set (b) is that the programs in (b) must halt only on all standard input, but in (d) they must halt on all input. So set (d) is contained in set (b). I would have to think harder whether the sets can differ. But the argument I gave above shows that (d) is never equal to (a) or (c). Sets (e) and (f) consists exactly of the finite sets of natural numbers. Any finite set of natural numbers has the property in (e) and (f), and conversely, if a set is infinite and definable in a nonstandard model $M$ of PA, then it must have nonstandard elements, since otherwise it would stand as a counterexample to induction in $M$. A modified version of (e) would be the property (e'), asserting that $X\subset\mathbb{N}$ and there is some program $e\in M$ that halts on all input in $M$, such that $\forall i\in \mathbb{N} (i\in X\iff \{e\}(i)\downarrow=1)$. This would be equivalent to (a) and (c).<|endoftext|> TITLE: What is the correct statement of Theorem 4.2 in Street's "Parity Complexes"? QUESTION [5 upvotes]: Ross Street's 1991 paper Parity Complexes (apologies; I don't know how to find DOI links for Cahiers papers) develops some very useful tools for working with free strict $\omega$-categories. There is a corrigenda to the paper. I find it a bit difficult to assemble together these two ingredients to be sure what are the correct statements of the main theorems in Street's paper. A useful later reference for this material is Buckley's Formalizing Parity Complexes. I am interested in understanding Theorem 4.2, which unfortunately is not covered in Buckley's treatment. The theorem states Theorem 4.2 The $\omega$-category $O(C)$ is freely generated by the atoms. Let's break that down: As indicated in the paper, the notion of "free generation" comes from Street's earlier The Algebra of Oriented Simplices. I believe this notion is to be read as-is without change from the corrigenda. I believe that as originally written (but see (5) below), $C$ was intended to be an arbitrary parity complex, a notion defined in Section 1 of the paper; I believe this definition is faithfully reproduced at the linked nlab page (the nlab's $<$ being Street's $\triangleleft$ and the nlab's $\prec$ being Street's $\blacktriangleleft$). $O(C)$ is the $\omega$-category defined at the beginning of Section 3. It is proven in Theorem 3.6 that for any parity complex $C$, $O(C)$ is an $\omega$-category. I believe that Theorem 3.6 is understood to be true as stated -- the corrigenda does not indicate that the definition of $O(C)$ (or the subsidiary notions of cells or well-formed subsets of $C$) need be changed, nor does it indicate that any additional hypothesis on the parity complex $C$ is needed to ensure that $O(C)$ is an $\omega$-category (Thm 3.6). The notion of an atom is as defined in Section 4 of the paper. I believe the corrigenda indicates that the statement of Thm 4.2 should be changed as follows. On p. 1 of the corrigenda, it is indicated that for every element $x \in C_p$ of the parity complex $C$, we need to assume throughout Section 4 (including, apparently, in the statement of Thm 4.2) that the sets $\mu(x)$ (defined at the beginning of Section 4, with the definition corrected at the beginning of the corrigenda) are tight in the sense defined further down p. 1 of the corrigenda. Therefore, I believe the correct statement of Theorem 4.2 is: Theorem 4.2, correcta: Let $C$ be a parity complex. Assume that for every $p \in \mathbb N$ and every $x \in C_p$, the sets $\mu(x)$ (as defined in the corrigendum, not as defined in the paper) are tight (as defined in the corrigendum). Then the $\omega$-category $O(C)$ is freely generated by the atoms. Question 1: Do I have that right? The corrigendum also defines a notion of globularity at the beginning of p.2, and I believe the corrigendum asserts that for every parity complex $C$, and for every relevant $x \in C_p$ (as defined in Section 4), the globularity condition holds for $x$. I believe that Prop 5.2 of the corrigendum gives a criterion ensuring that the globularity condition holds for all $x \in C_p$, not just for the relevant $x \in C_p$. This condition uses the corrected definition of $\mu(x)$ as well as the corrected definition of $\pi(x)$. Question 2: Can the corrected statement of Theorem 4.2 be simplified by assuming something about the "globularity condition" rather than explicitly assuming something about tightness? REPLY [4 votes]: I hope I can offer some quick answers to your questions without errors. Let's tackle the breaking down: As indicated in the paper, the notion of "free generation" comes from Street's earlier The Algebra of Oriented Simplices. I believe this notion is to be read as-is without change from the corrigenda. Yes, it is the notion of freeness of this article. Using some rephrasing, it means that there exists a polygraph/computad $P$ such that $O(C)$ is isomorphic to $P^*$, the free $\omega$-category on $P$. I believe that as originally written (but see (5) below), $C$ was intended to be an arbitrary parity complex, a notion defined in Section 1 of the paper; I believe this definition is faithfully reproduced at the linked nlab page (the nlab's $<$ being Street's $\triangleleft$ and the nlab's $\prec$ being Street's $\blacktriangleleft$). It seems that the well-formed condition of parity complexes is badly reproduced in the nlab (condition 2.). Indeed, it not only a condition on the 1-cells but also on higher cells. Moreover, nlab's $<$ is Street's $<$ and nlab's $\prec$ is Street's $\triangleleft$. $O(C)$ is the $\omega$-category defined at the beginning of Section 3. It is proven in Theorem 3.6 that for any parity complex $C$, $O(C)$ is an $\omega$-category. I believe that Theorem 3.6 is understood to be true as stated -- the corrigenda does not indicate that the definition of $O(C)$ (or the subsidiary notions of cells or well-formed subsets of $C$) need be changed, nor does it indicate that any additional hypothesis on the parity complex $C$ is needed to ensure that $O(C)$ is an $\omega$-category (Thm 3.6). If I remember correctly, yes, the additions of the corrigenda is not required in order to obtain an $\omega$-category. So one can start from any parity complex. The notion of an atom is as defined in Section 4 of the paper. Yes. I believe the corrigenda indicates that the statement of Thm 4.2 should be changed as follows. On p. 1 of the corrigenda, it is indicated that for every element $x \in C_p$ of the parity complex $C$, we need to assume throughout Section 4 (including, apparently, in the statement of Thm 4.2) that the sets $\mu(x)$ (defined at the beginning of Section 4, with the definition corrected at the beginning of the corrigenda) are tight in the sense defined further down p. 1 of the corrigenda. Yes, this correta seems correct to me, in the sense that it should be the one deduced from the corrigenda. Question 2: Can the corrected statement of Theorem 4.2 be simplified by assuming something about the "globularity condition" rather than explicitly assuming something about tightness? No, it can not be simplified to a globularity condition. The counter-example I gave in my article is a parity complex which satisfies the globularity condition. Still, Theorem 4.2 does not hold for this example. By the way, it is no coincidence that Theorem 4.2 of Street's paper is not covered by Buckley, since it does not hold in its full generality with or without the corrigenda (but the counter-examples, like the already cited one, are very peculiar, so that most if not all the examples which use parity complexes in the literature should be fine).<|endoftext|> TITLE: Does the spectrum of Morava E-theory depend only on height? QUESTION [10 upvotes]: I almost expect the answer to this question to be no, but I can't find it explicitly said anywhere. Given a formal group law $f$ of height $n$ over a perfect field $k$ of characteristic $p$, we can construct the Morava E-theory $E(n)$. The resulting cohomology theory is $$E(n)^m (X) = MP^m(X)\otimes _{MP(*)} R$$ where $R = W(k)[[v_1,...,v_{n-1}]]$ is the Lubin-Tate ring of $f$ and $MP$ is complex cobordism. Also $$ \pi_\bullet (E) = E(n)^\bullet (*) = R[\beta, \beta^{-1}]$$ so at least the homotopy groups of the spectrum are independent of our choice of formal group law. This suggests that maybe the spectrum is the same, and the difference is the multiplicative structure on it. I think my confusion also might come from the fact that people usually refer to it as "Morava E-theory" rather than "a Morava E-theory", somewhat implying uniqueness. However, the main problem I see with this is that the map $MP(*) \to R$ from the Lazard ring classifying the universal deformation of $f$ (which makes $R$ a $MP(*)$-module) will be different for different formal group laws, so I would expect different values (as a graded abelian group) of the cohomology theory depending on the formal group chosen. Can someone please link an appropriate reference? REPLY [11 votes]: Here's an argument that Eric Peterson and I came up with showing that the homotopy type of Morava $E$-theory only depends on the choice of perfect char $p$ field $k$ and the height $n<\infty$. $\textbf{Lemma}$: Let $R$ be a ring with two Landweber exact formal group laws $e,f:\text{Laz}\rightarrow R$ and let $E$ and $F$ be the spectra corresponding to these two formal group laws under the Landweber exact functor theorem. Then $E$ and $F$ have the same homotopy type if there is a ring extension $u: R\rightarrow S$ which is split as an $R$-module map, and such that the formal group laws $u\circ e$ and $u\circ f$ are isomorphic and Landweber exact. $\textbf{Proof}$: First we show that $E\simeq F$ if the map $\eta:E_*\rightarrow F_*E$ (induced by $1\wedge id: \mathbb{S}\wedge E\rightarrow F\wedge E$) is split (as a map of $F_*$-modules). Note that $E_*$ is canonically an $F_*$ module by the \textit{equality} $E_*=R=F_*$. We claim that $F_*E$ is flat over $F_*$. That boils down to interpreting Landweber exactness as flatness of maps to $\mathcal{M}_{fg}$, noting that flatness is preserved under pullback, and that (Spec of) $E_*$, $F_*$ $F_*E$ and $\mathcal{M}_{fg}$ fit into a pullback square (with $\mathcal{M}_{fg}$ in the bottom right and Spec of $F_*E$ in the top left). It follows that $F\wedge E$ represents the cohomology theory $X\mapsto F^*X\otimes_{F_*}F_*E$, and then Brown representability (and Strickland's result about the absence of phantom maps between Landweber exact spectra) produces a map $\phi: \text{Mod}_{F_*}(F_*E,F_*)\rightarrow F^*E$. Now suppose $s$ is a splitting of $\eta$. Then there is a map $\phi(s):E\rightarrow F$ which on homotopy groups (set $X=\text{pt}$ above) is the composite $s\circ\eta$ which is the identity map (recall that $E_*$ and $F_*$ are canonically isomorphic to $R$), so $E\simeq F$. Now we show that $\eta:E_*\rightarrow F_*E$ is split if there is a ring extension $u: R\rightarrow S$ which is split as an $R$-module map, and such that the formal group laws $u\circ e$ and $u\circ f$ are isomorphic and Landweber exact. Write $g:=u\circ e$ and $h:=u\circ f$. Assume they are isomorphic and Landweber exact. Let $G$ and $H$ be the spectra associated to $g$ and $h$. Then we have the following commutative diagram of $F_*(=R)$-modules $\require{AMScd}$ \begin{CD} S=G_* @>1\otimes 1\otimes id>> H_*G= S\otimes_{MU_*}MU_*MU\otimes_{MU_*}S\\ @A u A A @AA u\otimes id\otimes u A\\ R=E_* @>>1\otimes 1\otimes id> F_*E=R\otimes_{MU_*}MU_*MU\otimes_{MU_*}R \end{CD} Since the right vertical map is split (as a map of $F_*$-modules), splitting the bottom horizontal map (as a map of $F_*$-modules) is equivalent to splitting the diagonal map (as a map of $F_*$-modules). Moreover, the top horizontal map is split (as an $S=H_*$-algebra (and hence $F_*$-algebra) map in fact) by first choosing an isomorphism of $g$ and $h$, which induces an $H_*$-algebra isomorphism $H_*G\simeq G_*G$ and then using the multiplication map $G\wedge G\rightarrow G$ to split $G_*\rightarrow G_*G$. Therefore splitting the diagonal map is equivalent to splitting the left vertical map (as a map of $F_*$-modules). $\textbf{End of proof}.$ Now we apply the lemma to show that at height $n<\infty$ the homotopy type of Morava $E$-theory depends only on the choice of a perfect characteristic $p$ field. In other words, let $k$ be a perfect characteristic $p$ field and let $E_1(n)$ and $E_2(n)$ be two Morava $E$-theory spectra corresponding to two arbitrary height $n< \infty$ formal group laws over $k$. Then we'll show that as spectra, $E_1(n)\simeq E_2(n)$. It suffices to check the conditions of the lemma. Let $\overline{k}$ be the algebraic closure of $k$. Let $R$ and $S$ be the (2-periodic) Lubin-Tate deformation rings $W(k)[[u_1,...,u_{n-1}]][\beta^\pm]$ and $W(\overline{k})[[u_1,...,u_{n-1}]][\beta^\pm]$. Let $\tilde{e}$ and $\tilde{f}$ be formal group laws over $R$ universally deforming $e_1$ and $e_2$. Then $\tilde{e}_1$ and $\tilde{e}_2$ are Landweber exact and the corresponding spectra are the Morava $E$-theory spectra $E_1(n)$ and $E_2(n)$. Let $u:R\rightarrow S$ be the map that extends coefficients by $W(k)\rightarrow W(\overline{k})$, sends $u_i$ to $u_i$, and $\beta$ to $\beta$. Then the formal group laws $\tilde{g}:=u\circ \tilde{e}_1$ and $\tilde{h}:=u\circ \tilde{e}_2$ over $S$ are isomorphic and Landweber exact, since they universally deform the isomorphic formal group laws $g:u\circ e_1$ and $h:u\circ e_2$ over $\overline{k}$. Finally, the map $u:R\rightarrow S$ is split as an $R$-module map, because the map $W(k)\rightarrow W(\overline{k})$ is split as a $W(k)$-module map (See (ill Sawin's comments at Splitting the Witt vectors of $\overline{\mathbb{F}_p}$))<|endoftext|> TITLE: Stable smoothing of topological manifolds relative to an embedding QUESTION [5 upvotes]: Let $M$ be a topological manifold. We know that $M$ is stably smoothable if and only its tangent microbundle, up to stabilization, admits a reduction to vector bundle. Now I wonder if there is a relative version of this fact. Suppose $M$ and $N$ are topological manifolds and $f: M \to N$ is a locally flat topological embedding and assume it admits a normal microbundle. We also assume that the normal microbundle is equivalent to a vector bundle. Suppose $M$ and $N$ are both stably smoothable. I wonder if one can prove that there are smoothable stablizations $M = M \times {\mathbb R}^m$ and $N' = N \times {\mathbb R}^{m + n}$ such that the natural extension $f': M' \to N'$ (extending $f$ by the inclusion ${\mathbb R}^m \to {\mathbb R}^{m+n}$) is a smooth embedding. If the above speculation is true, I also wonder if such stable smoothings are in any sense unique. REPLY [4 votes]: I think the answer is yes, it follows from this: if $M$ is a triangulable manifold of dimension $n$ greater than 4, and if the total space $E$ of a vector bundle over $M$ is smoothable, then the smooth structure is concordant to one where the zero section is a smooth submanifold. To see why it follows, stabilize $M$ so that it has dimension at least $5$. Then the normal bundle of the embedding satisfies the requirements of our theorem since the tubular neighborhood is assumed to be a vector bundle and is an open subset of a smoothable space. By the above proposition and the concordance extension theorem, we can find a concordant smooth structure on $N$ which agrees with the smooth structure on the tubular neighborhood of $M$ which we have manufactured to have $M$ as a submanifold. So why is the first proposition true? Pick a triangulation of $M$. Over an n-simplex $\Delta$ we inductively apply the fact that any smooth structure on $\Delta \times \mathbb{R}^k$ which is a product structure when restricted to some set of faces $I$ is concordant relative $I \times \mathbb{R}^k$ to a product structure. This is a consequence of the product structure theorem. After this process is completed, the smooth structure on our vector bundle has the property that the restriction to every $n$-simplex in $M$ is a smooth submanifold. Hence, $M$ is a smooth submanifold. The vector bundle requirement ensures that any two trivializations over a shared face $\sigma$ induce the same smooth structure on $\sigma \times \mathbb{R}^k$, meaning the notion of a local smooth product structure on a vector bundle is well defined.<|endoftext|> TITLE: Non-isotopic homology spheres in $S^4$ with equal complements? QUESTION [9 upvotes]: Are there two diffeomorphic smoothly embedded homology 3-spheres $M_1^3, M_2^3 \subset S^4$ that have diffeomorphic complements but such that $M_1$ and $M_2$ are not isotopic? I would be interested in examples in other 4-manifolds as well. I am more interested in examples where $M_1$ and $M_2$ are not smoothly isotopic (I imagine there is a difference between smooth and topological isotopy of 3-manifolds in 4-manifolds, I guess I'd love to hear about references to that as well, specifically in $S^4$). On a more general note, what are some general ways people tend to tell apart isotopy classes of codimension one submanifolds? REPLY [5 votes]: There is an example of an Akbulut cork, proved to be a strong cork by Dai-Hedden-Mallick, $W= W(0,1)$ whose boundary $M$ is $+1$ surgery on the Stevedore knot, and has a mapping class group of order 4 generated by two involutions $S$ and $T$. If one doubles $W$ along $M$, then one gets $S^4=-W\cup W$, and hence an embedding of $M$ into $S^4$, since $W$ is a Mazur manifold (described first by Akbulut-Kirby). (Proposition 1 of Mazur proves that any such manifold has double diffeomorphic to $S^4$) Akbulut also shows that $-W \cup_S W$ obtained by gluing two copies by $S$ is diffeomorphic to $S^4$, giving another embedding of $M$ into $S^4$, and same for $-W\cup_T W$. Suppose the two embeddings are isotopic, in particular there is a diffeomorphism taking one embedding to the other. Then this diffeomorphism takes the $W$ on each side to each other. We may think of one $W$ as fixed, and the diffeomorphism extends over the other $W$. But the other two copies of $W$ are glued by the identity and $S$ respectively, so we see that the involution $S$ extends to a diffeomorphism of $W$, a contradiction. Thus we have two copies of $M$ embedded in $S^4$ which are not isotopic, and which have diffeomorphic complements. One comment: if the isotopy takes the $W$ on one side to the other copy of $W$ in such a way that the diffeomorphism is not isotopic to the identity on $M$, then it must be the involution $ST$ in the mapping class group of $M$, since $S$ and $T$ don’t extend. Then the two mapping classes will differ by $T$, which still doesn’t extend over the other copy of $W$.<|endoftext|> TITLE: Rank one adjoint operators on a Lie algebra QUESTION [6 upvotes]: Let $\mathfrak{g}$ be a (finite dimensional) semi-simple Lie algebra over a field $k$ and let $x \in \mathfrak{g}$. By definition, we have the equivalence: $$ \mathrm{rk}(\mathrm{ad}_x) = 0 \iff x = 0,$$ where $\mathrm{rk}(\mathrm{ad}_x)$ is the rank of $\mathrm{ad}_x$ seen as an element of $\mathrm{End}(\mathfrak{g})$. I would like to know if there is a classification of elements $x \in \mathfrak{g}$ such that $\mathrm{rk}(\mathrm{ad}_x) \leq 1$? I am primarily interested in the case where $k = \mathbb{C}$ and $\mathfrak{g}$ is simple of classical type. REPLY [8 votes]: $\DeclareMathOperator\ad{ad}\DeclareMathOperator\rk{rk}$I claim that it is impossible to have $\rk(\ad_x) = 1$. I'll assume that $\mathfrak g$ is $k$-split, which seems to be OK since you are interested in the case $k = \mathbb C$. (Or you could just tensor up to $\overline k$, which does not change the rank of the adjoint operator.) Let $\mathfrak h$ be a split Cartan subalgebra. Suppose that $x \in \mathfrak g$ satisfies $\rk(\ad_x) = 1$. If $x$ lies in $\mathfrak h$, $\beta$ is any root that does not vanish at $x$, and $E_\beta$ and $E_{-\beta}$ are non-$0$ vectors in the appropriate root spaces, then $\ad_x(E_\beta) = \beta(x)E_\beta$ and $\ad_x(E_{-\beta}) = -\beta(x)E_{-\beta}$ are non-$0$ elements of different root spaces, hence are linearly independent, which is a contradiction. Thus, $x$ does not lie in $\mathfrak h$. Let $x = \sum x_\alpha$ be the root-space decomposition with respect to $\mathfrak h$, where $\alpha$ runs over the roots of $\mathfrak h$ in $\mathfrak g$ and $0$, and choose a root $\alpha$ such that $x_\alpha$ is non-$0$. We have that $\ad_x(h)$ equals $-\sum \alpha(h)x_\alpha$ for all $h \in \mathfrak h$. Since all these elements lie on a common line, if $\beta \ne \pm\alpha$ is a root, then, since $\{\alpha, \beta\}$ and $\{\alpha, -\beta\}$ are linearly independent subsets of $\mathfrak h^*$, we have $x_\beta = 0$ and $x_{-\beta} = 0$. Let $h$ be an element of $\mathfrak h$ not annihilated by $\alpha$. We have that $\ad_x(h) = -\alpha(h)x_\alpha + \alpha(h)x_{-\alpha}$ and $\ad_x(x_\alpha)$ equals $\alpha(x_0)x_\alpha + \ad_{x_{-\alpha}}(x_\alpha)$. Noting that $\ad_{x_{-\alpha}}(x_\alpha)$ lies in $\mathfrak h$ and comparing root-space decompositions of these linearly dependent elements shows that $\alpha(h)x_{-\alpha}$ equals $0$, so that $x_{-\alpha}$ equals $0$. That is, $\ad_x(h)$ equals $-\alpha(h)x_\alpha$, hence is a non-$0$ element of the $\alpha$-root space. If $x_0$ is non-$0$, then we can choose a root $\beta \ne -\alpha$ (possibly $\beta = \alpha$) such that $\beta(x_0) \ne 0$. If $E_\beta$ is a non-$0$ element of the $\beta$-root space, then $\ad_x(E_\beta) = \beta(x_0)E_\beta + \ad_{x_\alpha}(E_\beta)$ has projection $\beta(x_0)E_\beta \ne 0$ on $\mathfrak h$, hence is linearly independent of $\ad_x(h)$, which is a contradiction. Thus $x$ lies in the $\alpha$-root space. If $E_{-\alpha}$ is a non-$0$ element of the $(-\alpha)$ root space, then $\ad_x(E_{-\alpha})$ is a non-$0$ element of $\mathfrak h$, hence is again linearly independent of $\ad_x(h)$, which is a contradiction. REPLY [6 votes]: Another approach. To show it's impossible (the rank can't be 1), it is enough to show this when the field (assumed of char 0) is algebraically closed, and in turn it's enough to show the result in case $\mathfrak{g}$ is simple. If $x$ has $\mathrm{ad}(x)$ of rank 1, $x$ has centralizer of codimension 1. It is known (see e.g. this MathSE answer) that $\mathfrak{g}$ has no subalgebra of codimension 1, unless $\mathfrak{g}$ is 3-dimensional. But for $\mathfrak{sl}_2$, the operator $\mathrm{ad}_x$ has rank 2 for every nonzero $x$ (alternatively, all 2-dimensional subalgebras have a trivial centralizer, so can't be centralizer of an element). A similar approach can be used (with a little further work) to classify which $\mathfrak{g}$ admit an $x$ with $\mathrm{ad}(x)$ of rank 2, and classify such elements $x$.<|endoftext|> TITLE: Theorems that are essentially impossible to guess by empirical observation QUESTION [99 upvotes]: There are many mathematical statements that, despite being supported by a massive amount of data, are currently unproven. A well-known example is the Goldbach conjecture, which has been shown to hold for all even integers up to $10^{18}$, but which is still, indeed, a conjecture. This question asks about examples of mathematical statements of the opposite kind, that is, statements that have been proved true (thus, theorems) but that have almost no data supporting them or, in other words, that are essentially impossible to guess by empirical observation. A first example is the Erdős–Kac theorem, which, informally, says that an appropriate normalization of the number of distinct prime factors of a positive integer converges to the standard normal distribution. However, convergence is so slow that testing it numerically is hopeless, especially because it would require to factorize many extremely large numbers. Examples should be theorems for which a concept of "empirical observation" makes sense. Therefore, for instance, theorems dealing with uncomputable structures are (trivially) excluded. REPLY [3 votes]: A slightly old question, but I've only just seen it... Lord Brouncker proved in 1654 that $$ \dfrac{1}{4}\pi = \cfrac{1}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\dotsb}}}} $$ This has very slow convergence. Apparently people didn't believe it initially. It takes nearly 50 terms for five decimal places of accuracy and nearly 120 for six! It converges in an oscillatory manner and the jumps are quite large for quite a long time. Of course, 120 terms isn't that many for modern computers and there would be no doubt about it these days. But, 120 is a lot to do by hand in the 1650s!<|endoftext|> TITLE: Finite presentability and elementary equivalence QUESTION [18 upvotes]: Do there exist two elementary equivalent finitely generated groups $G,H$ such that $G$ is finitely presented but $H$ is not finitely presentable? It seems reasonable to think that finite presentability is not preserved under elementary equivalence among finitely generated groups, so such groups probably exist. However, I do not know any example. The case where $G,H$ have the same universal theory, instead of having the same first order theory, would be already interesting. REPLY [2 votes]: The problem is known and still open but does not seem hopeless. I would start with considering Zilber's old example of two f.g. non-isomorphic nilpotent of class $2$ groups which are e.e.: B.I. Zilʹber, An example of two elementarily equivalent but not isomorphic finitely generated metabelian groups. Algebra i Logika 10 (1971), 309–315. Both of his groups were of course finitely presented. But I would guess that there are similar examples when one of the groups is not finitely presented.<|endoftext|> TITLE: Invariants of $\mathrm{GL}_n$ representations QUESTION [10 upvotes]: $\DeclareMathOperator\GL{GL}$Let $V=\mathbb C^n$ be the natural representation of $\GL_n(\mathbb C)$ and let $W=\operatorname{Sym}^2(V)$ be the symmetric square representation. Let $W^k$ denote the direct sum of $k$-copies of $W$. A version of the first fundamental theorem gives a generating set for $\mathbb C[V^k \oplus (V^*)^k]^{\GL_n(\mathbb C)}$. Is a generating set for the ring of invariants $\mathbb C[W^k \oplus (W^*)^k]^{\GL_n(\mathbb C)}$ known? REPLY [3 votes]: In classical terms, this is about joint invariants of $k$ quadratic forms $Q^{(1)}(x),\ldots,Q^{(k)}(x)$ in point coordinates $x=(x_1,\ldots,x_n)$ as well as $k$ quadratic forms $R^{(1)}(u),\ldots,R^{(k)}(u)$ in dual coordinates $u=(u_1,\ldots,u_n)$. By the FFT for $GL_n$, these invariants are linear combinations of graphs which tell you how to build invariant polynomials by contracting indices (along edges). Here the latter are made of two types of vertices of degree two (two incident edges), ones for the $Q$'s and ones for the $R$'s. Moreover all edges are $QR$, i.e., no $QQ$ or $RR$ edges. So basically these are disjoint collections of cycles of even length made by alternating $Q$'s and $R$'s. For a proof of the relevant FFT see my two answers to How to constructively/combinatorially prove Schur-Weyl duality? Another description of these invariants involves introducing the $k^2$ matrices $A^{(a,b)}$ with matrix elements $$ A^{(a,b)}_{ij}=\sum_{\ell=1}^{n} Q_{i\ell}^{(a)} R_{\ell j}^{(b)}\ . $$ Then the basic invariants are traces of arbitrary products or words made of $A$ matrices. In the $k=1$ case, there is only one matrix $A$ and the generators are simply the traces of the powers $A,A^2,\ldots,A^n$ and they are algebraically independent because one can take $R$ to be the identity, and $Q$ some generic diagonal matrix and the result follows from the algebraic independence of the elementary symmetric functions. So one recovers the result in Mark's answer. For general $k$ and $n$, finding a minimal set of (algebra) generators and a description of their relations, sounds to me like a hopeless task. Note that one may get some information via the relation to invariants under conjugation of $k^2$ matrices. This subject has been extensively investigated in work of Procesi, Rasmyslov, Formanek, etc.<|endoftext|> TITLE: Polynomials with roots on, outside but not inside the unit circle QUESTION [7 upvotes]: In relation to the (possible non-)existence of non-contracting ergodic endomorphisms on the torus, I have the following question on polynomials: If the polynomial is monic, with integer coefficients and has no root strictly inside the unit circle, is it then also true that roots on the unit circle are roots of unity? For polynomials up to degree 4 this seems to be the case, but I'd be happy with a higher degree counter-example. I am aware that Kronecker's Theorem says the answer is "yes" if all roots are on the unit circle. REPLY [12 votes]: Let $f(z)$ be a degree-$n$ irreducible polynomial with integer coefficients and roots on the unit circle but no roots strictly within it. Then $g(z) := z^nf(1/z)$ is also a degree-$n$ irreducible polynomial, whose roots on the unit circle are the same as those of $f$. Since $f$ is the minimal polynomial of those unit roots and $g$ has the same degree as $f$, $g$ must also have the same roots as $f$. Now since $f$ has no roots within the unit circle, $g$ has by construction no roots outside the unit circle, so all roots of $f$ lie on the unit circle. The rest follows from Kronecker's theorem.<|endoftext|> TITLE: Parallel lines containing a subset with even cardinality QUESTION [5 upvotes]: For each $\alpha \in \mathbf{R}\cup \{\infty\}$, let $\mathscr{L}_\alpha$ denote the collection of lines $\ell$ of $\mathbf{R}^2$ with slope $\alpha$. More explicitly: if $\alpha \in \mathbf{R}$, then $\ell \in \mathscr{L}_\alpha$ if and only if $\ell=\{(x,y) \in \mathbf{R}^2: y=\alpha x+\beta\}$, for some $\beta \in \mathbf{R}$; and $\ell\in \mathscr{L}_\infty$ if and only if $\ell=\{(x,y) \in \mathbf{R}^2: x=\gamma\}$, for some $\gamma \in \mathbf{R}$. Now, let us fix a finite subset $\mathscr{P}:=\{P_1,\dotsc,P_k\}\subseteq \mathbf{R}^2$, where $k$ is a given positive even integer. For each $\alpha \in \mathbf{R}\cup \{\infty\}$, let $N(\alpha, \mathscr{P})$ be the number of points contained in lines $\ell$ with slope $\alpha$ which contain at least two points of $\mathscr{P}$, that is, $$ N(\alpha, \mathscr{P}):=|\mathscr{P} \cap L(\alpha,\mathscr{P})|, \quad \text{ where }\quad L(\alpha,\mathscr{P}):=\{\ell\in \mathscr{L}_\alpha: |\,\ell \cap \mathscr{P}\,|\ge 2\}. $$ Question. Is it true that there exists $\alpha \in \mathbf{R}\cup \{\infty\}$ such that $N(\alpha,\mathscr{P})$ is an even positive integer? Some observations: if the points in $\mathscr{P}$ are collinear, the answer is clearly affirmative. Otherwise, it is also affirmative if $\mathscr{P}\subseteq \{a,b\}\times \mathbf{R}$ for some distinct $a,b$: indeed, in such case, there would be a maximal slope $\alpha \in \mathbf{R}$ connecting two points with different abscissas so that $N(\alpha,\mathscr{P})=2$. Moreover, it may be possible that $|L(\alpha,\mathscr{P})| \neq 1$ for every $\alpha$: for, let $\mathscr{P}$ be the grid $\{1,\dotsc,2n\}^2$, for some $n\ge 2$. Differently from the estimates on point-line incidences, here we are interested in the parity of the latter ones. REPLY [3 votes]: It was conjectured by Dumitrescu and Toth in "Distinct triangle areas in a planar point set" (see Problem 1 at the end) that for every large enough $k$ for any $\mathscr P$ there is an ordinary direction $\alpha$ such that for all $\ell\in L(\alpha,\mathscr P)$ we have $|\ell\cap \mathscr P|=2$. This would imply an affirmative answer to your question, as in this case $N(\alpha,\mathscr P)=2|L(\alpha,\mathscr P)|$. Though there is no implication in the reverse direction, I don't see how one could exploit parity, so I think that your problem is just as hard as theirs.<|endoftext|> TITLE: Silver-like forcing preserves p-points (reference request) QUESTION [5 upvotes]: A Silver forcing "below $2^n$" is defined e.g. in Definition 7.4.11 of [Tomek Bartoszyński and Haim Judah, Set Theory: on the structure of the real line, A. K. Peters, Wellesley, 1995.]. It is called Infinitely equal forcing EE there. In the same book, in Lemma 7.4.15 the authors show that EE preserves p-points. However, the proof of this lemma seems to be not complete/correct: the choice of conditions $p^{n+1}$ is not clear, as needed extensions of $p^n$ for various $r_n^j$ may "contradict" each other. (This would not be a problem if conditions were $2^n$-splitting trees, i.e., if the forcing were more like the Sacks rather than Silver.) Do you know another source/reference for the proof that EE preserves p-points? Or perhaps I am missing something and the proof in the book is actually complete? REPLY [6 votes]: David Chodounský and Osvaldo Guzmán showed in arXiv:1703.02082, There are no P-points in Silver extensions that There are no P-points in Silver extensions. They prove that after adding a Silver real no ultrafilter from the ground model can be extended to a P-point, and this remains to be the case in any further extension which has the Sacks property. In particular, any free ultrafilter from the ground model will not be a p-point, and hence cannot be an ultrafilter, as Silver forcing is proper. Silver is a complete subforcing of the forcing EE (right? In each entry of the generic, use the first bit — this will be a Silver generic) so also EE destroys all p-points. (Edit 2022: Published 2019 in the Israel Journal of Mathematics, vol 232 (2) 759–773, https://doi.org/10.1007/s11856-019-1886-2)<|endoftext|> TITLE: A tale of two maps into a Grassmannian QUESTION [5 upvotes]: I suspect that the answer to this question is well-known to the experts. However, I was not able to find it in the literature, so let me ask here. Setup. In the sequel, $X$ is a compact complex surface and $E$ is a rank $2$ vector bundle on $X$. We call $\pi \colon \mathbb{P}(E) \to X$ the projective bundle in the sense of Hartshorne. In my applications, $X$ is of general type and $E=\Omega_X^1$, but let me ask the question in generality. We assume that $S^nE$ is globally generated, namely, that the evaluation of sections $$H^0(X, \, S^nE) \otimes \mathcal{O}_X \to S^nE$$ is surjective. If $V$ is a vector space, we denote by $G(r, \, V)$ the Grassmannian of $r$-dimensional subspaces of $X$, and by $G(V, r)$ the Grassmannian of the $r$-dimensional quotients of $V$, so that $G(r, \, V) \simeq G(V^*, \, r)$. Moreover, we denote by $\mathbb{G}(r-1, \, \mathbb{P}(V))$ and $\mathbb{G}(\mathbb{P}(V), \, r-1)$ the corresponding projective Grassmannians. We are now going to construct two morphisms $X \to \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*)$. The first morphism. Take a point $x \in X$. The fibre $\pi^{-1}(x)$ is the projective line $\mathbb{P}(E(x))$, and $|\mathcal{O}_{\mathbb{P}(E)}(n)|$ cuts on it the complete linear system $\mathcal{O}(n)$. Identyfying the global sections of $\mathcal{O}_{\mathbb{P}(E)}(n)$ with those of $S^nE$, we see that $\pi^{-1}(x)$ is embedded in $\mathbb{P}H^0(X, \, S^nE)^*$ as a rational normal curve $C_x$ of degree $n$. There is exactly one $n$-plane $L_x$ containing $C_x$, and we get a morphism $$\mathscr{G} \colon X \to \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*), \quad x \mapsto L_x.$$ The second morphism. Take a point $x \in X$. There is a surjection $$H^0(X, \, S^nE) \to S^nE(x) \to 0,$$ where $S^nE(x)$ is the fibre of $S^nE$ over $x$, namely, a $n+1$-dimensional vector space. This provides by definition an element $$s_x \in G(H^0(X, \, S^nE), \, n+1) \simeq G(n+1, \, H^0(X, \, S^nE)^*) \simeq \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*),$$ and we can define a morphism $$\mathscr{H} \colon X \to \mathbb{G}(n, \, \mathbb{P}H^0(X, \, S^nE)^*), \quad x \mapsto s_x.$$ Question. How are the morphisms $\mathscr{G}$ and $\mathscr{H}$ related? Are they the same? Every answer and/or reference to the relevant literature will be greatly appreciated. REPLY [12 votes]: I guess, when you say "in the sense of Hartshorne" you mean the projective spectrum of $\oplus S^kE$. Yes, the morphisms are the same, and to see this just note that there is a natural (relative Veronese) embedding $$ \mathbb{P}_X(E) \to \mathbb{P}_X(S^nE), $$ such that every fiber of the (former) $\mathbb{P}^1$-bundle becomes a rational normal curve in the (latter) $\mathbb{P}^n$-bundle. Moroever, the composition $$ \mathbb{P}_X(E) \to \mathbb{P}_X(S^nE) \to \mathbb{P}(H^0(X,S^nE)^*) $$ (where the second arrow is induced by the dual of the evaluation morphism $H^0(X,S^nE) \otimes \mathcal{O}_X \to S^nE$) coincides with the morphism induced by the line bundle $\mathcal{O}_{\mathbb{P}_X(E)}(n)$. This shows that the two maps coincide.<|endoftext|> TITLE: Density of fake zeros of Zeta QUESTION [5 upvotes]: I am investigating whether or not there exist $\epsilon > 0$ such that $\zeta(s) \neq 0$ on the strip $1-\epsilon < \Re(s) \leq 1$. Suppose not. Then given $\delta > 0$ there exists a zero of zeta $\rho$ such that $1 -\delta < \Re(\rho) < 1 $. Hence, there exists a sequence of zeros $\{ z_n \}_{n=1}^\infty$ with increasing imaginary parts that have real part getting closer and closer to the line $\Re(z)=1$. Can we somehow estimate the density of this infinite set of "fake zeros." The idea is to compare the density of this set with the best known estimate for the density of zeros that are on the critical line. REPLY [11 votes]: To add to GH from MO's answer, Chapter 10 of Iwaniec and Kowalski's "Analytic Number Theory" is another good reference. A few of the classical papers are Ingham's "On the Estimation of $N(\sigma,T)$," Montgomery's "Mean and Large Values of Dirichlet Polynomials" and "Zeros of $L$-functions," and Huxley's "On the Difference Between Consecutive Primes." Define $$ N(\sigma,T) = |\left\{\rho=\beta+i\gamma: \zeta(\rho)=0,\ \beta\geq \sigma,\ 0\leq \gamma \leq T\right\}|, $$ so $N(\sigma,T)$ counts the zeros of $\zeta$ with real part at least $\sigma$ and imaginary part at most $T$. In general, zero density estimates for the Riemann zeta function take the form $$ \tag{1} N(\sigma,T) \ll T^{\lambda(\sigma)(1-\sigma)}(\log T)^B $$ for some relatively unimportant positive constant $B$ and some function $\lambda(\sigma)$. Note that the bound shrinks as $\sigma \to 1$. Ingham shows that (1) holds with $$ \lambda(\sigma) = \frac{3}{2-\sigma}. $$ Montgomery proves an analogous result for all Dirichlet $L$-functions and also proves that (1) holds with $$ \lambda(\sigma) = \frac{5}{2}, $$ which Huxley improves to $$ \lambda(\sigma) = \frac{12}{5}. $$ Both Montgomery and Huxley prove results slightly stronger than what I've stated. The density conjecture states that one can take $\lambda(\sigma)=2$, but this is only known in limited ranges of $\sigma$. For instance, $\sigma \geq \frac{5}{6}$ follows from Huxley's work. The full range would follow from the Riemann Hypothesis. Zero density estimates are connected to the distribution of primes in short intervals. Ingham, in a paper with the same name as Huxley's, shows that any estimate of the form (1) with $\lambda$ constant implies that any interval $[x,x+x^\theta]$ with $\theta > 1-\lambda^{-1}$ and $x$ sufficiently large contains a prime. The general strategy for proving zero density estimates is to construct "zero detecting polynomials," which are functions that take unusually large values at the zeros of $\zeta$. One then uses "mean and large value estimates" to show that these functions cannot be large too often.<|endoftext|> TITLE: Questions on the group $\mathrm{GL}(H)$ QUESTION [9 upvotes]: $\DeclareMathOperator\GL{GL}\DeclareMathOperator\U{U}$Let $H$ be an infinite dimensional complex Hilbert space. Consider the group $\GL(H)$ of bounded invertible operators on $H$. Question 1. I've heard that $\GL(H)$ is not a topological group regardless we take (the subspace topology of) the strong-$*$, strong, weak, ultra-strong-$*$, ultra-strong or ultra-weak topologies. It quickly becomes a tedious exercise to see how the continuity of multiplication or inverse fails in each situation, can anyone point out a quick proof or some references of these facts? Question 2. If $G$ is a subgroup of $\GL(H)$ that is bounded in norm, is it then a topological group with respect to any of the above topologies? Note that if $G$ is conjugate to a subgroup of the group $\U(H)$ of unitary operators, the answer is affirmative with respect to all of the above mentioned topologies, since they all coincide on $\U(H)$, and multiplication on either side by a fixed $a \in \mathcal{B}(H)$ is always continuous with respect to all of these topologies. Question 3. Is it true that the subgroup $G$ in question 2 is always a conjugate of a subgroup of $\U(H)$? I know questions 2 & 3 are related to Kadison's famous problem of whether every strongly continuous uniformly bounded representations of a locally compact group is equivalent to a unitary one, to which various counter-examples have been found. Though I am not familiar with these works, they still make me suspect that the answer to questions 2 and 3 are both likely to be negative, and it seems to me that Q3 might be easier. REPLY [3 votes]: I think it is possible to give a "nearly unified" approach to Q1 and Q2 (Q3 being answered in the comments), and also to show why a completely unified approach is not possible. I cannot answer one part of Q2 however. General comments: As observed in the OP, translation (on left or right) by a fixed operator is always continuous for any of the topologies. So it suffices to consider continuity at the identity. Question 1 We construct nets $(T_i), (S_i)$ which converge $\sigma$-strong-$^\ast$ to $1$ yet $(T_iS_i)$ does not converge strongly to $1$. This deals with all the "strong" topologies. Let $X$ be the collection of all families $(\xi_n)$ with $\sum_n \|\xi_n\|^2<\infty$, let $X_{<\infty}$ be the finite subsets of $X$, and let $I=X_{<\infty}\times\mathbb N$ with the obvious ordering. Fix a unit vector $\xi_0$. For $i=(F,m)\in I$ choose a unit vector $\eta_i$ with $(\eta_i|\xi_0)=0$ and with $\sum_n |(\eta_i|\xi_n)|^2 < m^{-3}$ for each $(\xi_n)\in F$. Define $T_i:\xi\mapsto m (\eta_i|\xi) \eta_i$ and $S_i:\xi\mapsto m^{-1} (\xi_0|\xi) \eta_i$, so $\|S_i\|=m^{-1}$ and hence $1+S_i\rightarrow 1$ in norm. As $S_i$ has small norm, $1+S_i$ is invertible, and as the spectrum of $T_i$ is $\{0,m\}$, $1+T_i$ is invertible. For any $(\xi_n)\in X$, $$ \sum_n \|T_i(\xi_n)\|^2 = \sum_n m^2 |(\eta_i|\xi_n)|^2 < m^{-1}, $$ so $1+T_i\rightarrow 1$ $\sigma$-strong$^\ast$ (as $T_i$ is self-adjoint). However, as $$ T_iS_i(\xi) = m^{-1} (\xi_0|\xi) T_i(\eta_i) = (\xi_0|\xi)\eta_i, $$ we see that the net $(T_iS_i(\xi_0))$ does not converge in norm, and so $(1+T_i)(1+S_i)$ does not converge strongly to $1$. We cannot extend this to the "weak" topologies, because: Theorem: Let $(T_i), (S_i)$ be nets converging strong$^*$ to $1$. Then $(T_iS_i)$ converges $\sigma$-weakly to $1$. Proof: It suffices to show weak convergence, the $\sigma$-weak case following by replacing $H$ by $H\otimes\ell^2$. Thus, let $\xi,\eta\in H$ can consider that $T_i^*\xi\rightarrow\xi$ and $S_i\eta\rightarrow\eta$ in norm, because of strong$^*$ convergence. Hence $$ (\xi|T_iS_i\eta) = (T_i^*\xi|S_i\eta) \rightarrow (\xi|\eta), $$ as required. Thus we seek a new counter-example for the "weak" topologies. Let $(e_n)_{n\geq 0}$ be an orthonormal sequence in $H$, and consider the operators $T_n:\xi\mapsto (e_n|\xi) e_0$ for $n\geq 1$. Again, the compact operator $T_n$ has spectrum $\{0\}$ and so $1+T_n$ is invertible. By Bessel's Inequality $T_n\rightarrow 0$ strongly, and $T_n^*\rightarrow 0$ weakly, hence $\sigma$-weakly, as $(T_n)$ is bounded. So $1+T_n, 1+T_n^*\rightarrow 1$ $\sigma$-weakly. However, $$ T_nT_n^*(\xi) = (e_0|\xi) T_n(e_n) = (e_0|\xi) e_0 $$ so $(1+T_n)(1+T_n^*) \rightarrow 1 + p_0$ weakly, where $p_0$ is the projection onto $\mathbb Ce_0$. (I do not know a reference. However, looking in e.g. Dixmier's book or Stratila-Zsido, there are exercises which motivate the 2nd counter-example. Once you have this, the 1st counter-example is not so hard to think of.) Question 2 Let $G$ be a bounded subgroup of $GL(H)$, say $K=\sup\{ \|T\| :T\in G \}$. This has an affirmative answer for the strong topologies. Let $(S_i), (T_i)$ be nets in $G$ converging strongly to $1$. For $\xi\in H$, the estimate $$ \| T_i S_i\xi - \xi\| \leq \|T_iS_i\xi - T_i\xi\| + \|T_i\xi-\xi\| \leq K \|S_i\xi - \xi\|+ \|T_i\xi-\xi\| $$ shows that $T_iS_i\rightarrow 1$ strongly. The estimate $$ \|T_i^{-1}\xi-\xi\| = \|T_i^{-1} (\xi-T_i\xi)\| \leq K \|T_i\xi-\xi\| $$ shows that $T_i^{-1}\rightarrow 1$ strongly. As $G^*:=\{T^*:T\in G \}$ is also a bounded subgroup of $GL(H)$, the same arguments applied to $G^*$ and $G$ at the same time show the results for the strong$^*$-topology. By replacing $H$ with $H\otimes\ell^2$ and letting $T\in G$ act as $T\otimes 1$, shows the result for the $\sigma$-strong and $\sigma$-strong$^*$-topologies. Above, we found bounded sequences $(S_n), (T_n)$ which converge $\sigma$-weakly to $1$, but with $S_nT_n\not\rightarrow 1$ weakly. However, a little thought will show that the subgroup these operators generate is not bounded. As I do not have a good source of examples of bounded (but not unitarizable) subgroups of $GL(H)$, I leave this open question: If $G\subseteq GL(H)$ is a bounded subgroup, must it necessarily be a topological group for the weak or $\sigma$-weak topologies?<|endoftext|> TITLE: What is the smallest set of real continuous functions generating all rational numbers by iteration? QUESTION [31 upvotes]: I recently came across this problem from USAMO 2005: "A calculator is broken so that the only keys that still work are the $\sin$, $\cos$, $\tan$, $\arcsin$, $\arccos$ and $\arctan$ buttons. The display initially shows $0$. Given any positive rational number $q$, show that pressing some finite sequence of buttons will yield $q$. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians." A surprising question whose ingenious solution actually shows how to generate the square root of any rational number. I'd like to pose the following questions related to this problem: What is the smallest set of real functions, continuous at all points of $\mathbb{R}$, which can be applied to $0$ to yield a sequence containing all the rational numbers? It's also interesting perhaps weaken this to allowing finite numbers of discontinuities so you can use the rational functions for example: What is the smallest set of real functions, continuous except at a finite set of points, which can be applied to $0$ to yield a sequence containing all the rational numbers? Note that these are slightly different questions to the one above in that we are asking not only to be able to produce any rational from $0$ but to produce all of them at some point after starting at $0$. In the case of the USAMO question you can generate a complete sequence of rationals as well as any given rational but this may not always be true. (See solution for details) For the second question note that from the theory of continued fractions of rational numbers the functions $f(x)=1/x$, $g(x)=x+1$ will generate any given rational starting from $0$. For example since $$\frac{355}{113} = 3+\cfrac{1}{7+\cfrac{1}{16}}$$ we have $\frac{355}{113}=g^{[3]}(f(g^{[7]}(f(g^{[16]}(0)))))$. If we also throw in $h(x)=x-1$ we again have every inverse included hence this set of three functions will generate all rationals. So we know that the smallest set must contain either $1$, $2$ or $3$ functions. In fact as pregunton noted in this related question the functions $f(x)=x+1$ and $g(x)= -1/x$ generate the modular group which acts transitively on $\mathbb{Q}$ and this gives an elegant example with only two functions. REPLY [32 votes]: It is enough with one continuous function. First, I'll give a simple example with one function which is discontinuous at one point. To do it, consider the function $$f:(0,\pi+1)\to(0,\pi+1)$$ with $$ f(x) = \begin{cases} x+1 &\text{if $x<\pi$,} \\ x-\pi &\text{if $x>\pi$,} \\ 1 &\text{if $x=\pi$.}\\ \end{cases} $$ Claim: The sequence $$1,f(1),f^2(1),\dots \tag{$*$}$$ is dense in $(0,\pi+1)$. To verify the claim, it is enough to see that the image is dense in the interval $(0,1)$, and that is true because for every $n$, the number $\lceil n\pi\rceil-n\pi$ is in the image, and the sequence of multiples of $\pi$ modulo 1 is dense in $(0,1)$ due to $\pi$ being irrational. Let $A$ denote the image of the sequence $(*)$. Since $A$ is dense in $(0,\pi+1)$, we can find an homeomorphism $h:(0,\pi+1)\to\mathbb{R}$ with $h(A)=\mathbb{Q}$ (using that $\mathbb{R}$ is countable dense homogeneous, see for example this reference). We can also suppose $h(1)=0$ changing $h$ by $h-h(1)$ if necessary. Then the function $F=hfh^{-1}$ does the trick, because $$F^n(0)=hf^nh^{-1}(0)=h(f^n(1)),$$ so $h(A)$, which is $\mathbb{Q}$, is the image of the sequence $0,F(0),F^2(0),\dots$ To prove that the problem can be solved with one continuous function, we can apply the same argument but taking instead of $f$ a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $0,g(0),g^2(0),\dots$ is dense in $\mathbb{R}$. As Martin M. W. noticed in his answer, those functions are known to exist (they are called transitive maps), this paper gives examples of them. REPLY [21 votes]: You only need one continuous function. There exists a continuous function $f: \mathbb{R} \to \mathbb{R}$ with a dense orbit, according to this MathOverflow answer. As in Saúl's construction, you can conjugate $f$ with a homeomorphism to get a map where the forward iterates of $0$ are $\mathbb{Q}$.<|endoftext|> TITLE: Completeness of the sequence $\left \{ \frac{1}{x+1},\frac{1}{x+2},\frac{1}{x+3}, \dots \right \}$ in $L^2[0,1]$ QUESTION [5 upvotes]: In the book R. M. Young, An introduction to non-harmonic Fourier series, I came across the following problem (page 18): Problem. Show that the sequence $\left \{ \frac{1}{x+1},\frac{1}{x+2},\frac{1}{x+3}, \dots \right \}$ is complete in $L^2[0,1]$. I tried to apply the Müntz-Szasz theorem but it didn't work out. Any ideas or approaches how to show completeness here? Thanks! REPLY [3 votes]: This is just a comment on the present responses which I am adding in the hope that it will be of interest--it will be too long for that format. Firstly, it treats completeness for more general functions of the form $$\frac1{x+\lambda_n}$$ where $(\lambda_n)$ is a sequence of positive numbers (even complex ones with positive real parts) and displays the connection with the Müntz-Szasz theorem. It also uses the Hahn Banach theorem to connect with uniqueness theorems for spaces of analytic functions as is done above. We begin by replacing the $L^2$-space with $C([0,1)$--anything complete in the latter remains so in the former. We now consider for any measure $\mu$ the function $$f(z)=\int_0^1 \frac{d\mu(x)}{x+z}.$$ This is bounded and analytic in the right half plane. We now use the fact that a sequence $(\lambda_n)$ there is the zero set of such a function if and only if $$\sum\left(1-\left|\frac{\lambda_n-1}{\lambda_n+1}\right|\right)<\infty.$$ This suffices to prove your result, together with a wide-ranging set of generalisations. The same proof, with the definition $$f(z)=\int_0^1 x^zd\mu(x),$$ can be used to prove the classic Müntz-Szasz theorem (which assumes that the sequence increases to infinity along the real line), but also variants which allow it to converge in more exotic manners to the boundary of the half-plane.<|endoftext|> TITLE: Existence of surjection vs injection over $\sf ZF$ QUESTION [5 upvotes]: Consider the following statements in $\sf ZF$: (S) If $A, B$ are nonempty sets, then there is a surjection $s:A \to B$, or there is a surjection $t:B\to A$. (I) If $A, B$ are sets, then there is an injection $i:A\to B$, or there is an injection $j:B\to A$. Note that (I) implies (S). Assuming $\sf AC$, both statements are true. Question. Is there a model of $\sf ZF$ in which (S) holds, but not (I)? REPLY [16 votes]: The answer is no. Both (I) and (S) are equivalent to AC over ZF. Indeed, for any set $S$ the class of ordinals $\alpha$ such that $\alpha$ injects into $S$ (resp. $S$ surjects onto $\alpha$) is a set, so there is some ordinal $\beta$ outside this set. Assuming (I) (resp. (S)), there is an injection $S\to\beta$ (resp. surjection $\beta\to S$). In the latter case, we also get an injection $S\to\beta$, by taking any element $S$ to the least element in its preimage. In either case, we see $S$ is well-orderable. Thus both (I) and (S) imply well-ordering theorem and hence AC. Since both (I) and (S) are equivalent to AC, they are also equivalent to each other.<|endoftext|> TITLE: When do binomial coefficients sum to a power of 2? QUESTION [53 upvotes]: Define the function $$S(N, n) = \sum_{k=0}^n \binom{N}{k}.$$ For what values of $N$ and $n$ does this function equal a power of 2? There are three classes of solutions: $n = 0$ or $n = N$, $N$ is odd and $n = (N-1)/2$, or $n = 1$ and $N$ is one less than a power of two. There are only two solutions $(N, n)$ outside of these three classes as far as I know: (23, 3) and (90, 2). These were discovered by Marcel Golay in 1949. There are no more solutions with $N < 30{,}000$. I've written more about this problem here. By the way, I looked in Concrete Mathematics hoping to find a nice closed form for $S(N, n)$ but the book specifically says there isn't a closed form for this sum. There is a sum in terms of the hypergeometric function $_2F_1$ but there's no nice closed form. REPLY [23 votes]: The case $n=2$ was settled by Nagell in 1948 and suspected (?) by Ramanujan in 1913, but in an equivalent form. As John points out in his growing blog post, the $n = 2$ case is a quadratic equation which, via the quadratic formula, requires that $2^n - 7 = x^2$ for some integer $x$. Motivated by who-knows-what, Ramanujan posted the following in 1913 (J. Indian Math.). Question 464. $2^n - 7$ is a perfect square for the values $3, 4, 5, 7, 15$ of $n$. Find other values. A posted "solution" just verified the values of $n$ he gave and did not address whether there are other solutions. The same problem was proposed by Ljunggren in a Norwegian journal in 1943; in 1948 Nagell proved that there are no other solutions, using a quadratic field with $\sqrt{-7}$ and focusing on values of $x$ rather than $n$. Skolem, Chowla, and Lewis (referencing Ramanujan but not aware of Nagell's solution) solved the problem using $p$-adic techniques in 1959, prompting Nagell to republish his easier 1948 proof in English. Meanwhile, in another part of the forest, error-correcting codes arose. With that motivation, Shapiro and Slotnick essentially reconstructed Nagell's approach in 1959. Their subsequent results make use of other error-correcting code structures; techniques in coding veer away from the binomial sum question. As van Lint explained in a 1975 survey, Although as far as perfect codes are concerned the problem has been settled, the purely number-theoretic problem of finding all solutions of (5.2) remains open. where (5.2) is the more general $\sum_{i=0}^e \binom{n}{i} (q-1)^i = q^k$ where $q$ is a power of a prime. Bringing Nagell into the error-correcting code literature occurred by 1964 (Cohen). The OEIS entries A215797, A060728, and A038198 address the problem from different viewpoints. There's one reference to another solution that I have not been able to track down. In a 1998 textbook on error correcting codes, John Baylis writes (p109) ...so $2+n+n^2$ must be a power of 2. It was shown in 1930 that $n = 1, 2, 5$ and 90 are the only positive integers for which this is true. Any idea what 1930 result he has in mind? References: Baylis, Error-Correcting Codes, Chapman & Hall, 1998. Berndt, Choi, Kang, The problems submitted by Ramanujan to the Journal of the Indian Mathematical Society, Contemporary Mathematics 236, 1999. Cohen, A note on double perfect error-correcting codes on $q$ symbols, Information and Control 7, 1964. Nagell, The diophantine equation $x^2 + 7 = 2^n$, Arkiv Math. 4, 1961 (English version of his 1948 article published in Norwegian). Shapiro, Slotnick, On the mathematical theory of error correcting codes, IBM Journal, January 1959 (available through IEEE). Skolem, Chowla, Lewis, The diophantine equation $2^{n+2} - 7 = x^2$ and related problems, Proc. AMS 10, 1959. van Lint, A survey of perfect codes, Rocky Mountain J. Math. 5, 1975.<|endoftext|> TITLE: Global symplectic (orthogonal) type of automorphic representation compels its type to all its local components? QUESTION [6 upvotes]: Let $F$ be a number field. For an irreducible cuspidal automorphic representation $\pi$ of $\operatorname{GL}_n(\mathbb{A}_F)$, we say that $\pi$ is symplectic (or orthogonal) if $L(s,\pi,\bigwedge^{2})$ (or $L(s,\pi,\operatorname{Sym}^2)$) has a pole at $s=1$. I am wondering whether if $\pi=\bigotimes \pi_v$ is symplectic (or orthogonal), then $\pi_v$’s are also symplectic (or orthogonal) for all places $v$? (Here, $\pi_v$ is symplectic (or orthogonal) means that its corresponding Weil–Deligne group representation by local Langlands correspondence is of such type.) Any comments are welcome! REPLY [4 votes]: Here is a proof of the claim using results from Arthur's monograph The Endoscopic Classification of Representations: Orthogonal and Symplectic Groups. Let $N = 2n$ be an even integer, and $\pi$ a cuspidal automorphic representation satisfying $\pi^\vee \cong \pi$. Since $\pi \cong \pi^\vee$, the Rankin-Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. We have the factorisation $L(\pi \times \pi, s) = L(\pi, S^2, s) L(\pi, \wedge^2, s)$ and both factors are non-vanishing at $s = 1$, so exactly one of them must have a pole; thus $\pi$ is orthogonal or symplectic, in the sense of the question, but never both. The discussion preceding Theorem 1.5.3 of Arthur shows that $\pi$ defines an element of his set $\tilde{\Phi}_{\mathrm{sim}}(N)$ of global parameters, and this lies in the subset $\tilde{\Phi}_{\mathrm{sim}}(G)$ for a uniquely determined quasi-split group $G$ whose Langlands dual $\widehat{G}$ is either $SO_{2n}$ or $Sp_{2n}$. Theorem 1.5.3 of Arthur shows that $\widehat{G}$ is symplectic if $\pi$ is of symplectic type, and $\widehat{G}$ is orthogonal if $\pi$ is of orthogonal type. (This is a very deep theorem, despite sounding like a tautology!) Theorem 1.4.2 of op.cit. now shows that there is a cuspidal automorphic representation $\sigma$ of $G$ such that, for every $v$, the Weil–Deligne representation associated to $\pi_v$ is the image in $GL_N$ of the ${}^L G$-valued parameter associated to $\sigma_v$. So, in particular, it is symplectic (resp. orthogonal) if $\pi$ is.<|endoftext|> TITLE: Literature on non-Archimedean analogues of basic complex analysis results QUESTION [8 upvotes]: It looks like there is some literature out there on what might be called 'non-Archimedean complex analysis' e.g. Benedetto - An Ahlfors Islands Theorem for non-archimedean meromorphic functions and Cherry - Lectures on Non-Archimedean Function Theory. I am mainly working on non-Archimedean functional analysis right now and need to become better acquainted with the non-Archimedean analogues of basic results that might be encountered in a first course on complex analysis, up to and including Liouville's Theorem e.g. Cauchy integral formulas, holomorphic functions etc. for a few spectral theory proofs. Of course I am well aware that with many results there will be no such analogue. What I would like to know is if there exists a good introduction to this area that I could look at, that starts with the fundamentals. For example, is there an analogue of 'holomorphic iff analytic'? Any advice much appreciated. This relates to my earlier question: Non-emptiness of spectrum σ(a) in non-Archimedean Banach algebras REPLY [2 votes]: You might like Nonarchimedean Functional Analysis by Peter Schneider; while it’s not directly concerned with $\mathbb{C}$, it does discuss properties of nonarchimedean Banach spaces and the like.<|endoftext|> TITLE: Does complexified isometry group act transitively on tangent bundle of compact Riemannian manifold? QUESTION [6 upvotes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$Let $ g $ be the round metric on the sphere $ S^n $. Since $ S^n $ is compact the isometry group $ \Iso(S^n,g) $ is also compact. And every compact group can be realized as the real points of some (reductive) linear algebraic group. Indeed, $ \Iso(S^n,g) = \O_{n+1}(\mathbb{R}) $. The complex points of this group are $ \O_{n+1}(\mathbb{C}) $. And $ \O_{n+1}(\mathbb{C}) $ acts transitively on the tangent bundle of the sphere $ T(S^n) $. Does this generalize from the round sphere to other compact homogeneous Riemannian manifolds? In other words, Let $ (M,g) $ be a compact Riemannian homogeneous space. Then $ \Iso(M,g) $ is a compact Lie group. So there exists some (reductive) linear algebraic group whose real points are isomorphic to $ \Iso(M,g) $. The question is, does there always exist a linear algebraic group $ G $ such that the real points of $ G $ are isomorphic to the isometry group $$ G_\mathbb{R} \cong \Iso(M,g) $$ and, in addition, the complex points of $ G $ act (transitively, smoothly) on the tangent bundle $ T(M) $? Note that this question is equivalent to a question which does not a priori in involve any geometry: The manifold $ G_\mathbb{C}/H_\mathbb{C} $ is the tangent bundle to $ G_\mathbb{R}/H_\mathbb{R} $ where $ G_\mathbb{R}$, $H_\mathbb{R} $ are compact real forms of $ G_\mathbb{C}$, $H_\mathbb{C} $ Note that while the action of $ G_\mathbb{R} $ is by isometries, the action of $ G_\mathbb{C} $ on $ T(M) $ can only be by isometries if $ M $ is parallelizable. So in particular the action of $ \O_{n+1}(\mathbb{C}) $ on $ T(S^n) $ can only be by isometries in the cases $ n=1,3,7 $. REPLY [2 votes]: Maybe these arguments are of interest to you. It is known that for any compact symmetric space $M$ the tangent bundle $TM$ possesses a canonical structure of a complex manifold. Multiplication by $-1$ on $TM$ is an antiholomorphic involution; its set of fixed points is $M$, when identified with the zero section of $TM$ (see e.g. Thm 2.5a of Szőke, R.: Complex structures on tangent bundles of Riemannian manifolds, Math. Ann. 291 (1991), 409-428). It follows that the action of the Lie group $G_{\mathbf R}$ on $M$ extends to an action on the complex manifold $TM$ by holomorphic automorphisms. One proves that it extends to a holomorphic action of the complexification $G_{\mathbf C}$ of $G_{\mathbf R}$ on $TM$. This can be done by considering the induced morphism from the real Lie algebra ${\mathfrak g}_{\mathbf R}$ of $G_{\mathbf R}$ into the complex Lie algebra ${\mathcal A}(TM)$ of holomorphic vector fields on $TM$. It naturally extends to the complexification ${\mathfrak g}_{\mathbf C}$ of ${\mathfrak g}_{\mathbf R}$, which gives rise to a holomorphic action of the complexification $G_{\mathbf C}^0$ of the neutral component $G_{\mathbf R}^0$ of the real Lie group $G_{\mathbf R}$. The latter action can easily be extended to a holomorphic action of $G_{\mathbf C}$ on the complex manifold $TM$. It is not hard to see that the action is transitive (for more details see e.g. Thm C of Szőke, R.: Automorphisms of certain Stein manifolds, Math. Z. 219 (1995), 357-385)<|endoftext|> TITLE: What is the rank of the period lattice of modular forms? QUESTION [8 upvotes]: Let $f$ be a weight $2$ cusp form for the group $\Gamma_0(N)$. I was experimenting with integrals of the form $$ \int_r^s f(z) \, dz$$ where $r, s \in \mathbf{P}^1(\mathbf{Q})$ and the integral above is over the geodesic in the upper-half plane connecting $r$ and $s$. When I plotted these period integrals, I noticed that the resulting values formed a rank $2$ lattice in $\mathbf{C}$. I have two questions about this: Why would the above period lattice have rank $2$? Since we are integrating $f(z) \, dz$ over paths in the relative homology group $H^1(X_0(N), \{\text{cusps}\}, \mathbf{Z})$, I would expect the rank of the period lattice to grow with the rank of this homology group. Why is the rank of the period lattice always $2$, even if the rank of the homology group is bigger? Inside the period lattice obtained from integrating over $H^1(X_0(N), \{\text{cusps}\}, \mathbf{Z})$, we can consider the sublattice from integrating over just the ordinary homology group $H^1(X_0(N), \mathbf{Z})$, omitting paths that are not closed loops in $X_0(N)$. (Equivalently, the sublattice is formed by only plotting $\int_r^s f(z) \, dz$ when $r$ and $s$ are $\Gamma_0(N)$-equivalent.) What is the index of this sublattice inside the full period lattice? Can one answer this in general? REPLY [3 votes]: My attempt. $f$ is an eigenform with integer coefficients $\in S_2(\Gamma_0(N))$, otherwise it is usually not true that its periods form a lattice in $\Bbb{C}$. As $f=\sum_{d|N} c_d \tilde{f}(dz)$ for some newform of lower level, we can restrict to the case that $f$ is a newform. Following Cremona, taking finitely many $\alpha\in \Bbb{Q}$ representing the finitely many cusps, for some prime $p\equiv 1\bmod N$ not dividing the numerator and denominator of any $\alpha$, $$(1+p-a_p(f))\int_\alpha^{i\infty} f(z)dz=\int_\alpha^{i\infty} (1+p-T_p) f(z)dz$$ $$=\int_\alpha^{p\alpha} f(z)dz+\sum_{k=0}^{p-1} \int_\alpha^{(\alpha+k)/p} f(z)dz\tag{1}$$ $p\alpha$ and $(\alpha+k)/p$ are $\Gamma_0(N)$ equivalent to $\alpha$, whence the RHS is equal to a sum of integrals over closed-loops in $X_0(N)$. From the convergence of the Rankin Selberg integral $\langle f E_2,f\rangle$ you'll get that $p+1-a_p(f)\ne 0$ for some $p\equiv 1\bmod N$ not dividing the numerator and denominator of the $\alpha$, obtaining that the subgroup of $\Bbb{C}$ generated by the $\int_\alpha^\beta f(z),\alpha,\beta\in \Bbb{Q}\cup i\infty$ is a lattice iff $\{\int_\gamma f(z),\gamma$ closed-loop in $X_0(N)\}$ is a lattice. Then the theorem we need is that with $g=\dim_\Bbb{C} S_2(\Gamma_0(N))$ then $ \pi_1(X_0(N))^{ab}\cong \Bbb{Z}^{2g}$. $S_2(\Gamma_0(N))$ has a $\Bbb{C}$-basis $f_1,\ldots,f_g$ of modular forms with integer coefficients. We choose it such that $f_1=f$. With $\gamma_1,\ldots,\gamma_{2g}$ a $\Bbb{Z}$-basis of $\pi_1(X_0(N))^{ab}$, and $\lambda_l=(\int_{\gamma_l} f_1(z)dz),\ldots,\int_{\gamma_l} f_g(z)dz))$, using that a harmonic function on $X_0(N)$ must be constant,you'll get that the columns of the $2g\times 2g$ matrix with rows $(\Re(\lambda_l),\Im(\lambda_l))$ are $\Bbb{R}$-linearly independent, whence so are its rows, so the $\lambda_l$ generate a lattice $\Lambda$ in $\Bbb{C}^g$. Let $\Bbb{T}$ be the $\Bbb{Z}$-algebra generated by the Hecke operators. Using that $f_1$ is a newform we can take an element $P\in \Bbb{T}$ such that $Pf_j=0$ except for $Pf_1=cf_1\ne 0$. Each $\int_\gamma Pf_j(z)dz$ is given by integrating $f_j$ on some curves from cusps to cusps, with the same $p$ as above and the second equality of $(1)$ you'll get that there is some $d$ such that for all closed-loop $\gamma$ in $X_0(N)$, $$(\int_\gamma (1+p-T_p)Pf_1(z)dz,\ldots,\int_\gamma (1+p-T_p)Pf_g(z)dz) \in \frac1d\Lambda$$ This proves that for any $\lambda\in \Lambda$, $((1+p-a_p(f))c\lambda_1,0\ldots,0) \in \frac1d\Lambda$, and hence $\{ \int_\gamma f_1(z)dz, \gamma \in \pi_1(X_0(N))\}$ is a lattice in $\Bbb{C}$, from which the $\int_\alpha^\beta f_1(z)dz,\alpha,\beta\in \Bbb{Q}\cup i\infty$ generate a lattice in $\Bbb{C}$.<|endoftext|> TITLE: How do pure mathematicians assess whether their research ambitions can be realistically achieved? QUESTION [60 upvotes]: I am an enthusiastic but ever-so-slightly naive PhD student and have been 'following my nose' a lot recently, seeing whether topics that I have studied can be generalised or translated in various ways into unfamiliar settings; exploring where the theory breaks down etc. When doing this, I have found it very difficult to assess whether it is going to 'work' in the more general sense of whether it could lead to a viable project for a PhD thesis or perhaps a short research paper. I guess it becomes easier to get a feel for these things as one gains experience and a better sense of perspective. Of course, one added complication over the past year has been that due to various lockdowns it been difficult to get to know other mathematicians and run ideas past them in the natural way that would have occurred in previous years. Suppose that a wise and experienced pure mathematician wishes to generalise a particular theory or shed some light on an open problem and will devote, say, at least 6 months to it. What reasonable steps should be taken to maximise the likelihood of this being a fruitful endeavour? My main concern personally would be (is?) a previously unforeseen obstacle rearing its ugly head only after a significant amount of time and energy has been invested that brings the whole thing crashing down. How can this scenario be avoided when exploring something brand new? EDIT: Although I have referred to my own circumstances above, my question relates primarily to the more general issue. REPLY [94 votes]: Over decades, and across multiple research fields, I've noticed a way to predict I'm on track to make progress. I discover something interesting, only to learn it is already known. As a student, this was incredibly discouraging, and in fact I stopped some lines of research for this very reason. But by now I'm used to it: I start looking at a new area, and have an insight. Arg, it turns out people knew it 10 years ago. I read some more, think some more, and have a new insight. Careful searching reveals a paper with that result from three years ago. Too bad—but that paper is fascinating, and I can deeply appreciate it and feel kinship with the author. Thinking about it leads to another insight, which I start writing up. Oof, then I see a preprint from a month ago which says the same thing. What I've learned over time is that this pattern of rediscovery, particularly if the dates of things I've been rediscovering get more and more recent, is a reliable sign I'm on a good path, and that I'm building my intuition in an area other people care about. So keep following your nose, check back with the literature regularly, and take any rediscoveries as a green light, not a red light. REPLY [32 votes]: The unfortunate fact of life, from your perspective, is that most mathematical questions are either intractable or known (or follow easily from known results). This is not to say that there aren't also a lot of interesting mathematical questions at the boundary between the two—otherwise doing mathematical research would be a hopeless enterprise—but it's not always so easy to locate that boundary. Roughly speaking, there are two cases to consider. Case 1 is that there is specific well-known open problem or conjecture that you're trying to attack. In this case, there is a reasonably good chance that someone has written a survey paper that outlines the partial progress that has been made. If you're lucky, someone may also have described some of the fundamental barriers to further progress. For example, in the case of the $\mathsf{P} = \mathsf{NP}$ problem, there are the so-called "naturalization" and "relativization" (or "algebrization") barriers. If you have what you think is a new idea, then chances are it's somewhat similar to something that has already been tried. The survey paper can help direct you to the most relevant literature. Often you'll find that your question has already been answered, or that there is a known barrier which means that a simple bare-hands approach is highly unlikely to succeed. But if you're lucky, you'll find some work that is very close to what you're thinking, yet does not answer your questions. This is usually a promising sign that you've found the elusive boundary that I alluded to above. Of course it's tough to be sure; perhaps you'll prove something that you think is new but which an expert can see is a very easy consequence of known results, or which turns out to be not very interesting or fruitful. But to some extent, that is an occupational hazard of all research. Case 2 is that you're not trying to solve a specific problem but are trying to develop some new theory. I have given my opinion about theory-building elsewhere on MO. Compared to Case 1, the advantage of Case 2 is that you have a somewhat better chance of coming up with something new. For example, if you're trying to generalize something, maybe nobody else has studied that generalization, not because it was intractable, but because they didn't see the point. One disadvantage is that it may be harder to figure out whether someone else has anticipated you, because there may not be a nice survey paper. Nevertheless, you should make a diligent effort to search for related work. If nothing else, the process will help you clarify in your mind how your ideas connect with other people's ideas. After all, if you're engaged in theory-building, clarifying your thoughts is the name of the game. Another disadvantage, especially when it comes to producing a research paper, is that if you're not making tangible progress toward answering a question that people care about, you may have trouble convincing people that you have made a meaningful contribution. But again, that is an occupational hazard of all theory-building. As for your worry that some unforeseen obstacle causes everything to come crashing down, I actually think that this is not the main thing to worry about. It can happen; for example, I know someone who wasted a year of his Ph.D. program because he was relying on a published "theorem" that was known (but not to him or his advisor) to be false. It can also happen if you foolishly decide to try to prove the Riemann hypothesis without reading any literature. But if you have done some due diligence searching the literature, and if you take the sensible approach of starting with small things that you can definitely prove and working your way up from there, then you're not likely to bang your head into the "intractability" side of the boundary. The larger risk is usually that you'll produce something that turns out to be known, or not interesting. So if you're not sure what you're doing, you shouldn't spend six months working on something without getting some feedback (ideally from your advisor, if you're a student) about whether what you're doing is worthwhile. On the other hand, for a shorter amount of time, just following your nose can be worthwhile even if it "amounts to nothing" in terms of publishable results, because it will probably give you a good handle on the subject that will serve you well in the future.<|endoftext|> TITLE: Looking for a "clever" argument for a $q$-series identity QUESTION [10 upvotes]: Consider the below $q$-series identity. One of the things I like about this expansion is how nicely the difference on the left hand side factors to the right hand side of the equation. $$\prod_{k\geq1}(1+q^k)^3-\prod_{k\geq1}(1+q^{3k}) =3q\prod_{n\geq1}(1+q^n)(1+q^{9n})^2(1+q^n+q^{2n}+\cdots+q^{8n}).$$ I have a "not-so-neat" proof, so QUESTION. Can you provide a "nifty" justification? Caveat: you decide what is "nifty". REPLY [16 votes]: Here's a proof that indicates a systematic method for proving such identities. Let $\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})$, with $q = e^{2 \pi i z}$. The identity you state in the equation is the same as $$ \frac{\eta^{3}(2z)}{\eta^{3}(z)} - \frac{\eta(6z)}{\eta(3z)} = 3 \frac{\eta(2z) \eta^{2}(18z)}{\eta^{2}(z) \eta(9z)}. $$ Replacing $z$ with $2z$ and multiplying by $\eta(2z) \eta(4z) \eta^{2}(6z)$ gives the equivalent $$ \frac{\eta^{4}(4z) \eta^{2}(6z)}{\eta^{2}(2z)} - \eta(2z) \eta(4z) \eta(6z) \eta(12z) = 3 \frac{\eta^{2}(4z) \eta^{2}(6z) \eta^{2}(36z)}{\eta(2z) \eta(18z)}.$$ Results about $\eta(z)$ in Gordon and Hughes - Multiplicative properties of $\eta$-products. II, and Ligozat - Courbes modulaires de genre 1 (MR) guarantee that each of the three terms in the above equation are in the space of weight $2$ cusp forms for the group $\Gamma_{0}(72)$. This space has dimension $5$, and if $f = \sum_{n=1}^{\infty} a_{n} q^{n}$ is a function in the space with $a_{1} = a_{2} = \dotsb = a_{7} = 0$, it follows that $f = 0$. It suffices to verify that the coefficients of $q^{1}$ through $q^{7}$ are the same for the two sides of the equivalent identity above.<|endoftext|> TITLE: A common combinatorial description for a certain type of recurrences QUESTION [5 upvotes]: For integer-valued sequences $(x_n)_{n=0}^\infty$, consider recurrences of the form $$x_n=ax_{n-1}+(bn+c)x_{n-2} \tag{$*$}\label{star}$$ for $n\ge2$, where $a,b,c$ are integers. There seem to be many such sequences in the OEIS — see e.g. A000085, A001475, A005425, and A000898 — with various combinatorial descriptions. Question: Is there a common combinatorial description for all integer-valued sequences $(x_n)_{n=0}^\infty$ satisfying recurrences of the form \eqref{star} — at least for natural $a$, $b$ ,$c$ and for some initial conditions (on $x_0$, $x_1$, depending on $a$, $b$, $c$)? REPLY [5 votes]: So, the Fibonacci numbers can be constructed from this recursion, so it is natural to look for generalizations of those (and combinatorial models for the Fibonacci numbers). The Fibonacci number $f_n$ (up to some index shift), can be seen as the number of integer compositions of $n$ (list of numbers summing to $n$), using only the numbers $1$ and $2$. Now, the recursion above, $x_n = a x_{n-1} + (bn+c)x_{n-2}$ hints that we also look at integer compositions using $1$ and $2$, but the $1$ comes in $a$ different colors. Moreover, the 2s have two types, either, one of $c$ colors, or it is a 'special' 2, with one of $b$ colors and a label. The label is between 1 and the total sum of entries up to that point. For example, $a=3$, $b=3$, $c=2$ $n=8$ has the object $$ 1_3, \; 2_{1, 1}, \; 2_{3, 5}, \; 2_{2}, \; 1_{1}. $$ The numbers sum to $n$, and subscripts of the $1$ is between $1$ and $a$. The $2's$ with only one subscript has the subscript between $1$ and $c$. Finally, each $2$ with two substripts, has the first subscript (color) between $1$ and $b$. The second subscript must be between $1$ and the sum of all numbers in the list up to that point. For example, for $2_{3, 5}$, the 5 is the maximum allowed since $1+2+2=5$. Below is Mathematica code for generating such compositions: a = 3; b = 3; c = 2; Clear[f]; f[0] := {{}}; f[1] := Table[{{1, j}}, {j, a}]; f[n_] := f[n] = Join[ Join @@ Table[ Append[f, {1, aa}], {f, f[n - 1]}, {aa, a}] , Join @@ Table[ Append[f, {2, bb}], {f, f[n - 2]}, {bb, b}] , Join @@ (Join @@ Table[ Append[f, {2, cc, j}], {f, f[n - 2]}, {cc, c}, {j, n}]) ]; The natural initial condition is $x_0 =0$ and $x_1 = a$. (In the code above, I use $f$ instead of $x$)<|endoftext|> TITLE: Mirror symmetry for K3 fibered Calabi-Yau threefolds QUESTION [6 upvotes]: By a K3 fibered Calabi-Yau threefold, I mean a smooth projective threefold $X$ with trivial canonical class and $h^{1,0}(X) =h^{2,0}(X) = 0$ that has a fibration $X \rightarrow \mathbb P^1$ whose generic fibers are smooth K3 surfaces (to be denoted by $S_X$) I gathered some examples of mirror pairs of those K3-fibered CY threefolds. Borcea--Voisin CY's, CY double coverings over some quasi-Fano threefolds in this paper. From these examples, I observed: In these examples of mirror pairs of K3-fibered CY threefolds X, Y. Their generic K3 fibers are K3 mirrors (in the definition of this paper). In concrete words, there are embeddings lattices $\phi_X: Pic(S_X) \rightarrow L$, $\phi_Y: Pic(S_Y) \rightarrow L$ to K3 lattice $L$ of rank 22 and signature $(3, 19)$ such that $L = im(\phi_X) \oplus im(\phi_Y) \oplus U$ ,where $U$ is a unimodular hyperbolic plane in $L$. My question is: Are there other examples of mirror pairs of K3-fibered CY threefolds that support or fail this observation. If other known examples also support this observation, Is this observation expected from any kind of theories of mirror symmetry of Calabi-Yau threefolds with K3 fibration? REPLY [5 votes]: Morally speaking, a K3 fibration on one side of the mirror correspondence should correspond to a K3 degeneration on the other side. In the physics literature, I think this observation goes back to studies of the F-theory/heterotic duality in the '90s; some coauthors and I wrote a paper with a more combinatorial take that describes how to understand and generate toric examples (though in the toric case, identifying the appropriate polarizing lattice can be subtle). Studies of Tyurin degeneration, including the Nam-Hoon Lee paper you linked and the work of Doran-Harder-Thompson, should give you another perspective on the degeneration/fibration duality.<|endoftext|> TITLE: Is Mazur's analogy between arithmetic and topology formal, in any sense? QUESTION [12 upvotes]: I preface my question by admitting I know no algebraic geometry nor algebraic number theory. I do know some algebraic topology. I'm a student. Recently I learned about sheaf cohomology. Then a little bit of etale cohomology, as much as I could stomach having never studied algebraic geometry. Then I came across Artin-Verdier duality, in particular the notion that $\mathcal{O}_K$ is 'like a 3-manifold.' This led me to the interesting area of arithmetic topology that wants to understand some larger 'arithmetic $\leftrightarrow$ topology dictionary.' Now I've done some reading to try to grasp the big picture of arithmetic topology. But I'm unclear on one point: Is the analogy pursued by this arithmetic $\leftrightarrow$ topology dictionary formal, in any sense? So far I've seen it said how this analogy gives a nice way of thinking about number-theoretic things with topology (e.g. prime ideals are like links, and their factors are like the constituent knots.) The words 'inspire' and 'motivate' are used a lot. And there are precise comparisons to be made between the objects on either side (e.g. the algebraic fundamental group of $\mathrm{Spec} ~\mathbb{Z}$ is isomorphic to the classical fundamental group of $S^3$.) But I'd like to know whether there is some larger framework that rigorously explains why this analogy exists. REPLY [13 votes]: For this analogy, like most analogies in mathematics, and indeed like most philosophical principles in mathematics, one can certainly make a part of it formal and rigorous, but I don't think any true formal statement could ever capture all of what we mean by the analogy. In particular, by well-chosen definitions, one can write down statements of the form "If X is either a 3-manifold or the ring of integers of a number field, then something is true about X", where "something" is expressed the same way in each case. But there's no reason to expect that there is a single statement that implies all true such statements. In particular, one can certainly not get an equivalence of categories between some category of 3-manifolds and some category of number fields (as Wojowu suggests in the comments), or any kind of correspondence between one 3-manifold and one number field, that respects the interesting structure like Artin-Verdier duality. (Thus I think the equivalence of fundamental groups between $S^3$ and $\mathbb Z$ is a red herring.) Note that in Verdier duality, a pretty fundamental concept is an orientation. Any 3-manifold is either oriented or has a double cover to be oriented. But from the form of Artin-Verdier duality, for a number field to be oriented, it would have to contain the $n$th roots of unity for all $n$, which is impossible. So the "oriented double cover" in this setting is actually a cover of infinite degree! Covering spaces and dualizing sheaves are some of the concepts we absolutely do want to match up, so I don't think there's any way to wriggle out of this.<|endoftext|> TITLE: On the endofunctor $\Sigma^\infty\Omega^\infty$ QUESTION [5 upvotes]: Consider the co-monad $M:=\Sigma^\infty \Omega^\infty$ on the category of spectra. It is clear that given a pointed space $X$, $M\Sigma^\infty X=\Sigma^\infty E(X)$,where $E(X)$ is the free unital $E_\infty$-algebra on $X$. As $\Sigma^\infty$ commutes with colimit, $M\Sigma^\infty X$ is equivalent to the free $E_\infty$-algebra on the spectrum $\Sigma^\infty X$, that is, to: $$ \lor_n (\wedge^n (\Sigma^\infty X)_{hS_n}).\tag{1} $$ Is this right? It looks like given an arbitrary spectrum $A$, there is no equivalence between $\Sigma^\infty\Omega^\infty A$ and $ \lor_n (\wedge^n (A)_{hS_n}) $ Is there any general condition for having such an equivalence? REPLY [4 votes]: For connected spectra $A$, there is an equivalence $$\Sigma^\infty \Omega^\infty A \simeq \bigvee_{n=1}^\infty A^{\wedge n}_{h\Sigma_n}$$ if and only if $A$ is a wedge summand of a suspension spectrum. This is Theorem 1.2 in N. Kuhn's paper Suspension Spectra and Homology Equivalences (TAMS, 1983). Kuhn calls spectra that are wedge summands of suspension spectra spacelike. I am guessing that if $A$ is not connected then such an equivalence can not exist. It is easy to see that there can not be an equivalence of ring spectra.<|endoftext|> TITLE: $E$-(co)homology of $BU(n)$ (Reference request) QUESTION [7 upvotes]: I am currently reading Lurie's notes on Chromatic Homotopy Theory (252x) and in Lecture 4 (https://www.math.ias.edu/~lurie/252xnotes/Lecture4.pdf), he skims through the calculation of $E^{\ast}(BU(n))$ and $E_\ast(BU(n))$ where $E^\ast(-)$ is a complex oriented cohomology theory. I am able to follow the calculation but am struggling with filling in the gaps. Could someone kindly refer me to a resource where these computations have been explicitly carried out? PS- I have tried having a look at Adams's blue book, I would appreciate it if there's a relatively modern reference/textbook. REPLY [9 votes]: I prefer an approach that does not go via ordinary cohomology. I'll assume you're happy that $E^*(BU(1))=E^*(\mathbb{C}P^\infty)=E^*[[x]]$. We can then use a Künneth theorem to get $E^*(BU(1)^d)=E^*[[x_1,\dotsc,x_d]]$. We have an inclusion $i\colon BU(1)^d\to BU(d)$, giving $$ i^*\colon E^*(BU(d))\to E^*(BU(1)^d)=E^*[[x_1,\dotsc,x_d]]. $$ This is permutation-invariant up to homotopy, and so gives a map from $E^*(BU(d))$ to the subring of symmetric functions in $x_1,\dotsc,x_d$. This subring is just $E^*[[c_{d1},\dotsc,c_{dd}]]$, where $c_{di}$ is the $i$'th elementary symmetric function in $x_1,\dotsc,x_d$. Our task is to prove that the resulting map $$ i^* \colon E^*(BU(d)) \to E^*[[c_{d1},\dotsc,c_{dd}]] $$ is an isomorphism. One approach is to use the Projective Bundle Theorem, as follows. Let $V$ be a complex bundle of dimension $d$ over a CW complex $X$. We then have an associated projective bundle $PV$ over $X$, with fibre $P(V_a)\simeq\mathbb{C}P^{d-1}$ over each point $a\in X$. There is then a tautological line bundle $L$ over $PV$, which is classified by a map $f\colon PV\to BU(1)$, giving an Euler class $f^*(x)\in E^2(PV)$. We will just write $x$ instead of $f^*(x)$. We now have a map $$ \phi\colon\bigoplus_{i=0}^{d-1} E^{*-2i}(X) \to E^*(PV) $$ given by $\phi(a_0,\dotsc,a_{d-1})=\sum_ia_ix^i$. More generally, we have a map $\phi_U\colon\bigoplus_iE^{*-2i}(U)\to E^*(P(V|_U))$ for each open subspace $U$ of $X$. If $V$ can be trivialised over $U$ then $P(V|_U)\simeq U\times\mathbb{C}P^{d-1}$ and it is easy to see that $\phi_U$ is an isomorphism. The maps $\phi_U$ are easily seen to be compatible with Mayer-Vietoris sequences, so if $\phi_U$, $\phi_W$ and $\phi_{U\cap W}$ are isomorphisms, then so is $\phi_{U\cup W}$ (by an application of the five lemma). If $X$ is compact then it can be written as the union of finitely many open sets over which $V$ can be trivialised, and it follows that $\phi_X$ is an isomorphism. We can now pass to limits to show that $\phi$ is an isomorphism in all cases. In particular, this means that the map $\pi^*\colon E^*(X)\to E^*(PV)$ is injective. Now take $V$ to be the tautological bundle over $BU(d)$, which we can think of as the Grassmannian of $d$-dimensional subspaces of $\mathbb{C}^\infty$. From this point of view it is not hard to identify $PV$ with $BU(1)\times BU(d-1)$, and we can assume inductively that $E^*(BU(d-1))$ is as expected. It follows that the ring $$ E^*(BU(1)\times BU(d-1)) = E^*[[x,c_{d-1,1},\dotsc,c_{d-1,d-1}]] $$ is freely generated by $\{1,x,\dotsc,x^{d-1}\}$ as a module over $E^*(BU(d))$. On the other hand, straightforward algebra shows that it is also freely generated by the same set as a module over $E^*[[c_{d1},\dotsc,c_{dd}]]$. The induction step can be deduced from this.<|endoftext|> TITLE: Is $\mathbb{Q}$ the orbit of a rational function under iteration? QUESTION [9 upvotes]: In this previous post I asked for the smallest set of continuous real functions that could generate $\mathbb Q$ by iteration starting from $0$. Surprisingly one continuous function suffices. In the question I gave the example of three rational functions that generate $\mathbb{Q}$, $f(x)=1/x$, $g(x)=x+1$ and $h(x)=x-1$. It would be interesting to know if this is best possible and in particular whether one rational function can generate all of $\mathbb{Q}$: Can $\mathbb{Q}$ be generated as the orbit of fewer than 3 rational functions? The question Orbits of rational functions asks a more general question but I don't think explicitly answers it for $\mathbb{Q}$ itself. REPLY [7 votes]: A rational function is as a self-map of $\mathbb P^1$. With that understanding, as was noted earlier, it is possible to generate all of the points $\mathbb P^1(\mathbb Q)$ by starting with the point $0$ and applying elements of the semi-group $\langle f_1,f_2\rangle$ generated by iteration using the two functions $f_1=x+1$ and $f_2=-1/x$. In this construction, both $f_1$ and $f_2$ are rational maps of degree $1$. However, if one instead uses sets of rational maps $f(z)\in\mathbb Q(z)$ of degree at least $2$, then no finitely generated semi-group of such rational maps has an orbit that contains all of $\mathbb P^1(\mathbb Q)$, and indeed, any such orbit will be fairly sparse. Here's a quick proof (shown to be by Wade Hindes). Let $\mathcal F=\langle f_1,\ldots,f_r\rangle$, where $f_i\in\mathbb Q(z)$ has degree $d_i\ge2$. Then we have the height estimate $$ h\bigl(f_i(P)\bigr) \ge d_i h(P) - C(f_i). $$ It follows that for each $i$, $$ f_i\bigl(\mathbb P^1(\mathbb Q)\bigr) := \bigl\{ f_i(Q) : Q \in \mathbb P^1(\mathbb Q) \bigr\} $$ has density $0$, where we use the height function to count points. But then for any starting point $P \in \mathbb P^1(\mathbb Q)$, the full orbit satisfies $$ \mathcal F(P) := \bigl\{ f(P) : f\in\mathcal F\bigr\} \subseteq \bigcup_{1\le i\le r} f_i\bigl(\mathbb P^1(\mathbb Q)\bigr). $$ Thus the orbit $\mathcal F(P)$ is the union of finitely many sets of density $0$, so the orbit $\mathcal F(P)$ has density $0$.<|endoftext|> TITLE: Random walk with decreasing steps QUESTION [5 upvotes]: I have a random walk $$R(t)= \sum_{n0$. I think that someone must have studied this before. I am interested in understanding the behavior of $R(t)$ for large $t$. For example can we estimate the probability of $R(t) \in [1, x)$? Obviously, $E(R(t))=0,$ and $Var(R)= \tfrac{1}{3}\sum_{n0$. Using the inequality $\dfrac{\sinh u}u0$, \begin{equation} P(R(t)\ge x)\le e^{-zx+z^2 B_{a,t}/6}. \end{equation} The latter bound on $P(R(t)\ge x)$ is minimized at $z=3x/B_{a,t}$. Thus, \begin{equation} P(R(t)\ge x)\le e^{-3x^2/(2B_{a,t})}. \end{equation}<|endoftext|> TITLE: When does $G\times G\times G$ admit a faithful group action on a set of size $|G|$? QUESTION [22 upvotes]: [Edited due to YCor's comment:] Given a finite group $G$, under what conditions does $G\times G\times G$ (the direct product of three copies of $G$) admit a faithful group action on a set of size $|G|$? This is equivalent to an injective group homomorphism from $G\times G\times G$ to $S_{|G|}$. For reference, for every group $G$ with a trivial center, $G\times G$ admits an easy faithful group action on $G$ by $(g_1, g_2): g \rightarrow g_1 g g_2^{-1}$. REPLY [8 votes]: This is more of an extended comment than an answer. I will determine all the abelian groups failing to act faithfully on at most $n/3$ points. Suppose $G$ is abelian, say $G = C_{q_1} \times C_{q_2} \times \cdots \times C_{q_k}$, where $q_1, q_2, \dots$ are (not necessarily distinct) prime powers. Then the minimal faithful permutation action of $G$ has $q_1 + q_2 + \cdots + q_k$ points. For a proof see https://mathoverflow.net/a/409831/20598. Hence $G$ does not act faithully on $|G|/3$ points iff $$q_1 + \cdots + q_k > q_1 \cdots q_k / 3.$$ It is maybe easier to think of this as $$ \sum_{i=1}^k q_i / (q_1 \cdots q_k) > 1/3 . $$ Each of the terms here is at most $1/2^{k-1}$, so $k / 2^{k-1} > 1/3$, so $k \leq 4$. By further analyzing $k=1,2,3,4$ we find all the solutions (where $q_1 \leq \cdots \leq q_k$): $$ (q_1),\\ (2, q_2), (3, q_2),\\ (4, 4), (4, 5), (4, 7), (4, 8), (4, 9), (4, 11), (5, 5), (5, 7),\\ (2, 2, q_3) \qquad (2 \leq q_3 \leq 11),\\ (2, 3, 3), (2, 3, 4), (2, 2, 2, 2), (2, 2, 2, 3). $$<|endoftext|> TITLE: Riemann-Hilbert approach to Selberg integral QUESTION [5 upvotes]: I am interested in matrix integrals, and I have seen many mentions to a certain Riemann-Hilbert approach that indicate that this is a very powerful tool to can be used in this area, when coupled with the theory of orthogonal polynomials. I would like to understand it, but the sources I found were hard to follow. It would be helpful if I could see it in action in a simple case, so I suggest what follows. Consider the matrix integral $$ f(a,N)=\int_{0}^\infty e^{-a\sum_{m=1}^\infty \frac{1}{m}{\rm Tr}(X^m)}|\Delta(X)|^2dX,$$ where $a>0$, $\Delta(X)$ is the Vandermonde and $X$ is diagonal of dimension $N$. I would like to know if this kind of integral is amenable to the R-H approach (with a non-polynomial potential). The series in the exponent is only finite if $0< X< 1$, in which case it gives $-{\rm Tr}\log(1-X)$ so using that $e^{-\infty}=0$ (I am not sure how rigorous this can be made): $$f(a,N)=\int_{0}^1 \det(1-X)^a|\Delta(X)|^2dX,$$ which is a particular case of the Selberg integral and there is an explicit solution to it, which is $$\prod_{j=0}^{N-1}\frac{\Gamma(a+j+1)j!(j+1)!}{\Gamma(a+N+j+1)}.$$ My question is: how can the R-H approach be applied to the first integral in order to produce the Selberg result? REPLY [5 votes]: Let me formulate the problem in a slightly more general way: We seek to evaluate the large-$N$ limit of the matrix integral $$\int e^{-\beta\,{\rm Tr}\,V(X)}|\Delta(X)|^\beta dX\equiv e^{-\beta N^2 F},$$ integrated over $N\times N$ Hermitian matrices $X$. In the OP the index $\beta=2$ and $V(X)=(a/2)\sum_m m^{-1}X^m$, which creates convergence problems, here I assume $V(\lambda)$ is a well defined function of the eigenvalues $\lambda$ of $X$. My goal here is to reduce the calculation of the matrix integral in the large-$N$ limit to the solution of a nonlinear integral equation which can be solved by the Riemann-Hilbert technique. Note for the OP: Selberg integrals are exact finite-$N$ results. The Riemann-Hilbert technique evaluates these integrals in the large-$N$ limit, by the method of stationary phase (or steepest descent). There is therefore no direct connection between the two techniques. Because of the invariance under unitary transformations the integral over the matrix elements can be reduced to an integral over the eigenvalues, $$e^{-\beta N^2 F}=\int_{-\infty}^\infty d\lambda_1\cdots \int_{-\infty}^\infty d\lambda_N \,e^{-\beta U(\lambda_1,\ldots\lambda_N)},$$ $$U(\lambda_1,\ldots\lambda_N)=\sum_{i=1}^N V(\lambda_i)-\sum_{i TITLE: Monotonic dependence on an angle of an integral over the $n$-sphere QUESTION [6 upvotes]: Let $v,w \in S^{n-1}$ be two $n$ dimensional real vectors on sphere. Consider the following integral: $$ \int_{x \in S^{n-1}} \big|\langle x,v \rangle\big|\cdot\big|\langle x,w \rangle\big|\; dx. $$ Since the integration is taking over the sphere, we have rotation invariance and the value of the integration only depends on the value of $\langle v,w \rangle$. Now the question is to show that the integral is a non-decreasing function w.r.t $\langle v,w \rangle$, for $\langle v,w \rangle > 0$. I believe that this question is connected to the problem of packing two pairs of two antipodal points on a sphere (i.e. four diametrically symmetric points), but I could not show the connection. Any help is much appreciated. REPLY [5 votes]: That is technically a 2D question. We can assume that $v=e^{-it}, w=e^{it}\in\mathbb R^2$ ($0 TITLE: Does iterating the derivative infinitely many times give a smooth function whenever it converges? QUESTION [69 upvotes]: I am a graduate student and I've been thinking about this fun but frustrating problem for some time. Let $d = \frac{d}{dx}$, and let $f \in C^{\infty}(\mathbb{R})$ be such that for every real $x$, $$g(x) := \lim_{n \to \infty} d^n f(x)$$ converges. A simple example for such an $f$ would be $ce^x + h(x)$ for any constant $c$ where $h(x)$ converges to $0$ everywhere under this iteration (in fact my hunch is that every such $f$ is of this form), eg. $h(x) = e^{x/2}$ or simply a polynomial, of course. I've been trying to show that $g$ is, in fact, differentiable, and thus is a fixed point of $d$. Whether this is true would provide many interesting properties from a dynamical systems point of view if one can generalize to arbitrary smooth linear differential operators, although they might be too good to be true. Perhaps this is a known result? If so I would greatly appreciate a reference. If not, and this has a trivial counterexample I've missed, please let me know. Otherwise, I've been dealing with some tricky double limit using tricks such as in this MSE answer, to no avail. Any help is kindly appreciated. $\textbf{EDIT}$: Here is a discussion of some nice consequences know that we now the answer is positive, which I hope can be generalized. Let $A$ be the set of fixed points of $d$ (in this case, just multiples of $e^x$ as we know), let $B$ be the set of functions that converge everywhere to zero under the above iteration. Let $C$ be the set of functions that converges to a smooth function with the above iteration. Then we have the following: $C$ = $A + B = \{ g + h : g\in A, h \in B \}$. Proof: Let $f \in C$. Let $g$ be what $d^n f$ converges to. Let $h = f-g$. Clearly $d^n h$ converges to $0$ since $g$ is fixed. Then we get $f = g+h$. Now take any $g\in A$ and $h \in B$, and set $f = g+h$. Since $d^n h$ converges to $0$ and $g$ is fixed, $d^n f$ converges to $g$, and we are done. Next, here I'm assuming the result of this thread holds for a general (possibly elliptic) smooth linear differential operator $d : C^\infty (\mathbb{R}) \to C^\infty (\mathbb{R}) $. A first note is that fixed points of one differential operator correspond to solutions of another, i.e. of a homogeneous PDE. Explicitly, if $d_1 g = g$, then setting $d_2 = d_1 - Id$, we get $d_2 g = 0$. This much is simple. So given $d$, finding $A$ from above amounts to finding the space of solutions of a PDE. I'm hoping that one can use techniques from dynamical systems to find the set $C$ and thus get $A$ after the iterations. But I'm approaching this naively and I do not know the difficulty or complexity of such an affair. One thing to note is that once we find some $g \in A$, we can set $h(x) = g(\varepsilon x)$ for small $\varepsilon$ and $h \in B$. Conversely, given $h \in B$, I'm wondering what happens when set set $f(x) = h(x/\varepsilon)$, and vary $\varepsilon$. It might not coincide with a fixed point of $d$, but could very well coincide with a fixed point of the new operator $d^k$ for some $k$. For example, take $h(x) = cos(x/2)$. The iteration converges to 0 everywhere, and multiplying the interior variable by $2$ we do NOT get a fixed point of $d = \frac{d}{dx}$ but we do for $d^4$. I'll leave it at this, let me know again if there is anything glaringly wrong I missed. REPLY [57 votes]: I was able to adapt the accepted answer to this MathOverflow post to positively answer the question. The point is that one can squeeze more out of Petrov's Baire category argument if one applies it to the "singular set" of the function, rather than to an interval. The key step is to establish Theorem 1. Let $f \in C^\infty({\bf R})$ be such that the quantity $M(x) := \sup_{m \geq 0} |f^{(m)}(x)|$ is finite for all $x$. Then $f$ is the restriction to ${\bf R}$ of an entire function (or equivalently, $f$ is real analytic with an infinite radius of convergence). Proof. Suppose this is not the case. Let $X$ denote the set of real numbers $x$ for which there does not exist any entire function that agrees with $f$ on a neighbourhood of $x$ (this is the "entire-singular set" of $f$). Then $X$ is non-empty (by analytic continuation) and closed. Next, let $S_n$ denote the set of all $x$ such that $M(x) \leq n$ for all $m$. As $M$ is lower semicontinuous, the $S_n$ are closed, and by hypothesis one has $\bigcup_{n=1}^\infty S_n = {\bf R}$. Hence, by the Baire category theorem applied to the complete non-empty metric space $X$, one of the sets $S_n \cap X$ contains a non-empty set $(a,b) \cap X$ for some $a < b$. Now let $(c,e)$ be a maximal interval in the open set $(a,b) \backslash X$, then (by analytic continuation) $f$ agrees with an entire function on $(c,e)$, and hence on $[c,e]$ by smoothness. On the other hand, at least one endpoint, say $c$, lies in $S_n$, thus $$ |f^{(m)}(c)| \leq n$$ for all $m$. By Taylor expansion of the entire function, we then have $$ |f^{(m)}(x)| \leq \sum_{j=0}^\infty \frac{|f^{(m+j)}(c)|}{j!} |x-c|^j$$ $$ \leq \sum_{j=0}^\infty \frac{n}{j!} (b-a)^j$$ $$ \leq n \exp(b-a)$$ for all $m$ and $x \in [c,e]$. Letting $(c,e)$ and $m$ vary, we conclude that the bound $$ M(x) \leq n \exp(b-a)$$ holds for all $x \in (a,b) \backslash X$. Since $(a,b) \cap X$ is contained in $S_n$, these bounds also hold on $(a,b) \cap X$, hence they hold on all of $(a,b)$. Now from Taylor's theorem with remainder we see that $f$ agrees on $(a,b)$ with an entire function (the Taylor expansion of $f$ around any point in $(a,b)$), and so $(a,b) \cap X$ is empty, giving the required contradiction. $\Box$ The function $f$ in the OP question obeys the hypotheses of Theorem 1. By Taylor expansion applied to the entire function that $f$ agrees with, and performing the same calculation used to prove the above theorem, we obtain the bounds $$ M(x) = \sup_{m \geq 0} |f^{(m)}(x)| \leq M(0) \exp(|x|)$$ for all $x \in {\bf R}$. We now have locally uniform bounds on all of the $f^{(m)}$ and the argument given by username (or the variant given in Pinelis's comment to that argument) applies to conclude.<|endoftext|> TITLE: Lagrange inversion for power-series with rational powers QUESTION [10 upvotes]: One can use Lagrange inversion to find the power series $F(x)$, which solves $F(x) = x(1+F(x)^p)$, where $p$ is a positive integer. Now, what if $p$ is not an integer, but rather a positive rational number, say $p=7/3$? As a concrete example, we are looking for a formal solution to $F(x) = x(1+F(x)^{7/3})$, but now, $F(x) \in \mathbb{C}[x^{1/3}]$. The Lagrange inversion formula still seem to work in this case, that is, the function $$ F(x) := \sum_{r>0} x^r \left( [t^{r-1}]\frac{1}{r} (1+t^p)^r \right) $$ is a solution to $F(x) = x(1+F(x)^p)$, but now we must have $F(x) \in \mathbb{C}[x^{1/d}]$, where $d$ is the denominator of $p$, and the sum ranges over all positive integer multiples of $1/d$. Is there some reference which proves this extension of Lagrange inversion? Edit: I think I managed to prove that Lagrange inversion generalizes to this setting, i.e, instead of having $f,g \in \mathbb{C}[x]$, we have $f,g \in \mathbb{C}[x^{1/d}]$, and we wish to express the coefficients of $g$, in terms of the coefficients of $f$, where $f(g(x))=x$. REPLY [4 votes]: I managed to write up a proof based on these lecture notes (which cites Stanley's EC2) The original proof uses residue calculus, and properties of analytic functions. This proof does not generalize to rational exponents, but the proof linked above, extends without any issues to "power series" with rational powers.<|endoftext|> TITLE: If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the conjecture proved? QUESTION [14 upvotes]: If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the Goldbach conjecture proved? If ZFC+CH implies Goldbach, and if the Goldbach turn out to be false, then it would mean that ZFC+CH is not consistent, but we know that ZFC+CH is consistent assuming that ZFC is consistent... What do you think? REPLY [32 votes]: Because the Goldbach conjecture is an arithmetic statement, it is absolute between any two models which agree on the natural numbers. Now, given any model of $\sf ZFC$, $M$, there is a forcing extension $M[G]$ with the same ordinals (and in particular, the same natural numbers, which are the just the finite ordinals), in which $\sf CH$ holds. Or, better yet, simply consider $L^M$, which is an inner model with the same ordinals (and, again, the same natural numbers), in which $\sf CH$ holds. Therefore, if you can prove Goldbach, Riemann, or the ABC Conjecture, assuming $\sf CH$, you may as well have proved it. Using $L$ will also tell you that using the Axiom of Choice was redundant, so in fact the proof is in $\sf ZF$ and not $\sf ZFC$. So, to sum this up, if you prove that $\sf ZFC+CH$ implies Goldbach's conjecture, and then you prove that Goldbach's conjecture is false, you've proved that $\sf ZF$ is inconsistent. Which, to my taste, is a far bigger result than Goldbach's conjecture (although others may disagree).<|endoftext|> TITLE: ℤ/18ℤ elliptic curves over cubic fields QUESTION [7 upvotes]: I am working on $\mathbb{Z}/18\mathbb{Z}$ elliptic curves over cubic fields. The curves are created using the formulas on p. 584 of D. Jeon, C. H. Kim, Y. Lee, Families of elliptic curves over cubic number fields with prescribed torsion subgroups, Mathematics of Computation, V. 80, 273, January 2011, p. 579-591, JSTOR: 41104715. My code snippet with the saved output for Magma Calculator online is available for download from MEGA. I observe that the following triples of rational $t$-values produce curves with similar characteristics: $$t_1=t$$ $$t_2=1-\frac{1}{t_1}=1-\frac{1}{t}$$ $$t_3=1-\frac{1}{t_2}=\frac{1}{1-t}$$ For a triple $t_1,t_2,t_3$, the three elliptic curves are different over three different cubic fields, with different discriminants and conductors. But the ranks, $j$-invariants, and heights of all generators are the same (e.g., for rank $2$ there will be height $h_1$ for generators $g_{11}, g_{12}, g_{13}$ and height $h_2$ for generators $g_{21}, g_{22}, g_{23}$, where $g_{ik}$ is the $i$-th generator for the curve created using $t_k$). It was also pretty straightforward to derive the formula for the $j$-invariant and see that it is always rational: $$j=\frac{(t^3-3t^2+1)^3(t^9-9t^8+27t^7-48t^6+54t^5-45t^4+27t^3-9t^2+1)^3}{(t^3-6t^2+3t+1)(t^2-t+1)^3(-1+t)^9t^9}$$ Magma is unable to check whether the curves are isomorphic, as they are defined over different cubic fields: >> IsIsomorphic(E1, E2); ^ Runtime error in 'IsIsomorphic': Curves must be defined over the same base ring Isogeneity check is unavailable over number fields (only over rationals or finite fields): >> IsIsogenous(E1, E2); ^ Runtime error in 'IsIsogenous': Bad argument types Argument types given: CrvEll[FldNum[FldRat]], CrvEll[FldNum[FldRat]] The curves seem to be essentially the same (not distinct in any real sense) to me, even though they might not be considered isomorphic and/or isogenous. Question 1: Does there exist a proper mathematical name for "essentially the same" used above, or the name for the observed connection between the curves? Question 2: Is it possible to map a generator discovered on one of the curves to the other two curves? The explicit expression for the map is not a priority yet. Question 3: If the height of the generator (but not the generator itself) is considered to be known, is it possible to speed up a search process for it? If so, how? Rationale for Questions 2 and 3: It is very easy to determine both generators (with heights $1.798$ and $11.652$, default Effort := 1 in $45$ seconds) for $t_1=\frac{1}{5}$, harder to do so for $t_2=-4$ (Effort := 1000 helps, takes much longer), and very hard to recover the second generator for $t_3=\frac{5}{4}$ (Effort := 1600 fails). REPLY [5 votes]: We already have an accepted answer, but since i had already started an answer and was at the half of the route beyond getting the essence of the structure, i completed it now, since it may be useful in similar contexts. On the mathematical side the situation is as follows, recalled for the convenience of the reader from the literature. Notations are as in the already cited paper: Families of Elliptic Curves over Cubic Number Fields with Prescribed Torsion Subgroups, Daeyeol Jeon, Chang Heon Kim, And Yoonjin Lee A further reference that should not be omitted is: Markus Reichert, Explicit Determination of Nontrivial Torsion Structures of Elliptic Curves Over Quadratic Number Fields In order to produce an example of a curve $E$ with torsion $\Bbb Z/18$, the Ansatz is to work with the Tate normal form, consider curves $E=E(b,c)$ parametrized by two algebraic numbers $b,c$ from a cubic number field $K$, $$ E = E(b, c)\ :\qquad y^2 + (1 − c)xy − by = x^3 − bx^2\ , $$ and arrange that the point $P = (0, 0)$ has order $18$. For this, pick two parameters $(U,V)$ satisfying the equation for $X_1(18)$: $$ \begin{aligned} X_1(18) \ :\qquad g_{18}(U,V) &= 0\ ,\qquad\text{ where} \\ g_{18}(U,V) &:=(U-1)^2 V^2 - (U^3 - U + 1)V + U^2(U - 1) \\ &= U^3(1-V) + U^2 (V^2 -1) + U(V-2V^2) + (V^2-V)\\ &\sim_{\Bbb Q(V)^\times} U^3 - U^2(V + 1) + \frac{2V^2-V}{V-1} -V\ . \ . \end{aligned} $$ Seen as a polynomial in $U$, it has degree $3$. We set $V=t$ to be a "suitable" rational number, and the polynomial $g_{18}(U,t)$ defines a cubic field $K=\Bbb Q(\alpha_t)$ generated by some $\alpha_t$. Let me plot the connection to $X_1(18)$ explicitly: $$ g_{18}(\alpha_t,t)=0\ . $$ Then the formulas for $b,c$ are given by one and the same rational function in $(U,V)=(\alpha_t,t)$. They are: $$ \begin{aligned} b(U,V) &= -\frac {V(U - V)(U^2 + V)(U^2 -UV + V)} {(U^2 -V^2+V)(U^2 + UV -V^2 + V)^2} \ , \\ c(U,V) &= -\frac {V(U - V)(U^2 -UV + V)} {(U^2 -V^2+V)(U^2 + UV -V^2 + V)} \ . \end{aligned} $$ Warming up. We proceed as follows in the given context from above. We fix some $t$. To have a concrete example, $t$ may be specialized to $t=t_1=1/5$, as the OP does it also. Let $t'$ be its cousin, $$ t'=t_3=\frac 1{ 1-t }\ . $$ We build the corresponding field $K=\Bbb Q(\alpha)$, where $\alpha =\alpha_t$ is a suitable root of the polynomial $g_{18}(U,t)$, seen as a polynomial in $U$. Let $K'=\Bbb Q(\alpha')$ be the cousin field, where $\alpha'$ is a specific root for $g_{18}(U, t')$. Question: Are $K$ and $K'$ isomorphic (for some good choice of $\alpha'$)? Answer: Yes, they are, take $\displaystyle \alpha'= 1-\frac 1\alpha$. To illustrate the situation, we consider first the sample case $t=1/5$. Sage gives this information as follows: def g18(U, V): return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V) R. = PolynomialRing(QQ) t1 = 1/5 a1 = g18(U, t1).roots(ring=QQbar, multiplicities=False)[0] t3 = 1/(1 - t1) a3 = 1 - 1/a1 print(f'g18(a3, t3) = {g18(a3, t3)}') The sage interpreter gives after a copy+paste of the above code, together with one more line to be sure we get a clean zero: g18(a3, t3) = 0.?e-17 sage: g18(a3, t3).minpoly() x Because of the rôle of $(\alpha,t)$ as a special value for $(U,V)$, i will use below rather $(u,v)$ pairs instead. Now the whole context can be explained structurally as follows. Proposition: Let $F$ be a field (of characteristic $\ne 2,3$). For two parameters $b,c\in F$, $b\ne 0$, let $E_T(b,c)$ be the elliptic curve in Tate normal form $$ E_T(b, c)\ :\qquad y^2 +(1-c)xy -by = x^3 bx^2\ , $$ so that $P=(0,0)$ is a rational point on it. For suitable ($\Delta(A,B)\ne 0$) parameters $A,B\in F$ let consider also the elliptic curve in short Weierstrass form $$ E_W(A,B)\ :\qquad y^2 = x^3 + Ax+B\ . $$ Fix $u,v$ în $F$, $u\ne 1$, so that the pair $(u,v)$ corresponds to a point on the moduli space $X_1(18)$ parametrized as mentioned above, i.e. it satisfies $$ g_{18}(u,v)=0\ ,\qquad\text{ where }\\ g_{18}(U,V)= (U-1)^2 V^2 -(U^3 -U + 1)V + U^2(U - 1)\ . $$ Then the pair $(u',v')$ with components $$ \begin{aligned} u' &= \frac 1{1-u}\ ,\\ v' &= 1-\frac 1v\ , \end{aligned} $$ is also defining a pointin the moduli space $X_1(18)$, i.e. $g_{18}(u',v')=0$. Let $\underline A$, $\underline B$ be the rational functions given by $$ \begin{aligned} \underline A(b,c) &= -\frac 1{48}\Big(\ ((c-1)^2 - 4b)^2 - 24b(c - 1)\ \Big)\ ,\\ \underline B(b,c) &= \frac 1{864}\Big(\ ((c-1)^2 - 4b)^3 - 36b(c-1)^3 + 72b^2(2c + 1)\ \Big)\ . \end{aligned} $$ Consider with a slight abuse of notation $b,c\in F$ and $b',c'\in F$, then $A,B\in F$ and $A',B'\in F$ as follows $$ \begin{aligned} b &= b(u,v)\ ,\qquad &b' &= b(u',v')\ ,\\ c &= c(u,v)\ ,\qquad &c' &= c(u',v')\ ,\\[2mm] A &= \underline A(b,c)\ ,\qquad &A' &=\underline A(b',c')\ ,\\ B &= \underline B(b,c)\ ,\qquad &B' &=\underline B(b',c')\ ,\\[2mm] &\qquad\text{ and consider the elliptic curves}\\[2mm] E_T &= E(b, c)\ , \qquad &E'_T &= E_T(b', c')\\ E_W &= E(A, B)\ , \qquad &E'_W &= E_W(A', B')\ . \end{aligned} $$ Then$$ \frac {A'}A = U^{12}\ ,\qquad \frac {B'}B = U^{18}\ , $$ so the elliptic curves $E_W$ and $E_W'$ are canonically isomorphic via a map $\Phi$, as shown in the diagram below. The functions $\underline A$, $\underline B$ were chosen to make $E_T(b,c)$ isomorphic $E_W(A,B)$. Then the following diagram is commutative: $\require{AMScd}$ $$ \begin{CD} E_T @>{\cong}>> E_W\\ @A{\cong} AA @A\cong A\Phi A\\ E'_T @>>\cong> E'_W \end{CD} $$ So we can compare the rational points $P=(0,0)\in E_T(F)$ and $P'=(0,0)\in E'_T(F)$ in one or any of the common worlds, e.g. in $E_W(F)$, and then $11P$ and $P'$ (or equivalently $P=5\cdot 11 P$ and $5P'$) correspond to one and the same torsion point of order (dividing) $18$. In a diagram: $\require{AMScd}$ $$ \begin{CD} P_T @>{\cong}>> P_W=5\Phi(P'_W)=\Phi(5P'_W)\\ @. @A\cong A\Phi A\\ 5P'_T @>>\cong> 5P'_W \end{CD} $$ Proof by computer. $\square$ Code for the proof. First let us define the needed functions, and needed objects. def bmap(U, V): return -(U^2 - U*V + V) * (U^2 + V) * (U - V) * V / (U^2 + U*V - V^2 + V)^2 / (U^2 - V^2 + V) def cmap(U, V): return -(U^2 - U*V + V) * (U - V) * V / (U^2 + U*V - V^2 + V) / (U^2 - V^2 + V) def Amap(b, c): return -1/48 * ( ((c-1)^2 - 4*b)^2 - 24*b*(c - 1) ) def Bmap(b, c): return 1/864 * ( ((c-1)^2 - 4*b)^3 - 36*b*(c-1)^3 + 72*b^2*(2*c + 1) ) def f(U, V): return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V) R. = PolynomialRing(QQ) Q = R.quotient( f(U, V) ) FR = R.fraction_field() FQ = Q.fraction_field() u1, v1 = FQ(U), FQ(V) u2, v2 = 1/(1 - u1), 1 - 1/v1 Now we can check: print(f'Is f(u2, v2) zero? {bool( f(u2, v2) == 0 )}' ) b , c = bmap(U , V ), cmap(U , V ) b1, c1 = bmap(u1, v1), cmap(u1, v1) b2, c2 = bmap(u2, v2), cmap(u2, v2) A , B = Amap(b , c ), Bmap(b , c ) A1, B1 = Amap(b1, c1), Bmap(b1, c1) A2, B2 = Amap(b2, c2), Bmap(b2, c2) print(f'Is A2/A1 = u1^12? {bool( A2/A1 == u1^12 )}') print(f'Is B2/B1 = u1^18? {bool( B2/B1 == u1^18 )}') ET = EllipticCurve(FR, [1 - c, -b, -b, 0, 0]) EW = EllipticCurve(FR, [A, B]) phi = ET.isomorphism_to( EW ) PT = ET.point( (0, 0, 1) ) PW = phi( PT ) five_PW = 5*PW x_PW , y_PW = PW.xy() x_five_PW, y_five_PW = five_PW.xy() x_PW.subs({U: u1, V:v1}) == x_five_PW.subs({U: u2, V: v2}) / u1^6 y_PW.subs({U: u1, V:v1}) == y_five_PW.subs({U: u2, V: v2}) / u1^9 This gives the needed confirmations: Is f(u2, v2) zero? True Is A2/A1 = u1^12? True Is B2/B1 = u1^18? True True True The last two True values confirm that the coordinates of $P=(0,0)=E_T(F)$ and $5P'$ where $P'=(0,0)\in E_T'$ are the same, when transported to $E_W(F)$. Note: Unfortunately, sage cannot build the needed curves over FQ.<|endoftext|> TITLE: Motivic cohomology with $\mathbb{Z}/2$ coefficients in positive characteristic QUESTION [5 upvotes]: In G. M. L. Powell's note 'Steenrod operations in motivic cohomology', he stated that if $\mathrm{char}(k)=0$, $$H^{*,*}(k,\mathbb{Z}/2)=K_*^M(k)/2[\tau]$$ where $\tau\in H^{0,1}$ is the unique nonzero element. I wonder whether this result holds when $char(k)>0$? REPLY [5 votes]: This holds if the characteristic of $k$ is not 2, and it follows from the Milnor conjecture proved by Voevodsky. Voevodsky ultimately proved the following (Theorem 6.17 in https://annals.math.princeton.edu/wp-content/uploads/annals-v174-n1-p11-s.pdf): If $m>0$ and $X$ is smooth over a field $k$ of characteristic prime to $m$, then the map $$ H^n(X, \mathbb Z/m(i)) \to H^n_{\mathrm{et}}(X,\mu_m^{\otimes i}) $$ is an isomorphism provided that $n\leq i$. When $X=\operatorname{Spec}(k)$ the right-hand side is $K^M_*(k)/m[\tau^{\pm 1}]$ if there is a primitive $m$th root of unity $\tau\in\mu_m(k)$ (this follows from the isomorphism for $n=i$), and the left-hand side is zero for $n>i$, so we know everything. When the characteristic is 2, or more generally when the characteristic is $p$ and the coefficients are $\mathbb Z/p$, we also know everything by Geisser and Levine: in this case the motivic cohomology vanishes when $n\neq i$, so there is only Milnor K-theory.<|endoftext|> TITLE: Subshifts with special property QUESTION [6 upvotes]: I am looking how to prove the following fact: If $ X \subseteq A^\mathbb{Z}$ is an infinite minimal subshift, then for any $N\ge 1$, $X$ is conjugate to a minimal subshift $Y\subseteq B^\mathbb{Z}$ such that for any $y\in Y$, $y(i)\neq y(j)$ if $|i-j|\le N , i\neq j$. (I have encountered this in a paper in which the author asserts this without a proof.) REPLY [8 votes]: Define $X(m)$ as the image of $X$ in $(A^m)^\mathbf{Z}$, mapping $(a_n)_{n\in\mathbf{Z}}$ to $((a_{n+k})_{0\le k TITLE: Hankel determinants for q-Catalan numbers where q is a root of unity? QUESTION [5 upvotes]: Let ${C_n}(q)$ be the weight of the Dyck paths of semilength $n$ where the upsteps have weight $1$ and the downsteps which end on height $i$ have weight $q^i$. They satisfy ${C_n}(q) = \sum\limits_{j = 0}^{n - 1} {{q^j}} {C_j}(q){C_{n - 1 - j}}(q)$ with ${C_0}(q) = 1.$ The first 3 Hankel determinants ${d_k}(n,q) = \det \left( {{C_{i + j + k}}(q)} \right)_{i,j = 0}^{n - 1}$ are ${d_0}(n,q) = {q^{\frac{{n(n - 1)(4n - 5)}}{6}}},$ ${d_1}(n,q) = {q^{\frac{{n(n - 1)(4n + 1)}}{6}}}$ and ${d_2}(n,q) = {q^{\frac{{n(n - 1)(4n + 7)}}{6}}}\frac{{1 - {q^{n + 1}}}}{{1 - q}}.$ For the higher Hankel determinants no simple formulas seem to be known. Is there anything known for $q$ a root of unity which generalizes the well-known formulas for $q=1?$ Edit Let me make my question more precise: For $q=1$ the determinants $d_{k}(n,1)$ satisfy $\frac{d_{k}(n,1)}{d_{k-1}(n,1)}=\frac{\binom{2n+2k-2}{k-1}}{\binom{2k-2}{k-1}}.$ For $q=-1$ analogous results hold for $d_{k}(2n,-1).$ Here we get $\frac{d_{2k}(2n,-1)}{d_{2k-1}(2n,-1)}=\frac{\binom{2n+2k-2}{k-1}}{\binom{2k-2}{k-1}}$ and $\frac{d_{2k+1}(2n,-1)}{d_{2k}(2n,-1)}=\frac{\binom{2n+2k}{k}}{\binom{2k}{k}}.$ So my question is: Let $\zeta_{m}$ be an $m-$th root of unity. Are there analogous results for $\frac{d_{mk+j}(mn, \zeta_{m})}{d_{mk+j-1}(mn, \zeta_{m})}?$ Computations suggest that this is true for $m=4.$ Here we get $\frac{d_{4k+j}(4n,i)}{d_{4k+j-1}(4n,i)}=\frac{\binom{2n+2k}{k}}{\binom{2k}{k}}$ for $j=1,2,3$ and $\frac{d_{4k}(4n,i)}{d_{4kj-1}(4n,i)}=\frac{\binom{2n+2k-1}{k}}{\binom{2k-1}{k}}.$ For $m=3$ we get $\frac{d_{3k+j}(3n,\rho)}{d_{3k+j-1}(3n,\rho)}=\frac{\binom{2n+2k}{k}}{\binom{2k}{k}}$ for $j=1,2.$ But I could not find a formula for $\frac{d_{3k}(3n,\rho)}{d_{3k-1}(3n,\rho)}$. I would be interested how to prove these conjectures and if there are analogous results for general $m.$ REPLY [5 votes]: This is more of an extended comment. It is in fact possible to guess formulas for larger $k$ using the FriCAS guessing package. At least for $k=3$ I obtain something reasonable: $$ q^{\frac{n(n-1)(4n+13)}{6}} \frac{(q^{n+3}+q^{n+1}-q-1)(q^{n+2}-1)(q^{n+1}-1)} {(q;q)_3} $$ The trick is to remove the power of $q$ first. Explicitly (I used FriCAS from within Sage, which makes things only slightly more complicated), I first define a few helpers: R. = QQ[] @cached_function def q_catalan(n): if n == 0: return 1 return sum(q_catalan(j)*q_catalan(n-1-j)*q^j for j in range(n)) @cached_function def hankel(k, n): return det(matrix([[q_catalan(i+j+k) for i in range(n)] for j in range(n)])) def get_q_power(poly): for f, e in poly.factor(): if f == q: return e return 0 Then, we find: sage: e = [get_q_power(hankel(3, n)) for n in range(10)]; e [0, 0, 7, 25, 58, 110, 185, 287, 420, 588] sage: fricas.guessRat(e)[0].sage().factor() 1/6*(4*n + 13)*(n - 1)*n sage: e = [get_q_power(hankel(4, n)) for n in range(10)]; e [0, 0, 9, 31, 70, 130, 215, 329, 476, 660] sage: fricas.guessRat(e)[0].sage().factor() 1/6*(4*n + 19)*(n - 1)*n from which we guess that we should take out a factor $q^{\frac{n(n-1)(4n+6k-5)}{6}}$: sage: hankel_reduced = lambda k, n: hankel(k, n)/q^(n*(n-1)*(4*n+6*k-5)/6) sage: k=3; fricas.guessRat(q)([hankel_reduced(k, n) for n in range(10)], []) 6 4 3 n 5 4 3 2 2 n 3 2 n (q + q )q + (- q - 2 q - 2 q - q )q + (2 q + 2 q + 2 q)q - q - 1 [-----------------------------------------------------------------------------] 6 5 4 2 q - q - q + q + q - 1 sage: k=4; fricas.guessRat(q)([hankel_reduced(k, n) for n in range(10)], []) [ 18 16 15 14 13 12 10 6 n (q + q + q + 2 q + q + q + q )q + 17 16 15 14 13 12 11 10 9 - q - 2 q - 4 q - 5 q - 8 q - 8 q - 8 q - 5 q - 4 q + 8 7 - 2 q - q * 5 n q + 15 14 13 12 11 10 9 8 2 q + 4 q + 10 q + 14 q + 20 q + 20 q + 20 q + 14 q + 7 6 5 10 q + 4 q + 2 q * 4 n q + 13 12 11 10 9 8 7 6 - q - 6 q - 11 q - 21 q - 26 q - 30 q - 26 q - 21 q + 5 4 3 - 11 q - 6 q - q * 3 n q + 10 9 8 7 6 5 4 3 2 2 n (3 q + 8 q + 16 q + 21 q + 24 q + 21 q + 16 q + 8 q + 3 q )q + 7 6 5 4 3 2 n 4 3 2 (- 3 q - 6 q - 10 q - 10 q - 10 q - 6 q - 3 q)q + q + 2 q + 2 q + 2 q + 1 / 18 17 16 15 14 13 12 11 10 9 8 7 q - q - q - q + q + 2 q + q + q - 2 q - 2 q - 2 q + q + 6 5 4 3 2 q + 2 q + q - q - q - q + 1 ] This indicates that we are actually expecting a $q$-polynomial (not a $q$-fraction) of $q$-degree $\binom{k}{2}$, which tells us the exact number of terms we need to provide to guess the polynomials. Next I tried to guess the denominator polynomials: sage: [fricas.guessRat(q)([hankel_reduced(k, n) for n in range(binomial(k, 2)+2)], [])[0].sage().denominator().factor() for k in range(1, 7)] [1, q - 1, (q^2 + q + 1)*(q + 1)*(q - 1)^3, (q^4 + q^3 + q^2 + q + 1)*(q^2 + q + 1)^2*(q^2 + 1)*(q + 1)^2*(q - 1)^6, (q^6 + q^5 + q^4 + q^3 + q^2 + q + 1)*(q^4 + q^3 + q^2 + q + 1)^2*(q^2 + q + 1)^3*(q^2 - q + 1)*(q^2 + 1)^2*(q + 1)^4*(q - 1)^10, (q + 1)^6 * (q - 1)^15 * (q^2 - q + 1)^2 * (q^2 + 1)^3 * (q^2 + q + 1)^5 * (q^4 + 1) * (q^4 + q^3 + q^2 + q + 1)^3 * (q^6 + q^3 + 1) * (q^6 + q^5 + q^4 + q^3 + q^2 + q + 1)^2] Taking out the factor $(q-1)^{\binom{k}{2}}$ and plugging in $q=1$, we obtain https://oeis.org/A086205, which has an obvious $q$-analogue, so we obtain $$ (q-1)^{\binom{k}{2}}\prod_{i=1}^{k-1}\frac{[2i-1]!}{[i-1]!} $$ for the denominator. Apparently, the numerator is not quite as easy to guess. However, it appears that $\prod_{i=1}^{k-1} (q^{n+i}-1)^{\min(i,k-i)}$ is a factor. This allows us to guess more terms, since the degree of what remains is just $\lfloor\frac{(k-1)^2}{4}\rfloor$. The first few, beginning with $k=2$, are [1, q^(n + 3) + q^(n + 1) - q - 1, q^4 + 2*q^3 + 2*q^2 + 2*q + q^(2*n + 10) + q^(2*n + 8) + q^(2*n + 7) + 2*q^(2*n + 6) + q^(2*n + 5) + q^(2*n + 4) + q^(2*n + 2) - 2*q^(n + 7) - 2*q^(n + 6) - 3*q^(n + 5) - 2*q^(n + 4) - 3*q^(n + 3) - 2*q^(n + 2) - 2*q^(n + 1) + 1, q^10 + 3*q^9 + 5*q^8 + 8*q^7 + 10*q^6 + 10*q^5 + 10*q^4 + 8*q^3 + 5*q^2 + q^(4*n + 25) + q^(4*n + 23) + q^(4*n + 22) + 3*q^(4*n + 21) + 2*q^(4*n + 20) + 5*q^(4*n + 19) + 4*q^(4*n + 18) + 7*q^(4*n + 17) + 5*q^(4*n + 16) + 6*q^(4*n + 15) + 5*q^(4*n + 14) + 7*q^(4*n + 13) + 4*q^(4*n + 12) + 5*q^(4*n + 11) + 2*q^(4*n + 10) + 3*q^(4*n + 9) + q^(4*n + 8) + q^(4*n + 7) + q^(4*n + 5) - q^(3*n + 22) - 3*q^(3*n + 21) - 4*q^(3*n + 20) - 6*q^(3*n + 19) - 10*q^(3*n + 18) - 14*q^(3*n + 17) - 18*q^(3*n + 16) - 21*q^(3*n + 15) - 24*q^(3*n + 14) - 27*q^(3*n + 13) - 27*q^(3*n + 12) - 24*q^(3*n + 11) - 21*q^(3*n + 10) - 18*q^(3*n + 9) - 14*q^(3*n + 8) - 10*q^(3*n + 7) - 6*q^(3*n + 6) - 4*q^(3*n + 5) - 3*q^(3*n + 4) - q^(3*n + 3) + 3*q^(2*n + 18) + 6*q^(2*n + 17) + 12*q^(2*n + 16) + 16*q^(2*n + 15) + 24*q^(2*n + 14) + 30*q^(2*n + 13) + 39*q^(2*n + 12) + 40*q^(2*n + 11) + 44*q^(2*n + 10) + 40*q^(2*n + 9) + 39*q^(2*n + 8) + 30*q^(2*n + 7) + 24*q^(2*n + 6) + 16*q^(2*n + 5) + 12*q^(2*n + 4) + 6*q^(2*n + 3) + 3*q^(2*n + 2) + 3*q - 3*q^(n + 14) - 7*q^(n + 13) - 13*q^(n + 12) - 19*q^(n + 11) - 25*q^(n + 10) - 29*q^(n + 9) - 32*q^(n + 8) - 32*q^(n + 7) - 29*q^(n + 6) - 25*q^(n + 5) - 19*q^(n + 4) - 13*q^(n + 3) - 7*q^(n + 2) - 3*q^(n + 1) + 1, q^20 + 4*q^19 + 9*q^18 + 18*q^17 + 31*q^16 + 46*q^15 + 64*q^14 + 82*q^13 + 96*q^12 + 106*q^11 + 110*q^10 + 106*q^9 + 96*q^8 + 82*q^7 + 64*q^6 + 46*q^5 + 31*q^4 + 18*q^3 + 9*q^2 + q^(6*n + 48) + q^(6*n + 46) + q^(6*n + 45) + 3*q^(6*n + 44) + 3*q^(6*n + 43) + 6*q^(6*n + 42) + 6*q^(6*n + 41) + 12*q^(6*n + 40) + 13*q^(6*n + 39) + 20*q^(6*n + 38) + 23*q^(6*n + 37) + 32*q^(6*n + 36) + 33*q^(6*n + 35) + 43*q^(6*n + 34) + 44*q^(6*n + 33) + 4*q + 56*q^(6*n + 32) + 56*q^(6*n + 31) + 64*q^(6*n + 30) + 61*q^(6*n + 29) + 68*q^(6*n + 28) + 61*q^(6*n + 27) + 64*q^(6*n + 26) + 56*q^(6*n + 25) + 56*q^(6*n + 24) + 44*q^(6*n + 23) + 43*q^(6*n + 22) + 33*q^(6*n + 21) + 32*q^(6*n + 20) + 23*q^(6*n + 19) + 20*q^(6*n + 18) + 13*q^(6*n + 17) + 12*q^(6*n + 16) + 6*q^(6*n + 15) + 6*q^(6*n + 14) + 3*q^(6*n + 13) + 3*q^(6*n + 12) + q^(6*n + 11) + q^(6*n + 10) + q^(6*n + 8) - 2*q^(5*n + 44) - 4*q^(5*n + 43) - 6*q^(5*n + 42) - 9*q^(5*n + 41) - 16*q^(5*n + 40) - 26*q^(5*n + 39) - 39*q^(5*n + 38) - 56*q^(5*n + 37) - 78*q^(5*n + 36) - 105*q^(5*n + 35) - 135*q^(5*n + 34) - 171*q^(5*n + 33) - 212*q^(5*n + 32) - 254*q^(5*n + 31) - 292*q^(5*n + 30) - 326*q^(5*n + 29) - 357*q^(5*n + 28) - 383*q^(5*n + 27) - 399*q^(5*n + 26) - 404*q^(5*n + 25) - 399*q^(5*n + 24) - 383*q^(5*n + 23) - 357*q^(5*n + 22) - 326*q^(5*n + 21) - 292*q^(5*n + 20) - 254*q^(5*n + 19) - 212*q^(5*n + 18) - 171*q^(5*n + 17) - 135*q^(5*n + 16) - 105*q^(5*n + 15) - 78*q^(5*n + 14) - 56*q^(5*n + 13) - 39*q^(5*n + 12) - 26*q^(5*n + 11) - 16*q^(5*n + 10) - 9*q^(5*n + 9) - 6*q^(5*n + 8) - 4*q^(5*n + 7) - 2*q^(5*n + 6) + q^(4*n + 40) + 8*q^(4*n + 39) + 15*q^(4*n + 38) + 29*q^(4*n + 37) + 45*q^(4*n + 36) + 77*q^(4*n + 35) + 113*q^(4*n + 34) + 172*q^(4*n + 33) + 234*q^(4*n + 32) + 325*q^(4*n + 31) + 412*q^(4*n + 30) + 523*q^(4*n + 29) + 623*q^(4*n + 28) + 745*q^(4*n + 27) + 841*q^(4*n + 26) + 942*q^(4*n + 25) + 997*q^(4*n + 24) + 1051*q^(4*n + 23) + 1054*q^(4*n + 22) + 1051*q^(4*n + 21) + 997*q^(4*n + 20) + 942*q^(4*n + 19) + 841*q^(4*n + 18) + 745*q^(4*n + 17) + 623*q^(4*n + 16) + 523*q^(4*n + 15) + 412*q^(4*n + 14) + 325*q^(4*n + 13) + 234*q^(4*n + 12) + 172*q^(4*n + 11) + 113*q^(4*n + 10) + 77*q^(4*n + 9) + 45*q^(4*n + 8) + 29*q^(4*n + 7) + 15*q^(4*n + 6) + 8*q^(4*n + 5) + q^(4*n + 4) - 4*q^(3*n + 35) - 16*q^(3*n + 34) - 37*q^(3*n + 33) - 68*q^(3*n + 32) - 116*q^(3*n + 31) - 182*q^(3*n + 30) - 271*q^(3*n + 29) - 383*q^(3*n + 28) - 518*q^(3*n + 27) - 673*q^(3*n + 26) - 841*q^(3*n + 25) - 1007*q^(3*n + 24) - 1163*q^(3*n + 23) - 1305*q^(3*n + 22) - 1415*q^(3*n + 21) - 1486*q^(3*n + 20) - 1510*q^(3*n + 19) - 1486*q^(3*n + 18) - 1415*q^(3*n + 17) - 1305*q^(3*n + 16) - 1163*q^(3*n + 15) - 1007*q^(3*n + 14) - 841*q^(3*n + 13) - 673*q^(3*n + 12) - 518*q^(3*n + 11) - 383*q^(3*n + 10) - 271*q^(3*n + 9) - 182*q^(3*n + 8) - 116*q^(3*n + 7) - 68*q^(3*n + 6) - 37*q^(3*n + 5) - 16*q^(3*n + 4) - 4*q^(3*n + 3) + 6*q^(2*n + 30) + 20*q^(2*n + 29) + 49*q^(2*n + 28) + 91*q^(2*n + 27) + 157*q^(2*n + 26) + 240*q^(2*n + 25) + 351*q^(2*n + 24) + 473*q^(2*n + 23) + 619*q^(2*n + 22) + 762*q^(2*n + 21) + 914*q^(2*n + 20) + 1036*q^(2*n + 19) + 1144*q^(2*n + 18) + 1202*q^(2*n + 17) + 1232*q^(2*n + 16) + 1202*q^(2*n + 15) + 1144*q^(2*n + 14) + 1036*q^(2*n + 13) + 914*q^(2*n + 12) + 762*q^(2*n + 11) + 619*q^(2*n + 10) + 473*q^(2*n + 9) + 351*q^(2*n + 8) + 240*q^(2*n + 7) + 157*q^(2*n + 6) + 91*q^(2*n + 5) + 49*q^(2*n + 4) + 20*q^(2*n + 3) + 6*q^(2*n + 2) - 4*q^(n + 25) - 14*q^(n + 24) - 33*q^(n + 23) - 64*q^(n + 22) - 109*q^(n + 21) - 166*q^(n + 20) - 234*q^(n + 19) - 308*q^(n + 18) - 381*q^(n + 17) - 448*q^(n + 16) - 501*q^(n + 15) - 536*q^(n + 14) - 548*q^(n + 13) - 536*q^(n + 12) - 501*q^(n + 11) - 448*q^(n + 10) - 381*q^(n + 9) - 308*q^(n + 8) - 234*q^(n + 7) - 166*q^(n + 6) - 109*q^(n + 5) - 64*q^(n + 4) - 33*q^(n + 3) - 14*q^(n + 2) - 4*q^(n + 1) + 1, -q^35 - 5*q^34 - 14*q^33 - 33*q^32 - 68*q^31 - 124*q^30 - 210*q^29 - 332*q^28 - 492*q^27 - 693*q^26 - 931*q^25 - 1196*q^24 - 1476*q^23 - 1754*q^22 - 2008*q^21 - 2220*q^20 - 2374*q^19 - 2453*q^18 - 2453*q^17 - 2374*q^16 - 2220*q^15 - 2008*q^14 - 1754*q^13 - 1476*q^12 - 1196*q^11 - 931*q^10 - 693*q^9 - 492*q^8 - 332*q^7 - 210*q^6 - 124*q^5 - 68*q^4 - 33*q^3 - 14*q^2 + q^(9*n + 84) + q^(9*n + 82) + q^(9*n + 81) + 3*q^(9*n + 80) + 3*q^(9*n + 79) + 7*q^(9*n + 78) + 7*q^(9*n + 77) + 14*q^(9*n + 76) + 16*q^(9*n + 75) + 28*q^(9*n + 74) + 33*q^(9*n + 73) + 53*q^(9*n + 72) + 63*q^(9*n + 71) + 94*q^(9*n + 70) + 113*q^(9*n + 69) + 157*q^(9*n + 68) + 187*q^(9*n + 67) + 247*q^(9*n + 66) + 288*q^(9*n + 65) + 368*q^(9*n + 64) + 424*q^(9*n + 63) + 523*q^(9*n + 62) + 591*q^(9*n + 61) + 704*q^(9*n + 60) + 776*q^(9*n + 59) + 900*q^(9*n + 58) + 972*q^(9*n + 57) + 1095*q^(9*n + 56) + 1158*q^(9*n + 55) + 1267*q^(9*n + 54) + 1307*q^(9*n + 53) + 1395*q^(9*n + 52) + 1405*q^(9*n + 51) + 1463*q^(9*n + 50) + 1440*q^(9*n + 49) + 1463*q^(9*n + 48) + 1405*q^(9*n + 47) + 1395*q^(9*n + 46) + 1307*q^(9*n + 45) + 1267*q^(9*n + 44) + 1158*q^(9*n + 43) + 1095*q^(9*n + 42) + 972*q^(9*n + 41) + 900*q^(9*n + 40) + 776*q^(9*n + 39) + 704*q^(9*n + 38) + 591*q^(9*n + 37) + 523*q^(9*n + 36) + 424*q^(9*n + 35) + 368*q^(9*n + 34) + 288*q^(9*n + 33) + 247*q^(9*n + 32) + 187*q^(9*n + 31) + 157*q^(9*n + 30) + 113*q^(9*n + 29) + 94*q^(9*n + 28) + 63*q^(9*n + 27) + 53*q^(9*n + 26) + 33*q^(9*n + 25) + 28*q^(9*n + 24) + 16*q^(9*n + 23) + 14*q^(9*n + 22) + 7*q^(9*n + 21) + 7*q^(9*n + 20) + 3*q^(9*n + 19) + 3*q^(9*n + 18) + q^(9*n + 17) + q^(9*n + 16) + q^(9*n + 14) - q^(8*n + 80) - 3*q^(8*n + 79) - 5*q - 6*q^(8*n + 78) - 9*q^(8*n + 77) - 15*q^(8*n + 76) - 25*q^(8*n + 75) - 42*q^(8*n + 74) - 66*q^(8*n + 73) - 100*q^(8*n + 72) - 148*q^(8*n + 71) - 217*q^(8*n + 70) - 309*q^(8*n + 69) - 433*q^(8*n + 68) - 595*q^(8*n + 67) - 803*q^(8*n + 66) - 1063*q^(8*n + 65) - 1385*q^(8*n + 64) - 1777*q^(8*n + 63) - 2246*q^(8*n + 62) - 2789*q^(8*n + 61) - 3409*q^(8*n + 60) - 4105*q^(8*n + 59) - 4877*q^(8*n + 58) - 5714*q^(8*n + 57) - 6605*q^(8*n + 56) - 7529*q^(8*n + 55) - 8466*q^(8*n + 54) - 9387*q^(8*n + 53) - 10272*q^(8*n + 52) - 11096*q^(8*n + 51) - 11835*q^(8*n + 50) - 12458*q^(8*n + 49) - 12944*q^(8*n + 48) - 13278*q^(8*n + 47) - 13449*q^(8*n + 46) - 13449*q^(8*n + 45) - 13278*q^(8*n + 44) - 12944*q^(8*n + 43) - 12458*q^(8*n + 42) - 11835*q^(8*n + 41) - 11096*q^(8*n + 40) - 10272*q^(8*n + 39) - 9387*q^(8*n + 38) - 8466*q^(8*n + 37) - 7529*q^(8*n + 36) - 6605*q^(8*n + 35) - 5714*q^(8*n + 34) - 4877*q^(8*n + 33) - 4105*q^(8*n + 32) - 3409*q^(8*n + 31) - 2789*q^(8*n + 30) - 2246*q^(8*n + 29) - 1777*q^(8*n + 28) - 1385*q^(8*n + 27) - 1063*q^(8*n + 26) - 803*q^(8*n + 25) - 595*q^(8*n + 24) - 433*q^(8*n + 23) - 309*q^(8*n + 22) - 217*q^(8*n + 21) - 148*q^(8*n + 20) - 100*q^(8*n + 19) - 66*q^(8*n + 18) - 42*q^(8*n + 17) - 25*q^(8*n + 16) - 15*q^(8*n + 15) - 9*q^(8*n + 14) - 6*q^(8*n + 13) - 3*q^(8*n + 12) - q^(8*n + 11) + 3*q^(7*n + 75) + 8*q^(7*n + 74) + 23*q^(7*n + 73) + 40*q^(7*n + 72) + 75*q^(7*n + 71) + 121*q^(7*n + 70) + 210*q^(7*n + 69) + 327*q^(7*n + 68) + 520*q^(7*n + 67) + 769*q^(7*n + 66) + 1147*q^(7*n + 65) + 1623*q^(7*n + 64) + 2300*q^(7*n + 63) + 3131*q^(7*n + 62) + 4250*q^(7*n + 61) + 5573*q^(7*n + 60) + 7263*q^(7*n + 59) + 9201*q^(7*n + 58) + 11578*q^(7*n + 57) + 14210*q^(7*n + 56) + 17289*q^(7*n + 55) + 20562*q^(7*n + 54) + 24230*q^(7*n + 53) + 27969*q^(7*n + 52) + 31982*q^(7*n + 51) + 35877*q^(7*n + 50) + 39864*q^(7*n + 49) + 43481*q^(7*n + 48) + 46959*q^(7*n + 47) + 49820*q^(7*n + 46) + 52343*q^(7*n + 45) + 54049*q^(7*n + 44) + 55260*q^(7*n + 43) + 55534*q^(7*n + 42) + 55260*q^(7*n + 41) + 54049*q^(7*n + 40) + 52343*q^(7*n + 39) + 49820*q^(7*n + 38) + 46959*q^(7*n + 37) + 43481*q^(7*n + 36) + 39864*q^(7*n + 35) + 35877*q^(7*n + 34) + 31982*q^(7*n + 33) + 27969*q^(7*n + 32) + 24230*q^(7*n + 31) + 20562*q^(7*n + 30) + 203662*q^(5*n + 37) + 94662*q^(5*n + 25) + 17289*q^(7*n + 29) + 14210*q^(7*n + 28) + 11578*q^(7*n + 27) + 9201*q^(7*n + 26) + 7263*q^(7*n + 25) + 5573*q^(7*n + 24) + 4250*q^(7*n + 23) + 3131*q^(7*n + 22) + 2300*q^(7*n + 21) + 1623*q^(7*n + 20) + 1147*q^(7*n + 19) + 769*q^(7*n + 18) + 520*q^(7*n + 17) + 327*q^(7*n + 16) + 210*q^(7*n + 15) + 121*q^(7*n + 14) + 75*q^(7*n + 13) + 40*q^(7*n + 12) + 23*q^(7*n + 11) + 8*q^(7*n + 10) + 3*q^(7*n + 9) - 3*q^(6*n + 70) - 16*q^(6*n + 69) - 43*q^(6*n + 68) - 99*q^(6*n + 67) - 191*q^(6*n + 66) - 344*q^(6*n + 65) - 587*q^(6*n + 64) - 965*q^(6*n + 63) - 1521*q^(6*n + 62) - 2330*q^(6*n + 61) - 3452*q^(6*n + 60) - 4981*q^(6*n + 59) - 6999*q^(6*n + 58) - 9618*q^(6*n + 57) - 12922*q^(6*n + 56) - 17016*q^(6*n + 55) - 21946*q^(6*n + 54) - 27776*q^(6*n + 53) - 34511*q^(6*n + 52) - 42139*q^(6*n + 51) - 50573*q^(6*n + 50) - 59706*q^(6*n + 49) - 69338*q^(6*n + 48) - 79259*q^(6*n + 47) - 89190*q^(6*n + 46) - 98845*q^(6*n + 45) - 107908*q^(6*n + 44) - 116073*q^(6*n + 43) - 123015*q^(6*n + 42) - 128478*q^(6*n + 41) - 132246*q^(6*n + 40) - 134166*q^(6*n + 39) - 134166*q^(6*n + 38) - 132246*q^(6*n + 37) - 128478*q^(6*n + 36) - 123015*q^(6*n + 35) - 116073*q^(6*n + 34) - 107908*q^(6*n + 33) - 98845*q^(6*n + 32) - 89190*q^(6*n + 31) - 79259*q^(6*n + 30) - 69338*q^(6*n + 29) - 59706*q^(6*n + 28) - 50573*q^(6*n + 27) - 42139*q^(6*n + 26) - 34511*q^(6*n + 25) - 27776*q^(6*n + 24) - 21946*q^(6*n + 23) - 17016*q^(6*n + 22) - 12922*q^(6*n + 21) - 9618*q^(6*n + 20) - 6999*q^(6*n + 19) - 4981*q^(6*n + 18) - 3452*q^(6*n + 17) - 2330*q^(6*n + 16) - 1521*q^(6*n + 15) - 965*q^(6*n + 14) - 587*q^(6*n + 13) - 344*q^(6*n + 12) - 191*q^(6*n + 11) - 99*q^(6*n + 10) - 43*q^(6*n + 9) - 16*q^(6*n + 8) - 3*q^(6*n + 7) + q^(5*n + 65) + 15*q^(5*n + 64) + 51*q^(5*n + 63) + 139*q^(5*n + 62) + 296*q^(5*n + 61) + 583*q^(5*n + 60) + 1036*q^(5*n + 59) + 1768*q^(5*n + 58) + 2840*q^(5*n + 57) + 4435*q^(5*n + 56) + 6634*q^(5*n + 55) + 9676*q^(5*n + 54) + 13636*q^(5*n + 53) + 18800*q^(5*n + 52) + 25198*q^(5*n + 51) + 33117*q^(5*n + 50) + 42470*q^(5*n + 49) + 53479*q^(5*n + 48) + 65875*q^(5*n + 47) + 79768*q^(5*n + 46) + 94662*q^(5*n + 45) + 110535*q^(5*n + 44) + 126661*q^(5*n + 43) + 142893*q^(5*n + 42) + 158337*q^(5*n + 41) + 172812*q^(5*n + 40) + 185382*q^(5*n + 39) + 195950*q^(5*n + 38) + 208606*q^(5*n + 36) + 210134*q^(5*n + 35) + 208606*q^(5*n + 34) + 203662*q^(5*n + 33) + 195950*q^(5*n + 32) + 185382*q^(5*n + 31) + 172812*q^(5*n + 30) + 158337*q^(5*n + 29) + 142893*q^(5*n + 28) + 126661*q^(5*n + 27) + 110535*q^(5*n + 26) + 79768*q^(5*n + 24) + 65875*q^(5*n + 23) + 53479*q^(5*n + 22) + 42470*q^(5*n + 21) + 33117*q^(5*n + 20) + 25198*q^(5*n + 19) + 18800*q^(5*n + 18) + 13636*q^(5*n + 17) + 9676*q^(5*n + 16) + 6634*q^(5*n + 15) + 4435*q^(5*n + 14) + 2840*q^(5*n + 13) + 1768*q^(5*n + 12) + 1036*q^(5*n + 11) + 583*q^(5*n + 10) + 296*q^(5*n + 9) + 139*q^(5*n + 8) + 51*q^(5*n + 7) + 15*q^(5*n + 6) + q^(5*n + 5) - 5*q^(4*n + 59) - 35*q^(4*n + 58) - 112*q^(4*n + 57) - 285*q^(4*n + 56) - 606*q^(4*n + 55) - 1168*q^(4*n + 54) - 2070*q^(4*n + 53) - 3465*q^(4*n + 52) - 5505*q^(4*n + 51) - 8404*q^(4*n + 50) - 12355*q^(4*n + 49) - 17592*q^(4*n + 48) - 24287*q^(4*n + 47) - 32639*q^(4*n + 46) - 42738*q^(4*n + 45) - 54657*q^(4*n + 44) - 68311*q^(4*n + 43) - 83570*q^(4*n + 42) - 100125*q^(4*n + 41) - 117609*q^(4*n + 40) - 135488*q^(4*n + 39) - 153203*q^(4*n + 38) - 170082*q^(4*n + 37) - 185475*q^(4*n + 36) - 198706*q^(4*n + 35) - 209203*q^(4*n + 34) - 216480*q^(4*n + 33) - 220209*q^(4*n + 32) - 220209*q^(4*n + 31) - 216480*q^(4*n + 30) - 209203*q^(4*n + 29) - 198706*q^(4*n + 28) - 185475*q^(4*n + 27) - 170082*q^(4*n + 26) - 153203*q^(4*n + 25) - 135488*q^(4*n + 24) - 117609*q^(4*n + 23) - 100125*q^(4*n + 22) - 83570*q^(4*n + 21) - 68311*q^(4*n + 20) - 54657*q^(4*n + 19) - 42738*q^(4*n + 18) - 32639*q^(4*n + 17) - 24287*q^(4*n + 16) - 17592*q^(4*n + 15) - 12355*q^(4*n + 14) - 8404*q^(4*n + 13) - 5505*q^(4*n + 12) - 3465*q^(4*n + 11) - 2070*q^(4*n + 10) - 1168*q^(4*n + 9) - 606*q^(4*n + 8) - 285*q^(4*n + 7) - 112*q^(4*n + 6) - 35*q^(4*n + 5) - 5*q^(4*n + 4) + 10*q^(3*n + 53) + 50*q^(3*n + 52) + 159*q^(3*n + 51) + 384*q^(3*n + 50) + 814*q^(3*n + 49) + 1538*q^(3*n + 48) + 2704*q^(3*n + 47) + 4445*q^(3*n + 46) + 6956*q^(3*n + 45) + 10391*q^(3*n + 44) + 14973*q^(3*n + 43) + 20800*q^(3*n + 42) + 28042*q^(3*n + 41) + 36688*q^(3*n + 40) + 46779*q^(3*n + 39) + 58103*q^(3*n + 38) + 70527*q^(3*n + 37) + 83608*q^(3*n + 36) + 97042*q^(3*n + 35) + 110202*q^(3*n + 34) + 122671*q^(3*n + 33) + 133762*q^(3*n + 32) + 143106*q^(3*n + 31) + 150074*q^(3*n + 30) + 154473*q^(3*n + 29) + 155910*q^(3*n + 28) + 154473*q^(3*n + 27) + 150074*q^(3*n + 26) + 143106*q^(3*n + 25) + 133762*q^(3*n + 24) + 122671*q^(3*n + 23) + 110202*q^(3*n + 22) + 97042*q^(3*n + 21) + 83608*q^(3*n + 20) + 70527*q^(3*n + 19) + 58103*q^(3*n + 18) + 46779*q^(3*n + 17) + 36688*q^(3*n + 16) + 28042*q^(3*n + 15) + 20800*q^(3*n + 14) + 14973*q^(3*n + 13) + 10391*q^(3*n + 12) + 6956*q^(3*n + 11) + 4445*q^(3*n + 10) + 2704*q^(3*n + 9) + 1538*q^(3*n + 8) + 814*q^(3*n + 7) + 384*q^(3*n + 6) + 159*q^(3*n + 5) + 50*q^(3*n + 4) + 10*q^(3*n + 3) - 10*q^(2*n + 47) - 45*q^(2*n + 46) - 138*q^(2*n + 45) - 328*q^(2*n + 44) - 684*q^(2*n + 43) - 1279*q^(2*n + 42) - 2218*q^(2*n + 41) - 3592*q^(2*n + 40) - 5519*q^(2*n + 39) - 8085*q^(2*n + 38) - 11381*q^(2*n + 37) - 15429*q^(2*n + 36) - 20242*q^(2*n + 35) - 25743*q^(2*n + 34) - 31827*q^(2*n + 33) - 38291*q^(2*n + 32) - 44912*q^(2*n + 31) - 51395*q^(2*n + 30) - 57447*q^(2*n + 29) - 62741*q^(2*n + 28) - 67003*q^(2*n + 27) - 69987*q^(2*n + 26) - 71528*q^(2*n + 25) - 71528*q^(2*n + 24) - 69987*q^(2*n + 23) - 67003*q^(2*n + 22) - 62741*q^(2*n + 21) - 57447*q^(2*n + 20) - 51395*q^(2*n + 19) - 44912*q^(2*n + 18) - 38291*q^(2*n + 17) - 31827*q^(2*n + 16) - 25743*q^(2*n + 15) - 20242*q^(2*n + 14) - 15429*q^(2*n + 13) - 11381*q^(2*n + 12) - 8085*q^(2*n + 11) - 5519*q^(2*n + 10) - 3592*q^(2*n + 9) - 2218*q^(2*n + 8) - 1279*q^(2*n + 7) - 684*q^(2*n + 6) - 328*q^(2*n + 5) - 138*q^(2*n + 4) - 45*q^(2*n + 3) - 10*q^(2*n + 2) + 5*q^(n + 41) + 23*q^(n + 40) + 67*q^(n + 39) + 159*q^(n + 38) + 327*q^(n + 37) + 606*q^(n + 36) + 1036*q^(n + 35) + 1656*q^(n + 34) + 2499*q^(n + 33) + 3589*q^(n + 32) + 4933*q^(n + 31) + 6517*q^(n + 30) + 8302*q^(n + 29) + 10227*q^(n + 28) + 12207*q^(n + 27) + 14143*q^(n + 26) + 15928*q^(n + 25) + 17453*q^(n + 24) + 18621*q^(n + 23) + 19355*q^(n + 22) + 19606*q^(n + 21) + 19355*q^(n + 20) + 18621*q^(n + 19) + 17453*q^(n + 18) + 15928*q^(n + 17) + 14143*q^(n + 16) + 12207*q^(n + 15) + 10227*q^(n + 14) + 8302*q^(n + 13) + 6517*q^(n + 12) + 4933*q^(n + 11) + 3589*q^(n + 10) + 2499*q^(n + 9) + 1656*q^(n + 8) + 1036*q^(n + 7) + 606*q^(n + 6) + 327*q^(n + 5) + 159*q^(n + 4) + 67*q^(n + 3) + 23*q^(n + 2) + 5*q^(n + 1) - 1] The constant term of these is easy to guess again, it is $\prod_{j=1}^{k-2} (1+q^j)^{k-1-j}$. The linear term also has some obvious factors, which are of somewhat similar form, but there is also a larger factor which I cannot explain.<|endoftext|> TITLE: Sequence A76132 eventually periodic modulo $2,3$ and $5$ QUESTION [6 upvotes]: Sequence A76132 starting as $1,1,2,4,10,36,218,\ldots$ of the OEIS is recursively defined by $a(1)=1$ and $a(n)=\sum_{k=1}^{n-1}a(n-k)^k$ for $n\geq 2$. It is eventually periodic of period 1,1 and 34 modulo $2,3$ and $5$ (the proof is essentially trivial: it is enough to check eventual periodicity on a big enough finite initial chunk). Curiously, this seems to stop: either the non-periodic part is very long (of length at least 100) or the period is of length longer than 5000 for $p=7$. Moreover, the sequence seems to be eventually periodic modulo $2^k$: up to $2^3=8$ the sequence is eventually constant modulo $2^k$ (for $k=1,2,3$) after that it seems to be eventually periodic with period-length $2^{k-3}$. Similarly, it seems to be eventually periodic with period length $3^k$ for $k\geq 2$. Curiously, it does not seem to be eventually periodic modulo $25$. Is there an easy explanation for this behaviour? Added: The generating series $\sum a(n)t^n$ is algebraic modulo every prime number. Eventual periodicity modulo $p$ boils down to showing that it is a rational series modulo $p$. Final addition. The remark of reuns implies that $\sum a(n)t^n$ is always rational modulo any integer $m$. The aperiodic part and the period length are however generally of order $O(m^m)$ which explains why they are not easily observable. REPLY [2 votes]: The comment from "reuns" is the real answer-- the sequence is periodic for all $M$. For prime $M$, we can slightly improve the implied bound as follows. We track the pair: $$\left(j, \sum_{k=1}^{n}(a_{n-k}){k+j}\right) \bmod M$$ for $j\in 0,1,..., \phi(M)-1$ (instead of $2\phi(M)$). We may end up detecting a multiple of a cycle, but the total size of the state space (and thus maximum cycle length) is smaller: $\phi(M)M^{\phi(M)}$. This is still not a great bound, e.g., the bound for M=11 is 259,374,246,010. In practice, though, the periods may be shorter. I'm including a little snippet of Python code for finding these. Running time is a fraction of a second. First, the results: Extending series to 1000000 places ### M=2 ### minimal period: start 1, length 1 ### M=3 ### minimal period: start 5, length 1 ### M=5 ### minimal period: start 3, length 37 ### M=7 ### minimal period: start 253, length 807 ### M=11 ### minimal period: start 303832, length 9 ### M=13 ### ... Period not found ... ### M=17 ### ... Period not found ... ### M=19 ### ... Period not found ... Next, the code: # Try all the values from 2 to 20 modulus_list = range(2,21) # "max_cycle_length" is optional. If set to 0, we detect cycles of # any length, but memory grows linearly as the experiment runs. If # it's >0, then memory is bounded and we can extend the cycle # indefinitely given enough time. max_cycle_length = 0 import numpy as np from math import gcd def phi(n): amount = 0 for k in range(1, n + 1): if gcd(n, k) == 1: amount += 1 return amount # Encode the historical state into a single, hashable number def encode(ar): out=0 multiplier=1 for a in ar: out+=a*multiplier multiplier*=M return((out, step % phiM)) num_trials = 1000000 print('Extending series to %d places' % num_trials) if max_cycle_length>0: print('Maximum detectable cycle length: %d' % max_cycle_length) for M in modulus_list: phiM = phi(M) # Restrict to primes for now if phiM < M-1: continue print("### M=%d ###" % M) state=np.ones(phiM) history={} a=np.zeros(num_trials) periodic = False for step in range(num_trials): new_value = state[step % phiM] a[step] = new_value powered = new_value for i in range(phiM): j = (1+i+step) % phiM state[j] = (powered+state[j]) % M powered = (powered*new_value) % M enc = encode(state) if enc in history: periodic = True break else: history[enc] = step if max_cycle_length > 0: if len(history) == max_cycle_length: history = {} if not periodic: print(' ... Period not found ...') continue # We may have a multiple of the fundamental period; minimize it. start = history[enc]+1 end = step if end <= start+1: print(' minimal period: start %d, length 1' % (start)) else: chunk = a[start:start+28] for i in range(start+2, end-len(chunk)+1): if np.array_equal(chunk, a[i:i+len(chunk)]): print(' minimal period: start %d, length %d' % (start, i)) break else: print(' minimal period: start %d, length %d' % (start, end-start)) Extending num_trials=1000000 did not turn up anything else, but a little more patience might reveal something<|endoftext|> TITLE: Distinct exponents in the factorization of the factorial, a problem of Erdős QUESTION [20 upvotes]: In the 1982 paper below, Paul Erdős proved that if $h(n)$ is the number of distinct exponents in the prime factorization of $n!$ then $$c_1\Big(\frac{n}{\log n}\Big)^{1/2} < h(n) < c_2\Big(\frac{n}{\log n}\Big)^{1/2}$$ for all sufficiently large $n$, where $c_1, c_2$ are positive constants. Then he said that "there is no doubt" that $$h(n) = (c + o(1))\Big(\frac{n}{\log n}\Big)^{1/2}$$ as $n$ goes to infinity, for some positive constant $c$, but that "we do not know enough about the difference of consecutive primes" to conclude so. I wonder if now, after 40 years, "we" know enough to prove Erdős conjecture (?) P. Erdős, Miscellaneous problems in number theory, Congressus Numerantium 34 (1982), 25-45. https://users.renyi.hu/~p_erdos/1982-08.pdf REPLY [14 votes]: The primes $p \leq \sqrt{n}$ can be ignored as the number of them is $$\approx \sqrt{n} / \log (\sqrt{n} )= 2 \sqrt{n}/\log n = o (1) \cdot \sqrt { n/\log n}$$ For primes $p> \sqrt{n}$, the exponent of $p$ is simply $\lfloor n/p \rfloor$. So an equivalent problem is to count the number of $1 \leq k \leq \lfloor \sqrt{n} \rfloor$ such that the interval $ (n/(k+1), n/k]$ contains a prime. When $k$ is small, these intervals are very long, and one can prove that all or almost all contain a prime. When $k$ is large, the intervals are very short, and one can prove that few of them contain two primes and then use the prime number theorem to count primes. The tricky case is the critical range when $k \approx \sqrt{n / \log n}$, say $ \sqrt{n/\log n}/2 < k < 2 \sqrt{n/\log n}$. In this range, the expected proportion of intervals that contain a prime is strictly between $0$ and $1$. To get the exact constant, we would need to estimate this proportion precisely, which seems quite difficult. Doing this by the moment method would require understanding the expected number of primes in the interval, the expected number of pairs of primes, triples, etc. and I don't think we have asymptotics for anything but the expected number and maybe the pairs. That said, the Cramér random model suggests that $c$ is exactly $$\int_{0}^{\infty} \left(1 - e^{ - 2/ x^2} \right) dx = \sqrt{2\pi},$$ if I did the calculation right, with the integrand at $x$ being the probability to have a prime in the interval associated to $k = x \sqrt{n/\log n}$.<|endoftext|> TITLE: How do the various homotopy 2-categories compare? QUESTION [6 upvotes]: There are various models of $\infty$-categories floating around, so there are as many models of the associated homotopy 1- and 2-categories. Because the relations between the former are worked out in great detail, it feels like the relations between the latter should be well-understood as well. But since I was unable to find much about the 2-dimensional case I'd like to ask for references here. Specifically I am interested in the following questions. Given a Kan-enriched category $\mathcal{C}$ we can apply the homotopy coherent nerve to get a quasicategory $N^\Delta(\cal{C})$. We can then apply [HTT Prop. 2.3.4.12] to pass to a simplicial set $h_2(N^\Delta(\cal{C}))$, which Lurie calls the underlying 2-category. On the other hand we can take $\cal{C}$ and apply hom-wise the homotopy-category functor and obtain a Grpd-enriched category, say $H_2(\cal{C})$. We can now apply the Duskin-nerve to get another simplicial set $N^D(H_2(\cal{C}))$, which ought to be a 2-category in the sense of HTT. Are both sSets equivalent in an appropriate sense (I guess it means Joyal-equivalent)? Even more important to me is the following variation: Note that $H_2(\cal{C})$ is a Grpd-enriched hence a strict (2,1)-category. In the light of Kerodon Rem. 2.3.6 it seems like the Duskin-nerve has a left adjoint $|-|^D:\mathsf{sSet} \rightarrow (2,1)\mathsf{Cat}_{str}$. So are the strict (2,1)-categories $H_2(\cal{C})$ and $|h_2(N^\Delta(\cal{C}))|^D$ equivalent as strict (2,1)-categories or at least as weak (2,1)-categories? Out of curiosity I'd like to add a third variation on the question: Using the Cordier-realization $|-|^C:\mathsf{sSet} \rightarrow \mathsf{sSet}\text{-}\mathsf{Cat}$ and the fibrant replacement of simplicial categories $R:\mathsf{sSet}\text{-}\mathsf{Cat} \rightarrow \mathsf{Kan}\text{-}\mathsf{Cat}$ we can turn a quasicategory $C$ into a Kan-enriched category $R(|C|^C)$. So how are $|h_2(C)|^D$ and $H_2(R(|C|^C))$ related? Thank you very much for your time and considerations! REPLY [6 votes]: The simplicial sets $h_2(N^\Delta(\mathcal{C}))$ and $N^D(H_2(\mathcal{C}))$ are isomorphic. To prove this, observe that the universal property of $h_2(N^\Delta(\mathcal{C}))$ applied to the image under $N^{\Delta}$ of the quotient simplicial functor $\mathcal{C} \to H_2(\mathcal{C})$ yields a map of simplicial sets $h_2(N^\Delta(\mathcal{C})) \to N^D(H_2(\mathcal{C}))$, which one can easily check is an isomorphism of simplicial sets (it suffices to check that it's a bijection on 0-, 1-, and 2-simplices). The strict (2,1)-categories $|h_2(N^\Delta(\mathcal{C}))|^{\mathcal{D}}$ and $H_2(\mathcal{C})$ are biequivalent; in fact, there is a strict 2-functor $|h_2(N^\Delta(\mathcal{C}))|^{\mathcal{D}} \to H_2(\mathcal{C})$ which is a bijective-on-objects biequivalence. To prove this, note that the composite of the Duskin nerve functor (for strict (2,1)-categories) with its left adjoint sends a strict (2,1)-category $\mathcal{A}$ to its "normal pseudofunctor classifier" $Q\mathcal{A}$, which is a strict (2,1)-category with the universal property that strict 2-functors $Q\mathcal{A} \to \mathcal{B}$ are in natural bijection with normal pseudofunctors $\mathcal{A} \to \mathcal{B}$, for $\mathcal{B}$ a strict (2,1)-category. Moreover, by this universal property, there is a "counit" strict 2-functor $Q\mathcal{A} \to \mathcal{A}$ which one can show is bijective on objects and an equivalence on hom-categories, and hence a biequivalence. The strict (2,1)-categories $|h_2(C)|^D$ and $H_2(R(|C|^C))$ are isomorphic (if you make a good choice of $R$, e.g. change-of-base along $Ex^\infty$). Indeed, the two functors $|h_2(-)|^D$ and $H_2(|-|^C)$ are naturally isomorphic, and the functor $H_2$ sends the "unit" map $\mathcal{E} \to R(\mathcal{E})$ to an isomorphism (for a good choice of $R$ as above). I don't know any references for these answers, but these are all straightforward and standard arguments. If you would like me to elaborate on any of these points, I would be happy to.<|endoftext|> TITLE: Is there a natural topology on the automorphism group of a topological group? QUESTION [6 upvotes]: $\DeclareMathOperator\TAut{TAut}\DeclareMathOperator\Homeo{Homeo}$Let $G$ be a topological group, and let $\TAut(G)$ denote the group of topological automorphisms of $G$ under composition (i.e. the group of maps $f \colon G \to G$ that are simultaneously group automorphisms and self-homeomorphisms). We wish to give $\TAut(G)$ a reasonable topology in the following sense: $\TAut(G)$ becomes a topological group with respect to this topology. This topology should interact with/depend on the topology of $G$ in some way, i.e. we can require that the natural action $\TAut(G) \times G \to G$ is continuous. In the case that $G$ is compact, it is known that giving $\TAut(G)$ the compact–open topology satisfies the above conditions, where the compact–open topology has as a subbasis sets of the form $$V(C, U) = \{f \colon G \to G \mid f(C) \subseteq U\},$$ where $C, U \subseteq G$ are compact and open, respectively. If $G$ is locally compact, we can instead give $\TAut(G)$ the $g$-topology, which has as a subbasis sets of the form $$V(K, W) = \{f \colon G \to G \mid f(K) \subseteq W\},$$ where either $K$ or $G \setminus W$ is compact. The cases where $G$ is compact or locally compact are discussed in Dijkstra - On Homeomorphism Groups and the Compact–Open Topology and Arens - Topologies for Homeomorphism Groups. In fact, these two papers discuss the group of self-homeomorphisms $\Homeo(X)$ for a space $X$ which is not necessarily a topological group, so the case for $G$ and $\TAut(G)$ follows from that (since $\TAut(G)$ is a subgroup of $\Homeo(G)$). My question is, for a general topological group $G$, is there a good way to describe a topology on $\TAut(G)$ satisfying the two conditions above? We can simply say: "give $\TAut(G)$ the coarsest topology such that it becomes a topological group and the action $\TAut(G) \times G \to G$ is continuous," but I am hoping for something more explicit than that. REPLY [8 votes]: $\DeclareMathOperator\Aut{Aut}$There is a recent paper Uniformly locally bounded spaces and the group of automorphisms of a topological group by Maxime Gheysens where he among other nice things systematically investigates the topologies on $\Aut(G)$ for any topological group $G$. On every topological group there are several natural uniform structures making translations become uniformly continuous: first of all, the left and right uniform structure, their supremum (the upper uniform structure) and their infimum (the lower or Roelcke uniform structure). Now, as it turns out, on $\Aut(G)$ one can usefully consider: the topology of uniform biconvergence on bounded sets with respect to the left, right, or upper uniform structure (they all give the same topology) or the topology of uniform biconvergence on bounded sets with respect to the lower uniform structure. In general, these two topologies are different, but they coincide for the so-called SIN groups (and even broader, for coarsely SIN groups), i.e. groups with admitting a basis of conjugation-invariant identity neighborhoods as well as for all locally compact groups. So the existence of two really different useful topologies on $\Aut(G)$ is purely a phenomenon in the world of “very big” groups.<|endoftext|> TITLE: Progress in robustifying mathematics - i.e. making mathematical theorems robust to small changes in hypotheses QUESTION [10 upvotes]: The idea of making a mathematical theorem robust to small changes in its hypotheses has been known for some time. In areas such as group theory reasonable progress has been made leading to the theory of approximate groups - see Terence Tao's comment here and related notes. It seems that Stanislav Ulam was the first to discuss this overall concept in reference to the stability of functional equations in a talk in 1940. In his "A Collection of Mathematical Problems", Chapter 6, Section 1 "Stability" he asks "When is it true that by changing a little the hypotheses of a theorem, one can still assert that the thesis of the theorem remains true or approximately true?" He gives the following example by way of illustration: "If $f(x)$ is a measurable real- valued function defined for all real $x$ satisfying the inequality $|f(x + y) - (f(x)+f(y))| TITLE: Fixpoints of $m\longmapsto \mathrm{rad}(\phi(m^2))$ under iteration QUESTION [10 upvotes]: Given a strictly positive integer $m$ let $\alpha(m)=\mathrm{rad}(m\phi(m))$ be the radical (product of all distinct prime divisors) of the product of $m$ and of Euler's totient function $\phi(m)=m\prod_{p\vert m} \left(1-\frac{1}{p}\right)$ (where the product is over all prime-divisors of $m$). (Equivalently, $\alpha(m)=\mathrm{rad}(\phi(m^2))$. Since $\alpha(m)$ is bounded above by the product of all primes at most equal to the largest prime-divisor of $m$, iterating $\alpha$ yields a fixed point $\alpha^\infty(m)=\alpha^k(m)=\alpha(F(m))$ for huge enough $k$ ($k=m$ certainly works). Curiously, the sequence $$1,2,6,10,30,34,42,78,102,110,\ldots$$ (given by enumerating in increasing order all elements of $\alpha^\infty(\mathbb N)$) of fixpoints for $\alpha$ is not recognized by the OEIS! If $m$ is square-free (i.e. if $m=\mathrm{rad}(m)$), we have the easy inequalities $\mathrm{loglog}(m)\leq \mathrm{loglog}(\alpha^\infty(m))\leq 2\mathrm{loglog}(m)$ (where $\mathrm{loglog}(x)=\log(\log(x))$) which imply the existence of a constant $$a=\limsup_{m\rightarrow \infty,\,\mathrm{rad(m)=m}}\frac{\mathrm{loglog}\,\alpha^\infty(m)}{\mathrm{loglog}\,m}\,.$$ ($\limsup$ is in fact attained over the set of prime-numbers.) Can the obvious bounds $1\leq a\leq 2$ be improved? The bound should be close to $2$ if there are very large 'towers' of Sophie Germain primes (primes such that iterating $p\longmapsto 2 p+1$ is also prime). The factor $2$ in $p\longmapsto 2 p+1$ can in fact be replaced by arbitrary small varying numbers (larger than $1$). Motivation I know of at least two motivations related to the map $\alpha^\infty$ considered above: (1) Prime certificates: Primality of a prime number $p$ is proven by exhibiting an element $g=g_p$ of order exactly $p-1$ in the cyclic group $(\mathbb Z/p\mathbb Z)^*$ of invertibles. This can be achieved by showing $g^{p-1}\equiv 1\pmod p$ and $g^{(p-1)/q}\not\equiv 1\pmod p$ for every prime divisor $q$ of $p-1$. (Finding such an integer $g$ should not be difficult: a positive proportion of random elements in $\{2,\ldots,p-2\}$ should work. Computing $g^r\pmod m$ is easy using fast exponentiation underlying for example the RSA cryptosystem). The constant $a$ above is thus related to the worst case of such prime certificates (which involve all primes dividing $\alpha^\infty(p)$). (2) Consider the orbit of an initial integer $s_0\geq 2$ under $x\longmapsto x+x^x$. The sequence $s_0,s_1=s_0+s_0^{s_0},s_2=s_1+s_1^{s_1},\ldots$ grows extraordinarily fast but is easily computable modulo $m$ (and eventually periodic modulo every integer $m$). Computations modulo $m$ involve however also the computation modulo suitable powers of all divisors of $A(m)$. REPLY [6 votes]: Your observation on towers of primes is spot on. To study the iterations of $\alpha$ at $m$, for each prime $p$ dividing $m$, one builds a tree whose descending branches are the primes that divide $p-1$, and repeats iteratively. These are known as Pratt trees; the primes that appear in (the union of) the tree for $m$ are those appearing in $\alpha^\infty (m)$. The current state of the art on general Pratt trees is Ford, Konyagin, and Luca, which still is a long way from the full structure of those trees. That the problem is likely difficult can be seen even by considering a single prime, as very large prime factors of $p-1$ are only known under Elliott-Halberstam. Unfortunately the extra structure coming from the full tree (as opposed to just going down one level) only comes into play when seeking for lower bounds, whereas upper bounds depend on constructions with specific primes for the worst-case scenario. What I can tell you is that in virtually all these situations, if you have an obvious bound like $1 \leq a(m) \leq 2$, one tipically has (search in the literature for "iterated totient") statements like: infinitely many $m$ come close to the upper bound (conditionally), while most $m$ are close to the lower bound; there are quantitative forms of the latter as well, i.e. that read like "all but $O(f(x))$ of $m \leq x$ have $a(m)<1+\varepsilon(m)$" for various explicit pairs of $f$, $\varepsilon$.<|endoftext|> TITLE: Explanation for $\chi(\operatorname{SL}_2(\mathbb{Z})) = -1/12$ with zeta function QUESTION [16 upvotes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}$One can compute the (group cohomological) Euler characteristic of $\SL_2(\mathbb{Z})$ via $$ \chi(\SL_2(\mathbb{Z})) = \chi(\mathbb{Z}/2) \cdot \chi(\PSL_2(\mathbb{Z})) = \frac{1}{2}\cdot \left(\frac{1}{2} + \frac{1}{3} - 1\right) = -\frac{1}{12} = \zeta(-1) $$ (since $ \PSL_2(\mathbb{Z}) \cong \mathbb{Z}/2 * \mathbb{Z}/3 $). Alternatively it follows as $\SL_2(\mathbb{Z})$ has $F_2$ as an index-$12$ subgroup. I was wondering whether the connection with $\zeta$ is coincidental. The only (far fetched) connection I could come up with is that the (representation theoretic) zeta function of $\SL_2(\mathbb{C})$ is the usual $\zeta$, as its irreducible representations have dimensions $1, 2, 3, \dotsc$. (Here $\zeta_G(s) = \sum_{V\in \operatorname{Irr}(G)} \dim(V)^{-s}$.) REPLY [14 votes]: (Expanding my comment into an answer) It is not a coincidence. Relating the Euler characteristic of certain arithmetic groups to the Zeta function is a theorem due to Harder [1] from 1971. It is expanded on in Brown's "Cohomology of Groups", Chapter IX.8. Taken from Gruenberg's AMS review of Brown's book is the following overview of the idea: Let $G$ be an algebraic subgroup of $\operatorname{GL}_n$ defined over $\mathbb{Q}$ and $\Gamma$ an arithmetic subgroup of $G(\mathbb{Q})$. Then $\Gamma$ is a discrete subgroup of the Lie group $G(\mathbb{R})$. If $K$ is a maximal compact subgroup of $G(\mathbb{R})$, then $X = G(\mathbb{R})/K$ is diffeomorphic to a euclidean space of dimension say $d$. [...] Number theory enters through the work of Harder (1971). The Gauss-Bonnet measure on $X$ lifts to a unique invariant measure $\mu$ on $G(\mathbb{R})$. Harder proved the deep theorem that $\chi(\Gamma) = \mu(G(\mathbb{R})/\Gamma)$. This leads to an explicit fromula [sic.] for $\chi(G(\mathbb{Z}))$ in terms of values of the zeta-function. For example, $\chi(\operatorname{SL}_2(\mathbb{Z})) = \zeta(-1)$ and since $\zeta(-1) = -1/12$, this gives a third way of arriving at the Euler characteristic of $\operatorname{SL}_2(\mathbb{Z})$. Edit: One can find more values using this method, of course. Some are given in Brown's book (p. 255-256). For example, we have $$ \chi(\operatorname{SL}_n(\mathbb{Z})) = \prod_{k=2}^n \zeta(1-k) $$ and $$ \chi(\operatorname{Sp}_{2n}(\mathbb{Z})) = \prod_{k=1}^n \zeta(1-2k). $$ Thus for example, we find $\chi(\operatorname{SL}_n(\mathbb{Z})) = 0$ for $n \geq 3$ and $\chi(\operatorname{Sp}_{4}(\mathbb{Z})) = \zeta(-1)\zeta(-3) = -\frac{1}{1440}$. [1] Harder, G., A Gauss-Bonnet formula for discrete arithmetically defined groups, Ann. Sci. Éc. Norm. Supér. (4) 4, 409-455 (1971). ZBL0232.20088.<|endoftext|> TITLE: What results about the topology of manifolds depend on the dimension mod 3? QUESTION [30 upvotes]: There are a lot of interesting results about the topology of manifolds that depend on the dimension of the manifold mod 2, mod 4, or mod 8. The simplest ones involve the cup product $$ \smile \colon H^{n/2}(M,R) \times H^{n/2}(M,R) \to H^{n}(M,R) \cong R$$ where $n$, the dimension of $M$, is even, $R$ is a commutative ring, and the last isomorphism is present if $M$ is compact and oriented (or more generally $R$-oriented). When $n$ is a multiple of $4$, this trick gives a nondegenerate symmetric bilinear form $$ \smile \colon H^{n/2}(M,\mathbb{R}) \times H^{n/2}(M,\mathbb{R}) \to \mathbb{R}$$ and the signature of this bilinear form is an important invariant called the signature of $M$. In this case we also get a lattice $L$ that's the image of $H^{n/2}(M,\mathbb{Z})$ in $H^{n/2}(M,\mathbb{R})$, and a unimodular symmetric bilinear form $$ \smile \colon L \times L \to \mathbb{Z}$$ which gives more refined information about $M$. This is very important in the classification of compact oriented 4-dimensional manifolds. When the dimension of $M$ equals 2 mod 4, we instead get a nondegenerate skew-symmetric bilinear form $$ \smile \colon H^{n/2}(M,\mathbb{R}) \times H^{n/2}(M,\mathbb{R}) \to \mathbb{R}$$ otherwise known as a symplectic structure, and a unimodular skew-symmetric bilinear form $$ \smile \colon L \times L \to \mathbb{Z}.$$ These data are not directly helpful in classifying manifolds because they're determined up to isomorphism by the dimension of $H^{n/2}(M,\mathbb{R})$, unlike in the case where $n$ is a multiple of $4$. But if $M$ is equipped a 'framing', then we can get an interesting invariant by taking $R = \mathbb{Z}/2$ and improving the bilinear form above to a quadratic form. This is called the Kervaire invariant. There are subtler tricks involving spinors that depend heavily on the dimension of the manifold mod 8. There's also a nice result that depends on the dimension of the manifold mod $2^n$ for all $n$: namely, any smooth compact $n$-dimensional manifold admits an immersion into Euclidean space of dimension $2n-H(n)$, where $H(n)$ is the number of 1's in the binary expansion of $n$. (See Cohen's paper The immersion conjecture for differentiable manifolds.) All this made me wonder if there are interesting results about manifolds that depend on their dimension mod 3 — or for that matter, any number that's not just a power of 2. Do you know any? If $M$ is oriented and its dimension is a multiple of 3, we can use the cup product to get a trilinear form $$ \smile \colon H^{n/3}(M,\mathbb{R}) \times H^{n/3}(M,\mathbb{R}) \times H^{n/3}(M,\mathbb{R}) \to \mathbb{R}$$ and if $L$ is the image of $H^{n/3}(M,\mathbb{Z})$ in $H^{n/3}(M,\mathbb{R})$ this restricts to give a trilinear form $$ \smile \colon H^{n/3}(M,\mathbb{Z}) \times H^{n/3}(M,\mathbb{Z}) \times H^{n/3}(M,\mathbb{Z}) \to \mathbb{Z}.$$ These are symmetric when $n$ is a multiple of 6 but skew-symmetric when $n$ equals 3 mod 6. Of course this already is a partial answer to my question, and we could easily generalize to numbers other than 3. But do these trilinear forms give nontrivial invariants of $M$? Are they used for anything interesting? In a quick attempt to search for this I bumped into a paper on cubic forms that are invariants of 12-dimensional manifolds: Fei Han, Ruizhi Huang, Kefeng Liu, Weiping Zhang, Cubic forms, anomaly cancellation and modularity. Also, Ahmet Beyaz has a paper A new construction of 6-manifolds that builds on a paper (Classification problems in differential topology. V. On certain 6-manifolds) where C.T.C. Wall classified simply-connected, compact oriented 6-manifolds with spin structure and torsion-free cohomology with the help of the trilinear form $$ \smile \colon H^{2}(M,\mathbb{Z}) \times H^{2}(M,\mathbb{Z}) \times H^{2}(M,\mathbb{Z}) \to \mathbb{Z}.$$ But I don't know if either of these is part of a bigger story that involves the manifold's dimension mod 3, or 6, or 12. REPLY [9 votes]: I learned recently that the 'topological modular form' cohomology theory TMF has $576$-fold periodicity. Not quite period $3$, but similar, since $576=3^2 \times 2^6$.<|endoftext|> TITLE: 3-periodic point implies positive topological entropy QUESTION [6 upvotes]: When I learn some basic ergodic theory, I encounter an interesting exercise. As we all know, 3-periodic point often means chaos. Therefore, when a continuous map has a 3-periodic point, it may have positive topological entropy. The following question exactly talks about it. Question: Let $ I=[0,1]$, and $f$ be a continuous map from $I$ to itself. Suppose $f$ have a 3-periodic point, where 3 means minimal period. Then try to prove the topological entropy of $f$ is strictly positive. Well, Sharkovsky's theorem tells us that $f$ has arbitrary periodic points, which means "chaos". But I do not know how to prove this exercise. My idea is to construct a symbolic system as a factor or sub-system by dividing the interval with periodic points, for Bernoulli shift has positive topological entropy. But I fail. Please help me! Thank you! REPLY [3 votes]: See Theorem 1 in The periodic points of maps of the disk and the interval. It states that if a continuous map $f:[0,1]\rightarrow [0,1]$ has a point of period $n$, and $n$ is not a power of two, then $h_{\rm{top}}(f)>\frac{1}{n}\log 2$. The proof establishes the lower bound via constructing a suitable $(n,\epsilon)$-separated subset for $f$. Also see Milnor and Thurston's paper on kneading theory which is concerned with interval maps with finitely many monotonic pieces. In sections 8-10, it relates the periodic points to the so called kneading invariant of $f$ which captures where in the interval the turning points (local extrema) go under iteration. Corollary 9.6 therein is the inverse of the result mentioned above: If $s:={\rm{e}}^{h_{\rm{top}}(f)}>1$, then $f$ has a periodic point whose period is not a power of two. Finally, the idea of constructing a Markov partition and studying the associated shift works if you further assume that the turning points of $f$ are eventually periodic. In that case, the turning points (local extrema) and their forward orbits cut the interval into finitely many subintervals. Each of the subintervals is mapped onto other via $f$. You can write down the transition matrix and compute the entropy as the logarithm of the leading eigenvalue of the matrix. For instance, if you have a unimodal map (a map with just one turning point), if the turning point is of period three, you can compute the entropy as $\log\left(\frac{1+\sqrt{5}}{2}\right)$.<|endoftext|> TITLE: Which finite dimensional Banach spaces can be represented isometrically as spaces of bounded operators on a finite dimensional Hilbert space? QUESTION [12 upvotes]: Background: It is known that every Banach space $X$ can be embedded isometrically as a subspace in the space $C(K)$ of continuous functions on a compact Hausdorff space $K$. Indeed, one can take $K$ equal to the closed unit ball in the dual $X^*$, and embed $X$ in $C((X^*)_1)$ using the duality mapping $x \mapsto \hat{x}\big|_{(X^*)_1} \in C((X^*)_1)$, where $\hat{x} \in X^{**}$ is given by $\hat{x}(f) = f(x)$, for $f \in X^*$. By the Gelfand-Naimark theorem we know that $C(K)$ can be faithfully represented as a sub-C*-algebra of the algebra $B(H)$ of bounded linear operators on some Hilbert space $H$, and so we conclude: Every Banach space can be represented isometrically as a closed subspace of $B(H)$ for some Hilbert space $H$. Question: Which (finite dimensional) Banach spaces can be represented isometrically as closed subspaces of $B(H)$ for a finite dimensional Hilbert space $H$? A related questions is: do you know any work related to this question? I usually have spaces of the complex numbers in mind, but I'll also be happy to hear about real spaces. Why did I ask myself this question? Nothing fancy. I was giving a lecture about operator systems to undergraduates and I started from operator spaces, but since they didn't all take a course in functional analysis I worked mostly in $M_n$. I made a point that every banach space is an operator space in $B(H)$ for some $H$, and showed them how $\ell^2_2$ and $\ell^\infty_2$ (the space $\mathbb{C}^2$ with the $\ell^2$ and $\ell^\infty$ norms, respectively) can be isometrically embedded in $M_2$. My older notes said that $\ell^1_2$ cannot be embedded in $M_n$ for finite $n$ (I forgot the complete details of the proof but I'll write the idea below). Speaking about this I got curious about which finite dimensional spaces can be embedded in $M_n$ for finite $n$. The simplest question is: Can $\ell^1_2$ be embedded in $M_n$? For this simple question I think that the answer is: no. The reason I think this is that $\ell^1_2$ has a unique operator space structure. We have an isometric representation as the operator subspace of the C-algebra $C(S^1)$ generated by the unitaries $1$ and $z$. Since unitaries are hyperrigid, any unital isometric map from $span\{1,z\}$ onto a subspace of $M_n$ would extend to $*$-isomorphism between the generated C-algebras, and this is impossible. Thus, $\ell^1_2$ is not a unital operator space. I haven't worked much in the nonunital setting, I think that in this particular case the technical difference between unital and nonunital won't change the end result. But then I got stuck (Gupta and Reza's linked paper below gives an elementary proof that $\ell^1_1$ cannot be embedded in $M_n$, and one of the steps is a nice trick that shows how indeed you can reduce to the case of unital operator space). Update: Bunyamin Sari, in a comment to Bill Johnson's answer, put a reference to a paper by Samya Kumar Ray showing that $\ell^p_n$ cannot be embedded isometrically as compact operators. That papers contains a reference to a paper by Gupta and Reza that provides an elementary proof that $\ell^1_2$ cannot be embedded in $M_n$. It follows, of course, that no $\ell^1_m$ can be embedded in $M_n$. Terry Tao commented below: "In order to embed for a finite dimensional Banach space to embed isometrically into some $M_n$ it is necessary that the unit ball be a semi-algebraic set." To see this, note that the unit ball of an operator space is the intersection of a linear subspace with the unit ball of $M_n$, which can be given as the set of matrices satisfying the matrix inequality $A^*A \leq I$. However, having a semi-algebraic unit ball is not a sufficient condition for embeddability as an operator subspace of $M_n$; for example consider $\ell^p_2$ for $p=4$ and see the paper by Ray above. (Thanks to Guy Salomon and Eli Shamovich for some offline discussions). REPLY [6 votes]: In Ray - On Isometric Embedding $\ell_p^m \to S_\infty^n$ and Unique operator space structure it is shown that for $p\in (2,\infty)\cup \{1\}$ and $n\ge 2$, $\ell_p^n$ does not isometrically embed into the space of compact operators on $\ell_2$. (This was previously in the comments, now deleted for housekeeping.)<|endoftext|> TITLE: Relaxation of notion of positive definite function QUESTION [9 upvotes]: A function $f:\mathbb{R}\to\mathbb{R}$ is called positive definite (in the semigroup sense) if for all $n\geq 1$ and $x_1,\ldots,x_n\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^n$ is positive definite. Such functions have a nice characterization by (a version of) Bochner's theorem. Now I am interested in the following relaxation of this notion: Fix $k\in\mathbb{N}$. We say that a function $f:\mathbb{R}\to\mathbb{R}$ is $k$-positive if for all $x_1,\ldots,x_k\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^k$ is positive definite. For instance $1$-positive means that $f(x)>0$ for all $x\in\mathbb{R}$. Being $2$-positive means additionally that $\log(f)$ is strictly midpoint convex. In general, it is clear that $k$-positive definite implies $(k-1)$-positive definite, and the above example shows that the converse is not always true. My question is the following: Is it true that for every $k\in\mathbb{N}$ there is a $k$-positive function which is not $(k+1)$-positive? REPLY [3 votes]: $\newcommand\R{\mathbb R}$For real $c$ and $x$, let $$f_c(x):=f(x)-c,$$ where $f(x):=e^{x^2/2}$. Note that $f$ is the moment generating function of the standard normal distribution and thus a mixture of exponential functions. Since the exponential functions are positive semidefinite (in the semigroup sense), $f$ is also positive semidefinite. With some further effort, one should be able to show that $f$ is positive definite. By Theorem 2.5 on p. 55, Theorem 5.3 on p. 65, and Theorem 8.1 on p. 78 of Karlin - Total positivity, vol. I, for $f_c$ to be $r$-positive semidefinite it is sufficient that the Hankel determinant $$d_{k,c}(x):=d_{f;k,c}(x):=\det((f_c^{(i+j)}(x))_{0\le i,j\le k-1})$$ be $>0$ for all $k\in[r]:=\{1,\dots,r\}$ and necessary that this determinant be $\ge0$ for all $k\in[r]$. Note that $$d_{k,c}(x)=d_{k,0}(x)-c\tilde d_k(x),$$ where $$\tilde d_k(x):=d_{f'';k-1,0}(x)=\det((f^{(i+j)}(x))_{1\le i,j\le k-1});$$ note also that, similarly to $d_{k,0}(x)\ge0$ for real $x$, we have $\tilde d_k(x)\ge0$ for real $x$ (and, likely, $\tilde d_k(x)>0$ for real $x$). Let $$c_k:=\sup\{c\colon d_{k,c}(x)\ge0\ \forall x\in\R\} =\inf_{x\in\R}\frac{d_{k,0}(x)}{\tilde d_k(x)}. $$ We have $$c_1=1>c_2=\frac{\sqrt e}2=0.82\ldots>c_3=\frac23=0.66\ldots.$$ Moreover, it appears that $c_k>c_{k+1}$ for all natural $k$. It will then follow that $f_c$ is $r$-positive semidefinite but not $(r+1)$-positive semidefinite if $c_{r+1}c_{k+1}$ for all natural $k$ and the above remark are illustrated by the following image of a Mathematica notebook (click on the image to magnify it):<|endoftext|> TITLE: When is a compact orbifold Riemann surface a global quotient of a Riemann surface QUESTION [6 upvotes]: While reading the paper Seifert Fibred Homology 3-Spheres and the Yang-Mills Equations on Riemann Surfaces with Marked points by M. Furuta and B. Steer, I stumbled upon the following statement: Any compact orbifold Riemann surface, with $n\geq 3$ singular points or $n=2$ and $\alpha_1=\alpha_2$ if the genus $g$ is zero, has the form $N/D$, where $N$ is a compact Riemann surface and $D$ is a finite group of diffeomorphisms. However, there is no comment why this should be true. The paper cites the article of P. Scott on the geometrization conjecture, but on that paper the only thing that is assured is that if this conditions are satisfied, then the orbifold is a global quotient, by some unspecified manifold. However, How can we deduce that this manifold must be a Riemann surface? This also leads me to ask, If $N$ is a compact Riemann surface and $D$ is a finite group of diffeomorphisms, is the quotient $N/D$ a compact Riemann orbifold surface? Thanks in advance for your answers. REPLY [8 votes]: Let $M$ be a your compact orbifold Riemann surface and let $S\subset M$ be the finite set of orbifold singularities. Theorem: The following are equivalent: $M$ is the quotient of a closed Riemann surface by a finite group $M$ admits a conformal metric of constant curvature (with the correct cone angles at the singularities) Proof: If $M = N/D$, $N$ admits a metric of constant curvature by Poincaré--Koebe'uniformization, which is essentially unique and thus invariant by $D$ (one has to be a bit careful in the positive curvature case). Conversely, a metric of constant curvature on $M$ gives rise to a holonomy representation $\rho$ of the fundamental group $\pi_1(M\backslash S)$ into $\mathrm{Isom}(X)$, where $X$ is the sphere/the euclidean plane/the hyperbolic plane depending on the sign of the curvature. The image of $\rho$ is a finitely generated linear group which admits a torsion-free subgroup of finite index $\Gamma$ (by Selberg's lemma). The preimage of $\Gamma$ is a normal finite index subgroup of $\pi_1(M\backslash S)$ which thus defines a finite ramified normal cover $N$ of $M$. One verifies that the ramifications have the correct degree at the singularities, so that $N$ is a smooth Riemann surface covering $M$. QED Now, by a theorem of Troyanov, every closed orbifold of non-positive Euler characteristic admits a conformal metric of constant non-positive curvature. This shows that every orbifold is covered by a smooth Riemann surface except for the few ones that have positive Euler characteristic, namely, spheres with 1 or 2 singularities, or 3 singularities of angles $\pi, 2\pi/3, 2\pi/3$ or $\pi, \pi, 2\pi/3$. These cases are not too hard to treat. For instance, the sphere with 3 singularities of angle $\pi, 2\pi/3, 2\pi/3$ is the double of a spherical triangle with angles $\pi/2, \pi/3, \pi/3$.<|endoftext|> TITLE: Improving Cauchy estimates? QUESTION [7 upvotes]: Consider an entire function $f:\mathbb C \to \mathbb C$ that is real on the real line and even. This function has a Taylor series of the form $$f(z) = \sum_{i=0}^{\infty} a_i z^{2i} \text{ with } a_i \in \mathbb R.$$ I have an estimate of the form $$\vert e^{z^2 \mu} f(z) \vert \le e^{\vert z \vert^4 \nu}$$ for all $z \in \mathbb C$ and $\mu, \nu >0.$ I could now bound the Taylor coefficients of this power series using Cauchy estimates, see for instance here, by using that $$\vert f(z) \vert \le e^{\vert z \vert^4\nu + \vert z \vert^2 \mu}.$$ However, I feel that I am wasting something here, as the initial estimate already implies that on the real line $f$ grows at most like $e^{z^4\nu}$ when multiplied by a Gaussian. So the bound I derived for my Cauchy estimate is very conservative on the real line. Of course I cannot immediately improve upon this estimate since when $z$ is purely imaginary, that bound may be optimal and Cauchy estimates are supposed to hold on a circle in the complex domain. However, since the function is real on the real line, the growth should really be determined by my initial estimate somehow. Therefore my question is: Can I do anything better here to bound the Taylor coefficients than applying Cauchy estimates to the estimate $$\vert f(z) \vert \le e^{\vert z \vert^4\nu + \vert z \vert^2 \mu}.$$ REPLY [8 votes]: Let $z=re^{i\theta}$. Then your original inequality implies $$|f(re^{i\theta})|\leq \exp(\nu r^4-\mu r^2\cos2\theta).$$ Then instead of Cauchy estimate you can use the exact formula for the coefficient: $$|a_{2n}|=\left|\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z^{2n+1}}dz\right|\leq\frac{1}{2\pi} \int_0^{2\pi}|f(re^{i\theta})|r^{-2n}d\theta$$ $$\leq r^{-2n}\exp(\nu r^4)\frac{1}{2\pi}\int_0^{2\pi}e^{-\mu r^2\cos2\theta}d\theta,$$ and now take the minimum, with respect to $r$. (The last integral can be expressed in terms of the modified Bessel function $I_0$, if needed.) However, this can give you an advantage only for small $n$, since for large $n$, the term with $r^2$ in the first estimate is negligible.<|endoftext|> TITLE: Does every infinite-dimensional Banach algebra contain an infinite-dimensional subalgebra with second-countable primitive ideal space? QUESTION [7 upvotes]: Let $A$ be an infinite dimensional Banach algebra. Even if separable the primitive ideal space of $A$ need not be second-countable when endowed with the hull-kernel topology. Can we at least find an infinite-dimensional subalgebra with this property? For C*-algebras, separability implies second-countable primitive ideal space. REPLY [3 votes]: The answer is no. Let $A=A(D)$ be the disk algebra. Every character on $A$ is a point evaluation at some $x\in D_1$, the closed unit disk of $\mathbb{C}$, i.e. the spectrum $\sigma(A)$ of $A$ is $D_1$. Let $D_r = \{z\in\mathbb{C}: |z|\leq r\}$ for $r<1$. Then, the restriction of the hull-kernel topology on $\sigma(A)$ to $D_r$ is cofinite topology. In fact, if $S\subseteq D_r$ is not finite, then $k(S) = \{0\}$, so $h(k(S)) = \sigma(A)$. Consequently, the hull-kernel topology on $\sigma(A)$ is not even first countable. Second, let $B\subseteq A$ be an infinite-dimensional closed subalgebra of $A$. Pick an $r<1$ and define an equivalence relation on $D_r$ via $$x\sim y \Leftrightarrow f(x) = f(y)\hspace{8mm} \forall f\in B.$$ By the basic properties of analytic functions, it is not difficult to show that the equivalence classes of $\sim$ are at most finite. Thus, $D_r/\sim$ is an uncountable set. Moreover, $D_r/\sim \subseteq\sigma(B)$. Proceeding the same as above, the restriction of the hull-kernel topology on $\sigma(B)$ to $D_r/\sim$ is the cofinite topology. We conclude that the hull-kernel topology on $\sigma(B)$ is not even first countable. On the positive side, if a commutative Banach algebra $A$ contains a separable regular (some sources use "completely regular" rather than "regular") subalgebra $B$, then the hull-kernel topology on $\sigma(B)$ is second countable. The reason for this is the fact that the Gelfand topology and the hull-kernel topology coincide on a spectrum of a regular Banach algebra, e.g. Theorem 3.2.10 in [Palmer, "Banach Algebras and the General Theory of *-Algebras I"]. Clearly, the Gelfand topology of a separable Banach algebra is second countable. Finally, I think the following related note might be useful although it is not asked in the question. Let $F=\{x_n:n\in\mathbb{N}\}$ be a countable subset of $D$ that has an accumulation point in $D$. Let $V=\{(f(x_n))_{n\in\mathbb{N}} : f\in A(D)\}$ with the norm adopted from $A(D)$. Then, the map $f\to(f(x_n))_{n\in\mathbb{N}}$ maps $A(D)$ isomorphically onto $V$, where $\sigma(V)$ is countable. Thus, the hull-kernel topology on $\sigma(V)$ is second countable. Paraphrasing, there exists countable subsets $F\subseteq\sigma{(A(D))}$, which are dense in $\sigma(A(D))$ in the hull-kernel topology of $\sigma(A(D))$, and for which the restriction of the hull-kernel topology to $F$ is second countable.<|endoftext|> TITLE: Strict graded commutativity of $\pi_*(\operatorname{THH}(A))$? QUESTION [5 upvotes]: $\DeclareMathOperator\THH{THH}\DeclareMathOperator\HH{HH}$A version of the strict graded commutativity (i.e. graded commutativity & $x^2=0$ for every homogeneous element $x$ of odd degree) of $\pi_*(\THH(A))$ seems to be used in Construction 6.8 of Bhatt–Morrow–Scholze, Topological Hochschild homology and integral $p$-adic Hodge theory to establish a form of HKR theorem due to Hesselholt (as indicated). Let me briefly summarize what is happening. Let $R$ be a perfectoid $\mathbb Z_p$-algebra (for example, a perfect field of characteristic $p$), and $A$ an $R$-algebra. The goal is to produce a map $(\Omega_{A/R}^*)_p^\wedge\to\pi_*(\THH(A;\mathbb Z_p))$ of graded $p$-complete $A$-module. The key step is to produce a map $\Omega_{A/\mathbb Z}^*\to\pi_*(\THH(A))$ of graded commutative $A$-algebras. The map $\THH(A)\to\HH(A):=\HH(A/\mathbb Z)$ becomes an equivalence after truncation $\tau_{\le2}$, and in particular, $\pi_1(\THH(A))\cong\pi_1(\HH(A))\cong\Omega_{A/\mathbb Z}^1$. Now they claim that, since $\pi_*(\HH(A))$ is strictly graded commutative as $\HH(A)$ is an animated $A$-algebra, the map $\Omega_{A/\mathbb Z}^1\to\pi_1(\THH(A))$ extends to a map $\Omega_{A/\mathbb Z}^*\to\pi_*(\THH(A))$ by the universal property of the exterior product. If I understand correctly, in order to construct such a map, one needs the strict commutativity of $\THH(A)$, not that of $\HH(A)$. However, $\THH(A)$ is no longer the underlying $\mathbb E_\infty$-ring of an animated ring in general. Update to clarify: no, my understanding was incorrect. The strict commutativity of $\THH$ is not needed to produce the map. Only that of $\HH$ is needed. See Tyler Lawson's answer. In retrospect, the text is clear and it is I who am stupid. In fact, by Bökstedt's periodicity, the graded ring $\pi_*(\THH(\mathbb F_p))$ is isomorphic to $\mathbb F_p[u]$ where $u$ is of degree $2$, but for animated rings, every element of $\pi_2$ has divided powers. In particular, if $\THH(\mathbb F_p)$ is the underlying $\mathbb E_\infty$-ring of an animated ring, this implies that $u^p=p!v$ for some $v\in\pi_{2p}(\THH(\mathbb F_p))$, which implies that $u^p=0$, contradiction (this argument also shows that $\THH(R;\mathbb Z_p)$ is not an animated ring, therefore neither is $\THH(R)$). Here are my questions: In this setting, is it true that $\pi_*(\THH(A))$ is strictly graded commutative? More generally, let $A$ be an animated ring. Is it true that $\pi_*(\THH(A))$ is strictly graded commutative? Of course, these questions are trivial except when $p=2$. REPLY [3 votes]: To my knowledge, there is no such result for THH of a commutative ring. (Could be?) However, we need less for this result; we only need degree 1 elements to square to zero. Every element in $\pi_1 THH(A)$ is a finite linear combination of elements $a \cdot \sigma(b)$ for $a, b$ in $A$, where $\sigma$ is the circle action. Therefore it suffices to show all elements in degree 1 square to zero in the case where $A$ is a finite polynomial algebra over $\Bbb Z$, or more generally a monoid algebra. If $M$ is a commutative monoid, then there is an equivalence of ring spectra $$ THH(\Bbb Z[M]) \simeq THH(\Bbb Z) \otimes Z(M)_+ $$ where $Z(M)$ is the cyclic bar construction, a simplicial commutative monoid. This agrees through degree 2 with $$ \Bbb Z \otimes Z(M)_+ \simeq HH(\Bbb Z[M]) $$ because $THH(\Bbb Z) \to \Bbb Z$ is a split map of ring spectra and an equivalence through degree 2. Therefore, it suffices to show that all elements in $HH_1$ square to zero, and in this case it is true due to coming from a simplicial commutative ring.<|endoftext|> TITLE: Chromatic number and vertex connectivity QUESTION [7 upvotes]: A conjecture of Mader implies that for any positive integer $n\geq2$, every graph with average degree at least $3n-4$ contains an $n$-connected subgraph. Mader himself proved this for $n=2,3,4,5,6,7$. In particular, any graph with minimum degree $5$ admits a $3$-connected subgraph, and so any graph with chromatic number at least $6$ admits a $3$-connected subgraph. Question: Does every graph with chromatic number $5$ already admit a $3$-connected subgraph? REPLY [4 votes]: Here are some observations that are slightly too long for a comment. First, note that the claim is false if we replace $5$ by $4$. Claim. There exists a graph $G$ with $\chi(G)=4$ such that $G$ does not contain a $3$-connected subgraph. Proof. The Moser Spindle is a well-known graph with chromatic number $4$ (since it is obtained via Hajó's construction applied to two copies of $K_4$). Moreover, it is easy to see that no subgraph of the Moser Spindle is $3$-connected. $\square$ On the other hand, every graph with chromatic number $4$ contains a subgraph which is 'almost' $3$-connected. Claim. Every graph $G$ with $\chi(G)=4$ contains a subgraph $A$ and $x,y \in V(A)$ such that $A+xy$ is $3$-connected. Proof. Let $G'$ be a subgraph of $G$ such that $\chi(G')=4$ and $G'$ is minimal (under the subgraph relation) with this property. If $G'$ is $3$-connected, we are done. Otherwise, let $(A,B)$ be a $(\leq 2)$-separation of $G'$ where $A$ corresponds to a leaf vertex in the SPQR tree of $G'$. By the minimality of $G'$, $\chi(A), \chi(B) \leq 3$. Thus, $(A,B)$ is a $2$-separation and the vertices $x,y \in V(A) \cap V(B)$ must be non-adjacent in $G'$ (otherwise $G'$ is $3$-colourable). Moreover, we must have $\chi(A)=\chi(B)=3$ (otherwise $G'$ is $3$-colourable). By the definition of the SPQR tree, $A+xy$ is either a cycle or is $3$-connected. However, $A+xy$ cannot be a cycle since $\chi(A)=3$. Thus, $A+xy$ is $3$-connected, as required. $\square$ Hopefully, for graphs $G$ with $\chi(G)=5$, it is not necessary to add the edge $xy$ to obtain a $3$-connected subgraph.<|endoftext|> TITLE: Rings s.t. each element has a power lying in the center (and their completely prime ideals) QUESTION [5 upvotes]: Let $R$ be a ring (throughout, all rings are associative and unital). We say $R$ satisfies condition (C) if, for every $a \in R$, there exists an integer $n \ge 1$ (depending on $a$) such that $a^n$ lies in the center $\mathcal Z(R)$ of $R$; and condition (C') if there exists an integer $n \ge 1$ such that $a^n \in \mathcal Z(R)$ for every $a \in R$ (so, the difference with condition (C) is that now $n$ does not depend on $a$). The following questions are motivated by those in a related thread here: Questions. (i) Is there any special/standard name for a ring satisfying either (C) or (C'), or any keyword I can use to read about them in the literature? (ii) Is it true that, if $R$ satisfies condition (C), then every non-unit of $R$ is contained in a completely prime${}^{(1)}$ (two-sided) ideal? (iii) If not, what about the case where $R$ satisfies condition (C')? Of course, (C') implies (C). Note also that (C') is equivalent to the existence of an integer $n \ge 1$ with the property that, for each $a \in R$, there is an integer $k$ between $1$ and $n$ (with $k$ depending on $a$) such that $a^k \in \mathcal Z(R)$. Edit 1. It's perhaps worth remarking that a ring satisfying (C) is Dedekind-finite${}^{(2)}$, which is a necessary condition for every non-unit of $R$ to lie in a proper (and, in particular, in a completely prime) ideal. In fact, assume $ab = 1_R$ for some $a, b \in R$; we need to check that $bu = 1_R$ for some $u \in R$. To begin, we have that $(ba)^k = b(ab)^{k-1} a = ba$ for every $k \in \mathbf N^+$. On the other hand, assuming $R$ satisfies (C) implies that $(ba)^n \in \mathcal Z(R)$ for a certain $n \in \mathbf N^+$. It follows that $$ ybax = y(ba)^n x = (ba)^n yx = bayx, \qquad \text{for all }x, y \in R, $$ which ultimately shows (by taking $x = b$ and $y = a$) that $ba^2b = (ab)^2 = 1_R$. [] Edit 2. In a comment to the OP, Benjamin Steinberg asked for non-commutative examples of rings satisfying, say, condition (C'). Luckily enough, the first idea that comes to mind seems to work just fine. Let $\mathscr F(X)$ be the free monoid on a non-empty set $X$, and let $F\langle X \rangle$ be the monoid ring of $\mathscr F(X)$ over the two-element field $F$ (i.e., $F\langle X \rangle$ is the free $F$-algebra on the basis $X$) and $\mathfrak S(X)$ be the group of permutations of $X$. We assume $X \subseteq \mathscr F(X)$ and use $\varepsilon$ for the identity of $\mathscr F(X)$ (namely, the empty $X$-word). Moreover, given $\mathfrak u \in \mathscr F(X)$, we denote by $\delta_{\mathfrak u}$ the "Kronecker delta" function $H \to F$ that maps $\mathfrak u$ to $1_F$ and every $X$-word $\mathfrak v \ne \mathfrak u$ to $0_F$. The quotient ring $R := F\langle X \rangle/\mathfrak i$ of $F\langle X \rangle$ by the (two-sided) ideal $\mathfrak i$ generated by the set $$ \bigcup_{\sigma \in \mathfrak S(X)} \{\delta_x \delta_y \delta_z - \delta_{\sigma(x)} \delta_{\sigma(y)} \delta_{\sigma(z)}: x, y, z \in X\} $$ is commutative if and only if $|X| = 1$; and we are going to show that $R$ satisfies condition (C') with $n = 4$. A couple of remarks are in order before proceeding: If $k$ is an integer $\ge 3$ and $\mathfrak z_1, \ldots, \mathfrak z_k$ are non-empty $X$-words, then it is straighforward from the definition of the ideal $\mathfrak i$ and the factoriality${}^{(3)}$ of $\mathscr F(X)$ that $\delta_{\mathfrak z_1} \cdots \delta_{\mathfrak z_k} \equiv \delta_{\mathfrak z_{\sigma(1)}} \cdots \delta_{\mathfrak z_{\sigma(k)}} \bmod \mathfrak i$ for every permutation $\sigma$ of the discrete interval $[\![1, k ]\!]$. By the previous remark, $\delta_\mathfrak{z} \bmod \mathfrak i \in \mathcal Z(R)$ for every $X$-word $\mathfrak z$ whose length is $\ge 3$. Now, pick $f \in F\langle X \rangle$; we need to check that $f^4 \bmod \mathfrak i \in \mathcal Z(R)$. For, note that, since $F$ has characteristic $2$ and $\delta_\varepsilon$ lies in the center of $F \langle X \rangle$, we have $(f \pm \delta_\varepsilon)^2 = f^2 + \delta_\varepsilon$ and hence $(f-\delta_\varepsilon)^4 = f^4 + \delta_\varepsilon$. It follows that $f^4 \bmod \mathfrak i \in \mathcal Z(R)$ if and only if $(f-\delta_\varepsilon)^4 \bmod \mathfrak i \in \mathcal Z(R)$, and we may therefore assume without loss of generality (as we do) that $f(\varepsilon) = 0_R$. In consequence, the support $\text{s}(f) := \mathscr F(X) \setminus f^{-1}(0_R)$ of $f$ is, by the very definition of $F\langle X \rangle$, a finite subset of $\mathscr F(X) \setminus \{\varepsilon\}$ with $f = \sum_{\mathfrak z \in \text{s}(f)} \delta_\mathfrak{z}$. Then $$ f^4 = \sum_{(\mathfrak z_1, \mathfrak z_2, \mathfrak z_3, \mathfrak z_4) \in \text{s}(f)^{\times 4}} \delta_{\mathfrak z_1} \delta_{\mathfrak z_2} \delta_{\mathfrak z_3} \delta_{\mathfrak z_4}; $$ and since $\varepsilon \notin \text{s}(f)$, we see that every $X$-word in the support of $f^4$ has length $\ge 4$. We thus conclude from the second remark above that $f^4 \bmod \mathfrak i \in \mathcal Z(R)$. [] Edit 3. A domain satisfying condition (C) is necessarily commutative. As noted by Benjamin Steinberg in a comment to the OP, this follows from the main theorem of I.N. Herstein, A Commutativity Theorem, J. Algebra 38 (1976), No. 1, 112-118. The result states that, if $R$ is a ring with the property that, for all $a, b \in R$, there exist positive integers $m$ and $n$ (depending on $a$ and $b$) such that $a^m b^n = b^n a^m$, then the commutator ideal of $R$ (that is, the two-sided ideal generated by the elements of the form $xy-yx$ with $x, y \in R$) is nil. But a domain has no non-zero nil ideals; and on the other hand, it is obvious that every ring satisfying condition (C) also satisfies the hypothesis of Herstein's theorem. Therefore, a domain satisfies condition (C) if and only if it is commutative. Notes. (1) An ideal $\mathfrak p$ of $R$ is completely prime if it is proper (in the sense that $\mathfrak p \subsetneq R$) and $ab \in \mathfrak p$ for some $a, b \in R$ implies $a \in \mathfrak p$ or $b \in \mathfrak p$; and is prime if it is proper and $aRb \subseteq \mathfrak p$ for some $a, b \in R$ implies $a \in \mathfrak p$ or $b \in \mathfrak p$ (cf. the article on prime ideals on Wiki.en). (2) $R$ is Dedekind-finite if every left- or right-invertible element is a unit (equivalently, if $ab = 1_R$ for some $a, b \in R$, then $ba = 1_R$). (3) I mean the fact that every non-empty $X$-word factors uniquely, in $\mathscr F(X)$, as a product of elements of the basis $X$. REPLY [3 votes]: Assume (C), so that for each $a\in R$, there exists some integer $n\geq 1$ (possibly depending on $a$) such that $a^n\in Z(R)$. The equivalence of conditions (1) and (3) in Theorem 12.11 from Lam's “A First Course in Noncommutative Rings” tells us that $R/J(R)$ is commutative; here, $J(R)$ is the Jacobson radical. [This result is attributed to Herstein and Kaplansky.] Thus, any maximal one-sided ideal of $R$ (which, of course contains $J(R)$) is two-sided, and is completely prime. This answers your second (and hence third) question. Regarding your first question, I'm not familiar with any terminology, probably because if $J(R)=0$, then the condition is equivalent to being commutative. I would recommend a literature search for those papers that cite Herstein's and Kaplansky's works, to see if the condition is studied more fully.<|endoftext|> TITLE: Bijective proof for a partition identity QUESTION [8 upvotes]: I came across the following cute fact about partitions: \begin{align} & |\{\lambda \vdash n \text{ with an even number of even parts}\}| \\[8pt] & {} - |\{ \lambda \vdash n \text{ with an odd number of even parts}\}| \\[8pt] = {} & |\{ \lambda \vdash n \text{ which are self-conjugate } (\text{i.e. } \lambda = \lambda^\top )\}| \end{align} I have a simple proof via the representation theory of $S_n$ -- or really the result just fell out of a calculation I was doing. I was wondering if anyone knew a purely combinatorial bijective proof or had a reference for one. REPLY [20 votes]: The answer is yes, there is a combinatorial proof, and both Sam's and Fёdor's proofs work. However, this is a really old result and a combinatorial proof is old. Here is the reference: H. Gupta, Combinatorial proof of a theorem on partitions into an even or odd number of parts, J. Combin. Theory Ser. A, 21 (1976), no. 1, 100–103. In general, you can find all of this in my survey. This particular problem is §3.2.4. This result goes back to Glaisher (1876), and the reference to Gupta is [79] given in the remarks at the end. You would need to also use an obvious bijection in §2.1.5 which is the first step in Sam's answer. According to Sylvester this is due to Durfee.<|endoftext|> TITLE: Is Heyting arithmetic sufficient to prove its own (formalized) existence property? QUESTION [9 upvotes]: Let $\mathsf{HA}$ denote first-order Heyting arithmetic (viꝫ., Peano axioms with unrestricted recursion scheme, in first-order intuitionistic logic). It is known (e.g., Troelstra & van Dalen, Constructivism in Mathematics (1988), 3.5.6; Beeson, Foundations of Constructive Mathematics (1985), VII.5.3 but concerning a different system; or Troelstra (ed), Metamathematical Investigations of Intuitionistic Arithmetic and Analysis (1973), 3.1.5) that $\mathsf{HA}$ has the number existence property: (NEP) If $\mathsf{HA} \vdash \exists n. P(n)$ for some closed formula $\exists n. P(n)$, then in fact there is a natural number $n$ such that $\mathsf{HA} \vdash P(\overline{n})$ (with $\overline{n}$ the obvious term that denotes $n$). Now this number existence property be formalized as a statement of arithmetic $\forall “P”.((\mathsf{HA} \vdash \exists n. P(n)) \Rightarrow \exists n. (\mathsf{HA} \vdash P(\overline{n})))$, beginning with a universal quantifier ranging over Gödel codes for formulas $P$. So we can ask whether $\mathsf{HA}$ proves this formalized number existence property. The techniques used to prove NEP in the references above all seem impossible to formalize in $\mathsf{HA}$, because they depend on something (like cut-elimination or formalization of arithmetical truth) that is beyond its power. Yet I also don't see how NEP would imply the consistency of $\mathsf{HA}$. So here are my questions: Questions: Does $\mathsf{HA}$ prove the (formalized) number existence property, $\forall “P”.((\mathsf{HA} \vdash \exists n. P(n)) \Rightarrow \exists n. (\mathsf{HA} \vdash P(\overline{n})))$, for $\mathsf{HA}$? If not, does it at least prove $((\mathsf{HA} \vdash \exists n. P(n)) \Rightarrow \exists n. (\mathsf{HA} \vdash P(\overline{n})))$ for each $P$? If a negative answer to (1), does $\forall “P”.((\mathsf{HA} \vdash \exists n. P(n)) \Rightarrow \exists n. (\mathsf{HA} \vdash P(\overline{n})))$ imply the consistency of $\mathsf{HA}$? If a negative answer to (2), does $((\mathsf{HA} \vdash \exists n. P(n)) \Rightarrow \exists n. (\mathsf{HA} \vdash P(\overline{n})))$ imply the consistency of $\mathsf{HA}$ for some judiciously chosen $P$? REPLY [10 votes]: The answers to all the questions are negative. First, none of the statements implies the consistency of HA, since if HA is inconsistent, all the statements trivially hold. On the other hand, considering the special case where $P$ is $\Delta_0$, the schema in 2 implies (classically, to make things simpler) “HA is inconsistent or $\Sigma_1$-sound”. More precisely, if HA proved 2, then PA + Con(PA) would prove the local $\Sigma_1$-reflection principle for HA, which implies the same principle for PA by (HA-verifiable) $\Sigma_1$-conservativity. Thus, PA + Con(PA) would prove its own consistency, contradicting Gödel’s theorem.<|endoftext|> TITLE: How should I think about 1-motives? QUESTION [11 upvotes]: By definition, a 1-motive over an algebraically closed field $k$ is the data $$ M = [X\stackrel{u}{\to}G] $$ where $X$ is a free abelian group of finite type, $G$ is a semi-abelian variety over $k$, and $u:X \to G(k)$ is a group morphism. As far as I understand, this was one of the earlier attempts of coming up with a workable notion of a mixed motive. As evidence, Deligne shows that 1-motives over $\mathbb{C}$ are equivalent to mixed Hodge structures (without torsion) of type $\{(0,0),(0,-1),(-1,0),(-1,-1)\}$. The problem is that I don't really understand the intuition behind this definition. I assume that Deligne had concrete examples in mind when he came up with this formulation. Pehaps some extensions of $H^1(A)$, where $A$ is an abelian variety? How should I interpret the data in the definition of a 1-motive? My naïve view on motives is that they should be compatible systems of realizations with "geometric origin", and extensions thereof. How to reconcile these two points of view? Perhaps what I lack is simply a bag of toy examples to help me think about it. REPLY [8 votes]: I don't fully understand the hostility to thinking of motives in terms of compatible systems. It's true that a motive is not just a compatible system of realizations (or else we would call it a "compatible system" and not a motive) but it's also true that studying compatible systems has always been a good way to get intuition for many aspects of the theory of motives. Maybe a better thing to say is that a motive should be a geometric recipe which produces a realization in any existing or future cohomology theory. You asked how to reconcile this perspective on 1-motives with the definition of 1-motives. I guess this means how to derive a compatible system of realizations from a map $X \to G$. The realization of the 1-motive $X \to G$ in any cohomology theory should be the extension of $H^1(G)$ by $H^1(X^\vee)$. (More properly I guess this is the dual of the realization). You ask if this is an extension of $H^1(A)$. The group $G$ is always an extension of an abelian variety $A$ by a torus $T$, and $H^1(G)$ is an extension of $H^1(T)$ by $H^1(A)$. So we see that we are potentially extending $H^1(A)$ in both directions. Which extension? The idea is we should morally get the $H^1$ of the quotient space $G/X$. Since this space is not very well-behaved, we can simulate this, in an arbitrary cohomology theory, as follows: We observe that any class in $H^1(G)$ (ideally, completely represented by a torsor) is multiplicative in the sense that its pullback to $G \times G$ under the multiplication map is equal to the sum of its pullbacks under the two projections. For torsors, we can fix an isomorphism of torsors realizing this. An element of $H^1( X \to G)$ is a class in $H^1(G)$ together with a trivialization of its pullback to $H^1(X)$ which is multiplicative in the sense that the pullback of this trivialization to $X \times X$ along the multiplication is equal to the sum of its pullbacks under the two projections (using the above isomorphism to define this sum). You can follow this recipe in any theory and get a concrete description of the extension. Furthermore, in every reasonable cohomology theory you will get a clue to the idea that this construction produces all the extensions with "geometric origin". For example, in $\ell$-adic cohomology theories, you will see that extensions of $H^1(G)$ by $X^\vee$ correspond to a Selmer group, and extensions arising from this construction are the image of the map from the group of rational points to the Selmer group. So as long as the Tate-Shafarevich group is finite, there will not be room to fit any more extensions.<|endoftext|> TITLE: Compactness theorem and topos theory QUESTION [10 upvotes]: The theory of classifying topoi due to Makkai, Reyes, Hakim, and Grothendieck supplies a bijection between geometric theories (up to Morita equivalence) and Grothendieck topoi, by assigning to each geometric theory $\mathbb T$ the unique Grothendieck topos $\mathcal E$ such that $$\hom(\mathcal F, \mathcal E)\cong \text{models of $\mathbb T$ in $\mathcal F$}$$ natural in $\mathcal F$. In particular, the points of the Grothendieck topos associated to $\mathbb T$ are precisely the models of $\mathbb T$ in the category of sets. This correspondence identifies Deligne's theorem (SGA4, Exposé VI, Section 9), which gives a sufficient condition for a topos to have points, with Gödel's completeness theorem from mathematical logic. A great explanation of that phenomenon has been given by Torsten Ekedahl (MO/68335). Question: Another fundamental theorem from mathematical logic is the compactness theorem. Can the compactness theorem be phrased as a statement about Grothendieck topoi, similar to how Gödel's completeness theorem can be phrased as a statement about Grothendieck topoi? REPLY [9 votes]: The following is an analogy with the compactness theorem. "Compactness" in logic refers to the fact that the first-order theory you're working with is finitary. In topos theory, I believe the closest approximation to this idea is restricting to coherent theories (Johnstone in "Topos Theory" calls them finitary theories). On the topos side, this means restricting to coherent toposes. The localic coherent toposes are those toposes of the form $\mathbf{Sh}(X)$ for $X$ a spectral space. Take a coherent theory with $\mathbf{Sh}(X)$ as classifying topos. Adding axioms changes the classifying topos, more precisely the new classifying topos is a subtopos $\mathbf{Sh}(Y) \subseteq \mathbf{Sh}(X)$ corresponding to a sublocale $Y \subseteq X$. If you added only finitary/coherent axioms, then $\mathbf{Sh}(Y) \subseteq \mathbf{Sh}(X)$ is a coherent subtopos, and these correspond precisely to the subsets $Y \subseteq X$ that are closed sets for the "patch topology". So adding a sequence of coherent axioms corresponds to a chain of closed sets $$X \supseteq Y_1 \supseteq Y_2 \supseteq Y_3 \supseteq \dots$$ for the patch topology. The compactness theorem can then be seen as the statement: $$\forall i \in I,~ Y_i \neq \varnothing ~\Rightarrow~ \bigcap_{i \in \mathbb{N}} Y_i \neq \varnothing$$ This statement follows from the fact that $X$ is compact for the patch topology (by looking at the complements of these closed sets). So an analogue of the compactness theorem for toposes would be: if $\mathcal{E}$ is a coherent topos, and $\mathcal{E} \supseteq \mathcal{E}_1 \supseteq \mathcal{E}_2 \supseteq \mathcal{E}_3 \supseteq \dots$ is a sequence of non-empty coherent subtoposes, then the intersection is again non-empty coherent. I don't know a proof of this in the non-localic case.<|endoftext|> TITLE: Can we define Whitney stratification algebraically? QUESTION [8 upvotes]: For a subset $S$ of a smooth manifold $M$, a locally finite decomposition $$S = \bigsqcup_{\alpha} S_\alpha$$ into smooth submanifolds (strata) is called a Whitney stratification of $S$ if each pair $(S_\alpha, S_\beta)$ satisfies Whitney's condition (b). Whitney's condition (b) is usually defined using local coordinates and specifically the notion of convergence in the usual Euclidean topology. It turns out that if $M$ is a nonsingular complex algebraic variety and $S$ is a subvariety, then there is a canonical Whitney stratification of $S$ whose strata are nonsingular algebraic sets. My question is: in the complex algebraic category, can one define the notion of Whitney stratifications using only commutative algebra? REPLY [2 votes]: In the case that the variety is a hypersurface, it seems to me that an answer is provided by the Zariski equisingular stratification. If $X = (f) \subset \Bbb C^{n+1}$ is a hypersurface, one defines the dimensionality type of $X$ at a point $x \in X$ inductively by $\mathrm{dt.}(X, x) = -1$ if $x \notin X$, and $\mathrm{dt.}(X, x) = 1 + \mathrm{dt.}(\Delta_{\pi}, 0)$ where $\pi : (X, x) \to (\Bbb C^n, 0)$ is a generic finite map provided by Noether normalization, and $\Delta_\pi$ is the discriminant, consisting of the hypersurface in $\Bbb C^n$ of points over which the map $\pi$ is not etale. Hironaka proved that the function $\mathrm{dt.}$ is upper semicontinuous on $X$, so as a consequence the sets $\{x \in X : \mathrm{dt.}(x) \geq i\}$ are closed subvarieties of $X$. One defines a partition of $X$ by "strata" as the connected components of the fibers $X_i \setminus X_{i-1}$ of $\mathrm{dt.}$; Zariski showed that this is indeed a topological stratification, in the sense that boundary of a "stratum" is a collection "strata". A proof of $(b)$-regularity may be found in Speder "Equisingularite et conditions de Whitney".<|endoftext|> TITLE: Diagonalizing the ‘restricted’ Hilbert transform on $L^2(0,1)$, $f(z_1) \mapsto \mathrm{p.v.} \int_0^1 \frac{i}{z_1-z_2}f(z_2) dz_2$ QUESTION [6 upvotes]: Consider the following operator on functions $\mathcal{T}: L^2(0,1) \to L^2(0,1)$ over the complex numbers. \begin{equation} (\mathcal{T} f)(z_1) = \mathrm{p.v.} \int_0^1 \frac{i}{z_1-z_2}f(z_2) dz_2 \end{equation} where $\mathrm{p.v.}$ means the Cauchy principal value. This is an analogue of the Hilbert Transform except restricted to functions on an interval. Note that this operator is Hermitian, so there should be an orthogonal basis of eigenfunctions. Is there a known description of such an orthogonal basis of the eigenfunctions and eigenvalues of this operator and a relevant 'Fourier inversion' formula? REPLY [9 votes]: This was carried out in Koppelman, W.; Pincus, J. D., Spectral representations for finite Hilbert transformations, Math. Z. 71, 399-407 (1959). ZBL0085.31701. The spectrum is purely continuous on $[-\pi,\pi]$ (with the OP's normalisation) and Theorem 3.1 gives the explicit Fourier-Plancherel theorem that diagonalises this operator (see Theorem 3.2). Note that the Hilbert transform should not be expected to be compact (it is not an integral operator due to the presence of the principal value) and so the fact that there are no eigenfunctions should not be surprising (this can presumably also be shown by some complex analysis argument, though I haven't worked out the details). It is also observed that the same Fourier-type transform diagonalises the self-adjoint differential operator $i \sqrt{x(1-x)} \frac{d}{dx} \sqrt{x(1-x)}$ in addition to the finite Hilbert transform (in particular, this differential operator commutes with that transform, and in fact can be expressed using the functional calculus of that transform).<|endoftext|> TITLE: A limit problem QUESTION [6 upvotes]: Let $f$ be a bounded and continuous function, $0 TITLE: Existence of orbifold vertex algebras – current status? QUESTION [5 upvotes]: Let the finite group $G$ act on a vertex algebra $V$. It is expected that there are certain vector spaces $V_g$ (with the structure of $g$ twisted $V$ modules), with $V_1=V^G$, and $$V/G\ :=\ \bigoplus_{g\in G}V_g$$ has a vertex algebra structure. I think the idea is that, if $V$ is a chiral quantisation of the jet space $J_\infty X$ of a scheme $X$, then $V/G$ is the chiral quantisation of the quotient stack $X/G$, $J_\infty(X/G)$. Question: What is the current status of the construction of $V/G$? When is it expected to exist? E.g. in Frenkel-Ben Zvi 5.7.1, it is claimed that there is a construction when $V$ is holomorphic and $G$ is cyclic. But I am wondering if there have been advances since then. REPLY [3 votes]: The question of constructing $G$-orbifold vertex algebras amounts to the problem of producing a suitable multiplication operation on a sum of modules for the fixed-point vertex algebra $V^G$. The parts of the orbifold vertex algebra are not twisted modules for $V$, but instead are pieces of twisted modules (like fixed-point submodules for some group action). When $V$ is holomorphic, we have a classification for all finite groups $G$, given in Evans-Gannon, under the assumption that $V^G$ is regular (which is unconditional when $G$ is solvable, by C-Miyamoto). In the cyclic case, this was a conjecture when Frenkel-Ben-Zvi was written, but it was solved by van Ekeren-Möller-Scheithauer. When $V$ is not holomorphic, it is not clear what sort of object you expect to get. You are basically trying to make a commutative ring in a braided category of $V^G$-modules, but I don't know what constraints are natural.<|endoftext|> TITLE: How to formulate the univalence axiom without universes? QUESTION [11 upvotes]: The standard formulation of the univalence axiom for a universe type $U$ is that, for all $X : U$ and $Y : U$, the canonical map $(X =_U Y) \to (X \simeq Y)$ is an equivalence. As we (usually) cannot form the type of universe types, the usual univalence axiom is actually an axiom scheme, consisting of an instance of the univalence axiom for each universe type. We can generalise (the form of) the univalence axiom as follows. Given a type $E (b)$ depending on $b : B$, we might say that $E (b)$ is univalent over $b : B$ if the canonical map $(b_0 =_B b_1) \to (E (b_0) \simeq E (b_1))$ is an equivalence. Of course, not all dependent types are univalent, but the univalence axiom for $U$ is precisely the condition that $X$ is univalent over $X : U$. Question. Without mentioning universes (or otherwise internalising the condition of being a universe type), could we formulate an axiom scheme or inference rule that is equivalent to the usual univalence axiom scheme in the presence of universe types? One consequence of the usual univalence axiom scheme is that, for every type $E (b)$ depending on $b : B$, there is a type $E' (b')$ univalent over $b' : B'$ and a map $\chi : B \to B'$ such that $E (b) \equiv E' (\chi (b))$. Indeed, given a univalent universe $U$ such that, (for every $b : B$) $E (b) : U$, we may take $B' \equiv U$, $E' (b') \equiv b'$, and $\chi \equiv (\lambda b : B . E (b))$. Anyway, we can take the image factorisation of $\chi$ to replace $B'$ with something smaller, but I'm not sure if this can be done respecting the judgemental equality $E (b) \equiv E' (\chi (b))$. If it could, it seems to me we would have a higher inductive type characterisation of $B'$ and $E'$. Applying this to a possibly non-univalent universe type would yield a univalent universe type, so by replacing "universe" with "univalent universe" in the appropriate places we would be able to interpret univalent type theory. Does this work? REPLY [7 votes]: One possibility along these lines is large eliminations for higher inductive types. For instance, here is a large elimination rule for the higher inductive interval type $\mathsf{I}$ with $0,1:\mathsf{I}$ and $\mathsf{seg}:\mathsf{Id}_{\mathsf{I}}(0,1)$: $$ \frac{\vdash A \,\mathsf{type} \quad \vdash B \,\mathsf{type} \quad \vdash e : \mathsf{Equiv}(A,B)}{x:\mathsf{I}\vdash E(x)\,\mathsf{type} \quad E(0)\equiv A \quad E(1) \equiv B \quad \mathsf{trans}^E_{\mathsf{seg}}= e} $$ (Everything in some ambient context $\Gamma$, of course.) Note that this can be stated without any universe, using only the "is a type" judgment that we always have in dependent type theory. (I'm assuming function extensionality here, so that it doesn't matter what kind of equality or homotopy we have in the final law $\mathsf{trans}^E_{\mathsf{seg}}= e$.) Now suppose we have an interval type with this rule, and a universe type $\mathsf{U}$. Suppose moreover that the large elimination rule relativizes to $\mathsf{U}$, i.e. if $A$ and $B$ belong to $\mathsf{U}$ then so does $E$; this seems like a reasonable thing to include when "adding a universe type" to a theory that already had large eliminations. We will prove that $\mathsf{U}$ is univalent, using the encode-decode method. Given $A,B:\mathsf{U}$, let $\mathsf{encode} : \mathsf{Id}_{\mathsf{U}}(A,B) \to \mathsf{Equiv}(A,B)$ be defined by transport or path induction; this is the map we want to show to be an equivalence. To define $\mathsf{decode}$ in the other direction, suppose given $e:\mathsf{Equiv}(A,B)$. Then by large elimination we have $E:\mathsf{I} \to \mathsf{U}$ with $E(0)\equiv A$ and $E(1)\equiv B$ and $\mathsf{trans}^E_{\mathsf{seg}}= e$. Now we can define $\mathsf{decode}(e) \equiv \mathsf{ap}_{E}(\mathsf{seg}) : \mathsf{Id}_{\mathsf{U}}(A,B)$. Then $\mathsf{encode}(\mathsf{decode}(e)) = \mathsf{trans}^E_{\mathsf{seg}}= e$. On the other hand, by path induction it's easy to prove $\mathsf{decode}(\mathsf{encode}(p)) = p$ for any $p:\mathsf{Id}_{\mathsf{U}}(A,B)$. Thus, $\mathsf{decode}$ is an equivalence, as desired. To my knowledge, this was first observed in this post to the homotopy type theory mailing list in 2014, and I don't know of another reference. That discussion also considered "equivalence induction" (Corollary 5.8.5 in the HoTT Book) as a form of universe-free univalence, although as Peter noted in 2011 that can't be phrased without a universe type unless you add some other kind of polymorphism.<|endoftext|> TITLE: Let $X$ be a manifold. Is it true that $\beta X\cong \operatorname{Specm}(C^\infty(X))$? QUESTION [8 upvotes]: Let $X$ be a (smooth) manifold. It's well known that its Stone-Cech compactification $\beta X$ is homeomorphic to $\operatorname{Specm}(C(X))$, with its Zariski topology. Is $\beta X$ also homeomorphic to $\operatorname{Specm}(C^\infty(X))$? REPLY [8 votes]: Indeed, $\operatorname{Specm}(C^\infty(X))$ is homeomorphic to $\beta X$ (through an explicit homeomorphism that will be described below). Essentially all the theory described in Gillman & Jerison's classic book Rings of Continuous Functions (1960) applies: for completeness of MathOverflow, let me recall how this works. If $f$ is a continuous real-valued function on $X$, its zero-set is the set $Z(f) := \{x\in X : f(x)=0\}$. The zero-set of a continuous function is closed, and conversely, as $X$ is a metric space, every closed set is the zero-set of a continuous function. But in fact, even more is true: every closed set of $X$ is the zero-set of a smooth function (as follows, e.g., from here). Because of this, the notions of “z-filter” and “z-ultrafilter” below do not depend on whether we are talking about zero-sets of continuous functions or of smooth functions (or, indeed, closed sets). Now let us make the following definitions: A set $\mathscr{F}$ of zero sets is said to be a z-filter iff it contains $X$, and is closed under enlargement (i.e., if $Z \subseteq Z'$ are zero-sets and $Z\in \mathscr{F}$ then $Z'\in \mathscr{F}$) and finite intersection. A z-filter is said to be a z-ultrafilter iff it is proper (i.e., does not contain $\varnothing$), and maximal for inclusion among proper z-filters. Now let $R$ be either the ring $C(X)$ of continuous real-valued functions on $X$ or the ring $C^\infty(X)$ of smooth real-valued functions on $X$. Define two maps $\mathscr{Z}\colon \{\text{ideals of $R$}\} \to \{\text{z-filters}\}$ and $\mathscr{I}\colon \{\text{z-filters}\} \to \{\text{ideals of $R$}\}$ as follows: If $I$ is an ideal of $R$, then $\mathscr{Z}(I) := \{Z(f) : f\in I\}$. This is indeed a z-filter, as is easy to check (for enlargement, note that if $Z(f) \subseteq Z(f')$ with $f\in I$ then $Z(f') = Z(ff')$; and for finite intersection, note that $Z(f_1) \cap Z(f_2) = Z(f_1^2 + f_2^2)$). If $\mathscr{F}$ is a z-filter, then $\mathscr{I}(\mathscr{F}) := \{f\in R : Z(f)\in \mathscr{F}\}$. This is indeed an ideal as is easy to check (note that $Z(g_1 f_1+\cdots+g_r f_r) \supseteq Z(f_1)\cap \cdots \cap Z(f_r)$ if $f_1,\ldots,f_r$ are in $\mathscr{I}(\mathscr{F})$). These functions are not bijections, but still, they preserve inclusions and we trivially have $\mathscr{Z}(I) \subseteq \mathscr{F}$ iff $I \subseteq \mathscr{I}(\mathscr{F})$ (when $I$ is an ideal of $R$ and $\mathscr{F}$ a z-filter); from this follows $\mathscr{Z}(\mathscr{I}(\mathscr{F})) \subseteq \mathscr{F}$ for any z-filter $\mathscr{F}$, and $\mathscr{I}(\mathscr{Z}(I)) \supseteq I$ for any ideal $I$ of $R$, but in fact it is clear that the former is easily seen to be an equality: $\mathscr{Z}(\mathscr{I}(\mathscr{F})) = \mathscr{F}$ (the latter inclusion, $\mathscr{I}(\mathscr{Z}(I)) \supseteq I$ can be proper in general as evidenced by the ideal of functions vanishing to order $2$ at a point). Also note that an ideal $I$ is the unit ideal $R$ iff $\mathscr{Z}(I)$ is the improper z-filter $\mathscr{Z}(R)$ (the one consisting of every zero-set) because $Z(f)$ is empty iff $f$ is invertible in $R$. At this point, it is easy to check that: if $M$ is a maximal ideal of $R$ then $\mathscr{Z}(M)$ is a z-ultrafilter, and if $\mathscr{U}$ is a z-ultrafilter then $\mathscr{I}(\mathscr{U})$ is a maximal ideal. So $\mathscr{Z}$ and $\mathscr{I}$ restrict to bijections between maximal ideals of $R$ and z-ultrafilters. But because the above worked just as well for $R$ being the ring $C(X)$ of continuous real-valued or the ring $C^\infty(X)$ of smooth real-valued functions, we can conclude that their maximal ideals are in bijection by the compositions of the corresponding $\mathscr{Z}$ and $\mathscr{I}$ functions, and in fact since $\mathscr{I}_{C^\infty(X)}(\mathscr{F}) = \mathscr{I}_{C(X)}(\mathscr{F}) \cap C^\infty(X)$ the bijection from maximal ideals of $C^\infty(X)$ to those of $C(X)$ can more simply be described as $M \mapsto M \cap C^\infty(X)$. Furthermore, the Zariski topologies correspond under this bijection, because (since every z-filter is of the form $\mathscr{Z}(I)$) they both correspond to the Zariski topology on the set of z-ultrafilters whose closed sets are given by the sets of ultrafilters containing a given z-filter. (It might be worth while examining exactly which properties are required for a ring $R$ of “functions” on $X$ to make the above reasoning ensure that $\operatorname{Specm}(R) = \beta X$.) PS: It might also be interesting to consider what happens for bounded functions. The maximal ideals of the ring $C^*(X)$ of bounded continuous functions on $X$ are also naturally in bijection with the Stone-Čech compactification, but the bijection between ideals of $C(X)$ and $C^*(X)$ is not the naïve $M \mapsto M \cap C^*(X)$. So I'm not sure what happens to bounded smooth functions (I didn't give it thought).<|endoftext|> TITLE: Moments of the Riemann zeta function QUESTION [8 upvotes]: Is it possible to get an upper bound better than $\ll_\sigma T^{3/2-\sigma}$ for $$\int_{0}^{T}|\zeta (\sigma +it)|\,dt,\qquad 0<\sigma<1/2\,?$$ REPLY [13 votes]: This answer is based on Lucia's remark, and is included for completeness. By (8.111) in Ivić's book "The theory of the Riemann zeta function with applications", we have $$\int_T^{2T}|\zeta(\sigma+it)|\,dt\asymp_\sigma T,\qquad T\geq 1,\quad 1/2<\sigma<1.$$ Hence, by the functional equation for $\zeta(s)$ and Stirling's approximation, we also have $$\int_T^{2T}|\zeta(\sigma+it)|\,dt\asymp_\sigma T^{3/2-\sigma},\qquad T\geq 1,\quad 0<\sigma<1/2.$$ In particular, the answer to the original question is negative.<|endoftext|> TITLE: What is an unstable dual-Steenrod comodule? QUESTION [9 upvotes]: $\newcommand\Sq{\mathit{Sq}}$Recall that a (graded) module $V^\ast$ over the Steenrod algebra $\mathcal A^\ast$ is said to be unstable if $\Sq^i v = 0$ for $i > |v|$. The motivating example, of course, is that if $V^\ast = H^\ast(X)$ for a space $X$ with its natural $\mathcal A^\ast$ structure, then $V^\ast$ is unstable. The category of $\mathcal A^\ast$-modules which are finite-dimensional over $\mathbb F_p$ is dual to the category of $\mathcal A_\ast$-comodules which are finite-dimensional over $\mathbb F_p$, where $\mathcal A_\ast$ is the dual Steenrod algebra. So the instability condition should be expressible from this dual perspective. Question 1: Let $V_\ast$ be a finite-dimensional graded $\mathbb F_p$-vector space equipped with the structure of an $\mathcal A_\ast$-comodule. Under what conditions is the dual vector space $V^\ast$ unstable (with its natural $\mathcal A^\ast$-module structure)? Ideally, the condition would be expressed in terms of Milnor's $\{\xi_m, \tau_n\}$ generators. In principle it should be straightforward to do the translation, but I am a bit intimidated by the Adem relations. I am also stuck already because even before dualizing, I don't know how to express the instability condition in terms of the generating set $\{\Sq^{2^k} \Sq^{2^{k-1}}\dotsm \Sq^1\}$, which I suppose leads to a subsidiary question: Question 2: Let $V^\ast$ be a finite-dimensional $\mathcal A^\ast$-module. In terms of the generating set $\{\Sq^{2^k} \Sq^{2^{k-1}} \dotsm \Sq^1\}$, when is $V^\ast$ unstable? I'd be happy to know the answer for $p=2$, $p>2$, or both. REPLY [4 votes]: Your question seems to be equivalent to asking for a description of the unstable condition in terms of the Milnor basis. This is easy to do: Given $r = (r_1,\dots, r_s)$, let $P(r)$ be dual to $\xi_1^{r_1}\cdots \xi_s^{r_s}$, and let $e(r)=r_1+\dots +r_s$. Then an $A$--module $M$ is unstable iff $P(r)x = 0$ if $e(r)>|x|$. This is pretty standard stuff, I think. This would be in Lionel Schwartz' book on unstable modules, and very possibly also in Margolis' earlier book. I use this characterization in section 6 of my first generic representation theory paper [Amer.J.Math. 116 (1994)]. If you are curious about such things, you should be aware that in the 1980's and 90's, the algebra related to the unstable condition was studied quite completely, with application to both the Sullivan conjecture (and then numerous other remarkable topological consequences), and also to representation theory. Schwartz' book is a good place to start learning about this.<|endoftext|> TITLE: Interesting and surprising applications of the Ising Model QUESTION [38 upvotes]: One of the most famous models in physics is the Ising model, invented by Wilhelm Lenz as a PhD problem to his student Ernst Ising. The one-dimensional version of it was solved in Ising's thesis in 1924; later, in 1944, Lars Onsager solved the two-dimensional case in the absense of external magnetic field and in a square lattice. Although primarily a physical model, it is quite fair to say that the model became part of the mathematics literature, since its descriptions and formulations involve many interesting tools from graph theory, combinatorics, measure theory, convex analysis and so on. Physically, the Ising model can me thought as a system of many little magnets in which case the spins $\pm 1$ represent a magnetic moment. It can also represent a lattice gas, where the spins now represent whether a site is occupied by a particle or not. I've heard before that the Ising model has a vast number of applications, some of them really interesting and curious. But after coming across the paper Social applications of two-dimensional Ising models, in which the authors use the Ising model to study socio-economic opinions, urban segregation and language change, I got really curious on what else can be done using it. So, the title says it all: what are other interesting and/or surprising applications of the Ising model? REPLY [2 votes]: The study of the Ising model (and Potts model in general) has led to the construction of a particular dependent percolation model, known as the random cluster model. Formally speaking, this is not quite an "application" of the Ising model but rather a representation/generalization of it (among others like cluster expansion, dimer, currents, etc). But nowadays, the random cluster model is studied extensively as a percolation model by itself, and offered some deep insights into duality and criticality of dependent percolation models.<|endoftext|> TITLE: Can GCH fail everywhere every way? QUESTION [10 upvotes]: The following question is about if it is compatible to add to $\sf ZF$ an axiom asserting the existence of a countable transitive model of $\sf ZF$ such that for every strictly increasing function $f$ on the ordinals, we have a transitive countable model of $\sf ZF $ satisfying: $$\forall\alpha>0:\beth_\alpha = \aleph_{f(\alpha)}$$ Formally: $ \exists M: M \equiv \operatorname {CTM}(\mathsf {ZF}) \land \forall f \subseteq M \ \big{(}\\f: \operatorname {Ord}^M \to \operatorname {Ord}^M \land \forall \alpha \forall \beta \, ( \beta > \alpha \to f(\beta) > f(\alpha) ) \\ \implies \\ \exists N : N \equiv \operatorname {CTM}(\mathsf {ZF}) \land \operatorname {Ord}^N =\operatorname {Ord}^M \land (N \models \forall \alpha > 0 : \beth_\alpha=\aleph_{f(\alpha)})\big{)} $ Where "$\equiv\operatorname {CTM}(\mathsf {ZF})$" means "is a countable transitive model of ZF" So this is to say that the generalized continuum hypothesis can fail everywhere and in everyway. REPLY [11 votes]: When working in ZF, one can have more freedom. See An Easton-like Theorem for Zermelo-Fraenkel Set Theory with the Axiom of Dependent Choice and An Easton-like theorem for Zermelo-Fraenkel Set Theory without Choice.<|endoftext|> TITLE: Applications of complex exponential QUESTION [19 upvotes]: In calculus we learn about many applications of real exponentials like $e^x$ for bacteria growth, radioactive decay, compound interest, etc. These are very simple and direct applications. My question is are there any similar applications of the complex exponential $e^z$? In other words, are there any phenomena in the natural world (physics, biology, etc.) which are modeled by the complex exponential? I am aware that it surfaces in electromagnetism and signal processing, although it seems to be buried in the equations and therefore indirect. REPLY [4 votes]: Currently, the very popular feature engineering technique in deep audio processing [audio classification, speaker recognition, noise suppression, speech enhancement] is to feed neural networks with spectrograms instead of raw waveforms. Given an audio waveform $x[n]$, the spectrogram is obtained by applying Discrete Short Time Fourier Transform (STFT) to $x$ by the formula $$\mathbf{STFT}\{x[n]\}(m,\omega)\equiv X(m,\omega) = \sum_{n=-\infty}^{\infty} x[n]w[n-m]e^{-j \omega n},$$ where $w$ (not $\omega$) is some window function. In the essence, it is a new complex valued signal in time-frequency domain. This channels (real and complex, or absolute/power) are then fed to neural networks (usually 2D CNNs)<|endoftext|> TITLE: For which $n$ is $\sum_{k=1}^n 1 / \varphi(k)$ an integer? QUESTION [18 upvotes]: For which positive integers $n$ is the sum $\sum_{k=1}^n 1 / \varphi(k)$ an integer? Here $\varphi$ is the Euler totient function. The question is a "totient-analog" of the well-known result that $\sum_{k=1}^n 1/k$ is an integer only for $n=1$, which can be proved either using Bertrand's postulate or considering the $2$-adic valuation. REPLY [7 votes]: Here is some computational evidence that $n\in\{1,2,4\}$ are the only $n$ that deliver an integer sum. For an odd prime $p$, let $a_p < b_p$ denote two smallest even positive integers such that $a_pp+1$ and $b_pp+1$ are prime. Then $n\in \big[a_pp+1,\min\{b_pp+1,p^2\}\big)$ cannot deliver an integer sum as its denominator will necessarily contain $p$. Here are these intervals for primes $p<100$: 3 [7, 9) 5 [11, 25) 7 [29, 43) 11 [23, 67) 13 [53, 79) 17 [103, 137) 19 [191, 229) 23 [47, 139) 29 [59, 233) 31 [311, 373) 37 [149, 223) 41 [83, 739) 43 [173, 431) 47 [283, 659) 53 [107, 743) 59 [709, 827) 61 [367, 733) 67 [269, 1609) 71 [569, 853) 73 [293, 439) 79 [317, 1423) 83 [167, 499) 89 [179, 1069) 97 [389, 971) As we can see primes 5, 11, 29, 41, 67 alone cover the interval $[11, 1609)$, while others provide extensive backup support. Such a pattern will likely continue to cover all integers above $11$. However I'm not sure whether this observation can be justified theoretically without relying on unproved conjectures. ADDED. Primes 5, 11, 29, 41, 67, 743, 7823, 165293, 6215171 cover the interval $[11,1131161123)$.<|endoftext|> TITLE: Primes between $x$ and $x+x^\theta$ QUESTION [9 upvotes]: Iwaniec [1] proved that $$ \pi(x+x^\theta)-\pi(x) < \frac{(2+\varepsilon)x^\theta}{\eta(\theta)\log x},\ x>x_0(\varepsilon,\theta). $$ with $$ \eta(\theta)=\frac{15\theta-2}{9}. $$ (Actually, he proves that a function $\eta(\theta)>\theta$ exists, and that this is an admissible choice. This choice gives nontrivial information for $\theta>1/3.$ He gives others like $\eta_1(\theta)=(1+\theta)/2$ for $\theta>1/2.$) Two other questions have asked about the primes in this interval Prime Power Gaps Prime powers between $x$ and $x+x^\theta$ but neither asks about upper bounds, nor do answers give information. Fundamentally, I'd like information on $$ f(\theta) := \limsup_{x\to\infty} \frac{\pi(x+x^\theta)-\pi(x)}{x^\theta/\log x} $$ What modern results are available? The result above is $f(\theta) \le 18/(15\theta-2)$ for $1/3<\theta<1.$ [1] Henryk Iwaniec. On the Brun–Titchmarsh theorem. Journal of the Mathematical Society of Japan, 34:1, pp. 95–123, 1982. REPLY [6 votes]: Montgomery (1) gives a list of 40 exponent pairs $(\kappa,\lambda)$ which can be plugged into Iwaniec's formula $$ \eta(\theta)=\left(1+\frac{1-\lambda+2\kappa}{3-\lambda-\kappa/2}\right)\theta - \frac{\kappa}{3-\lambda-\kappa/2} $$ to yield bounds for $0<\theta\le1/2.$ Of these, 36 are optimal in some interval; adding the zeta function value for $\theta>1/2$ yields $$ \eta(\theta)=\begin{cases} \theta,&\text{ if }0<\theta\le1/9\\ \frac{1047\theta-2}{1029},&\text{ if }1/9\le\theta\le514/3597\\ \frac{531\theta-2}{515},&\text{ if }514/3597\le\theta\le322/2061\\ \frac{369\theta-2}{354},&\text{ if }322/2061\le\theta\le194/1101\\ \frac{444\theta-4}{417},&\text{ if }194/1101\le\theta\le130/669\\ \frac{189\theta-2}{176},&\text{ if }130/669\le\theta\le46/219\\ \frac{498\theta-8}{451},&\text{ if }46/219\le\theta\le2/9\\ \frac{228\theta-4}{205},&\text{ if }2/9\le\theta\le362/1569\\ \frac{1161\theta-22}{1037},&\text{ if }362/1569\le\theta\le182/743\\ \frac{1167\theta-26}{1027},&\text{ if }182/743\le\theta\le52/201\\ \frac{921\theta-22}{805},&\text{ if }52/201\le\theta\le342/1291\\ \frac{254\theta-8}{215},&\text{ if }342/1291\le\theta\le226/833\\ \frac{669\theta-22}{563},&\text{ if }226/833\le\theta\le118/415\\ \frac{116\theta-4}{97},&\text{ if }118/415\le\theta\le186/641\\ \frac{589\theta-22}{487},&\text{ if }186/641\le\theta\le94/303\\ \frac{587\theta-26}{473},&\text{ if }94/303\le\theta\le310/959\\ \frac{1439\theta-66}{1153},&\text{ if }310/959\le\theta\le286/855\\ \frac{1327\theta-66}{1049},&\text{ if }286/855\le\theta\le127/376\\ \frac{351\theta-22}{265},&\text{ if }127/376\le\theta\le74/217\\ \frac{393\theta-26}{293},&\text{ if }74/217\le\theta\le45/128\\ \frac{325\theta-22}{241},&\text{ if }45/128\le\theta\le62/171\\ \frac{347\theta-26}{251},&\text{ if }62/171\le\theta\le37/98\\ \frac{281\theta-22}{201},&\text{ if }37/98\le\theta\le66/173\\ \frac{823\theta-66}{585},&\text{ if }66/173\le\theta\le426/1093\\ \frac{317\theta-26}{224},&\text{ if }426/1093\le\theta\le682/1733\\ \frac{1243\theta-114}{851},&\text{ if }682/1733\le\theta\le438/1103\\ \frac{1601\theta-150}{1089},&\text{ if }438/1103\le\theta\le36/89\\ \frac{1651\theta-162}{1107},&\text{ if }36/89\le\theta\le486/1181\\ \frac{659\theta-66}{439},&\text{ if }486/1181\le\theta\le474/1141\\ \frac{1465\theta-150}{969},&\text{ if }474/1141\le\theta\le1626/3865\\ \frac{595\theta-66}{383},&\text{ if }1626/3865\le\theta\le120/281\\ \frac{1427\theta-162}{911},&\text{ if }120/281\le\theta\le422/973\\ \frac{1249\theta-150}{781},&\text{ if }422/973\le\theta\le846/1921\\ \frac{923\theta-114}{571},&\text{ if }846/1921\le\theta\le1542/3469\\ \frac{470\theta-60}{287},&\text{ if }1542/3469\le\theta\le34/75\\ \frac{15\theta-2}{9},&\text{ if }34/75\le\theta\le1/2\\ \frac{\theta+1}{2},&\text{ if }1/2<\theta\le7/12\\ 2,&\text{ if }7/12<\theta<1. \end{cases} $$ This is hardly a satisfactory answer, giving only 20-year-old results and neglecting Vinogradov's method for small $\theta,$ but it may be useful so I'll leave it here. I’ve added Huxley’s result $$ \pi(x)-\pi(x-y) \sim \frac{y}{\log x}\text{ for }x^\theta \le y \le x/2 $$ with $\theta>7/12$. Hugh L. Montgomery, Harmonic Analysis as found in Analytic Number Theory, in Twentieth Century Harmonic Analysis—A Celebration, Springer, 2001 M. N. Huxley, On the difference between consecutive primes, Invent. Math. 15 (1972), pp. 164-170.<|endoftext|> TITLE: Would efficient factoring have any *other* useful applications? QUESTION [12 upvotes]: This question is certainly somewhat opinion-based, but hopefully not hopelessly so. The granddaddy of all applications for an efficient period finding or factoring capability (e.g. Shor's algorithm) is obviously breaking the public-key cryptography systems that currently encrypt Internet traffic. But if we could eventually implement Shor's algorithm on a large-scale quantum computer, would it have any other useful applications, whether for pure math research or out in the "real world"? This question is inspired by an essay by Boaz Barak, in which he mentions that he doesn't know of any reason why efficient factoring is inherently interesting other than for cryptography. Two scope clarifications: I know that the quantum Fourier transform at the heart of Shor's algorithm has other potential practical applications, e.g. for solving large linear systems via the HHL algorithm, but I'm specifically wondering about period finding and factoring, and not general ultra-fast Fourier transforms. I know that the existence of an efficient quantum factoring algorithm has huge implications for computational complexity theory: it provides perhaps the strongest evidence we have (a) that BPP != BQP, (b) against the extended Church-Turing thesis, and (c) depending on your philosophical beliefs, perhaps against the real-world feasibility of building a fault-tolerant quantum computer. But I'm wondering about actually executing such an algorithm, not about its existence or properties. REPLY [9 votes]: I'm not sure about "real world", but studies around multiplicative functions (e.g., aliquot sequences) will definitely benefit from the availability of a fast factorization method. At very least it will allow to verify, refine, or refute existing conjectures with extensive computational evidence. Also, there is a list of OEIS sequences "needing factors", which gives other examples of topics unrelated to cryptography but relying on integer factorization.<|endoftext|> TITLE: Is Joyal's category $\Theta_n$ the "only" Reedy category which is dense in $n$-categories? QUESTION [8 upvotes]: Let $Cat_n$ denote the category of $n$-categories. Then Joyal's category $\Theta_n$ is (1) a full dense subcategory of $Cat_n$ which (2) is also a Reedy category. Question: Is Joyal's category $\Theta_n$ somehow uniquely characterized by (1) and (2)? Notes: I've been ambiguous about what "category" and "$n$-category" mean -- the above statements are true whether "category" means "1-category" or "$(\infty,1)$-category". In the latter case, "$n$-category" can mean either "strict $n$-category" or "weak $(\infty,n)$-category" or various things in between. I'm interested in understanding the story for any $n \in \mathbb N \cup \{\infty\}$. One nice story which makes Joyal's category $\Theta_n$ look very "canonical" is the theory of generalized nerves. I'd be interested, for example, if there were something to say in that framework about getting one's nerves to be indexed by Reedy categories. I'd also be interested if $\Theta_n$ didn't turn out to be literally "unique" with respect to (1) and (2), but at least "minimal" with respect to (1) and (2), or something like that. When $n=1$, the question asks whether the simplex category $\Delta = \Theta_1$ is the "unique" full subcategory of $Cat$ which is a Reedy category. It sounds like the answer is probably "no" in this case, but I don't know a good counterexample. REPLY [2 votes]: EDIT: As $\Theta_n \subseteq Cat_n$ is not "irredunant" in the sense below ((3) fails), the following doesn't really answer the question. If the Claim below could be upgraded to work with "isomorphism" in (3) rather than "split epimorphism", then we'd be in business. Here's an answer which really surprises me: I claim that $\Theta_n$ is the unique idempotent-complete Reedy category $\mathcal R$ appearing as a dense full subcategory of $Cat_n$ (where $Cat_n$ means the weak $(\infty,1)$-category of weak $(\infty,n)$-categories) which is irredundant in the following sense: Definition: Let $\mathcal C$ be an $\infty$-category. Say that $\mathcal R \subseteq \mathcal C$ is a basis if it is a dense, idempotent-complete, full subcategory. Say that a basis $\mathcal R \subseteq \mathcal C$ is irredundant if the following conditions are satisfied: $\mathcal R$ is a Reedy category Every morphism of $\mathcal R^-$ is a split epimorphism. For every $R \in \mathcal R$, the reflection in $\mathcal C \subseteq Psh(\mathcal R)$ of the inclusion $\partial R \to R$ is not a split epimorphism. Claim: Irredundant bases are unique and minimal. In other words, let $\mathcal C$ be an $\infty$-category with an irredundant basis $\mathcal R$. Then for any other basis $\mathcal R' \subseteq \mathcal C$, we have $\mathcal R \subseteq \mathcal R'$ up to repletion in $\mathcal C$. Proof: Let $R \in \mathcal R$, and let us show that $R \in \mathcal R'$. Consider the map of $\mathcal R$-presheaves $$(\ast) \qquad \varinjlim_{\mathcal R' \ni R' \to R} R' \to R$$ The reflection of $(\ast)$ in $\mathcal C$ is an isomorphism. If $(\ast)$ is not split epi, then by the Lemma below, it factors through $\partial R \to R$. Upon reflection in $\mathcal C$, we obtain a splitting of the $\mathcal C$-reflection of $\partial R \to R$, contradicting irredundancy of $\mathcal C$. So $(\ast)$ is split epi. This means that there exists a split epimorphism $R' \to R$ with $R' \in \mathcal R'$. Since $\mathcal R'$ is closed under idempotent splitting, this implies that $R \in \mathcal R'$. **UPSHOT:** The main question is given an affirmative answer of some form by the following: Corollary: Since $\Theta_n$ is an irredundant basis of $Cat_n$, it follows that any other basis of $Cat_n$ contains $\Theta_n$, and no other basis of $Cat_n$ is irredundant. We have used the following Lemma. Thanks to Chris Schommer-Pries for pointing out the necessity of something like this in the comments below. Lemma: Let $\mathcal C$ be an $\infty$-category with a bases $\mathcal R, \mathcal R'$. Suppose that $\mathcal R$ satisfies (1) and (2) above. Let $R \in \mathcal R$. If the map $(\ast)$ is not a split epimorphism of $\mathcal R$-presheaves, then it factors through the map $\partial R \to R$. Proof: By Yoneda, $(\ast)$ is split epi iff $(\ast)(R)$ (that's the map of spaces obtained by evaluating the map of presheaves $(\ast)$ at the object $R$) is split epi, if and only if $id_R$ is in the image of $(\ast)(R)$. Note that the codomain of $(\ast)(R)$ is the discrete space $R(R)$. Moreover, $\partial R(R) \subset R(R)$ is the complement of $\{id_R\} \subseteq R(R)$. So if $(\ast)$ is not split epi, then the image of $(\ast)(R)$ is contained in $\partial R(R)$. Now for any $T \in R$, the inclusion of discrete spaces $\partial R(T) \subseteq R(T)$ has complement given by the maps $f : T \twoheadrightarrow R$ which lie in $\mathcal R^-$. Suppose for contradiction that such an $f$ is in the image of $(\ast)(T)$ evaluated at $T$. Then $f$ factors through some $R' \in \mathcal R'$. By condition (2) in the definition of irredundancy, $f$ has a section $g$, and now $id_R = fg$ factors through $R' \in \mathcal R$, which means that $id_R$ is in the image of $(\ast)(R)$, contrary to what we concluded in the previous paragraph. Therefore $(\ast)(T)$ factors through $\partial R(T)$ for each $T \in \mathcal R$. Since $\partial R(T) \to R(T)$ is always a monomorphism of discrete spaces, this means that $\ast$ factors through $\partial R \to R$. Notes: Conditions (1) and (2) in the definition of irredundancy are implied by the stronger assumption that $\mathcal R$ is an elegant Reedy category, which is indeed the case when $\mathcal R = \Theta_n$. $\Theta_n$ is not an irredundant basis of the category of strict $n$-categories (cf. Zhen Lin's comment above), because condition (3) fails in this case. For irredundancy, it's important that we put ourselves in the weak setting. Another way of stating condition (3) in the definition of irredundancy is this: for every $R \in \mathcal R$, it is not the case that every $C \in \mathcal C$ is weakly right orthogonal to the inclusion $\partial R \to R$. I suspect that the Claim will remain true if condition (3) in the definition of irredundancy is weakened to stipulate only that $\partial R \to R$ does not reflect to an isomorphism (this corresponds, in the previous bullet, to replacing weak orthogonality with strong orthogonality). I was almost able to modify the proof of the above Claim to verify this, at the cost of introducing into the structure of the proof an induction on the degree of $R$, but there were some coherence issues I was unsure of (I didn't push terribly hard, though).<|endoftext|> TITLE: The Wiener measure of an open set QUESTION [6 upvotes]: There is so much written about the Brownian motion and I suspect the answers to the questions below are hidden in somewhere in the literature but I cannot find them Denote by $E$ the Banach space of continuous functions $\newcommand{\bR}{\mathbb{R}}\newcommand{\ve}{\varepsilon}\newcommand{\bsW}{\boldsymbol{W}}$ $f:[0,1]\to\bR$ such that $f(0)=0$. The norm on $E$ is the sup-norm. The Wiener measure defines a Borel measure $\bsW$ on $E$. Let $f_0\in E_0$ and $\ve>0$. We set $$ w(f_0,\ve):=\bsW\Big[\;\big\{\; f\in E;\;\;\Vert f-f_0\Vert<\ve\,\big\}\;\Big]. $$ Question 1. Is it true that $w(f_0,\ve)>0$ for all $f_0\in E$ and $\ve>0$? Question 2. Can one produce an explicit positive lower bound for $w(f_0,\ve)$ in terms of $\ve>0$ and the modulus of continuity of $f$? In particular, if $f_0$ is Lipschitz, can one produce a lower bound in terms of $\ve>0$ and the Lipschitz constant of $f_0$? For Question 1 I have an argument based on Cameron-Martin formula? Is there any other more "elementary" argument? ${{}}$ REPLY [10 votes]: This is known as the support theorem for Brownian motion. Besides the proof in the answer of Iosif Pinelis and the proof in Exercise 1.8 of [1], there is also a proof on page 59 of [2]. Generalizations are discussed in [3]-[5]. [1] Mörters, Peter, and Yuval Peres. Brownian motion. Vol. 30. Cambridge University Press, 2010. https://yuvalperes.com/brownian-motion/ [2] R. Bass, Probabilistic Techniques in Analysis, Springer, New York (1995). MR1329542 [3] https://fabricebaudoin.wordpress.com/2013/04/10/lecture-30-the-stroock-varadhan-support-theorem/ [4] http://www.numdam.org/article/SPS_1994__28__36_0.pdf [5] Stroock, Daniel W.; Varadhan, S. R. S. On the support of diffusion processes with applications to the strong maximum principle. Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability (Univ. California, Berkeley, Calif., 1970/1971), Vol. III: Probability theory, pp. 333–359. Univ. California Press, Berkeley, Calif., 1972. REPLY [7 votes]: $\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\om}{\omega}$Let $g:=f_0$. There is some real $\de>0$ such that \begin{equation*} \om(g,\de):=\max\{|g(y)-g(x)|\colon x,y\in[0,1], |y-x|\le\de\}<\ep/3<\ep/2. \end{equation*} Take now any $t_0,\dots,t_n$ such that $0=t_0<\dots\ep/2$ and hence the length of the interval $J_k$ is $>\ep/6>0$. Also, for each $k\in[n]$ the shortest distance from any point $x\in J_{k-1}$ to the set $\{g(t_k)-\ep/2,g(t_k)+\ep/2\}$ of the endpoints of the interval $I_k$ is $>\ep/6>0$. So, $p_k>0$ for all $k\in[n]$, and hence $p>0$. Moreover, one can give an explicit expression of each $p_k$ in terms of $\de_k$, $I_k$, $J_k$, $J_{k-1}$ -- cf. e.g. Proposition 6.10.6, p. 533. Thus, one can explicitly express the lower bound $p>0$ on $w(g,\ep)$ in terms of $\ep$, $t_1,\dots,t_n$, and $g(t_1),\dots,g(t_n)$. For $g(t)=t (5 - 18 t + 12 t^2)$, $\ep=5/2$, $n=4$, $\de_k=1/4$ for $k\in[4]$, the picture below shows the graphs $\{(t,g(t))\colon t\in[0,1]\}$ (blue), $\{(t,g_-(t))\colon t\in[0,1]\}$ (gold), $\{(t,g_+(t))\colon t\in[0,1]\}$ (green), and a path of the Wiener process (gray) belonging to the event \begin{equation*} \begin{aligned} B&:=\{g_-(t)0$.<|endoftext|> TITLE: When are the zero sets of two continuous functions in the Stone-Čech compactification included in one another? QUESTION [8 upvotes]: Let $X$ be a topological space (feel free to add some simplifying assumptions here, like “completely regular” provided at least the case of finite-dimensional manifolds is covered). Let $f,g \in C^*(X)$ where $C^*(X)$ denotes the ring of bounded continuous real-valued functions on $X$. Denote $f^\beta, g^\beta \colon \beta X\to \mathbb{R}$ the continuous functions corresponding to $f,g\colon X\to \mathbb{R}$ under the canonical isomorphism $C(\beta X) \cong C^*(X)$ where $\beta X$ is the Stone-Čech compactification of $X$, and let $Z(f^\beta),Z(g^\beta)$ be their zero-sets, i.e., $Z(f^\beta) = \{p\in\beta X : f^\beta(p)=0\}$. Question: how can we tell whether $Z(f^\beta) \subseteq Z(g^\beta)$ merely by looking at $f,g$ (without looking at the Stone-Čech compactification)? To give an idea of the flavor of criteria I'm looking for, let me observe that: Fact: $Z(f^\beta) = \varnothing$ iff $f$ is bounded away from $0$ on $X$ (i.e. $\exists \varepsilon>0. |f|\geq\varepsilon$). (Proof: if $|f|\geq\varepsilon$ then clearly $|f^\beta|\geq \varepsilon$ so $Z(f^\beta)=\varnothing$. But conversely, if $Z(f^\beta)=\varnothing$, since $\beta X$ is compact, the image of $|f^\beta|$ is a compact subset of $\mathbb{R}_{>0}$, so it is lower bounded by some $\varepsilon>0$.) Equivalently, this means that $f$ is invertible in $C^*(X) \cong C(\beta X)$. Based on the above fact, I first thought that $Z(f^\beta) \subseteq Z(g^\beta)$ meant $|g|\leq C|f|$ on $X$ for some constant $C$, but this is not correct (take $X=\mathbb{R}$ and $f\colon x\mapsto \min(1,x^2)$ and $g\colon x\mapsto \min(1,|x|)$: then $f^\beta$ and $g^\beta$ both vanish only at $0$ but we don't have $|g|\leq C|f|$). Maybe something like “$Z(f)\subseteq Z(g)$ and there exists $K\subseteq X$ compact and $C$ such that $|g|\leq C|f|$ outside $K$”? Motivation: thinking about the PS in this answer made me ask whether, for $X$ a manifold and $f$ bounded continuous on $X$, we can find $g$ bounded and smooth on $X$ such that $Z(f^\beta) = Z(g^\beta)$ (and to find a criterion for the latter, looking at $Z(f^\beta) \subseteq Z(g^\beta)$ first seems natural). REPLY [7 votes]: Lemma: Suppose that $X$ is a compact Hausdorff space. Let $f,g:X\rightarrow[0,\infty)$ be continuous functions. Then the following are equivalent: $Z(f)\subseteq Z(g)$. For all $\epsilon>0$, there exists a $\delta>0$ where $f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$. There exists a function $u:[0,\infty)\rightarrow[0,\infty)$ that is continuous at the point $0$ with $u(0)=0$ and where $g\leq u\circ f$. There exists a continuous function $v:[0,\infty)\rightarrow[0,\infty)$ with $v(0)=0$ and where $g\leq v\circ f$. There exists a continuous bijection $w:[0,\infty)\rightarrow[0,\infty)$ with $w(0)=0$ and where $g\leq w\circ f$ (the mapping $w$ is necessarily a homeomorphism and increasing). Proof: $5\rightarrow 4,4\rightarrow 3.$ These are trivial. $4\rightarrow 5.$ Suppose that $v:[0,\infty)\rightarrow[0,\infty)$ is a bijection with $v(0)=0$ and $g\leq v\circ f$. Then let $w:[0,\infty)\rightarrow[0,\infty)$ be the function defined by letting $w(x)=v(x)+x$. Then $g\leq w\circ f$ and $w(0)=0$. $3\rightarrow 4.$ There is some continuous $v:[0,\infty)\rightarrow[0,\infty)$ where $v(0)=0$ and where either $\min(\max(g),u(y))\leq v(y)$ for all $y$. Therefore, for all $x$, we have $g(x)\leq u(f(x))$, so $g(x)\leq\min(\max(g),u(f(x)))\leq v(f(x)).$ $5\rightarrow 2$. Suppose that $\epsilon>0$. Then set $\delta=w^{-1}(\epsilon)$. If $x\in f^{-1}[0,\delta]$, then $f(x)\leq\delta$, so $g(x)\leq w(f(x))\leq w(\delta)=\epsilon.$ Therefore $x\in g^{-1}[0,\epsilon]$. $2\rightarrow 1$. If $\epsilon>0$, then there is a $\delta>0$ where $Z(f)\subseteq f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$. Therefore, $Z(f)\subseteq\bigcap_{n=1}^{\infty}g^{-1}[0,1/n]=Z(g)$. $1\rightarrow 3.$ Suppose that $Z(f)\subseteq Z(g)$. If $Z(f)=\emptyset$, then $\min(f)>0$, so just select a suitable function $u$ with $u(y)\geq\max(g)$ whenever $y\geq\min(f)$. Now, assume $Z(f)\neq\emptyset$. Let $u:[0,\infty)\rightarrow[0,\infty)$ be the mapping such that $u(y)=\max\{g(x)\mid f(x)\leq y\}=\max g[f^{-1}(-\infty,y]].$ Then $u(0)=\max\{g(x)\mid f(x)\leq 0\}=0$. Furthermore, if $x_{0}\in X$, then $u(f(x_{0}))=\max\{g(x)\mid f(x)\leq f(x_{0})\}$. Therefore, $g(x_{0})\leq u(f(x_{0}))$. I claim that $u$ is upper semicontinuous (and therefore continuous at $0$). Suppose that $y\in u^{-1}(-\infty,c)$. Then $\max g[f^{-1}(-\infty,y]]=\max\{g(x)\mid f(x)\leq y\}=u(y)y$ where $f^{-1}(-\infty,z]\subseteq g^{-1}(-\infty,c)$. However, if $s0$, there exists a $\delta>0$ where $f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$. There exists a function $u:[0,\infty)\rightarrow[0,\infty]$ that is continuous at the point $0$ with $u(0)=0$ and where $g\leq u\circ f$. There exists a continuous function $v:[0,\infty)\rightarrow[0,\infty)$ with $v(0)=0$ and where $g\leq v\circ f.$ There exists a continuous bijection $w:[0,\infty)\rightarrow[0,\infty)$ with $w(0)=0$ and where $g\leq w\circ f$. Now suppose that $X$ is a locally compact regular space with compactification $C$, and suppose that $f,g:X\rightarrow[0,\infty)$ and $f,g$ extend to mappings $\overline{f},\overline{g}:C\rightarrow[0,\infty)$. The following result characterizes when $Z(\overline{f}|_{C\setminus X})\subseteq Z(\overline{g}|_{C\setminus X})$, so one can use the following result to produce more characterizations of when $Z(\overline{f})\subseteq Z(\overline{g})$ when $X$ is locally compact. Theorem: Suppose that $X$ is a non-compact locally compact regular space with compactification $C$. Let $f,g:X\rightarrow[0,\infty)$ be bounded continuous functions, and let $\overline{f},\overline{g}:C\rightarrow[0,\infty)$ be the continuous extensions of $f,g$ to the domain $C$. Then the following are equivalent. $Z(\overline{f}|_{C\setminus X})\subseteq Z(\overline{g}|_{C\setminus X})$. For each $\epsilon>0$, there exists a $\delta>0$ and a compact $K\subseteq X$ such that if $x\in X\setminus K$, then $f(x)<\delta\rightarrow g(x)<\epsilon.$ There exists a continuous bijection $u:[0,\infty)\rightarrow[0,\infty)$ with $u(0)=0$ and a function $A:X\rightarrow[0,\infty)$ where $A^{-1}[\epsilon,\infty)$ is compact for each $\epsilon>0$ and where $g\leq A+(u\circ f)$. Proof: $2\rightarrow 1$. Suppose that $c_{0}\in Z(f|_{C\setminus X})$. Therefore, let $(x_{d})_{d\in D}$ be a net that converges to $c_{0}$. Then for each $\epsilon>0$, there is some $\delta>0$ and compact set $K\subseteq X$ where if $x\in X\setminus K$ and $f(x)\leq\delta$, then $g(x)\leq\epsilon$. Then there is some $d_{0}\in D$ where if $d\leq d_{0}$, then $x_{d}\not\in K$ and where $f(x_{d})\leq\delta$. In this case, we have $g(x_{d})\leq\epsilon$. Therefore, we conclude that $g(x_{d})_{d\in D}\rightarrow 0$, so since $g$ is continuous, we know that $g(c_{0})=0$, and therefore $c_{0}\in Z(g|_{C\setminus X})$. Thus, $Z(f|_{C\setminus X})\subseteq Z(g|_{C\setminus X})$. $1\rightarrow 3$. By the above results, we know that there is a continuous mapping $u:[0,\infty)\rightarrow[0,\infty)$ such that $\overline{g}|_{C\setminus X}\leq u\circ\overline{f}|_{C\setminus X}$. Therefore, let $A:X\rightarrow[0,\infty)$ be the mapping defined by $A=\max(0,g-(u\circ f))$. Then $g=g-(u\circ f)+(u\circ f)\leq A+(u\circ f),$ and $\overline{A}(c)=0$ whenever $c\in C\setminus X$, so $A^{-1}[\epsilon,\infty)$ is a compact subset of $X$ whenever $\epsilon>0$. $3\rightarrow 2$. Suppose that $g\leq A+(u\circ f)$. Then for all $\epsilon>0$, there is a $\delta>0$ with $u(\delta)<\epsilon$. Therefore, suppose that $c\in C\setminus X$, and $\overline{f}(c)\leq\delta$. Then $$\overline{g}(c)\leq\overline{A}(c)+(u\circ\overline{f})(c)=u(\overline{f}(c))\leq u(\delta)<\epsilon.$$ In particular, there is no $c\in C\setminus X$ with $\overline{f}(c)\leq\delta$ and $\overline{g}(c)\geq\epsilon$. Therefore, if we set $K=\overline{f}^{-1}[0,\delta]\cap\overline{g}[\epsilon,\infty)$, then $K$ is a closed subset of $C$, so $K$ is compact, but $K$ is also a subset of $X$. Therefore, if $x\in X\setminus K$, then $f(x)\leq\delta\rightarrow g(x)<\epsilon$. Q.E.D. REPLY [3 votes]: I'm reasonably convinced (but don't have time to check it carefully right now) that $Z(f^\beta)\subseteq Z(g^\beta)$ if and only if, for every $\varepsilon>0$, $f$ is bounded away from zero on $\{x\in X:|g(x)|\geq\varepsilon\}$. (Note that this agrees with the Fact in the question, by taking $g$ to be identically $1$.)<|endoftext|> TITLE: Scattered hereditarily separable does not imply countable in ZFC QUESTION [7 upvotes]: Recall that a topological space $X$ is scattered if and only if every non-empty subset $Y$ of $X$ contains at least one point which is isolated in $Y$. Consider the statement: "Every scattered hereditarily separable space is countable". In this book it is mentioned that, under $\mathsf{CH}$, K. Kunen constructed a compact, scattered, hereditarily separable space of size $\aleph_1$. My question is if a "real" example of a scattered, hereditarily separable, uncountable space is known, or if this statement is independent of $\mathsf{ZFC}$. REPLY [8 votes]: As clarified in the comments, the existence of a regular S-space is independent of ZFC, but a "real" $T_2$ example can be constructed by taking a well-ordering of a set of reals in order type $\omega_1$ and refining the Euclidean topology by declaring initial segments open. This example is $T_2$ since it refines a $T_2$ topology and is still hereditarily separable (any subset has an initial segment that is dense in the Euclidean topology and this initial segment is still dense in the stronger topology). It is not Lindelof since the cover by initial segments has no countable subcover, and it is scattered since any subset has a minimal element which is isolated in that subspace. This example is described in Mary Ellen Rudin's excellent book "Lectures on Set Theoretic Topology" published by the AMS.<|endoftext|> TITLE: Set theory determined by $V_\alpha$ for limit ordinals $\alpha>\omega$ QUESTION [9 upvotes]: Von Neumann hierarchy has a critical role in set theory. It is well-known that $V_\alpha$ is a model of $\mathsf{ZC}$ if $\alpha$ is a limit ordinal. Furthermore, $V_\alpha$ satisfies the cumulative hierarchy axiom ($\mathsf{CHA}$), an axiom claiming the existence of sequence $\langle V_\eta\mid \eta<\xi\rangle$ for each ordinal $\xi$ with that every set is contained in some $V_\eta$, which is apparently not a theorem of $\mathsf{ZC}$. (Caution. $\mathsf{CHA}$ is not equivalent to the claim that $V_\xi$ exists for all ordinal $\xi$. See Hamkins' answer.) However, $V_\alpha$ does not satisfy Replacement even for $\Delta_1$-formulas. Theorem. $V_{\omega+\omega}$ does not satisfy Replacement for $\Delta_1$-formulas. Proof. Consider \begin{align} \phi(x,y) :\equiv \exists n<\omega\exists f [x\omega$? (According to Mathias's The Strength of Mac Lane Set Theory, $\mathsf{ZC}$ proves Replacement for stratified formulas.) Can we characterize the theory of $V_\alpha$ for limit $\alpha>\omega$? The second question needs some clarification. Assume that $\mathsf{ZFC}$ is consistent. My question is we can characterize the theory $T$ defined by A sentence $\sigma$ is a member of $T$ if and only if, for any models $M\models \mathsf{ZFC}$ and $\alpha\in M$ such that $M\models \alpha>\omega\text{ is a limit ordinal}$, $M\models (V_\alpha\models \sigma)$. Obviously $T\supseteq \mathsf{ZC+CHA}$. Furthermore, if we assume $\mathsf{ZFC}$ + there is a worldly cardinal is consistent, then $T\nvdash\lnot\sigma$ for axioms $\sigma$ of $\mathsf{ZFC}$: If $M$ is a model of $\mathsf{ZFC}$ and $M\models \kappa \text{ is worldly}$, then $V^M_\kappa\models \mathsf{ZFC}$. If $\mathsf{ZFC}$ proves we can force $\sigma$, then $T\nvdash\lnot\sigma$. Again, let $M$ be a countable model of $\mathsf{ZFC}$ and $M\models \kappa \text{ is worldly}$. By the assumption, we can find a forcing poset $\mathbb{P}\in V^M_\kappa$ such that $V^M_\kappa$ thinks $\mathbb{P}$ forces $\sigma$. Now consider the $\mathbb{P}$-generic extension $M[G]$ of $M$. Then $V^{M[G]}_\kappa = V^M_\kappa[G]$ and $V^M_\kappa[G]\models \mathsf{ZFC}+\sigma$. These results provide some upper bound for $T$. Could we find a better upper bound for $T$? REPLY [8 votes]: This answer concerns the second question (which asks whether there is a characterization of the theory of models of the form $V_{\alpha}$, where $\alpha$ ranges over limit ordinals), and its elaboration which asks whether there is a characterization of the theory $T$ defined by: A sentence $\sigma$ is a member of $T$ if and only if, for any models $M\models \mathsf{ZFC}$ and $\alpha\in M$ such that $M\models \alpha>\omega\text{ is a limit ordinal}$, $M\models (V_\alpha\models \sigma)$. Let $T^{+}$ be the theory of models of the form $V_{\alpha}$, where $\alpha$ ranges over limit ordinals. The observations below make it clear that $T \neq T^{+}$. Observation 1. $T$ coincides with the collection of set-theoretical sentences $\sigma$ such that the implication "For all $\alpha$, if $\alpha$ is a limit ordinal greater than $\omega$, then $\sigma^{V_{\alpha}}$" is provable in $\mathsf{ZFC}$. Proof: this is a simple consequence of the completeness theorem for first order logic. Observation 2. $T$ is recursively enumerable. Proof: This is an immediate corollary of Observation 1. Observation 3. $T^{+}$ is not arithmetically definable (and indeed, $T^{+}$ is not definable in $n$-order arithmetic for any $n \in \omega$). Proof: Given an arithmetical sentence $\varphi$, $\mathbb{N}\models \varphi$ iff the set-theoretical sentence expressing [if $\omega$ exists, then $\mathbb{N}\models \varphi$] is a member of $T^{+}$. So by Tarski's undefinability of truth theorem, $T^{+}$ is not arithmetically definable (the parenthetical clause of Observation 3 has a similar proof). Finally, looking at the definition of $T^{+}$, $T^{+}$ is definable by a $\Pi_2$-formula (in the Levy hierarchy). Based on a "back-of-the-envelope-proof", this calibration is optimal. The proof is based on the "folklore" characterization of $\Sigma_2$-statements, as in Lemma 2 (page 4) of the paper The universal finite set of Hamkins and Woodin (note that $\sigma \in T^{+}$ iff the implication "if there is no last ordinal, then $\sigma$" is true in all structures of the form $V_{\alpha}$).<|endoftext|> TITLE: What is known about the least cardinal where $\kappa$ fails to be supercompact? QUESTION [6 upvotes]: Assume $\kappa$ is $\lambda$-supercompact for some $\lambda$ but not fully supercompact. Are there any known restrictions (or provably non-restrictions) on the least $\delta$ such that $\kappa$ is not $\delta$-supercompact (e.g. $\text{cf}(\delta)\geq\kappa$ or $\text{lim}(\delta)$)? REPLY [5 votes]: I'm not sure if this gets to the heart of the question, but here are some observations. For a successor ordinal $\alpha$, let $\kappa_\alpha$ be the least $\kappa>\alpha$ which is $\kappa^{+\alpha}$-supercompact. Then $\kappa_\alpha$ is not $2^{\kappa_\alpha^{+\alpha}}$-supercompact. This is because an embedding $j : V \to M$ with critical point $\kappa_\alpha$ witnessing $2^{\kappa_\alpha^{+\alpha}}$-supercompactness would have that $M$ has a measure witnessing $\kappa_\alpha$ is $\kappa_\alpha^{+\alpha}$-supercompact. (Here, we need to invoke the theorem of Solovay that if $\lambda$ is regular and $\kappa$ is $\lambda$-supercompact, then $\lambda^{<\kappa}=\lambda$.) Since $j(\kappa_\alpha) > \kappa_\alpha$, there is $\kappa<\kappa_\alpha$ that is $\kappa^{+\alpha}$-supercompact, contradicting the definition of $\kappa_\alpha$. On the other hand, if $\kappa$ is $\kappa^{+\alpha}$-supercompact, where $\alpha<\kappa$ is a limit ordinal, then $\kappa$ is $\kappa^{+\alpha+1}$-supercompact, since a $\lambda$-supercompactness measure for $\kappa$ witnesses $\lambda^{<\kappa}$-supercompactness. See Solovay-Reinhardt-Kanamori. I suppose an interesting specialization of the question is, if $\kappa$ is $\kappa^{+}$-supercompact and $2^{\kappa^+} > \kappa^{++}$, is it $\kappa^{++}$-supercompact? We can preserve the $\kappa^+$-supercompactness and kill the $\kappa^{++}$-supercompactness by adding a nonreflecting stationary subset of $\kappa^{++}$, but I don't know how to do it without collapsing $2^{\kappa^+}$ to $\kappa^{++}$ simultaneously. REPLY [4 votes]: One can get fairly close bounds by observing that if $\kappa$ is $\lambda$-supercompact, then by using a supercompactness measure of least Mitchell rank, there is a $\lambda$-supercompactness embedding $j:V\to M$ for which $\kappa$ is not $\lambda$-supercompact in $M$. But $\kappa$ will be $\theta$-supercompact in $M$ whenever $2^{\theta^{<\kappa}}\leq\lambda^{<\kappa}$. So from the point of view from $M$, we have a good understanding of the exact point of failure of supercompactness of $\kappa$. Indeed, under GCH, these bounds are nearly tight.<|endoftext|> TITLE: Are nearby subalgebras of matrix algebras conjugate? QUESTION [7 upvotes]: Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are conjugate in $M_n(k)$, i.e., there exists a matrix $X \in M_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$. It we do not assume that $A$ is simple, then this result is generally false (see e.g. the example given by Denis Serre at Conjugation between commutative subalgebras of a matrix algebra?). My question is: Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate? Here by "close" I mean that for some small $\varepsilon>0$, we have $\lVert f(a) - g(a)\rVert \leq \varepsilon \lVert a\rVert$, for some submultiplicative norm on $A$. REPLY [8 votes]: Here's another example. It's quite distinct from my previous answer since in my previous answer the homomorphisms $f_t$ are non-injective and have images that are not isomorphic (and hence non-conjugate). Here I provide an example with injective homomorphisms $f_t$ (so that their images are isomorphic). I'll not be explicit, using a dimensional argument "there are too many injective homomorphisms for all of them to be conjugate". Namely, $A=A_n$ is the unital (commutative associative) algebra of dimension $n+1$, with basis $(I,X_1,\dots,X_n)$ with $I$ unit and all other products being zero. Hence, a representation of $A$ is just the data of $n$ matrices squaring to zero and with pairwise zero product. And I'll assume $n=8$ although passing to $n\ge 8$ should be immediate (while decreasing to smaller $n$ should be possible, although I'm not really sure about $n=2,3$). Let $R_n$ be the nilradical of $A_n$, that is (with basis $(X_1,\dots,X_n)$). So a representation of the unital algebra $A_n$ is the same as a representation of the (non-unital) algebra $R_n$. Consider in $M_8(K)$ the matrices by $4+4$ blocks $\begin{pmatrix}0 & *\\ 0 & 0\end{pmatrix}$. These form a 16-dimensional subspace of $M_8(K)$. Its 8-Grassmanian has dimension 64. Each element in this 8-Grassmanian corresponds to a faithful representation of $R_8$, and hence of $A_8$. But the conjugacy action is that of $\mathrm{PGL}_8(K)$ and $\mathrm{PGL}_8$ has dimension 63. This implies, for $K$ being $\mathbf{R}$ or $\mathbf{C}$, that the conjugation action cannot be locally transitive. Hence every algebra in this family has arbitrary close neighbors in the same family, which are not conjugate to it.<|endoftext|> TITLE: Can $\mathcal{L}_{\omega_1,\omega}$ detect $\mathcal{L}_{\omega_1,\omega}$-equivalence? QUESTION [11 upvotes]: Roughly speaking, say that a logic $\mathcal{L}$ is self-equivalence-defining (SED) iff for each finite signature $\Sigma$ there is a larger signature $\Sigma'\supseteq\Sigma\sqcup\{A,B\}$ with $A,B$ unary relation symbols and an $\mathcal{L}[\Sigma']$-sentence $\eta$ such that the following are equivalent for all $\Sigma$-structures $\mathfrak{A},\mathfrak{B}$: $\mathfrak{A}\equiv_\mathcal{L}\mathfrak{B}$. There is a $\Sigma'$-structure $\mathfrak{S}$ such that $\mathfrak{S}\models \eta$, $A^\mathfrak{S}\upharpoonright\Sigma\cong\mathfrak{A}$, and $B^\mathfrak{S}\upharpoonright\Sigma\cong\mathfrak{B}$. For example, Fraisse showed that $\mathsf{FOL}$ is SED (this is used crucially in the proof of Lindstrom's theorem - it's a very happy construction). That same argument gives as a corollary that $\mathsf{SOL}$ is also SED, roughly because $(i)$ "$X$ is the powerset of $Y$" is second-order expressible and $(ii)$ $\mathsf{SOL}$-elementary equivalence between two structures amounts to $\mathsf{FOL}$-elementary equivalence between their "power-structures." The nicest logic whose SED status I don't know is $\mathcal{L}_{\omega_1,\omega}$. This logic isn't powerful enough to perform the same sort of "cheat" as $\mathsf{SOL}$; on the other hand, the direct game-theoretic attack analogizing the situation for $\mathsf{FOL}$ results in a game which is a bit too complicated for $\mathcal{L}_{\omega_1,\omega}$ to handle appropriately (see here, and note that this was also an issue in this earlier question of mine). So my question is: Is $\mathcal{L}_{\omega_1,\omega}$ SED? I'm separately interested in the general situation of which infinitary logics are SED. It's not hard to show that for $\kappa$ satisfying appropriate large cardinal properties we have that $\mathcal{L}_{\kappa,\omega}$ is SED, but I don't see how to extract large cardinal strength from SEDness. Note (per Peter LeFanu Lumsdaine's comment below) that looking at a single sentence $\eta$, as opposed to a theory $E$, is needed to avoid every (regular) logic being trivially SED; I messed this up in the original version of this question. REPLY [11 votes]: (Working in ZFC.) No (re $\mathcal{L}_{\omega_1,\omega}$). Suppose it is. Consider the signature $\Sigma$ with just one binary relation symbol $<$. Let $\Sigma',\eta$ witness SED-ness for $\Sigma$. Let $\pi:M\to V_\eta$ be elementary,with $\eta$ some sufficiently large limit ordinal, $M$ transitive, $M$ of cardinality $\kappa=2^{\aleph_0}$, with $\mathbb{R},2^{\aleph_0}\subseteq M$. So $\mathrm{crit}(\pi)=\kappa^{+M}<\kappa^+$ and $\pi(\kappa^{+M})=\kappa^+$. Let $\mathfrak{A}=(\kappa^{+M},{\in}\upharpoonright\kappa^{+M})$ and $\mathfrak{B}=(\kappa^+,{\in}\upharpoonright\kappa^+)$. Note that $\mathfrak{A}\equiv_{\mathcal{L}}\mathfrak{B}$, since $\pi$ is elementary and all the sentences in $\mathcal{L}$ are in $M$, since $\mathbb{R}\subseteq M$. So by hypothesis, there is some $\Sigma'$-structure $\mathfrak{G}$ such that $\mathfrak{G}\models\eta$, $A^{\mathfrak{G}}\upharpoonright\Sigma\approx\mathfrak{A}$ and $B^{\mathfrak{G}}\upharpoonright\Sigma\approx\mathfrak{B}$. Now let $G$ be $V$-generic for $\mathrm{Coll}(\omega,\max(\kappa^{+},|\mathfrak{G}|))$. In $V[G]$, there is an $\mathcal{L}_{\omega_1,\omega}$ sentence $\varphi$ in the signature $\Sigma$ asserting that the model is (isomorphic to) $\kappa^{+M}$. So in $V[G]$, $\mathfrak{A}\models\varphi$ but $\mathfrak{B}\models\neg\varphi$. So $V[G]$ models the statement "there are countable structures $\mathfrak{A}_0,\mathfrak{B}_0$ in the signature $\Sigma$ and an $\mathcal{L}_{\omega_1,\omega}$-sentence $\varphi_0$ such that $\mathfrak{A}_0\models\varphi_0$ and $\mathfrak{B}_0\models\neg\varphi_0$ and there is a countable structure $\mathfrak{G}_0$ in the signature $\Sigma'$ such that $\mathfrak{G}_0\models\eta$ and $A^{\mathfrak{G}}\upharpoonright\Sigma\approx\mathfrak{A}_0$ and $B^{\mathfrak{G}}\upharpoonright\Sigma\approx\mathfrak{B}_0$". But this statement is $\Sigma^1_2$ in a real coding $\eta$. (I don't see that it is $\Sigma^1_1$ in such a real, because to assert that $\varphi_0$ is really a sentence of $\mathcal{L}_{\omega_1,\omega}$ involves saying that it is built along a real ordinal.) Since $\eta\in V$ and by Shoenfield absoluteness, $V$ models the same statement. But this contradicts our assumptions.<|endoftext|> TITLE: closest equidistant point to N points in M dimensions QUESTION [5 upvotes]: Is there a formula/algorithm/etc. to find the closest equidistant point (assuming it exists) to a set of points, allowing that the number of dimensions of the space is independent of the number of points? Any help would be greatly appreciated. REPLY [4 votes]: If such a point exists, then it is the center of a sphere on which all $N$ points lie, so I think your question is equivalent to asking, do all $N$ points lie on some sphere? Since $M+1$ points determine a sphere in $\mathbb{R}^M$, I think it would be enough to first use $M+1$ points to determine that sphere (for example, as described in a/the answer here: https://math.stackexchange.com/questions/2611326/how-many-points-define-a-sphere-of-unknown-radius), and then check whether the remaining $N-(M+1)$ points are located at the given distance from the given center. In short: the formula/algorithm merely involves solving a linear system of $M+1$ equations, and then checking distances of the remaining points $N-(M+1)$.<|endoftext|> TITLE: Is rationality a deformation invariant property for smooth threefolds? QUESTION [7 upvotes]: Let $X\to S$ be a family of smooth projective complex threefolds over a connected base $S$, could it happen that for some $a, b\in S(\mathbb{C})$, the fiber $X_a$ is birational to $\mathbb{P}^3_{\mathbb{C}}$, while the fiber $X_b$ is not birational to $\mathbb{P}^3_{\mathbb{C}}$? REPLY [6 votes]: I believe this is still an open question in dimension $3$. There is an example due to Hassett, Kresch and Tschinkel (Stable rationality in smooth families of threefolds) of a smooth projective family of threefolds over a connected base where some fibers are stably rational and others are not. Also in dimension at least $4$, there are several examples of families over a connected base where $X_a$ is rational and $X_b$ is not even stably rational (you can find references in the linked paper).<|endoftext|> TITLE: L'Hopital rule for upper and lower limit? QUESTION [11 upvotes]: I am reading the following paper 1998(H.Hudzik) P.574 It reads using L'Hopital rule$$\liminf_{u\to\infty} \frac{1/\varphi(1/u)}{\psi(u)}=\liminf_{u\to\infty}\frac{\varphi'(u)}{\psi'(u)u^2[\varphi(1/u)]^2}.$$ That means we can apply L'Hopital for lower limits i.e. $$\liminf_{u\to\infty} \frac{f(u)}{g(u)}=\liminf_{u\to\infty}\frac{f'(u)}{g'(u)}?$$ But I only know the classical one. Is there someone can give me some reference to check this formula? Or if possible someone can give a proof? REPLY [23 votes]: The full L'Hopital rule says that $$\liminf \frac{f'}{g'}\leq\liminf\frac{f}{g}\leq\limsup\frac{f}{g}\leq\limsup\frac{f'}{g'}.$$ So in the special case when the limit of $f'/g'$, exists, the limit of $f/g$ also exists and is equal to the limit of $f'/g'$. This general rule is proved by integration.<|endoftext|> TITLE: Smooth complete intersections QUESTION [6 upvotes]: Let $X_{2,3}\subset\mathbb{P}^n$, with $n\geq 5$, be a complete intersection of a quadric $X_2$ and a cubic $X_3$ containing a $2$-plane $H$. Assume $X_2$ and $X_3$ to be general among the hypersurface of the same degree containing $H$. In particular $X_2$ and $X_3$ are smooth. Question: If $n = 5$ is $X_{2,3}$ necessarily singular? Is $X_{2,3}$ smooth for $n\geq 6$? REPLY [8 votes]: If $X \subset \mathbb{P}^n$ is a non-degenerate, smooth complete intersection variety of dimension at least $3$, then the restriction map $$\operatorname{Pic}(\mathbb{P}^n) \to \operatorname{Pic}(X)$$ is an isomorphism by Grothendieck-Lefschetz theorem. So, we get $\operatorname{Pic}(X) = \mathbb{Z}$, generated by the hyperplane section. This implies that $X$ contains no linear spaces of codimension $1$; in particular, the threefold $X_{2,3} \subset \mathbb{P}^5$ is necessarily singular. REPLY [6 votes]: If $n=5$ then let $\mathbb P^5$ have coordinates $x_0,\ldots,x_5$ and suppose the plane is $H=\mathbb P^2_{(x_0:x_1:x_2)}$. The two equations of $X$ are necessarily of the form $$ \begin{pmatrix} A_1 & B_1 & C_1 \\ D_2 & E_2 & F_2 \end{pmatrix}\begin{pmatrix} x_3 \\ x_4 \\ x_5 \end{pmatrix} = 0 $$ for some polynomials $A_1,\ldots,F_2\in\mathbb C[x_i]$ of the indicated degree. The scheme $Z\subset \mathbb P^5$ defined by the $2\times 2$ minors of the $2\times 3$ matrix has codimension 2, and thus $Z\cap H$ is nonempty (generically given by some points). Since both equations have order of vanishing $\geq2$ along $(Z\cap H)\subset X$, the 3-fold $X$ must be singular there. In dimension $n\geq6$ the scheme $Z$ (obtained by the analogous argument) has codimension $n-3\geq3$ so in the general case $Z\cap H=\emptyset$ and the $n$-fold $X$ is smooth (along $H$ at least). Essentially, since the matrix never drops rank we can always use the equations to eliminate two of the variables $x_3,\ldots,x_n$, showing that $X$ is smooth of dimension $n$ at every point along $H$.<|endoftext|> TITLE: Some special sequence in $C(\mathbb{R})$ QUESTION [6 upvotes]: Let us consider $C(\mathbb{R})$, the space of continuous functions on the reals. Q. Does there exist a sequence $\{f_n\}$ in $C(\mathbb{R})$ such that for every $f\in C(\mathbb{R})$ one may find a subsequence $\{f_{n_k}\}$ of $\{f_n\}$ pointwise converging to $f$, i.e., $f(x)=\lim f_{n_k}(x)$ for all $x$? REPLY [11 votes]: Yes. Take the countable set $P$ of all polynomials with rational coefficients and enumerate it somehow so that $P=\{p_1,p_2,\dots\}$. Given any $f\in C(\mathbb R)$, for each natural $k$ there exists some natural $n_k$ such that $\sup_{x\in[-k,k]}|p_{n_k}(x)-f(x)|<1/k$. Here without loss of generality we may assume that $n_1 TITLE: Why are quasi-categories better than simplicial categories? QUESTION [20 upvotes]: There are many models for $(\infty,1)$-categories: simplicial categories, Segal categories, complete Segal spaces, and quasi-categories. Doubtlessly the model most used to do higher category theory in is the model of quasi-categories, due to the work of Lurie (Higher Topos Theory, Kerodon), who calls them $\infty$-categories. Just browsing through these books, I noticed that however, simplicial categories are used too, for instance to construct examples of quasi-categories, as in the construction of the quasi-category of spaces: there is a simplicial category of Kan complexes, and to get the quasi-category of Kan complexes we take the homotopy coherent nerve (also called simplicial nerve in HTT I think) of that. Question: If simplicial categories are a more practical model for actually constructing examples, why are quasi-categories used for most of the theory? That is, what advantages do quasi-categories have over simplicial categories that outweigh the complications of constructing quasi-categories directly? REPLY [28 votes]: As a preface, I think that this question should be viewed as analogous to "what are the advantages of ZFC over type theory" or vice versa. We're talking about foundations -- in principle, it doesn't matter what foundations you use; you end up with an equivalent model-independent theory of $\infty$-categories. The paradigm is "use simplicial categories for examples; use quasicategories for general theorems". There are a lot of constructions which are simpler in quasicategories. I tend to think that a lot of the difference is visible at the model category level: the Joyal model structure on $sSet$ is just much nicer to work with than the Bergner model structure on $sCat$ (of course, the latter is still theoretically very important, at the very least for the purposes of importing examples which start life as simplicial categories). Some of these differences are: The Joyal model structure is defined on a presheaf category. The Joyal model structure is cartesian, making it much easier to talk about functor categories. In the Joyal model structure, every object is cofibrant. There's a synergy between (1) and (2) -- if $X$ is a quasicategory and $A$ is any simplicial set, then the mapping simplicial set $Map(A,X)$ gives a correct model for the functor category from $A$ to $X$ -- you don't need to do any kind of cofibrant replacement of $A$. This is nicely explained in Justin Hilburn's answer. (2) is quite convenient. For example, in general frameworks like Riehl and Verity's $\infty$-cosmoi, a lot of headaches are avoided by assuming something like (2). Here are a few examples of some things which are easier in quasicategories -- I'd be curious to hear other examples folks might mention! The join functor is very nice. Consequently (in combination with the nice mapping spaces), limits and colimits can be defined pretty cleanly. An example of a theorem proven in HTT using quasicategories which I imagine would be hard to prove (maybe even to formulate) directly in simplicial categories is the theorem that an $\infty$-category with products and pullbacks has all limits. The proof uses the fact that the nerve of the poset $\omega$ is equivalent to a 1-skeletal (non-fibrant) simplicial set, and relies on knowing how to compute co/limits indexed by non-fibrant simplicial sets like this. The theory of cofinality is very nice, arising from the (left adodyne, left fibration) weak factorization system on the underlying category -- I imagine it would be quite complicated with simplicial categories. Roughly at this point in the theory, though, one starts to have enough categorical infrastructure available that it becomes more possible to think "model-independently", and the differences start to matter less. Here's a few more: When you take the maximal sub-$\infty$-groupoid of a quasicategory, it is literally a Kan complex, ready and waiting for you to do simpicial homotopy with. This is especially nice when you take the maximal sub-$\infty$-groupoid of a mapping object -- which doesn't quite make the model structure simplicial, but it's kind of "close". The theory of fibrations is pretty nice in quasicategories -- just like in ordinary categories, left, right, cartesian, and cocartesian fibrations are "slightly-too-strict" notions which are very useful and have nice properties like literally being stable under pullback. I don't know what the theory of such fibrations looks like in simplicial categories. The fact that $sSet$ is locally cartesian closed is sneakily useful. Even though $Cat_\infty$ is not locally cartesian closed, there's a pretty good supply of exponentiable functors, and it's not uncommon to define various quasicategories using the right adjoint to pullback of simplicial sets. (Rule of thumb: in HTT, when Lurie starts describing a simplicial set by describing its maps in from simplices over a base, 90% of the time he's secretly describing the local internal hom of simplicial sets.) REPLY [6 votes]: In order to compute the correct mapping space $\text{Hom}(x,y)$ in a model category you have to replace $x$ with a cofibrant object and $y$ with a fibrant object. In the Joyal model structure on simplicial sets everything is cofibrant and quasi-categories are fibrant. In the Bergner model structure on simplicial categories most naturally occurring objects are not fibrant or cofibrant.<|endoftext|> TITLE: Given $f$ continuous, find $g$ smooth such that $g→0$ when $f→0$ and vice versa QUESTION [11 upvotes]: Let $X$ be a smooth manifold (where, throughout, “smooth” means “$C^\infty$”). Consider the following statement: For all $f\colon X\to\mathbb{R}$ continuous there exists $g\colon X\to\mathbb{R}$ smooth such that both of the following hold: $\forall \varepsilon>0.\; \exists \delta>0.\; \forall x\in X.\; (|f(x)|<\delta \Rightarrow |g(x)|<\varepsilon)$ $\forall \varepsilon>0.\; \exists \delta>0.\; \forall x\in X.\; (|g(x)|<\delta \Rightarrow |f(x)|<\varepsilon)$ Question: Is this statement true? If so, what is a proof? Is it a standard fact? If so, what is a reference? Comments: The statement is of the form $\forall f. \exists g. \forall \varepsilon$. This is not an error. (I state this explicitly, because everyone I asked initially seems to want to read it as $\forall f. \forall \varepsilon. \exists g$ instead.) With a slight abuse of notation/terminology, we might rephrase the question as: “for every continuous $f$, there exists $g$ smooth such that both $g\to 0$ when $f\to 0$ and conversely”. This explains the title of this question and I hope helps to make it seem more natural. For the significance of the conditions on $g$, see this answer (note that we can assume w.l.o.g. that $f$ is bounded, and we can demand w.l.o.g. that $g$ is: then they say $Z(\tilde f) \subseteq Z(\tilde g)$ and $Z(\tilde f) \supseteq Z(\tilde g)$ respectively, where $\tilde f,\tilde g$ are the continuous extensions of $f,g$ to the Stone-Čech compactification $\beta X$ of $X$). Two special cases of the above statement are worth noting (and are indeed proven as follows): If $X$ is compact, then the conditions on $g$ simply state that $g$ vanishes exactly when $f$ does (viꝫ. $Z(g) = Z(f)$), i.e., the statement is that every zero-set the zero-set of a smooth function. This is indeed the case: see here and here (I don't know a printed reference even for this particular case, so if you know one, please suggest). If $f$ does not vanish, then the statement can be proved as follows: we can assume w.l.o.g. that $f>0$ (by considering the clopen sets $\{f>0\}$ and $\{f<0\}$ separately). By a standard approximation theorem (Hirsch, Differential Geometry (1976, Springer GTM 33), chapter 2, theorem 2.2 on page 44), there exists a smooth function $v$ such that $|u-v|<1$ on $X$, where $u := \log f$: then $0 < C_1 f < g < C_2 f$ for some constants $00\exists\delta>0\forall x\in X(|f(x)|<\delta\rightarrow|g(x)<\epsilon|)\wedge (|g(x)|<\delta\rightarrow|f(x)<\epsilon|).$ See my answer to the previous question for an explanation of this notation. I claim that for every paracompact smooth manifold $X$ and continuous function $f:X\rightarrow\mathbb{R}$, there is a smooth function $h:X\rightarrow\mathbb{R}$ such that $Z(\overline{f})=Z(\overline{h}).$ Suppose that $X$ is a manifold, and $f:X\rightarrow[0,\infty)$ is continuous (we can assume $0\leq f(x)$ everywhere because we can just replace $f$ with $x\mapsto\min(1,|f(x)|)$). Then whenever $c_{1}0$, and multi-index $\alpha$, the function $D^{\alpha}G$ is bounded on the set $\{c\in C\mid d(c,C\cap\partial Z(G))\geq\delta\}$. Therefore, let $M(\iota,C,\alpha,\delta)$ be the maximum of $|D^{\alpha}G|$ on the set $\{c\in C\mid d(c,C\cap\partial Z(G))\geq\delta\}$. For each multi-index $\alpha$, there are polynomials $P_{\alpha,0},\dots,P_{\alpha,k}$ with non-negative coefficients where $D^{\alpha}H\circ G(\mathbf{x})=\sum_{k=0}^{|\alpha|}H^{(k)}(G(\mathbf{x}))P_{\alpha,k}((G_{\beta}(\mathbf{x}))_{\beta\leq\alpha})$ (a more explicit computation of these polynomials is obtained from the Faà di Bruno's formula). Therefore, $$|D^{\alpha}H\circ G(\mathbf{x})|\leq \sum_{k=0}^{|\alpha|}|H^{(k)}(G(\mathbf{x}))|\cdot P_{\alpha,k}((M(\iota,C,\beta,d(x,C\cap Z(G))))_{\beta\leq\alpha}).$$ Now, let $\omega:[0,\infty)\rightarrow[0,\infty)$ be a continuous increasing function such that $|G(x)-G(y)|\leq\omega(\|x-y\|)$ whenever $x,y\in C$. In particular, if $\|x-c\|=d(x,C\cap Z(G))$, then $$|G(x)|=|G(x)-G(c)|\leq\omega(\|x-c\|)=\omega(d(x,C\cap Z(G))).$$ Suppose now that for each $n$, there is an open set $U_{n}\subseteq\mathbb{R}^{n}$, a compact set $C_{n}$ and a chart $\iota_{n}:U_{n}\rightarrow X$ such that $X=\bigcup_{n}\iota[C_{n}^{\circ}]$. For simplicity, assume that $\iota[C_{n}^{\circ}]\cap Z(g)\neq\emptyset$ for all $n$. For each $n$, let $\omega_{n}:[0,\infty)\rightarrow[0,\infty)$ be a continuous increasing function with $\omega_{n}(0)=0$ ​and where $|(g\circ\iota_{n})(x)|\leq\omega_{n}(d(x,C_{n}\cap Z(g\circ\iota_{n}))$ whenever $x\in C_{n}$. Therefore, $$|D^{\alpha}H\circ g\circ\iota_{n}(x)|\leq \sum_{k=0}^{|\alpha|}|H^{(k)}\circ g\circ\iota_{n}(x)|\cdot P_{\alpha,k}((M(\iota_{n},C_{n},\beta,d(x,C_{n}\cap Z(g\circ\iota_{n}))))_{\beta\leq\alpha}).$$ By the above lemma, one can choose the function $H$ so that $$\lim_{x\rightarrow 0^{+},y\rightarrow 0,|y|\leq\omega_{n}(|x|)}\sum_{k=0}^{|\alpha|}|H^{(k)}(y)|\cdot P_{\alpha,k}(M(\iota_{n},C_{n},\beta,x)))|x|^{-1}=0.$$ In this case, if $x_{0}\in C_{n}\cap Z(g\circ\iota_{n})$, then $$\lim_{x\rightarrow x_{0}}\sum_{k=0}^{|\alpha|}|H^{(k)}\circ g\circ\iota_{n}(x)|\cdot P_{\alpha,k}((M(\iota_{n},C_{n},\beta,d(x,C_{n}\cap Z(g\circ\iota_{n}))))_{\beta\leq\alpha})\cdot\|x-x_{0}\|^{-1}=0.$$ Therefore, $$\lim_{x\rightarrow x_{0}}|D^{\alpha}(H\circ g\circ\iota_{n})(x)|\cdot\|x-x_{0}\|^{-1}=0.$$ We conclude that $H\circ g\circ\iota_{n}$ is smooth on $C_{n}^{\circ}\cap Z(G)$. Therefore, $H\circ g\circ\iota_{n}$ is smooth on $C_{n}^{\circ}$, so $H\circ g$ is smooth on $\iota[C_{n}^{\circ}]$. We conclude that $H\circ g$ is smooth everywhere. Q.E.D.<|endoftext|> TITLE: Pólya–Vinogradov inequality for Eisenstein integers QUESTION [6 upvotes]: The Pólya–Vinogradov inequality asserts that a non-principal Dirichlet character $\chi$ with modulus equal to $q$ satisfies $$\displaystyle \left \lvert \sum_{N < n < N+M} \chi(n) \right \rvert = O \left(\sqrt{q} \log q \right)$$ for all positive integers $N$, $M$. My question concerns changing the summation condition to run over Eisenstein integers having bounded norm, and the character to a primitive cubic character given by the cubic residue symbol of some element $w \in \mathbb{Z}[\zeta_3]$, where $\zeta_3$ is a primitive third root of unity. Specifically, I want to understand the sum $$\displaystyle \left \lvert \sum_{A < N(z) < A + B} \left(\frac{z}{w} \right)_3 \right \rvert$$ where $\left(\frac{\cdot}{\cdot} \right)_3$ is the cubic residue symbol over the Eisenstein integers and $N(\cdot)$ is the norm on $\mathbb{Z}[\zeta_3]$. Can a bound of $O(\sqrt{N(w)} \log N(w))$ be obtained as in the case of running over rational integers? REPLY [10 votes]: No. Such a bound would imply a similar bound on $$\displaystyle \left \lvert \sum_{N(z) = M} \left(\frac{z}{w} \right)_3 \right \rvert.$$ If $M$ is a product of distinct primes $p_1,\dots p_n$ congruent to $1$ mod $3$, the norms of primes $\pi_1,\dots,\pi_n$ then $$\sum_{N(z) = M} \left(\frac{z}{w} \right)_3 = \left( \left(\frac{1}{w} \right)_3 + \left(\frac{-1 }{w} \right)_3 \right) \left( \left(\frac{1}{w} \right)_3 + \left(\frac{\omega }{w} \right)_3 + \left(\frac{\omega^2 }{w} \right)_3 \right) \prod_{i=1}^n \left( \left(\frac{\pi_i}{w} \right)_3 + \left(\frac{\overline{\pi_i} }{w} \right)_3 \right)$$ so if we choose $w$ such that $\left(\frac{-1 }{w} \right)_3 =\left(\frac{\omega }{w} \right)_3 =1$ and then choose $n$ different primes $\pi$ such that $ \left(\frac{\pi_i}{w} \right)_3 =\left(\frac{\overline{\pi_i} }{w}\right)_3$ then this will have size $6 \cdot 2^n$. Taking $n$ sufficiently large, we contradict any bound like the one you request. To sketch a proof of a bound, let's first understand $$\left \lvert \sum_{z} \phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) \left(\frac{z}{w} \right)_3 \right \rvert$$ where $\phi$ is a smooth bump function. The trivial bound is $O(RT)$ as we are summing over an annulus of radius $R$ and thickness $T$. Another bound is provided by integrating the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ against the Fourier transform of $\left(\frac{z}{w} \right)_3 $. Since the Fourier transform of $\left(\frac{z}{w} \right)_3 $ is a bunch of Gauss sums, the same at every point, the bound we get this way will be $\sqrt{N(w)}$ times the $L^1$ norm of the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $. For $T$ small, the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ will look like the integral of a phase over a circle, i.e. a Bessel function, thus, viewed as a function of a complex variable $x$, approximately constant for $|x| < R^{-1}$ and proportial to $1/\sqrt{x}$ for larger values of $x$. The bump function will give us a rapidly decreasing cutoff at a radius of $1/T$. Since the value at $0$ should be $RT$, the Fourier transform will take the value $\approx RT$ for $|x| TITLE: Determinant connection between Schur polynomials and power sum polynomials QUESTION [9 upvotes]: Let $f_i=f_i(x_1,x_2,\ldots, x_n),i=0,1,2, \ldots $ be a family of symmetric polynomials. For the partition $\lambda=(\lambda_1,\lambda_2, \ldots, \lambda_n)$ consider the determinant $$ D_\lambda(f)=\left | \begin{array}{lllll} f_{\lambda_1} & f_{\lambda_1+1} & f_{\lambda_1+2} & \ldots & f_{\lambda_1+n-1}\\ f_{\lambda_2-1} & f_{\lambda_2} & f_{\lambda_2+1} & \ldots & f_{\lambda_2+n-2}\\ \vdots & \vdots & \vdots & \ldots & \vdots \\ f_{\lambda_n-(n-1)} & f_{\lambda_n-(n-2)} & f_{\lambda_n-(n-3)} & \ldots & f_{\lambda_n} \end{array} \right|. $$ It is well known (Jacobi−Trudi formulas) that for the elementary symmetric polynomials $e_i=e_i(x_1,x_2,\ldots, x_n)$ and for the complete homogeneous symmetric polynomials $h_i=h_i(x_1,x_2,\ldots, x_n)$ we have $$ D_\lambda(h)=s_{\lambda}(x_1,x_2,\ldots, x_n) \text{ and } D_\lambda(e)=s_{\lambda'}(x_1,x_2,\ldots, x_n), $$ where $s_{\lambda}(x_1,x_2,\ldots, x_n)$ is the Schur polynomial and $\lambda'$ is the conjugate partition. Question. Is there a similar expression for $D_\lambda(p)$ where $p_i=p_i(x_1,x_2,\ldots, x_n)$ is the power sum symmetric polynomials? By direct calculation for $n=2, \lambda_2\geq 1$ I got that $$D_{(\lambda_1,\lambda_2)}(p)=-s_{(\lambda_1-1,\lambda_2-1)} V(x_1,x_2)^2$$ and for $n=3,\lambda_3\geq 2$ $$D_{(\lambda_1,\lambda_2,\lambda_3)}(p)=-\frac{s_{(\lambda_1-1,\lambda_2-1,\lambda_3-1)}}{s_{(1,1,1)}} V(x_1,x_2,x_3)^2$$ and so on. Here $V$ is the Vandermonde determinant. I hope that must be a nice formula and for any $n$. REPLY [3 votes]: I don't know a fully general result, but your pattern for partitions $\lambda$ of length $\leq n$ with $n$-th entry $\lambda_{n}\geq n-1$ and with $n$ indeterminates persists: Theorem 1. Let $n$ be a positive integer. Let $\lambda=\left( \lambda _{1},\lambda_{2},\ldots,\lambda_{n}\right) $ be an integer partition with at most $n$ parts. Assume that $\lambda_{n}\geq n-1$. Consider polynomials in $n$ indeterminates $x_{1},x_{2},\ldots,x_{n}$. For each nonnegative integer $k$, we set \begin{align*} p_{k}:=x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}. \end{align*} (This is the $k$-th power-sum symmetric polynomial in $x_{1},x_{2} ,\ldots,x_{n}$ when $k>0$. We have $p_{0}=n$.) Define the $n\times n$-matrix \begin{align*} P:=\left( p_{\lambda_{i}-i+j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}=\left( \begin{array} [c]{cccc} p_{\lambda_{1}} & p_{\lambda_{1}+1} & \cdots & p_{\lambda_{1}+n-1}\\ p_{\lambda_{2}-1} & p_{\lambda_{2}} & \cdots & p_{\lambda_{2}+n-2}\\ \vdots & \vdots & \ddots & \vdots\\ p_{\lambda_{n}-n+1} & p_{\lambda_{n}-n+2} & \cdots & p_{\lambda_{n}} \end{array} \right) . \end{align*} Let $\mu=\left( \mu_{1},\mu_{2},\ldots,\mu_{n}\right) $ be the partition defined by \begin{align*} \mu_{i}=\lambda_{i}-\left( n-1\right) \ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{ 1,2,\ldots,n\right\} . \end{align*} (This is indeed a partition, since $\mu_{n}=\underbrace{\lambda_{n}}_{\geq n-1}-\left( n-1\right) \geq0$.) Let $s_{\mu}$ be the corresponding Schur polynomial in the $n$ indeterminates $x_{1},x_{2},\ldots,x_{n}$. Furthermore, let \begin{align*} V_{n}:=\prod_{1\leq i TITLE: Are there highly composite prime gaps? QUESTION [6 upvotes]: Definition: Highly composite prime gap The three composite numbers between the consecutive primes $643$ and $647$ each have at least three distinct prime factors. This is the first occurrence of prime gap of length $> 1$ where each composite number in the gap has at least $3$ distinct prime factors. We call prime gap between $643$ and $647$ as the highly composite prime gap of order $3$. We have the highly composite prime gaps for of order $k$ for $k = 3,4,5,6$ and $7$ as follows: $k = 3; p = 643$ $k = 4; p = 51427$ $k = 5; p = 8083633$ $k = 6; p = 1077940147$ $k = 7; p = 75582271489$ Question 1: Are there infinitely many highly composite prime gaps of order $k \ge 3$? Question 2: Given $k$ is there always a highly composite gap of order $k$? An ordinary linear regression between $k$ and $\log p$ gives a surprisingly strong fit with $R^2 \approx 0.99915$. Although it is based on only six data points, this suggests a relationship of the form $p \sim ab^k$ forsome fixed $a$ and $b$. Definition: Maximal highly composite gap The maximal highly composite gap is defined as a prime gap which is longer than any previous gap and each composite in the gap has at least $3$ distinct prime factors. The longest such gap I have found is of $75$ consecutive composite between the primes $535473480007$ and $535473480083$. Question 3: Are there arbitrarily long prime gaps in which each composite number in the gap has at least three distinct prime factors? Note: This question was posted in MSE six months ago; it got many votes but not answer hence posting in MO. REPLY [11 votes]: Assuming the prime tuples conjecture, all of these questions have affirmative answers. For instance, one can use the Chinese remainder theorem to find $a,b$ such that the tuple $$ an+b, \frac{an+b+1}{2^2 \times 5}, \frac{an+b+2}{3 \times 7}, \frac{an+b+3}{2 \times 11}, an+b+4$$ (for instance) are an admissible tuple of linear forms (in that they have integer coefficients, and for each prime $p$ there exist a choice of $n$ in which all forms are simultaneously coprime to $p$); concretely, one can take $a=2^2\times 3 \times 5 \times 7 \times 11 = 4620$ and $b=19$. The tuples conjecture then shows that there are infinitely many $n$ in which these forms are all simultaneously prime, giving infinitely many gaps of order $3$ and length $4$. Similarly for other gap orders and lengths. On the other hand, unconditionally not much can be said. I think at our current level of understanding we cannot even rule out the ridiculous scenario in which every sufficiently large prime gap $(p_n,p_{n+1})$ contains a semiprime. (Given that semiprimes are denser than the primes, it is likely that most prime gaps contain a semiprime, but it is highly unlikely (and inconsistent with the prime tuples conjecture) that all of them do.) The "bounded gaps between primes" technology of Goldston-Yildirim-Pintz, Zhang, Maynard, and Polymath does provide prime gaps of short length, but the method also inevitably produces several almost primes in the vicinity of such gaps, and with our current techniques we cannot prevent one of these almost primes being a semiprime inside the prime gap.<|endoftext|> TITLE: Analogue of Bockstein for crossed module extensions and higher Steenrod square QUESTION [8 upvotes]: It is well known that in $\mathbb{Z}_2$-valued simplicial cohomology (and other cohomologies) $$ Sq^1 = \beta\;,$$ where $Sq^1$ is the first Steenrod square and $\beta$ is the Bockstein homomorphism for the short exact sequence $$ \mathbb{Z}_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}_2\;.$$ I was wondering whether there is a way to also interpret $Sq^2$ or even higher Steenrod squares in a similar fashion. One could think that instead of $$\beta = g^{-1} d h^{-1}$$ for a short exact sequence of abelian groups $$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C$$ one could try something like $$g^{-1}dh^{-1}di^{-1}$$ for an exact sequence $$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C\overset{i}{\rightarrow} D\;.$$ However, if I'm not mistaken, the long exact sequence splits into two short exact sequences, and the above is just the product of the two corresponding Bockstein homomorphisms. E.g., when applied to the long exact sequence $$ \mathbb{Z}_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}_2$$ we get $\beta^2$ for the short exact sequence given in the beginning, which is trivial. Crossed module extensions are a more non-trivial analogue to ordinary group extensions which also come with an action of $C$ on $B$. It is known that crossed module extensions give rise to group 3-cocycles in $H^3(BD, A)$ just as ordinary extensions give rise to 2-cocycles. If $D$ acts trivially on $A$, the group 3-cocycle can be used to map a (simplicial) 1-cocycle to a 3-cocycle, and the "higher Bockstein" should generalize this to a map from $i$-cocycles to $i+2$-cocycles. However, I'm not sure how to use the action of $C$ on $B$ to define sich a higher Bockstein. So I basically have two questions: 1) Is there a non-trivial analogue of a Bockstein operation for some sort of longer exact sequences/crossed module extensions? 2) If yes, is any of these operations equal to $Sq^2$ (in general, or maybe at least when applied to a specific degree)? REPLY [11 votes]: $\DeclareMathOperator{\Sq}{Sq}\newcommand{\Z}{\mathbb{Z}}$The short version is that every cohomology operation can be interpreted as a Bockstein operator for an "exact sequence" (read: fiber sequence) of grouplike $E_\infty$ spaces. Any cohomology operation $\delta:H^*(-;A)\Rightarrow H^{*+k}(-;B)$ such as $\Sq^i$ gives a morphism of Eilenberg-MacLane spectra $f_\delta:HA\to \Sigma^k HB$. You can take the fiber of $f_\delta$ to obtain a spectrum $F$ which defines a generalized cohomology theory $E_F^*:X\mapsto \pi_{-*} F^X$, and there is a resulting long exact sequence $$ \dots\to E_F^i(X)\to H^i(X;A)\xrightarrow{\delta}H^{i+k}(X;B)\to E_F^{i+1}(X)\to\dots $$ If $\delta = \Sq^1$, the fiber is again an Eilenberg-MacLane spectrum (namely $F\simeq H\Z/4$). If the degree of $\delta$ is bigger than $1$, it will have two non-zero homotopy groups, namely $A$ in degree $0$ and $B$ in degree $k-1$. The connection to the crossed modules you mention is that applying the functor $\Omega^{\infty-1}$ to the fiber of $\Sq^2$ gives rise to a $2$-group, i.e. a homotopy type $X$ whose homotopy groups vanish outside degrees $1$ and $2$. As you mention, these are classified by the two groups $A = \pi_1(X),B = \pi_2(X)$, the action of the former on the latter, and a $k$-invariant in $H^3(A;B)$, which together can be packaged into the datum of a crossed module. However, these deloop once if and only if the action is trivial and the $k$-invariant vanishes. It is still possible to find reasonably easy algebraic models for spaces whose homotopy groups vanish outside degrees $k$ and $k+1$ (given by braided ($k=2$) and symmetric ($k\ge 3$) monoidal Picard (every object has a tensor inverse) groupoids), although I do not know a definition of $\Sq^2$ in this language. Questions in the comments You already discuss the resulting cohomology operation in the case that the action of $A$ on $D$ is trivial. For the general case, one first construts a natural transformation from $H^1(-;A)$ to local systems: $H^1(-;A)$ are isomorphism classes of $A$-principal bundles, and this natural transformation sends a principal bundle $P$ to $P\times_A D$. A cohomology class in $H^k(A;D)$ then gives a natural transformation from $H^1(-;A)$ to $H^k$ of this local system, by the same construction as in the trivial case. Yes, for a $(2,3)$-type $X$ the $2$-group $\Omega X$ is split, i.e. the action of $\pi_1(\Omega X)\cong \pi_2(X)$ on $\pi_2(\Omega X)\cong \pi_3(X)$ is trivial (this can be shown by the Eckmann-Hilton argument) and the $k$-invariant vanishes. For degree $2$ cohomology operations, there is in fact a complete classification (compare (Co)homology of the Eilenberg-MacLane spaces K(G,n) and the cited references): operations $H^2(-;A)\to H^4(-;B)$ are given by quadratic functions $q:A\to B$ (i.e. such that $q(x+y) - q(x) - q(y)$ is bilinear and $q(kx) = k^2q(x)$), and the resulting cohomology operation is given by a suitable version of the Pontryagin square. For $k\ge 3$, operations are in bijection with linear maps from $A\otimes \Z/2$ to $B$ (observe that such a linear map is also quadratic by the "freshman's dream"), and the resulting cohomology operation is the composition $$ H^*(-;A)\to H^*(-;A\otimes Z/2)\xrightarrow{\Sq^2} H^{*+2}(-;A\otimes\Z/2)\to H^{*+2}(-;B) $$ The relation to Picard groupoids is a consequence of the Homotopy hypothesis, and given a braided Picard groupoid $C$, you can associate to it its abelian group $\pi_0 C$ of isomorphism classes and the (abelian!) group $\pi_1 C$ of automorphisms of the unit $1$, together with the map $q: \pi_0 C\to \pi_1 C$ which sends $x$ to the composition $$ 1\cong x\otimes x^{-1}\xrightarrow{\beta_{x,x^{-1}}} x^{-1}\otimes x\cong 1 $$ It's a fun exercise to show that $q$ is quadratic in the above sense. It is not straightforward to give an inverse to this construction, i.e. construct the braided Picard groupoid from the quadratic map; for a reference in the symmetric setting, see Cegarra, A. M.; Khmaladze, E., Homotopy classification of graded Picard categories, Adv. Math. 213 (2007). A chain level representative of $\Sq^2$ (and higher Steenrod squares) can be found in Ralph M. Kaufmann, Anibal M. Medina-Mardones. Cochain level May-Steenrod operations.<|endoftext|> TITLE: Is $\prod_{i=1}^\infty (1-\frac{1}{2^{(2^i)}})$ transcendental? QUESTION [15 upvotes]: Motivation. In a coin game, a player flips all their coins every turn, starting with just one coin. If the coins all land heads then the game stops; otherwise, the number of coins is doubled for the following turn. While the game may clearly terminate on any turn, there is in fact a positive probability that it will never terminate: this is the infinite product $$p = \prod_{i=1}^\infty \left(1-\frac{1}{2^{(2^i)}}\right).$$ Questions. Do we have $p \in \mathbb{R}\setminus \mathbb{Q}$? Is $p$ transcendental? REPLY [27 votes]: This number is irrational. By Euler's pentagonal number theorem, we have $$\prod_{i=1}^{\infty} (1-2^{-i}) = 1 + \sum_{k=1}^{\infty} (-1)^k \left( 2^{-k(3k+1)/2} +2^{-k(3k-1)/2} \right).$$ This shows that the binary expansion is clearly not periodic. I have no idea whether this number is transcendental. Your number is closely related to the value of various modular forms like the Dedekind eta function at $\tau = \tfrac{\log 2}{2 \pi} i$, so $q = \tfrac{1}{2}$. Chapter 27 of Transcendental numbers, by Murty and Rath, discusses evaluating modular forms when $\tau$ is an algebraic number, but a quick search didn't show me anything about evaluating modular forms when $q$ is algebraic. REPLY [18 votes]: I can prove that the number actually described by the word problem, which is $$ \prod_{i=0}^{\infty} \left( 1- \frac{1}{2^{2^i}} \right),$$ is irrational, by a method similar to David Speyer's. Expanding out the product, we get $$\sum_{j=0}^{\infty} \frac{(-1)^{t_j} }{2^j}= 1+\sum_{j=1}^{\infty} \frac{(-1)^{t_j} }{2^j} = \sum_{j=1}^{\infty} \frac{1}{ 2^j} + \sum_{j=1}^{\infty} \frac{(-1)^{t_j} }{2^j} = \sum_{j=1}^{\infty} \frac{1+ (-1)^{t_j} }{2^j} = \sum_{j=1}^{\infty} \frac{1+ (-1)^{t_j} }{2} \cdot 2^{1-j} $$ where $t_j$ is the number of $1$s in the binary expansion of $j$. Thus the binary digit in the $j-1$th place is $1$ if $t_j$ is even and $0$ if $t_j$ is odd. Since this sequence is not periodic, the number is irrational. REPLY [9 votes]: Q: Is $\prod_{i=1}^\infty(1-2^{-i})$ transcendental? The question as posed originally by the OP. A: This is the Euler function $\phi(1/2)$, see https://oeis.org/A048651 --- I don't think it is known whether it is transcendental or not [as is the case for other instances of $\phi(q)$].<|endoftext|> TITLE: Ultrafilters of closed sets QUESTION [5 upvotes]: The following definition should be standard, but let me state it just in case there is some ambiguity: If $\mathscr{L}$ is a set of subsets of a set $X$ that is closed under finite unions and intersections and contains $\varnothing,X$ (or more generally, if $\mathscr{L}$ is a distributive lattice with top and bottom elements), let us say that $\mathscr{F} \subseteq \mathscr{L}$ is a filter in $\mathscr{L}$ provided it contains $X$ (the top element), is closed under finite intersections (meets), and is an up-set (meaning that if $A \subseteq B$ with $A \in \mathscr{F}$ and $B \in \mathscr{L}$ then in fact $B \in \mathscr{F}$). Let us say that such a filter is proper iff it does not contain $\varnothing$; and that it is an ultrafilter iff it is proper and maximal for inclusion among proper filters. If $\operatorname{Ult}(\mathscr{L})$ denotes the set of ultrafilters in $\mathscr{L}$, then for every $A\in\mathscr{L}$ we define a set $Z(A) := \{\mathscr{U}\in\operatorname{Ult}(\mathscr{L}) : \mathscr{U} \ni A\}$ of ultrafilters containing $A$ as an element. It is almost trivial that $Z(A\cap B) = Z(A) \cap Z(B)$, and it is also true, though slightly less trivial, that $Z(A\cup B) = Z(A) \cup Z(B)$ (sketch of proof: the inclusion $\supseteq$ is clear, so let us see $\subseteq$: if $\mathscr{U} \ni A\cup B$ and $\mathscr{U} \not\ni A$ then the filter generated by $\mathscr{U}$ and $\{A\}$ contains $\varnothing$, so $A\cap U = \varnothing$ for some $U\in\mathscr{U}$, and then $\mathscr{U} \ni (A\cup B)\cap U = B\cap U \subseteq B$ so $\mathscr{U}\ni B$). In particular, the $Z(A)$ form a basis of closed sets for a topology on $\operatorname{Ult}(\mathscr{L})$ called the Zariski topology. (Digression: note that they also form a basis of open sets for another topology, the Stone topology; I mention this in passing, because this confused the hell out of me: part of the confusion comes from the fact that, if $\mathscr{L}$ is actually a Boolean algebra, — e.g., if we consider all subsets of $X$, or more generally the clopen subsets of a topological space $X$, — then the complement of $Z(A)$ is $Z(X\setminus A)$ so the two topologies coincide.) As an example, if $X$ is a topological space and $\mathscr{Z}$ is the lattice of zero-sets of continuous real-valued functions on $X$ (“z-sets”), then $\operatorname{Ult}(\mathscr{Z})$, with its Zariski topology, is the Stone-Čech compactification of $X$ (Gillman & Jerison, Rings of Continuous Functions (1960), points (a) and (b) in the proof of theorem 6.5). This example has long bothered me because z-sets seem to make a fairly incongruous appearance, and I wondered why we don't take closed sets instead, which seem more “fundamental”, and what happens if we do. Let me ask precisely that: Question: if $X$ is a topological space, $\mathscr{C}$ is the lattice of closed sets of $X$, can we better describe the space $\operatorname{Ult}(\mathscr{C})$ of ultrafilters in $\mathscr{C}$, endowed with its Zariski topology? Does it have a name? What is “wrong” with it that makes it less interesting than the Stone-Čech compactification? (Note that if $X$ is metric, then closed sets and z-sets coincide, so the above space is the Stone-Čech compactification.) Maybe assume that $X$ is $T_1$, which ensures that we have a continuous map $X \to \operatorname{Ult}(\mathscr{C})$ taking $x\in X$ to the set of $F \subseteq X$ closed such that $F\ni x$. Remark: For the description of the space of ultrafilters of open sets (with the Stone topology), see this answer on MSE, where I prove that it is the “Gleason space” of $X$. REPLY [4 votes]: The construction you describe when $\mathscr{C}$ consists of all closed sets of $X$ is known as the Wallman compactification of $X$. I'll denote if $\omega(X)$. It is due to Wallman; Lattices and topological spaces, Ann. Math. 39 (1938) 112-126. Of course some sort of techincal assumption is required. Let $X$ be a $T_1$ space. Then $\omega(X)$ is a compact $T_1$ space containing $X$ as a dense subspace. Moreover it has the property that every continuous map $X\rightarrow K$ into a compact Hausdorff space $K$ extends over $\omega(X)$. The space $\omega(X)$ is Hausdorff if and only if $X$ is normal, and in this case $\omega(X)\cong\beta(X)$. Shanin later generalised Wallman's construction; On special extensions of topological spaces, Dokl. SSSR 38 (1943) 6-9, On separation in topological spaces, Dokl. SSSR 38 (1943) 110-113, On the theory of bicompact extensions of topological spaces, Dokl. SSSR 38 (1943) 154-156. The compactifications that Shanin constructed allowed for the ultrafilters to come from more general lattices $\mathscr{L}$ of closed subsets of $X$. Of course at the expense of added assumptions: $\mathscr{L}$ is required to be a so-called $T_1$-base for the closed subsets of $X$. Denote by $\omega(X;\mathscr{L})$ the Wallman-Shanin compactification built using the $T_1$-base $\mathscr{L}$. Here are some examples to convince you that these compactifications are interesting. $X$ is locally compact $T_2$ and $\mathscr{L}$ consists of all $(i)$ compact subsets of $X$, and $(ii)$ all closed subsets $A\subseteq X$ for which there is a compact $K\subseteq X$ with $A\cup K=X$. Then $\omega(X;\mathscr{L})$ is the Alexandroff compactification of $X$. $X$ is Tychonoff and $\mathscr{L}=\mathscr{Z}(X)$ is the collection of zero sets. Then $\omega(X;\mathscr{L})\cong\beta(X)$, as you have recognised. $X$ is rim-compact $T_2$ and $\mathscr{L}$ is the set of all finite intersections of regularly closed sets with compact boundaries. Then $\omega(X;\mathscr{L})=\mathfrak{f}(X)$ is the Freudenthal compactification of $X$.<|endoftext|> TITLE: Alternating subgroups of $\mathrm{SU}_n $ QUESTION [11 upvotes]: $\DeclareMathOperator\PSU{PSU}$Let $ \PSU_n $ be the projective unitary group. Let $ A_m $ be the alternating group on $ m $ letters. $ \PSU_2 $ contains a subgroup isomorphic to $ A_5 $. It is the largest of the exceptional finite subgroups of $ \PSU_2 $ and it is maximal (the only closed subgroup containing it is the whole group). The references in The finite subgroups of SU(n) show that $ \PSU_3 $ contains a subgroup isomorphic to $ A_6 $ and that again this subgroup is the largest of the exceptional finite subgroups. Again it is maximal. EDIT: The bolded text is from the original question but is wrong: "The reference https://arxiv.org/abs/hep-th/9905212 from the same MO question shows that $ A_7 $ is a subgroup of $ \PSU_4 $. It seems that $ A_7 $ is maximal in $ \PSU_4 $ as well. That leads me to ask: Is $ A_{n+3} $ always a maximal closed subgroup of $ \PSU_n $?" The following text is correct: The reference https://arxiv.org/abs/hep-th/9905212 from the same MO question shows that $ A_7 $ is a maximal subgroup of $ SU_4/C_2\cong SO_6(\mathbb{R}) $. However it seems that $ A_7 $ is not a subgroup of $ PSU_4=SU_4/C_4 $ since the corresponding subgroup lifted to $ SU_4 $ does not contain the full center $ C_4 $. So the pattern already seems to fail for $ n=4 $ REPLY [15 votes]: The comment of @abx is part of a general picture. For $n >7$ the alternating group $A_{n}$ has no non-trivial complex irreducible character of degree less than $n-1$, so that for $n > 7$, $A_{n}$ is not isomorphic to any subgroup of ${\rm GL}(n-3, \mathbb{C}).$ Furthermore, if $n > 7$ is also even, then the double cover of $A_{n}$ (which, by a theorem of I. Schur is a maximal perfect central extension of $A_{n}$) has no faithful complex irreducible character of odd degree, so certainly not of degree $n-3$ ( as an involution in its centre would be represented by a matrix of determinant $-1$). Hence the double cover of $A_{n}$ has no non-trivial irreducible complex representation of degree $n-3$, faithful or not. Thus $A_{n}$ does not embed as an irreducible subgroup of ${\rm PSU}(n-3, \mathbb{C})$ (and it is clear that a maximal closed subgroup has to be irreducible). (A few points for clarity: for finite groups, all finite dimensional complex representations are equivalent to unitary ones. Also, there is no real distinction between projective representations (in Schur's sense) and genuine representations of covering groups. The fact that the minimal degree of an non-triival irreducible complex representation of $A_{n}$ is $n-1$ ( for $n > 7$) can be found in almost any text on representation theory of $S_{n}$ ( eg that of G.D. James)). Later edit: I am not sure of the smallest degree of a faithful projective (in Schur's sense) complex representation of $A_{n}$, though I believe this must be well-known to symmetric group experts. However, results such as Zsygmondy's theorem provide a lower bound. If $p$ is a prime less than or equal to $m-2$, then $A_{m}$ contains a $p$-cycle which is conjugate to all its non-identity powers, from which it follows that $A_{m}$ can have no non-trivial complex projective representation of degree less than $p-1.$ Hence if $n$ is an even integer such that there is a prime $p$ with $\frac{n}{2} < p < n-2$, we see that $A_{n}$ has no complex irreducible projective representation of degree less than $\frac{n}{2}$ (and we may find similar bounds for odd $n$).<|endoftext|> TITLE: Étale fundamental group of rigid analytification QUESTION [7 upvotes]: Let $X$ be a quasi-projective variety over a $p$-adic field. Denote by $X^{an}$ its rigid analytification. Does $\pi_1^{et}(X)=\pi_1^{et}(X^{an})$? REPLY [11 votes]: Yes, if $X$ is proper (by GAGA). Otherwise, this is false for covers of degree divisible by $p$. See Example 51 in Ducros's survey "Étale Cohomology of Schemes and Analytic Spaces", though unfortunately he does not give a reference. For an example of this failure in equal characteristic $p$, see section 7.4 in de Jong–van der Put "Etale cohomology of rigid analytic spaces" (Doc. Math. 1995). They construct an example of a $\mathbf{Z}/p$-covering of $\mathbf{A}^{1, \rm an}$ which is not the analytification of a finite etale covering of the line. In mixed characteristic, almost the same argument should work if we replace Artin-Schreier coverings with Kummer coverings of degree $p$ (i.e. replace equations $T^p-T=f$ with $T^p=f$), but I didn't check the details.<|endoftext|> TITLE: Generalized Stokes' theorem QUESTION [18 upvotes]: In the Wikipedia article on Stokes' theorem the following claim is advanced without any references given: The main challenge in a precise statement of Stokes' theorem is in defining the notion of a boundary. Surfaces such as the Koch snowflake, for example, are well-known not to exhibit a Riemann-integrable boundary, and the notion of surface measure in Lebesgue theory cannot be defined for a non-Lipschitz surface. One (advanced) technique is to pass to a weak formulation and then apply the machinery of geometric measure theory; for that approach see the coarea formula. Can someone explain what is the weak formulation announced here and how exactly can we obtain the general Stokes' theorem in $\mathbb{R}^{3}$ for surfaces with non-smooth boundaries? REPLY [20 votes]: When I wrote that Wikipedia paragraph, I think I had in mind Theorem 5.16 in Evans & Gariepy, Measure Theory and the Fine Properties of Functions, which essentially proves Stokes' theorem for level sets of functions with bounded variation. Unfortunately, the proof given in the book is just citations to Lemmas 5.2 and 5.5, Theorem 5.15, and "the foregoing theory", so I can't be confident (via a cursory review) that "weak formulation+coarea" is the correct summary.<|endoftext|> TITLE: What is the right notion of a functor from an internal topological category to topological spaces? QUESTION [8 upvotes]: Let $\mathcal C=(\mathcal O, \mathcal M)$ be a category internal to topological spaces. Thus $\mathcal O$ and $\mathcal M$ are topological spaces: the space of objects and the space of morphisms respectively. These spaces come endowed with structure maps: $i \colon \mathcal O \to \mathcal M$, $s, t\colon \mathcal M\to \mathcal O$, and $c\colon \mathcal M \times_{\mathcal O} \mathcal M \to \mathcal M$, which satisfy well-known identities. Here $\mathcal M \times_{\mathcal O} \mathcal M$ is the pullback $\mathcal M\xrightarrow{s} \mathcal O \xleftarrow{t} \mathcal M$. Let Top be the category of topological spaces. Feel free to use your favorite ``convenient'' category of topological spaces. My question is What is the correct notion of a topological functor $F\colon \mathcal C\to $Top ? Intuitively I feel that the following is a reasonable notion: Proposed definition: A functor $F\colon \mathcal C\to $Top consists of a space over $\mathcal O$, that I will denote by $F\to \mathcal O$. This map is not necessarily a fibration. The fiber at a point $x\in \mathcal O$ corresponds to the value of $F$ at $x$. Furthermore, there has to be a structure map $$F\times_{\mathcal O} \mathcal M \to F$$ which satisfies certain more or less evident relations. Is this notion of a topological functor in the literature? Does it have a name? Where can I read about it? Disclosure: I have actually used this definition in a couple of papers, but in an ad hoc manner. I want to know if other people used it, and if it has been developed systematically. The following question is not of immediate practical consequence to me, but it is presumably important for the general picture: Is it possible to reinterpret this definition as an internal functor from $\mathcal C$ to Top, perhaps using the language of higher categories? What I really want to know is whether people have studied the homotopy theory of such functors. Is there a projective model structure on the category of functors $\mathcal C \to $Top, where weak equivalences/fibrations are fiber homotopy equivalences/fibrations over $\mathcal O$? Has anyone studied homotopical notions, such as homotopy limits and colimits, derived Kan extensions, and so forth, in this setting? REPLY [12 votes]: The definition you propose is that of a $\mathsf{Top}$-internal diagram in $\mathcal{C}$. It comes from viewing categories as many-object monoids, and functors/presheaves on categories as the many-object generalisation of left or right modules over a monoid (and then internalising these notions to $\mathsf{Top}$). Similarly, the bimodule version of this is called an "internal profunctor". You can find more about internal diagrams in the following references: Andrade, From manifolds to invariants of Eₙ-algebras, Section 2.6. Borceux, Handbook of Categorical Algebra, Volume I, Section 8.2. Jacobs, Categorical Logic and Type Theory, Section 7.4. Johnstone, Sketches of an Elephant, Volume 2, Section B.2.3. Johnstone, Topos Theory, Section 2.2. Tomašić, A topos-theoretic view of difference algebra, Section I.4. Wong, The Grothendieck Construction in Enriched, Internal and ∞-Category Theory, Section 4.4. (This one treats the non-Cartesian case). Wong, Smash Products for Non-cartesian Internal Prestacks, paper version of the above. MO 199237 and MO 263927. (Incidentally, a second possible (non-standard) such definition would be via locally internal categories, the "locally small version" of internal categories: we could pick a nice category of spaces $\mathsf{Spc}$ (one that is locally Cartesian closed; in particular that of compactly generated Hausdorff spaces isn't; see also this nLab page), and then view it as a locally $\mathsf{Spc}$-internal category, via self-internalisation. Then a topological functor from $\mathcal{C}$ to $\mathsf{Spc}$ would be a locally $\mathsf{Spc}$-internal functor from the externalisation of $\mathcal{C}$ (i.e. $\mathcal{C}$ viewed as a locally internal category) to this locally $\mathsf{Spc}$-internal version of $\mathsf{Spc}$. I don't think these two definitions agree (Edit: They actually do! See the comments below), though in any case this second definition is definitely a hassle to work with! :/)<|endoftext|> TITLE: Homomorphisms between Oort–Tate group schemes QUESTION [10 upvotes]: Let $R$ be a complete local $\mathbf{Z}_p$-algebra, for some prime $p$. In the 1970 paper Group schemes of prime order by Oort and Tate, they write down an explicit finite flat group scheme $G_R(a, b)$ of rank $p$, for each pair of elements $a,b \in R$ satisfying $ab = p$, and show that every FFGS over $R$ of rank $p$ is isomorphic to one of these. What are the homomorphisms (of $R$-group schemes) from $G_R(a, b)$ to $G_R(a', b')$? The underlying scheme of $G_R(a, b)$ is $\{ X : X^p - aX = 0\}$. If we look for homomorphisms which are "linear", $X \mapsto \lambda X$ for some $\lambda$, then (after a bit of unravelling) we conclude that $\lambda$ has to be a point of the $R$-scheme $$\{ Z : Z(aZ^{p-1} - a') = Z(b - b'Z^{p-1}) = 0\}.$$ (This is a group scheme, but neither finite nor flat over $R$ in general, although it is a FFGS if $\{a, b'\}$ generate the unit ideal of $R$.) Are these all the homomorphisms $G_R(a, b) \to G_R(a', b')$? (The answer is clearly "yes" for $p = 2$. It is also "yes" for $p = 3$ via a deeply nasty polynomial computation.) REPLY [8 votes]: Yes. The Tate-Oort description is an equivalence of categories between finite flat group schemes (over a $\Lambda$-scheme $S$ where $\Lambda$ is a certain ring decribed in the paper) and the category of triples $(L , a, b)$ where $L$ is an invertible sheaf over $S$ and $a\in \Gamma(S,L^{\otimes (p−1)})$, $b\in\Gamma(S,L^{\otimes (1− p)})$ satisfy $a\otimes b=w_p\cdot 1$. The morphisms between $(L,a,b)$ and $(L',a',b')$ are the morphisms of invertible sheaves $f:L\to L'$, viewed as global sections of $L^{\otimes -1}\otimes L'$, such that $a\otimes f^{\otimes p}=f\otimes a'$ and $b'\otimes f^{\otimes p}=f\otimes b$. In the case where the base is a local scheme, this is exactly what you wrote. All I wrote is almost verbatim from the Tate-Oort paper.<|endoftext|> TITLE: Finite simple groups and $ \operatorname{SU}_n $ QUESTION [5 upvotes]: A follow-up question to Alternating subgroups of $\mathrm{SU}_n $. $\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\PSL{PSL}$Let $ \PSU_m $ be the projective unitary group, a compact simple adjoint Lie group corresponding to the root system $ A_{m-1} $. Let $ \PSL_n(q) $ be the finite simple group of Lie type $ A_{n-1}(q) $ given by taking the special linear group with entries from the field with $ q $ elements and modding out by the center. Let $ \PSU_n(q^2) $ be the finite simple group of Lie type $ ^2 A_{n-1}(q^2) $ given by taking the special unitary group with entries from the field with $ q^2 $ elements and modding out by the center. $ \PSU_2 $ contains a $ 60 $ element subgroup isomorphic to $ \PSL_2(4) \cong \PSL_2(5) $. It is the largest of the primitive finite subgroups of $ \PSU_2 $ and it is maximal (the only closed subgroup containing it is the whole group). The references in The finite subgroups of SU(n) show that $ \PSU_3 $ contains a subgroup of order $ 360 $ isomorphic to $ \PSL_2(9) $ and that again this subgroup is the largest of the primitive finite subgroups. Again it is maximal. Also $ \PSU_3 $ contains a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $ which seems like it should be maximal. There is also a 60 element $ \PSL_2(4) \cong \PSL_2(5) $ subgroup of $ \PSU_3 $ but that is already in $ \PSU_2 $ so it's probably not maximal. The reference Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4) from the same MO question shows that $ \PSU_4 $ contains a subgroup of order $ 25{,}920 $ isomorphic to $ \PSU_4(4) $ and that again this subgroup is the largest of the primitive finite subgroups. Again it is maximal. Also $ \PSU_4 $ contains a group of order $ 360 $ isomorphic to $ \PSL_2(9) $, and a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $, but both of these are already in $ \PSU_3 $ so probably not maximal. Finally it contains a group of order 60 isomorphic to $ \PSL_2(4) \cong \PSL_2(5) $ but that is already in $ \PSU_2 $ so almost certainly not maximal. That leads me to ask: Does $ \PSU_m $ always have a maximal $ \PSL_n(q) $ or $ \PSU_n(q^2) $? REPLY [7 votes]: One can consult the tables of Bray-Holt-Roney Dougal to work out the subgroups of $\mathrm{PSU}_m$. For $m=5$, we have copies of $PSL_2(11)$ and $PSU_4(4)$. For $m=6$ we have $PSL_2(11)$, $PSL_3(4)$ and $PSU_4(9)$. For $m=7$ we have $PSU_3(9)$. For $m=8$ we have $PSL_3(4)$. For $m=9$ we have $PSL_2(19)$. For $m=10$ we have $PSL_2(19)$ and $PSL_3(4)$. For $m=11$ we have $PSL_2(23)$ and $PSU_5(4)$. For $m=12$ we have $PSL_2(23)$ and $PSL_2(9)$. So far, so good. Bray-Holt-Roney Dougal stops at dimension 12. For dimensions 13 to 15 one consults tables in Anna Schroeder's PhD thesis (St Andrews, 2015) and there are examples in those dimensions. For dimensions 16 and 17 one consults the thesis of Daniel Rogers (Warwick, 2017). For 16 there is an example, but there are no interesting subgroups (i.e., finite subgroups not contained in a closed positive-dimensional subgroup) of $\mathrm{PSU}_{17}$ at all, never mind linear or unitary ones. So it seems $m=17$ is the first case where things go wrong.<|endoftext|> TITLE: What's the point of a point-free locale? QUESTION [8 upvotes]: In [1, example C.1.2.8], a locale $Y$ (dense in another locale $X$) without any point is given. I fail to understand the point of such point-less locale - Why can't we identify those as the trivial locales, and what's so great about considering locales that have no points? Anyway, here's the construction of $X$ and $Y$ (taken from [1]). Let $A$ be an uncountable nonempty set (e.g. $\mathbb{R}$) (equipped with the discrete topology), and let $X$ be the set of all functions $\mathbb{N} \to A$, equipped with the Tychonoff topology. For each $a \in A$, let $X_a$ be the subspace $\{f \in X \,|\, a \in im(f)\}$, and let $$ Y = \bigcap_{a\in A} X_{a}.$$ Now the point set $Y_p$ of $Y$ is empty because there is no onto map from $\mathbb{N}$ to $\mathbb{R}$. In [2, section 5], Johnstone demonstrates why considering such locales could be useful. The main argument is that topoi are nice things to consider. However, at the point of writing, the (external) applications of topos theory seem lacking. Hopefully the situation has changed in mathematics in recent years. Thus the second question: How does the consideration of pointless locales help topos theory, and how does that in turn applies (externally) to mathematics? Reference [1] Sketches of an Elephant: A Topos Theory Compendium [Peter T. Johnstone] [2] The point of pointless topology-[Peter T. Johnstone] REPLY [14 votes]: A good answer to both questions is provided by the following variant of the Gelfand duality for commutative von Neumann algebras, which shows that the following categories are equivalent: The category CSLEMS of compact strictly localizable enhanced measurable spaces; The category HStonean of hyperstonean spaces and open maps. The category HStoneanLoc of hyperstonean locales and open maps. The category MLoc of measurable locales, defined as the full subcategory of the category of locales consisting of complete Boolean algebras that admit sufficiently many continuous valuations. The opposite category CVNA^op of commutative von Neumann algebras, whose morphisms are normal *-homomorphisms of algebras in the opposite direction. The first category, despite the rather complicated name, is essentially the correct category for measure theory: it incorporates equality almost everywhere, a (generalized) σ-finiteness property, and an abstract variant of the Radon measure property, which eliminate pathological measurable (and measure) spaces for which some of the most basic theorem of measure theory (such as the Riesz representation theorem or the Radon–Nikodym theorem) fail. Of particular interest is the fourth category MLoc of mesurable locales. It is a full subcategory of the category of locales, which quite interesting: it demonstrates that both point-set general topology (as implemented by the category of topological spaces) and point-set measure theory (as implemented by the above category CSLEMS) are a part of pointfree general topology, implemented by full subcategory of the category of locales. These parts (i.e., general topology and measure theory) are almost disjoint: locales corresponding to topological spaces are spatial, i.e., have enough points. On the other hand, points in a measurable locale are in a bijective correspondence with atoms in the original measure space. In particular, atomless measure spaces (i.e., what is typically used in practice) correspond to locales that have no points at all. Returning to topos theory: working in the topos of sheaves of sets on a measurable locale amounts to doing ordinary mathematics in measurable families over a measurable space. For example, doing internal linear algebra in such a topos corresponds to working with measurable vector bundles, etc.<|endoftext|> TITLE: Tanglegrams and functional equations of M. Somos QUESTION [6 upvotes]: Recent references on the matter at hand include, a lecture slide The Konvalinka-Amdeberhan conjecture and plethystic inverses and a preprint on Counting tanglegrams with species by I. Gessel; the initial work On the enumeration of tanglegrams and tangled chains by S. Billey et al. For each prime $p$, let's consider the family of (integral) sequences $$a_p(n):=\sum_{\lambda\vdash n}\frac{1}{z_\lambda}\prod_{i=2}^{\ell(\lambda)}(p\lambda_i+\cdots+p\lambda_{\ell(\lambda)}+1);$$ where the sum runs through all $p$-ary partitions $\lambda$, of $n$, and $\ell(\lambda)$ is the length of the partition and the numbers $z_{\lambda}$ are well-known since the number of permutations in $\mathfrak{S}_n$ with cycle type $\lambda$ is computed by $\frac{n!}{z_{\lambda}}$. On the other hand, Michael Somos proposed the functional equations $$ x\cdot A_p(x)^p=\frac{x\cdot A_p(x^p)}{1-p\cdot x\cdot A_p(x^p)} $$ and a couple of these are listed on OEIS as A085748 and A091190. However, there are not enough interpretations attached to $A_p(x)$ or its coefficients on OEIS. After some experimentation, I wish to ask: QUESTION. Can the following be justified or refuted? $$A_p(x)=\sum_{n\geq0}a_p(n)x^n.$$ Remark. Observe that neither $A_p(x)$ nor $a_p(n)$ have been found to output integers, unless $p$ is a prime. REPLY [6 votes]: In my lecture slides that Tewodros cites, I studied symmetric functions that I called $g_m$, where $m$ is a positive integer, that have constant term 1 and satisfy $$ -L_m[g_m] = p_1,$$ where $L_m$ is the Lyndon symmetric function given by $$L_m = \frac{1}{m} \sum_{d\mid m} \mu(d) p_d^{m/d}.$$ Here $\mu$ is the Möbius function, the $p_i$ are power sum symmetric functions, and $L_m[g_m]$ is the plethysm of $L_m$ and $g_m$. I showed that $g_m$ is an integral symmetric function; i.e., its coefficients in the underlying variables are integers, and that if $m$ is a power of the prime $q$ then for any $\alpha$, \begin{multline*} \quad g_m^{-\alpha} = 1+ \sum_{n=1}^\infty \sum_{\lambda}\frac{p_{\lambda}}{z_\lambda} \times \alpha\prod_{j=2}^{l(\lambda)} (m\lambda_j+m\lambda_{j+1}+\cdots +m\lambda_{l(\lambda)}+\alpha),\quad\tag{1} \end{multline*} where the sum on $\lambda$ is over all partitions of $n$ in which every part is a power of $q$. (I'm using $q$ here instead of $p$ to avoid confusion with the power sum symmetric functions.) This shows that if $\alpha$ is positive and $m$ is a power of a prime then all coefficients of $g_m^{-\alpha}$ are positive. I don't know that this is true if $m$ is not a power of a prime, though I suspect that it is. Now let us specialize the symmetric functions (in the variables $x_1, x_2,\dots$) by setting $x_1=x$ and $x_i=0$ for $i>0$, or equivalently, $p_i =x^i$ for all $i$. Let $G_m(x)$ be the image of $g_m$ under this specialization. Then $G_m(x)$ satisfies the functional equation $$-\frac{1}{m}\sum_{d\mid m}\mu(d) G_m(x^d)^{m/d} = x.$$ If $m$ is a power of the prime $p$, this may be written $$G_m(x^p)^{m/p} -G_m(x)^m = mx$$ and it's easy to check that if $m=p$ then $A_p(x) =1/G_p(x)$, where $A_p(x)$ is as in the original question. Moreover, it follows from $(1)$ that if $m$ is a power of the prime $p$ then for any $\alpha$ we have \begin{multline*} \quad G_m(x)^{-\alpha} = 1+ \sum_{n=1}^\infty x^n \sum_{\lambda}\frac{\alpha}{z_\lambda}\, \times\prod_{j=2}^{l(\lambda)} (m\lambda_j+m\lambda_{j+1}+\cdots +m\lambda_{l(\lambda)}+\alpha),\quad\tag{2} \end{multline*} where the sum on $\lambda$ is over all $p$-ary partitions of $n$. The case $m=p$, $\alpha=1$ answers the OP's questions in the affirmative. In all cases $G_m(x)$ has integer coefficients, but I can only prove that $G_m(x)^{-1}$ has positive coefficients when $m$ is a prime power, when it follows from $(2)$, though this is likely true in all cases. The stronger statement that $1-G_m(x)$ has positive coefficients also follows from $(2)$ when $m$ is prime power. I am working on a paper with detailed proofs of these formulas, but it will take a while.<|endoftext|> TITLE: More precise bound for the maximin number of arrows pointing to a square when square grid is filled with arrows QUESTION [13 upvotes]: The problem is as follows: Suppose we fill each cell of a $n\times n$ square with one of the four arrows $\uparrow$, $\downarrow$, $\gets$, $\to$, like the following $\begin{array}{|c|c|c|} \hline \downarrow & \downarrow & \gets\\ \hline \to& \to& \uparrow\\ \hline \to& \uparrow& \uparrow\\ \hline \end{array}$ We define that entry $A$ (in $i$th column $j$th row) points to entry $B$ (in $k$th column $l$th row) if one of the four happens: $A$ is filled with $\uparrow$ and $j> l$ and $i=k$ $A$ is filled with $\downarrow$ and $j< l$ and $i=k$ $A$ is filled with $\gets$ and $j= l$ and $ik$ (This is straightforward if you really fill them with arrows...) For the example above, only the entry in first row, third column, points to the left upper corner. The right bottom corner entry points to the entry on the first row, third column and the entry on the second row, third column. Let $r(n)$ be the maximum number such that there is a filling method such that every entry is pointed to by at least $r(n)$ other entries. Now we have a bound $\frac{2}{3}n+O(1)\le r(n)\le \frac{5}{6}n+O(1)$ (update: now $ r(n)\le \frac{3}{4}n+O(1)$) (update2: now $ r(n)\le (\sqrt{3}-1)n+O(1)$) (update3: now $ r(n)\le n/\sqrt{2}+O(1)$) The lower bound is obtained by the following construction: We first split the grid into the central $n/3\times n/3$ square surrounded by four regions as drawn. The red region is filled with $\to$, the green region is filled with $\downarrow$, The blue region is filled with $\gets$, the purple region is filled with $\uparrow$. The central square can be filled with anything. One can prove that every entry is pointed to by $\frac{2}{3}n+O(1)$ other entries. On the other hand, counting the total number of entries one entry can point to, there are at most $\frac{5}{6}n^3+O(n^2)$ pairs (entry 1, entry 2) such that entry 1 points to entry 2. So, in average, an entry is pointed to by at most $5/6\; n+O(1)$ other entries. Now, the question is: can the lower bound $2/3$ be raised and $5/6$ the upper bound be made smaller? I believe $2/3$ can be raised a little bit (for example, $0.01$) since the central $n/3\times n/3$ is not filled. But can it be substantially increased? Any improvement is welcome and appreciated, even minor improvement with $2/3$ is welcome. Update: a friend of mine gave me an $3/4\; n+O(1)$ upper bound. He considered the square connected the midpoints, where there are $2n$ entries. The entries inside the square point to at most one entry on the square, and the entries outside the square point to at most two. Thus, the number of entry points to the other entries is at most $1/2n^2+2\times 1/2n^2+O(n)=3/2n^2+O(n)$, and since there are $3n$ entries, averagely, one of the entries has at most $3n/4+O(1)$ entries points to it. Update 2: That friend suggested another way: choose these entries in green The green line four corners are of length $(\sqrt{3}-1)n/2$ entries. Using the similar argument above we can get $(\sqrt{3}-1)n+O(1)$ upper bound. Update 3: That friend, again, suggested a way: choose these entries in green, the lengths of the green lines in the four corners are $\frac{\sqrt{2}-1}{2}n$ (they are the diagonal of the square of side length $\frac{\sqrt{2}-1}{2}n$) and the central green square has diagonal length $(2-\sqrt{2})n$ These entries on the central square we give them a weight of $1/2$. So all the entries in the outside four triangles will point to weight two entries, the entries inside the central square will point to weight $1/2$ (one entry on the central square) and the rest will point to weight $1$ (two entries on the corner green lines, or two entries on the central square.) So the total weight of pointing will be $n^2+O(n)$. The total weight of green entries is $\sqrt{2} n+O(1)$, and by average, the number of entries pointed to a certain entry is at most $n/\sqrt{2}+O(1)$, so there must be one entry pointed by at most other $n/\sqrt{2}+O(1)$ entries. REPLY [10 votes]: I used integer linear programming (in SAS) to compute the exact values for $n\in\{2,\dots,50\}$, and they match https://oeis.org/A248183. Let $C_{i,j,d} \subseteq [n] \times [n]$ be the set of cells that can point to cell $(i,j)$ via direction $d$. Let binary decision variable $x_{i,j,d}$ indicate whether the arrow in cell $(i,j)$ points in direction $d$, and let $z$ represent the smallest number of arrows pointing to a cell. The problem is to maximize $z$ subject to \begin{align} \sum_d x_{i,j,d} &= 1 &&\text{for all $(i,j)$} \tag1\\ z &\le \sum_{(u,v) \in C_{i,j,d}} x_{u,v,d} && \text{for all $(i,j)$} \tag2 \end{align} Constraint $(1)$ chooses exactly one arrow per cell. Constraint $(2)$ enforces the maximin objective. Figure 1 shows an optimal solution for $n=50$, with $r(50)=34$. Maybe it will suggest a pattern for general $n$. Restricting U to the bottom half and D to the top half does not change the optimal values for $n\le 50$. For example, see Figure 2: Nor does further restricting L to the right half and R to the left half. For example, see Figure 3: For even $n \le 50$, the optimal values still do not change if you further restrict each direction to appear $n^2/4$ times. For example, see Figure 4: Forcing horizontal and vertical symmetry, and forcing triangle with side length $10$ ($>10$ loses optimality), as suggested by @Wolfgang, but not forcing $n^2/4$ arrows per direction. See Figure 5: Forcing diagonal symmetry loses optimality. I also tried forcing horizontal and vertical symmetry, $r(50)=34$, and minimizing the number of D (hence U). The minimum is at most $192$, as shown in Figure 6: With horizontal and vertical symmetry and $r(64)=44$, the minimum number of D (hence U) is $328$, as shown in Figure 7: I solved the LP relaxation for $n=50$ with no restrictions and got objective value $34+6/7$. The resulting optimal solution is not unique. Figure 8 shows a plot where the color is determined by a fractional value $\ge 1/2$ (gray if the largest fractional value is $< 1/2$): Figure 9 shows a heat map of the fractional assignments of cell to direction: Figure 10 shows a heat map of the number of arrows pointing to each cell for an optimal LP solution with no restrictions: Figure 11 shows a heat map of the number of arrows pointing to each cell for an optimal ILP solution with no restrictions: Figure 12 shows a solution for $n=50$ where each cell has the minimum number $34$ of arrows pointing to it:<|endoftext|> TITLE: Transversality and $C^l$, $C^{\infty}$ spaces of almost complex structures QUESTION [5 upvotes]: Recently I have been trying to get a grip on transversality results in Floer homology. That is suppose we the section $\partial_{J,H}: W^{1,p}(u^*(TM))\rightarrow L^{p}(u^*(TM))$ and we want to prove that there exists a generic set of almost complex structures $\mathcal{J}_\text{reg}$ for which its derivative is surjective. To prove this the idea is to consider an universal moduli space, prove this is a banach manifold, using the fact that set of regular points $R(u)$ is dense. Then one considers the projection $\pi: \mathcal{M}_\text{univ}\rightarrow \mathcal{J}$ and proves this is a Fredholm map, and then the desired result will follow from the Sard–Smale theorem. Now there a few technicalities here. First of all this direct approach only works in the $C^l$-topology of almost complex structures and to get a result on the $C^{\infty}$-topology one needs to use a method due to Taubes. This is one thing I would like to understand. Why is the completion of $\mathcal{J}$ in the $C^l$-topology a banach manifold and not in the $C^{\infty}$-topology? I am not looking for all the analytical details regarding the proof of this, but I would like to have some kind of intuition on why this happens. Any insight is appreciated. REPLY [6 votes]: There are several different issues happening here. The space of $C^\infty$ is not a Banach space because the topology is generated by a countable family of semi-norms not by one norm. You might try to think about why you can’t generate the topology by some magical norm. OK so why do we care about $C^\infty$ anyway. The point is applying Sard Smale. This theorem (like Sard’s Thm) requires more regularity as the formal dimension ( index) goes up so of you want one perturbation to rule all modulo spaces no matter the index you need it to be smooth. How do you marry these two issues. The idea is to find Banach spaces contained in $C^\infty$. The trick appears first to my knowledge in Floer’s paper “The unregularized gradient flow of the symplectic action”. Perhaps he mentions that he learned the idea from Taubes. What you do is pick a rapidly decreasing sequence $\epsilon_k$ and take the norm $$\sum_k \epsilon_k \|\cdot\|_{C^k}. $$ Floer shows in that paper that for suitable sequences there are plenty (enough to get transversality) of perturbations with this norm finite but this is not all of $C^\infty$ . Thus there are Banach spaces of perturbations contained the smooth perturbations. We can now apply Sard Smale to a countable family of moduli space problems to get still a dense set of perturbations that are regular for all of these.<|endoftext|> TITLE: A question about colorings of the vertices of a p-agon, where p is prime QUESTION [5 upvotes]: Let $V$ be the vertices of a regular $p$-agon in the plane, where $p$ is prime, and let $C$ be a set. Given two maps $f,g:V\rightarrow C,$ I believe it is true that either (a) $f=g\circ r$ for some rotation $r:V\rightarrow V,$ or (b) there is $h:V\rightarrow C$ such that for all $v\in V,$ $h(v)\neq f(v),$ but for all rotations $r:V\rightarrow V,$ there exists $v\in V$ (depending on $r$) such that $h(v)=g\circ r(v).$ Note these are mutually exclusive. (Aside: Calling the maps "colorings" of $V,$ I am saying if $f$ and $g$ are colorings, which are not just rotations of each other, then there is another coloring $h$ that doesn't match $f$ anywhere, but matches every rotation of $g$ somewhere. Perhaps this sounds more intuitive.) My proof is of course elementary but involves some case by case analysis, so I would like to ask if it follows from known results, or has an easy proof that I have missed. I don't know if it's essential that $p$ be prime, but I need that for my proof. REPLY [5 votes]: We may identify $V$ with the field $\mathbb{F}_p$ of size $p$, then denoting $g_k(x):=g(x+k)$ for $k\in \mathbb{F}_p$ the questions reads as: if $f(x)\not\equiv g_k(x)$ for each $k\in \mathbb{F}_p$, prove that there exists $h(x)$ such that $h(x)\ne f(x)$ for all $x$; but for each $k$ there exists $x(k)$ for which $h(x(k))=g_k(x(k))$. If $g\equiv c$ is a constant coloring, we find a vertex $v\in \mathbb{F}_p$ for which $f(v)\ne c$ and put $h(v)=c$, and put $h(x)\in C\setminus \{f(x)\}$ for $v\ne x$ arbitrarily (this is possible since $|C|\geqslant |\{c,f(v)\}|=2$.) If $g$ is not constant, we additionally require that $x(k)\colon \mathbb{F}_p\to \mathbb{F}_p$ is a bijection. Then we need $f(x(k))\ne g_k(x(k))$ for all $k$ (and $h$ is automatically defined by $h(x(k))=g_k(x(k))$). Consider the $p\times p$ matrix with both rows and columns indexed by $\mathbb{F}_p$ and put a star at the entry $(x,k)$ iff $f(x)\ne g(x+k)$. We want $p$ non-attacking rooks on the starred entries. Assume the contrary, then by Hall lemma (or Koenig theorem if you prefer) there exist sets $A,B\subset \mathbb{F}_p$ such that $|A|+|B|=p+1$ and $f(x)=g(x+k)$ for $x\in A$, $k\in B$. If $|B|=1$, then $|A|=\mathbb{F}_p$ and $f(x)=g(x+k)$ for all $x$ where $B=\{k\}$. This contradicts to our assumption. If $|A|=1$, then $B=\mathbb{F}_p$ and we get that $g$ is constant, which also is not the case. So, assume hereafter that $1<|A|,|B|d_i$ there exist $a_i\in A_i$ such that $f(a_1,\ldots,a_k)\ne 0$.<|endoftext|> TITLE: Are there infinitely many positive integer solutions to $(3+3k+l)^2=m\,(k\,l-k^3-1)$? QUESTION [15 upvotes]: I usually work in the field of differential geometry, but I have encountered the following problem in my research: Are there infinitely many positive integers $k,l,m\in\mathbb N^{>0}$ such that $$(3+3k+l)^2=m\,(k\,l-k^3-1)\,?$$ Obviously, taking $l=k^2$ and $m=-(3+3k+l)^2$ gives infinitely many integer solutions, but $m<0$ is negative. As a non-expert, I imagine that there is either a simple answer to this question, or the problem is not so simple to solve. Of course, I've played around with the equations a bit, but other than finding numerous examples, I haven't made any progress. I would appreciate an existence or non-existence statement for infinitely many positive integer solutions, but also some hints that the problem is most likely hard to solve would help me. Background: I am looking for certain integer representations of a surface group, and I can show that integer solutions to this diophantine equation actually give rise to integer representations. The condition that $k,l,m$ are positive is equivalent to the condition that the corresponding representation is contained in a higher Teichmüller component (which is important for my differential geometric application). REPLY [4 votes]: Here is another way maybe simpler as it reveals a certain Pell-Fermat equation. I don't know the relation with the other answer but i did many restrictions to get the following infinite sequence of solutions: For $n\ge 1$, $$\begin{cases}q_n=-7q_{n-1}+24d_{n-1}\\d_n=2q_{n-1}-7d_{n-1}\end{cases}$$ with $q_0=3$ and $d_0=1$, (solutions of $q^2-12d^2=-3).$ Set $j=2|d_n|$, $i=4|d_n|+|q_n|$, $r=3(i+j)$, so that $m=r^2$, $k=ij-1$ and $x:=3+3k+l=\dfrac{mk+r|j^3-i^3|}{2}$. This implies the positive integer solutions $(m,k,l)$ are infinite. The starting equation to solve in $ x $ over $\mathbb{N}$ was $x^2-mkx+m(k+1)^3$. Its discriminant is a square $X^2=(r\cdot y)^2$ as $\Delta=X^2=m^2k^2-4m(k+1)^3$, say $m=r^2$ is a perfect square, then $(rk-y)(rk+y)=4(k+1)^3$ and set $(k+1)^3=i^3j^3$, ($k=ij-1$). $$\begin{cases}rk-y=2i^3\\rk+y=2j^3\end{cases}.$$ So $r=(i+j)(-1+\dfrac{i^2+j^2-1}{ij-1})$ and $y=j^3-i^3$. Let $\dfrac{i^2+j^2-1}{ij-1}=s$ an integer say $s=4$. Finally we solve in $i$, $i^2+j^2-1=4ij-4$, taking again its discriminant to be a perfect (even) square $p=2q$, $j=2d$ we solve $p^2-12j^2=-12$ as $q^2-12d^2=-3$.<|endoftext|> TITLE: Expected value of non-negative iid random variables QUESTION [5 upvotes]: I want to show that, for $n$ iid non-negative random variables $x_1, \dots, x_n$, we have $$ \mathbb{E} \max_{i \in [n]} x_i \lesssim (\mathbb{E} [x_i^{\log n}])^{\frac{1}{\log n }}. $$ I can only go so far as to show $\mathbb{E} \max_{i \in [n]} x_i \leq n (\mathbb{E} [x_i^{\log n}])^{\frac{1}{\log n }}$ for $n \geq 3$, using Jensen's inequality and the inequality $\max_i x_i \leq \sum_i x_i$, which seems a bit wasteful (and responsible for the factor of $n$). Any tips for getting rid of the factor of $n$ would be greatly appreciated! REPLY [2 votes]: Here is another proof that the constant factor $e$ in mathworker21's nice proof is the best possible. Suppose that $1-P(x_1=0)=P(x_1=1)=p:=p_n:=\dfrac{\ln n}n$. Then $$E\max_{i\in[n]}x_i=1-(1-p)^n\ge1-e^{-np}=1-1/n\to1$$ and $$(Ex_1^{\ln n})^{1/\ln n}=p^{1/\ln n} =\exp\frac{\ln\ln n-\ln n}{\ln n} \to e^{-1}.$$ So, the constant factor $e$ is the best possible.<|endoftext|> TITLE: Average number of tiles of a (0,1)-matrix? QUESTION [5 upvotes]: Given a (0,1)-matrix $A$, I'll denote by $\mu(A)$ the number of maximal monochromatic polyominoes in $A$ (i.e., the number of connected polyominoes contained in $A$ each of which is either all 0 or all 1 such that each polyomino is as large as possible - so they thus tile $A$). I do not know of any literature on $\mu(A)$, so I apologize if there is already some notation for the quantity that I am not using. So for example if $A$ is the $n \times n$ identity matrix, $\mu(A) = n + 2$. For an $n \times n$ matrix $A$, we have $\mu(A) \leq n^2$, with equality if and only if $A$ is one of the "checkerboard matrices". I would like to understand what $\mu(A)$ is on average, as we allow the size of the matrix $A$ to tend toward infinity. In particular, is it true that $$ \lim_{n \to \infty} \frac{\text{Average value of $\mu$ over all $n\times n$ (0,1)-matrices}}{n^2} = 1 $$ REPLY [3 votes]: On the 2004 Putnam, Problem A5, it was essentially asked to show that the value of your limit is greater than 0.125. One way to achieve this which yields better lower bounds than Noam's answer proceeds by coloring one square at a time and is as follows: First color the top and left edges of a square of side length $n$. This is just the one dimensional case with $2n - 1$ tiles, so the expected number of components in what has been colored so far is $n$. Now color the squares in the interior one at a time, always choosing the topmost and leftmost square, and examine how the expected number of components can change. If the two already colored squares adjacent to the chosen square are part of different components, the expected change to the number of components is 0. Thus we care only about the probability that the two adjacent squares belong to the same component. This occurs slightly more than $\frac{1}{4}$th of the time, with the most common case being when the two squares have a common neighbor of the same color as them. From this we see that we expect the number of components to increase by slightly more than $\frac{1}{8}$th each time, giving that the expected number of components is greater than $$\frac{(n-1)^2}{8} + n.$$ This yields a lower bound on the limit of $0.125$, which can be strengthened slightly by considering more complicated "loops" which would allow the two squares adjacent to the chosen square to belong to the same component. For example, with probability $\frac{2}{2^8}$ (ignoring concerns about the edge), the two squares will already be joined by a path which walks around the outside of a $3$-by-$3$ square, but not by their common neighbor. This improves the lower bound to $\frac{33}{2^8}$, and its not hard to get better lower bounds by considering longer loops. By using the first eight such loops I could think of (those which can be contained in a 4x4 square), I obtained a lower bound of $\frac{4261}{32768} \approx 0.13$. Forgive me for this answer which maybe should have been a comment - I did not have enough reputation to just comment a mention of the Putnam problem.<|endoftext|> TITLE: Is it possible to realize the Moebius strip as a linear group orbit? QUESTION [22 upvotes]: On MSE this got 5 upvotes but no answers not even a comment so I figured it was time to cross-post it on MO: Is the Moebius strip a linear group orbit? In other words: Does there exists a Lie group $ G $ a representation $ \pi: G \to \operatorname{Aut}(V) $ and a vector $ v \in V $ such that the orbit $$ \mathcal{O}_v=\{ \pi(g)v: g\in G \} $$ is diffeomorphic to the Moebius strip? My thoughts so far: The only two obstructions I know for being a linear group orbit is that the manifold (1) must be smooth homogeneous (shown below for the the group $ \operatorname{SE}_2 $) and (2) must be a vector bundle over a compact Riemannian homogeneous manifold (here the base is the circle $ S^1 $). The Moebius strip is homogeneous for the special Euclidean group of the plane $$ \operatorname{SE}_2= \left \{ \ \begin{bmatrix} a & b & x \\ -b & a & y \\ 0 & 0 & 1 \end{bmatrix} : a^2+b^2=1 \right \}. $$ There is a connected group $ V $ of translations up each vertical line $$ V= \left \{ \ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix} : y \in \mathbb{R} \right \}. $$ Now if we include the rotation by 180 degrees $$ \tau:=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ then $ \langle V, \tau \rangle$ has two connected components and $$ \operatorname{SE}_2/\langle V, \tau \rangle $$ is the Moebius strip. REPLY [18 votes]: Here is another solution, using the special Euclidean group $\operatorname{SE}(2) := \operatorname{SO}(2) \ltimes \mathbb{R}^2$ instead of Robert Bryant's solution which uses $\operatorname{SO}(2,1)$. Let $\operatorname{SE}(2)$ act on $\mathbb{R}^2$ in the usual way. Let $V$ be the vector space of (inhomogenous) polynomials of degree $\leq 2$ on $\mathbb{R}^2$, so $\operatorname{SE}(2)$ acts on $V$. Then the orbit of the polynomial $x^2$ is in bijection with the set of lines. (Namely, the zero locus of each such polynomial is a line, and, given a line $\ell$, the function $d(\ell,(x,y))^2$ is a quadratic polynomial in $(x,y)$.) So the orbit of $x^2$ is the space of lines in $\mathbb{R}^2$, which is a Mobius strip. Here is the way I would think about this. Let $M$ be a smooth manifold and let $G$ be a group acting on $M$. Let $W$ be any finite dimensional subspace of $C^{\infty}(M)$. Then sending $x \in M$ to the "evaluation at $x$" gives a smooth map $M \to W^{\vee}$. If $W$ is $G$-invariant, then $W^{\vee}$ inherits a $G$-action and the map is $G$-equivariant. Unless we are very unlucky, the map is an embedding. So I would start by thinking about a group $G$ acting on the Mobius strip $M$, take some function $f \in C^{\infty}(M)$ and see if the $G$-orbit of $f$ spans a finite dimensional vector space. That is how I found the above example, thinking about functions on the space of lines like "slope" and "distance from the origin", until I discovered that "square of distance to the origin" worked.<|endoftext|> TITLE: Is every topos a sheaf topos with values in a well-pointed one? QUESTION [10 upvotes]: Here's a mix of heuristic and precise questions as I try to grapple with topos theory. I try to think of topoi as two notions of "$1$" being glued at the hip. One is the "building block" $1$, generating the naturals, the ordinals, the cardinals... with all its usual arithmetic properties. This generates the set theory of a topos. The other is the "all-encompassing" $1$, the highest truth value, the top element of a Heyting/Boolean algebra, whose subobjects form the algebra of truth values in the topos. This generates the logic of a topos. Hence the set-theoretical properties (existence of an NNO, choice, CH, large cardinals...) happen "above $1$" and the logical properties (being Boolean, two-valued, well-pointed...) happen "below $1$" (whatever this means, this is just my heuristic). Now Grothendieck topoi are defined as sheaf topoi with values in a certain well-pointed topos, namely $\textbf{Set}$. I can construct topoi that are not Grothendieck by taking a $\textbf{Set}$-like topos that is not $\textbf{Set}$, for example $\textbf{FinSet}$, and then considering sheaves valued in it. My first question is: are all topoi generated like this? Is every topos equivalent to a topos of sheaves on a site (its logic) with values a well-pointed topos (its set theory)? Taking the functor $(-)^1$ of global elements seems to suggest the affirmative, as it is a logical functor to its image $\mathcal{I}$, which is well-pointed. Then the topos is, I assume, equivalent to the $\mathcal{I}$-valued sheaves on its Heyting algebra. My second suspicion was that the logic and the set theory of a topos are independent of each other, but upon reviewing forcing, this doesn't seem to be the case. For example, following Mac Lane and Moerdijk, I can start with $\textbf{Set}$, value presheaves on a forcing poset $P$ in it ($\widehat{P}$), make it Boolean ($\widehat{P}_{\neg\neg}$) and mod out an ultrafilter ($\widehat{P}_{\neg\neg}/\mathcal{U}$). But then I've changed the set theory of a topos by only tinkering with its logic. In this light, forcing seems really unexpected and counterintuitive, by creating an interplay between what's happening "above $1$" and "below $1$". [I suppose this is exactly what the forcing theorem in material set theory says, though.] So my second question is: To what extent does forcing measure the interdependence of the logic and the set theory of a topos? Is there a theorem describing this? It also surprises me that the two constructions $(-)^1$ and $/\mathcal{U}$ to collapse a topos to a well-pointed one don't coincide. For example, $\textbf{Set} \cong (\widehat{P}_{\neg\neg})^1$ and $\widehat{P}_{\neg\neg}/\mathcal{U}$ generally differ. Is there a deep reason for this? Is it perhaps because the ultrafilter-quotient construction is not as well-behaved and purely logical as I think? REPLY [4 votes]: A solution to your first question is given in "Sketches of an Elephant" by Johnstone, Example A4.4.2(d). If a topos $\mathcal{E}$ is the topos of sheaves on some internal site in a topos $\mathcal{S}$, then there is a natural geometric morphism $p : \mathcal{E} \to \mathcal{S}$. It turns out that there are toposes that do not admit geometric morphisms to Boolean toposes whatsoever. Two examples given by Johnstone are the effective topos and the topos of triples $(A,B,f)$ where $A$ is a set, $B$ is a finite set, and $f : A \to B$ is a function. So these toposes cannot be written as a topos of sheaves over a Boolean topos (so also not over a well-pointed topos, because well-pointed toposes are Boolean). Below is the argument that a geometric morphism $p : \mathcal{E} \to \mathcal{S}$ does not exist, for $\mathcal{E}$ the topos of triples $(A,B,f)$ as above and $\mathcal{S}$ a Boolean topos. I follow the Elephant, Example A4.5.24. Because subtoposes of Boolean toposes are again Boolean, we can assume that $p$ is surjective, so $p^*$ is faithful. Further, in $\mathcal{S}$ all objects are decidable, i.e. the diagonal embeddings $X \to X \times X$ have a complement. So the same holds for objects of the form $(A,B,f) = p^*(X)$. This condition implies that $f$ is injective, and as a result both $A$ and $B$ are finite. We conclude that there are only finitely many morphisms $p^*(X) \to p^*(Y)$, and because $p^*$ is faithful, it follows that there are only finitely many morphisms $X \to Y$, for $X,Y$ arbitrary. However, for $(A,1,f)$ with $A$ infinite, we get that there are infinitely many morphisms $1 \to p_*(A)$, so this gives a contradiction.<|endoftext|> TITLE: Does ZF(C) prove countable reflection? QUESTION [5 upvotes]: Is the following a theorem of $\sf ZF(C)$? Countable reflection: If $\phi$ is a sentence in which $W$ is not free, then: $\phi \to \exists W: |W|= \omega \land \operatorname {Transitive}(W) \land \phi^W$ Where $\phi^W$ is the sentence obtained by merely bounding all quantifiers in $\phi$ by $ \in W$. That is: every true sentence is reflected upon a countable transitive set. So all true sentences are reflected upon some elements of $V_{\omega_1}$ (or $V_{\omega_2}$ in case of $\sf ZF$). REPLY [18 votes]: As Monroe Eskew points out in his comment to the question, the positive answer is well-known for ZFC, thanks to the ZF reflection theorem and the Löwenheim-Skolem Theorem (in the form: every model in a countable language has a countable elementary submodel). See, e.g., part (iii) of Theorem 12.14 in Jech's canonical textbook Set theory, third millenial edition. Note that ZFC can be weakened in the above to ZF+ DC (DC = dependent choice) since it is well-known that in the presence of ZF, DC implies (and indeed is equivalent to, see, e.g., Asaf Karagila's note) the statement that every structure in a countable language has a countable elementary submodel. The point of this answer is to point out that even DC can be eliminated from the proof of countable reflection. Theorem. ZF proves every instance of countable reflection. Proof: In light of the fact that the Mostowski collapses can be carried out in ZF, and the constructible universe L satsifies ZFC (and therefore L satisfies "every model in a countable language has a countable elementary submodel), the proof is complete once we note: $(*)$ For any set-theoretical sentence $\phi$, if $\phi$ has a transitive model, then the sentence "$\phi$ has a transtive model" holds in the constructible universe L. The above statement, as far as I know, was first noted in a (famous) paper of Barwise and Fisher entitled "The Shoenfield Absoluteness Lemma" (Israel J. Math. 8 1970, pp.329-339) in which a fine-tuning of the Shoenfield Absoluteness Lemma is presented (in particular, it is shown that the Lemma can be proved without invoking DC). More specifically, $(*)$ follows from Theorem 1b (page 336) of the Barwise-Fisher paper, which states that if $\phi$ has a transitive model $A$ with $\rho(A)=\alpha$ (where $\rho$ is the usual ordinal-valued rank function on sets), then there is a transitive model of $\phi$ in the $\kappa^{+}(\alpha)$-th level $L_{\kappa^{+}(\alpha)}$ of the constructible universe $L$, where $\kappa^{+}(\alpha)$ is the next admissible ordinal after the least admissible ordinal above $\alpha$.<|endoftext|> TITLE: Unlinked interlocking planar polygons QUESTION [15 upvotes]: Let $P$ and $Q$ be the boundary segments of two planar simple polygons. View these boundaries as rigid wires. Fix $Q$ in, say, the $xy$-plane, and imagine $P$ arranged in $\mathbb{R}^3$ so that $P$ and $Q$ are disjoint and not topologically linked. Say that $P$ and $Q$ are interlocked if it is impossible to separate $P$ to infinity via rigid motions that at all times avoids $P$ touching $Q$. The polygons cannot flex or distort; each is rigid. Two unlinked but interlocked polygons are shown below ($Q$ red):       My question is: Q. What is the fewest number of vertices $|P|+|Q|$ that achieves unlinked interlocking? Reducing the red rectangle $Q$ to a triangle, and reducing $P$ to a hexagon, achieves $9$ vertices. Can interlocking be achieved with $8$ total vertices? REPLY [6 votes]: It is not possible with 7 (i.e., with a triangle $T$ and a quadrilateral $Q$). I write a rough proof. First, any quadrilateral $Q$ lying in a plane $\pi$ can be partitioned in two triangles $Q_1$ and $Q_2$, whose common edge is a diagonal $d$ of $Q$. Now the intersection of the triangle $T$ with $\pi$ consists of two points (otherwise they are coplanar and the conclusion is trivial). Since the polygons are not linked, there are two cases: either both points lie outside $Q$, or they both lie inside $Q$. Case 1: both lie outside. An inspection of the cases shows that we can just translate $T$ in direction parallel to one of the bisectors of the triangles $Q_1$ or $Q_2$ starting from one of the vertices not belonging to $d$. Case 2: both lie inside. We will consider a modified problem: given a quadrilateral $Q$ and two initial points $x(0),y(0)$ inside it, find two continuous curves $x(t),y(t)$ such that the distance between the two is decreasing in time to 0, and they never lie in $Q$. The solution is, e.g., a linear homotopy that sends both points to the midpoint of $d$. Now we only have to realize $x(t)$ and $y(t)$ as the intersections of $T$ and the plane $\pi$. One can convince themselves that this is possible by pulling $T$ in direction orthogonal to $\pi$ while translating it.<|endoftext|> TITLE: $ \mathbb{R}P^n $ bundles over the circle QUESTION [11 upvotes]: Is every $ \mathbb{R}P^{2n} $ bundle over the circle trivial? Are there exactly two $ \mathbb{R}P^{2n+1} $ bundles over the circle? This is a cross-post of (part of) my MSE question https://math.stackexchange.com/questions/4349052/diffeomorphisms-of-spheres-and-real-projective-spaces which has been up for a couple weeks and got 8 upvotes and some nice comments but no answers. My intuition for thinking both answer are yes is that there are exactly 2 sphere bundles over the circle. The trivial one and then the non-trivial (and non-orientable) one which can be realized as the mapping torus of an orientation reversing map of the sphere. So importing that intuition to projective spaces then the orientable $ \mathbb{R}P^{2n+1} $ should have a nontrivial (and non orientable) bundle over the circle while the non orientable $ \mathbb{R}P^{2n} $ should have only the trivial bundle. For $ n=1 $ this checks out since that projective space is orientable and thus we have exactly two bundles over the circle (the trivial one=the 2 torus and the nontrivial one=the Klein bottle). REPLY [8 votes]: No. Every smooth bundle over $S^1$ with fiber $M$ is the mapping torus of some diffeomorphism $f:M\to M$. Isomorphism classes of bundles correspond to conjugacy classes in the group of isotopy classes of diffeomorphisms. There are already surprises when $M$ is $S^n$. For example, an exotic $7$-sphere can be realized as the union of two copies of $D^7$ glued by a diffeomorphism of $S^6$ that is not isotopic to the identity, and this leads to examples. There are related examples with $M=P^n$. REPLY [8 votes]: Your answer is correct if appropriately understood, but it's a little subtle. Here I should note that I'm interpreting your question as a purely homotopy theoretic one (in particular ignoring smooth structure), and that by "bundle" you mean "Serre bundle". If you care about smooth bundles, see Tom Goodwillie's answer. The way I know to do this involves a little algebraic topology. First, it's always the case that $X$-bundles over a circle (for $X$ some topological space) are classified by the group $\pi_0(\mathrm{Aut}(X)),$ where $\mathrm{Aut}(X)$ is the group of maps $X\to X$ which are invertible up to homotopy. Thus a good technique to carry out this classification is to first find all self-maps $X\to X$ up to homotopy, then see which ones are invertible. Now we want to specialize to the case $X = RP^n,$ for some $n$. However it turns out that it's much easier to classify maps not into $RP^n$ itself, but rather into $RP^\infty.$ Namely, maps from some space $X$ to $RP^\infty$ up to homotopy are always classified by $H^1(X, \mathbb{Z}/2)$ (equivalently, this is group homomorphisms from $\pi_1(X)$ into $\mathbb{Z}/2$). Now the difference between $RP^n$ and $RP^\infty$ isn't actually too terrible. Namely, the CW approximation theorem tells us that any map $RP^n\to RP^\infty$ actually can be chosen up to homotopy such that it factorizes through $RP^n\subset RP^\infty.$ So we can compute $$\pi_0\mathrm{Maps}(RP^n, RP^\infty) \cong H^1(RP^n, \mathbb{Z}/2)\cong \mathbb{Z}/2.$$ The element $0\in \mathbb{Z}/2$ corresponds to the trivial map $$*:RP^n\to RP^n$$ mapping everything to a basepoint in $RP^\infty$ and the nontrivial element $1\in \mathbb{Z}/2$ corresponds to the identity map, $$id:RP^n\to RP^n\subset RP^\infty.$$ And any other map $RP^n\to RP^n$ will be homotopy equivalent to one of these as an element of $\mathrm{Maps}(RP^n, RP^\infty)$. However there's a catch: while any map $RP^n\to RP^n$ is homotopy equivalent to one of the maps $id, *$ as a map to $RP^\infty$, the homotopy between the two maps might not live in $RP^n$. So a priori, there can be multiple homotopy classes of self-maps of $RP^n$ homotopic to one of these maps in $RP^\infty$ (and indeed, sometimes there are). To get a handle on how badly maps to $RP^\infty$ are undercounting, you can apply the CW approximation theorem again to see that any homotopy between two maps $RP^n\to RP^\infty$ will factorize, up to homotopy, through $$RP^{n+1}\subset RP^\infty.$$ The "error" of such a homotopy existing in $RP^n$ will be classified by a map to the quotient, $$RP^n\times [0,1]\to RP^{n+1}/RP^n\cong S^{n+1},$$ taking both $RP^n\times \{0\}, RP^n\times \{1\}$ to the basepoint, in other words, a based map from the space $$RP^n_+\wedge S^1 = \Sigma(RP^n_+)$$ (you can think of this as the ordinary suspension $\Sigma(RP^n)$ with the two suspension points identified) to $S^{n+1}.$ Let's write $$D: = \pi_0\text{Maps}(RP^n_+\wedge S^1, S^n)$$ for the set of possible such "defects" of a homotopy in $RP^{n+1}$ restricting to $RP^n$. Based maps from a closed $n+1$-dimensional manifold to $S^{n+1}$ are classified by ordinary $H^{n+1},$ and so we have $$D \cong H^{n+1}(\Sigma(RP^n_+)) \cong H^n(RP^n) \cong \begin{cases} \mathbb{Z}/2, & n\text{ even}\\ \mathbb{Z}, & n\text{ odd}. \end{cases}$$ Thus for each of the maps $*, id: RP^n\to RP^n,$ there can be at worst $D$ worth of distinct other homotopy classes of self-maps $RP^n\to RP^n$ homotopic to it as maps to $CP^{\infty}.$ Since a map homotopic to $*$ cannot be an automorphism (it would have to induce the trivial map on $H^1$ which cannot come from an automorphism), we can restrict our attention to maps homotopic in $CP^\infty$ to $id:RP^n\to RP^n.$ A priori, there could have been elements of $D$ which are not realizable as "defects" of homotopies between maps $RP^n\to RP^n$, but in this case we don't run into this problem: indeed, every element of $D$ occurs as the defect of some homotopy $RP^n\times [0,1]\to RP^{n+1}$ between the identity $id:RP^n\to RP^n$ and another map. Namely, recall that we have realized $D \cong H^n(RP^n)$ as a cyclic group, either $\mathbb{Z}$ or $\mathbb{Z}/2$ (depending on parity). Let $\alpha$ be a generator of this group. Then every element of $D$ can be written $k\alpha$ for some $k\in \mathbb{Z}$. By doing a calculation, you can see that each element $k\alpha\in D$ is realized as the defect of the homotopy $$[0,1]\cdot RP^n\to RP^{n+1}$$ induced by the map $[0,1]\times S^n\to S^{n+1}$ given by rotating $S^n$ in a circle around some ($n-1$-dimensional) axis inside $S^{n+1},$ by an angle of $$k\cdot \pi.$$ Now if $n$ is even, we see the resulting "new" map $RP^n\to RP^n$ is once again the identity. If $n$ is odd, the new map is induced from the "reflection" map given by $$\sigma:(x_1,x_2,\dots, x_n)\mapsto (-x_1,x_2, \dots, x_n)$$ (in some coordinates). Thus from what we've done so far, there can be at most two homotopy invertible self-maps up to homotopy $$RP^n\to RP^n$$ for any $n$, namely $id$ and $\sigma.$ It remains to check whether the induced two self-maps $RP^n\to RP^n$ are homotopic to each other. When $n$ is odd, they cannot be homotopic to each other since $RP^n$ is orientable, and $\sigma$ reverses orientation (so $\sigma$ can be distinguished from $id$ by looking at action on $H^n$). But when $n$ is even, the map $\sigma$ is homotopy equivalent via sphere rotations to the map $(x_1,x_2\dots, x_n)\mapsto (-x_1,-x_2\dots, -x_n),$ and this induces a homotopy between $\sigma$ and $id$ as maps $RP^n\to RP^n.$ Thus we have $$\pi_0(\mathrm{Aut}(RP^n)) \cong \begin{cases} \{id\}, & n \text{ even}\\ \{id, \sigma\}, & n \text{ odd}. \end{cases} $$ As mentioned, $RP^n$-bundles on $S^1$ are classified by the same data, so your guess is correct.<|endoftext|> TITLE: Are there infinitely many primes of the form $\lfloor e x\rfloor$ for $x\in\mathbb{Z}^+$? QUESTION [7 upvotes]: Construct a function $f(x)=\lfloor e x\rfloor$. For each positive integer $x$, $f(x)$ will be a positive integer. Among these integers $f(x)$, are there an infinite number of primes? REPLY [25 votes]: Something is known in considerably greater generality. Let $\lVert x\rVert=\min_{n\in\mathbb{Z}}|x-n|$. For $\gamma\in\mathbb{R}-\mathbb{Q}$, define $$\tau(\gamma) = \sup\Big\{\rho\in\mathbb{R}\colon \liminf_{n\to\infty} n^{\rho}\lVert\gamma n\rVert=0\Big\},$$ where the limit inferior is as $n\to\infty$ along the integers. Dirichlet's approximation theorem implies that for all irrational $\gamma$, we have $\tau(\gamma)\geq 1$. Khinchin and Roth proved that $\tau(\gamma)=1$ for almost all real numbers $\gamma$ (in the sense of Lebesgue measure), and $\tau(\gamma)=1$ for all (real) irrational algebraic integers $\gamma$. If $\gamma>1$ and $\tau(\gamma)<\infty$, then there is an asymptotic prime number theorem that counts the primes of the form $p=\lfloor \gamma n+\beta\rfloor$, where $n\geq 1$ is an integer and $\beta\in\mathbb{R}$ is fixed. A standard source for this is the paper Prime numbers with Beatty sequences by Banks and Shparlinski.<|endoftext|> TITLE: Logics detecting their own equivalence notions, take two: $\mathcal{L}_{\omega_2,\omega}$ QUESTION [6 upvotes]: This question is a follow-up to another question of mine, with different language - see the link below. Say that an infinite regular cardinal $\kappa$ is Fraissean iff the logic $\mathcal{L}_{\kappa,\omega}$ has the following property (called "SED" in the below-linked question): For every finite signature $\Sigma$ there is a larger signature $\Sigma'$ containing $\Sigma$ and two new unary predicate symbols $A,B$ - and possibly more symbols besides - and an $\mathcal{L}_{\kappa,\omega}[\Sigma]$-sentence $\eta$ such that, for every pair of $\Sigma$-structures $\mathfrak{A},\mathfrak{B}$, we have $\mathfrak{A}\equiv_{\kappa,\omega}\mathfrak{B}$ iff there is an $\mathfrak{M}\models\eta$ with $A^\mathfrak{M}\upharpoonright\Sigma\cong\mathfrak{A}$ and $B^\mathfrak{M}\upharpoonright\Sigma\cong\mathfrak{B}$. (That $\omega$ is Fraissean is an immediate consequence of Fraisse's characterization of elementary equivalence in terms of Ehrenfeucht-Fraisse games, hence the name.) Farmer S showed that $\omega_1$ is not Fraissean; that argument, however, does not seem to immediately generalize to higher cardinals, the issue being that $\mathcal{L}_{\kappa,\omega}$-sentences are not generally coded by reals and so even in the presence of large cardinals we lose a necessary absoluteness result. My question is: what can we say, in $\mathsf{ZFC}$ alone, about the situation re: $\omega_2$? Neither possibility has an obvious consistency proof to me; I would tentatively hazard a guess that $L$ thinks $\omega_2$ is not Fraissean and that the tree property at $\omega_2$ implies that $\omega_2$ is Fraissean, but both of these are essentially just free association. REPLY [10 votes]: (Working in ZFC.) $\omega_2$ is not Fraissean. In fact, it is not Fraissean with respect to $\Sigma$, where $\Sigma$ is the signature with a single binary relation $<$. To see this we use a variant of the argument you linked in the question. Suppose otherwise, and let $\Sigma'$ and $\eta$ witness this. Let $\gamma$ be a large enough ordinal and let $\pi:M\to V_\eta$ be elementary, with $M$ transitive and $M^{\omega_1}\subseteq M$ and $\mathrm{crit}(\pi)=\kappa$ exists. Let $\mathfrak{A}=(\kappa,{{\in}\upharpoonright\kappa})$ and $\mathfrak{B}=(\pi(\kappa),{{\in}\upharpoonright\pi(\kappa)})$. Since $M^{\omega_1}\subseteq M$, $M$ is correct about $\mathcal{L}_{\omega_2,\omega}$-truth, and also by elementarity of $\pi$, therefore $\mathfrak{A}\equiv_{\mathcal{L}_{\omega_2,\omega}}\mathfrak{B}$. So let $\mathfrak{M}$ witness the choice of $\Sigma',\eta$ with respect to $\mathfrak{A},\mathfrak{B}$. Let $G$ be $V$-generic for $\mathbb{P}=\mathrm{Coll}(\omega_1,\theta)$ where $\theta=\max(\pi(\kappa),\mathrm{card}(\mathfrak{M}))$ (collapsing $\theta$ to size $\aleph_1$ with countable conditions). Then $V[G]\models$"There are structures $\mathfrak{A}',\mathfrak{B}',\mathfrak{M}'$, each having universe $\omega_1$, such that $\mathfrak{A}',\mathfrak{B}'$ are in signature $\Sigma$, and $\mathfrak{M}'$ in signature $\Sigma'$, and $A^{\mathfrak{M}}\upharpoonright\Sigma\approx\mathfrak{A}'$ and $B^{\mathfrak{M}}\upharpoonright\Sigma\approx\mathfrak{B}'$, as witnessed by isomorphisms $\sigma,\tau$, and $\mathfrak{M}'\models\eta$ and there is a sentence $\varphi$ of $\mathcal{L}_{\omega_2^{V[G]},\omega}$, coded by a set $X\subseteq\omega_1$, such that $\mathfrak{A}'\models\varphi$ but $\mathfrak{B}'\models\neg\varphi$" (consider the natural sentence specifying the ordertype of $\kappa$). Fix names $\dot{\mathfrak{A}}',\dot{\mathfrak{B}}',\dot{\mathfrak{M}}',\dot{\sigma},\dot{\tau},\dot{X}$ for such objects (we may assume the empty condition forces the above things to hold for these names). So these are all basically names for subsets of $\omega_1$. Now working in $V$, given an $\aleph_1$-sized family $\mathscr{F}$ of dense subsets $D\subseteq\mathbb{P}$, we can build an $\mathscr{F}$-generic filter $G$, because $\mathbb{P}$ is countably closed. We claim that by picking $\mathscr{F}$ appropriately, and $G$ be $\mathscr{F}$-generic, then letting $\mathfrak{A}'$, etc, be the interpretations $\dot{\mathfrak{A}}'_G$, and $\varphi$ the sentence in $\mathcal{L}_{\omega_2,\omega}$ the sentence coded by $X'$, then the sentence of the previous paragraph which held in the generic extension, holds in $V$ about these objects, which contradicts our assumptions. To arrange $\mathscr{F}$: First, it is straightforward to arrange $\aleph_1$-many dense sets which arrange that $\dot{\sigma}_G$ and $\dot{\tau}_G$ will truly be isomorphisms; the main thing is to arrange the that the domain and codomain are the right sets. To arrange that $\mathfrak{M}'\models\eta$, consider not just $\eta$, but the set $S$ of all subformulas thereof. Now here that all formulas in $S$ have only finitely many distinct free variables (if $\psi$ has infinitely many distinct free variables, proceed by induction on the rank of formulas which have it as a subformula, to see that none of them are sentences). Since $\eta$ is in $\mathcal{H}_{\omega_2}$, we can fix a surjection $\pi:\omega_1\to S^+$, where $S^+$ is the set of pairs $(\psi,\vec{\alpha})$, where $\psi\in S$ and $\vec{\alpha}$ is an assignment of the (finitely many) free variables of $\psi$ to ordinals ${<\omega_1}$. Now the basic point is that if $(\psi,\vec{\alpha})\in S^+$ and $\psi$ is a disjunction $\bigvee_{\gamma<\omega_1}\psi_\gamma$, and $p\in\mathbb{P}$ forces $\dot{\mathfrak{M}}'\models\psi(\vec{\alpha})$, then for each $q\leq p$ we can pick some $\gamma<\omega_1$ and $r\leq p$ such that $r$ forces that $\dot{\mathfrak{M}}'\models\psi_\gamma(\vec{\alpha})$. Thus, we can include a dense set $D_{\psi,\vec{\alpha}}$ consisting of those conditions $p$ such that either $p$ forces that $\dot{\mathfrak{M}}'\models\neg\psi(\vec{\alpha})$ or there is $\gamma<\omega_1$ such that $p$ forces that $\dot{\mathfrak{M}}'\models\psi_\gamma(\vec{\alpha})$. Similarly, if $\psi$ is of form $\exists x\varrho$, then we can include the dense set of conditions $p$ either forcing that $\dot{\mathfrak{M}}'\models\neg\psi(\vec{\alpha})$, or such that there is $\alpha<\omega_1$ such that $p$ forces $\dot{\mathfrak{M}}'\models\varrho(\alpha,\vec{\alpha})$. And naturally, for each $(\psi,\vec{\alpha})\in S^+$, we include the dense set of $p$ deciding whether $\dot{\mathfrak{M}}'\models\psi(\vec{\alpha})$. There are also natural dense sets forcing the atomic formulas the correspond to the structure of $\dot{\mathfrak{M}}'$. For $X'$ and $\varphi'$ it is similar, but we no longer have $\varphi$ fixed in advance. But we can fix a name $\dot{S}_{\dot{\varphi}}^+$ for the set of all pairs $(\psi,\vec{\alpha})$ such that $\psi$ is a subformula of $\varphi$ (note in an actual forcing extension, we could have $\psi\notin V$), and $\vec{\alpha}$ an assignment of the free variables of $\psi$ to elements ${\in\omega_1}$. Fix a name for a surjection $\dot{f}:\omega_1\to \dot{S}^+_{\dot{\varphi}}$. Write $\dot{\psi}_\gamma$ for the name for the first component of $\dot{f}(\gamma)$, and $\vec{\alpha}_\gamma$ for the name for the second. Then for each $\gamma<\omega_1$, we can use the dense set deciding what sort of formula $\dot{\psi}_\gamma$ is (in particular, whether it is a disjunction, and whether it is existential), and the dense set of all conditions $p$ such that (i) there is $\vec{\alpha}$ such that $p$ forces "$\dot{\vec{\alpha}}_\gamma=\vec{\alpha}$", (ii) for some $\beta<\omega_1$, either $p$ forces that $\dot{\psi}_\gamma$ is not a disjunction, or $p$ forces that $\dot{\mathfrak{A}}'\models\neg{\dot{\psi}}_\gamma(\vec{\alpha})$, or $p$ forces "$\dot{\psi}_\gamma$ is a disjunction, $\dot{\psi}_\beta$ is one of the disjuncts of $\dot{\psi}_\gamma$, $\dot{\vec{\alpha}}_\beta=\vec{\alpha}$ and $\dot{\mathfrak{A}}'\models\dot{\psi}_\beta(\vec{\alpha})$", and (iii) there are $\alpha,\beta<\omega_1$ such that either $p$ forces that $\dot{\psi}_\gamma$ is not existential, or $p$ forces that $\dot{\mathfrak{A}}'\models\neg\dot{\psi}_\gamma$, or $p$ forces "letting $\varrho=\dot{\psi}_\beta$, then $\dot{\psi}_\gamma$ is the formula $\exists x\ \varrho$, and $\vec{\alpha}_\beta=\vec{\alpha}\frown(\alpha)$, and $\dot{\mathfrak{A}}'\models\varrho(\vec{\alpha},\alpha)$". Also include dense sets deciding whether $\dot{\mathfrak{A}}'\models\dot{\psi}_\gamma(\dot{\vec{\alpha}}_\gamma)$ holds, for each $\gamma$. Likewise for $\dot{\mathfrak{B}}'$. Also for each pair of ordinals $\alpha,\beta<\omega_1$, include the dense set of conditions $p$ such that either $p$ forces that $\dot{\psi}_\alpha$ is not a disjunction/conjunction, or $p$ forces that $\dot{\psi}_\alpha$ is a disjunction/conjunction, and $\dot{\psi}_\beta$ is one of the disjuncts/conjuncts, or $p$ forces that $\dot{\psi}_\alpha$ is a disjunction/conjunction, and $\dot{\psi}_\beta$ is not one of the disjuncts/conjuncts. Also for each $\alpha<\omega_1$, the dense set of conditions deciding the number (finitely many) of free variables of $\dot{\psi}_\alpha$, and arranging that $\dot{f}_G$ truly enumerates all of the relevant pairs $(\psi,\vec{\alpha})$. (I think this is now about enough dense sets.) Now let $G$ be $\mathscr{F}$-generic. Let $X'=\dot{X}'_G$, etc. It is straightforward to see that $\mathfrak{M}'\models\eta$ and $\mathfrak{A}'\approx A^{\mathfrak{M}'}\upharpoonright\Sigma$ (as witnessed by $\sigma'$) and likewise for $\mathfrak{B}'$. So by our assumptions, $\mathfrak{A}'\equiv_{\mathcal{L}_{\omega_2,\omega}}\mathfrak{B}'$. But $X'$ does really code a sentence $\varphi'$ of $\mathcal{L}_{\omega_2,\omega}$, because the ordertype along which it is built is wellfounded, because by the $\sigma$-closure of $\mathbb{P}$, otherwise we can find a condition $p$ which forces something to be an illfounded ordinal. And $\dot{S}^+{\dot{\varphi}}_G$ is the set of pairs $(\psi,\vec{\alpha})$ such that $\psi$ is a subformula of $\varphi'$, and $\vec{\alpha}$ an assignment of its free variables, and this set is enumerated by $\dot{f}_G$. And note that the names for the satisfaction relations amongst these formulas evaluate to the correct satisfaction relation. Thus, $\mathfrak{A}'\models\varphi'$ but $\mathfrak{B}'\models\neg\varphi'$, a contradiction.<|endoftext|> TITLE: How to integrate the multinomial over a ball in $\mathbb{R}^{n}$? QUESTION [5 upvotes]: I got an interesting question. Consider this integral: $$ \int_{B(0,1)}\bigg(\sum_{j=1}^{n}a_{j}x_{j}^2\bigg)^m \mbox{d}x, \quad m,n\in \mathbb{N}, \ a_{i}>0, \ i=1,2,\ldots,n.$$ It is clear that if we choose all $a_{j}=1$, then it will be simple! Futhermore, I want to know if the answer is connected with some special functions like hyperbolic functions, please do not use a finite summation in the result. If we just expand this multinomial directly, that is, $$ \bigg(\sum_{j=1}^{n}a_{j}x_{j}^2\bigg)^m =\sum_{\lambda_{1}+\lambda_{2}+\ldots+\lambda_{n}=m}\frac{m!}{\lambda_{1}!\lambda_{2}\ldots\lambda_{n}!}a_{1}^{\lambda_{1}}a_{2}^{\lambda_{2}}\ldots a_{n}^{\lambda_{n}}x_{1}^{2\lambda_{1}}x_{2}^{2\lambda_{2}}\ldots x_{n}^{2\lambda_{n}}, $$ then, after a simple calculation, in Table of Integrals I found a related formula, $$\frac{\bigg(\sum_{i=1}^{r}a_{i}^{2}\bigg)^{\frac{n}{2}}}{n!}H_{n}\left(\frac{\sum_{i=1}^{r}a_{i}x_{i}}{\sqrt{\sum_{i=1}^{r}a_{i}^{2}}}\right)=\sum_{\substack{m_{1}+\ldots+m_{r}=n,\\{m_{i}\geq 0}}}\prod_{k=1}^{r}\bigg\{\frac{a_{k}^{m_{k}}}{m_{k}!}H_{m_{k}}(x_{k})\bigg\}, $$ where $H_{n}(x)$ denotes the Hermite polynomials; I don't know if it works. If this result is a little hard, one can just find the asymptotic representation as $m\leq n$ and $m\rightarrow \infty$. REPLY [4 votes]: $\newcommand\R{\mathbb R}\newcommand{\Ga}{\Gamma}$The integral in question is \begin{equation*} \begin{aligned} \int_0^1 dr\,\int_{rS_{n-1}}dx\,f(x) &=\int_0^1 dr\,r^{2m+n-1}\int_{S_{n-1}}du\,f(u) \\ &=\frac1{2m+n}\,|S_{n-1}|Ef(U_n), \end{aligned} \end{equation*} where \begin{equation*} f(x):=\Big(\sum_{j=1}^n a_j x_j^2\Big)^m \end{equation*} (so that $f$ is $2m$-homogeneous), $|S_{n-1}|$ is the surface area of the unit sphere $S_{n-1}$ in $\R^n$, and $U_n$ is a random vector uniformly distributed on $S_{n-1}$. So, the problem reduces to finding the asymptotics of $Ef(U_n)$. Without loss of generality, \begin{equation*} U_n=\frac G{\|G\|}, \end{equation*} where $G=(G_1,\dots,G_n)$ is a standard Gaussian vector in $\R^n$ and $\|\cdot\|$ is the Euclidean norm. Let us normalize the $a_j$'s by assuming that \begin{equation*} \sum_{j=1}^n a_j=1 \tag{0} \end{equation*} and consider the following two extreme cases: Case 1: $a_1=1$, $a_2=\cdots=a_n=0$ and Case 2: $a_1=\cdots=a_n=1/n$. In Case 1, \begin{equation*} f(U_n)=\Big(\frac{G_1^2}{\|G\|^2}\Big)^m, \end{equation*} and $\dfrac{G_1^2}{\|G\|^2}$ has the beta distribution with parameters $1/2,(n-1)/2$. So, here \begin{equation*} Ef(U_n)=\frac{\Ga(n/2)\Ga(m+1/2)}{\Ga(1/2)\Ga(m+n/2)}. \end{equation*} So, if e.g. $m=n\to\infty$, then the integral in question is \begin{equation*} \sim\frac1{3n}\,\sqrt6\,\Big(\frac2{3\sqrt3}\Big)^n\,|S_{n-1}|. \tag{1} \end{equation*} So, here the coefficient of $|S_{n-1}|$ decreases exponentially. In the trivial Case 2, $f(U_n)=1/n^m$, and hence for $m=n\to\infty$ the integral in question is \begin{equation*} =\frac1{3n^{n+1}}\,|S_{n-1}|, \end{equation*} which is very different from the asymptotics for Case 1 given in (1). Between these two extreme cases, any intermediate asymptotics (or absense of any asymptotics) should be possible. As was noted in comments, the asymptotics in general (if any) will very much depend on how much the $a_j$'s differ from one another. Addendum 1: To be more specific, consider the following setting, intermediate between Cases 1 and 2: $a_1=\cdots=a_k=1/k$, $a_{k+1}=\cdots=a_n=0$ for natural $k\in[2,n-1]$, where $k$ is allowed to vary with $n$ and $m$. Then \begin{equation*} f(U_n)=\frac1{k^m}\,\Big(\frac{G_1^2+\cdots+G_k^2}{\|G\|^2}\Big)^m, \end{equation*} and $\dfrac{G_1^2+\cdots+G_k^2}{\|G\|^2}$ has the beta distribution with parameters $k/2,(n-k)/2$. So, here \begin{equation*} Ef(U_n)=\frac1{k^m}\,\prod_{r=0}^{m-1}\frac{k/2+r}{n/2+r}. \tag{2} \end{equation*} So, if $k/m^2\to\infty$, then \begin{equation*} Ef(U_n)\sim\frac1{n^m}, \tag{3} \end{equation*} which is the same asymptotics as for $m$ fixed. However, if $n/m^2\to\infty$ and $k\sim cm^2$ for some $c\in(0,\infty)$, then \begin{equation*} Ef(U_n)\sim\frac{e^{1/c}}{n^m}, \end{equation*} now with the additional factor $e^{1/c}$ as compared to (3). If we now let $n/m^2\to\infty$ and $k\sim cm$ for some $c\in(0,\infty)$, then an additional exponentially growing (with $m$) factor (as compared to (3)) will appear. Addendum 2: Letting $g(a_1,\dots,a_n):=Ef(U_n)$, we see that the function $g$ is symmetric and convex, and hence Schur convex -- see e.g. Theorem A on p. 258. So, in view of condition (0), \begin{equation*} Ef(U_n)\ge g(1/n,\dots,1/n)=\frac1{n^m}. \tag{$\clubsuit$} \end{equation*} That is, the smallest value of $Ef(U_n)$ is attained in Case 2 ($a_1=\cdots=a_n=1/n$), considered above. Addendum 3: For completeness, consider also the case when $m$ is fixed (even though $m\to\infty$ in the OP). Then, by (2), \begin{equation*} Ef(U_n)\sim\frac1{n^m}\,\prod_{r=0}^{m-1}\Big(1+\frac{2r}k\Big) \end{equation*} if $a_1=\cdots=a_k=1/k$ and $a_{k+1}=\cdots=a_n=0$ for a fixed natural $k\in[2,n-1]$ and $n\to\infty$. We see that, even when $m$ is fixed, the asymptotics depends on $k$ and, more generally, on how much the $a_j$'s differ from one another. Addendum 4: Here we complement the lower bound on $Ef(U_n)$ given by ($\clubsuit$) in Addendum 2 by providing a matching upper bound on $Ef(U_n)$ that implies the following: If the $a_i$'s are uniformly small in the sense that \begin{equation*} a:=\max_{i=1}^n a_i\to0 \end{equation*} and, moreover, $m$ is at most moderately large in the sense that \begin{equation*} m^3 a\to0, \tag{4} \end{equation*} then \begin{equation*} Ef(U_n)\sim\frac1{n^m} \tag{$\heartsuit$} \end{equation*} (as $n\to\infty$). Indeed, the random point $(x_1^2,\dots,x_n^2)$ has the Dirichlet distribution with parameters $1/n,\dots,1/n$, and the Dirichlet distribution has the negative association (NA) property. So, by (say) Theorem 2, \begin{equation*} Ef(U_n)\le E\Big(\sum_{j=1}^n a_j Y_j\Big)^m, \tag{5} \end{equation*} where the $Y_j$'s are iid random variables each with the beta distribution with parameters $1/2,(n-1)/2$. Denoting now the $L^m$ norm by $\|\cdot\|_m$ and using Minkowski's inequality, we get \begin{equation*} (Ef(U_n))^{1/m}\le\Big\|\sum_{j=1}^n a_j Y_j\Big\|_m \le \Big\|\sum_{j=1}^n a_j EY_j\Big\|_m +\Big\|\sum_{j=1}^n a_j Z_j\Big\|_m, \end{equation*} where $Z_j:=Y_j-EY_j$. Since $EY_j=1/n$, we have \begin{equation*} \Big\|\sum_{j=1}^n a_j EY_j\Big\|_m = \sum_{j=1}^n a_j EY_j=\frac1n. \tag{6} \end{equation*} Note also that $Var\,Y_j\sim2/n^2$ and $\|Z_j\|_m\le2\|Y_j\|_m\ll m/n$; here and in what follows, $A\ll B$ means $A\le CB$ for some universal real constant $C>0$. Using now an appropriate version of Rosenthal's inequality (see e.g. Theorem 6.1), we get \begin{equation*} \begin{aligned} \Big\|\sum_{j=1}^n a_j Z_j\Big\|_m &\ll\frac1n\,(m^2 a^{1-1/m}+m^{1/2} a^{1/2}) \\ &=\frac1n\,\frac1m\,((m^3 a)^{1-1/m}m^{3/m}+(m^3 a)^{1/2}) =o\Big(\frac1n\,\frac1m\Big), \end{aligned} \end{equation*} by (4). So, by (5) and (6), \begin{equation*} (Ef(U_n))\le\frac1{n^m}\Big(1+\frac{o(1)}m\Big)^m =\frac{1+o(1)}{n^m}. \end{equation*} Now ($\heartsuit$) follows, in view of ($\clubsuit$).<|endoftext|> TITLE: The site of extremally disconnected sets QUESTION [5 upvotes]: In proposition 2.7. of the condensed notes of professors Scholze and Clausen it is said that the category of extremally disconnected sets is a site, but in the definition of a site in the Stacks Project (https://stacks.math.columbia.edu/tag/00VH) it is necessary for a site to have fibre products sometimes (Axiom (3) in the definition) and extremally disconnected sets don't have all fibre products. The category of extremally disconnected sets is a site using the definition from the Stacks Project? REPLY [8 votes]: Usage varies. Let's at least stipulate that "site" is synonymous with "category equipped with a Grothendieck topology". Some, but not all, authors, require a site to have pullbacks, because this assumption simplifies the definition a bit. But e.g. the nlab gives the definition which doesn't assume one has pullbacks. Apparently this version of the definition goes back at least to SGA 4.<|endoftext|> TITLE: M. M. Artyukhov / М. М. Артюхов QUESTION [20 upvotes]: Does anybody know any biographical information about М. М. Артюхов (e.g., first name, affiliation)? It seems he discovered a criterion for primality equivalent to the Solovay–Strassen one in 1966, in this paper: Некоторые критерии простоты чисел, связанные с малой теоремой Ферма. (Keith Conrad discusses the story a little bit here: The Miller–Rabin test.) I'm just curious. :-) REPLY [13 votes]: In the books [1], pp. 41-42 and [2], p. 67, it is stated that his name (and patronymic) is Mikhail Mikhailovich, confirming Anatoly Kochubei's comment to Kostya_I's answer. References also state that he was born on the 26 of April 1910 in Moscow, laureated from (the then called) Leningrad University in 1936, become "aspirant" (aspirant in science) until 1939, then "kandidat fiz.-matem. nauk" (candidate in physics and mathematical sciences) and "docent" (professor) from 1945 onward. After 1954 he worked at the North Caucasus (Ordzhonikidze) Metallurgical Institute ([1], p. 41), thus Kostya_I's answer is further confirmed. In [1], p. 42, there's a list of six of his works, the first being dated 1934 and the latest one being dated 1957: in [2], p. 67, the list is continued with four more works, dated from 1958 to 1963. References [1] Fomin, S. V.; Shilov, G. E., eds. (1970), Математика в СССР 1958–1967 [Mathematics in the USSR 1958–1967] (in Russian), Том второй: Биобиблиография выпуск второй М–Я, Москва: Издательство "Наука", p. 762, MR 0250816, Zbl 0199.28501. [2] Kurosh, A. G.; Vityushkov, V. I.; Boltyanskii, V. G.; Dynkin, E. B.; Shilov, G. E.; Yushkevich, A. P., eds. (1959), Математика в СССР за сорок лет 1917–1957 [Mathematics in the USSR 1917–1957] (in Russian), Том второй: Биобиблиография, Москва: Государственное Издательство Физико–Математическои Литературы, p. 819, MR 0115874, Zbl 0191.27501.<|endoftext|> TITLE: Regarding the surgery construction in "A procedure for killing homotopy groups of differentiable manifolds" by Milnor QUESTION [8 upvotes]: In the first section of "A procedure for killing homotopy groups of differentiable manifolds", Milnor gives the surgery construction as follows. Let $W$ be an $n=p+q+1$ dimensional manifold. Given a smooth, orientation preserving embedding: $$f:S^p\times D^{q+1}\to W$$ we may obtain a new manifold as the disjoint sum $$(W-f(S^p\times 0))\cup (D^{p+1}\times S^q)$$ modulo some equivalence relation. My question is: why in this construction is Milnor deleting $S^p\times 0$ from $W$ and not $S^p\times D^{q+1}$? I expected this disjoint sum to be $$(W-f(S^p\times D^{q+1}))\cup (D^{p+1}\times S^q).$$ REPLY [8 votes]: There is no error in the paper as far as I can tell. One thinks of the surgery as taking out the interior of the image of $S^p\times D^{q+1}$, and glueing in a copy of $D^{p+1}\times S^q$ along the common boundary $S^p\times S^q$. However, that does only define a topological space and not an orientable differentiable manifold. That's why one takes $W'=W\setminus f(S^p\times\{0\})$ and glues in $D^{p+1}\times S^q$ on the overlap $$ f(S^p\times (D^{q+1}\setminus\{0\}))\cong (D^{p+1}\setminus\{0\})\times S^q $$ the identification being $f(u,\theta v)\leftrightarrow (\theta u,v)$, for $(u,v)\in S^p\times S^q$ and $0<\theta\leq 1$.<|endoftext|> TITLE: Hochschild cohomology of group ring of a free group QUESTION [6 upvotes]: Let $G$ be a free group of finite rank. Consider any commutative ring $R$ containing $\mathbb{Z}$. Consider the group ring $RG$. Q) What can we say about the Hochschild cohomology groups of $RG$ with coefficients in $R$? I can find the $HH^1(RG)$ in terms of outer derivations but again I don't know the complete description of derivations on a free group. Also, I could not find a single element of $HH^n(RG)$ for $n\geq 2$. Any comment, reference, or suggestion will be extremely helpful. Thanks in advance. REPLY [8 votes]: Hochschild cohomology of group rings is reducible to ordinary group cohomology. (Not really if you're interested in multiplicative structure, but that's another story). Everyting works over any commutative base ring, you can assume it's $\Bbb Z$ if you'd like. Namely, there's an adjunction between $kG-bimod$ and $kG-mod$ (I mean left modules): forgetful right adjoint functor $\rho$ sends a bimodule $A$ to the module $\rho A$ where action is conjugation, and left adjoint $\lambda$ is $M \mapsto \lambda M := M \otimes_k kG$. Right action is multiplication in $kG$, and left action is $h \cdot \sum_G m \otimes g = \sum_G hm \otimes hg$. So, now we can get a string of identifications $$HH^*(kG, A) = Ext_{bimod}(kG, A) = Ext_{bimod}(\lambda k, A) = Ext_{mod}(k, \rho A) = H^*(G, \rho A)$$ In particular, Hochschild cohomological dimension of a group ring is equal to comohomological dimension of a group itself, and $HH^*(kG, k) = H^*(G, k)$. (so if your group is free, there's no second cohomology at all) Everything above is pretty much copied verbatim from chapter X of Homological algebra by Cartan and Eilenberg. I'd suggest looking at Dan Burghelea and Sarah Witherspoons papers for further reading, but cannot provide explicit references for a while (because searching old papers from a phone is somewhat difficult).<|endoftext|> TITLE: Functional equation with Fourier transform and $\frac{1}{x} f(\frac{1}{x}) $ QUESTION [8 upvotes]: What are the continuous functions $f$ such that on $\mathbb{R}^{+*}$, they satisfy following functional equation: $$\int_0^\infty f(t) e^{-itx} \, dt =\lambda \frac{1}{x} f\left(\frac{1}{x}\right)$$ $\lambda$ is a constant. The functions $f(x)=x^{\alpha}$ with $-1<\operatorname{Re}(\alpha)<0$ are solution, but can we find other solutions to this equation ? Any method to solve this problem ? I tried to transform it to find a differential equation but did not succeed ... REPLY [2 votes]: To start with: this functional equation is a "Fredholm integral equation of second kind". We use the Mellin transform to find solutions. (See page 657 of "Handbook of integral equations"). Lets make Mellin transform on both side: $$\int_0^\infty \int_0^\infty f(t) e^{-itx} \, dt \, x^{s-1} dx=\lambda \int_0^\infty \frac{1}{x} f\left(\frac{1}{x}\right) x^{s-1} dx$$ $$\int_0^\infty \int_0^\infty f(t) t^{-s}e^{-ix} \, x^{s-1} \, dt \,dx=\lambda \int_0^\infty f\left(x\right) x^{-s} dx$$ We note $\mathcal{M}(f)(s)$ the Mellin transform of $f$ $$\mathcal{M}(e^{-ix})(s)\cdot \mathcal{M}(f)(1-s) =\lambda \,\mathcal{M}(f)(1-s)$$ $$[\mathcal{M}(e^{-ix})(s) - \lambda] \mathcal{M}(f)(1-s)=0$$ $$\left[\cos\left(\frac{\pi s}{2}\right) + i\sin\left(\frac{\pi s}{2}\right) - \lambda\right] \mathcal{M}(f)(1-s)=0$$ Under this form, we see either $\mathcal{M}(f)(1-s)=0$ either we have $\lambda$ such that there exist $\alpha$ such that $\cos(\frac{\pi \alpha}{2}) + i\sin(\frac{\pi \alpha}{2})=\lambda$ and $\mathcal{M}(f)(1-s)=\delta(s-\alpha)$. So taking Mellin inverse of $\delta(s-\alpha)$ we see the functional equation has only solutions of the form $x^{a}$.<|endoftext|> TITLE: Computation of modified Gauss sums QUESTION [11 upvotes]: Let $\chi$ be a primitive Dirichlet character of conductor $q$. I want to compute numerically $$G(k)=\sum_{n\bmod q}\chi(n)e^{2\pi i n(n-k)/(2q)}$$ for all $k$ with $0\le k<2q$ with $k\equiv q\pmod2$ (thanks to this last condition the sum $G(k)$ is well defined). I have two questions: For now I compute these values in a naive way, so using $q*q$ steps (of course precomputing the $2q$th roots of unity). Is there a better method (even reducing to $q^2/2$ steps would be nice)? Imagine $q=10^5$ or $q=10^6$. When $q$ is not prime, $G(k)$ is often equal to $0$. For instance if $p$ is a prime congruent to $3$ mod 4 dividing $q$ and $k$, it seems that $G(k)=0$, at least for quadratic characters (but I am interested in all characters). My question is: give a necessary and sufficient condition for $G(k)=0$ (sufficient would already be nice). REPLY [3 votes]: I might be missing something, but I believe the below gives $O(q \log q)$ to compute $G(k)$ for all $k$. For a FFT-type solution to the first question, we write $k = 2k' + (q\bmod 2)$ (where we in particular mean the residue of $q\bmod 2$ in $\{0,1\}$), and note that $$\exp\left(-\frac{2\pi i n(2k'+(q\bmod 2))}{2q}\right) = \exp\left(-\frac{2\pi i nk'}{q}\right)\exp\left(-\frac{2\pi i n(q\bmod 2)}{2q}\right) = \zeta_q^{-nk'}\zeta_{2q}^{-n(q\bmod 2)}.$$ We can therefore write $$G(2k'+(q\bmod 2)) = \sum_{n\bmod q} \chi(n)\zeta_{2q}^{n^2-n(q\bmod 2)}\zeta_q^{-nk'}$$ Let $a_n = \chi(n)\zeta_{2q}^{n^2-n(q\bmod 2)}$. Then computing $G(k)$ for all $k\in [0, 2q)$ such that $k\equiv q\bmod 2$ reduces to evaluating the polynomial $$A(x) = \sum_{n = 0}^{q-1}a_nx^n$$ at the points $\zeta_q^{-k'}$ for $k'\in [0,q)$. This can be done with the standard complex DFT. To see this, we use the exposition of the complex DFT in terms of matrix multiplications, namely that evaluating $A(x)$ on $q$ points $x_0,\dots, x_{q-1}$ is equivalent to multiplying the vector $(a_0,\dots, a_{q-1})$ by the Vandermonde matrix associated with those points, i.e. computing the product $$ \begin{pmatrix} 1 & 1 & 1&\dots & 1\\ 1 & \zeta_q^{-1} & \zeta_q^{-2} & \dots & \zeta_q^{-(q-1)}\\ 1 & \zeta_q^{-2} & \zeta_q^{-4}&\dots & \zeta_q^{-2(q-1)} \\ \vdots & & & \ddots&\vdots\\ 1 & \zeta_q^{-(q-1)} & \zeta_q^{-2(q-1)} & \dots & \zeta_q^{-(q-1)(q-1)} \end{pmatrix}\begin{pmatrix} a_0\\ a_1\\ \vdots\\ a_{q-1} \end{pmatrix} $$ This product can be computed in $O(q\log q)$ time using standard FFT techniques. Note that the $i$th coordinate of the output is precisely $\sum_{n = 0}^{q-1} a_n \zeta_q^{-in} = \sum_{n = 0}^{q-1} \chi(n)\zeta_{2q}^{n^2-n(q\bmod 2)}\zeta_q^{-ni} = \sum_{n = 0}^{q-1} \chi(n)\zeta_{2q}^{n(n-2i-(q\bmod 2))}$, i.e. is precisely $G(2i+(q\bmod 2)) = G(k)$ for some $k\in[0, 2q)$ with $k\equiv q\bmod 2$.<|endoftext|> TITLE: Do pretopoi have cohomology and homotopy groups? QUESTION [13 upvotes]: Grothendieck topoi have cohomology: the abelian category of abelian group objects in a topos has enough injectives, hence one can consider the right derived functors of the global sections functor from abelian group objects to abelian groups. (For details see Chapter 8 in Johnstone's book Topos theory.) Also, Grothendieck topoi have homotopy groups. (I think a reference is Artin and Mazur's Etale homotopy.) Question: Can some of these algebraic invariants (or other "topological properties") of Grothendieck topoi be generalized to pretopoi? REPLY [5 votes]: There's a long story that can be told here but I will try to be brief. In one sense, the answer is yes – you can certainly define cohomology and homotopy groups and so on for pretoposes and have them coincide with the classical definitions for Grothendieck toposes – but in another sense the answer is no – because you are essentially just embedding the pretopos into a suitable Grothendieck topos and reducing to that case. Let $\mathcal{E}$ be a pretopos. That means $\mathcal{E}$ is a category with a terminal object, pullbacks, finitary coproducts, and coequalisers of internal equivalence relations, such that finitary coproducts are disjoint and preserved by pullback, and coequalisers of internal equivalence relations are effective and preserved by pullback. In short, $\mathcal{E}$ satisfies the exactness part of the Giraud axioms, with finitary coproducts eplacing infinitary coproducts. That in itself should be a powerful reason to believe that any finitary constructions that can be carried out in a Grothendieck topos can also be carried out in $\mathcal{E}$ with the same results. Indeed: Proposition. Assuming $\mathcal{E}$ is small, there is a fully faithful embedding of $\mathcal{E}$ into a Grothendieck topos where the embedding preserves finite limits, finitary coproducts, and coequalisers of internal equivalence relations. Proof. Regard $\mathcal{E}$ as a site where the covering sieves are those that contain a sieve generated by a finite family that is jointly strongly epimorphic, and take the topos of sheaves on this site.　◼ (If $\mathcal{E}$ is not small then go up to a universe where it is, or find a subpretopos that is small and contains the objects and morphisms you care about.) Concretely, the category $\textbf{Ab} (\mathcal{E})$ of internal abelian groups in $\mathcal{E}$ is an abelian category (but not necessarily AB4 or AB4*, let alone AB5). So you can go on to define the category $\textbf{Ch} (\mathcal{E})$ of chain complexes in $\textbf{Ab} (\mathcal{E})$ and then the (unbounded) derived category $\mathbf{D} (\mathcal{E})$. What you do not get is the existence of enough injectives in $\textbf{Ab} (\mathcal{E})$ itself. Nonetheless, the definition of derived functors as (absolute, or at least pointwise) Kan extensions makes sense, and some derived functors can be constructed without injective resolutions. For example, although $\mathbf{R} \textrm{Hom}_{\textbf{Ch} (\mathcal{E})} (A, -) : \mathbf{D} (\mathcal{E}) \to \mathbf{D} (\textbf{Ab})$ itself does not have an obvious construction, $H_0 \mathbf{R} \textrm{Hom}_{\textbf{Ch} (\mathcal{E})} (A, -) : \mathbf{D} (\mathcal{E}) \to \textbf{Ab}$ always exists: you can directly check that $\textrm{Hom}_{\mathbf{D} (\mathcal{E})} (A, -) : \mathbf{D} (\mathcal{E}) \to \textbf{Ab}$ works. Also, if $\mathcal{E}$ is small, then any functor $\textbf{Ch} (\mathcal{E}) \to \textbf{Ab}$ whatsoever admits a pointwise left Kan extension along $\textbf{Ch} (\mathcal{E}) \to \mathbf{D} (\mathcal{E})$... but it is unclear to me whether this is consistent with what $(\infty, 1)$-category theory would give. Similarly (in some sense...), the category $\textbf{Kan} (\mathcal{E})$ of internal Kan complexes in $\mathcal{E}$ is a category of fibrant objects (in the sense of Brown) where the fibrations are the internal Kan fibrations and the weak equivalences are the internal weak homotopy equivalences. Thus the homotopy category $\mathbf{H} (\mathcal{E})$, obtained by localising $\textbf{Kan} (\mathcal{E})$ with respect to internal weak homotopy equivalences, is reasonable in the sense that there is a nice-ish formula for its hom-sets. The category $\textbf{Set}_\textrm{fin}$ of finite sets is the initial pretopos, so we get an induced functor $L : \mathbf{H} (\textbf{Set}_\textrm{fin}) \to \mathbf{H} (\mathcal{E})$. The homotopy type of $\mathcal{E}$ is "morally" a representing object for the functor $\textrm{Hom}_{\mathbf{H} (\mathcal{E})} (1, L {-}) : \mathbf{H} (\textbf{Set}_\textrm{fin}) \to \textbf{Set}$, where $1$ is the terminal object, but in practice this functor is rarely representable (even if $\mathcal{E}$ is a Grothendieck topos) so we are forced to make various tweaks like replacing $\textbf{Set}_\textrm{fin}$ with $\textbf{Set}_{< \kappa}$ (if $\mathcal{E}$ has coproducts of families of size $< \kappa$), or allowing more generalised notions of representability (e.g. pro-representability), or both. (Actually, this phenomenon can already be seen at homotopy level 0 for Grothendieck toposes, so in some sense the difficulty does not (only) come from trying to work in higher homotopy levels or with general pretoposes instead of Grothendieck toposes.)<|endoftext|> TITLE: Powers in finite extensions of the p-adics QUESTION [5 upvotes]: Fix a prime $p$. A p-adic field is a finite extension of $\mathbb{Q}_p$. Question 1: Let $K$ be a $p$-adic field and fix $n$. Is there $m$ such that if $\alpha \in \mathbb{Q}_p$ is an $m$th power in $K$ then $\alpha$ is an $n$th power in $\mathbb{Q}_p$? $\quad$ (Probably $m$ is a multiple of $n$.) I think that you'd approach this by decomposing $K/\mathbb{Q}_p$ into a tower of extensions of some sort and then go up the tower. So we probably want to actually prove the following. Question 2: Suppose that $K/L$ is an extension of $p$-adic fields and fix $n$. Is there $m$ such that if $\alpha \in K$ is an $m$th power in $L$ then $\alpha$ is an $n$th power in $K$? I can prove Question 1 in the case when $K/\mathbb{Q}_p$ is unramified, but I don't know about the general case. I think that one might be able to use class field theory, or maybe it's already well-known, or maybe I am just missing something elementary. I have good motivation for this question, but it would take a bit of work to explain, the motivation comes from logic. REPLY [4 votes]: Yes. I'll solve the more general question 2. Let $v_p(n)$ be the highest power of $p$ dividing $n$. Then if $\alpha \in L$ is congruent to $1$ modulo $p^{v_p(n)+1}$, then $\alpha$ is an $n$th power in $L$. Every $\alpha \in L$ can be written as $\pi^j u$ where $j\in \mathbb Z$ and $u$ is a unit in $L$. It suffices to find $m$ such that if $\pi^j u$ is an $m$th power in $K$ then $j$ is divisible by $n$ and $u$ is congruent to $1$ modulo $p^{v_p(n)+1}$. Let $e$ be the ramification degree of $K$ over $L$, i.e. the greatest $e$ such that there is an element of $K$ whose $e$'th power is $\pi$ times a unit and let $k$ be any natural number. Then if $\pi^j u$ is a $ke$'th power in $K$ then $j$ is divisible by $k$, and hence $u = \pi^j u / (\pi^{j/k})^k$ is also a $k$th power in $K$. So it suffices to find $k$ such that if $u$ is a unit that is a $k$th power in $K$ then $u$ is congruent to $1$ mod $p^{v_p(n)+1}$. Let $q$ be the order of the residue field in $k$. If $u$ is a $k$th power, then it is a $k$th power of some unit $v$, and $v^{q-1}$ is congruent to $1$ modulo the uniformizer $\pi'$ of $K$. So $$ v^{ (q-1) p^r} = (1 + (v^{q-1}-1))^{p^r} = \sum_{i=0}^{P^r} \binom{p^r}{i} (v^{q-1}-1)^i$$ where the $i$th term is divisible by $\binom{p^r}{i} \pi'^{i}$. Taking $r$ sufficiently large, we can ensure that $\binom{p^r}{i} \pi'^{i}$ is divisible by $p^{ v_p(n)+1}$ for all $i>0$, so $v^{ (q-1) p^r}$ is congruent to $1$ mod $p^{ v_p(n)+1}$, so we may take $k= (q-1) p^r$.<|endoftext|> TITLE: Scattering amplitudes and the Riemann zeta function QUESTION [9 upvotes]: I'm reading Amplitudes and the Riemann Zeta Function, which recently appeared in Physical Review Letters. It's received some publicity, including my own campus' PR operation. From the abstract (adapting the notation) "Physical properties of scattering amplitudes are mapped to the Riemann zeta function. Specifically, a closed-form amplitude is constructed, describing the tree-level exchange of a tower with masses $m_n^2=\gamma_n^2$, where $\zeta(1/2+i\gamma_n)=0$. Requiring real masses corresponds to the Riemann hypothesis..." NB: I'm skeptical that this will lead to any progress on the Riemann Hypothesis, and not interested in that aspect. I'm trying to determine the new (if any) mathematical content. The author develops the identity $$-4+\frac{\pi^2}{8}+G+\frac{\zeta^{\prime\prime}(1/2)}{2\zeta(1/2)}-\frac{1}{8}\left(C+\frac{\pi}{2}+\log8\pi\right)^2 =2\sum_{n=1}^\infty\frac{1}{\gamma_n^2}.$$ Here $G$ is the Catalan constant and $C$ denotes the Euler constant. (I can check this via the logarithmic derivative of the Hadamard product for $\Xi(\sqrt{s})$, and differentiating the functional equation for $\zeta(s)$ twice.) More generally, with $$\zeta_n(s)=\frac{\zeta^{(n)}(s)}{\zeta(s)}, \qquad\zeta_n^k=\zeta_n(1/2)^k,$$ he has identities for $2\sum_{n=1}^\infty 1/\gamma_n^{2k}$ for odd $k$, for example $$ 2\sum_{n=1}^\infty \frac{1}{\gamma_n^6}=-128+\frac{1}{7680}\Psi^{(5)}(1/4)-\zeta_1^6+3\zeta_1^4\zeta_2-\frac{9}{4}\zeta_1^2\zeta_2^2+\frac{1}{4}\zeta_2^3-\zeta_1^3\zeta_3+\zeta_1\zeta_2\zeta_3-\frac{1}{12}\zeta_3^2+\frac{1}{4}\zeta_1^2\zeta_4-\frac{1}{8}\zeta_2\zeta_4-\frac{1}{20}\zeta_1\zeta_5+\frac{1}{120}\zeta_6 $$ The author writes "[These] can be proven exactly albeit laboriously, without appeal to our amplitude, using repeated differentiation of the functional equation and the Hadamard product form of the zeta function, as well as various polygamma identities... what is remarkable is that our amplitude construction allows for much simpler, physical derivations of these identities." Are such identities new? Are they interesting? Regarding the latter, the author in an online talk describes such sums as moments of the $\{\gamma_n\}$, and they look like moments to me. REPLY [9 votes]: Thanks to @reuns for the answer in the comments. I've asked him to post as an answer, and I will accept it if he does. Meanwhile, his comments encouraged me to look again, and here is another approach, quite easy (Lemma 4.11 in Equivalents of the Riemann Hypothesis vol. II) Let $\Xi(s)=\xi(1/2+is)$, so $\Xi(-s)=\Xi(s)$. Let $\gamma_n$ be the zeros with positive real part, so the Hadamard product looks like $$ \Xi(s)=\xi(1/2)\prod_{n=1}^\infty\left(1-\frac{s^2}{\gamma_n^2}\right) $$ Expanding $\Xi(s)$ as a power series about $s=0$ $$ \sum_{k=0}^\infty \frac{(-1)^k\xi^{(2k)}(1/2)}{2k!}s^{2k}=\Xi(s)=\xi(1/2)\left(1-s^2\left(\sum_n \gamma_n^{-2}\right)+s^4\left(\sum_{m,n}\gamma_m^{-2}\gamma_n^{-2}\right)-\ldots\right) $$ Equating coefficients and the Newton Identities gives the moments as polynomials in the $\xi^{(2k)}(1/2)$, for example $$ \sum_n\frac{1}{\gamma_n^6}=\frac{\xi^{(6)}(1/2)}{240\,\xi(1/2)}-\frac{\xi^{(2)}(1/2)\xi^{(4)}(1/2)}{16\,\xi(1/2)}+\frac{\xi^{(2)}(1/2)^3}{8\,\xi(1/2)}. $$<|endoftext|> TITLE: Characterize this subspace of the bounded operators on $ L^2(\mathbb{R}) $ QUESTION [6 upvotes]: I posted this on MSE a couple months ago and it got three upvotes but no answers or even comments so I decided to cross-post it here: For every pair $ a,b $ of real numbers define the operator $ U_{a,b} $ on $ L^2(\mathbb{R}) $ sending $ \psi \in L^2(\mathbb{R}) $ to $ U_{a,b}\psi $ defined by the equation $$ [U_{a,b}\psi](x)=e^{ibx}\psi(x+a) $$ Consider the set of operators $$ \mathcal{B}:=\{ U_{a,b}:a,b \in \mathbb{R} \} $$ Let $ V $ be the closure in the operator norm topology of the span of the set $ \mathcal{B} $. Does anyone have a good idea for a nice characterization of what sort of operators are and are not in $ V $? Does $ V $ include all trace class operators? All compact operators? All unitary operators? This is a follow up question to my question: https://math.stackexchange.com/questions/4303824/is-this-a-basis-for-the-bounded-operators-on-l2-mathbbr REPLY [7 votes]: Any linear combination $L$ of $U_{a,b}$'s can be written $(L\psi)(x) = \sum_{k=1}^n \alpha_ke^{ib_kx}\psi^{\to a_k}(x)$, where $\psi^{\to a_k}(x) = \psi(x + a_k)$. Fix $L$. Let $N \in \mathbb{N}$ be such that $Nb_k$ is close to an integer multiple of $2\pi$, for all $k$. Then $$(L\psi^{\to N})(x) = \sum \alpha_k e^{ib_kx}\psi^{\to a_k + N}(x) = (\sum \alpha_k e^{ib_k(x - N)}\psi^{\to a_k}(x))^{\to N} = (\sum \alpha_k'e^{ib_kx}\psi^{\to a_k}(x))^{\to N}$$ where each $\alpha_k'$ is close to $\alpha_k$. Okay, now find a sequence $N_j \to \infty$ such that as $j \to \infty$ the multiples $N_jb_k$ get arbitrarily close to integer multiples of $2\pi$. Then $(L\psi^{\to N_j})^{-N_j} \to L\psi$ in $L^2(\mathbb{R})$, for every $\psi$. But any compact operator $T$ satisfies $T\psi^{\to N_j} \to 0$ in $L^2(\mathbb{R})$. Taking $\psi$ with $\|\psi\|_2 = 1$ and $\|L\psi\|_2$ close to $\|L\|$, we have $\|T\psi^{\to N_j}\|_2 \to 0$ but $\|L\psi^{\to N_j}\|_2 \to \|L\psi\|_2 \cong \|L\|$. This shows you that $\|T - L\| \geq \|L\|$; that is, the distance from $L$ to the compact operators is $\|L\|$. Every element of $V$ will have the same property, so in particular $V$ contains no compact operators besides $0$. Every operator is a linear combination of four unitaries, so if $V$ contained every unitary then it would be all of $B(L^2(\mathbb{R}))$, which we've just seen is not the case. On the other hand, the WOT closure of $V$ does equal $B(L^2(\mathbb{R}))$; this follows from the double commutant theorem, since any operator that commutes with $U_{0,b}$ for all $b$ must be a multiplication operator and hence won't commute with $U_{a,0}$ for all $a$ unless it is a scalar. That is, $V' = \mathbb{C}\cdot I$, so $V'' = B(L^2(\mathbb{R}))$.<|endoftext|> TITLE: Locating generic filters in the Lévy collapse QUESTION [5 upvotes]: Let $\operatorname{Col}(\omega,<\kappa)$ denote the Lévy collapse of an inaccessible cardinal $\kappa$. A variant of the Factor Lemma is as follows: Lemma. Suppose that $\kappa$ is an inaccessible cardinal and that $\mathbb{P}$ is a poset of size $<\kappa$. Let $G$ be $\operatorname{Col}(\omega,<\kappa$)-generic over $V$. If in $V[G]$ there is a filter $h \subseteq \mathbb{P}$ that is $\mathbb{P}$-generic over $V$, then there is $G^* \in V[G]$ that is $\operatorname{Col}(\omega,<\kappa)$-generic over $V[h]$ and such that $V[h][G^*] = V[G]$. For those interested, I took this variant from the paper Happy and mad families in $L(\mathbb{R})$, Lemma 17 of Page 10. This lemma shows that, in particular, $h \in V[G]$. Is it true that there must exist some ordinal $\beta < \kappa$ such that $h \in V[G\upharpoonright\beta]$? REPLY [10 votes]: Note that the lemma doesn't show that $h$ is in $V[G]$, it assumes this. But yes, if $h\in V[G]$ is a subset of a set $X\in V$ such that $|X| < \kappa$, then for some $\beta < \kappa$, $h\in V[G\restriction \beta]$. The argument to follow is by no means original, but I don't remember where I saw it. Fix a name $\dot h$ such that $h = \dot h_G$, and for each $x\in X$, let $A_x\subseteq \text{Col}(\omega,{<}\kappa)$ be a maximal antichain consisting of conditions that force either $\check{x}\in \dot h$ or its negation. Recall that $\text{Col}(\omega,{<}\kappa)$ has the $\kappa$-cc, so $|A_x| < \kappa$. Let $A = \bigcup_{x\in X} A_x$. Since $|X| < \kappa$, $|A| < \kappa$, and therefore there is some $\beta < \kappa$ be such that $A\subseteq \text{Col}(\omega,{<}\beta)$. We have $h\in V[G\restriction \beta]$ since $h = \{x\in X : \exists q\in G\restriction \beta\, (q\Vdash \check{x} \in \dot h)\}$.<|endoftext|> TITLE: Integral surgeries on $3$-manifolds QUESTION [5 upvotes]: Let $K$ be a knot in $S^3$. Let $N(K)$ be a tubular neighborhood of $K$, a solid torus. On $\partial N(K)$, we may specify a preferred longitude $\lambda$, i.e., a simple closed curve whose linking number with $K$ is $0$. Also, we can choose a canonical meridian $\mu$ whose linking number with $K$ is $1$. A $p/q$-surgery on $S^3$ along $K$ is a closed oriented $3$-manifold given by $$S^3_{p/q} (K) = (S^3 - \mathrm{int}(N(K))) \cup_\varphi (D^2 \times S^1)$$ where $\varphi: S^1 \times S^1 \to S^1 \times S^1$ is a homeomorphism that sends $\partial D^2 \times \{ 0 \}$ to $p \mu + q \ell$. I want to understand the generalization of this process to an arbitrary closed oriented $3$-manifold $M$. The concept of integral surgery makes sense for any $M$? If yes, how? Here, the choice of $\lambda$ is not obvious. How about the rational surgeries? P.S. I checked several reference books about the knot theory. I couldn't find a precise approach for this generalization. Any reading advise will be appreciated. REPLY [5 votes]: Two places in which this is discussed are Gompf and Stipsicz's book 4-manifolds and Kirby calculus (Sections 5.2 and 5.3) and Ozbagci and Stipsicz's Surgery on contact 3-manifolds and Stein surfaces (Chapter 2). The reason this is not discussed in many knot theory references is that this is more of a 3.5-dimensional issue (that is, in between dimension 3 and dimension 4) than a knot-theoretical one. Yes, this makes sense in arbitrary 3-manifolds, since the fact that $q=1$ is independent of your choice of longitude. The meridian is always well-defined, and any two longitudes differ by a multiple of the meridian, so having an expression of the form $p\mu + \lambda$ is independent of which $\lambda$ you choose. (The $p$ will vary, of course.) A different, arguably "better" perspective is that this corresponds to attaching a 4-dimensional 2-handle to $M \times I$ along $K$, which is what the two references I gave you are mostly about. I'm not sure what you mean by "how about", so I'll at least tell you that the construction you have for knots in $S^3$ is completely general, and it goes by the name of "Dehn surgery". You can do it for any knot in any 3-manifold, and the datum you need is just the choice of (the homology class of) a simple closed curve in $\partial N(K)$, also called the slope (which is your $(p,q)$ or $p/q$). The only drawback is that if $K$ and $M$ are arbitrary you don't have a canonical way to translate this slope into a rational number, unless $K$ is null-homologous in $M$ (see below). About 2., let me add that the choice of $\lambda$ is not only non-obvious, but also not possible in general. The only instance in which there is a canonical choice of $\lambda$ is when the knot $K$ is null-homologous in $M$, i.e. $[K] \in H_1(M;\mathbb{Z})$ vanishes. Then you have a preferred longitude (still called the Seifert longitude), which is the only curve on $\partial N(K)$ (up to isotopy) whose homology class dies in $H_1(M\setminus K; \mathbb{Z})$. In general, whenever you have a knot $K \subset M$, you always have a rank-1 subgroup of $H_1(\partial N(K); \mathbb{Z})$ which dies in $H_1(M \setminus K; \mathbb{Z})$. This group needs not be generated by a primitive element, so there might be no simple closed curve on $\partial N(K)$ that bounds a surface in $M \setminus {\rm int}(N(K))$. It will be generated by a longitude if and only if $K$ is null-homologous in $M$.<|endoftext|> TITLE: Order ideals of positive root systems and avoiding group elements in the Weyl group QUESTION [6 upvotes]: Let $X$ be the poset of positive roots of a finite root system of Dynkin type $Q$. Question 1: In Dynkin type $A_n$, is it true that the poset of order ideals of $X$ is isomorphic to the poset of [2,1,3]-avoiding permutations under strong Bruhat order? I tested this experimentally in Sage, for example: n=4 S=Permutations(n) W=[p for p in S if p.avoids( [2,1,3])] B = Poset([W, lambda x,y: x.bruhat_lequal(y)]) O=LatticePoset(B).join_irreducibles_poset() display(B) display(O) Question 2: Is there a generalisation to the other Dynkin types so that the poset of order ideals of $X$ is isomorphic to the poset of $\{g_i \}$-avoiding permutations under the strong Bruhat order of the Weyl group for some group elements $g_i$? REPLY [7 votes]: Regarding Q1: I believe this is proved in "A Distributive Lattice Structure Connecting Dyck Paths, Noncrossing Partitions and 312-avoiding Permutations" by Barcucci et al. https://doi.org/10.1007/s11083-005-9021-x. (Note that 213-avoiding and 312-avoiding permutations are in bijection via inversion, a poset automorphism of the Bruhat order.) Regarding Q2: I doubt this coincidence will extend in a straightforward way to other types. One reason is because in other types the number of order ideals of the root poset (the "W-Catalan number") is not the same as the number of fully commutative elements of the Weyl group. In Type A, fully commutative elements are 321-avoiding: not exactly the same as 213-avoiding, but equinumerous with them. EDIT: By the way, even to define what pattern avoidance means in other types is a nontrivial task, but there is such a notion, due to Billey and Postnikov https://doi.org/10.1016/j.aam.2004.08.003.<|endoftext|> TITLE: Who introduced the notation for $\beth$ numbers and when? QUESTION [12 upvotes]: Georg Cantor, when developing the basics of set theory, noted that there are two ways to increase cardinality: power sets and successors (or, in modern terms, the Hartogs operation).1 Eventually the notation for the successors became the $\aleph$ numbers. The power set operation eventually became what we now know as the $\beth$ numbers: $\beth_0(\kappa)=\kappa$, $\beth_{\alpha+1}(\kappa)=2^{\beth_\alpha(\kappa)}$, and $\beth_\alpha(\kappa)=\sup\{\beth_\xi\mid\xi<\alpha\}$ for limit steps. If $\kappa=\aleph_0$, we simply omit it from the notation and we get the $\beth$ numbers. What I am trying to find is the origin of the notation for $\beth$ numbers. Kanamori's article in the Handbook only mentions this in relation to the Erdős–Rado paper from 1956, where the notation does not appear. Jech, which normally has a reasonably thorough historical overview in the 3rd Millennium Edition of "Set Theory", has no mention as to who came up with the notation. The notation does appear in the 1978 first edition (p. 72), but there is no mention of its origin; the notation is also missing from The Axiom of Choice (written in 1973). So, who came up with the notation and when? Footnotes We also have the Lindenbaum operator, similar to Hartogs but with surjections, which can grow "quicker" than the Hartogs and differently from the power set, at least in the absence of choice. But we're not here to discuss $\sf ZF$. REPLY [14 votes]: Charles Sanders Peirce is credited with the beth notation ℶ, first introduced in a December 1900 letter to Cantor. Apparently, this was then forgotten for half a century. I reproduce the relevant text from Gregory Moore's Early history of the generalized continuum hypothesis.<|endoftext|> TITLE: Importance of third homology of $\operatorname{SL}_{2}$ over a field QUESTION [8 upvotes]: $\DeclareMathOperator\SL{SL}$I am reading some papers about the third homology of linear groups. In particular for the $\SL_{2}$ over a field. Why is it important to study these homologies? I have been trying to read some papers and books (for example, Knudson - Homology of linear groups, Hutchinson - A Bloch–Wigner complex for $\operatorname{SL}_2$, Hutchinson - A refined Bloch group and the third homology of $\operatorname{SL}_2$ of a field, Mirzaii - Third homology of general linear groups) and they only mention homology stabilization theorems and the Theorem of Hurewicz but they did not go deep on this. I only found on the paper Scissor Congruences II by Johan L. Dupont and Chih Han Sah that it explored the properties of the Bloch group due to its connection to their study of the Third Hilbert problem in Hyperbolic 3-space and for the case $F=\mathbb{C}$ they proved the Bloch–Wigner Theorem that states that there is a short exact sequence $$0\rightarrow\mu_{\mathbb{C}}\rightarrow H_{3}(\SL_{2}(\mathbb{C}),\mathbb{Z})\rightarrow \mathcal{B}(\mathbb{C}) \rightarrow 0,$$ where $\mathcal{B}(F)$ is the Bloch group of a field $F$. Also when $F=\mathbb{C}$ (and more generally $F^{\times}=(F^{\times})^{2}$) it is proved that $K_{3}^{\operatorname{Ind}}(\mathbb{C})=H_{3}(\SL_{2}(\mathbb{C}),\mathbb{Z})$. Andrei Suslin proved for any infinite field $F$ in his paper $K_{3}$ of a field and the Bloch group that there is a natural exact sequence $$0\rightarrow \widetilde{\operatorname{Tor}_{1}^{\mathbb{Z}}(\mu_{F},\mu_{F})}\rightarrow K_{3}^{\operatorname{Ind}}(F)\rightarrow \mathcal{B}(F)\rightarrow0$$ where $\widetilde{\operatorname{Tor}_{1}^{\mathbb{Z}}(\mu_{F},\mu_{F})}$ is the unique nontrivial extension of $\operatorname{Tor}_{1}^{\mathbb{Z}}(\mu_{F},\mu_{F})$ by $\mathbb{Z}/2$ when $\operatorname{Char}(F)\ne2$ (and $\widetilde{\operatorname{Tor}_{1}^{\mathbb{Z}}(\mu_{F},\mu_{F})}=\operatorname{Tor}_{1}^{\mathbb{Z}}(\mu_{F},\mu_{F})$ if $\operatorname{Char}(F)=2$). So we have a relationship through $K$-theory and the Bloch group via the homology of linear groups. But I think there is more things for the studies of the homology of the linear groups. I would appreciate any hint or indication of where I can read about it. REPLY [5 votes]: I don't know if this is quite the answer you're looking for, but elements of $H_3(\operatorname{PSL}_2(\mathbb{C}),\mathbb{Z})$ can be identified as fundamental classes of finite-volume hyperbolic $3$-manifolds. Neumann [1] showed how to construct the Cheeger–Chern–Simons class as a map $\mathfrak{c} : H_3(\operatorname{PSL}_2(\mathbb{C}),\mathbb{Z}) \to \mathbb{C}/4\pi^2\mathbb{Z}$. Evaluating $i\mathfrak{c}([M])$ on a manifold computes the complex volume, whose real part is the volume of the hyperbolic metric and whose imaginary part is the Chern–Simons invariant. To effectively apply this formula (you need some extra combinatorial data called a flattening) it is helpful to use a particular coordinate system on the space of $\operatorname{PSL}_2(\mathbb{C})$-bundles over the manifold, the Ptolemy coordinates [2]. I think that these were originally discovered when thinking about things from the perspective of group homology: see [2, Sections 2 and 3]. [1] Extended Bloch group and the Cheeger–Chern–Simons class, Walter Neumann. [2] The volume and Chern–Simons invariant of a representation, Christian K. Zickert.<|endoftext|> TITLE: Group-like elements in quotients of group rings QUESTION [7 upvotes]: $\DeclareMathOperator\Gr{Gr}$Let $R$ be a local ring, let $A$ be a finite abelian group, and let $I$ be a Hopf ideal of the ring $R[A]$. The quotient $R[A]\twoheadrightarrow R[A]/I$ induces a map on group-like elements $f\colon \Gr(R[A])\to \Gr(R[A]/I)$. Is the map $f$ surjective? The answer is yes if the order of $A$ is invertible in $R$, so let's assume it's not. (By $R[A]$ I just mean the usual ring with generators $T_a$ for $a\in A$ and relations $T_0=1$ and $T_aT_b=T_{a+b}$). REPLY [2 votes]: I found a counter-example, so the answer is no. Let $B=\mathbb{F}_2[a,b,c]/J$ where $J=(a,b,c)^3+(ab+ac-bc)$ and put $s=ab$ and $t=ac$. Let $A=(\mathbb{Z}/2\mathbb{Z})^2$ and define $R = B[x_{10},x_{01},x_{11}]/J'$ where $$ J'=(x_{10}^2-s, x_{01}^2-t, x_{11}^2-(s+t), x_{10}x_{01}-ax_{11}, x_{10}x_{11}-bx_{01}, x_{01}x_{11}-cx_{10})\,. $$ Then $R[A]\cong R[T_{10},T_{01}]/(T_{10}^2-1, T_{01}^2-1)$ with the usual Hopf-algebra structure. Define $$ I=(x_{10}(T_{10}-1), x_{01}(T_{01}-1), x_{11}(T_{10}T_{01}-1))\,. $$ I claim that element $g=1+t(T_{10}-1)$ is not equal to 1 but group-like modulo $I$. To see that $g$ is group-like, we must show that $\varepsilon(g)=1$ and $\Delta(g)=g\otimes g$, where $\varepsilon$ denotes the counit and $\Delta$ denotes the coaction. It is clear that $\varepsilon(g)=1$ so we will show that $\Delta(g)=g\otimes g$. Since $$ T_{\lambda}-1=(T_{\lambda-\lambda'}-1)(T_{\lambda'}-1)+(T_{\lambda-\lambda'}-1)+(T_{\lambda'}-1)\,, $$ we have $x_{\lambda'}(T_\lambda-1)=x_{\lambda'}(T_{\lambda-\lambda'}-1)$ modulo $I$ for all $\lambda,\lambda'\in A$. This implies that, modulo $I\otimes R[A]+R[A]\otimes I$, we have $$ \begin{split} \Delta(g)-g\otimes g & = \Delta(1+t(T_{10}-1))-(1+t(T_{10}-1))\otimes (1+t(T_{10}-1)) \\ & = t(T_{10}-1)\otimes (T_{10}-1) \\ & = (s+t)(T_{10}-1)\otimes (T_{10}-1) \\ & = (s+t)(T_{01}-1)\otimes (T_{10}-1) \\ & = t(T_{01}-1)\otimes (T_{10}-1)+(T_{01}-1)\otimes s(T_{10}-1) \\ & = 0\,. \end{split} $$ Hence $g$ is a group-like element which is not the image of a group-like element in $R[A]$.<|endoftext|> TITLE: What conditions are sufficient for the Leray-Hirsch theorem to be a Künneth formula? QUESTION [6 upvotes]: This was originally posted on MSE, and since it didn't receive much attention, I'll try here. Let me know if this is not the appropriate place. Given a fiber bundle $F \to E \to B$ over a paracompact base $B$, assume its cohomology satisfies all the required properties for the Leray-Hirsch theorem to hold. This tells us that, as groups, $$ H^n (E,R) \cong \bigoplus_{p+q=n} H^p (F,R) \otimes H^q (B, R). $$ But this does NOT tell us that, as rings, we have a ring isomorphism $$ H^* (E,R) \cong H^* (F,R) \otimes H^* (B,R). $$ I'm wondering when this stronger isomorphism actually does hold. What extra structure on the fiber bundle is sufficient, without it being a trivial bundle? I know of the classic counterexample with the bundle $\mathbb{C}P^3$ over $S^4$ with fiber $S^2$. I know that principle bundles can have the module isomorphism of Leray-Hirsch upgraded to the ring isomorphism. What else do we know? Any references on this subject would be highly appreciated. EDIT: How about the following rather strong condition on our fiber bundle. Say both the integral cohomology of $F$ and $B$ lie in even degrees only, and that the bundle is torsion-free (one of the two cohomology rings are free and finitely generated, I believe is the criteria). Then, when applying the Serre spectral sequence, we would immediately get that it collapses for degree reasons. I only have a basic understanding of the SSS, but would this be enough to show that $H^* (E, \mathbb{Z})$ splits as a tensor products of the two other cohomology rings? REPLY [3 votes]: (Not really an answer, but too long for a comment, bear with me.) I'd say that those conditions will be pretty subtle and I doubt anything both universal and meaninful can be said. For example, let's restrict attention to sphere bundles which come as unit sphere bundles of vector ones of rank $r$, so we have fibration $S^{r-1} \to E \to B$. If cohomology coefficients are 2-divisible, then top Stiefel-Whitney class determines $H^*(E)$. It's more or less folklore, usually attributet to R. Thom afair. But, if you look at integral cohomology, then you need top three Stiefel-Whitney classes to vanish for cohomology to split as a product. (I'm not going to write a proof here, because the one I'm able to produce is lengthy and is based on lengthy and totally unilluminating diagram chasing in spectral sequence associated to the universal bundle over $BSO(r)$ on cochain level)<|endoftext|> TITLE: Examples of non-adjoint equivalences QUESTION [7 upvotes]: What are some examples of equivalences whose canonical unit/counit fail to satisfy the triangle identities? It is common knowledge that not all equivalences satisfy the triangle identities, but that any equivalence can be refined by swapping out its unit (or counit) with a different one to form an adjoint equivalence which does satisfy the triangles while leaving both functors intact, so all functors that are part of an equivalence are also part of an adjoint equivalence. I'm curious about equivalences where the canonical unit and counit do not satisfy the triangle identities -- the meaning of canonical here is hopefully canonical, but to be more precise I mean that the unit and counit that are 'obvious' to write down do not satisfy the triangles and need to be modified using the refinement to adjoint equivalences to do so. REPLY [4 votes]: Let's start with a functor $U:{\mathcal A}\to{\mathcal X}$ that's full and faithful. Bypassing questions of Choice, it's also essentially surjective on objects in the sense that there is a function $F:{\mathsf ob}{\mathcal X}\to{\mathsf ob}{\mathcal A}$ along with an assignment of an isomporhism $\eta_X:X\to U F X$ to each object $X$. From these data, for each morphism $f:X\to Y$ of $\mathcal X$, we define $F f:F X\to F Y$ as the unique $\mathcal A$-map such that $$ \eta_X^{-1};f;\eta_Y = U(F f), $$ in other words such that $\eta$ is natural. It is easy to check that $F$ preserves identity and composition. It remains to define $\epsilon$. For one of the triangle laws we require $U\epsilon_A=\eta_{U A}^{-1}$, which uniquely defines $\epsilon_A$ since $U$ is full and faithful. Naturality of $\epsilon$ follows from that of $\eta$. The other triangle law is $\eta_{U A};U\epsilon_A={\mathsf id}$, for which it suffices that this hold with $U$ applied, since that's full and faithful. By naturality of $\eta$ and the first triangle law, we have $$ \eta_X;U F\eta_X; U\epsilon_{F X} = \eta_X;\eta_{U F X}; U\epsilon_{F X} = \eta_X $$ but $\eta_X$ is invertible an $U$ is full and faithful, so the other triangle law holds. In other words, the obvious data for "full, faithful and essentially surjective on objects" yield an adjoint equivalence. So what other kind of equivalence is there? If the isomorphism $\eta$ in "essential surjectivity" is to be natural, it can only be as above. However, the other isomorphism $\epsilon'$ could come from somewhere else. Nevertheless, $\eta_{U A};U\epsilon'_A$ is still a natural automorphism of $U A$, which must be $U$ applied to a natural isomorphism of $A$. In other words, a non-adjoint equivalence is given by arbitrary natural isomorphisms applied to an adjoint equivalence. This is a situation that can easily be realised with group isomorphisms, yielding the counterexample that was requested.<|endoftext|> TITLE: A question about the adjoint of the Adams operations on representation rings QUESTION [11 upvotes]: Let $G$ be a finite group, and $R(G)$ its representation ring over $\mathbb{C}$. We have the Adams operations $\psi^k:R(G)\rightarrow R(G)$, given on the level of characters by: $$\chi_{\psi^k{V}}(g)=\chi_V(g^k).$$ Since the power sums can be expressed as polynomials in the homogenous symmetric functions $h_n$, these $\psi^kV$ correspond to virtual representations given by integer polynomials in the symmetric powers of $V$. By using these same polynomials in the symmetric powers, we can define Adams operations on the level of Burnside rings $\psi^k:A(G)\rightarrow A(G)$, such that these operations intertwine the natural map $A(G)\rightarrow R(G)$ induced by taking the free (virtual) vector space on the (virtual) $G$ set. Since $R(G)$ has a nondegenerate quadratic form, we can take the adjoint of the $\psi^k$ map, call it $\nu^k$. Since it is an adjoint, this map preserves characters/virtual representations. On characters, this map is given by:$$\chi_{\nu^k V}(g)=\sum_{h^k=g}\chi_V(h).$$ My question is then, does a natural lift of $\nu^k$ to the Burnside ring exist? It is natural condition for such a $\nu^k$ to commute with induction from subgroups, so it suffices to define $\nu^k(\ast)$ for the trivial $G$ set $\ast$. So to show that such a Burnside ring version of this map isn't possible, it would suffice to show that $\nu^k(\mathbb{1})$ isn't in the image of the natural map $A(G)\rightarrow R(G)$, but I'm struggling to find a counterexample to this claim. REPLY [6 votes]: If I understand the question correctly, then $\nu^{2}(1)$ isn't in the image of the natural map $A(G) \to R(G)$ when $G = Q_{8},$ the quaternion group of order $8$. The number of square roots of the identity in $G$ is $2$, the number of square roots of the central involution $z$ in $G$ is $6$, and the number of square roots of each element of order $4$ is $0$. Hence we have $\nu^{2}(1) = \lambda_{1} + \lambda_{2}+ \lambda_{3} + \lambda_{4} - \chi$, where the $\lambda_{i}$ are the linear characters of $G$ and $\chi$ is the unique irreducible character of $G$ of degree $2$. Now I claim that this virtual character is not a difference of permutation characters. Indeed, it is not even a difference of characters afforded by $\mathbb{R}G$-modules. For the irreducible character $\chi$ has real Schur index $2$, so occurs with even multiplicity in any character of $G$ afforded by an $\mathbb{R}G$-module. Hence $\chi$ occurs with even multiplicity in any difference of permutation characters, so that $\nu^{2}(1)$ is not expressible as a difference of permutation characters. More generally, if $G$ is any finite group of even order which has an irreducible character $\chi$ with Frobenius-Schur indicator $-1$, then $\nu^{2}(1)$ is not expressible as a difference of permutation characters of $G$. An induction theorem of G. Segal may be relevant to trying to characterize exceptions for which $\nu^{2}(1)$ is not a virtual permutation character of $G$.<|endoftext|> TITLE: In surreal numbers, what is $\ln \omega$? QUESTION [8 upvotes]: Can this number $\ln \omega$ be written in $\{L|R\}$ form? What's its birthday? REPLY [12 votes]: In general this is taken to mean the value of Gonshor's logarithm at $\omega$. This was defined in the tenth chapter of his 1986 book An introduction to the theory of surreal numbers where you can find a justification for my answer. In an informal way, the function $\ln$ is the "simplest" function that is eventually smaller than each power function $x\mapsto x^r$ for $r \in \mathbb{R}^{>0}$ but eventually greater than any constant function. So if $\ln(\omega)$ could be the simplest number that is greater than each real number but smaller than each power $\omega^r$ for $r \in \mathbb{R}^{>0}$, that would be nice. Indeed $\ln(\omega)=\{ \mathbb{R} \ | \ \omega^{r}: r \in \mathbb{R}^{>0}\}$. In Conway normal form, this is a monomial $\omega^{\omega^{-1}}$. You can also write $\ln(\omega)=\{ \mathbb{N} \ | \ \omega^{2^{-n}}: n \in \mathbb{N}\}$, then the difference is that the elements in brackets are simpler than $\ln(\omega)$ in the sense of the simplicity relation on surreal numbers. re-edit: my past answer for the birth day was wrong. In fact each $\omega^{2^{-n}}$ has birth day $\omega+\omega^2.n$, so the birth day of $\ln(\omega)$ is actually $\omega^3$.<|endoftext|> TITLE: Polygamma function in mathematical physics QUESTION [5 upvotes]: Are there situations in which the polygamma pops up naturally in a mathematical physics context? In particular: are there examples of potentials having some interest for which the dependence on the distance is expressed in terms of $\psi^{(n)}$? Update: While Carlo Beenakker's answer is clearly useful, the references therein still don't contain exactly what I'm looking for in the second part of the question. REPLY [6 votes]: Abou-Salem, L. I., A study on baryons spectroscopy using digamma-function as interacting potential, https://arxiv.org/abs/1311.6743 studies using the digamma function as an interaction potential (for quarks in baryons).<|endoftext|> TITLE: Why are exponential sums so bad at solving this very easy problem? QUESTION [8 upvotes]: Exponential sums are a powerful tool in additive combinatorics and number theory. In my understanding, when it comes to estimate the cardinality of a certain set, exponential sums are (essentially) used in this way: (1) the indicator function of the set is replaced by an appropriate exponential sum; (2) the sum of the indicator function and the sum of the exponential sum are swapped; (3) the main term is extracted; 4) the error term is bounded. However, recently I stumble upon some simple additive problems for which the exponential-sums approach seems to fail miserably. My (somehow philosophical) question is: Am I applying exponential sums in the wrong way? (And, in such a case, how should apply them to solve such problems?) Or are exponential sums are not good for these problems? (And, in such a case, what is the reason it is so?) Below an extremely simple example in which exponential sums seem to fails. Of course, I do not really care about solving such problem with exponential sums, but it is just for the sake of example. Let $1 \leq h \leq m$ be fixed integers. We want to estimate the number $C$ of $(x, y) \in \{0,\dots,h-1\}^2$ such that $x \equiv y \bmod m$. Obviously, $C = h$. Proceeding with exponential sums, we have: $$C = \sum_{0 \leq x < h} \sum_{0 \leq y < h} \begin{cases} 1 & \text{ if } x \equiv y \bmod m \\ 0 & \text{ if not} \end{cases}$$ $$\stackrel{(1)}{=} \sum_{0 \leq x < h} \sum_{0 \leq y < h} \frac1{m} \sum_{0 \leq k < m} \exp\Big(\frac{2 \pi i (x - y)k}{m}\Big)$$ $$\stackrel{(2)}{=} \frac1{m} \sum_{0 \leq k < m} \sum_{0 \leq x < h} \sum_{0 \leq y < h} \exp\Big(\frac{2 \pi i (x - y)k}{m}\Big)$$ $$\stackrel{(3)}{=} ???$$ In step (3) comes the problem. It is not clear what the main term of $\sum_{0 \leq k TITLE: On the chromatic number of an analytic graph QUESTION [5 upvotes]: Let $X$ be a Polish space and let $G\in\mathbf{\Sigma}^1_1(X^2)$ be a graph on $X$, that is an irreflexive and symmetric relation on $X$. Given a cardinal $\kappa$ we say that $G$ has chromatic number $\kappa$, in symbols $\chi(G)=\kappa$, if there is a function $\varphi\colon X\to Y$ for some Polish space $Y$ such that $|\varphi(X)|=\kappa$ and for every $y\in\varphi(X)$, we have that $\varphi^{-1}(y)$ is $G$-independent, meaning that $G\cap(\varphi^{-1}(y))^2=\varnothing$, and $\kappa$ is the least cardinal with this property. We say that $G$ has Borel chromatic number $\kappa$, in symbols $\chi_B(G)=\kappa$ if we additionally require $\varphi$ to be Borel (of course the fact that $Y$ is Polish is completely irrelevant in the definition of $\chi(G)$ and I'm only phrasing it this way to define $\chi_B(G)$ analogously). Question: Does $\mathsf{ZFC}$ prove that $\chi(G)\in\{1,2,\ldots,\aleph_0,2^{\aleph_0}\}$? Of course the answer is positive under $\mathsf{CH}$ but that is hardly interesting. Remarks: If $\mathsf{CH}$ fails and there are no regularity assumption on $G$ then there are trivial counterexamples, fix $A\subseteq\Bbb R$ with $|A|=\aleph_1$ and let $G=A^2\setminus\Delta_\Bbb R$. Then $\chi(G)=\aleph_1$ but $G$ is not analytic. The same question for $\chi_B(G)$ (or even for Baire measurable colourings) has a positive answer by the Kechris-Solecki-Todorcevic $G_0$-dichotomy. $\chi(G)$ and $\chi_B(G)$ can be wildly different, there are examples with $\chi(G)=2$ ($G$ can even be taken to be acyclic) and $\chi_B(G)=2^{\aleph_0}$. REPLY [6 votes]: There is a counterexample due to Todorcevic. See Proposition 9.2 in Kechris, Solecki, Todorcevic, Borel Chromatic Numbers, Advances in Mathematics, Vol. 141, Issue 1, 1999, Pages 1-44<|endoftext|> TITLE: What is the homotopy type of the poset of nontrivial decompositions of $\mathbf{R}^n$? QUESTION [19 upvotes]: Consider the following partial order. The objects are unordered tuples $\{V_1,\ldots,V_m\}$, where each $V_i \subseteq \mathbf{R}^n$ is a nontrivial linear subspace and $V_1 \oplus \cdots \oplus V_m = \mathbf{R}^n$. We say that $\{V_1,\ldots,V_m\} \geq \{W_1,\ldots,W_i\}$ if for each $j=1,\ldots,m$ there exists $1 \leq k \leq i$ such that $V_j \subseteq W_k$. (I.e. if $\{V_1,\ldots,V_m\}$ is a finer decomposition than $\{W_1,\ldots,W_i\}$.) This partial order has a minimal element, $\{\mathbf{R}^n\}$, which is the trivial decomposition. If we remove it, what is the homotopy type of this partial order? Some things that are known about similar problems. If we're working over $\mathbb{F}_1$, i.e. with subsets of $\{1,\ldots,n\}$, then there is both a minimal and a maximal object. If we remove both of these then the homotopy type of the partial order is a wedge of $S^{n-3}$'s. If instead we work with ordered decompositions, then this is a barycentric subdivision of Ruth Charney's split building, so it is a wedge of $S^{n-2}$'s. Edit: clarity and notation. REPLY [10 votes]: Let me write $V$ for a finite-dimensional vector space over some field (the field will not play a role), and $\mathsf{P}(V)$ for the poset described in the question, which I consider as a category. Let me rephrase the order relation: $\{V_1, \ldots, V_n\} \geq \{W_1, \ldots, W_m\}$ if and only if each $W_j$ is a direct sum of $V_i$'s. The nerve of this poset is $(\mathrm{dim}(V)-2)$-dimensional, so to see that it is a wedge of spheres one must show that it is $(\mathrm{dim}(V)-3)$-connected. I think this is true. Let $\mathsf{S}(V)$ denote the following category. It has objects given by pairs $([n], f : [n] \to Sub(V))$ consisting of a standard set $[n] := \{1,2,\ldots, n\}$ and a function from $[n]$ to the set of proper vector subspaces of $V$, such that $\bigoplus_{i \in [n]} f(i) = V$. A morphism in $\mathsf{S}(V)$ from $([n], f)$ to $([n'], f')$ is given by a surjection $e : [n] \to [n']$ such that $f'(i) = \bigoplus_{j \in e^{-1}(i)} f(j)$. There is a functor $F : \mathsf{S}(V) \to \mathsf{P}(V)$ given by sending $([n], f)$ to the unordered collection $\{f(i)\}_{i \in [n]}$. Now let me do something a bit odd: choose a total order $\prec$ on the set of vector subspaces of $V$. Then we can consider objects of $\mathsf{P}(V)$ as given by lists of subspaces $(V_1, \ldots, V_n)$ with $V_1 \prec V_2 \prec \cdots \prec V_n$, and such that $\bigoplus_{i=1}^n V_i = V$. Attempt to define a functor $G : \mathsf{P}(V) \to \mathsf{S}(V)$ by $$G(V_1, \ldots, V_n) := ([n], i \mapsto V_i)$$ on objects. If $(V_1, \ldots, V_n) \geq (W_1, \ldots, W_m)$ in $\mathsf{P}(V)$ then each $W_j$ is a direct sum of $V_i$'s. Define a function $e : [n] \to [m]$ by $e(i) = j$ if $V_i \subset W_j$; this gives a morphism in $\mathsf{S}(V)$ from $([n], i \mapsto V_i)$ to $([n], j \mapsto W_j)$. To check that this is indeed a functor suppose that $(W_1, \ldots, W_m) \geq (U_1, \ldots, U_\ell)$, with $e'(j) = k$ when $W_j \subset U_k$. Then we indeed have $e' e(i) = k$ when $V_i \subset W_{e(i)} \subset U_k$, so $G$ is indeeed a functor. The conclusion of this discussion is that $\mathsf{P}(V)$ is a retract of $\mathsf{S}(V)$, so it suffices to show that $\mathsf{S}(V)$ is $(\mathrm{dim}(V)-3)$-connected. The full name of $\mathsf{S}(V)$ is $\mathsf{S}^{E_\infty}(V)$, the $E_\infty$-splitting category defined by Galatius, Kupers, and I in Definition 17.17 of Cellular $E_k$-algebras (applied to the symmetric monoidal groupoid of finite-dimensional vector spaces; the way I have described it here is not identical to that definition, but is an equivalent category). It has a cousin $\mathsf{S}^{E_1}(V)$ which turns out to be precisely (the category of simplices of) Charney's split building. In particular $\mathsf{S}^{E_1}(V)$ is $(\mathrm{dim}(V)-3)$-connected by Charney's theorem. It would take too long to explain all the details here, but the theory developed in that paper shows that the collection of all $\Sigma^2 \mathsf{S}^{E_\infty}(V)$'s can be obtained from the collection of all $\Sigma^2 \mathsf{S}^{E_1}(V)$'s by an iterated bar construction, and in particular given Charney's connectivity result (for all $V$) it follows that the $\Sigma^2 \mathsf{S}^{E_\infty}(V)$ are also $(\mathrm{dim}(V)-1)$-connected, so the $\mathsf{S}^{E_\infty}(V)$ are homologically $(\mathrm{dim}(V)-3)$-connected. (It should not be hard to show it is simply-connected, by hand.) I'm happy to discuss the details by e-mail, if you like. (Also, now that I believe it is true I expect there must be a more elementary way to deduce it from Charney's theorem.) Edit: As @inna says in the comments, the functor $F : \mathsf{S}(V) \to \mathsf{P}(V)$ is actually an equivalence, because there is a natural isomorphism $\mathrm{Id} \Rightarrow F \circ G$ given by applying the unique permutation necessary to put things in the order given by $\prec$. Let me try to give some references for what I said in the final paragraph. All references are to Cellular $E_k$-algebras: Writing $\mathsf{G}$ for the symmetric monoidal category of finite-dimensional vector spaces and linear isomorphisms, we work in the category $\mathsf{sSet}_*^\mathsf{G}$ of functors from $\mathsf{G}$ to pointed simplicial sets. This is again symmetric monoidal by Day convolution, and the functor $$\mathbb{t}(V) = \begin{cases} S^0 & V \neq 0\\ * & V=0 \end{cases}$$ is a nonnital commutative monoid, and hence also a nonunital $E_\infty$-algebra. (In the paper this object is called $\underline{*}_{>0}$.) There is another character involved: the derived $E_k$-indecomposables $Q^{E_k}_\mathbb{L}(\mathbb{t})$ for $1 \leq k \leq \infty$, which are again objects of $\mathsf{sSet}_*^\mathsf{G}$. For $V$ an object of $\mathsf{G}$ we write $H_{V,d}^{E_k}(\mathbb{t}) := H_d(Q^{E_k}_\mathbb{L}(\mathbb{t})(V))$. The ingredients I have in mind are now: a. Combining Proposition 17.4 and Lemma 17.10 shows that $$\Sigma Q^{E_1}_\mathbb{L}(\mathbb{t})(V) \simeq \Sigma^2 \mathsf{S}^{E_1}(V)$$ (in fact by Section 17.5 this can be desuspended once) and hence Charney's theorem shows that $H_{V,d}^{E_1}(\mathbb{t})=0$ for $d < \dim(V)-1$. b. Theorem 14.4 (this is the application of the bar construction result, but is packaged so one doesn't explicitly have to think about that) applied with $\rho(V) := \mathrm{dim}(V)$ shows that $H_{V,d}^{E_\infty}(\mathbb{t})=0$ for $d < \dim(V)-1$ too. c. Corollary 17.23 shows that $Q^{E_\infty}_\mathbb{L}(\mathbb{t})(V) \simeq \Sigma \mathsf{S}^{E_\infty}(V)$ so the above translates to $\tilde{H}_*(\mathsf{S}^{E_\infty}(V))=0$ for $* \leq \mathrm{dim}(V)-3$.<|endoftext|> TITLE: Quasi-isometry groups of metric spaces QUESTION [13 upvotes]: Given a metric space $(X, d)$, we can consider the set of all quasi-isometries $f: X \to X$, and quotient out by the equivalence relation identifying $f$ and $g$ if $\sup_{x \in X}d(f(x), g(x))$ is finite. Doing so, we obtain a set of equivalence classes $\mathcal{QI}(X)$ that is a group under composition. In the same spirit as the questions Every group is a fundamental group and Is every group the automorphism group of a group?, we can ask: for which groups $G$ does there exist a metric space $X$ such that $\mathcal{QI}(X) \cong G?$ Surprisingly, someone told me today that basically nothing is known about this question. According to them, we do not even know how to construct a metric space $X$ such that $\mathcal{QI}(X)$ is a finite cyclic group. Given this, my question is: what do we know about the quasi-isometry groups of metric spaces? For example, what are some metric spaces $X$ for which $\mathcal{QI}(X)$ has been computed? Do we know of any groups $G$ which are not isomorphic to $\mathcal{QI}(X)$ for any $X$? REPLY [8 votes]: Regarding the question of spaces $X$ for which $QI(X)$ is known, a good keyword is quasi-isometric rigidity. One reference would be the survey in Chapter 25 of the Druţu–Kapovich book "Geometric group theory". A space $X$ is called strongly QI rigid if the map $\operatorname{Isom}(X) \to QI(X)$ is surjective. Often $X$ has no non-trivial isometries that are bounded distance from the identity, so we get an isomorphism. For example, Pansu showed that quaternionic hyperbolic spaces are strongly QI rigid. Pansu, Pierre, Carnot-Carathéodory metrics and quasiisometries of symmetric spaces of rank 1, Ann. Math. (2) 129, No. 1, 1-60 (1989). ZBL0678.53042..<|endoftext|> TITLE: What flavor of set theory is used in model theory? QUESTION [12 upvotes]: When I read statements like ‘first order theories can’t control cardinalities of their models’ I wonder, what flavor of set theory is used in a (meta)model theory? (I hope not a naïve set theory, lol). Can I use, say, New Foundations theory with a universal set of all sets, talking about ZFC models, or vice versa? Can I ask if any first order theory can have a model with a cardinality in between countable and continuum, thus asking how in a meta-set theory, used in model theory, the continuum hypothesis is settled? REPLY [9 votes]: As many have pointed out by now, model theory, as usually practiced nowadays, doesn't require a lot of attention to foundations; the answers to questions in tame model theory like neostability theory are usually absolute. But I wanted to mention a few examples where moving beyond ZFC was important to bring core model-theoretic phenomena to light. Shelah's 'dividing lines' philosophy, which has had a huge impact on the development of model theory, suggests that it is interesting to find properties that divide the space of first order theories into tame and wild. Good dividing lines have the property that you can prove interesting structure theorems on the tame side and interesting nonstructure theorems on the wild side. But they also have the property that they can be characterized by both 'inside' (i.e. syntactic) and 'outside' (i.e. semantic or set-theoretic) criteria. The outside characterization usually has a set-theoretic flavor and this sometimes requires working outside of ZFC to obtain the equivalence. Here are some examples: 1. NIP - A theory $T$ is NIP if no formula has the independence property, i.e. there is no $\varphi(x;y)$ and sequence $(a_{i})_{i \in \mathbb{N}}$ in a model of $T$ such that $\{\varphi(x;a_{i}) : i \in X\} \cup \{\neg \varphi(x;a_{i}) : i \not\in X\}$ is consistent for all $X \subseteq \mathbb{N}$. This is equivalent to an outside condition: define $\mathrm{ded}(\kappa)$ to be the supremum of cardinals $\lambda$ such that there is a linear order of size $\lambda$ with a dense subset of size $\kappa$. Shelah proved that the number of types over a set of size $\kappa$ in a countable NIP theory can be bounded above by the cardinal $\mathrm{ded}(\kappa)^{\aleph_{0}}$, while the number of types over a set of size $\kappa$ in any theory with the independence property will be $2^{\kappa}$. So one can characterize the (countable) NIP theories by their type counting function, provided $\mathrm{ded}(\kappa )^{\aleph_{0}} < 2^{\kappa}$ for some $\kappa$. This can be arranged by forcing results of Mitchell. 2. Simple theories - A theory $T$ is simple if no formula has the tree property, i.e. there is no $\varphi(x;y)$, tree of tuples $(a_{\eta})_{\eta \in \omega^{<\omega}}$ in a model of $T$, and $k < \omega$ such that (a) For all $\eta \in \omega^{\omega}$, $\{\varphi(x;a_{\eta | i} : i < \omega\}$ is consistent. (b) For all $\eta \in \omega^{<\omega}$, $\{\varphi(x;a_{\eta^{\frown}\langle i \rangle}) : i < \omega\}$ is $k$-inconsistent. This is equivalent to an outside condition. Define the saturation spectrum $\mathrm{SP}(T)$ to be the set of pairs of cardinals $\kappa \leq \lambda$ such that every model of $T$ of size at most $\lambda$ has a $\kappa$-saturated elementary extension of size $\lambda$. It follows from a standard argument that if $\lambda = \lambda^{<\kappa}$, then $(\kappa,\lambda)$ is in the saturation spectrum of any $T$ (in a countable language, say) and Shelah proves that, modulo a forcing axiom he calls $\mathrm{Ax}_{0}\mu$, a theory $T$ is simple if and only if there is some pair $(\lambda, \kappa)$ in the saturation spectrum of $T$ which does not satisfy $\lambda = \lambda^{<\kappa}$. Shelah proves the consistency of this forcing axiom by a class forcing argument. 3. NSOP$_{2}$ A theory $T$ is said to be NSOP$_{2}$ if there is no formula $\varphi(x;y)$ and tree of tuples $(a_{\eta})_{\eta \in \omega^{<\omega}}$ in a model of $T$ such that (a) For all $\eta \in \omega^{\omega}$, $\{\varphi(x;a_{\eta | i}) : i < \omega\}$ is consistent. (b) For all $\eta \perp \nu$ in $\omega^{<\omega}$, $\{\varphi(x;a_{\eta}), \varphi(x;a_{\nu})\}$ is inconsistent. This has an outside characterization as well, via something called the interpretability order (sometimes also called the `triangle star order'). Suppose $T_{1},T_{2}$ are countable theories and $\lambda$ is a cardinal. Say $T_{1} \unlhd^{*}_{\lambda} T_{2}$ if there is a theory $\tilde{T}$ in a language of size $< \lambda$ that interprets both $T_{1}$ and $T_{2}$ and, moreover, has the property that, in any model of $\tilde{T}$, if the interpreted model of $T_{2}$ is $\lambda$-saturated, then the interpreted model of $T_{1}$ is $\lambda$-saturated. Then say $T_{1} \unlhd^{*} T_{2}$ if $T_{1} \unlhd^{*}_{\lambda} T_{2}$ for all sufficiently large $\lambda$. Assuming GCH, a countable theory is maximal in this pre-order $\unlhd^{*}$ on theories if and only if that theory has SOP$_{2}$. This is really three theorems: Džamonja-Shelah proved that maximality implies a property called SOP$_{2}''$ assuming GCH. Shelah-Usvyatsov proved that SOP$_{2}''$ and SOP$_{2}$ are equivalent for theories, and then Malliaris-Shelah proved recently that SOP$_{2}$ implies maximality. The point of these examples is that set-theoretic characterizations have been established as part of the method for recognizing dividing lines in model theory and often these characterizations require making assumptions that go beyond what can be established in ZFC alone. It is no doubt true that most questions about simple or NIP theories are absolute, but the fact that they can be given this sort of abstract set-theoretic description is part of what contributes to the sense that they mark meaningful notions of complexity.<|endoftext|> TITLE: Classification of homogeneous Einstein manifolds QUESTION [11 upvotes]: In Besse's "Einstein manifolds", p. 177, he states that, until that moment, no general classification of homogeneous Einstein manifolds was know, even in the compact case. More specifically, he poses a problem: classify the compact simply connected homogeneous manifolds $M=G/K$ which admit a $G$-invariant Einstein metric. Does that question remain open to this day? REPLY [13 votes]: Yes, the question is still open. I suggest to read this quite recent paper by Kerr and Böhm. It reviews some of the most important advances in the problem and includes several open problems.<|endoftext|> TITLE: Group of isometries of Banach spaces a topological group? QUESTION [9 upvotes]: Let $X$ be a Banach space and let $\mathrm{Iso}(X)$ be its group of isometries, i.e., the set of surjective linear maps $T: X \to X$ with $\|Tx\| = \|x\|$. Q: Is $\mathrm{Iso}(X)$ a topological group under the strong topology? While it is easy to show that multiplication is continuous, it is not clear to me how to show that inversion is continuous. I did not find a reference in the literature for this statement. If $X = H$ is a separable Hilbert space, then $\mathrm{Iso(X)} = \mathrm{U}(H)$, the unitary group of $H$, and the statement that this is a topological group is well-established. However, the proof (that I know) uses the fact that the weak and the strong operator topology agree on $\mathrm{U}(H)$ and that the inverse is just given by $u \mapsto u^*$, which is continuous in the weak operator topology. REPLY [10 votes]: That the inverse is continuous for the strong topology would actually be true for any bounded subgroup of $GL(X)$, the invertible operators on $X$. Firstly, as translation is continuous, it suffices to consider continuity at the identity. Now let $(T_i)$ be a bounded net of invertibles converging strong to $I$, and with $(T_i^{-1})$ also bounded, say a common bound of $K$. Then, for $x\in X$, $$ \|T_i^{-1}(x) - x\| = \|T_i^{-1}(x - T_i(x))\| \leq K \|x-T_i(x)\| \rightarrow 0 $$ by assumption that $T_i\rightarrow I$ strongly. Thus also $T_i^{-1}\rightarrow I$ strongly.<|endoftext|> TITLE: Solve in integers: $y(x^2+1)=z^2+1$ QUESTION [9 upvotes]: Find all integer solutions to the equation $$ y(x^2+1)=z^2+1. $$ There is, for example, an infinite family of solutions $x=u$, $y=(uv\pm1)^2+v^2$, $z=(u^2+1)v \pm u$, $u,v \in {\mathbb Z}$, but there are also solutions outside of this family, e.g. $(x,y,z)=(8,5,18)$ or $(x,y,z)=(12,2,17)$. The question is to describe all integer solutions. Any reasonable description is ok. An algorithm generating all solutions is also ok, provided that it does not involve any search by trial and error (otherwise there is a trivial algorithm that tries all triples $(x,y,z)$ in some order). Using parametric expressions, recurrence relations, or something like ``start with this solution and apply these operations in any order'' (like generating Markov numbers via Markov tree) would be ideal, but more complicated algorithms are also possible. For example, for equation $yz=x^3+1$, there is an obvious algorithm "let $x$ be an arbitrary integer, let $y$ be any divisor of $x^3+1$, and then let $z=(x^3+1)/y$", which I think is acceptable. The equation in question is one of the smallest/simplest ones for which I do not see any reasonable method/algorithm to describe all solutions, hence the question. Remark: If we define, for any integer $u$, set $S(u)$ as a set of integers $0\leq r < u^2+1$ such that $\frac{r^2+1}{u^2+1}$ is an integer, then all the solutions to the equation are in the form $x=u$, $z=(u^2+1)v \pm r$ and $y=(u^2+1)v^2 \pm 2vr + \frac{r^2+1}{u^2+1}$ for $u,v \in {\mathbb Z}$ and $r \in S(u)$, but we need trial and error to construct set $S(u)$, so I do not think this is an acceptable answer. REPLY [3 votes]: This answer (which I made Community Wiki) attempts an explicit summary of how to ``list" explicitly the solutions of the given equation, given the answer of Robert Israel and the comment of David Speyer which followed it. It's convenient to deal with the prime $2$ first, but this is easy, because the only condition imposed on $z$ to get the right power of $2$ dividing $z^{2}+1$ is that if $x$ is odd, we need $z$ odd. If $p$ is an odd prime such that $p^{n}$ is the exact power of $p$ dividing $x^{2}+1$, then we just require that $p^{n}$ divides $z^{2} -x^{2}$, so we need $ z \equiv \pm x$ (mod $p^{n}$) since $p$ is odd. Notice then that, in all cases, there are $2^{k}$ possible congruences for $z$ (mod $x^{2}+1$), where $k$ is the number of distinct odd primes dividing $x^{2}+1$. To be more explicit, if $x$ is even, let the prime factorization of $x^{2}+1$ be $x^{2}+1 = \prod_{i=1}^{k}p_{i}^{n_{i}}$, where the $p_{i}$ are distinct primes, and if $x$ is odd, let the prime factorization of $\frac{x^{2}+1}{2}$ be $\frac{x^{2}+1}{2} = \prod_{i=1}^{k}p_{i}^{n_{i}}$. For any ordered $k$-tuple of signs $(\epsilon_{1}, \ldots, \epsilon_{k})$, solve (simultaneously) $z \equiv \epsilon_{i} x$ (mod $p_{i}^{n_{i}}$) for each $i$. If $x$ is odd, also impose the condition that $z$ is odd. In either case, this gives $2^{k}$ distinct possibilities for the congruence of $z$ (mod $x^{2}+1$) which yield $x^{2} + 1 | z^{2}+1$, and there are no other valid congruences (where, as noted earlier, $k$ is the number of distinct odd prime divisors of $x^{2}+1$). For each integer $r$ and each integer $m \neq 0$ let $S(m,r)$ be the set of integers $0\leq w<|m|$ such that $w^2 \equiv r \,(\text{mod } m)$. We have just explicitly described the set $S(x^2+1,-1)$ for any integer $x$. Using it, the complete solution set to our equation is given by $x=u$, $z=(u^2+1)v + r$ and $y=(u^2+1)v^2 + 2vr + \frac{r^2+1}{u^2+1}$ for $u,v \in {\mathbb Z}$ and $r \in S(u^2+1,-1)$.<|endoftext|> TITLE: Irreducible 3-manifold with boundary of genus greater than 1 QUESTION [5 upvotes]: Let $M$ be an irreducible 3-manifold with incompressible boundary of genus > 1. When is $M$ homotopy equivalent to an Eilenberg-MacLane space? Or it is never true? REPLY [11 votes]: M is always aspherical, and hence a Eilenberg-Maclane space. This is because $\pi_1(\partial M)$ embeds to $\pi_1(M)$, by the incompressibility condition. If the genus of the boundary is no less than 1, then $M$ has infinite fundamental group. This forces the universal cover $\tilde M$ to be a non-compact 3-manifold, so $H_i(\tilde M)=0,\ i\geqslant 3$. By sphere theorem, $M$ is irreducible implies that $\pi_2(\tilde M)=\pi_2(M)=\{e\}$. Note that $\pi_1(\tilde M)=\{e\}$, then by Hurewicz theorem we have $H_i(\tilde M)=0,\ i=1,2.$ Then $\tilde M$ has trivial integral homology and is simply connected, so $\tilde M$ is contractible by Whitehead theorem.<|endoftext|> TITLE: Are all degree-1 cohomology operations Bocksteins? QUESTION [10 upvotes]: I'm interested in cohomology operations (in ordinary cohomology) $$H^i(-, G)\rightarrow H^{i+1}(-, H)\;,$$ that is, elements of $$H^{i+1}(K(G, i), H)\;.$$ I know that $K(G, 1)=BG$, so for $i=1$, those cohomology operations are in $H^2(BG, H)$, and therefore given by the Bocksteins of the corresponding central extensions of $G$ by $H$. Also, for $G=H=\mathbb{Z}_p$, the stable cohomology operations are given by the Steenrod algebra, and the only degree-1 elements are Bocksteins. However, I don't know how it is for unstable cohomology operations, or groups other than $\mathbb{Z}_p$. I'm mostly interested in simple groups, such as finitely generated or $\mathbb{R}/\mathbb{Z}$. REPLY [16 votes]: Yes. For $i\ge1$ you can build $K(G,i)$ from the Moore space $M(G,i)$ by adding cells of dimension $\ge i+2$, so $H_i(K(G,i); Z) = G$ and $H_{i+1}(K(G,i); Z) = 0$. Hence $Ext(G, H) \cong H^{i+1}(K(G,i); H)$ by the UCT. The elements of $H^{i+1}(K(G,i); H)$ represent the cohomology operations $H^i(-;G) \to H^{i+1}(-;H)$, and the elements of $Ext(G,H)$ correspond to the Bockstein operations. The case $i=0$ may require special attention.<|endoftext|> TITLE: What is the most general form of Hurwitz's theorem in complex analysis? QUESTION [10 upvotes]: In complex analysis, Hurwitz's  theorem  roughly states that, under certain conditions, if a sequence of holomorphic functions converges uniformly to a holomorphic function on compact sets, then after a while  (above a certain rank ) those functions and the limit function have the same number of zeros  in any open disk. I read somewhere that the result is valid more generally on bounded convex sets , from what I remember, but no references were given. Question. Where can I find this more general statement of Hurwitz's theorem, and where can I find some useful references in this direction? Thank you. REPLY [11 votes]: This has nothing to do with convexity. The exact formulation is this: Let $\Omega$ be an arbitrary region, and $f_n\to f$ is a sequence of holomorphic functions converging uniformly on compacts in $\Omega$. If $f\neq 0$ (this is an important condition!), and $D$ is a region, such that $\overline{D}\subset \Omega$, and $f(z)\neq 0$ on $\partial D_1$, then there exists $N$ (dependng on $D$) such that for $n\geq N$, $f$ and $f_n$ have the same number of zeros in $D$. For the proof, you find an intermediate region $D_1$ such that $D\subset D_1\subset\overline{D_1}\subset\Omega$, and $\partial D_1$ is piecewise smooth, and the number of zeros of $f$ in $\overline{D_1}$ is the same as the number of zeros in $D$. Such region exists since $\overline{D}\subset\Omega$, and since the zeros of $f$ are isolated and $f(z)\neq 0,\; z\in\partial D$. Then the number of zeros $f$ in $D$ is the same as in $D_1$ and is equal to $$\frac{1}{2\pi i}\int_{\partial D_1} \frac{df}{f},$$ and the number of zeros of $f_n$ in $D_1$ is $$\frac{1}{2\pi i}\int_{\partial D_1} \frac{df_n}{f_n}.$$ Since $f_n\to f$ uniformly on the compact set $K:=\overline{D_1}\backslash D$, and $f(z)\neq 0$ on $K$, we conclude that for $n\geq N$, $f_n$ has the same number of zeros in $D$ and $D_1$. Since the integrals for $f_n^\prime/f_n$ converge to the integral for $f'/f$, and all these integrals are integers, we conclude that they are equal for $n\geq N$. Refs. This is stated in full generality in the book A. I. Markushevich, Theory of functions of a complex variable, vol. I, Ch. IV, Sect 3 (p. 426 of the Russian original). Special cases are in: Ahlfors, Complex Analysis, p. 178, Theorem 2, Titchmarsh, The theory of functions, sect. 3.45, Marshall, Complex Analysis, Theorem 8.8, But all these special cases have the same proof as the general theorem whose complete proof I wrote.<|endoftext|> TITLE: Eigenvalues and eigenfunctions of the Laplace operator on entire plane QUESTION [6 upvotes]: According to the answers in the the following questions: How to prove the spectrum of the Laplace operator? and What is spectrum for Laplacian in $\mathbb{R}^n$ , the spectrum of the Laplace operator $\Delta :H^2(\mathbb{R}^2)\subset L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$ is in fact $\sigma(\Delta)=(-\infty,0].$ However, I was not able to find a discussion on the eigenvalues of $\Delta$. The set of eigenvalues $\sigma_p(\Delta)$ (also called point spectrum) is known to be contained in $\sigma(\Delta)$ and one can have $\sigma_p(\Delta)\subsetneq \sigma(\Delta)$. Indeed, by taking the Fourier transform $\mathcal{F}:L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$ of the eigenvalue problem one has $$\Delta u(x) = \lambda u(x),\;\;\forall x\in \Bbb R^2 \;\;\;\overset{\mathcal F}{\longrightarrow}\;\;\;\;-4\pi^2|\xi|^2\hat u(\xi) =\lambda \hat u(\xi), \;\;\;\forall\xi\in \Bbb R^2,$$ and this can only be satisfied by $\hat u=u=0$. This means that the only eigenvalue-eigenvector pair in this setting is $(\lambda,u)=(0,0)$ . Also, the same argument applies when $\Delta$ is seen as $\Delta:W^{m,p}(\mathbb{R}^2)\subset L^p(\mathbb{R}^2)\to L^p(\mathbb{R}^2)$ with $p\in [1,2)$ and $\mathcal F:L^p(\mathbb{R}^2)\to L^{p^*}(\mathbb{R}^2)$ with $1/p+1/p^*=1$. Question 1. What happens when $p>2$ and the Fourier transform becomes distribution valued, so that the above elementary argument cannot be applied directly? It seems if $u\in C^2(\mathbb{R}^2)$ is in fact an eigenvalue of $\Delta$, then it cannot be in $L^p(\mathbb{R}^2)$ for any $p\in [1,2]$. Depending on the answer to Q1, this might also hold for $p>2$. In any case, it seems that the $L^p$ framework is not suitable for this problem. Question 2. On what space(s) could one define the domain of $\Delta$ to obtain non-trivial eigenvalues? Edit. The crossed out sentence should be replaced by: "The point spectrum $\sigma_p(\Delta)$ is therefore empty." REPLY [7 votes]: The point spectrum coincides with the spectrum minus 0 if $p>2n/(n-1)$ and it is empty in the remaining cases ($n$ is the dimension). This is proved in G. Talenti: "Spectrum of the Laplace operator acting in $L^p(R^n)$", Indam, Symposia Mathematica vol VII, Academic Press 1971.<|endoftext|> TITLE: Is this a Brownian motion? QUESTION [6 upvotes]: I am building a 2D stochastic process as follows. I start with a point $P_0=(0,0)$. Then $P_k=(X_k,Y_k)$ is defined as follows, for $k>0$: \begin{align} X_k & =X_{k-1}+R_k \cos(2\pi\theta_k) \\ Y_k & =Y_{k-1}+R_k \sin(2\pi\theta_k) \end{align} where the $\theta_k$'s are uniformly and independently distributed (iid) on $[0,1]$, the $R_k$'s are also iid, and independent from the $\theta_k$'s. The $R_k$'s are generated as follows: $$R_k=\frac{1}{\lambda}\Big[-\log(1-U_k)\Big]^c,$$ where $\lambda>0$ and $U_k$ is uniform on $[0,1]$. If $c=1$, $R_k$ has an exponential distribution of parameter $\lambda$. If $c>0$, $R_k$ has a Weibull distribution; if $c<0$, $R_k$ has a Fréchet distribution. In all cases, $E[R_k]=\Gamma(1+c)/\lambda$ if $c>-1$, otherwise the expectation is infinite. It is possible to rescale the process, in the same way a random walk is rescaled to become at the limit, a Brownian motion. My question is this: will my simulation always result in a Brownian motion regardless of the parameter values? My hope is that the answer is sometimes yes, sometimes no. For standardization purposes, assume that $\lambda=\Gamma(1+c)$ if $c>-1$. I am interested in finding parameter values such that the resulting process is not Brownian. What if $c=4$? Below is a realization with $10^4$ points and $c=4$. It does not look Brownian, it consists of well separated clusters, typical for a large value of $c$. And my goal is to illustrate clustering techniques (in particular, identifying the number of clusters) for processes that are Brownian-related, but exhibiting a much stronger cluster structure with well separated clusters. REPLY [6 votes]: I vote for Mateusz Kwaśnicki. The condition for whether the random walk you generate this way scales towards Brownian motion under taking long times and rescaling is whether or not the variance is finite. You already know how to take the expectation, your formula is correct. $$ \int_0^1 \left(-\ln(1-x)\right)^c\, dx\, =\, \Gamma(1+c)\, . $$ So if $\lambda=1/\Gamma(1+c)$, then $\mathbf{E}[R_k]=1$. Note that you still have $$ \mathbf{E}[X_k]\, =\, \mathbf{E}[X_{k-1}]\, ,\ \mathbf{E}[Y_k]\, =\, \mathbf{E}[Y_{k-1}]\, , $$ because of the $\theta_k$ randomization. Then by the same integral formula $$ \mathbf{E}[R_k^2]\, =\, \frac{1}{\lambda^2}\, \int_0^1 \left(-\ln(1-x)\right)^{2c}\, dx\, =\, \frac{\Gamma(1+2c)}{\Gamma(1+c)^2}\, . $$ So, as long as $-1/2 TITLE: Early successes of Schwartz distribution theory QUESTION [10 upvotes]: What are the early successes of Schwartz distributions theory? What are the hard theorems that became simple and what open problems were solved with this new tool soon after Laurent Schwartz released his book in the fifties? REPLY [15 votes]: Following the citation for the 1950 Fields medal I would argue that putting the Dirac delta function on a firm ground was the early success of the theory of distributions.$^\ast$ An extensive list of (later) applications is at Nice applications for Schwartz distributions $^\ast$ Incidentally, this is Dyson's take on applications of the theory of distributions in physics [source] The theory of generalized functions (alias distributions) is the only part of post-war pure mathematics which has turned out to have a genuine usefulness in physics. In fashionable field-theoretic circles a paper can now hardly be submitted for publication without at least a reference to Schwartz's two-volume Théorie des Distributions. In some recent papers one can find evidence that Schwartz's work has not only been quoted but has even been read. Applications of distribution theory are being found all over physics, in nonlinear mechanics and fluid dynamics as well as in field theory. REPLY [9 votes]: This is a very broad topic. If you want a nice little book on the use of distributions in mathematical physics, I suggest this one by Demidov. In the Preface and in Chapter 1, Section 1, the author deals a bit with the history of the concept of function and the role there played by generalized functions (a particular case of which, in his terminology, are Schartz distributions). The book proceeds by showing how some classical PDEs of mathematical physics admit a much more natural and explicative solution if considered in weak form, and how this is further generalized in the theory of distributions. I also quite like the final chapter on the (underestimated, in my opinion) concept of "generalized function according to Egorov". So all in all I think this may be a useful read for your aim. But please notice that, as usual in the Russian school, quite a bit of work, including some key proofs, is left to the reader in the problems.<|endoftext|> TITLE: Philosophy of forcing and ctm QUESTION [9 upvotes]: I asked a similar question on SE before and received an answer. Not completely convinced, I decided to ask it here with some modifications. Note: I understand how forcing works and how it proves relative consistency statement in the metatheory; this question is more about the philosophical interpretation, hence the tag "soft-question". There are mainly two approaches to forcing as far as I know: the internal approach, usually with a complete Boolean algebra, and the external approach, using a countable transitive model (ctm). While the internal approach is elegant and intuitive in my opinion, it only creates a Boolean-valued model instead of a usual transitive model. Moreover my impression is that this is not how people usually think about forcing; I feel that when people do forcing they really imagine that some new set is created out of thin air and thrown into the current universe. Now for the ctm approach. The problem (I feel) is that it only creates a ctm. All kinds of stange stuff can happen for ctm: (Hamkins) A ctm can be pointwise definable. (Hamkins) Every ctm embeds into its own constructible universe. There exists a statement $\phi$ such that ZFC+"there is a countable transitive model of ZFC$+\phi$" is consistent (relative to some reasonable hypothesis), while it becomes inconsistent if the word countable is changed to uncountable. An example of $\phi$ is "there does not exist ctm of ZFC" (consider the minimal model). Is there a more interesting example? I think if a statement has no uncountable transitive model then it "obviously should not be true", even if there is a consistency proof via forcing or whatever. Edit: seems like such a $\phi$ cannot be too interesting, because as long as $\phi$ is consistent with "$Ord$ is Mahlo", then by reflection there exists inaccessible $\kappa$ such that $V_\kappa\models\phi$. Basically, my question is why should one believe that forcing is meaningful? Yes forcing shows there is a ctm of $\lnot$CH, so $\lnot$CH is consistent, but could that be merely a pathology about ctm, just like 1-3 above? I mean there is also a model of ZFC+$\lnot$Con(ZFC), but it has to be ill-founded so nobody believes this should actually be the case; I don't feel much better about ctm than ill-founded model. For the particular case of $\lnot$CH, another doubt of mine is that the von-Neumann hierarchy picture is often used to justify the consistency of ZFC, but the idea of forcing seems to say that we actually never reach the "true" power set of a set, so the hierarchy cannot proceed in the first place. Of course large cardinals directly imply the consistency of $\lnot$CH, but that is another story. Here is another fun thought (to add more vagueness to this question): set theory people often start a sentence with "collapse $X$ to a countable set". If we take this seriously, does it mean actually all sets are countable, and uncountability is just an illusion? Dana Scott seems to share a similar view in his foreword to the book Set Theory: Boolean-Valued Models and Independence Proofs: ......Perhaps we would be pushed in the end to say that all sets are countable (and that the continuum is not even a set) when at last all cardinals are absolutely destroyed. But really pleasant axioms have not been produced...... REPLY [3 votes]: A little while ago, I wrote a paper addressing some linked worries to what you have here: Barton, N. Forcing and the Universe of Sets: Must We Lose Insight?. J Philos Logic 49, 575–612 (2020). https://doi.org/10.1007/s10992-019-09530-y I'll add a very brief summary of what I argue there: There's pressure to want to provide "nice" interpretations of forcing, since forcing is more than just a tool for proving relative consistency. We can also formulate axioms about uncountable sets using forcing (e.g. remarkable cardinals) and prove theorems about uncountable sets in the universe using forcing (cf. Todorčević and Farah's book, Malliaris and Shelah's result that $\mathfrak{p}=\mathfrak{t}$). You can get "nice" interpretations of forcing within the universe, where the model you use is very "close" to $V$ in certain senses. In this sense, studying forcing can be differentiated from the case of $\sf ZFC$ + $\neg Con(\mathsf{ZFC})$ that you point to.<|endoftext|> TITLE: Anti-delta function? QUESTION [23 upvotes]: Did anyone ever consider a "function" or "distribution" $F(x)$ with the following property: its integral $\int_a^b F(x)\,dx=0$ for any finite interval $(a,b)$ but $\int_{-\infty}^\infty F(x)\,dx=1$? It definitely can be seen as a limit of smooth functions, and its Fourier transform will be a finction, equal to $1$ at $0$ and otherwise $0$. I think such object for sure should be considered and named by someone. REPLY [9 votes]: This is more a long comment following the ones of Mateusz Kwasnicki and Anixx and the nice answers given by Michael Hardy, Gro-Tsen, et al. As stated by Mateusz, such a mathematical objiect cannot be a function nor a distribution: indeed in this last case, without a compactification of the real line (as suggested by Gro-Tsen'asnswer), there's no $\varphi\in C_0^\infty$ whose support is compact and includes the point at infinity, so the space of functionals on function spaces made of this functions is necessarily empty. However, this limitation is not inherited by hyperfunctions: there exist hyperfunctions, notably Fourier hyperfunctions, "supported" (it is better to say "carried", since the support of a distribution and the carrier of a hyperfunction are not exactly the same) at $x=+\infty$. The first explicit example of such kind of generalized functions was constructed by Morimoto and Yoshino [2], and was lated simplified and extended by Akira Kaneko in [1] (who also deals with the higher dimensional case). The method (as explained by Kaneko [1] pp. 99-100) consists in choosing a defining function for the sought for hyperfunction which grows very fast on the real axis, for example $e^{e^{z^2}}=\exp\big(e^{z^2}\big)$, and a path $\gamma$ to $+\infty$ along which it rapidly decreases: then the holomorphic function $F(z)$ defined as $$\DeclareMathOperator{\dmu}{\operatorname{d}\!} F(z) = \frac{1}{2\pi i}\int\limits_\gamma \frac{\exp\big(e^{\zeta^2}\big)}{\zeta-z} \dmu\zeta $$ is such that its "boundary value" (i.e. the defined hyperfunction) $f(x) =F(x+i0)-F(x-i0)$ is $0$ for all finite $x\in\Bbb R$, as it can be proved by deforming the integration path and using the standard Cauchy integral theorem. But nevertheless it is $$ F(x)= \exp\big(e^{x^2}\big) + \frac{1}{2\pi i}\int\limits_\gamma \frac{\exp\big(e^{\zeta^2}\big)}{\zeta -x} \dmu\zeta $$ thus the "boundary value" of $F$ produces a non trivial Fourier hyperfunction carried at $+\infty$. Bibliography [1] Akira Kaneko, "Explicit construction of Fourier hyperfunctions supported at infinity" (English), Kawai, Takahiro (ed.) et al., Microlocal analysis and complex Fourier analysis, River Edge, NJ: World Scientific Publishers, ISBN 981-238-161-9/hbk, pp. 99-114 (2002), MR2068531, Zbl 1046.46031. [2] Mitsuo Morimoto, Kunio Yoshino, "Some examples of analytic functionals with carrier at the infinity" (English),Proceedings of the Japan Academy, Series A 56, 357-361 (1980), MR0596004, Zbl 0471.46024.<|endoftext|> TITLE: Is $\mathbb{Q}$ the orbit of a continuous function that is computable when restricted to $\mathbb{Q}$? QUESTION [5 upvotes]: In the previous post What is the smallest set of real continuous functions generating all rational numbers by iteration? I asked for the smallest set of continuous real functions that could generate $\mathbb Q$ by iteration starting from $0$. Surprisingly one continuous function suffices and this can also be Lipschitz and possibly analytic. Given that we are applying the function only to rational numbers to generate our sequence the question arises whether it is possible that the function restricted to the rationals is also computable: Is there a continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ that generates $\mathbb{Q}$ by iteration where $f$ restricted to $\mathbb{Q}$ is computable? REPLY [6 votes]: Yes. This answer is based on the answers to your previous question. Start with a computable ergodic map $T$ (D. Thomine constructs an example here). For every basic open neighborhood $B_i$, the set $D_i = \bigcup_n T^{-n}(B_i)$ is a dense effectively open set, uniformly in $i$. So some computable real $p$ meets all the $D_i$, meaning $X = \{ T^n(p) : n \in \omega\}$ is dense. There is a computable, order-preserving bijection $g: X \to \mathbb{Q}$ (back-and-forth argument), and we may assume that $g(p) = 0$. $g$ induces a (bi-computable) homeomorphism $G: \mathbb{R} \to \mathbb{R}$. Define $f = G\circ T\circ G^{-1}$. Constructing the bijection: We'll build a computable sequence of rationals $(q_n)_{n \in \omega}$ and define $g(T^n(p)) = q_n$. Begin with $q_0 = 0$. Then compute enough of $p$ and $T(p)$ to determine how they are ordered ($p < T(p)$ or $T(p) < p$). If $p < T(p)$, choose $q_1$ to be a rational greater than 0; otherwise, choose $q_1$ to be a rational less than 0. In either case, choose $q_1$ to be the appropriate rational with the smallest Gödel number. Then compute enough of $p$, $T(p)$ and $T^2(p)$ to determine how they are ordered, and choose $q_2$ to be a rational in the same relative position to $q_0$ and $q_1$. Again, choose the appropriate rational of smallest Gödel number. Etc. This is all a computable process, so $g$ is computable. It's a total order-preserving injection by construction. Surjectivity is by induction on Gödel number. For a rational $r$, by the inductive hypothesis all rationals of smaller Gödel number are in the range of $g$. So fix an $n$ such that all rationals of smaller Gödel number occur in $q_0, \dots, q_{n-1}$, and assume $r$ does not occur in this list, as otherwise we are done. Define $C = \{i < n : q_i < r\}$. By the density of $X$, there is an $m \ge n$ with $T^i(p) < T^m(p)$ for all $i \in C$, and $T^m(p) < T^i(p)$ for all $i < n$ with $i \not \in C$. Fix the least such $m$. Then by construction, $q_m = r$.<|endoftext|> TITLE: Must uncountable standard models of ZFC satisfy CH? QUESTION [17 upvotes]: In Cohen's article, The Discovery of Forcing, he says that "one cannot prove the existence of any uncountable standard model in which AC holds, and CH is false," and offers the following proof. If $M$ is an uncountable standard model in which AC holds, it is easy to see that $M$ contains all countable ordinals. If the axiom of constructibility is assumed, this means that all the real numbers are in $M$ and constructible in $M$. Hence CH holds. But this argument, on the surface of it, invokes $V = L$. Can we eliminate the use of $V = L$? The discussion in a related MO question seems to come close to answering this question, but doesn't directly address it. REPLY [5 votes]: Here is a dichotomy which is close to what you want. Given a model $V$ of $\mathrm{ZFC} + \mathrm{CH}$. If $\omega_1^V$ is inaccessible to reals then every real number is contained in an uncountable standard model of $\mathrm{ZFC}+\lnot\mathrm{CH}$. If $\omega_1^V$ is not inaccessible to reals then there is a real $x$ such that every uncountable standard model of $\mathrm{ZFC}$ containing $x$ satisfies $\mathrm{CH}$ (perhaps vacuously). For 1, if $x$ is any real then $\omega^V_1$ is inaccessible in $L[x]$. Then $L_{\omega^V_1}[x]$ is an uncountable standard model of $\mathrm{ZFC}$ and we can force to add $\kappa$ Cohen reals over $L_{\omega^V_1}[x]$ for any $\omega_2^{L[x]} \leq \kappa < \omega_1^V$ to obtain an uncountable standard model of $\mathrm{ZFC}+\lnot\mathrm{CH}$. For 2, there must be a real $x$ such that $\omega^V_1 = \omega_1^{L[x]}$. Then we can mimic Cohen's argument to see that every uncountable standard model $M$ of $\mathrm{ZFC}$ that contains $x$ in must satisfy $\mathrm{CH}$ since such a model must have $\omega_1^M = \omega_1^V$.<|endoftext|> TITLE: The center of a representation von Neumann algebra, and finite index subgroups QUESTION [5 upvotes]: Consider a (countable) group $G$, a subgroup $H\leq G$ of finite index, and a unitary representation $\pi:G\to \mathcal{U}(\mathcal{H})$. If the center of the von Neumann algebra $\pi(H)''$ is finite dimensional, does this imply that the center of $\pi(G)''$ is finite dimensional as well? Is it at least true assuming that the representation $\pi$ is induced from a representation of $H$? Remarks: As far as I'm concerned, $H$ can be assumed to be normal if it makes things simpler. If $\pi$ is of type I, then the question is whether $\pi$ decomposes to finitely many irreducible representations provided that the restriction $\mathrm{Res}_H^G\pi$ does. This is obviously true. REPLY [7 votes]: The answer is YES and it follows from the general theory of finite-index inclusions of von Neumann subalgebras (a la Jones, Pimsner, Popa...), which says that finite-dimensionality of the center is preserved under taking a finite-index subalgebra/extension. This is perhaps an overkill and so I will try a layman's proof. Definition. An inclusion $M\subset N$ of von Neumann algebras is of finite-index if there is a (necessarily faithful normal) conditional expectation $E$ from $N$ onto $M$ such that $E\geq\lambda\,\mathrm{id}_N$ for some $\lambda>0$. Note that $E$ maps $Z(N)$ into $Z(M)$ and that a von Neumann algebra $A$ is finite-dimensional iff $\mathbb{C}1\subset A$ is of finite-index. Thus if $M\subset N$ is of finite-index, then $$\dim Z(M)<\infty\Rightarrow\dim Z(N)<\infty.$$ Example. If $H\le G$ is of finite-index, then the inclusion $\pi(G)'\subset\pi(H)'$ is of finite-index with the conditional expectation given by $$E(x)=\frac{1}{[G:H]}\sum_{g\in G/H}\pi(g)x\pi(g)^*,$$ which satisfies $E\geq\frac{1}{[G:H]}$. Hence $\dim Z(\pi(G)'') <\infty\Rightarrow\dim Z(\pi(H)'') <\infty$. We want to prove the converse. As I said in the beginning, one way is to use von Neumann algebra machinery (the standard form and the basic construction) to "flip" the inclusion. Here is an alternative way. By passing to a finite-index subgroup, we may assume that $H$ is normal. Take a natural realization (which I'm too lazy to write) of the induced representation $\mathrm{Ind}_H^G\,\pi$ on $\ell_2(G/H)\otimes\mathcal{H}$ that satisfies $$(\mathrm{Ind}_H^G\,\pi)(H)''\subset(\mathrm{Ind}_H^G\,\pi)(G)''\subset B(\ell_2(G/H)) \otimes\pi(H)''.$$ Then one can show the commutant inclusion ${\mathbb C}1\otimes\pi(H)'\subset (\mathrm{Ind}_H^G\,\pi)(H)'$ is of finite-index (via the conditional expectation coming from the trace on $B(\ell_2(G/H))$) and so the intermediate inclusion ${\mathbb C}1\otimes\pi(H)'\subset (\mathrm{Ind}_H^G\,\pi)(G)'$ is also of finite-index. From this and the fact that $\pi\subset\mathrm{Ind}_H^G\,\pi$, one obtains $$\dim Z(\pi(H)'') <\infty\Rightarrow\dim Z(\pi(G)'') <\infty.$$<|endoftext|> TITLE: Can this probability be obtained by a combinatorial/symmetry argument? QUESTION [16 upvotes]: Suppose that $a_1,\dots,a_n,b_1,\dots,b_n$ are iid random variables each with a symmetric non-atomic distribution. Let $p$ denote the probability that there is some real $t$ such that $t a_i \ge b_i$ for all $i$. It was shown that $$p=\frac{n+1}{2^n}.$$ Can this be proved by a combinatorial/symmetry argument? REPLY [24 votes]: If I understand correctly, $c_i := b_i/a_i$ should also be symmetric and non-atomic. Then the result holds if there exists $t$ so that for all $i$ $t \geq c_i$ if $a_i > 0$; $c_i \geq t$ if $a_i < 0$. Reorder the indices so that $|c_1| < |c_2| < \dots < |c_n|$. There are $2^n$ ways to assign signs to $a_i$ for each $i \in [n]$. The only assignments for which an appropriate $t$ exists are those where the signs of $a_1,\dots,a_k$ are positive and the signs of $a_{k+1},\dots,a_n$ are negative with $k \in \{0,1,2,\dots,n\}$. There are $n+1$ such assignments, so the desired probability is $(n+1)/2^n$.<|endoftext|> TITLE: Classification of simple Lie algebras over finite fields QUESTION [5 upvotes]: Classification of simple (or simple-restricted) Lie algebras over algebraically closed fields in positive characteristic is studied for a long time. Today, we know all finite-dimensional simple (or simple-restricted) Lie algebras over algebraically closed fields of characteristic $p \ge 5$. But, how about over finite fields? I don't know either the results of classifications over finite fields or some examples. Does anyone know some references? REPLY [4 votes]: The classification of simple finite-dimensional Lie algebras over finite fields is a very hard task and only few results are known. However, I suggest to have a look at the following paper by Bettina Eick and references therein: Some new simple Lie algebras in characteristic 2.<|endoftext|> TITLE: A fibration equivalent to having a terminal object QUESTION [7 upvotes]: It is well known that the codomain functor $$cod:\mathcal{C}^\to\to\mathcal{C}$$ from the arrow category of a category $\mathcal{C}$ to itself is a fibration iff $\mathcal{C}$ has binary pullbacks. Is there another functor that always exists for which promotion to a fibration is equivalent to $\mathcal{C}$ having a terminal object? I naively tried $!:\mathcal{C}\to{\bf 1}$ and $\{X\}:{\bf 1}\to\mathcal{C}$ without any luck. This would be of interest for the obvious reason -- having binary pullbacks and a terminal object is equivalent to being finitely complete, so a functor answering the above question positively would allow for a 'fully fibrational' characterization of finite completeness. Further, the construction below yields a functor that always exists such that promotion to a fibration is equivalent to having all pullbacks -- this together with Alexanders answer below offers a fully fibrational characterization of completeness, as originally desired. Thanks to Alexander and Andrej for pointing out in the comments that $cod$ as above being a fibration only yields binary pullbacks, not arbitrary ones. EDIT: I was able to fix the following construction to actually yield a functor that always exists such that promoting this functor to a fibration is equivalent to $\mathcal{C}$ having all pullbacks. (The first correction worked; we want them all to be the same commutative square universally, so that all paths from the pullback object into the object we're pulling back over are equal.) For a category $\mathcal{C}$, consider the category $Sink(\mathcal{C})$ whose objects are sinks in $\mathcal{C}$ indexed over ordinals $\alpha\in{\bf O_n}$ $$\{f_i:X_i\to X\}_{i<\alpha}$$ and whose arrows are only defined from sinks indexed over $1$ to arbitrary sinks $$\hat h:\{f:A\to B\}\to\{g_i:C_i\to D\}_{i<\alpha}$$ given by ordered pairs $$\hat h=\big(\{h_i:A\to C_i\}_{i<\alpha},h:B\to D\big)$$ such that all the obvious squares commute, so $$h\circ f=g_i\circ h_i$$ for all $i<\alpha$. We have an obvious functor $$cod:Sink(\mathcal{C})\to\mathcal{C}$$ $$\{f_i:X_i\to X\}_{i<\alpha}\mapsto X,$$ $$\hat h\mapsto h$$ and this functor is a fibration iff $\mathcal{C}$ has all pullbacks. We can identify $\mathcal{C}^\to$ with the subcategory of $Sink(\mathcal{C})$ indexed over $1$. REPLY [10 votes]: A category $C$ has a terminal object if and only if the canonical functor $C \star \mathbf{1} \to \mathbf{1} \star \mathbf{1} = \mathbf{2}$ is a fibration.<|endoftext|> TITLE: Is any abelian category a subcategory of $\mathrm{Ab}^I$? QUESTION [6 upvotes]: Motivation: define a concrete Abelian category as a category with a univalent and injective functor in $\mathrm{Ab}^I$ (such that all homological concepts in it coincide with simple set-theoretic concepts $\mathrm{Ab}^I$) Is it true that any abelian category is isomorphic to a subcategory $\mathrm{Ab}^I$ for some small category $I$? (I know Mitchell's embedding, but I don't see if it's useful here). The subcategory is closed with respect to taking kernels and cokernels. The isomorphism must be additive (as far as I understand, it follows from this that it preserves kernels, cokernels, exact sequences). If yes, then is it true that any Abelian category is isomorphic to a full subcategory $(R-\mathrm{Mod})^I$ for some ring $R$ and some small category $I$ (but this seems much more doubtful to me). REPLY [5 votes]: Observe that $\bigoplus_I : \textbf{Ab}^I \to \textbf{Ab}$ is a conservative exact functor: it is right exact by general nonsense, it preserves monomorphisms (because e.g. $\bigoplus_{i \in I} A_i$ is naturally a subgroup of $\prod_{i \in I} A_i$), and it is conservative because $\bigoplus_{i \in I} A_i \cong 0$ implies each $A_i \cong 0$. It seems to me that you are looking for a conservative exact functor $\mathcal{A} \to \textbf{Ab}$, so this observation implies that allowing $I$ with more than one element does not add any generality. For small abelian categories $\mathcal{A}$, the Freyd–Mitchell embedding theorem gives a fully faithful exact functor $\mathcal{A} \to R\textbf{-Mod}$ (where $R$ is a not necessarily commutative ring), so every small abelian category admits a conservative exact functor $\mathcal{A} \to \textbf{Ab}$. In particular, $\mathcal{A}$ will be isomorphic to a (not necessarily full) abelian subcategory of $\textbf{Ab}$, by which I mean a subcategory that is closed under finite direct sums/products, kernels, and cokernels. If $\mathcal{A}$ is not required to be small then one has to do more work. The existence of a conservative exact functor $\mathcal{A} \to \textbf{Ab}$ is itself a size restriction on $\mathcal{A}$. For example: Proposition. If there is a conservative exact functor $\mathcal{A} \to \textbf{Ab}$, then $\mathcal{A}$ is locally small and wellpowered. Proof. A conservative exact functor is automatically faithful, and $\textbf{Ab}$ is locally small, so $\mathcal{A}$ must also be locally small. Similarly, a conservative exact functor induces embeddings of subobject lattices, and $\textbf{Ab}$ is wellpowered, so $\mathcal{A}$ must also be wellpowered.　◼ Maybe you don't believe in categories that are not locally small and wellpowered. Even so, I am not aware of any embedding theorems that work for arbitrary locally small and wellpowered abelian categories. (I am also not aware of counterexamples. Perhaps there is some large cardinal axiom that implies it can be done.) In practice, it is not necessary to embed the entire category to prove the theorems you want – you can usually find a small abelian subcategory containing all the objects and morphisms you need for your theorem and then you can embed that subcategory. Asking for an embedding of the whole category at once is being greedy – like asking for a global holomorphic chart of a complex manifold when local charts suffice. Here are some embedding theorems I know for non-small abelian categories. Theorem. If $\mathcal{A}$ is a Grothendieck abelian category then there is a conservative exact functor $\mathcal{A}^\textrm{op} \to \textbf{Ab}$ that has a left adjoint. Proof. It is a well-known theorem that Grothendieck abelian categories have injective cogenerators. But an injective cogenerator of $\mathcal{A}$ is precisely an object $I$ such that $\textrm{Hom}_\mathcal{A} (-, I) : \mathcal{A}^\textrm{op} \to \textbf{Ab}$ is a conservative exact functor. Furthermore, representable functors in cocomplete categories automatically have a left adjoint, and $\mathcal{A}^\textrm{op}$ is indeed cocomplete.　◼ Maybe contravariance is jarring. But $\textbf{Ab}$ is itself a Grothendieck abelian category, so the theorem (or Pontryagin duality) gives us a conservative exact functor $\textbf{Ab}^\textrm{op} \to \textbf{Ab}$, and composing them yields a conservative exact functor $\mathcal{A} \to \textbf{Ab}$. The following is a small generalisation. Theorem. If $\mathcal{A}$ is a locally small abelian category and has a small generating set, then there is a conservative exact functor $\mathcal{A}^\textrm{op} \to \textbf{Ab}$. Proof. The hypothesis implies there is a small full abelian subcategory $\mathcal{B}$ containing the given small generating set of $\mathcal{A}$. We get a fully faithful functor $\mathcal{A} \to [\mathcal{B}^\textrm{op}, \textbf{Ab}]$, but in any case it is not automatically exact. Let $\textbf{Lex} (\mathcal{B}, \textbf{Ab})$ be the full subcategory of left exact functors $\mathcal{B}^\textrm{op} \to \textbf{Ab}$. Then, the earlier functor factors as a fully faithful exact functor $\mathcal{A} \to \textbf{Lex} (\mathcal{B}, \textbf{Ab})$. But $\textbf{Lex} (\mathcal{B}, \textbf{Ab})$ is a Grothendieck abelian category, so we may apply the earlier theorem to conclude.　◼<|endoftext|> TITLE: R-module hom a direct summand of Z-module hom? QUESTION [15 upvotes]: $\DeclareMathOperator\Hom{Hom}$Fix a commutative ring $R$. For $R$-modules $M$ and $N$, there is an inclusion of abelian groups $\Hom_R(M,N) \to \Hom_{\mathbb{Z}}(M,N).$ Are there conditions on $R$ that will ensure this is an inclusion of a direct summand (preferably with a section natural in $M$ and $N$)? (I have tried googling this question, but all the words are too common. Pointers to the right terminology would also be appreciated!) REPLY [20 votes]: There exists a natural section if and only if $R$ is separable over $\mathbb Z$ (more generally over $k$, if you have a $k$-algebra $R$ and are asking about the morphism $\hom_R(M,N)\to \hom_k(M,N)$). If you're only asking for a section, no naturality, then I don't know any reasonable condition, but maybe someone can chime in and say something about it. Here's a proof of my claim about separability (I'll do it over a base commutative ring $k$, because $\mathbb Z$ adds nothing specific here): $\hom_k(M,N)\cong \hom_R(R\otimes_k M, N)$ and the forgetful map $\hom_R(M,N)\to \hom_k(M,N)$ corresponds to precomposition by the action map $R\otimes_k M\to M$ under this equivalence. So by the Yoneda lemma, a natural section $\hom_R(R\otimes_k M, {-}) \to \hom_R(M,{-})$ is the same thing as a natural section $M\to R\otimes_k M$ to the action map $R\otimes_k M\to M$ (note that we use naturality in $N$ to get a section $M\to R\otimes_k M$, and naturality in $M$ to get that this section is natural). We now apply this to $M = R$, we get a section $R\to R\otimes_k R$ to the multiplication. Now, this is very weak as is, but this section is natural in endomorphisms of $R$ as a left $R$-module, i.e. in multiplications on the right by elements of $R$. If you unravel what this naturality means, it will imply that $R\to R\otimes_k R$ is a section of $R\otimes_k R^\text{op}$-modules ($R$-bimodules over $k$), which is what is called a "separability idempotent" for $R$. Recall: Definition : A $k$-algebra $R$ is called separable (over $k$) if the multiplication map $R\otimes_k R^\text{op}\to R$ admits a $R\otimes_k R^\text{op}$-module section. The converse is not hard to show: if $R$ is separable, then the action map has a natural section $M = R\otimes_R M \to (R\otimes_k R)\otimes_R M = R\otimes_k M$, and so $\hom_R(M,N)\to \hom_R(R\otimes_k M, N) = \hom_k(M,N)$ has a natural section too. Examples of separable algebras: Finite étale extensions of $k$ are separable. Matrix rings over commutative rings are separable over their base (they all "look like this" étale-locally). A group ring $k[G]$ is separable if and only if $\lvert G\rvert\in k^\times$. Finally, separable algebras are closed under many familiar operations: taking the center, finite products, taking tensor products (over the same base or over a different base).<|endoftext|> TITLE: Unifying two definitions of $L^\infty$ QUESTION [5 upvotes]: Let $X$ be a locally compact Hausdorff space and $\mu$ a Radon measure on $X$. Definitions: A subset $E\subseteq X$ is called locally Borel if $F \cap E$ is Borel for every Borel set $F\subseteq X$ with $\mu(F) < \infty.$ The locally Borel subset $E$ is called locally null if $\mu(E \cap F)= 0$ for every $F\subseteq X$ with $\mu(F) < \infty$. We say that a property of the points of $X$ holds locally almost everywhere (= a.e.) if the set of points where the property fails is locally null. A function $f: X \to \mathbb{C}$ is called locally measurable if $f^{-1}(A)$ is locally Borel for every Borel subset $A \subseteq \mathbb{C}$, i.e. $f$ is measurable w.r.t. the larger $\sigma$-algebra of locally Borel sets on the domain. Construction $1$: [As e.g. defined in Folland's book "A course in abstract harmonic analysis", p50.] Define $\mathscr{L}_1^\infty$ to be the set of locally measurable functions $f: X \to \mathbb{C}$ with the following property: there exists $\alpha \in (0, \infty)$ such that $$\{x \in X: \lvert f(x)\rvert > \alpha\}$$ is a locally null set. The vector space $\mathscr{L}_1^\infty$ has the seminorm $$\lVert f\rVert_\infty:= \inf\{\alpha \ge 0: \{x \in X: \lvert f(x)\rvert> \alpha\} \text{ is locally null}\}.$$ Let $\mathscr{N}_1$ be the subspace of locally measurable functions that are $0$ locally a.e.. We then define the quotient vector space $$L_1^\infty:= \mathscr{L}_1^\infty/\mathscr{N}_1$$ on which the seminorm $\lVert\cdot\rVert_\infty$ becomes a norm. The standard argument proves that this is a Banach space. Construction 2: [As e.g. defined in "Abstract harmonic analysis" by Hewitt and Ross, p151, or as e.g. defined in "Measure theory" by Cohn, section 3.3.] Let $\mathscr{L}_2^\infty$ be the set of bounded measurable functions (with the usual supremum norm) and let $\mathscr{N}_2$ be the set of measurable bounded functions that are $0$ locally almost everywhere. Then we define the quotient vector space $$L_2^\infty:= \mathscr{L}_2^\infty/\mathscr{N}_2$$ (together with the quotient norm). Question: Do these two constructions give the same Banach space, i.e. is there a canonical isometric isomorphism $$\Phi: L_2^\infty \to L_1^\infty?$$ There is an obvious well-defined linear isometry $$\Phi: L_2^\infty \to L_1^\infty: f + \mathscr{N}_2 \mapsto f + \mathscr{N}_1.$$ However, is it surjective? I.e. can we choose a representative of an element of $L_1^\infty$ to be measurable, instead of only locally measurable? There is a possibility that this doesn't need any topological assumptions and that the above is true for general measure spaces and measures. Context question: I'm reading the book "A course in abstract harmonic analysis" by Folland. On p50, it is claimed that the canonical map $L^\infty_1 \to (L^1)^*$ is an isometric isomorphism, but for the surjectivity Folland refers to Hewitt and Ross, where the other definition is used, so I'm trying to unify them. In this context, it would be desirable that the map $\Phi$ above is surjective. REPLY [3 votes]: I think one does need to be careful (c.f. the comments). There are Radon measures for which locally null and null are different. The canonical example is to take $X=\mathbb R^2$ with the topology that $U$ is open if and only if $U_x = \{ y : (x,y)\in U \}$ is open, for each $x$. Then define a Radon measure by the functional $$ C_{00}(X)\rightarrow\mathbb C; \quad f\mapsto \sum_x \int f(x,y) \ dy $$ where we use Lebesgue measure. Then $\{(x,0)\}$ is locally null, but not null. See exercises 3.3.6 and 7.2.4 in Cohn's book. I think the resolution to the problem asked is to realise that locally compact groups are not entirely arbitrary locally compact spaces. The "trick" which Folland uses in his Harmonic Analysis book is to realise that for any $G$ we can always find an open and closed, $\sigma$-compact subgroup $H$. Then $G/H$ has the discrete topology, and being $\sigma$-compact, everything works fine on $H$. Then you can reconstruct the results you want by working piecewise on each coset of $H$. (See page 51 of Folland). This resolution works quite adequately with "construction 2". Indeed, see Cohn, Theorem 9.4.8 which shows exactly that $L^\infty$ is the dual of $L^1$ for any regular Borel measure on $G$. However, beware of the subtle point in the proof, that you cannot exactly just work on each coset of $H$: to get a Borel function, some trick using Lusin's theorem (to approximate by continuous functions) is needed. (An alternative approach to "fix" the duality issue is to work with a larger $\sigma$-algebra than the Borel sets; compare Cohn Section 7.5. If I understand things right, this is the same notion of "measurability" which Hewitt+Ross uses. Here, Exercise 7.5.5 in Cohn is interesting: $L^p$ for this $\sigma$-algebra, or $L^p$ for Borel sets, are isometrically isomorphic, so long as $p$ is finite.) In conclusion, the two constructions asked about are in general different. But in this special case (locally compact groups) they agree. A meta-question is: which construction to use? I guess I don't know (and it doesn't matter, if you believe this answer!) From my experience, the literature in abstract harmonic analysis almost always uses construction 1, and the "locally" language. Furthermore, in most cases, it doesn't really matter exactly which $\sigma$-algebra one works with. More pragmatically, the vast majority of examples are $\sigma$-compact, and so there is relatively little to be lost by just assuming that $G$ is $\sigma$-compact. Then it's clear that both constructions agree. (Related to the often seen "In this paper we assume that all Hilbert spaces are separable" caveats).<|endoftext|> TITLE: Are there finitely many sieves on each object in the distributive lattice cube category? QUESTION [10 upvotes]: Define the distributive lattice cube category or Dedekind cube category $\square_{\land\lor}$ to be the full subcategory of the category of posets and monotone maps consisting of objects of the form $[1]^n \triangleq \{0 < 1\}^n$ for $n \ge 0$. Morphisms $[1]^m \to [1]^n$ in this category correspond to $n$-tuples of terms in $m$ variables in the theory of bounded distributive lattices, hence the name. (In terms of generators, every morphism can be written as a composite of face maps, degeneracies, permutations, diagonals, and min- and max-connections─this fact appears on the logical side as the disjunctive/conjunctive normal form presentations.) The presheaf category $\mathrm{PSh}(\square_{\land\lor})$ is then one of many variations on "cubical sets". I'd like to know whether the following finiteness property is satisfied by the representable cubes in this category: Question: In the presheaf category $\mathrm{PSh}(\square_{\land\lor})$, does every representable object have finitely many subobjects up to isomorphism? Equivalently, are there finitely many sieves on every object of $\square_{\land\lor}$? In general, say that a category $\mathcal{C}$ has the finite sieve property if there are finitely many sieves on each object of $\mathcal{C}$. Recall that a sieve on an object $A$ in category $\mathcal{C}$ is a set $\mathcal{S}$ of arrows into $A$ such that if $g : B \to A$ is in $\mathcal{S}$ and $f : C \to B$ is an arbitrary arrow, then $g \circ f$ is in $\mathcal{S}$. To show that the set of sieves on some $A$ is finite, it suffices to show there are finitely many principal sieves, that is, sieves of the form $\langle g \rangle \triangleq \{g \circ f \mid f : C \to B\}$ for some $g : B \to A$. Using this and some specifics of $\square_{\land\lor}$, we can thus give the following more concrete reformulations: Equivalent Question: Fix $k \ge 0$. Does there exist $\ell \ge 0$ such that every $f : [1]^n \to [1]^k$ generates the same principal sieve as some $f' : [1]^\ell \to [1]^k$? Equivalent Question 2: Fix $k \ge 0$. Does there exist $\ell \ge 0$ such that for every $f : [1]^n \to [1]^k$, there are maps $[1]^n \overset{g}\to [1]^\ell \overset{h}\to [1]^n$ with $f \circ h \circ g = f$? (If this is so, then $\langle f \rangle = \langle f \circ h \rangle$.) I'm interested in this cube category because of its relevance to modelling cubical type theories. In work with Christian Sattler we're in the process of writing up, we've found that finite product categories with the finite sieve property can be embedded in generalized Reedy categories in a useful way. We used this to characterize a certain model structure on $\square_\lor$ (the subcategory with just one connection, see remarks below), but don't know if any of the arguments apply with $\square_{\land\lor}$. Some partial and related answers The remainder of this post just collects special cases and related results that might or might not be helpful. Remark: Any Eilenberg-Zilber category $\mathcal{C}$ in which there are finitely many face maps into any object has the finite sieve property. Proof: Any $f : B \to A$ generates the same sieve as the degree-raising map of its Reedy factorization, as the degree-lowering map is a split epimorphism. $\blacksquare$ In particular, the simplex category and cartesian cube category (the wide subcategory of $\square_{\land\lor}$ generated by face maps, degeneracies, permutations, and diagonals) have the finite sieve property. If we add one connection to the latter, we get a category that is not generalized Reedy but still has the finite sieve property. Proposition: The wide subcategory $\square_\lor$ of $\square_{\land\lor}$ consisting of maps generated by face maps, degeneracies, permutations, diagonals, and the max-connection $\lor$ has the finite sieve property. (Note: this is equivalently the wide subcategory of join-preserving maps.) Proof: Let $f : [1]^n \to [1]^k$. Such a morphism corresponds to a tuple of $k$ terms in $n$ variables in the theory of bounded join-semilattices. We can assume without loss of generality that no component $f_j : [1]^n \to [1]$ for $1 \le j \le k$ is constant $1$. Then for each $f_j$, we have a normal form $f_j(x_1,\ldots,x_n) = \bigvee S_j$ for some $S_j \subseteq \{x_1,\ldots,x_n\}$. For $1 \le i \le n$, define $T_i = \{j \mid x_i \in S_j\} \subseteq \{1,\ldots,k\}$. If $n > 2^k$, then we must have $T_i = T_{i'}$ for some $i,i'$. Without loss of generality, say $T_1 = T_2$. Then $x_1$ and $x_2$ appear as disjuncts in the same components, so we have $f \circ d \circ c = f$ where $c : [1]^n \to [1]^{n-1}$ maps $(x_1,\ldots,x_n) \mapsto (x_1 \lor x_2, x_3, \ldots, x_n)$ and $d : [1]^{n-1} \to [1]^n$ maps $(y_1,\ldots,y_{n-1}) \mapsto (y_1,y_1,y_2, \ldots, y_{n-1})$. Thus $\langle f \rangle = \langle f \circ d \rangle$. By repeatedly reducing in this way, we find that $\langle f \rangle = \langle f' \rangle$ for some $f' : [1]^{2^k} \to [1]^k$. $\blacksquare$ Lemma: The finite sieve property holds for the objects $[1]^k$ with $k \le 2$ in $\square_{\land\lor}$. Proof: The $k = 0$ case is obvious. For $k = 1$, any $f : [1]^n \to [1]$ is either constant $0$, constant $1$, or (we can see from the disjunctive normal form) satisfies $f \circ \Delta \circ f = f$ where $\Delta : [1] \to [1]^n$ is the diagonal $x \mapsto (x,\ldots,x)$. Now $k = 2$, a bit sketchily. Suppose we have $f : [1]^n \to [1]^2$ and neither component is constant $0$ or $1$ (in which case we could reduce to the $k = 1$ case). We can write each component in disjunctive normal form, $f_j(x_1,\ldots,x_n) = \bigvee_{i \in I_j} \bigwedge C^j_i$ where $I_j$ is nonempty and each $C^j_i$ is a nonempty set of variables. If there exists some $C^1_i$ which is not a superset of any $C^2_{i'}$, choose one and call this $K_1$; otherwise choose an arbitrary $C^1_i$ to be $K_1$. (Note that in the latter case $f_1 \le f_2$.) Likewise, choose $K_2$ among the $C^2_i$ which is not a superset of any $C^1_{i'}$ if possible. Define $g : [1]^2 \to [1]^n$ by $$ g(u_1,u_2)_i \triangleq \bigvee \{u_j \mid x_i \in K_j \} $$ Then $f \circ g \circ f = f$. $\blacksquare$ In the $k \le 2$ case, we always get $f \circ g \circ f = f$ for some $g$. But this does not happen in general, as demonstrated by the following lower bound: Proposition: For any $k \ge 1$, there exists some $f : [1]^n \to [1]^k$ such that $\langle f \rangle \neq \langle g \rangle$ for every $g : [1]^\ell \to [1]^k$ with ${\ell \choose {\lfloor \ell/2 \rfloor}} < 2^{k-1}$. Proof: For any $v \in [1]^k$, write $\#_v \in [1]^{\lvert [1]^k \rvert}$ for the vector taking value $1$ in all coordinates except the $v$th. We define a monotone map $f : [1]^{\lvert [1]^k \rvert} \to [1]^k$. $$ \begin{align*} f(\top) &\triangleq \top \\ f(\#_v) &\triangleq v \\ f(\_) &\triangleq \bot &\text{otherwise} \end{align*} $$ Suppose we have $\langle f \rangle = \langle g \rangle$ for some $g : [1]^\ell \to [1]^k$, so there exist $h : [1]^n \to [1]^\ell$ and $t : [1]^\ell \to [1]^n$ with $f \circ t \circ h = f$ (and $g = f \circ t$). Then because $f^{-1}(v) = \{\#_v\}$ for $v \in [1]^k\setminus\{\bot,\top\}$, we must have $(t \circ h)(\#_v) = \#_v$ for these $v$. As $\#_v$ and $\#_w$ are incomparable for $v \neq w$, so must be $h(\#_v)$ and $h(\#_w)$. Thus $[1]^\ell$ must contain a set of at least $\lvert [1]^k\setminus\{\bot,\top\} \rvert = 2^{k-1}$ pairwise-incomparable vectors. The maximal antichain in $[1]^\ell$ has ${\ell \choose {\lfloor \ell/2 \rfloor}}$ elements, so ${\ell \choose {\lfloor \ell/2 \rfloor}} \ge 2^{k-1}$. $\blacksquare$ In particular, there is a principal sieve on $[1]^3$ not generated by any $f : [1]^3 \to [1]^3$. REPLY [4 votes]: $\newcommand{\Pos}{\mathbf{Pos}}$ $\square_{\land\lor}$ does not have the finite sieve property. The following paper provided the key lemma: Imrich, Wilfried; Kalinowski, Rafał; Lehner, Florian; Pilśniak, Monika, Endomorphism breaking in graphs, Electron. J. Comb. 21, No. 1, Research Paper P1.16, 13 p. (2014). ZBL1300.05100. Write $C_n$ for the cycle graph on vertices $\{0,\ldots,n-1\}$. We can regard $C_{2n}$ as a bipartite graph, with the even numbers in one part and the odd numbers in the other, and thereby as a poset with $i < j$ iff $i$ is even, $j$ is odd, and $i-j = \pm 1 \bmod 2n$. From the paper above, I just want the following (simple) fact, which is in their proof of Lemma 2. Proposition 1. For any $n \ge 3$, there exists a 2-coloring of $C_{2n}$ such that the only color-preserving graph endomorphism of $C_{2n}$ is the identity function. Corollary 2. For any $n \ge 3$, there exists a poset map $f_n \colon C_{2n} \to C_4$ such that the only endomorphism of $f_n \in \Pos/C_4$ is the identity function. Proof. Choose a 2-coloring $c \colon C_{2n} \to \{0,1\}$ as in the proposition. Define $f \colon C_{2n} \to C_4$ by $f(i) \triangleq 2c(i) + i \mathbin{\%} 2$. We have an embedding of posets $\iota \colon C_4 \hookrightarrow [1]^4$ defined as follows. \begin{align*} 0 &\mapsto 0001 & 1 &\mapsto 0111 \\ 2 &\mapsto 0010 & 3 &\mapsto 1011 \end{align*} Given $i \in \mathbb{N}$, write $\chi_i \in [1]^n$ for the element with $1$ at index $i \bmod n$ and $0$ elsewhere. For any $n \ge 3$, we have a poset embedding $z_n \colon C_{2n} \hookrightarrow [1]^n$ defined as follows. \begin{align*} z(i) &\triangleq \left\{ \begin{array}{ll} \chi_{i/2} &\text{if $i$ is even} \\ \chi_{\lfloor i/2 \rfloor} \vee \chi_{\lceil i/2 \rceil} &\text{if $i$ is odd} \end{array} \right. \end{align*} For each $n \ge 3$, we define a monotone map $g_n \colon [1]^n \to [1]^4$ by cases as follows. \begin{align*} g_n(v) &\triangleq \left\{ \begin{array}{ll} \iota(f_n(i)), &\text{if $\exists i < 2n.\ v = z_n(i)$} \\ \bot, &\text{if $v = \bot$} \\ \top, &\text{otherwise} \end{array} \right. \end{align*} By construction, we have $g_n \circ z_n = \iota \circ f_n$ as well as $g_n^{-1}(\iota(C_4)) = z_n(C_{2n})$. Now, suppose we are given an endomorphism $h : g_n \to g_n$ in $\Pos/[1]^4$. Because $h(z_n(C_{2n})) = h(g_n^{-1}(\iota(C_4))) \subseteq g_n^{-1}(\iota(C_4)) = z_n(C_{2n})$, there is a restriction $h \upharpoonright z_n : C_{2n} \to C_{2n}$ fitting into the following commuting diagram in $\Pos$. $\require{AMScd}$ \begin{CD} C_{2n} @>{h \upharpoonright z_n}>> C_{2n} \\ @V{z_n}VV @V{z_n}VV \\ {[1]^n} @>h>> {[1]^n} \\ @V{g_n}VV @V{g_n}VV \\ {[1]^4} @= {[1]^4} \end{CD} It follows that we have $h \upharpoonright z_n \colon f_n \to f_n$ in $\Pos/C_4$, thus that $h \upharpoonright z_n = \mathrm{id}$. Thus the image of any such $h \colon g_n \to g_n$ has cardinality at least $2n$; if $h$ factors through some $[1]^\ell$, we must therefore have $\ell \ge \log_2 n + 1$.<|endoftext|> TITLE: Seeking concrete examples of "generic" elliptic fibrations of K3 surfaces QUESTION [8 upvotes]: For me a K3 surface will be a smooth complex projective variety of dimension 2 that is simply-connected and has trivial canonical bundle. Given a K3 surface $X$, an elliptic fibration $\pi \colon X \to \mathbb{C}P^1$ is a proper morphism with connected fibers such that all but finitely many fibers are smooth curves of genus 1. I've learned a little about these from here: Daniel Huybrechts, Lectures on K3 surfaces, Chapter 11: Elliptic K3 surfaces. Generically a K3 surface admits no elliptic fibration, but among those that do, generically the fiber of $\pi$ is a smooth curve of genus 1 at all but 24 points, where the fiber is a rational curve with a single double point. Huybrecht also catalogues the less generic cases where $\pi$ has fewer singular (i.e. non-smooth) fibers, but with correspondingly worse singularities. On MathOverflow there's a nice easy example: the Fermat quartic surface admits an elliptic fibration with 6 singular fibers, each of which has 4 double points. But I'd like to see concrete easy examples of elliptic fibrations with 24 singular fibers. By 'concrete easy examples' I mean that ideally I would like there to be simple explicit formulas for the K3 surface $X$, the hyper-Kähler structure on $X$, the elliptic fibration $\pi$, the 24 points on $\mathbb{C}\mathrm{P}^1$ with singular fibers, the double points on these fibers, and also the points of $X$ where $d\pi$ is not injective. But of course I'll settle for whatever I can get! REPLY [2 votes]: One can get examples by taking a double cover of the Segre-Hirzebruch surface $\mathbb F_4$. Write $C^-$ for the negative section and $C^+$ for a zero section and $F$ for a fiber of the projection $\mathbb F_4\to \mathbb P^1$, so that $C^+$ is linearly equivalent to $C^-+4F$. Set $L=2C^-+6F$ and choose $D$ in the linear system $|3C^+|$. Then there is a double cover $f\colon X \to \mathbb F_4$ branched on $C^-+D$ and such that $f_*\mathcal O_X=\mathcal O_{\mathbb F_4}\oplus L^{-1}$; if $D$ is smooth, the surface $X$ is also smooth. Standard formulae for double covers give $h^1(\mathcal O_Y)=0$ and $K_Y=0$, so $Y$ is a K3. The map $X\to \mathbb F_4\to \mathbb P^1$ is an elliptic fibration. The singular points of the elliptic fibers lie above the points where $D$ is tangent to the ruling of $\mathbb F_4$. If the tangency is simple the singularity of the elliptic fiber is a node. A dimension count shows that for a general choice of $D$ all the tangency points of $D$ to a ruling are simple. I think that using computer algebra it should not be hard to find explicit examples of $D$ where this condition holds.<|endoftext|> TITLE: Proven chaos in logistic maps QUESTION [7 upvotes]: For a logistic map using $f_r(x)=rx(1-x)$, what values of $r$ and starting $x$ are guaranteed (i.e., with an accepted proof) to be chaotic? I mean "chaotic" in the loosest sense: The limiting sequence of $x$ does not tend to a finite number of points. I am currently using $r=3.8$ and starting $x=0.501234567890123456789$, but have only tested through 10,000 iterations. What is the probability that I am chaotic? EDIT: Below are new results (with 4,000,000 bits of precision to avoid any rounding problems) for 2,000,000 iterations (showing matches to "0.72224", the end's most significant digits). So, I believe it is fair to say that there are 3 possible cases: There is no limit cycle (through infinity), There is a limit cycle of at least 1,105,578 points, or There is a smaller limit cycle but any "two points chosen from the first 2,000,000 points" are not both within one limit point's attractor zone. #2 seems the most unlikely. #3 seems unlikely simply because I chose such round numbers from the start. According to answers here, however, it does seem like the probability for #1 is not 100%. Maybe someone can put my statements here into proper mathematical language and clarify this better. n: x_n ----------------------- 53951: 0.7222489331 66539: 0.7222408270 68976: 0.7222441979 75138: 0.7222495664 120428: 0.7222473699 134963: 0.7222441673 235912: 0.7222411119 395643: 0.7222459509 closest greater value 417062: 0.7222404139 462528: 0.7222468852 472142: 0.7222408308 645137: 0.7222474275 679584: 0.7222492244 731458: 0.7222410420 761284: 0.7222468048 891274: 0.7222442328 894423: 0.7222448046 closest lower value 935412: 0.7222498698 1110025: 0.7222446506 1220483: 0.7222447341 1222255: 0.7222485044 1269796: 0.7222407187 1301786: 0.7222439936 1422147: 0.7222488714 1431959: 0.7222457998 1503338: 0.7222445272 1509878: 0.7222404127 1568206: 0.7222447453 1569439: 0.7222415020 1612039: 0.7222497768 1634269: 0.7222406207 1642044: 0.7222450907 1791569: 0.7222487370 1865739: 0.7222420900 1879844: 0.7222427753 1902889: 0.7222493257 2000000: 0.7222453893 end REPLY [7 votes]: There is a nice article by Mikhail Lyubich in the October 2000 edition of Notices of the AMS, "The Quadratic Family as a Qualitatively Solvable Model of Chaos" that provides a nice summary of rigorous results regarding logistic maps. The main results relevant to your question are the following (some of these are in Lyubich's paper, others are not). Let $S$ denote the set of parameter values $r$ for which the map is "stochastic" in the sense that there is an invariant probability measure that is an absolutely continuous measure with respect to Lebesgue. Then the Lebesgue measure of $S$ is positive (so a positive proportion of parameter values lead to stochastic behavior); this was first proved by Jakobson (Comm. Math. Phys. 1981), see also work by Collet and Eckmann (Comm. Math. Phys. 1980, ETDS 1983) and by Benedicks and Carleson (Annals 1985). For $r\in S$, Lebesgue-a.e. $x$ has an orbit that is distributed according to this measure, and in particular its $\omega$-limit set is a union of intervals. Let $R$ denote the set of parameter values $r$ for which the map is "regular" in the sense that there is an attracting periodic orbit. Then $R$ is open and dense in parameter space, and $R\cup S$ has full Lebesgue measure: Lebesgue-almost every $r$ gives a map that is either regular or stochastic in the above senses. Regarding specific values of $r$, say that the parameter $r$ is a Misiurewicz point if the critical point for the map is pre-periodic but not periodic. Such parameter values are stochastic in the sense above (Misiurewicz 1981, Pub Math IHES). Note that in particular $r=4$ is a Misiurewicz point. If I recall correctly, the proof that $S$ has positive Lebesgue measure involves a process called "parameter exclusion" near Misiurewicz points and actually shows that each Misiurewicz point is a Lebesgue density point for $S$. It is an interesting question to get rigorous bounds on the Lebesgue measure of $S$. I don't know the full story here, but there is a paper by Tucker and Wilczak (Physica D 2009) and a more recent one by Golmakani, Koudjinan, Luzzatto, Pilarczyk (Chaos 2020) with some results.<|endoftext|> TITLE: Do combinatory logic bases need a function of 3 variables? QUESTION [17 upvotes]: All the known bases of combinatory logic, such as $\{S,K\}$, or $\{K,W,B,C\}$, have one or more combinators using 3 variables: \begin{align*} S ={} & \lambda x\lambda y\lambda z. x z(y z), \\ B ={} & \lambda x\lambda y\lambda z. x (y z), \\ C ={} & \lambda x\lambda y\lambda z. x z y. \end{align*} This raises the question whether we can make a basis with functions of only 2 variables. Surely if such a thing were possible, it would be a most interesting result and it would be well-known. But as far as I can tell, no such thing is known. Thus it seems nobody believes such a thing to be possible. But has this been proven anywhere? How do we know that $\{K,W,2,O,T,D\}$ is not a basis, where \begin{align*} K ={} & \lambda x\lambda y. x, \\ W ={} & \lambda x\lambda y. x y y, \\ 2 ={} & \lambda f\lambda x. f(f x), \\ O ={} & \lambda x\lambda y. y(x y), \\ T ={} & \lambda x\lambda y. y x, \\ D ={} & \lambda x. x x? \end{align*} REPLY [18 votes]: @PeterTaylor’s excellent answer points to exactly the result wanted. But the proof there can be streamlined a bit, and may be paywalled for some readers, so I’ll write it out here for accessibility. Theorem. Let $\newcommand{\B}{\mathbf{B}}\B$ be any set of combinators of rank ≤2. Then there is no $\B$-expression $T$ such that $\newcommand{\reddto}{\twoheadrightarrow}Tabc \reddto cab$. Proof. Fix such $\B$ throughout. The idea is (unsurprisingly) to find an invariant of $Tabc$ that is sufficiently closed under reduction, and not satisfied by $cab$. Say that an expression is bad if it is of the form $E[F[a,b],c]$, where $E$ and $F$ are each $\B$-expressions of 2 variables. Then we have: Lemma. Any bad expression is either normal, or else has a reduction chain to another bad expression, including at least one leftmost reduction. Proof of lemma. If a bad expression $E[F[a,b],c]$ is not stuck, consider its leftmost redex. There are three possible cases: The head combinator of the redex is not in an occurrence of $F$. Then each occurrence of $F$ is either outside the redex, or inside an argument of the head combinator; so the reduct is of the form $E'[F[a,b],c]$. The whole redex is inside an occurrence of $F$. Then by reducing it in every occurrence of $F$, we have $E[F[a,b],c] \reddto E[F'[a,b],c]$, and this includes the leftmost redex. The head combinator $H$ is iside an occurrence of $F$, but not the whole redex. Then $H$ must be of rank ≥1. If $H$ is of rank 1, then $F$ is just $H$, so $x,y$ don’t occur, and the whole expression (both before and after reduction) is of the form $E'[c]$, so is still bad. If the head combinator is of rank 2, then we must have $F[a,b] \equiv HF'[a,b]$, so by re-parsing as $E[F[a,b],c] \equiv E'[F'[a,b],c]$, case 1 applies. This proves the lemma. It follows that any bad expression either reduces to a bad normal form, or has an infinite quasi-leftmost reduction sequence. So by the quasi-leftmost-reduction theorem (Hindley–Seldin 2008, Thm 3.22), any normal form of a bad expression is bad; and so the bad expression $Tabc$ cannot reduce to the normal, non-bad expression $cab$. □ References: Rémi Legrand, A Basis Result in Combinatory Logic, J. Symb. Logic 1988; jstor full text J. Roger Hindley, Jonathan P. Seldin, Lambda-Calculus and Combinators, an Introduction, CUP 2008<|endoftext|> TITLE: Oscillatory integrals with a decaying factor in the integrand QUESTION [6 upvotes]: Lemma 4.5 of Titchmarsh's book The Theory of the Riemann Zeta function says (slightly rephrased): Let $F$ be a twice differentiable real function such that $ F''(x) \geq r > 0$ for all $x$ in $[a,b]$ or $ F''(x) \leq -r < 0$ for all $x$ in $[a,b]$. Let $G$ be a real function such that $G(x)/F'(x)$ is monotonic and $|G(x)| \leq M$ for all $x$ in $[a,b]$. Then we have $$ \left| \int_a^b G(x) e^{iF(x)} dx \right| \leq \frac{8M}{\sqrt{r}} .$$ I am considering the case $$ I_\alpha(T) = \left| \int_1^T x^\alpha e^{iF(x)} dx \right| $$ where $ F''(x) \geq 1/T$ and $x^\alpha/F'(x)$ is monotonic for all $x$ in $[1, T]$ and for all $\alpha > -1$. Titchmarsh's Lemma (for $\alpha > -1/2$) and the trivial bound (for $-1/2 \geq \alpha > -1$) give $$ I_\alpha(T) \leq \begin{cases} 8T^{\alpha + 1/2} \ &\text{if} \ \alpha \geq 0 \\ 8T^{1/2} \ &\text{if} \ 0 >\alpha > -\frac{1}{2} \\ \frac{T^{\alpha + 1}}{\alpha+1} &\text{if} -\frac{1}{2} \geq \alpha > -1 \end{cases} $$ I am wondering if one can do better (asymptotically in $T$) in the cases when $0 > \alpha > -1$ by combining the oscillatory cancellation from $e^{iF(x)}$ with the decay from $x^{\alpha}$. REPLY [3 votes]: $\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}$If a function $h\colon\R\to\R$ is monotonic and does not change sign on an interval $[a,b]$, then it is easy to see (make a picture) that for all real $s$ \begin{equation*} \Big|\int_{[a,b]}h(u)\sin(u+s)\,du\Big|\le\sup_{c\in\R}\int_{[a,b]\cap[c,c+\pi]}|h(u)|\,du, \end{equation*} whence \begin{equation*} \Big|\int_{[a,b]}h(u)e^{iu}\,du\Big|\le\sqrt2\,\sup_{c\in\R}\int_{[a,b]\cap[c,c+\pi]}|h(u)|\,du. \end{equation*} Since $x^\al/F'(x)$ is monotonic in $x\in[1,T]$, $F'$ cannot change its (nonzero) sign on $[1,T]$. Making now the substitution $u=F(x)$, letting here \begin{equation*} h(u):=\frac{g(u)^\al}{F'(g(u))} \text{ with } g:=F^{-1}, \end{equation*} and then making the inverse substitution $x=g(u)$, we get \begin{equation*} \begin{aligned} I_\al(T)= &\Big|\int_{F([1,T])}h(u)e^{iu}\,du\Big| \\ &\le\sqrt2\,\sup_{c\in\R}\int_{F([1,T])\cap[c,c+\pi]}|h(u)|\,du \\ &=\sqrt2\,\sup_{c\in\R}\int_{[1,T]\cap g([c,c+\pi])}x^\al\,dx. \end{aligned} \tag{1} \end{equation*} Recall that $F'$ cannot change its (nonzero) sign on $[1,T]$. If $F'>0$ on $[1,T]$, then for any $x,y$ in $[1,T]$ such that $x\le y$ \begin{equation*} |F(y)-F(x)|=F(y)-F(x)\ge F'(x)(y-x)+\frac{(y-x)^2}{2T}\ge\frac{(y-x)^2}{2T}, \end{equation*} since $F'>0$ and $F''\ge1/T$. Similarly, if $F'<0$ on $[1,T]$, then for any $x,y$ in $[1,T]$ such that $x\le y$ \begin{equation*} |F(y)-F(x)|=F(x)-F(y)\ge F'(y)(x-y)+\frac{(x-y)^2}{2T}\ge\frac{(y-x)^2}{2T}, \end{equation*} since $F'<0$ and $F''\ge1/T$. So, in either case, $|y-x|\le\sqrt{2T|F(y)-F(x)|}$ for any $x,y$ in $[1,T]$. So, denoting by $l$ the length of the interval $[1,T]\cap g([c,c+\pi])$, we have $l\le\sqrt{2\pi T}$. Hence, by (1), \begin{equation*} I_\al(T) \le\sqrt2\,\int_0^{\sqrt{2\pi T}}x^\al\,dx \le\frac C{1+\al}\, T^{(1+\al)/2}, \end{equation*} whence $C$ is a universal positive real constant. The latter bound is indeed an improvement of the corresponding bound in your post. Morever, the latter bound is optimal, as it is asymptotically attained (as $T\to\infty$, up to a positive real constant factor depending only on $\al$) if $F(x)=\dfrac{x^2}{2T}$:<|endoftext|> TITLE: What are abelian categories enriched over themselves? QUESTION [8 upvotes]: As far as I understand, an arbitrary abelian category is not enriched over itself, for example, $\mathrm{ChainComplex}(\mathrm{Ab})$ is, right? On the other hand, the categories $\mathrm{Mod}(R)$ (in particular, $\mathrm{Ab}$) are enriched over themselves. What other examples, necessary and sufficient conditions are there for abelian categories enriched over themselves? Maybe they were researched somewhere? The question arose because the dualization of a chain complex (leading to cohomology with values ​​in the same category) is definable in $\mathrm{Mod}(R)$, but apparently undefinable in an arbitrary abelian category. Update: (thanks to a discussion in the comments) I realized that I start initially from an abelian category $\mathcal{A}$ with a univalent functor to the category $\mathrm{Ab}$ and look for an enrichment of $\mathcal{A}$ over itself consistent with the standard enrichment of $\mathcal{A}$ over $\mathrm{Ab}$ in the sense of this functor (as far as I understand, this guarantees , that, in accordance with my motivation, the cohomology will not change, but will acquire an additional structure of objects $\mathcal{A}$). For abelian categories of the form "a subcategory of $\mathrm{Func}(I, \mathcal{A})$, where $\mathcal{A}$ is an abelian category with a standard forgetful functor in $\mathrm{Ab}$ and $I$ is small" we define a forgetful functor in $\mathrm{Ab}$ as $F \mapsto \bigoplus\limits_{i \in I} F(i)$ (and, accordingly, on morphisms). Is the category of chain complexes of $R$-modules consistently enriched over itself? And the category of sheafs? REPLY [12 votes]: To make sense of enrichment over a category $V$, you want $V$ to have a monoidal structure. Indeed, you want to be able to compose morphisms so you need a way to go from "something in $\hom(a,b)$ and something in $\hom(b,c)$ to something in $\hom(a,c)$", and this is neatly encoded by a monoidal structure where an enrichment is, among other things, a morphism $\hom(a,b)\otimes \hom(b,c)\to \hom(a,c)$ (if you're only monoidal and not symmetric monoidal you have to be careful about what kind of hom's you take, and also in what order you take this tensor, but let me not worry about it here, and pretend I'm in a symmetric monoidal category). Now, a monoidal category $V$ has a "canonical candidate" for self-enrichment : one would want the internal hom's (which I'll denote $\mathcal{H}om$) to have the following universal property (which reflects the usual cases): $\hom(v, \mathcal Hom(v_0,v_1))\cong \hom(v\otimes v_0, v_1)$. In other words, you want $-\otimes v_0$ to have a right adjoint $\mathcal Hom(v_0,-)$, for all $v_0$. It turns out that this is enough: Suppose $V$ is a symmetric monoidal category in which $-\otimes v_0$ has a right adjoint for each $v_0\in V$ [NB : $V$ is called 'closed (symmetric) monoidal' in this case]. Then there is a canonical enrichment of $V$ over itself which satisfies the above property. (again, you can say similar things without "symmetric", but it gets a bit more subtle so I won't state them here) Note that this is the "usual story" for enrichment and self-enrichment. I don't know of any other way to tell those stories, and they are the ones that tend to come up. In your examples, $Mod(R)$ has a tensor product $\otimes_R$ when $R$ is commutative, and so it is self-enriched - note that this doesn't work if $R$ is not commutative ! If you want to make it work, you have to move to bimodules or something, and then you have choices to make to decide what the enrichment is. Similarly, chain complexes over a (cocomplete) symmetric monoidal abelian category have a usual monoidal structure, and if they're sufficiently complete, the right adjoints from above exist. Here is one condition that is often met in practice and guarantees the existence of the right adjoints : If $V$ is symmetric monoidal, presentable and the tensor product $-\otimes v$ preserves colimits as a functor $V\to V$, then it has a right adjoint. Here the power is hidden in "presentable", but this is actually often met in practice, despite being a very strong assumption. (note that this is unrelated to abelian-ness of your category : for a general category $C$ it doesn't make sense to speak about enrichment in $C$, the structure that makes this avaiable is a monoidal structure, and if it is closed, then in fact $C$ can itself be enriched over itself. In the abelian context, you'll often want the tensor product to be at the very least nicely behaved, e.g. right exact in each variable)<|endoftext|> TITLE: Compact flat orientable 3 manifolds and mapping tori QUESTION [9 upvotes]: There are 10 compact flat 3 manifolds up to diffeomorphism, 6 orientable and 4 non orientable. I am looking to better understand how to construct the orientable ones. The six orientable ones are determined by their holonomy groups $$ C_1,C_2,C_3,C_4,C_6 $$ and $$ C_2 \times C_2 $$ The five with cyclic holonomy all arise as the mapping torus of a mapping class of $ T^2 $ with the corresponding order: 1,2,3,4, or 6. These five Euclidean manifolds with cyclic holonomy can even be constructed as a quotient of the special Euclidean group $ SE_2 $ by a cocompact lattice constructed as the semidirect product of a lattice in $ \mathbb{R}^2 $ and a finite cyclic subgroup of $ SL_2(\mathbb{Z}) $ preserving that lattice. For example $ C_1 $ corresponds to the three torus $ T^3 $. I am very curious about the compact flat orientable 3 manifold with holonomy $ C_2 \times C_2 $ (known as the Hantzsche-Wendt manifold). It is not a mapping torus of $ T^2 $ like the other five, but perhaps it is a mapping torus of the Klein bottle $ K $? REPLY [4 votes]: The remaining orientable manifold is called the Hantzsche-Wendt manifold $M^{HW}$, and is not a mapping torus over the Klein bottle. It has first homology $H_1(M^{HW}, \mathbb{Z}) = \mathbb{Z}_4 \times \mathbb{Z}_4$. Any mapping torus $MT(M, f)$ of a manifold $M$ via the map $f: M \to M$ has $\pi_1(MT(M, f)) \cong \pi_1(M_0) \rtimes_{f_*} \mathbb{Z}$. Abelianizing and using Hurewicz's theorem yields that if $M$ is a mapping torus of the Klein bottle, $H_1(M, \mathbb{Z})$ must be of the form $\mathbb{Z} \times H$, where $H$ is a quotient of $H_1(K, \mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}_2$, and hence cannot be $\mathbb{Z}_4 \times \mathbb{Z}_4$. It is worth mentioning that the $3$-torus is a normal covering space of all flat compact $3$-manifolds. The Hantzsche-Wendt Manifold has a nice description as a branched cover of the complement of the Borromean rings, which is described in Zimmerman's paper On the Hantzsche-Wendt Manifold. The standard reference for a classification of compact flat 3-manifolds is J. Wolf's book Spaces of Constant Curvature, which gives a classification of them and explicit constructions as quotients of $\mathbb{R}^3$ by isometries. A resource to develop visual intuition about these manifolds is Jeffrey Weeks' program Curved Spaces, which simulates how it would look to "fly around inside of" manifolds and has a number of flat 3 manifolds as pre-built examples, including the Hantzsche-Wendt manifold.<|endoftext|> TITLE: Only odd primes? QUESTION [18 upvotes]: For $k \ge 2$, let $$u = \{\lfloor{(k - \sqrt{k})n}\rfloor : n \ge 1\}$$ $$v = \{\lfloor{(k + \sqrt{k})n}\rfloor : n \ge 1\}.$$ My computer suggests that $u$ and $v$ are disjoint if and only if $k$ is an odd prime. Can someone give a reference, proof, or counterexample? REPLY [32 votes]: If $k$ is odd and not a perfect square, then the sets are disjoint. In particular, if $\alpha = \frac{k - \sqrt{k}}{\frac{k-1}{2}}$ and $\beta = \frac{k + \sqrt{k}}{\frac{k-1}{2}}$, then $\alpha$ and $\beta$ are irrational and $\frac{1}{\alpha} + \frac{1}{\beta} = 1$. Therefore, by Beatty's theorem, $A = \{ \lfloor \alpha n \rfloor : n \geq 1 \}$ and $B = \{ \lfloor \beta n \rfloor : n \geq 1 \}$ form a partition of the positive integers. The sets $u$ and $v$ are subsets of $A$ and $B$ respectively (in particular those with $\frac{k-1}{2} \mid n$). In particular, the sets $u$ and $v$ are disjoint for $k = 15$, which is the smallest non-prime odd number that is not a perfect square. REPLY [20 votes]: This is a supplement to Jeremy Rouse's nice answer. Let $\alpha$ and $\beta$ be positive irrational numbers. Skolem proved in 1957 (see Theorem 8 in On certain distributions of integers in pairs with given differences) that the Beatty sequences $[\alpha n]$ and $[\beta n]$ are disjoint if and only if $a/\alpha+b/\beta=1$ holds for some positive integers $a$ and $b$. It follows that the OP's sets $u$ and $v$ are disjoint if and only if $k$ is odd and not a perfect square.<|endoftext|> TITLE: Realizing spherical complexes as convex polytope QUESTION [5 upvotes]: A spherical polytope is the intersection of some closed hemispheres which is non-empty and does not contain a pair of antipodal points. A spherical complex is a tiling of the whole (d−1)-dimensional sphere by spherical polytopes. Equivalently, it is the complex obtained by intersecting a complete polyhedral fan with the a (d−1)-dimensional sphere centered at the vertex of the fan. We know that every combinatorial information of a spherical complex can be realized as a convex polytope in dimension 3 by considering the so-called canonical polytope. The proof from Wikipedia seems to depend on the circle packing theorem and the midsphere theorem. So my question is: Is any spherical polytope of dimension $n$ isomorphic to a convex polytope as abstract polytope for $n > 3$? According to This question the midsphere theorem is still unknown for higher dimension, so I suspect there is a counterexample. Any idea or reference is appreciated. EDIT: (1) The definition of spherical polytope is edited several times (see the discussion on comment) (2) I just saw this wiki where it is stated that "convex polytopes in p-space are equivalent to tilings of the (p−1)-sphere." which seems to give a positive answer of my question, but no reference is provided there. REPLY [6 votes]: Another way to phrase your question is "whether every complete fan is (combinatorially equivalent to) the face fan (or the normal fan) of a convex polytopes". The answer is No in dimension $n\ge 4$ and I will provide an example derived from the following non-polytopal 4-diagram: This picture is taken from Example 5.10 in Günter Ziegler's book "Lectures on Polytopes". The book explains why this diagram is not the Schlegel diagram of any 4-dimensional polytope. It is important to note that the "outer cell" of the diagram is a simplex. Now, consider this diagram $D$ embedded into a 3-dimensional affine subspace of $\Bbb R^4$ that does not pass though the origin. Let $|D|$ be the simplex that forms the "outer cell" of $D$. Let $x\in\Bbb R^4$ be a point for which the convex hull $\Delta:=\mathrm{conv}(|D|\cup\{x\})$ contains the origin in its interior. This convex-hull is a 4-simplex and $|D|$ is one of its facets. Take the face fan of the simplex $\Delta$ and subdivide the cone over the facet $|D|$ into the cones over the cells of $D$. This is a complete fan in which no cone contains antipodal points. You can intersect it with a sphere to obtain a spherical complex. Suppose now that $P$ is a polytope combinatorially equivalent to this complex. Let $y\in P$ be the vertex that corresponds to the vertex $x$ of the complex. Its four neighbors in $P$ span a simplex $\Delta'$ (that corresponds to $|D|$ in the fan above). But deleting $y$ from $P$ (i.e. taking the convex hull of all vertices of $P$ except for $y$) leaves a polytope for which $\Delta'$ is a facet (here we use that $\Delta'$ resp. $|D|$ is a simplex) and whose Schlegel diagram based at $\Delta'$ is exactly $D$. This is a contradiction.<|endoftext|> TITLE: How are Sheffer polynomials related to Lie theory? QUESTION [6 upvotes]: Sheffer polynomials are orthogonal polynomial sets $\{P_n(x)\}$ having generating function $P(x,t) = \sum_{n=0}^{\infty}P_n(x)t^n=A(t)e^{xu(t)}$. This form reminds me of the Lie group–Lie algebra correspondence. Is there any connection to Lie theory, obvious or not? Here are some articles on Sheffer polynomials: Daniel Galiffa, Tanya Riston. An elementary approach to characterizing Sheffer A-type 0 orthogonal polynomial sequences. 2015. Dongsu Kim, Jiang Zeng. A Combinatorial Formula for the Linearization Coefficients of General Sheffer Polynomials. 2001. REPLY [9 votes]: First, a general set of Sheffer polynomials is not orthogonal with respect to some weight function; for example, the prototypical sequence $p_n(x) = x^n$, which belongs to both special sub-groups of Sheffer polynomials—the Appell and binomial Sheffer polynomials—with the e.g.f. $e^{xt}$, is not an orthogonal set—neither is the celebrated Bernoulli Appell Sheffer sequence. (Usually the Sheffer polynomials are defined by $A(t) e^{xB(t)} = \sum_{n\geq 0} p_n(x) \frac{t^n}{n!}= e^{tp_\cdot(x)}$. The normalization $P_n(x)=n!p_n(x)$ often serves to give integer coefficients.) However, there are connections to constructs associated to Lie theory: 1) The binomial Sheffer sequences have e.g.f.s of the form $e^{xu(t)}$, where $u(0)=0$ and $u'(0) \neq 0$. A dual sequence, its umbral compositional inverse, has the e.g.f. $e^{xu^{(-1)}(t)}$ defined by the compositional inverse function. The Graves–Lie infinitesimal generator $g(z)\partial_z$ with $g(z) = 1/u'(z)$ enters the picture via $$e^{tg(z)\partial_z}z \rvert_{z=0}= u^{(-1)}(t),$$ so $$e^{xu^{(-1)}(t)}= e^{tg(z)\partial_z} e^{zx} \;\rvert_{z=0}.$$ 2) Each Sheffer sequence has a pair of ladder ops—the raising/creation and lowering/destruction/annihilation ops defined by $L \; p_n(x) = n \; p_{n-1}(x)$ and $R \; p_n(x) = p_{n+1}(x)$—satisfying the Graves–Lie bracket of vector fields, the commutator, relation $$[L,R] = 1,$$ from which the Graves–Pincherle derivative $$[f(L),R] = f'(L)$$ follows. There is also the commutator $$[(g(z)\partial_z)^n,u(z)] = n \; (g(z)\partial_z)^{n-1}$$ for the powers of the Lie derivative / infinitesimal generator for the Scherk-Comtet partition polynomials of A139605, forming an underlying calculus for the binomial Sheffer sequences. 3) A general Sheffer sequence (a semidirect product of the Appell and binomial Sheffer sequences) is associated with a generalized Lie derivative via $$e^{t(q(z)+g(z)\partial_z)} \; e^{xz} |_{z=0} = A(t) e^{xf^{(-1)}(t)}$$ with $q(z) = \partial_t \ln(A(t)) \;|_{t=f(z)}$ and $g(z) = 1/f'(z)$. I have an extensive set of posts at my web blog and links to numerous OEIS entries and MO-Q&As and MSE-Q&As related to this topic. All the symmetric polynomials/functions--complete, elementary, power, Faber--are related to Sheffer Appell sequences with their related ladder ops. The Faa di Bruno/Bell and cycle index polynomials of the symmetric groups, a.k.a the refined Stirling partition polynomials of the second and first kinds, and other compositional partition polynomials are all Appell Sheffer sequences in a distinguished indeterminate. The closely related sets of Lagrange inversion partition polynomials, including the refined Euler characteristic partition polynomials of the associahedra, all have the raising op $g(z)\partial_z$, as shown above--one set, OEIS A134264, is an Appell sequence and is related to free probability, inversion of Laurent series, and characterization of Kac-Schwarz operators related to Heisenberg-Virasoro groups. Multiplicative inversion is intimately bound with the refined Euler characteristic partition polynomials of the permutahedra and the operational (differential/matrix) calculus of Appell Sheffer polynomials. The Bernoulli Appell polynomials are of course related to the BCHD theorem, exponential mappings of the Lie commutator, topology, Todd operator and class of Hirzebruch, the Hurwitz zeta function, and more. Formal group laws (and generalized local Lie groups) are related to the linearization coefficients of products of binomial Sheffer polynomials. The list goes on. The orthogonal Sheffer sequences such as the associated Laguerre polynomials have a lot of underlying group theory and vector fields associated with them. Vilenkin with "Special Functions and the Theory of Group Representations", Miller, Feinsilver, and many others have written books on the underlying group theory. (None covers every aspect.) One group of researchers has also written extensively on this topic. See, e.g., "One-parameter groups and combinatorial physics" by Duchamp, Penson, Solomon, Horzela, and Blasiak (although quite negligent of reffing the OEIS--tribal instincts...sigh...What can you do?).<|endoftext|> TITLE: Does the (1-)topos structure on simplicial sets have any homotopy-theoretic significance? QUESTION [8 upvotes]: To give an example of a peculiar feature of simplicial sets that I cannot remember encountering anywhere in the context of homotopy theory: every simplicial set $X$ possesses partial map classifier $X\rightarrowtail\widetilde X$: the $n$-simplices of $\widetilde X$ are partial simplices of $X$, i. e. maps from all kinds of simplicial subsets of the standard $n$-simplex $\Delta[n]$ to $X$. This $\widetilde X$ is a contractible Kan complex, in strongest possible way: it is an injective object, i. e. any $Y\leftarrowtail Y'\to\widetilde X$ extends to $Y$. It is thus some sort of a fibrant cone for $X$. This construction is functorial, in fact, part of a monad structure (sometimes called "lift monad" or "maybe monad"). Is not existence of such a thing useful for homotopy-theoretic purposes? Note that this is just one example, there surely are many other features that can be extracted from the topos structure. I found a related question Internal logic of the topos of simplicial sets but it is rather about peculiarities of simplicial sets as a particular topos than peculiarities of this topos as a particular homotopy-theoretic universe. REPLY [7 votes]: To answer the title question: Yes, the fact that the 1-category $sSet$ is a 1-topos has homotopy-theoretic significance. It is closely related to the fact that the $\infty$-category $Spaces$ is an $\infty$-topos! To a large degree, constructions using the topos structure on $sSet$ can be recast model-independently as constructions using the $\infty$-topos structure on $Spaces$. If you like, I suppose you can see the fact that $Spaces$ is an $\infty$-topos as a consequence of the fact that $sSet$ is a 1-topos using Rezk's model topos machinery. On the other hand, I'm not so sure about the significance of the partial map classifier in particular. The "$\infty$-partial map classifier" exists in $Spaces$, but there are not so many partial maps in the $\infty$-topos $Spaces$ as in the 1-topos $sSet$ (since a monomorphism is just a coproduct inclusion in $Spaces$). Probably this construction has some homotopy-theoretic significance in some model-dependent context, but it will depend on what the other model-dependent particulars of the situation are, I suspect. On the gripping hand, an $\infty$-topos like $Spaces$ actually has object classifiers which are arguably even "better" than subobject classifiers.<|endoftext|> TITLE: Modern results that are widely known, yet which at the time were ignored, not accepted or criticized QUESTION [51 upvotes]: What is your favorite example of a celebrated mathematical fact that had a hard time to become accepted by the community, but after overcoming some initial "resistance" quickly took on? It can be a theorem, a proof method, an algorithm or a definition, that is widely known and very useful in the present day, less than 99 years old; this is in order to avoid examples from the very distant past, such the difficulties Grassmann's work had being accepted but which at the time of its inception was not appreciated, misunderstood or ignored by the mathematical community, before it became mainstream or inspired other research which in turn became mainstream. This is a partial converse question to this one that asks for mathematical facts that were quickly accepted but then discarded by the community. This and this question are somewhat related, but former focuses on people (resp. their entire works, see Grassmann) not being accepted, rather then individual results, whereas the latter solely on famous articles rejected by journal; also, the results that are being mentioned in these links are often rather old and do not fit this question. Example. Numerical optimization: The first quasi-Newton algorithm was discovered in 1959 and "was not accepted for publication; it remained as a technical report for more than thirty years until it appeared in the first issue of the SIAM Journal on Optimization in 1991" (Nocedal & Wright, Numerical Optimization). But the algorithm inspired a slew of other variants, has been cited over 2000 times to this day and quasi-Newton type algorithms are still state-of-the-art in for certain optimization problems. REPLY [9 votes]: Grothendieck's inequality, now a fundamental result in functional analysis, with connections to computer science and quantum physics, had a difficult birth. It was proved by Grothendieck in the paper Résumé de la théorie métrique des produits tensoriels topologiques, published in French in 1953 in an obscure Brazilian journal, only in very few copies, making it almost impossible to find. The paper was almost completely ignored by the community, until it was rediscovered in 1968 by Lindenstrauss and Pelczynski who realized that in particular it contained answers to questions raised after its publication. The story is explained in the first pages of the survey article by Gilles Pisier, Grothendieck's Theorem, past and present<|endoftext|> TITLE: Breen-Deligne packages and the Liquid Tensor Experiment QUESTION [17 upvotes]: I am trying to understand some things about Condensed Mathematics and the Liquid Tensor Experiment. The aim of the LTE is to provide a formalised proof of Theorem 9.4 in Scholze's paper Lectures on Analytic Geometry (describing joint work with Clausen). Part of the LTE is a blueprint of the general approach. Theorem 9.4 is stated for a rather general class of chain complexes, but the blueprint works with a more restricted class which is presumably sufficient for the intended application; I have not yet understood this reduction. Specifically, the blueprint considers complexes defined with the aid of a structure called a Breen-Deligne package (Definition 2.11 in the blueprint). As pointed out in Definition 2.13 of the blueprint, there is a default example of a Breen-Deligne package. My question is: is it sufficient to consider this default example, or is there a real need to consider all possible examples? If it is sufficient to consider the default example, then it seems to me that the whole proof can be substantially simplified. (However, I am just beginning to get to grips with these ideas, so I could easily be mistaken.) REPLY [26 votes]: The comments have already given the answers, but let me assemble them here with my account of the story. When Scholze first posted the Liquid Tensor Experiment, it was quickly identified (by both Peter and Reid, somewhat independently I think) that Breen–Deligne resolutions would be the largest input in terms of prerequisites that needed to be formalised. Back then, I had no idea what these resolutions were, or how to prove that they existed. But they were needed for the statement of Theorem 9.4, which seemed to be a very natural first milestone to aim for. So I set out to formalise the statement of the existence of Breen–Deligne resolutions, with the expectation that we would never prove the result in Lean, but just assume it as a black box. Let me recalll the statement: Theorem (Breen–Deligne). For an abelian group $A$, there exists a resolution of the form $$ \dots → \bigoplus_{j = 0}^{n_i} ℤ[A^{r_{i,j}}] → \dots → ℤ[A^2] → ℤ[A] → A → 0 $$ that is functorial in $A$. On the proof. See appendix to Section IV of Condensed.pdf. As Reid remarked in the comments, the proof relies on the fact that stable homotopy groups of spheres are finitely generated. ∎ (Aside: for the proof of Theorem 9.4, we really need a version that applies to condensed abelian groups $A$, so in practice we want to abstract to abelian sheaves.) In the rest of the Lecture notes, Scholze often uses a somewhat different form of the above resolution, by assuming $n_i = 1$ and effectively dropping all the $\bigoplus$'s. I think this was done mostly for presentational reasons. I did the same thing when I axiomatized Breen–Deligne resolutions in Lean. So really, they weren't axiomatized at all. I don't know of any reason to expect that there exists a resolution with the property that $n_i = 1$ for all $i$, but I also don't see any reason why there shouldn't. Anyway, I needed a name for the axioms that I did put into Lean, and I chose Breen–Deligne package for that. So what is that exactly? Well, a functorial map $ℤ[A^m] → ℤ[A^n]$ is just a formal sum of matrices with coefficients in $ℤ$. So as a first approximation, we record the natural numbers $r_j = r_{1,j}$ (since $n_i = 1$); for every $j$, a formal sum of $(r_{j+1}, r_j)$-matrices with coefficients in $ℤ$. But we need one more property of Breen–Deligne resolutions: if $C(A)$ denotes the complex, then there are two maps induced by addition. There is the map $σ \colon C(A^2) → C(A)$ that comes from the functoriality of $C$ applied to the addition map $A^2 → A$. But there is also the map $π \colon C(A^2) → C(A)$ that comes from addition in the objects of the complex. (All objects are of the form $ℤ[A^k]$ and we can simply add elements of these free abelian groups.) The final axiom of a Breen–Deligne package is there is a functorial homotopy between $σ$ and $π$. While playing around with the axioms, I noticed that I could write down inductively a somewhat non-trivial example of such a package. When I discussed this example with Peter, he suggested that it might in fact be suitable as a replacement for Breen–Deligne resolutions in all applications in his lecture notes so far. Several months later we found out that I had rediscovered MacLane's $Q'$-construction. So, let's denote by $Q'$ the complex corresponding to the example package. The following result is, as far as I know, original: Lemma. Let $A$ and $B$ be (condensed) abelian groups. If $\text{Ext}^i(Q'(A),B) = 0$ for all $i ≥ 0$, then $\text{Ext}^i(A,B) = 0$ for all $i ≥ 0$. Proof. I've written up a proof sketch on Zulip (public archive of that thread). ∎ We are in the process of formalizing the necessary homological algebra to verify the proof in Lean. It's the last major milestone left to complete the challenge in Peter's original blogpost. After that, we mostly need some glue. See this blogpost for an update on the formalisation effort. (Edit: see also https://math.commelin.net/files/LTE.pdf for a more precise roadmap of what remains to be done to complete the challange.) Once everything is done, it should all be written up in some paper. So now, let me turn to the question is it sufficient to consider this default example, or is there a real need to consider all possible examples? It is indeed sufficient to consider this default example. The reasons for working with the abstract concept of Breen–Deligne packages are historical: I had formalised the statement of Theorem 9.4 and some other material in terms of BD packages before I realised that this default example was indeed sufficient for our purposes. practical: as Remy points out in the comments, it is (i) helpful to abstract away concrete details into a conceptual object, and (ii) there is a chance that there might be better examples leading to better constants. What would be really interesting is an example of a BD package that gives resolutions instead of merely functorial complexes. It is known that the $Q'$ construction is not a resolution. Indeed, one can easily show inductively that $H_i(Q'(ℤ))$ contains a copy of $ℤ^{2^i}$. With more work (relying again on abstract homotopy theory), Peter gave a proof that $Q'(A) ≅ Q'(A)^{⊕2}[-1] ⊕ ℤ ⊗_{\mathbb S} A$ on Zulip (public archive).<|endoftext|> TITLE: Variance of the norm of a random variable under finite-moment assumptions QUESTION [5 upvotes]: There is the following exercise in Vershynin's book on High-Dimensional Probability. Exercise 3.1.6: Let $X = (X_1, \dots, X_n) \in \mathbb{R}^n$ be a random vector with independent coordinates $X_i$ that satisfy $\mathbb{E}[X_i^2] = 1$ and $\mathbb{E}[X_i^4] \leq K^4$. Show that $\mathrm{Var}(\|X\|_2) \leq CK^4$. What if we remove the condition that the coordinates are independent? Suppose instead that we have only the condition that the coordinates are uncorrelated (i.e., $X$ has covariance $I$), and we additionally have some bound on the fourth moment (e.g., $\mathbb{E} [ \langle X, v\rangle^4]\leq K^4$ for every unit vector $v$). What is the best bound we can prove on $\mathrm{Var}(\|X\|_2)$? REPLY [4 votes]: $\newcommand\R{\mathbb R} \renewcommand{\P}{\operatorname{\mathsf P}}\newcommand\E{\operatorname{\mathsf{E}}}\newcommand\Var{\operatorname{\mathsf{Var}}}$The best bound on $\Var \|X\|_2$ is about $n$. Indeed, \begin{equation*} \Var \|X\|_2\le \E \|X\|_2^2=\sum_{i=1}^n \E X_i^2=n. \tag{1}\label{1} \end{equation*} On the other hand, suppose that \begin{equation*} \mu_X=q\,\mu_{aG}+p\,\mu_{bG}, \end{equation*} where $\mu_Y$ denotes the distribution of a random vector $Y$, $G$ is a standard Gaussian random vector in $\R^n$, and \begin{equation*} a:=0,\quad b\in[1,\infty),\quad p:=\frac1{b^2},\quad q:=1-p. \end{equation*} Then the covariance matrix of $X$ is $pb^2I=I$. Also, for all unit vectors $v\in\R^n$, \begin{equation*} \E (X\cdot v)^4=pb^4\E (G\cdot v)^4=3b^2=K^4 \end{equation*} if $b=K^2/\sqrt3$ (assuming that $K^4\ge3$). So, all your conditions hold. However, $\E\|X\|_2^2=pb^2\E\|G\|_2^2=n$ and $\E\|X\|_2=pb\E\|G\|_2\le pb\sqrt{\E\|G\|_2^2}=pb\sqrt n=\sqrt n/b$ and hence \begin{equation*} \Var \|X\|_2=\E\|X\|_2^2-(\E\|X\|_2)^2\ge\Big(1-\frac1{b^2}\Big)n=\Big(1-\frac3{K^4}\Big)n; \tag{2}\label{2} \end{equation*} the inequality $\Var \|X\|_2\ge\big(1-\frac3{K^4}\big)n$ trivially holds if $K^4\not\ge3$. Conclusion 1: The trivial upper bound $n$ on $\Var \|X\|_2$ in \eqref{1} is optimal, in the sense that it cannot be replaced by $cn$ for any real $c<1$, if $K$ is allowed to be large enough. Nonetheless, the upper bound $n$ on $\Var \|X\|_2$ in \eqref{1} can be "second-order" improved as follows. Note that \begin{equation*} \E\|X\|_2^4=\E\Big(\sum_{i=1}^n X_i^2\Big)^2=\sum_{i,j=1}^n \E X_i^2 X_j^2 \le\sum_{i,j=1}^n \frac{\E X_i^4+\E X_j^4}2 =n\sum_{i=1}^n \E X_i^4\le n^2K^4. \end{equation*} So, by the log-convexity of $\E\|X\|_2^p$ in $p>0$, \begin{equation*} (\E\|X\|_2)^2\ge\frac{(\E\|X\|_2^2)^3}{\E\|X\|_2^4}\ge\frac{n^3}{n^2K^4} =\frac n{K^4}. \end{equation*} So, \begin{equation*} \Var \|X\|_2=\E\|X\|_2^2-(\E\|X\|_2)^2\le\Big(1-\frac1{K^4}\Big)n, \tag{3}\label{3} \end{equation*} which is an improvement of \eqref{1}. Conclusion 2: In view of \eqref{2} and \eqref{3}, the best upper bound on $\Var \|X\|_2$ is of the form \begin{equation*} \Big(1-\frac C{K^4}\Big)n \end{equation*} for some $C\in[1,3]$. This refines Conclusion 1.<|endoftext|> TITLE: Is there a version of the Poincaré–Hopf theorem for manifold with corners? QUESTION [7 upvotes]: As we know, the square $S=[0,1]\times[0,1]$ is not a manifold with boundary. Instead, it's a manifold with corners. For a tangent vector field on a compact manifold with boundary, we have the Poincaré–Hopf index theorem. Then here is my question: is there a similar index theorem for manifolds with corners (e.g., the square, the cube) which relates the indices of a tangent vector field to the Euler characteristics of the manifold? If affirmative, what is the statement of this theorem? Btw, I find an economics textbook (Vives, X., 1999. Oligopoly pricing: old ideas and new tools. MIT Press) which states that the Poincaré–Hopf theorem can apply to a compact cube (a manifold with corners). The statement (on page 362) is as below: This "version" is frequently used in economics papers. Is it correct? REPLY [10 votes]: With regards to the updated question: Note that the quoted statement is that the vector field points inward at the boundary. In particular this means that there are no singularities at the corners (nor at the flat parts of the boundary). Thus you can simply replace the manifold by a slightly smaller one, for which the corners are "rounded off". This then is a manifold with boundary, to which the usual Poincaré-Hopf theorem applies. That being said, as discussed in the comments, the situation gets a bit more interesting when there are singularities at the boundary. There, the theorem still holds, provided the notion of index is correctly defined for those singularities. I think for this, one can simply copy the usual proof of the Poincaré-Hopf theorem which cuts out small balls around the singularities and then simply define the index as whatever is needed to make this work. Toying around with the Gauss-map, I would expect something like $$ \operatorname{ind}(x_0) = \lim_{\epsilon \to 0} \frac{1}{|S^{n-1}|} \int_{S\cap \partial B_\epsilon(x_0)} \left(\frac{u}{|u|}\right)^* \omega_{S^{n-1}}$$ but for corners there might be situations where this limit is not entirely well defined.<|endoftext|> TITLE: Topos-theoretic Galois theory QUESTION [12 upvotes]: This page is an overview of some of the types of "Galois theories" there are. One of the most basic type is the fundamental theorem of covering spaces, which says, roughly, that for each topological space $X$, there is an equivalence of categories $$\mathrm{Cov}(X)\simeq \pi_1(X)\mathbf{Set}.$$ Grothendieck proved an analogue of that statement for schemes $X$: $$\mathrm{EtCov}(X)\simeq \pi_1(X)\mathbf{Set}.$$ (This is again just a very rough formulation and omits some of the assumptions, but you know what I mean.) I am interested in "topos-theoretic Galois theory". Unfortunately, this section of the nLab page isn't filled out ("(...)"), but I guess that the topos-theoretic formulation of Galois theory states, roughly, that for each topos $\mathcal E$, $$\mathrm{Gal}(\mathcal E)\simeq \pi_1(\mathcal E)\mathbf{Set},\qquad (\ast)$$ where $\mathrm{Gal}(\mathcal E)$ is the full subcategory of $\mathcal E$ consisting of locally constant objects in $\mathcal E$, and $\pi_1(\mathcal E)$ is the fundamental group of $\mathcal E$. (This is suggested by the nLab section "Reformulation of classical Galois theory".) Question: Is there a reference for $(\ast)$ (and the definition of the fundamental group of a topos which is used here)? Is this in SGA 4? The linked nLab page fundamental group of a topos refers to (and is mostly copy-pasted from) Porter's paper Abstract Homotopy Theory: The interaction of category theory and homotopy theory, which contains a section called "The fundamental group of a topos", which in turn refers to SGA 1. This is weird, because SGA 1 doesn't discuss topoi, so in particular not the fundamental group of a topos! The nLab also refers to SGA 4 Exposé IV Exercice 2.7.5 for the definition of the fundamental group and SGA 4 Exposé VIII Proposition 2.1 for, I guess, $(\ast)$ in the special case that $\mathcal E$ is the étale topos of the scheme $X=\mathrm{Spec}(k)$ for some field $k$. (But this is really just a guess - I can't read French. So correct me if I'm wrong.) Is there more of "topos-theoretic Galois theory" in SGA 4 or are these the only two paragraphs about that topic? Concerning the definition of the fundamental group of a topos, there is a construction in Moerdijk's Classifying Spaces and Classifying Topoi, in which he nevertheless remarks: The profinite fundamental group is discussed in SGA1. This suggests there are two version of the fundamental group of a topos: the one he discusses and the "profinite" version. However, as I said, topoi don't occur in SGA 1, so I wonder where I can find the definition of the "profinite" fundamental group, if that's the notion that should be used in $(\ast)$. (The definition used in $(\ast)$ should of course have the property that if $\mathcal E$ is the étale topos of a scheme $X$, then $\pi_1(\mathcal E)$ is isomorphic to the étale fundamental group of $X$.) REPLY [6 votes]: Maybe you would like to see the thesis of O. Leroy, Groupoïde fondamental et théorème de Van Kampen en théorie des topos, available from https://plmbox.math.cnrs.fr/f/7ffa366379144dd4bacc/?dl=1 — the password is : groupoide (a project of transcription) or the thesis of V. Zoonekynd, La Tour de Teichmuller-Grothendieck, available from https://tel.archives-ouvertes.fr/tel-00001140/document REPLY [6 votes]: I believe that chapter $5$ of Galois Theories by Borceux and Janelidze contains what you're looking for -- they develop a general 'categorical Galois theorem' invoking toposes. Chapter 7 gets even more general, presenting a 'non-Galoisian Galois theorem' that holds for any descent morphism in a category -- a morphism $f$ is a descent morphism iff the 'pullback along $f$' functor is monadic. There is also the classic paper by Joyal and Tierney, An Extension of the Galois theory of Grothendieck, where they prove that each Grothendieck topos is equivalent to the category of equivariant sheaves on a groupoid internal to the category of locales. This paper by Christopher Townsend might be of interest to your specific question; he re-proves Joyal and Tierney’s result on the representation of Grothendieck toposes as localic groupoids using a simplified case of the aforementioned categorical Galois theorem, then proceeds to actually prove the whole theorem using this trivial case as a key ingredient.<|endoftext|> TITLE: Latin squares with one cycle type? QUESTION [8 upvotes]: Cross posting from MSE, where this question received no answers. The following Latin square $$\begin{bmatrix} 1&2&3&4&5&6&7&8\\ 2&1&4&5&6&7&8&3\\ 3&4&1&6&2&8&5&7\\ 4&3&2&8&7&1&6&5\\ 5&6&7&1&8&4&3&2\\ 6&5&8&7&3&2&4&1\\ 7&8&5&2&4&3&1&6\\ 8&7&6&3&1&5&2&4 \end{bmatrix}$$ has the property that for all pairs of two different rows $a$ and $b$, the permutations $ab^{-1}$ have the same cycle type (one 2-cycle and one 6-cycle). What is known about Latin squares with the property that all $ab^{-1}$ have the same cycle type (where $a$ and $b$ are different rows)? For example, do they have a particular structure, for which cycle types do they exist, are there any infinite families known, do they have a name, etc? The only example of an infinite family I'm aware of are powers of a single cyclic permutation when $n$ is prime, for example: $$\begin{bmatrix} 1&2&3&4&5\\ 2&3&4&5&1\\ 3&4&5&1&2\\ 4&5&1&2&3\\ 5&1&2&3&4 \end{bmatrix}$$ REPLY [6 votes]: One way to achieve the required property is to construct a Latin square whose autotopism group acts transitively on unordered pairs of rows. This can be achieved for orders that are a prime power congruent to 3 mod 4, by means of the quadratic orthomorphism method, described in this paper. The focus of that paper is the atomic squares I mentioned in other comments, but the construction will always achieve the property that each pair of rows produces the same cycle structure. For prime powers that are 1 mod 4, quadratic orthomorphisms will build Latin squares with at most two types of cycle structures formed by pairs of rows.<|endoftext|> TITLE: From a physicist: How do I show certain superelliptic curves are also hyperelliptic? QUESTION [16 upvotes]: As the title suggests, I am a physicist and have a question about how to show certain superelliptic curves are also hyperelliptic. The superelliptic Riemann surfaces in question has the form $$w^n = \prod_{\alpha=1}^N(z-u_\alpha)(z-v_\alpha)^{n-1},\quad u_12$ then we must have $f(z)=z$. If $f(z)=z$, then the hyperelliptic involution sends $w$ to another $n$th root of $w^n$. This must be obtained by multiplying $w$ by an $n$th root of unity. Since the hyperelliptic involution is an involution, this must be multiplying $w$ by $-1$, so $n$ must be even and the quotient by the hyperelliptic involution is given by $$ w^{n/2}= \frac{ \prod_{\alpha=1}^N(z- u_\alpha )} {\prod_{\alpha=1}^N (z- v_\alpha)}.$$ This is a smooth curve of bidegree $(n/2,N)$ in $\mathbb P^1 \times \mathbb P^1$, and must have genus $0$, so we must have $n/2 \leq 1$ or $N \leq 1$. Again the case $N \leq 1$ is trivial, so we must have $n=2$. So the only hyperelliptic cases are $n=2$ and $N=2$.<|endoftext|> TITLE: Does $A-A=\mathbb Q$ hold for $A=\{x^4+y^4:\ x,y\in\mathbb Q\}$? QUESTION [14 upvotes]: Let $A=\{x^4+y^4:\ x,y\in\mathbb Q\}$. Then $$A-A:=\{a-b:\ a,b\in A\}=\{u^4+v^4-x^4-y^4:\ u,v,x,y\in\mathbb Q\}.$$ Motivated by Question 415482, here I ask the following question. Question. Is it true that $A-A=\mathbb Q$? Any effective way to approach it？ By my computation, $A-A$ at least contains $0,1,\ldots,562$. For example, $$248=\left(\frac{95}{28}\right)^4+\left(\frac{135}{14}\right)^4-\left(\frac{13}7\right)^4-\left(\frac{269}{28}\right)^4\in A-A.$$ From the viewpoint of additive combinatorics, the question looks interesting. I guess that it should have a positive answer. Any ideas to solve it? REPLY [19 votes]: According to Tito Piezas's website $x^4+y^4-(z^4+t^4) = N$, There is an identity $((2a+b)c^3d)^4 + (2ac^4-bd^4)^4 - (2ac^4+bd^4)^4 - ((2a-b)c^3d)^4 = a(2bcd)^4$ where $b = c^8-d^8$, for arbitrary {$a,c,d$}.<|endoftext|> TITLE: Ramanujan-Petersson conjecture at various cusps QUESTION [6 upvotes]: Suppose that $f \in S_k(\Gamma_0(N)) $ be a Hecke eigenform whose Fourier expansion at $ i\infty $ is given by $$ f(z) = \sum_{n=1}^{\infty} \lambda(n) n^{\frac{k-1}{2}} \exp(2\pi i n z), $$ normalized so that $\lambda(1)=1$. In this setting the Ramanujan-Petersson conjecture states that $ |\lambda(n)| \leq d(n) $ the number of divisors of $n$ (for all $ n $ coprime to the level $ N $). Does the same bound hold if I consider the Fourier expansion of $f$ at some other cusp? REPLY [9 votes]: The estimate $|\lambda(n)| \leq C_N d(n)$ remains valid at all cusps, but $C_N$ cannot in general be taken independent of $N$. See Remark 3.14 of this paper (arxiv link), where it is noted that for certain $(N,f,n)$, with $f$ a newform, one can have $\lambda(n) \gg n^{1/4}$ at some cusp. (One can take for $n$ a suitable power of a prime $p$ for which $p^2 | N$.)<|endoftext|> TITLE: De Rham via topoi QUESTION [16 upvotes]: Étale cohomology of schemes $X$ is constructed as follows: one associates to $X$ the so-called étale topos of $X$, and then one just takes the sheaf cohomology of that topos. Is it possible to associate to each smooth manifold $M$ a "de Rham topos of $M$", whose sheaf cohomology yields the de Rham cohomology of $M$? REPLY [13 votes]: One can define an analogue of the crystalline topos for smooth manifolds. This is known as the de Rham stack of $M$. One of the easiest constructions of the de Rham stack embeds smooth manifolds fully faithfully (using the Yoneda embedding) into the category of ∞-sheaves on affine smooth loci, the latter being defined as the opposite category of $\def\Ci{{\rm C}^∞} \Ci$-rings satisfying certain properties. In this language, the de Rham stack of an ∞-sheaf $F$ is the ∞-sheaf $\def\dR{{\rm dR}} \dR(F)$ defined by $\def\Spec{\mathop{\rm Spec}} \def\red{{\rm red}} \dR(F)(\Spec A)=\dR(F)(\Spec(\red(A))$, where $\Spec A$ denotes the spectrum of a $\Ci$-ring (defined purely formally in this context) and $\red(A)$ denotes the quotient of $A$ by its ideal of nilpotent elements. One can then prove that the commutative differential graded algebra of smooth functions on $\dR(M)$ is precisely the de Rham algebra of $M$. The de Rham stack has other exciting properties: vector bundles (and, more generally, sheaves) on $\dR(M)$ can be identified with D-modules, etc. The cited nLab article has the relevant pointers to the literature. The de Rham stack is also closely related to the definition of the differential graded algebra of differential forms in synthetic differential geometry as the differential graded algebra of infinitesimal smooth singular cochains equipped with the cup product. See the nLab article differential forms in synthetic differential geometry for further pointers to the literature.<|endoftext|> TITLE: Questions about SGA 4 QUESTION [8 upvotes]: What is Deligne's motivation in Appendice 9 of Exposé VI to prove that every coherent topos has enough points? For instance, does that have applications in étale cohomology (or other parts of algebraic geometry)? Which topics are discussed in the Exposés Vbis, VI, VIII, and IX? An answer to the question should provide either buzzwords or English literature covering these topics. What role do topoi play in these Exposés? Here Joyal writes the following. What could he mean by that? About half of the topos theory of SGA4 is devoted to categorical generalities. They are now subsumed by the modern theory of (locally) presentable categories. REPLY [10 votes]: I'm not aware of any applications of Deligne's result to algebraic geometry at the time. It does formally imply that certain topologies (e.g. fppf) have enough points, but this wasn't very useful because the actual points in many well-known topologies were only described by Gabber–Kelly and Schröer in 2014. On the other hand, points in the étale topology are just geometric points, so those are very useful in the theory, but you don't need Deligne's completeness theorem for this. Deligne's theorem does have formal consequences to the systematic study of coherent topoi, such as in Barwick–Glasman–Haine's Exodromy, where it is used in the proof of a base change theorem for oriented fibre products of bounded coherent $\infty$-topoi (Thm. 7.1.7). This is many questions in one, but let me say some words. V$^{\text{bis}}$ is the origin of hypercoverings and cohomological descent. This is a powerful tool that is used frequently in algebraic geometry, especially in combination with resolution of singularities or alterations to build up the cohomology of a variety from the cohomology of smooth projective varieties. This happens for example in Deligne's Hodge III paper, where he uses proper hypercoverings to construct the mixed Hodge structure on a singular or open variety. One alternative source for the material from V$^{\text{bis}}$ is the chapter on hypercoverings in the Stacks project, but I imagine to a category theorist this might look a bit dated. Descent with respect to arbitrary hypercoverings is also a topic that comes up in $\infty$-topoi, where this property is called hyperdescent. The relation between hypercompleteness and hyperdescent is explained in sections 6.5.3 of Higher Topos Theory, and §6.5.4 makes the case that you do not want to impose this as an axiom on an $\infty$-topos. Because hypercompleteness is related to Whitehead's theorem (which holds on the $\infty$-topos of a point), an $\infty$-topos can only have enough points (in the sense of question 1) if it is hypercomplete! VI is about coherent topoi and is purely topos-theoretic in nature (although the definitions of quasi-compact and quasi-separated are inspired by the spaces occurring in algebraic geometry). I don't know secondary references for this, nor is it incredibly clear to me where it is used, so I will leave this to someone else to answer. VIII is the first place where some actual computations in the étale cohomology happen. For example, it is explained that the étale site of a field $k$ with absolute Galois group $\Gamma_k$ is a $B\Gamma_k$ (at least if you understand $\Gamma_k$ as a profinite group and interpret $B\Gamma_k$ in that light). This material is all specific to the étale site in algebraic geometry, and strays away from general topos theory. Most of the results can be found in the Stacks Project, or Milne's book or notes on Étale Cohomology, or the book by Freitag and Kiehl, or ... IX is about constructible sheaves and the cohomology on a curve. Constructible sheaves also exist in topology, but in algebraic geometry there are pervasive finiteness hypotheses for everything to work. More interesting is the computation of the cohomology of a curve. For example, you want to know that étale cohomology with finite coefficients vanishes above dimension $2$. In algebraic geometry, the only way to do this is to show that the Galois group of a function field of a curve has cohomological dimension $1$, and then use the Leray spectral sequence for $\operatorname{Spec} K(C) \hookrightarrow C$. The rest of the computation of the cohomology on a curve is based on its Picard group (or Jacobian). This has been addressed by Tim Campion in the comments. As I noted, schemes don't start coming into SGA 4 until V$^{\text{bis}}$ (a little bit) and really starting at VII. All the material prior to IV (and even some of the material after that) really is general category theory, a substantial part of which consists of putting size bounds on everything to make sure all categories are small or at least locally small. This is exactly the study of accessible categories. The classical reference is probably Adámek–Rosicky's Locally presentable and accessible categories, and I suspect SGA 4 would be substantially shortened if one already has access to all that.<|endoftext|> TITLE: Littlewood-Richardson coefficients in terms of Specht modules QUESTION [10 upvotes]: Littlewood-Richardson coefficients $c_{\nu\mu}^{\lambda}$ (where $\nu$,$\mu$ and $\lambda$ are integer partitions such that $|\nu| + |\mu| = |\lambda|$) are well-known coefficients appearing in various contexts. In terms of Schur functions, which form a basis of the symmetric functions, the coefficients are the multiplicative constants, meaning that $$ s_{\nu}s_{\mu} = \displaystyle \sum_{\lambda} c_{\nu\mu}^{\lambda} s_{\lambda}. $$ This is the context in which the rule was first stated. As Schur functions are the characters of irreducible representations of $\text{GL}_n$, it means that if we denote $V_{\lambda}$ for the irreducible representation of $\text{GL}_n$ associated to $\lambda$, then: $$ V_{\nu} \otimes V_{\mu} \cong \displaystyle \bigoplus_{\lambda} (V_{\lambda})^{\oplus c_{\nu\mu}^{\lambda}} $$ The transition between those two contexts is relatively easy. There is also another context closely related to the two above, which is the representations the symmetric group. Via the Schur-Weyl duality, the above results implies that if $S^{\lambda}$ is the Specht module associated to $\lambda$ (which are the irreducible representations of $\mathfrak{S}_{|\lambda|}$), then $$ \left( S^{\nu} \otimes S^{\mu} \right) \big\uparrow_{\mathfrak{S}_{|\nu|} \times \mathfrak{S}_{|\mu|}}^{\mathfrak{S}_{|\nu|+|\mu|}} \cong \bigoplus_{\lambda} \left( S^{\lambda} \right)^{\oplus c_{\nu\mu}^{\lambda}} $$ Using Schur's lemma, the latter means that $$ \text{dim} \left( \text{Hom}_{\mathfrak{S}_{|\nu|+|\mu|}} \left( S^{\lambda}, \left( S^{\nu} \otimes S^{\mu} \right) \big\uparrow_{\mathfrak{S}_{|\nu|} \times \mathfrak{S}_{|\mu|}}^{\mathfrak{S}_{|\nu|+|\mu|}} \right) \right) = c_{\nu\mu}^{\lambda} $$ My question is: is there any construction of the Littlewood-Richardson coefficients which is stated only in terms of the symmetric group and that gives the last equality? For example, Specht modules is the span of certain elements of the symmetric group algebra, called polytabloids. Can we calculate the Littlewood-Richardson coefficients using those? REPLY [4 votes]: Gordon James gave a fairly simple proof entirely within the representation theory of the symmetric groups: James, G. D. A characteristic-free approach to the representation theory of $_n$. J. Algebra 46 (1977), no. 2, 430–450.<|endoftext|> TITLE: A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$? QUESTION [16 upvotes]: Let $f(x)$ be sufficiently regular (e.g. a smooth function or a formal power series in characteristic 0 etc.). In my research the following recursion made a surprising entrance $$ f_1(x) = f(x),\ f_{n+1}(x) = f(x) f'_{n}(x) $$ Thus I would like to understand the sequence $$ \left(f(x) \frac{d}{dx} \right)^n f(x) $$ This looks like a classical question that must have been already studied in dynamical systems, or Weyl algebras (say for $f(x) \in R[x]$, $R$ a commutative ring of characteristic $0$, and the derivation $\partial := f(x) \frac{d}{dx} \in A_1(R)$), or generating functions in combinatorics. But I have been unable to pinpoint it. My question is this: Is there a known formula in the spirit of the general Leibniz formula which expresses $$ \left(f(x) \frac{d}{dx} \right)^n f(x) $$ in terms of $f$ and its derivatives $f',f'',\dots, f^{(n)}$? Any references would be also very much appreciated! REPLY [3 votes]: This, too, is not an answer, just a preview of a paper I probably won't be writing for a while (one of only 9 papers on my immediate to-do list). It so happens I derived a related formula a few weeks ago when proving an apocryphal identity of Hochschild. The formula needs some preliminary definitions, and unfortunately my notations are different from the OP's, but I hope it is of some use. Note that, while I am able to prove everything I claim, I am not fluent in its combinatorial background (umbral calculus, in particular). The general setup will be as follows: We fix a commutative ring $R$. Let $d:R\rightarrow R$ be a derivation (i.e., a $\mathbb{Z}$-linear map satisfying $d\left( ab\right) =d\left( a\right) \cdot b+a\cdot d\left( b\right) $ for all $a,b\in R$). Let $a\in R$ be an element. Let $L_{a}$ be the map $R\rightarrow R,\ r\mapsto ar$ (known as "left multiplication by $a$", but can just as well be called "right multiplication by $a$" since $R$ is commutative). Note that $L_{a}\circ d$ is a derivation (commonly denoted by $ad$), but $d\circ L_{a}$ is not (in general). For each nonnegative integer $n$, we define the following: Let $\left[ n\right] $ be the set $\left\{ 1,2,\ldots,n\right\} $. A set partition of $\left[ n\right] $ means a set $P$ of subsets of $\left[ n\right] $ (called the blocks of $P$) such that each element of $\left[ n\right] $ belongs to exactly one block of $P$. Let $\operatorname*{SP}\left( n\right) $ denote the set of all set partitions of $\left[ n\right] $. For instance, \begin{align*} \operatorname*{SP}\left( 3\right) & ={\Large \{}\left\{ \left\{ 1,2,3\right\} \right\} ,\ \ \ \left\{ \left\{ 1\right\} ,\left\{ 2,3\right\} \right\} ,\ \ \ \left\{ \left\{ 2\right\} ,\left\{ 1,3\right\} \right\} ,\\ & \ \ \ \ \ \ \left\{ \left\{ 3\right\} ,\left\{ 1,2\right\} \right\} ,\ \ \ \left\{ \left\{ 1\right\} ,\left\{ 2\right\} ,\left\{ 3\right\} \right\} {\Large \}}. \end{align*} Note that $\left\vert \operatorname*{SP}\left( n\right) \right\vert $ is the $n$-th Bell number $B_{n}$. If $P=\left\{ P_{1},P_{2},\ldots,P_{k}\right\} $ is a set partition of $\left[ n\right] $ (with $P_{1},P_{2},\ldots,P_{k}$ being distinct), then we set $\ell\left( P\right) :=k$ and \begin{align*} b_{P}:=\prod_{i=1}^{k}\left( d\circ L_{a}\right) ^{\left\vert P_{i} \right\vert -1}\left( 1\right) . \end{align*} For example, if $n=5$ and $P=\left\{ \left\{ 1,4\right\} ,\left\{ 2,3,5\right\} \right\} $, then \begin{align*} b_{P}=\underbrace{\left( d\circ L_{a}\right) ^{2-1}\left( 1\right) }_{\substack{=\left( d\circ L_{a}\right) \left( 1\right) \\=d\left( a\right) }}\cdot\underbrace{\left( d\circ L_{a}\right) ^{3-1}\left( 1\right) }_{\substack{=\left( d\circ L_{a}\right) ^{2}\left( 1\right) \\=d\left( ad\left( a\right) \right) }}=d\left( a\right) \cdot d\left( ad\left( a\right) \right) . \end{align*} Of course, $b_{P}$ depends only on the sizes of the blocks of $P$. Now, my formulas claim that each integer $m\geq0$ and each $w\in R$ satisfy \begin{equation} \left( d\circ L_{a}\right) ^{m}\left( w\right) =\sum_{P\in \operatorname*{SP}\left( m+1\right) }b_{P}\cdot a^{\ell\left( P\right) -1}\cdot d^{\ell\left( P\right) -1}\left( w\right) \label{darij1.eq.damw} \tag{1} \end{equation} and \begin{equation} \left( L_{a}\circ d\right) ^{m}\left( w\right) =\sum_{P\in \operatorname*{SP}\left( m\right) }b_{P}\cdot a^{\ell\left( P\right) }\cdot d^{\ell\left( P\right) }\left( w\right) . \label{darij1.eq.admw} \tag{2} \end{equation} For instance, for $m=2$, these two formulas become \begin{align*} \left( d\circ L_{a}\right) ^{2}\left( w\right) & =\underbrace{b_{\left\{ \left\{ 1,2,3\right\} \right\} }}_{=\left( d\circ L_{a}\right) ^{2}\left( 1\right) }\cdot\underbrace{a^{1-1}}_{=1} \cdot\underbrace{d^{1-1}\left( w\right) }_{=w}\\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{ \left\{ 1\right\} ,\left\{ 2,3\right\} \right\} }}_{\substack{=\left( d\circ L_{a}\right) ^{0}\left( 1\right) \cdot\left( d\circ L_{a}\right) ^{1}\left( 1\right) \\=\left( d\circ L_{a}\right) ^{1}\left( 1\right) }}\cdot\underbrace{a^{2-1}} _{=a}\cdot\underbrace{d^{2-1}}_{=d}\left( w\right) \\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{ \left\{ 2\right\} ,\left\{ 1,3\right\} \right\} }}_{\substack{=\left( d\circ L_{a}\right) ^{0}\left( 1\right) \cdot\left( d\circ L_{a}\right) ^{1}\left( 1\right) \\=\left( d\circ L_{a}\right) ^{1}\left( 1\right) }}\cdot\underbrace{a^{2-1}} _{=a}\cdot\underbrace{d^{2-1}}_{=d}\left( w\right) \\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{ \left\{ 3\right\} ,\left\{ 1,2\right\} \right\} }}_{\substack{=\left( d\circ L_{a}\right) ^{0}\left( 1\right) \cdot\left( d\circ L_{a}\right) ^{1}\left( 1\right) \\=\left( d\circ L_{a}\right) ^{1}\left( 1\right) }}\cdot\underbrace{a^{2-1}} _{=a}\cdot\underbrace{d^{2-1}}_{=d}\left( w\right) \\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{ \left\{ 1\right\} ,\left\{ 2\right\} ,\left\{ 3\right\} \right\} }}_{\substack{=\left( d\circ L_{a}\right) ^{0}\left( 1\right) \cdot\left( d\circ L_{a}\right) ^{0}\left( 1\right) \cdot\left( d\circ L_{a}\right) ^{0}\left( 1\right) \\=1}}\cdot\underbrace{a^{3-1}}_{=a^{2}}\cdot\underbrace{d^{3-1}}_{=d^{2} }\left( w\right) \\ & =\left( d\circ L_{a}\right) ^{2}\left( 1\right) \cdot w+3\cdot\left( d\circ L_{a}\right) ^{1}\left( 1\right) \cdot ad\left( w\right) +a^{2}\cdot d^{2}\left( w\right) \end{align*} and \begin{align*} \left( L_{a}\circ d\right) ^{2}\left( w\right) & =b_{\left\{ \left\{ 1,2\right\} \right\} }\cdot a^{1-1}\cdot d^{1-1}\left( w\right) +b_{\left\{ \left\{ 1\right\} ,\left\{ 2\right\} \right\} }\cdot a^{2-1}\cdot d^{2-1}\left( w\right) \\ & =\left( d\circ L_{a}\right) \left( 1\right) +ad\left( w\right) ; \end{align*} both are easily verified. The right hand sides of the formulas \eqref{darij1.eq.damw} and \eqref{darij1.eq.admw} can be rewritten in terms of exponential Bell polynomials. As a result, the formulas take the forms (I hope I got them right) \begin{equation} \left( d\circ L_{a}\right) ^{m}\left( w\right) =\sum_{k=0}^{m} B_{m+1,k+1}\left( u_{0},u_{1},u_{2},\ldots\right) \cdot a^{k}\cdot d^{k}\left( w\right) \nonumber \end{equation} and \begin{align*} \left( L_{a}\circ d\right) ^{m}\left( w\right) =\sum_{k=0}^{m} B_{m,k}\left( u_{0},u_{1},u_{2},\ldots\right) \cdot a^{k}\cdot d^{k}\left( w\right) , \end{align*} where \begin{align*} u_{i}:=\left( d\circ L_{a}\right) ^{i}\left( 1\right) \ \ \ \ \ \ \ \ \ \ \text{for each integer }i\geq0. \end{align*} Proving the formulas is fairly easy. You can show \eqref{darij1.eq.damw} by induction on $m$ (using the Leibniz rule and the easy observation that a set partition $P\in\operatorname*{SP}\left( m\right) $ with $\ell\left( P\right) =k$ can be obtained from exactly $k+1$ different set partitions $P^{\prime}\in\operatorname*{SP}\left( m+1\right) $ by removing the element $m+1$). Then, \eqref{darij1.eq.admw} follows by applying \eqref{darij1.eq.damw} to $m-1$ and $d\left( w\right) $ instead of $m$ and $w$ (and multiplying the result by $a$). There are probably some more essential ways to prove the formulas -- in particular, the uncanny similarity to Faa di Bruno's formula cries out for an explanation -- but I am happy enough with the induction. To see why this all is related to the OP, let $R$ be the ring of functions, and let $a=f\left( x\right) $ and $d=\dfrac{d}{dx}$. Not sure if the results are of much use, though. Now, what is the connection to Hochschild? Let $p$ be a prime number, and let $w\in R$ be such that $pw=0$. (Usually, one just considers the case when $p=0$ in $R$, but all we need is $pw=0$.) Applying \eqref{darij1.eq.damw} to $m=p-1$, we obtain \begin{equation} \left( d\circ L_{a}\right) ^{p-1}\left( w\right) =\sum_{P\in \operatorname*{SP}\left( p\right) }b_{P}\cdot a^{\ell\left( P\right) -1}\cdot d^{\ell\left( P\right) -1}\left( w\right) . \label{darij1.eq.p-1} \tag{3} \end{equation} However, the cyclic group $\mathbb{Z}/p$ acts on the set $\operatorname*{SP} \left( p\right) $ (by cyclically rotating each number in each block -- i.e., the element $\overline{1}\in\mathbb{Z}/p$ sends a set partition $\left\{ \left\{ a_{1},a_{2},\ldots,a_{k}\right\} ,\left\{ b_{1},b_{2} ,\ldots,b_{\ell}\right\} ,\ldots,\left\{ g_{1},g_{2},\ldots,g_{r}\right\} \right\} $ to $\left\{ \left\{ a_{1}+1,a_{2}+1,\ldots,a_{k}+1\right\} ,\left\{ b_{1}+1,b_{2}+1,\ldots,b_{\ell}+1\right\} ,\ldots,\left\{ g_{1}+1,g_{2}+1,\ldots,g_{r}+1\right\} \right\} $, where addition happens modulo $p$). This action splits $\operatorname*{SP}\left( p\right) $ into orbits of sizes $1$ and $p$. Each orbit of size $p$ contributes a total of $0$ to the right hand side of \eqref{darij1.eq.p-1}, because the $p$ addends corresponding to the entries of this orbit are all equal and contain a $w$ (so summing them $p$ times yields $0$ since $pw=0$). What remains are the addends corresponding to the orbits of size $1$. These orbits are the ones of the set partitions $\left\{ \left\{ 1,2,\ldots,p\right\} \right\} $ and $\left\{ \left\{ 1\right\} ,\left\{ 2\right\} ,\ldots,\left\{ p\right\} \right\} $. The corresponding addends are $\left( d\circ L_{a}\right) ^{p-1}\left( 1\right) \cdot a^{1-1}\cdot d^{1-1}\left( w\right) =\left( d\circ L_{a}\right) ^{p-1}\left( 1\right) \cdot w$ and $\underbrace{\left( \left( d\circ L_{a}\right) ^{1-1}\right) ^{p}}_{=1^{p}=1}\cdot a^{p-1}\cdot d^{p-1}\left( w\right) =a^{p-1}\cdot d^{p-1}\left( w\right) $. Thus, \eqref{darij1.eq.p-1} simplifies to \begin{equation} \left( d\circ L_{a}\right) ^{p-1}\left( w\right) =\left( d\circ L_{a}\right) ^{p-1}\left( 1\right) \cdot w+a^{p-1}\cdot d^{p-1}\left( w\right) . \label{darij1.eq.hoch1} \tag{4} \end{equation} If we apply this to $d\left( w\right) $ instead of $w$, and multiply the result by $a$, we obtain \begin{equation} \left( L_{a}\circ d\right) ^{p}\left( w\right) =\left( d\circ L_{a}\right) ^{p-1}\left( 1\right) \cdot ad\left( w\right) +a^{p}\cdot d^{p}\left( w\right) . \label{darij1.eq.hoch2} \tag{5} \end{equation} In particular, if $p=0$ in $R$, then we thus have \begin{align*} \left( L_{a}\circ d\right) ^{p} & =\underbrace{\left( d\circ L_{a}\right) ^{p-1}\left( 1\right) \cdot a}_{=\left( L_{a}\circ d\right) ^{p-1}\left( a\right) }d+a^{p}\cdot d^{p}\\ & =\left( L_{a}\circ d\right) ^{p-1}\left( a\right) \cdot d+a^{p}\cdot d^{p} \end{align*} as maps $R\rightarrow R$. This is Proposition 1.1 in Andrzej Nowicki, Integral derivations, Journal of Algebra 110(1):262-276 (another scan at https://doi.org/10.1016/0021-8693(87)90045-7 ), where it is credited to Hochschild. Something similar does indeed appear as Lemma 1 in a 1955 paper by Hochschild, with a vague sketch of a proof I never decided whether to trust; the modern statement seems to have been first made by Cartier in Questions de rationalité des diviseurs en géométrie algébrique (see (43) therein). I am not sure if a complete proof without holes has ever appeared in the literature. Note that this is not the same as that other Hochschild identity, which is (42) in Cartier's op.cit. and doesn't seem to appear in Hochschild's work at all. But the latter identity, too, can be proved combinatorially using a $\mathbb{Z}/p$-action :)<|endoftext|> TITLE: Are lacunary functions still lacunary over rings larger than $\mathbb{C}$? QUESTION [5 upvotes]: Take a lacunary function of your choice ex: $$ f(z) = z + z^2 + z^4 + \cdots = \sum_{k=0}^\infty z^{2^n} $$ Obviously this cannot really be analytically or meromorphically continued outside the unit disk over $\mathbb{C}$. But suppose now that $z \in \mathbb{C}^{2 \times 2}$ the set of $2\times 2$ complex matrices or some other bigger ring of your choice which contains a copy of $\mathbb{C}$ We could define some notion of distance on this space (I like the matrix $2$-norm). And now using our norm define open balls of the form $|z - z_0| \le k$ and now that we have open sets we can build sheaves and attempt to do a form of analytic continuation in this higher space. So now the question is: does our function still resist continuation, or is it possible to extend this over the entire higher space? Have people looked at problems like this before? REPLY [5 votes]: $f:M_k(\Bbb{C})\to M_k(\Bbb{C})$ is analytic on $\{ A\in M_k(\Bbb{C}),\sigma(A)<1\}$ and not beyond, where $\sigma(A)=\sup |\lambda_j(A)|$ is the spectral radius. If $\sigma(A)< 1$ then write the Jordan normal form $A=P (D+L)P^{-1}$ where $D$ is diagonal, $\|D\|_\infty=\sigma(A), DL=LD, L^k=0$ so that $A^n = P \sum_{m=0}^{k-1} {n\choose m} L^m D^{n-m} P^{-1}, \|A^n\|_\infty = O(n^k \sigma(A)^n)$ and hence the power series converges absolutely. If $\sigma(A)\ge 1$ then we are assuming that $f$ is analytic on a curve $0\to A$ so it will be analytic on a small disk around some $B$ such that $\sigma(B)=1$. In this disk there is some $C$ with $\sigma(C)=1$ and an eigenvalue $\exp(2i\pi s/2^t)$ with eigenvector $v$ so that $\lim_{r\to 1^-} \|f(r C) v\|=\infty$ contradicting that $f$ extends continuously to $C$. No idea for more general infinite dimensional (Banach) $\Bbb{C}$-algebras.<|endoftext|> TITLE: Cutting a spherical surface into mutually non-congruent pieces of equal area QUESTION [7 upvotes]: Question: For what values of integer $n$ can the surface of a sphere be partitioned into $n$ convex and mutually non-congruent pieces of same area? (convexity could be viewed as geodesic convexity). If such a partition exists for some $n$'s, one can ask if one can achieve a 'fair' convex partition - pieces with same area and same perimeter - for some of those n's. Note: one can replace sphere by general ellipsoids or tori or polyhedrons. Speculation: Even a negative result like: "the spherical surface (or maybe even any ellipsoid or say, toroid) cannot be cut into any number of mutually non-congruent convex pieces of same area (or for that matter, 'same area and same perimeter')" could be interesting. REPLY [5 votes]: There is no solution for $n=4$, by the following result. Proposition. If $ABC$ is a spherical triangle of area $\pi$, then there is a partition of the sphere into four copies of $ABC$, and that is the unique partition into four triangles of area $\pi$ with $ABC$ as one of the triangles. Proof of existence. By the relation of area and excess, the angles within $ABC$ sum to $2\pi$. So consider a point $P$ with $\angle ABP= \angle BCA$ and $BP=AC$. Then \begin{align} \angle PBC &= 2\pi - \angle ABC - \angle ABP \text{, since the angles at }B\text{ sum to }2\pi\\ &= 2\pi - \angle ABC - \angle BCA \text{, by construction}\\ &= \angle CAB,\phantom{2\pi - \angle ABC - }\text{ since the angles in }ABC\text{ sum to }2\pi. \end{align} From this equality we can prove others to show that the triangles $ABC$, $APB$, $BPC$, $CPA$ are congruent, and therefore all of area $\pi$. The final result has the following diagram —- where, because this is a sphere, the area outside the triangle is also a triangle. Proof of uniqueness. Suppose $ABC$, $AOB$, $BOC$, $COA$ are all of area $\pi$. If $AOB \subset APB$ or $BOC \subset BPC$ or $COA \subset CPA$ then the triangle with $O$ would have area of less than $\pi$. Since this is impossible, $O$ must avoid all those inclusions, and therefore $O=P$. $\square$<|endoftext|> TITLE: Mathematical journals that accept long papers (up to 100 pages) QUESTION [25 upvotes]: I have a problem with finding a suitable mathematical journal (maybe not of the highest level) that accepts longer paper (say 100 pages). I know only two such journals: Memoirs of AMS and Dissert.Math. But both journals now are in Q1 quartile, so are very selective, without any chances to be accepted there. In fact, "Dissertationes Mathematicae" was created in 1951 (as "Rozprawy Matematyczne") with the purpose of publishing good Ph.D. Theses and indeed some time ago it was quite real to be accepted there. But since 2019 when they entered to the Q1 quartile this journal accepts only papers of exceptional quality (so far higher than an average Ph.D. Thesis). So, what would be your suggestion concerning journals for longer papers (at the moment I would not like to consider splitting the long paper into smaller pieces). Thanks for the advise (I suspect that many mathematicians had such problems in mind, so the collective experience had to elaborate some helpful answer here). REPLY [2 votes]: I recall here my answer to a similar question: a journal which accepts long papers and 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". I recall the statements one can read on the homepage of the Journal: 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.<|endoftext|> TITLE: Are hyperbolic spaces actually better for embedding trees than Euclidean spaces? QUESTION [10 upvotes]: There is a folklore in the empirical computer-science literature that, given a tree $(X,d)$, one can find a bi-Lipschitz embedding into a hyperbolic space $\mathbb{H}^n$ and that $n$ is "much smaller" than the smallest dimension of a Euclidean space in which $(X,d)$ can be bi-Lipschitz embedded with similar distortion. Question A: Is there any theoretical grounding to this claim? Namely, can one prove that $(X,d)$ (where $\# X = n\in\mathbb{N}_+$) admits a bi-Lipschitz embedding into some $\mathbb{H}^n$ with: distortion strictly less that $O(\log(n))$ $n0$ does there exist a finite metric space $(X,d)$, which don't admit a bi-Lipschitz (resp. possibly uniform embedding) into $\mathbb{H}^n$ with distortion at-most $D$? Relevant Definition (For completeness) A bi-Lipschitz embedding $f:X\rightarrow \mathbb{H}^n$ of a metric space $(X,d)$ into $\mathbb{H}^n$ with distortion $D>0$ is a Lipschitz homeomorphism $f:X\rightarrow \mathbb{H}^n$ Lipschitz inverse $f^{-1}$ such that there is some $s>0$ satisfying $$ sd(x_1,x_2) \leq d_{\mathbb{H}^n}(f(x_1),f(x_2)) \leq sDd(x_1,x_2) \qquad (1) $$ for every $x_1,x_2\in X$. Here, $d_{\mathbb{H}^n}$ is the usual geodesic distance on the $n$-dimensional hyperbolic space. Some Relevant posts: Hyperbolic embeddings References to embedding into hyperbolic spaces Representability of finite metric spaces Flat Embeddings Problem with embedding expanders into "flat" spaces Characterizing finite metric spaces which embed into Euclidean space Uniform Embeddings Notes on coarse and uniform embeddings REPLY [3 votes]: Here is a trivial example for question B (in the $s=1$ case): The discrete metric space on $N$ points with distance 1 between every two distinct points has the minimal distortion of an embedding into $\mathbb H^n$ going to $\infty$ as $N$ goes to $\infty$ with respect to $n$. Indeed, for an embedding of distortion $D$, any two points in the embedding must have distance at least $1$, so the balls of radius $1/2$ around these points must be disjoint. But every two points have distance at most $D$, so the balls of radius $1/2$ around these points must be contained in the ball of radius $D+1/2$ around one point. Thus, for an embedding to exist, the volume of the ball of radius $D+1/2$ must be at least $N$ times the volume of the ball of radius $1/2$. Thus $D$ must go to $\infty$ if $N$ goes to $\infty$ with fixed $n$.<|endoftext|> TITLE: Norm of contragredient of unitary representations of compact quantum groups QUESTION [6 upvotes]: Maybe the answer to the following question is known but I am unable to find it in the literature. Anyway, let me begin my question by fixing some notations and terms. Let $G = (A, \Delta)$ be a compact quantum group in the sense of Woronowicz. For a finite dimensional unitary representation $U \in B(H) \otimes A$ of $G$, the contragredient $U^c$ of $U$ is the representation $(j \otimes \operatorname{Id})(U^*) \in B\bigl(\overline{H}\bigr) \otimes A$ of $G$, where the carrier space $\overline{H}$ is the conjugate Hilbert space of the finite dimensional Hilbert space $H$, and $j : B(H) \to B\left(\overline{H}\right)$ the $\ast$-anti-isomorphism sending $T$ to $\overline{T^*}$, the latter denotes the operator $\overline{\xi} \mapsto \overline{T^* \xi}$ on $\overline{H}$. Define $c(G)$ to be the supremum of all $\| U^c \|$, where $U$ runs through all finite dimensional unitary representations (of course, irreducible ones suffice) of $G$. It is clear that if $G$ is of Kac-type, then $c(G) = 1$, as $U^c$ in the above is always unitary. With some effort, I can prove that $c(G) = +\infty$ for $G = SU_q(2)$ with $-1 < q \ne 0 < 1$, but whether $c(G) = +\infty$ for general non-Kac type $G$ seems more delicate, which prompts me into asking the following Question. Does $c(G) < +\infty$ imply $G$ being of Kac type? As there are already many results concerning characterization of Kac type compact quantum groups, it may well be possible that this question is already settled, in which case, I appreciate a reference to the literature. REPLY [6 votes]: Yes, $c(G) < +\infty$ implies that $G$ is of Kac type. As far as I know, this result is not in the literature, but can be proven as follows. Let $h$ be the Haar state. For every irreducible unitary representation $u \in M_n(\mathbb{C}) \otimes A$, there is a unique positive invertible $Q_u \in M_n(\mathbb{C})$ with $\operatorname{Tr}(Q_u) = \operatorname{Tr}(Q_u^{-1})$ and giving the orthogonality relations $$h(u_{ij} u_{kl}^*) = \delta_{i,k} \, (Q_u)_{jl} \, \operatorname{Tr}(Q_u)^{-1} \; .$$ Note that $G$ is of Kac type if and only if $Q_u = 1$ for all irreducible unitary representations $u$. This positive $Q_u$ also unitarizes $u^c$. First defining the matrix $u^c \in M_n(\mathbb{C}) \otimes A$ by $(u^c)_{ij} = (u_{ij})^*$, we get that $(Q_u^{1/2} \otimes 1) u^c (Q_u^{-1/2} \otimes 1)$ is a unitary representation that we denote as $\overline{u}$. This is the unitary contragredient of $u$. Proposition. The following statements are equivalent. $G$ is of Kac type. There exists a $C > 0$ such that $\|Q_u\| \leq C$ for every irreducible unitary representation $u$. There exists a $C > 0$ such that $\|u^c\| \leq C$ for every irreducible unitary representation $u$. Proof. When $u$ is an arbitrary finite dimensional unitary representation, we can still canonically define $Q_u$ by decomposing $u$ as a sum of irreducibles. We have $Q_{u \otimes v} = Q_u \otimes Q_v$. Also, $Q_{\overline{u}} = Q_u^{-1}$. For every unitary representation $u \in M_n(\mathbb{C}) \otimes A$, we denote $\dim u = n$ and $\dim_q u = \operatorname{Tr}(Q_u)$. The orthogonality relations then say that $$(\mathord{\text{id}} \otimes h)((u^c)^* u^c) = \frac{\dim u}{\dim_q u} \, Q_u \; .$$ 1 $\Rightarrow$ 2 is trivial. 2 $\Rightarrow$ 3. Since $Q_{\overline{u}} = Q_u^{-1}$, also $\|Q_u^{-1}\| \leq C$. By definition, $\|u^c\| \leq C$ for every irreducible unitary representation $u$. 3 $\Rightarrow$ 1. Assume that $G$ is not of Kac type. Fix $C > 0$. We construct an irreducible unitary representation $u$ with $\|u^c\| > C$. Since $G$ is not of Kac type, we can choose an $n$-dimensional irreducible unitary representation $v$ such that $Q_v \neq 1$. Define $q_{\max}$ as the largest eigenvalue of $Q_v$. Since $Q_v \neq 1$, we get that $\operatorname{Tr}(Q_v) < q_{\max} \, n$. Take an integer $k \geq 1$ such that $(\operatorname{Tr}(Q_v)^{-1} \, q_{\max} \, n)^k > C^2$. Define $w$ as the $k$-fold tensor power of $v$. Then, $$(\mathord{\text{id}} \otimes h)((w^c)^* w^c) = \frac{\dim w}{\dim_q w} \, Q_w = \Bigl(\frac{\dim v}{\dim_q v}\Bigr)^k \, Q_v^{\otimes k} \; .$$ By construction, the operator norm of the right hand side is strictly larger than $C^2$. The operator norm of the left hand side is the maximum of $\| (\mathord{\text{id}} \otimes h) ((u^c)^* u^c)\|$, where $u$ runs through the irreducible subrepresentations of $w$. So, there exists an irreducible $u$ such that $\| (\mathord{\text{id}} \otimes h) ((u^c)^* u^c)\| > C^2$. It follows that $\|u^c\| > C$.<|endoftext|> TITLE: Equivalence between geometric theories and frames internal to the free topos QUESTION [9 upvotes]: What is a reference for "the equivalence between geometric theories and frames internal to the free topos"? This seems to be an extremely interesting theorem. REPLY [9 votes]: Simon essentially answered the question already, but I will expand some of the parts that may not be clear to the experts. Sketches of an elephant is a good reference for everything I am going to say. I will take a bit of a different narrative to Simon's, implicitly assuming a good understanding of classifying topoi, blurring the distinction between a topos and the geometric theory it classifies, $$\mathcal{E} \simeq \text{Set}[\mathbb{T}] .$$ Def (Prebounds). Let $\mathcal{E}$ be a topos. A prebound $e \in \mathcal{E}$ is an object such that the subobjects of its finite powers $m: a \to e^n$ are a generator for the topos. Such an object always exist and can be obtained by manipulating a generator (or a site). Construction (From prebounds to localic geometric morphisms). Given a couple $(\mathcal{E},e)$ where $\mathcal{E}$ is a topos and $e$ is a prebound, we can construct a localic geometric morphism $$f_e: \mathcal{E} \to \text{Set}[\mathbb{O}]. $$ Of course, this is the same of a cocontinuous left exact functor $f_e^*: \text{Set}[\mathbb{O}] \to \mathcal{E}$, which is the same of a lex functor $\text{Fin}^\circ \to \mathcal{E}$, $$\text{Topoi}(\mathcal{E}, \text{Set}[\mathbb{O}]) \simeq \text{Cocontlex}( \text{Set}[\mathbb{O}], \mathcal{E}) \simeq \text{Cocontlex}( \text{Set}^{\text{Fin}}, \mathcal{E})\simeq \text{Lex}(\text{Fin}^\circ, \mathcal{E}). $$ The latter, is given by sending $n \mapsto e^n$. The geometric morphism obtained in this way is localic by definition of prebound. This construction appeared for the first time in Freyd's All topoi are localic. Remark. If you think about it, I am just spelling out in categorical terms what Simon suggested in somewhat mystical language. Remark (Morita-like phenomena). Notice that each prebound (and we can construct a prebound from any site) gives a different localic morphism, thus we have many localic representation for the same topos! Remark (Topoi are geometric theories, generators are their presentation). Following Thm 2.1.1 in Caramello's Theories, Sites, Toposes, we see that a generator, or a site, is essentially the same of a linguistic presentation of the geometric theory classified by the topos. Theorem (Internal locales are localic geometric morphisms). There is a biequivalence of categories between the $2$-category of internal locales in $\text{Set}[\mathbb{O}]$ and the $2$-category of localic geometric morphisms over $\text{Set}[\mathbb{O}]$, $$\text{Loc}(\text{Set}[\mathbb{O}]) \leftrightarrows \text{Topoi}_{\text{loc} / \text{Set}[\mathbb{O}]}. $$ Proof. Lemma 1.2 in Johnstone, Factorization theorems for geometric morphisms. Cahiers, 22, no1 (1981) Remark (On the emergence of Lawvere-like doctrines). When one spells out what a locale internal to $ \text{Set}[\mathbb{O}]$ is, one discovers that it is nothing but a functor $$\mathbb{P}: \text{Fin} \to \text{Heyt}$$ verifying the Beck-Chevalley condition and Frobenius reciprocity (see Lemma C.1.6.9 and Cor. C.1.6.10 in Sketches of an Elephant). Suddenly we see how doctrine-like objects emerge in the representation of theories! That's beautiful in my opinion. $\text{Fin}$ acts as a fact as a set of variables, while $\mathbb{P}(n)$ gives us the set of formulas on those $n$-variables. Def (Well presented topoi). The $2$-category WTopoi of well presented topoi has objects $(\mathcal{E},e)$ where $\mathcal{E}$ is a topos and $e$ is a prebound and morphism geometric morphisms whose left adjoint preserve the prebuound. Remark. This notion does not appear in the literature (to my knowledge), I just need it as an intermediate notion. A good intuition for it is that the topos is specified together with a precise language generator of the geometric theory it classifies. WTopoi is really much more a $2$-category of sites, together with a relational notion of morphism of sites, rather than a $2$-category of topoi. Remark (Every topos can be well presented). Of course, the WTopoi is not the same of Topoi but the forgetful functor $$\mathsf{U}: \text{WTopoi} \to \text{Topoi} $$ is essentially surjective on objects, and on morphisms (!). Theorem (Internal locales are well presented topoi and vice versa). There is a biequivalence of categories $$\text{Loc}(\text{Set}[\mathbb{O}]) \leftrightarrows \text{Topoi}_{\text{loc} / \text{Set}[\mathbb{O}]} \leftrightarrows \text{WTopoi}.$$<|endoftext|> TITLE: Quadrisecants of knots QUESTION [9 upvotes]: Recall that a quadrisecant of a knot is a line that passes thru four points on it. If the points appear on the line in the order $a$, $b$, $c$, $d$ and on the knot in the order $a$, $c$, $b$, $d$, then the quadrisecant is called alternated. It is well known that any nontrivial knot has an (alternated) quadrisecant. The statement looks very elementary, but all proofs I saw are quite involved, and it surprises me. Is there an elementary proof that every knot has an (alternated) quadrisecant? REPLY [9 votes]: Here is how I like to think of these kinds of geometric problems. Rather than working in $\Bbb R^3$, let's work in $S^3$. Given a knot $K$ in $S^3$, there is the configuration space of points in $K$. Let's consider $5$ points in $K$, i.e. $C_5 K$. This is a 5-manifold, diffeomorphic to a trivial 4-ball bundle over $K$. In $C_5 S^3$ there is the subspace where all five points sit on a round circle. By "round circle" I mean the intersection of a 2-dimensional hyperplane in $\Bbb R^4$ with $S^3$. Let's call this the Round Circle Subspace, $RSS \subset C_5 S^3$. $RSS$ has dimension $11$, or co-dimension $4$. $RSS$ has a subspace that I call the pentacle subspace. The idea is that the five points have a natural cyclic ordering $1 \to 2 \to 3 \to 4 \to 5 \to 1$. But five points on a circle also have a cyclic ordering. So we take the path-component of $RSS$ where adjacent points in the natural (or "label") ordering are not adjacent in the circle ordering. i.e. if you draw lines from $p_i$ to $p_{i+1}$ for all $i$ (index mod $5$) then you get a pentacle, provided you also draw the circle they are on. Given a knot, generically the pentacles in $C_5 K$ are a $1$-dimensional submanifold. If you stereographically project at a point of $S^3$ that is a point of a pentacle, that pentacle is converted into an alternating quadrisecant for the projected knot -- which will now be a long knot, i.e. running off to infinity. In my opinion this setup makes a lot of formal sense. If you can prove non-trivial long knots have alternating quadrisecants, it follows for closed knots by a closure operation. A long knot you could view as a smooth embedding $\mathbb R \to \mathbb R^3$ that has a fixed axis as an asymptote. Define the collinear triple subspace of $\mathbb R^3$ as the space of triples that sit on a common straight line, as a subspace of $C_3 \mathbb R^3$. This space has three path components, usually denoted $Col_i$ where $p_i$ is in the middle of the linear ordering of the line. Given a long knot $f : \mathbb R \to \mathbb R^3$ one takes the induced map $f_* : C_3 \mathbb R \to C_3 \mathbb R^3$ and the preimages $f_*^{-1}(Col_1)$ and $f_*^{-1}(Col_3)$. $C_3 \mathbb R$ is basically a copy of $\mathbb R^3$ and the alternating quadrisecants correspond to double points of $f_*^{-1}(Col_1)$ passing under $f_*^{-1}(Col_3)$, when forgetting the first coordinate of $C_3 \mathbb R$. You can also do it as a crossing count forgetting the third coordinate. This does not answer your question, but it gives some context. If you don't have an alternating quadrisecant, it means $f_*^{-1}(Col_1)$ is purely "over top of" $f_*^{-1}(Col_3)$, from the point of view of projecting out the first coordinate of $C_3 \mathbb R$. To me this is the primary geometric meaning associated to not having alternating quadrisecants. edit: I should emphasize one other point. Non-trivial knots having alternating quadrisecants is equivalent to stating that non-trivial knots in $S^3$ have pentacles. The space of pentacles in $C_5 K$ is generically an oriented $1$-manifold. If you consider the forgetful map $C_5 K \to K$ that forgets all but one point, then the degree of this map, restricted to the pentacle subspace, is the type-$2$ Vassiliev invariant of the knot, i.e. the coefficient of $z^2$ in the Conway normalization of the Alexander polynomial.<|endoftext|> TITLE: Can a countable union of two-element sets be uncountable? QUESTION [26 upvotes]: I am thinking about the Axiom of Choice and I am trying to understand the Axiom with some but a little progress. Many questions are arising in my head. So, I know that there exists a model of ZF set theory in which the set of real numbers, which is provably uncountable, is a countable union of countable sets. Question: does there exist a model of ZF set theory for which there exists a collection $A_n$, $n\in\mathbb{N}$, of pairwise disjoint two-element sets such that their union is not countable? Some thoughts. Let $A_n$, $n\in\mathbb{N}$, be a collection of pairwise disjoint two-element sets. Then for every $n\in\mathbb{N}$ there exists a bijection $f:\{1,2\}\to A_n$. But when we want to prove that $\bigcup_{n\in\mathbb{N}}A_n$ is countable, we have to choose a countable number of bijections $f_n:\{1,2\}\to A_n$, $n\in\mathbb{N}$, at once (simultaneously). After this we plainly define the bijection $f:\mathbb{N}\to\bigcup_{n\in\mathbb{N}}A_n$ by $f(1):=f_1(1)$, $f(2):=f_1(2)$, $f(3):=f_2(1)$, $f(4):= f_2(2)$, and so on. Rigorously, we write $f(k)=f_l(1)$ if $k=2l-1$ and $f(k)=f_l(2)$ if $k=2l$. Clearly, $f$ is a bijection and we are done. But without the Axiom of Countable Choice we can not choose $f_n$, $n\in\mathbb{N}$, simultaneously and the argument does not work. It is worth mentioning that if $A_n$ are subsets of $\mathbb{R}$, then we can choose $f_n$, $n\in\mathbb{N}$, simultaneously. Indeed, we can define $f_n(1):=\min A_n$ and $f_n(2):=\max A_n$, $n\in\mathbb{N}$, and the natural proof given above works. So if a counterexample exists, the sets $A_n$, $n\in\mathbb{N}$, have to be "abstract", say pairs of socks. REPLY [10 votes]: The point of this answer is to draw attention to the easy proposition below and the historical remark that follows it. In this answer "countable" means countably infinite (the finite case is trivial since within $\mathrm{ZF}$-set theory, a simple induction on the cardinality of $F$, where $F$ is a finite set of finite sets, shows that the union of $F$ is finite). Let $\mathrm{CUPC}$ denote the statement: "every countable disjoint union of pairs (two-element sets) is countable", and consider the weak form of the axiom of choice that is often denoted $\mathrm{C}^{\omega}_{2}$, which states: "every countable disjoint family of 2-element sets has a choice function". Note that it is precisely $\mathrm{C}^{\omega}_{2}$ that is alluded to in Russell's infinitely many pairs of socks set-up. Proposition. Within $\mathrm{ZF}$-set theory, $\mathrm{C}^{\omega}_{2}$ is equivalent to $\mathrm{CUPC}$. Proof. Suppose $P$ is a countable disjoint family of pairs (two-element sets), thus each $p\in P$ has two elements, and there is a bijection $f:\omega \to P$. We will show that $P$ has a choice function iff the union $\cup_{n \in \omega} f(n)$ of members of $P$ form a countable set. Suppose $\mathrm{C}^{\omega}_{2}$ holds. Then there is (choice) function $C:\omega \to \cup_{n \in \omega} f(n)$ such that $C(n) \in f(n)$ for each $n \in \omega$. To see that $\cup_{n \in \omega} f(n)$ is countable we simply note that the function $g:\omega \to \cup_{n \in \omega} f(n)$ is a bijection, where $g(2n) = C(n)$ and $g(2n+1) = b$, where $f(n)\setminus \{C(n)\} = \{b\}$ (i.e., $b$ is the element of $f(n)$ that is not chosen by the choice function $C$). Now suppose $\mathrm{CUPC}$. Then there is a bijection $g:\omega \to \cup_{n \in \omega} f(n)$. This allows us to define the desired choice function $C:\omega \to \cup_{n \in \omega} f(n)$ via $C(n)=g(k)$, where $k$ is the least $m\in \omega$ such that $g(m)\in f(n)$. Historical Remark. The independence of $\mathrm{C}^{\omega}_{2}$ from $\mathrm{ZFA}$, i.e., set theory with atoms/urelements was first established by Abraham Fraenkel in the 1920s, his technique was extended by Andrzej Mostwoski in the 1930s, in the form of what is nowadays known as Fraenkel-Mostowski permutation models (the modern expositions of this technique employ the further conceptual machinery of "filters" introduced by Ernst Specker in the 1950s). Several decades later, in the early 1960s, Paul Cohen invented the method of forcing, and was able to transplant the Fraenkel-Mostwoski independence to the $\mathrm{ZF}$-setting. In "Cohen's second model", there is a countable family $P$ of 2-element sets $p$ such that each member of $p$ is a subset $\mathbb{R}$ with the property that $P$ has no no choice function, i.e., each "Russell sock" is a collection of real numbers. Postscript: It is well-known that the assumption of disjointness can be removed from the choice principle discussed here, and other similar situations, by the trick of replacing a family of sets $\mathcal{A}$ which might have intersecting members with the family of disjoint sets $\mathcal{A}^{*}$ that results from replacing each $A \in \mathcal{A}$ with $A \times \{A\}$, and noting that $\mathcal{A}^{*}$ has a choice function iff $\mathcal{A}$ does.<|endoftext|> TITLE: Ways of proving normal distribution (with a view towards Selberg's central limit theorem) QUESTION [8 upvotes]: Given an random variable $Y:\Omega \to \mathbb{R}$ with finite mean $\mu$ and finite, positive variance $\sigma^2$, let $X = \frac{Y-\mu}{\sigma}$ be the renormalization with mean $0$ and variance $1$. what are some general techniques for showing that $Y$ has a normal distribution? That is, $$P(X\leqslant a) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^a e^{-t^2/2}\,dt.$$ The standard technique I know is to compute the moments or cumulants and then use the fact that the normal distribution is characterized by its moments/cumulants. Are there any other general techniques, and what are their advantages and disadvantages? The motivation for this question is number theoretic (as with this related question), hence the number theory tags. Specifically, the motivating theorem is Selberg's central limit theorem (first published in Tsang's thesis, see also this article of Radziwill-Soundararajan) which states that for large $T$, the real valued random variable on $[T,2T]$ given by $t \mapsto \log|\zeta(\tfrac12 + it)|$ is approximately normally distributed with mean $0$ and variance $\frac{1}{2}{\log\log T}$. Both the proofs I know of (Selberg's original proof, and that of Radziwill-Soundararajan) use the method of moments. Morally speaking, the analytic-number-theoretic input goes into showing that the contributions from the zeros of zeta can be controlled, and hence at least for the distributional question with $t \in [T,2T]$, $$\log|\zeta(\tfrac12 + it)| \simeq \Re\sum_{p\leqslant T^{o(1)}} \frac{1}{p^{1/2 + it}}.$$ One can then compute the moments of the right hand side and show that as $T \to \infty$, the moments appropriately normalized converge to the moments of a standard Gaussian. The hope is to see if there's a way to prove Selberg's CLT in a situation where the moments are harder to compute, and so the method of moments may not be tractable. REPLY [2 votes]: There is also the information-theoretic method; see Linnik (1959), Barron (1986), etc. Oliver Johnson has written an entire book on this topic.<|endoftext|> TITLE: Is a "separable" algebra over a field finite-dimensional? QUESTION [10 upvotes]: Let $k$ be a field and $A$ a unital associative (possibly non-commutative) $k$-algebra, and let $A^e$ denote the enveloping algebra of $A$, namely, $A^e = A \otimes_k A^{op}$. It seems that there are two typical definitions for separable $k$-algebras. Consider the following two conditions. $A$ is projective as an $A^e$-module. For any field extension $K$ of $k$, the algebra $A \otimes_k K$ is semsimple. Also consider the following condition. $A$ is finite-dimensional over $k$ and the condition 2 is satisfied. In some literature (e.g. Corollary 10.6 in Pierce's Associative Algebra), it is shown that 1 is equivalent to 3. I wonder whether 2 is actually equivalent to 1 and 3, that is, Question: Suppose that the condition 2 is satisfied. Then is $A$ finite-dimensional over $k$? I found some people say that this question is true, and even adopt the condition 2 as the definition of separable algebras: e.g. nLab's article on separable algebra, and Lemma 3.3 in Reyes-Rogalski's Graded twisted Calabi-Yau algebras are generalized Artin-Schelter regular. I looked up some cited references, but it seems that in the proofs, $A$ is assumed to be finite-dimensional. I'm not sure whether the question is true even when $A$ is commutative, or $A$ is a filed. REPLY [5 votes]: I claim that no infinite dimensional algebra $A$ over $k$ satisfies 2. I am indebted to the comment of @UriyaFirst for the idea of the main step in the proof. First I claim we may assume that $k$ is algebraically closed. Indeed, if $[A:k]=\infty$ and satisfies 2, then for an algebraic closure $\overline k$ of $k$ we have that $A'=A\otimes_k \overline k$ is infinite dimensional over $\overline k$ and satisfies 2 over $\overline k$ by transitivity of extension of scalars. So assume $k$ is algebraically closed and $[A:k]=\infty$. We show that $A$ does not satisfy $2$. If $A$ is not semisimple, then it fails 2, so we may assume that $A$ is a direct product of matrix algebras over division algebras and at least one of these division algebras $D$ is infinite dimensional over $k$. Since extension of scalars commutes with direct product and matrix algebras, it suffices to show that $D\otimes_k K$ is not semisimple for some field extension $K/k$. The key observation is that if $F/k$ is an infinite extension, then $F\otimes_k F$ is either not Artinian or not Noetherian. If $F/k$ is not finitely generated, then $F\otimes_k F$ is not Noetherian by Theorem 11 of P. Vámos, On the minimal prime ideals of a tensor product of two fields, Mathematical Proceedings of the Cambridge Philosophical Society, 84 (1978), pp. 25-35. If $F/k$ is finitely generated but infinite dimensional then it has positive transcendence degree. The Krull dimension of $F\otimes_k F$ is the transcendence degree of $F/k$, by Rodney Y. Sharp, The Dimension of the Tensor Product of Two Field Extensions, Bulletin of the London Mathematical Society 9 Issue 1 (1977) pp 42–48. Thus $F\otimes_k F$ is not Artinian in this case. There are two cases. If $[Z(D):k]=\infty$, then $D\otimes_k Z(D)$ has center $Z(D)\otimes_k Z(D)$ by general facts on centers of tensor products over a field, and so the center of $D\otimes_k Z(D)$ is not a finite direct product of fields by the above observation (since it fails either the Artinian or Noetherian condition). But the center of a semisimple ring is a direct product of fields by Wedderburn-Artin, contradiction. So assume $[Z(D):k]<\infty$. Then $Z(D)=k$ because $k$ is algebraically closed. Let $\alpha\in D\setminus k$. Then $\alpha$ is transcendental over $Z(D)=k$ and so $k\leq K=k(\alpha)$ with $K/k$ a purely transcendental extension. Now using @UriyaFirst comment (but with a descending chain argument), we have that $K\otimes_k K$ is not Artinian by the above and so has an strictly descending chain of ideals $I_1\supsetneq I_2\supsetneq \cdots$. Now $D\otimes_k K$ is a free right $K\otimes_k K$-module (since $D$ is a free $K$-module) and so the $(D\otimes_k K) I_k$ form a strictly descending chain of left ideals in $D\otimes_k K$. Thus $D\otimes_k K$ is not left Artinian and hence not semisimple.<|endoftext|> TITLE: Double ($p$-fold) coverings of $B^4$ along ribbon/slice disks QUESTION [6 upvotes]: I have two questions that seem to be related. I wonder if there is a user-friendly algorithm (starting from ribbon/slice presentation of knots/disks) for the construction of double (in general $p$-fold) coverings of $B^4$ along ribbon/slice disks. Using the handle decomposition of ribbon/slice disks can we control the handle decomposition of double (in general $p$-fold) coverings of $B^4$ along these disks? REPLY [3 votes]: I would like to popularize the lecture notes of Brendan Owens about the double branched cover of knots in $S^3$ and of ribbon/slice disks in $B^4$. He both discussed constructions and obstructions. Also, it includes the classical references listed by Professor Ruberman. Owens, Brendan. "Knots and 4-manifolds." Winter Braids Lecture Notes 6 (2019): 1-26. It is avaliable online here.<|endoftext|> TITLE: Quotient of a quotient stack: interesting examples? QUESTION [5 upvotes]: Let $X$ be a scheme acted on by an algebraic group $G$. Also, let $H$ be an algebraic group acting on the quotient stack $X/G$, for the definition of "act", see Romagny - Group Actions on Stacks and Applications. I have for a long time been extremely confused about the 2-stack $(X/G)/H$, e.g. when it is a quotient stack, and if it isn't what it can look like. I think to not get bogged down I'll ask my question in a simple case. Question: Take $X=V$ a vector space acted on linearly by $G\to\operatorname{GL}(V)$, and $H$ an algebraic group. What are examples of $H$ acting on $V/G$ (1a. acting weakly but not strongly) (1b. acting strongly)? What is $(V/G)/H$ in these examples? In general, when is this double quotient a quotient stack, i.e. of the form $V/G'$? REPLY [7 votes]: [Edit: removed some nonsense in first version where I conflated $J$ and $BJ$.] There's an important example in representation theory of something the situation you're asking about, i.e., a quotient stack $V/G$ with a symmetry group[-scheme] $J$ over $V$ acting inertially (i.e., mapping equivariantly to the family of stabilizer groups, or to the inertia group-scheme of the quotient stack), so that automorphisms of points in $V$ themselves have automorphisms. This comes (by ``looping") from an action of a 2-group (or group-stack) $BJ$ on the stack $V/G$, and the quotient is an honest 2-stack. This example is due to Ngô as part of his proof of the Fundamental Lemma https://arxiv.org/abs/0801.0446 (the 2-stacky formulation I learned from a lecture of Drinfeld, http://math.uchicago.edu/~drinfeld/langlands/Regular_centralizers.pdf - which says it more clearly than I could). In this example $V=\mathfrak g$ is a reductive Lie algebra (eg $\mathfrak sl_2$), $G$ is a corresponding group (e.g. $SL_2$), but $H=J$ is not quite a group, it's a family of groups, i.e. a group-scheme over $V$ -- namely the group-scheme $J$ of regular centralizers. It's defined as follows: for any $x\in \mathfrak g$ take the centralizer of a regular element with the same characteristic polynomial as $x$ (here regular means the dimension of this centralizer is the smallest possible, ie the rank of $\mathfrak g$, or that this centralizer is commutative, or for matrices that minimal polynomial=char.polynomial). Turns out this is a (well-defined) family of commutative groups over $\mathfrak g$. For $GL_n$ this is a simple classical construction -- $J_x$ is just invertible functions on the spectrum of $x$. Ngô showed that $J$ acts inertially on $\mathfrak g / G$ (and that the action deloops to an action of the group-stack $BJ$ on $\mathfrak g/G$). More concretely, he showed the tautological isomorphism between $J_x$ and the inertia of the $G$-action (the centralizer $I_x\subset G$ of $x$) for $x$ regular extends uniquely (by Hartogs' theorem) to a map from $J$ to the inertia group-scheme $I$ of $G$ acting on $\mathfrak g$. Moreover this action deloops to an action of $BJ$ on the stack $\mathfrak g/G$. For $GL_n$ again this is something fairly classical. The inertial action is the statement that invertible functions on the spectrum of a matrix give (by "functional calculus") operators commuting with that matrix. Its delooped version is the statement that you can tensor vector spaces with endomorphisms by invertible modules for functions on the spectrum. (What was important for Ngô is that the fibers of $J$ are in general not connected, as can be seen already for $x=0$ in $SL_2$, and those component groups are crucial for the geometric interpretation of endoscopy.) Anyway this action gives an interesting quotient 2-stack --- eg on the regular locus the quotient $(\mathfrak g/G)/BJ$ just looks like the affine variety $\mathfrak g//G\simeq \mathfrak h//W$ of invariant polynomials (equivalently, generically $\mathfrak g/G$ is isomorphic to $BJ$), but eg for $x=0$ this action is trivial, so we find $J_0$ (the centralizer of a regular nilpotent element in $\mathfrak g$) acting trivially as automorphisms of the stabilizer of $0$ (i.e. $G$ itself), so producing genuine 2-stackiness.<|endoftext|> TITLE: Status of the Hopf-Thurston sign conjecture in dimension 4 QUESTION [5 upvotes]: A famous conjecture in topology asserts: The Euler characteristic of a closed aspherical $2n$-manifold $M$ satisfies $(-1)^n\chi(M) \geq 0$. This was conjectured by Hopf for manifolds with non-positive sectional curvature and (much later) by Thurston for all aspherical manifolds. By the classification of surfaces, it holds in dimension $2$. I am curious about the status of this conjecture in dimension 4. It seems as if the special case for manifolds with negative sectional curvature was settled by Milnor, as explained in this paper by Chern. Undoubtedly, the more general conjecture by Thurston is much younger. I wonder if the general version has some kind of reformulation in less manifoldy and more algebraic terms. More precisely, I am curious about the following. For which fundamental groups of aspherical $4$-manifolds is the $4$-dimensional Thurston conjecture settled? One approach to the Thurston conjecture that I am aware of is the employment of $\ell^2$ invariants: the Singer conjecture implies the Hopf conjecture in general. However, the Singer conjecture is open. REPLY [8 votes]: There has been a lot of work on cases of this conjecture connected to Coxeter groups. M. Davis and R. Charney made a conjecture that comes from these cases in 1995 in The Euler characteristic of a non-positively curved piecewise Euclidean manifold. Also have a look at the article Vanishing theorems and conjectures for the $\ell^2$-homology of right-angled Coxeter groups by M Davis and B Okun. They prove the case when $M^4$ has a non-positively curved piecewise Euclidean cubical structure. This is a far larger class than it sounds: every flag triangulation of the 3-sphere gives rise to such a manifold via the Davis construction. Davis-Okun also give a conjecture for the $\ell^2$-homology of groups that act properly cocompactly on contractible manifolds that extends the Singer conjecture, which is equivalent to the case when the group acts freely properly cocompactly.<|endoftext|> TITLE: Whitney's approximation theorem for Lipschitz manifolds QUESTION [8 upvotes]: In the smooth setting, Whitney's approximation theorem says the following: If $M,N$ are smooth manifolds and $f,g:M\to N$ are smooth functions that are continuously homotopic (ie there is a continuous homotopy $H:M\times [0,1]\to N$ between them) then they are also smoothly homotopic (ie there exists a smooth homotopy $\tilde{H}$ between them). Does something similar hold for Lipschitz functions between Lipschitz manifolds? More precisely, if $M,N$ are two Lipschitz manifolds and $f,g:M\to N$ are Lipschitz functions that are continuously homotopic, is it true that they are also Lipschitz homotopic (ie there exists a Lipschitz homotopy between them)? REPLY [5 votes]: For compact Lipschitz manifolds this follows from the main result of the paper by Liu, Luofei, Yu, Hanfu, Liu, Ye "Converting uniform homotopies into Lipschitz homotopies via moduli of continuity." Topology Appl. 285 (2020) But the above paper is surely an overkill. It gives quantitative bounds on the homotopy and qualitative result should be known much earlier. I expect it's known that for a compact Lipschitz manifold $N$, for some Lipschitz embedding of $N$ into some $\mathbb R^n$ a small neighbourhood of $N$ Lipschitz retracts onto $N$. This would easily imply the result. It's likely even true that $N$ is a Lipschitz ANR. It's also clear that there is an appropriate version of this for noncompact manifolds too. Edit: I found a reference to the fact that any Lipschitz manifold is a Lipschitz ANR. This immediately follows from Theorem 5.1 in this paper. Therefore what I said above works. It is enough to show that any two sufficiently $C^0$ close Lipschitz maps $f,g:M\to N$ are Lipschitz homotopic. Embed $N$ by a Lipschitz map into a Euclidean space $\mathbb R^n$, connect $f$ to $g$ by straight line homotopy in $\mathbb R^n$ and then use Lipschitz neighborhood retraction to push this homotopy into $N$.<|endoftext|> TITLE: Euclidean norms of matrices QUESTION [5 upvotes]: Let us consider the euclidean norm on $\mathbf{R}^2$. After some computations, I have obtained the following expression for the associated operator norm on 2 by 2 matrices. $$ \left\lVert\pmatrix{a&b\cr c&d\cr}\right\rVert^2 = {1\over 2} \Bigl(\lvert a+ib\rvert^2+\lvert c+id\rvert^2+\lvert(a+ib)^2+(c+id)^2\rvert\Bigr). $$ This expression is new to me and I am wondering if there is a conceptual explanation for such a formula. Also, is there an analogous formula for higher dimensional matrices? REPLY [3 votes]: In essence, the answer is: no. $\|A\|^2$ is always equal to the largest eigenvalue of $A^tA$. If $A$ is already symmetric, $\|A\|$ is equal to the (absolute value of) the largest eigenvalue of $A$. The characteristic polynomial of a matrix can be essentially anything, so the matrix norm can be essentially any algebraic number. This means a nice expression with only basic arithmetic is impossible, and any somewhat nice expression is unlikely. Otherwise, linear algebra software would not go throught all the trouble of using complicated iterative numerical algorithms to approximate eigenvalues. But it is possible to extend (slightly) beyond $2\times2$. For example for a symmetric $3\times 3$ or $4\times 4$ matrix, you can compute the characteristic polynomial, and plug its coefficients into the formulas for the roots of these polynomials (i.e. Cardano or Ferrari), and then pick out the largest of the solutions. Note that this is actually done in practice, as it can be faster than the iterative methods that are otherwise used to compute eigenvalues of (large) matrices. See for example here for the special case of $3\times3$ matrices. Though already in this case, the formulas are rather cumbersome. If you are looking for bounds, the simplest ones are \begin{align} \frac{1}{\sqrt{n}}\|A\|_F \le \|A\| \le \|A\|_F \end{align} where \begin{align} \|A\|_F^2 = \sum_{i,j=1}^n a_{ij}^2 \end{align} is the "Frobenius norm" of the matrix. More precise bounds can be obtained using the Gershgorin circle theorem for example As a side note: because of all these problems, in numerical linear algebra, one rarely uses the $\ell_2$-induced matrix norm in actual calculations. For example the matrix-norms induces by the $\ell_1$ and $\ell_\infty$ norms have very simple expressions, so they are preferable in practical calculations.<|endoftext|> TITLE: Adèlic points and algebraic closure QUESTION [6 upvotes]: Consider $\mathcal{X}$ a projective and flat scheme over $\text{Spec}(\mathcal{O}_K)$, with $\mathcal{O}_K$ the ring of integers of a number field $K$. Let $F/K$ vary over all finite Galois number field extensions and define $\mathbf{A} := \mathbf{A}_{\overline{K}}$ as the direct limit of the topological rings $\mathbf{A}_F$ of the adèles of each $F$. Question 1 Is there a good and intrinsic definition of $\mathcal{X}(\mathbf{A})$ and does it agree with the direct limit topological space $\varinjlim_{F/K}\mathcal{X}(\mathbf{A}_F)$? (or is this latter the definition usually given?) Question 2 Is $\mathcal{X}(\mathbf{A})$ compact? The ring $\mathbf{A}$ must probably be replaced with a restricted product of countably many copies of $\mathbf{C}$, $\mathbf{R}$, $(\mathbf{C}_p,\mathcal{O}_{\mathbf{C}_p})$ for infinitely many $p$. Even so, $\mathbf{C}_p$ is not locally compact, so I expect the answer to question 2 is "no", no matter how we put it. REPLY [6 votes]: Since $\mathcal X$ is projective, a section is given by finitely many coordinates. If the $i$'th coordinate lies in $\mathbf A_{F_i}$ for some extension $F_i$ of $K$, then all the coordinates lie in $\mathbf A_{F}$ for $F$ the composition of the $F_i$. So $\mathcal X(\mathbf A_{\overline {K}})$ agrees with the direct limit of $\mathcal X(\mathbf A_{\overline {F}})$ - at least as sets: I didn't check the topology but I imagine it's fine. I'm not sure what you mean by an intrinsic definition. Both these definitions seem fine to me. $\mathcal X(\mathbf A_{\overline {K}})$ is not compact. Choose a $p$-adic valuation on $\overline{K}$, We have maps $$\mathcal X ( \mathbf A_{\overline{K}}) \to \mathcal X (\overline{K}_p) = \mathcal X( \overline{\mathbb Q}_p) = \mathcal X( \overline{\mathbb Z}_p) \to \mathcal X ( \overline{\mathbb F}_p)$$ and the composition is continuous with the discrete topology on $\mathcal X ( \overline{\mathbb F}_p)$, but $\mathcal X ( \overline{\mathbb F}_p)$ is infinite. Restricted to each individual $\mathcal X( \mathbf A_{F})$, we're restricting the valuation to $F$, mapping to the appropriate completion, clearing denominators (because we are working in a projective variety), and reducing modulo the uniformizer.<|endoftext|> TITLE: Easiest proof of computability of homotopy groups of spheres QUESTION [16 upvotes]: Has it gotten easier to prove all homotopy groups of spheres are computable? I don’t care if the computation is inefficient, what’s the easiest proof? Are we still stuck doing Postnikov towers? REPLY [13 votes]: It may not be any "easier" than Brown's proof, but the work of people like Francis Sergeraert and others has created a nice conceptual framework for these types of questions. See especially the paper, Effective homotopy of fibrations by Romero and Sergeraert. They have also implemented their algorithms in Kenzo. See also An algorithm computing homotopy groups by Pedro Real, which uses the ideas of effective homology to sketch an algorithm for the homotopy groups of spheres specifically.<|endoftext|> TITLE: Is there an eigenvalue of modulus larger than 1? QUESTION [12 upvotes]: Given a matrix $A\in \operatorname{SL}_d(\mathbb{Z})$ (the special linear group) satisfying the two conditions: (1) no eigenvalue of $A$ is a root of unity, (2) the characteristic polynomial of $A$ is irreducible over $\mathbb{Q}$. QUESTION. Does it follow that at least one eigenvalue $\lambda$ of $A$ fulfills $\vert\lambda\vert>1$? Assuming true, it seems that there must be a theorem of a sort here, but I couldn't recall. It would also be nice if one can relax the conditions to gain the same conclusion, if possible. UPDATE. After exploring papers by Kronecker (thanks Terry Tao) and others, I realized that we don't quite need "irreducibility" but "monic" is enough. REPLY [22 votes]: Since the characteristic polynomial is irreducible, eigenvalues are simple and hence the matrix is $\mathbf{C}$-diagonalizable. If all were on the unit cercle, it would follow that $\{ A^n:n\in\mathbf{Z}\}$ is bounded. But since it is contained in the set of integral points, this would force it to be finite, and hence all eigenvalues would be roots of unity, contradiction. Hence one is not on the unit circle. Since the product of eigenvalues is $\pm 1$ (the determinant), it follows that some eigenvalue has modulus $>1$. PS: Here's the straightforward way to extend this the general case (without irreducibility). Let $A\in\mathrm{M}_d(\mathbf{Z})$ be a matrix whose eigenvalues are not only among $0$ and roots of unity. As any integral matrix, we can block-triangulate it (over $\mathbf{Z}$) so that all diagonal blocks are $\mathbf{Q}$-irreducible. Hence at least one diagonal block, with characteristic polynomial $P=P(t)$, also satisfies the assumption. So $P$ is $\mathbf{Q}$-irreducible, and not equal to $t$. If $|P(0)|\ge 2$ then clearly some root has modulus $>1$. Otherwise, $P(0)\in\{\pm 1\}$ and we are in the previous (main) case.<|endoftext|> TITLE: How many tensor products of chain complexes are there? QUESTION [12 upvotes]: Let $Ch$ be the category of nonnegatively-(homologically-)graded chain complexes of abelian groups. Suppose that $(Ch,\boxtimes)$ is a monoidal biclosed structure. Assume that the forgetful functor $(Ch, \boxtimes) \to (Gr,\otimes)$ is strong monoidal, where $Gr$ is the category of nonnegatively-graded abelian groups and $\otimes$ its usual tensor product. Thus $(A\boxtimes B)_n = \oplus_{i+j = n} A_i \otimes B_j$, equipped with some differential, in an associative and unital way. Question 1: Can one classify all such monoidal structures $(Ch,\boxtimes)$, up to monoidal equivalence over $(Gr,\otimes)$? Notes: The usual tensor product $\otimes$ of chain complexes is an example, with $d(x\otimes y) = d x \otimes y + (-1)^{|x|} x \otimes d y$ but not the only one -- e.g. one might define the differential as $d'(x \otimes y) = (-1)^{|y|} d x \otimes y + x \otimes d y$ for a different example. If it's easier to work over a field instead of $\mathbb Z$, that would be an interesting start. I've singled out nonnegatively-graded chain complexes, but if the situation changes when passing to unbounded chain complexes, that would be interesting. Question 2: Are all such monoidal structures $\boxtimes$ symmetric monoidal? In how many ways? REPLY [3 votes]: Okay, here's a solution when we allow unbounded chain complexes. First, note that the automorphism group of $Ch$ over $Gr$ is $C_2^\omega$ (or rather $C_2^{\mathbb Z}$ since we've switched to the unbounded case), because swapping the sign of the $n$th differential on all objects of $Ch$ affects neither the underlying graded abelian groups, nor the notion of chain map (and similarly for swapping the sign of $d_n$ for each $n \in S$, where $S \subseteq \mathbb Z$ is an arbitrary subset). And indeed, the two examples of monoidal structures which I gave are related by such an automorphism -- the one where you swap the sign of $d_n : A_n \to A_{n-1}$ whenever $n$ is odd. Claim: All the monoidal biclosed structures on $Ch$ over $Gr$ are related to the standard one by such an automorphism. An easy exercise shows that the only nonidentity automorphism of the standard tensor product over $Gr$ is the one which flips the signs of all the differentials, so most of these monoidal structures are different from one another. And apparently they are all symmetric monoidal (in exactly as many ways as the standard monoidal structure -- however many that is). Proof: Consider the short exact sequence $S(0) \to D(1) \to S(1)$ (here and in the following, $D(n)$ is the degree $n$ disk and $S(n)$ is the degree $n$ sphere; we take advantage in the following of the fact that these objects have canonical bases which we sometimes silently identify). Tensoring the exact sequence against itself, we get an exact $3\times 3$ grid. Unitality of the monoidal structure tells us that the differentials on $S(0) \boxtimes D(1)$ and $D(1) \boxtimes S(0)$ are canonically isomorphic to the differential on $D(1)$. By looking at the inclusion maps from these objects to $D(1)^{\boxtimes 2}$, we see that the $d_1$ differential on $D(1)^{\boxtimes 2}$ is given by the codiagonal $[1,1]: D(1)_1 \otimes D(0)_0 \oplus D(0)_0 \otimes D(0)_1 \to D(0)_0 \otimes D(0)_0$. This implies that the $d_2$ differential is of the form $[a,-a] : D(1)_1 \otimes D(1)_1 \to D(1)_1 \otimes D(0)_0 \oplus D(0)_0 \otimes D(0)_1$ for some $a \in \mathbb Z$. Conjugating by an automorphism of $Ch$ over $Gr$, we may assume that $a \geq 0$. Quotienting along the maps to $D(1) \boxtimes S(1)$ and $S(1) \boxtimes D(1)$, we see that the differential on the former is given by $a$ and on the latter the differential is $-a$. I claim that $a = 1$. For if $a = 0$, then $S(1) \boxtimes D(1)$ has a null differential. Tensoring with $S(-1)$, it follows that $D(1)$ has a null differential, a contradiction. On the other hand, if $a > 1$, then there is an exact sequence $0 \to D(2) \to S(1) \boxtimes D(1) \to S(1)/a \to 0$. Again tensoring with $S(-1)$, we obtain an exact sequence $0 \to S(-1) \boxtimes D(2) \to D(1) \to S(0)/a \to 0$. We know all the groups involved here, and because the differential on $D(1)$ is a unit, the differential on $S(-1) \boxtimes D(2)$ would have to be $1/a$, which is of course impossible. Now just as we used knowledge of the differential on $S(0) \boxtimes D(1)$ to determine the differential on $S(1) \boxtimes D(1)$, we can work our way up to work out that the differential on $S(n) \boxtimes D(1)$ is standard for all $n \geq 0$. A dual argument allows us to work out negative $n$'s as well. At this point we know the differential on $D(1) \boxtimes D(1)$. It’s the standard one, which splits as a sum $D(2) \oplus D(1)$. We also have canonical isomorphisms $S(n-1) \boxtimes D(1) = D(n)$ and dually $D(1) \boxtimes S(m-1) = D(m)$. Together we can now determine the differential on $D(n) \boxtimes D(m)$. By compatibility with colimits, this determines the entire monoidal structure. If we stick to just nonnegatively graded complexes, then I have difficulty ruling out the cases $a=0$ or $a > 1$ in the above argument. I would be interested to see the argument pushed through to completion in this case, but I think I have run out of energy to do it myself right now.<|endoftext|> TITLE: Symmetric polynomials that detect positivity QUESTION [9 upvotes]: Imagine there are numbers $a_1,\ldots,a_n \in \mathbb R$ and you want to know whether they are all positive. You cannot access the numbers themselves, but you can choose any symmetric polynomials you like to evaluate at $(a_1,\ldots,a_n)$. Actually, suppose you don't even get the results of those evaluations, but only the signs. What is a good family $s_{i,n}$ of symmetric polynomials that allows you to do this? To restate the problem, we want (for each $n=1,2,...$) a family $s_{1,n},s_{2,n},\ldots,s_{k,n}$ of symmetric polynomials in the variables $x_1,\ldots,x_n$ with the property that these are equivalent: $a_j > 0$ for $j=1,\ldots,n$. $s_{i,n}(a_1,\ldots,a_n) > 0$ for $i=1,\ldots,k$. For extra credit, forget about $a_1,\ldots,a_n$ being real, and replace $a_j > 0$ with $\mathfrak{R}(a_j) > 0$. How do I know this is even possible? It's because a related question has been treated extensively, namely that of determining, given the coefficients of a polynomial, whether all of the roots of the polynomial are in the right (or left or whatever) half-plane. Since the coefficients are (the elementary) symmetric polynomials applied to the roots of the polynomial, results like the Routh–Hurwitz stability criterion, which is usually expressed in terms of the signs of certain polynomials in the coefficients, can be rewritten in terms of symmetric polynomials applied to the roots. But what if you don't care about going through the coefficients? What if you are happy enough to write down symmetric polynomials in terms of the roots themselves? Then what is the “best” or “most natural” family of symmetric polynomials that will do the job? REPLY [12 votes]: The elementary symmetric functions $1=e_0,e_1,\dots,e_n$ will suffice. Clearly $e_j>0$ if each $a_i>0$. Conversely, let $P(x)= \prod_{i=1}^n(x+a_i)$. If some $a_i<0$ and all $e_j>0$ then $0=P(-a_i)=\sum_j e_j (-a_i)^{n-j}>0$, a contradiction. Incidentally, if we assume only that $a_i\in\mathbb{C}$ we can still determine whether each $a_i$ is a positive real number using $n-1$ additional symmetric functions. See Gantmacher, Matrix Theory, vol. 2, Chapter 15, Section 9.1.<|endoftext|> TITLE: Is there an "anti-choice axiom"? QUESTION [20 upvotes]: Edit: I have now rewritten the question in terms of well-ordering. Before, had I used the axiom of choice directly, which made the question unclear. The anti-foundation axiom is a nice thing: It not only denies the axiom of foundation, but also gives an interesting structure to the non-founded sets. I now want to know whether there is a similar axiom that constructively denies the axiom of choice, in the form that for each set a well-ordering exist. In greater detail: If the axiom of choice is not true, then there are usually some sets that still can be well-ordered, for example the countable sets. All sets that can be mapped injectively into a well-ordered set can also be well-ordered. Or in terms of cardinality: All sets smaller than a well-ordered set are also well-orderable. So in a set theory without the axiom of choice, we have two kinds of sets: Small ones, which have a well-ordering, and big ones, which do not have one. A good anti-choice axiom would therefore provide a "natural" division between small and big sets and give the big sets some additional properties that they cannot have under the usual (ZFC) set theory. So the question is: What can I get instead if I cannot well-order some sets? REPLY [18 votes]: There are two ways to understand this. Well. Three. If $X$ is a set, then $\mathcal P(X)$ admits a choice function.1 This is equivalent to the statement that $X$ can be well-ordered. This means that all the ordinals are "small", and since in $\sf ZF$ the axiom of choice is equivalent to saying that $\mathcal P^2(\alpha)$ admits a choice function for all ordinals, we know that the failure of choice is just saying that for some $\alpha$, $\mathcal P(\alpha)$ is not small anymore. If you want this ordinal to be explicit, we can introduce many various axioms that make it so. Most famously, Determinacy implies that $\omega$ already fails that. As do many other axioms. If $X$ is a set, then we can think of $X$ as small if any family of sets indexed by $X$ admits a choice function. Here we have that finite sets are small; $\sf AC_\omega$ states that countable sets are small; and we can have $\sf AC_X$ for non well-orderable sets $X$ just as well. Again, we can make this more explicit, as $\sf AC$ fails, by stating that $\sf AC_\omega$ fails, or even that for any infinite set $X$, there is a family indexed by $X$ which does not admit a choice function. We can improve upon this by also providing limitations on how big the sets inside the family are allowed to be, e.g. "every countable family of finite sets admits a choice function", but not so for anything larger: either in index or in content. We can mix those two. We can think of a set as being small if any well-ordered family of subsets admits a choice function, for example. Or any well-ordered family of subsets by a small ordinal (in the sense of (1), that is) admits a choice function. Etc. etc. etc. There's a lot of variety here. And it gets weirder and weirder, and more and more explicit as we dive into this rabbit hole. At this point, you might argue, none of these are "constructive negations of choice". Insofar that none of them really pinpoint the exact failure. This is why more structural axioms, such as $\sf AD$, do offer a modicum of success here. Since $\sf AD$ provides us with a fairly rich theory of how badly things get. If you couple it with $V=L(\Bbb R)$, we can say even more about how terrible things can get. But we can turn our heads to a different path. For example, Monro proved in Monro, G. P., Decomposable cardinals, Fundam. Math. 80, 101-104 (1973). ZBL0272.02085. That it is consistent that an infinite set is well-orderable if and only if it cannot be written as the union of two sets of strictly smaller cardinality. In the Cohen model, which Monro studies, this is true, and we have a relatively straightforward description of this odd partition. There are many axioms like that, which infuse the universe of sets with a sense of chaos, either in the structure of "what type of families have a choice function" or "what kind of sets can be well-ordered". It seems that you're not entirely clear as to what you're looking for exactly, but there's a lot of these things in the literature. I suggest starting with Herrlich's "Axiom of Choice" book to read about "disasters" and go from there. Footnotes. Yes, $\varnothing$ is an issue. But we can just agree that a family of sets $A$ admits a choice function if $\prod (A\setminus\{\varnothing\})$ is non-empty.<|endoftext|> TITLE: Fraction of $S_n$ reachable by using every transposition once as $n\to\infty$? QUESTION [17 upvotes]: For $n\in \mathbb{N}$ let $S_n$ denote the set of permutations (bijections) $\pi: \{0,\ldots,n-1\}\to \{0,\ldots,n-1\}$. A transposition swaps exactly $2$ elements and is often denoted by $(i \; k)$ if $i\neq k\in\{0,\ldots,n-1\}$ are the elements being swapped. For $n\in\mathbb{N}$ let $E_n$ be the number of elements of $S_n$ that can be obtained by a composition of all the transposition, such that every transposition is used exactly once. What is the value of $\lim\sup\frac{E_n}{n!}$? REPLY [25 votes]: The value is $1/2$. The problem can be reformulated as follows: How to sort a non-sorted list $(a_1,...,a_n)$ that is a permutation of $\{1 .. .n\}$ with the parity the same as $n \choose 2$ by exactly one transposition between every pair of indices? In this answer the phrase "unsorted element" means an element $a_p$ that is not equal to $p$. This is solved by a recursive manner: Suppose $n\geq 4$ and $a_p\neq p$. Then there's a sequence of transpositions of $a_p$ with all the other elements such that $a_p=p$ after the transpositions and there exists an unsorted element in the list. Thus, we can proceed to the subproblem of finding such a sequence on the list with $a_p$ removed. This recursion stops at $n=3$, but by the parity assumption and the unsorted element assumption, there must be exactly two unsorted elements left, and it's easy to sort them by three transpositions. Now we only need to prove the existence of such transpositions. We may assume that $a_1 \neq 1$, $a_n=1$ and we need to achieve $a_1=1$ after the transpositions. Let $(1,2), (1,3), ..., (1,n)$ be the sequence of transpositions. Then $a_1$ will be equal to $1$ after applying the transpositions in order. If there are unsorted elements, we are done; otherwise the sequence $(1,2), (1,3), ..., (1,n-1), (1,n-2), (1,n)$ will give an unsorted element.<|endoftext|> TITLE: Volume of conic bundles QUESTION [5 upvotes]: Consider a smooth conic bundle $X\rightarrow \mathbb{P}^1$ with discriminant of degree $d$ (the locus of $\mathbb{P}^1$ over which the fibers are reducible conics). There is a formula for $(-K_X)^2$ that reads $(-K_X)^2 = 8-d$. So that the volume depends only on the degree of the discriminant. Now, consider conic bundles $Y\rightarrow\mathbb{P}^2$. For instance, if $Y\subset\mathbb{P}^2\times\mathbb{P}^2$ is a divisor of bidegree $(4,2)$ I got $(-K_Y)^3 = -6$. Note that in this case $d = 12$. If $Y\subset\mathbb{P}(\mathcal{O}_{\mathbb{P}^2}(2)\oplus \mathcal{O}_{\mathbb{P}^2}(1)\oplus \mathcal{O}_{\mathbb{P}^2})$ given by an equation of the form $$ a_0x^2+a_1xy+a_2xz+a_3y^2+a_4yz+a_5z^2 = 0 $$ where the $a_i$ are polynomials on the base $\mathbb{P}^2$ of degree respectively $6,5,4,4,3,2$. Note that also in this case $d = 12$. So it seems that unlike the case of surfaces in higher dimension the is no formula for $(-K_Y)^n$ where $n$ is the dimension of the conic bundle $Y$. Is this correct and if so is there a conceptual explanation for this fact? REPLY [3 votes]: If I understand your question, then there is no such formula. Since there are examples where the degree of the discriminant curve is $5$ but with different $(-K_{X})^3$. The blow up of a smooth cubic $3$-fold in a line is a conic bundle over $\mathbb{P}^2$ with discriminant curve of degree $5$. It is a Fano $3$-fold with $(-K)^3 =18$. On the other hand Panin showed that the blow-up of $\mathbb{P}^3$ in a smooth curve of degree $7$ and genus $5$ is a conic bundle over $\mathbb{P}^2$ with discriminant curve of degree $5$. It is a Fano $3$-fold, it has $(-K)^3 =16$. I learned both these examples from https://arxiv.org/pdf/1712.05564.pdf, (example 3.4.2 and example 3.4.3 respectively). There is many more examples of conic bundles over $\mathbb{P}^2$ given there, and I guess many more counter examples to the formula.<|endoftext|> TITLE: Groups in which Computational Diffie Hellman is in $P$ but Discrete Logarithm is not known to be in $P$ QUESTION [6 upvotes]: The Computational Diffie Hellman (CDH) problem is to compute $g^{XY}$ given $g^X$ and $g^Y$ where $g$ generates the group. The Discrete Logarithm (DLOG) problem is to compute $X$ given $g^X$. The latter in $P$ implies the former in $P$. Are there groups where CDH problem is easy and in $P$ while DLOG is difficult and not known to be in $P$? REPLY [2 votes]: Some background for this answer: A computation is non-uniform if, in addition to an input $x$, one gets an "advice string" that depends solely on the length of $x$. One can view this as precomputation, although this advice string need not be computable (this should not be relevant here though). In this answer, non-uniformity is used to ensure one has access to parameters of certain elliptic curves, and potentially a factorization of a certain number. The generic group model is an abstract model in cryptography where, rather working directly with group elements, one is given oracles for the relevant group operations. Algorithms in the generic group model therefore work for all groups with efficiently-computable operations, and do not exploit the particular structure of any single group. There are lower bounds in the generic group model --- solving DLOG in $\Theta(\exp(\sqrt{\log |G|}))$ operations is provably optimal iirc. The only "mainstream" groups used within cryptography that are plausibly generic are Elliptic Curve groups ($\mathbb{F}_p^\times$ and related groups admit non-generic algorithms that go broadly by the name of "index calculus" algorithms). The paper you quote there requires $|G|$ to be $\mathsf{polylog}|G|$ smooth according to abstract. Are you sure it applies to cyclic multiplicative group mod p a prime? Please post full answer for this group. It does not, as $\mathbb{F}_p^\times$ is not thought to be a "generic" group --- one can speed up DLog via appealing to index calculus methods. The paper works in the generic group model, so is mainly applicable to groups we think are generic, e.g. elliptic curve groups. Still, I will include two of the papers results, as your understanding of the paper from the abstract is wrong, and it can handle the case of $|G| = p$ a large prime. Here is corollary 5 of the paper Corollary 5: If the smoothness assumption is true, then there exists a polynomial-time generic algorithm computing discrete logarithms in cyclic groups of order $n$, making calls to a DH oracle for the same group, if and only if all the multiple prime factors of $n$ are of order $(\log n)^{O(1)}$. This result is non-uniform, and requires a certain number theoretic assumption. It clearly handles the case of $|G| = p$ though. Removing the smoothness assumption (and I believe the non-uniformity assumption, but I have not carefully read), corollary 7 states that Corollary 7. Let $P$ be a fixed polynomial, let $G$ be a cyclic group with generator $g$, and let $B := P (\log |G|)$. Then there exists a list of expressions $A(p)$ in $p$ with the following property: if every prime factor $p$ of $|G|$ greater than $B$ is single and if for every such prime factor at least one of the expressions $A(p)$ is $B$-smooth, then breaking the DH protocol in $G$ with respect to $g$ is polynomial-time equivalent to computing discrete logarithms in $G$ to the base $g$. The list contains the following expressions: $p − 1 , p + 1 , p + 1 \pm 2a$, if $p \equiv 1 (\mod 4)$, where $p = a^2 + b^2$, $p + 1 \pm 2a, p + 1 \mp a \pm 2b , p + 1 \pm(a + b)$, if $p \equiv 1 \bmod 3$, where $p = a^2 − ab + b^2, a \equiv 2 \bmod 3$, and $b \equiv 0 \bmod 3$, $\frac{(p^k)^l − 1}{p^k − 1} = (p^k)^{l−1} + · · · + p^k + 1$, where $k, l = (\log p)^{O(1)}$, and $f(p)$, where $f(x) \in \mathbb{Z}[x]$ is a nonconstant polynomial dividing $x^n − 1$ for some $n = O(1)$. these technical conditions are related to constructions of what the authors call "auxillary groups", which are the non-uniform information the authors need for their reduction. This is to say that the result can definitely be improved, for example removing the smoothness assumption for cor. 4, or relaxing the technical conditions with respect to the construction of auxillary groups, Of main interest to you though, the work is in the generic group model, and the example you have mentioned in the comments is not a generic group.<|endoftext|> TITLE: Analytical solution for coupled ODE system QUESTION [5 upvotes]: I came across this coupled system of ODE in my research (related to foulant in reaction diffusion system): $$\frac{d^2y}{dx^2} - A\frac{yz}{x} = 0$$ $$\frac{d^2z}{dx^2} - B\frac{yz}{x} = 0$$ where $A$ and $B$ are constants. Can it be solved analytically? REPLY [5 votes]: Mathematica cannot solve this system of ODEs, even after the decoupling suggested by user LeechLattice, even with $\alpha=\beta=0$ and $A=1$. This means that the existence of an explicit solution is highly unlikely. Here is the image of the corresponding Mathematica notebook:<|endoftext|> TITLE: Why is $\operatorname{U}(n,\mathbb{H})\subset \operatorname{SL}(n,\mathbb{H}) $? QUESTION [5 upvotes]: This question is inspired by Maximal compact subgroup of $\mathrm{SL}(2,\mathbb{H})$. Consider the embedding $\operatorname{U}(n,\mathbb{H})\subset \operatorname{GL}(n,\mathbb{H}) $. Since $\operatorname{U}(n,\mathbb{H})\cong \operatorname{Sp}(n) $ is almost simple, we know that it is actually contained in $\operatorname{SL}(n,\mathbb{H}) $, the kernel of the Dieudonné determinant $\ \det: \operatorname{GL}(n,\mathbb{H}) \rightarrow \mathbb{R}^*_{+}\,$. Is there a direct way to prove $\det A=1$ for $A\in \operatorname{U}(n,\mathbb{H}) $? (Recall that $\ \det\, {}^{t}\overline{A}\neq \det A\ $ for $A$ general in $\operatorname{GL}(n,\mathbb{H}) $.) REPLY [3 votes]: Yes, relying on the fact that elements of the symplectic group (over $\mathbf{C}$, and hence over $\mathbf{R}$) have determinant 1. Indeed, an element $g$ of $\mathrm{U}(n,\mathbf{H})$ preserves the canonical Hermitian form $b$. Let $b'$ be any imaginary component of $b$. Then $b'$ is a $g$-invariant real-valued symplectic form on the $4n$-dimensional real space $\mathbf{H}^n$. So the determinant of $g$ as $4n$-dimensional matrix (which is the Dieudonné determinant) is $1$. Edit (abx): actually the determinant $\det_{\mathbb{R}}(g)$ of $g$ as a real $4n$-dimensional matrix is $(\det_{D}( g))^4$, where $\det_D$ is the Dieudonné determinant — since $\det_D(g)\in\mathbb{R}^*_+$, this implies $\det_D(g)=1$ as required. Indeed $\det_{\mathbb{R}}$ is a homomorphism from $\operatorname{GL}(n,\mathbb{H}) $ to $\mathbb{R}^*$; such a homomorphism is trivial on the derived subgroup $\operatorname{SL}(n,\mathbb{H}) $. Since $\det_D$ induces an isomorphism of $\operatorname{GL}(n,\mathbb{H}) /\operatorname{SL}(n,\mathbb{H}) $ onto $\mathbb{R}^*_+$, we have $\det_{\mathbb{R}}=\varphi \circ \det_{D}$, where $\varphi :\mathbb{R}^*_+\rightarrow \mathbb{R}^*$ is a continuous homomorphism. Such a homomorphism is of the form $x\mapsto x^{a}$ for some $a\in\mathbb{R}$; thus $\det_{\mathbb{R}}= (\det_{D})^{a}$. Taking $g= tI_n$ gives $a=4$.<|endoftext|> TITLE: Computation of the torsion of K-groups related to elliptic curves QUESTION [5 upvotes]: Let $E$ be an elliptic curve over $\mathbb Q$. Let $F$ be the rational function field of $E$. The $K_2$ group of $F$ may be described by elements in $F^\times ⊗_\mathbb{Z} F^\times$ quotiented by the relations $\langle f ⊗ (1 − f)\rangle (f ∈ F^\times, f\neq 0,1)$. For $P ∈ C(\bar {\mathbb Q})$ and $\{g, h\}\in K_2(F)$, the tame symbol is defined as $$T_P (\{g, h\}) = (−1)^{\text{ord}_P(g)\text{ord}_P(h)}\frac{g^{\text{ord}_P (h)}}{h^{\text{ord}_P (g)}}(P)$$ The tame kernel $K_2^T(E)$ is the subgroup of $K_2(F)$ whose elements $\{g,h\}$ satisfies $T_p(\{g,h\})=1$ for every $P ∈ E(\bar {\mathbb Q})$. Question: How to compute* the torsion of $K_2(E)$, or $K_2^T(E)/K_2(\mathbb Q)$? *It's not known whether the torsion of the groups above are finitely generated. So it could be possible that there is no algorithm to compute the torsion because the torsion is infinite. Having this in mind, the word "compute" should be interpreted as follows: Is there an algorithm to determine whether a symbol $\{g,h\}$ is trivial in one of the K-groups? What is the most efficient way to find as many as possible torsion elements in one of the K-groups? REPLY [5 votes]: The most efficient way I know to detect whether an element of $K_2(E)$ is torsion is to use the elliptic dilogarithm. This relies however on a conjecture, and it is not an exact method, in the sense that it uses floating-point arithmetic. Consider the map \begin{align*} \beta : F^\times \otimes F^\times & \to \mathbb{Z}[E(\overline{\mathbb{Q}})] \\ g \otimes h & \mapsto \sum_{P,Q \in E(\overline{\mathbb{Q}})} \mathrm{ord}_P(f) \mathrm{ord}_Q(g) [P-Q]. \end{align*} Bloch has defined a regulator map \begin{equation*} \mathrm{reg}_\infty \colon K_2(E) \to \mathbb{R}, \end{equation*} which can be computed using the elliptic dilogarithm $D_E : E(\mathbb{C}) \to \mathbb{R}$. Namely, given $x \in K_2(E)$, whose image in $K_2^T(E)$ is written as $\sum_i \{g_i,h_i\}$, we have (up to a constant factor) \begin{equation*} \mathrm{reg}_\infty(x) = \sum_i D_E(\beta(g_i \otimes h_i)) \end{equation*} where $D_E$ is extended by linearity by defining $D_E(\sum n_P [P]) = \sum n_P D_E(P)$. The elliptic dilogarithm can be computed very rapidly: writing $E(\mathbb{C}) = \mathbb{C}^\times/q^{\mathbb{Z}}$, one has $D_E([x]) = \sum_{n \in \mathbb{Z}} D(xq^n)$. Here $D$ is the Bloch-Wigner dilogarithm, implemented e.g. in PARI/GP as $\texttt{polylog(2,x,2)}$. Since $|q|<1$, the series for $D_E$ converges exponentially fast. One should also take into account the bad places of $E$. For each prime $p$, there is a residue map $K_2(E) \to K'_1(\mathcal{E}_p)$, where $\mathcal{E}_p$ is the fiber at $p$ of the minimal regular model of $E$ over $\mathbb{Z}$. It turns out that $V_p := K'_1(\mathcal{E}_p) \otimes \mathbb{Q}$ is nonzero precisely when $E$ has split multiplicative reduction at $p$, in which case $\operatorname{dim}(V_p)=1$. It is also possible to compute the residue map, see Bloch, Grayson, $K_2$ and $L$-functions of elliptic curves - Computer calculations and Rolshausen, Schappacher, On the second $K$-group of an elliptic curve. Now, it is conjectured that the extended regulator map \begin{equation*} \mathrm{reg} \colon K_2(E) \to \mathbb{R} \oplus \bigoplus_p (V_p \otimes \mathbb{R}) \end{equation*} is an isomorphism after tensoring $K_2(E)$ with $\mathbb{R}$. In particular, and conjecturally, an element $x$ of $K_2(E)$ is torsion precisely when it is in the kernel of the extended regulator map. If the image of $x$ appears numerically to be 0, then one may try to ascertain that $x$ is torsion in $K_2^T(E)$ by finding Steinberg relations (this becomes a linear algebra problem in the group $F^\times \otimes F^\times$). On the other hand, if the image appears to be nonzero, this can in principle be proved by computing with enough accuracy. Regarding the torsion in $K_2(E)$ and $K_2^T(E)/ K_2(\mathbb{Q})$, it is known that $K_2(\mathbb{Q})$ embeds in $K_2(E)$ by means of the structural morphism $E \to \operatorname{Spec} \mathbb{Q}$, since this morphism has a section. The group $K_2(\mathbb{Q})$ is infinitely generated (for a description, see Milnor's book Introduction to algebraic $K$-theory). So the torsion of $K_2(E)$ is infinite. I don't know about the torsion of $K_2^T(E)/ K_2(\mathbb{Q})$ in general, but Goncharov and Levin have given a complex computing $K_2^T(E)/K_2(k)$ for an elliptic curve $E/k$, at least when $k$ is algebraically closed, see Theorem 1.5 in Zagier's conjecture on $L(E,2)$. For $k=\overline{\mathbb{Q}}$, this shows (if my computation is correct) that $K_2^T(E)/K_2(\overline{\mathbb{Q}})$ contains a copy of the group $E(\overline{\mathbb{Q}})_{\mathrm{tors}}$.<|endoftext|> TITLE: Projections of compact real algebraic sets QUESTION [8 upvotes]: Suppose that $M$ is a compact, real algebraic subset of $\mathbb R^n$ and $f:\mathbb R^n \to \mathbb R^m$ is the projection to the first $m$ coordinates. If $f$ maps $M$ bijectively unto its image $f(M)$, is it true that $f(M)$ is algebraic? REPLY [9 votes]: The answer is no. (Although the previous example I gave was bad.) Let $C$ be the curve $y^2 = x^2 (x-1)(2-x)$, so $C$ has a smooth component with $1 \leq x \leq 2$, and also a node at $(0,0)$. Let $M$ be the normalization of $C$; explicitly, $$M = \{ (x,y,z) : z^2 = (x-1)(2-x),\ y=xz \}.$$ Then the projection $M \to C$ is $1$-to-$1$ over the smooth component but misses the node. However, such examples can only miss a set of lower dimension than $M$, and cannot occur if all connected components of the Zariski closure of $f(M)$ have the same dimension! This follows from Bialynicki-Birula, A.; Rosenlicht, M., Injective morphisms of real algebraic varieties, Proc. Am. Math. Soc. 13, 200-203 (1962). ZBL0107.14602. At the start of section 2, they prove the following result: Let $V$ and $W$ be real algebraic sets and let $f: V \to W$ be an injective morphism such that $f(V)$ is Zarsiki-dense in $W$. Then $f(V)$ contains a Zariski-open Zariski-dense subset of $W$. Note that this is very false for $f$ non-injective; consider the map $x \mapsto x^2$ from $\mathbb{R}$ to itself. In our setting, take $V = M$ and let $W$ be the Zariski-closure of $f(M)$. For simplicity, let $M$ be irreducible, so $W$ will be as well. Bialynicki-Birula and Rosenlicht's result shows that $f(M)$ must contain a Zariski-open Zariski-dense subset $U$ of $W$, and thus $W \setminus U$ must be a proper Zariski closed subset of $W$. In particular, $W \setminus U$ must have dimension lower than $\dim W = \dim M$. If we now assume that $W$ is irreducible and all its connected components have the same dimension, then $U$ must be dense in $M$ for the analytic topology. But, also, $f(M)$ is closed in the analytic topology since $M$ is compact. We have shown that $f(M)$ is closed for the analytic topology and contains a dense set (namely $U$) for the analytic topology, so $f(M) = W$.<|endoftext|> TITLE: Prime numbers from permutation QUESTION [9 upvotes]: Let $P(n)$ of a sequence $s(1),s(2),s(3),...$ be obtained by leaving $s(1),...,s(n)$ fixed and reverse-cyclically permuting every $n$ consecutive terms thereafter; apply $P(2)$ to $1,2,3,...$ to get $PS(2)$, then apply $P(3)$ to $PS(2)$ to get $PS(3)$, then apply $P(4)$ to $PS(3)$, etc. The limit of $PS(n)$ is $a(n)$ (A057063). The sequence begins $$1, 2, 4, 6, 3, 10, 12, 7, 16, 18, 11, 22, 13, 5, 28$$ Some examples: $$1,2,(4,3),(6,5),(8,7),(10,9),(12,11),(14,13),(16,15),(18,17)$$ $$1,2,4,(6,5,3),(7,10,8),(12,11,9),(13,16,14),(18,17,15)$$ $$1,2,4,6,(3,7,10,5),(12,11,9,8),(16,14,18,13)$$ $$1,2,4,6,3,(10,5,12,11,7),(8,16,14,18,9)$$ I conjecture that $a(n)+1$ is prime if and only if $a(n)=2(n-1)$. Is there a way to prove it? REPLY [9 votes]: 1. Answer to the question. Claim 1. $a(n)=2(n-1)$ iff the number $d=a(n)$ moves only leftwards (while it moves). Proof. At each move, every moving number moves either to the right or $1$ left. The number $d=a(n)$ came to the $n$th position during $P(n-1)$, moving $1$ left. If $d$ moved leftwards through all $P(2),\dots, P(n-1)$, then it started at position $n+(n-2)=2(n-1)$ (so $d=2(n-1)$). Otherwise it shifted leftwards by a smaller distance, so it was smaller than $2(n-1)$. $\quad\square$ Claim 2. A number $d$ moves only to the left (while it moves) iff $d+1$ is prime. Proof. While $d$ moves to the left, it appears at position $d-i+2$ before $P(i)$. Then, at $P(i)$, it moves to the right iff $d-i+2>i$ and $d-i+2\equiv 1\pmod i$, that is, if $d\equiv -1\pmod i$ and $d+1\geq 2i$, or in other wirds, $i\mid d+1$ and $(d+1)/i>1$. So, if $p\leq d/2$ is the least prime divisor of $d+1$, then at $P(p)$ the number $d-p+2=(d+1)-(p-1)>p$ shifts to the right (if it did not shift earier --- in fact, it did not). Otherwise, $d$ is prime, and it cannot shift right. $\quad\square$ The two claims yield the result. 2. Proof that the resulting arrangement is a permutation. We fix a number $d$ and describe how it moves. We aim at proving that it shifts rightwards only finitely many times; this clearly yields that $d$ will eventually stop. Let $a_n$ denote the position of $d$ before $P(n)$, and put $b_n=a_n+n-1$. Informally speaking, $b_n-1$ denotes a position from which $d$ would come to its current position moving only to the left. If $d$ moves at $P(n)$, we have $a_{n+1}=a_n-1$ if $a_n\not\equiv 1\pmod n$ and $a_{n+1}=a_n+n-1$ otherwise. Therefore, $b_{n+1}=b_n$ if $n\nmid b_n$, and $b_{n+1}=b_n+n$ otherwise (in the latter case we call $n$ a crucial index); we want to show that there are finitely many crucial indices. Let $n1$, we have $b_{n+1}=n(c_n+1)$, and hence $c_n+1>b_{n+1}/(n+1)\geq b_{n+1}/m=b_m/m=c_m$. Since $c_m$ is an integer, we conclude that $c_m\leq c_n$. Number all crucial indices as $n_11$, we have $B_{i+1}=B_i\cdot \frac{k+1}k$. This may happen only finitely many times if $k>1$, since all the $B_i$ are integers. Thus, the (non-increasing) sequence $C_i$ cannot preserve any value $k>1$ indefinitely, so it decreases from time to time, and eventually it reaches $1$, as desired.<|endoftext|> TITLE: Non-finitely presented FP groups with cohomological dimension $2$ QUESTION [8 upvotes]: In this recent preprint, the authors construct a certain uncountable family of non-finitely presented FP groups. Recall that group is an FP group if the trivial $\mathbb Z[G]$-module $\mathbb Z$ has a finite projective resolution by finitely generated $\mathbb Z[G]$-projectives. In fact, if I understand correctly, the family that they construct is a family of groups of cohomological dimension $2$; but this specific point does not seem to be the focus of their paper. Furthermore, theirs is not the first example of a non-finitely presented FP group, and so it leads me to my question: Were there earlier examples of non finitely-presented FP groups of cohomological dimension $2$ ? I want to be liberal with the meaning of the word "example": my question is specifically whether there was an earlier proof that such things existed, whether or not some explicit examples were given. (NB : this is not quite my field of research, so maybe this is (very?) classical : any pointers to classical literature on the topic where this question is discussed would be helpful) REPLY [11 votes]: The Bestvina-Brady construction of non-finitely presented groups of type FP produces groups of cohomological dimension two. Bestvina-Brady groups are parametrized by finite flag simplicial complexes. The Bestvina-Brady group is of type FP iff the flag complex is acyclic (=has the same ordinary homology as a point), is finitely generated iff the flag complex is connected, is finitely presented iff the flag complex is simply connected. For an acyclic flag complex the cohomological dimension of the Bestvina-Brady group is equal to the dimension of the flag complex. So if you apply the Bestvina-Brady construction to any non-simply connected 2-dimensional acyclic complex you get a group having the properties that you asked for. My `generalized Bestvina-Brady groups' allow you to construct uncountable families of groups of type FP, each having cohomological dimension two. The main novelty in my article with Tom Brown that you referenced in your question is that we use very different methods, whereas previous constructions all used CAT(0) cubical techniques. As for some references: for a discussion of the question before it was answered, look at K S Brown's book Cohomology of Groups, Chapter VIII, especially VIII.5-VIII.8. The article by Bestvina and Brady is the best place to look for the construction of the examples.<|endoftext|> TITLE: Asymptotic for the average of $|d(n)-\log n|$? QUESTION [10 upvotes]: Let $d(n)$ be the number of positive integers that divide $n$. It is well known that $d(n)$ is on average $\log n$. However, it is also well known that for most $n$ the number $d(n)$ is rather close to $(\log n)^{\log 2}$. What explains the anomaly is that the values of $d(n)$ that are larger than its typical value are typically way larger, hence, they dominate the first moment. The average of $d^2(n)$ is $\gg (\log n)^3$, which is just another manifestation of the same anomaly. To understand these things a little better I was interested in estimating the average $$ R(x):= \sum_{1\leq n \leq x } | d(n)-\log n|.$$ Has anyone seen an asymptotic for this in the literature? As Alexander Kalmynin wrote in the comments one has $$R(x)=\sum_{1\leq n \leq x } |d(n)-\log n| \ll \sum_{n\leq x } d(n) +x\log x\ll x\log x .$$ Furthermore, for each $\epsilon>0$ there exists a subset $A_\epsilon\subset \mathbb N$ of density $1$ with $d(n)\leq (\log n)^{\log 2+\epsilon}$ for all $n\in A_\epsilon$, we get $$ R(x) \geq \sum_{n\in A_\epsilon\cap[1,x]} ((\log n)- (\log n)^{9/10}) \gg x \log x.$$ Hence, $$ x\log x\ll R(x) \ll x \log x.$$ REPLY [7 votes]: Here is a variation of Lucia's nice argument. Writing $$|d(n)-\log n|=d(n)+\log n-2\min(d(n),\log n),$$ we see that $$R(x)=2x\log x+O(x)-2\sum_{n\leq x}\min(d(n),\log n).$$ Now let $\kappa\in(0,1)$ be fixed, and let us use that $\min(d(n),\log n)\leq d(n)^\kappa(\log n)^{1-\kappa}$. Then, by Theorem 2 in Chapter II.6 of Tenenbaum: Introduction to analytic and probabilistic number theory, we get $$\sum_{n\leq x}\min(d(n),\log n)\leq(\log x)^{1-\kappa}\sum_{n\leq x}2^{\kappa\Omega(n)}\ll_\kappa x(\log x)^{2^\kappa-\kappa}.$$ The value $\kappa:=-\log\log 2/\log 2\approx 0.529$ yields the minimal exponent $$2^\kappa-\kappa=(1+\log\log 2)/\log 2\approx 0.914,$$ whence $$R(x)=2x\log x + O\left(x(\log x)^{(1+\log\log 2)/\log 2}\right).$$<|endoftext|> TITLE: Has the "partial Sophomore's Dream function" been studied before? QUESTION [6 upvotes]: We can consider the generalized Harmonic numbers $$H_{n,m} := \sum_{k=1}^{n} \frac{1}{k^{m}} $$ as a partial version of the Riemann zeta function, because $$\lim_{n \to \infty} H_{n,m} = \zeta(m). $$ We could also define the "partial Sophomore's Dream function" $$S_{r} := \sum_{q=1}^{r}\frac{1}{q^{q}} ,$$ as we have $$\lim_{r \to \infty} S_{r} := S= \int_{0}^{1}x^{-x}dx ,$$ where the integral on the right is equal to the first Sophomore's Dream constant. Question: while studying the series $$A := \sum_{r=1}^{\infty}(S-S_{r}) \approx 0.3371877158, $$ I was wondering whether the function $S_{r}$ has already been studied, and if someone has coined a name for it that is hopefully less awkward than mine. I am looking for alternative representations of $S_{r}$ that might help finding a closed form of $A$. Note: this question was previously asked on MSE. REPLY [6 votes]: We have that $S=\sum_{n=1}^\infty \frac 1 {n^n}$ and $S_r=\sum_{n=1}^r \frac 1 {n^n}$. Thus, $A=\sum_{r=1}^\infty \sum_{n=r+1}^\infty \frac 1 {n^n} = \sum_{n=2}^\infty \sum_{r=1}^{n-1} \frac 1 {n^n}$ But then, $A=\sum_{n=2}^\infty \frac {n-1} {n^n}$. To evaluate this, consider the integral $\int_0^1 t^{-tx} dt = \sum_{n=1}^\infty \frac {x^{n-1}} {n^n}$. Differentiating with respect to $x$, we get $\sum_{n=2}^\infty \frac {(n-1)x^{n-2}} {n^n} = \int_0^1 -t\ln t \cdot t^{-xt} dt$. And substituting $x=1$ finally gives $A=\int_0^1 -\ln t \cdot t^{1-t} dt$. Note that a WolframAlpha calculation gives 0.3371877 as the value of the integral, in excellent agreement with the value you obtained. REPLY [3 votes]: I don't know about an alternative name, but your query for "alternative representations of $S_r$" can be answered by the identity $$\sum_{q=1}^r \frac{1}{q^q}=\int_0^1 \frac{\Gamma(r,-x\ln x)}{x^x\, \Gamma(r)}dx.$$ Since $\lim_{r\rightarrow\infty} \Gamma(r,a)/\Gamma(r)=1$, in that limit you recover the sophomore's dream.<|endoftext|> TITLE: Singular curves of genus 1 QUESTION [7 upvotes]: Let $C$ be an irreducible curve of arithmetic genus $1$ over a field $k$ and with a double $k$-point $p\in C$. Is $C$ rational over $k$? If $C$ is a plane cubic the answer is positive since we can parametrize $C$ with the lines through $p$. In general I would embed $C$ in some projective space $\mathbb{P}^n$ and I would project $C$ until I get a plane cubic with a node. I think this should be ok if $k$ is infinite but if $k$ is finite I do not know how to make sure that $C\subset \mathbb{P}^n$ can be birationally projected onto a plane nodal cubic. Thank you. REPLY [10 votes]: There's no problem over finite fields, but there is a problem over fields that have a nontrivial Brauer class. If you take a genus $0$ curve that's not rational (say a plane quadric), it will always have points over a degree $2$ extension (intersect with a line), and then you can glue two of them together to get a nodal curve of arithmetic genus $1$. This curve will not be a nodal cubic, but you can embed it as a degree $4$ curve in $\mathbb P^3$. For example, let $a$ be a quadratic nonresidue mod $p$, consider the curve in $\mathbb P^3$ with coordinates $x,y,z,w$ given by the equations $x^2 -a y^2 - p z^2$ and $xy = zw$ over $\mathbb Q_p$. The only rational point of this curve is $(x:y:z:w)=(0:0:0:1)$, which is a node, so the curve has arithmetic genus $1$, has a node, and isn't rational.<|endoftext|> TITLE: Fixed points of a function $z\mapsto\overline{P(z)}$ of a complex variable QUESTION [6 upvotes]: The equation $z^2=\overline{z}$ has four zeros and this example motivates us to generalize the problem to this form; How many zeros does the equation $P(z)=\overline{z}$ have if $P(z)$ is a polynomial of degree $n>1?$ Can we find the bound for the number of zeros of this problem? The example motivate us to conjecture that it may be at most $2n,$ if not at most $2n+n-2=3n-2.$ I am suggesting mere by intuition! May I request you to share your thoughts on this? REPLY [9 votes]: Function $z\mapsto\overline{P(z)}$ has at most $3d-2$ fixed points, where $d\geq 2$ is the degree of $P$, and this is best possible. This remarkable result is due to Khavinson and Świa̧tek, MR1933331 Khavinson, Dmitry, Świa̧tek, Grzegorz , On the number of zeros of certain harmonic polynomials, Proc. Amer. Math. Soc. 131 (2003), no. 2, 409–414. and it was later generalized to rational functions, and to some transcendental functions. There is a survey of related results: D. Khavinson and G. Neumann, From the fundamental theorem of algebra to astrophysics: a “harmonious” path, Notices Amer. Math. Soc. 55 (2008), no. 6, 666–675. Let me mention a major unsolved problem: let $p$, $q$ be polynomials of degrees $m>n$. How many solutions can the equation $$\overline{p(z)}=q(z)$$ have? Can one do better than the Bézout estimate $mn$?<|endoftext|> TITLE: Specht modules for symmetric group $S_{\infty}$ QUESTION [7 upvotes]: Specht modules of $S_n$, the symmetric group on n symbols is well-known. Is there an analogue of these modules for $S_{\infty}$, the set of all permutations of $\mathbb N$? Also, please share some references to learn the representation theory of $S_{\infty}$. Thank you. REPLY [13 votes]: There are at least two possible analogues here, depending on which features of Specht modules you consider to be most essential. However, I would like to emphasise that in both cases, they are representations of the group of all "finitary" permutations of $\mathbb{N}$ (permutations that fix all but finitely many elements), rather than the group of arbitrary permutations of $\mathbb{N}$. This finitary version can be described as $\varinjlim S_n$, with respect to the standard inclusions of symmetric groups (if $m \leq n$, $S_m$ is the subgroup of $S_n$ fixing all numbers larger than $m$). So it is closely tied to the finite symmetric groups, and in this answer, I will only consider this finitary version. There is also the theory of Deligne categories (usually written $\underline{Rep}(S_t)$), which interpolate the Specht modules in a way that is similar to what is described in the first analogue below. Since the theory is more abstract (doesn't directly have modules or characters), I leave it out. Analogue 1: Stable Representation Theory of Sam and Snowden. Let's consider the simplest non-trivial Specht module $S^{(n-1,1)}$. If $\mathbb{C}^n$ is the permutation representation of $S_n$, it contains an invariant subspace $ \mathbb{C}(1,1,\ldots,1)$. A complement to this invariant subspace is given by mean-zero vectors: $\{(x_1, x_2, \ldots, x_n) \mid \sum_i x_i = 0\}$; the condition $\sum_i x_i = 0$ is preserved under permutation of the coordinates. This subspace is the Specht module $S^{(n-1,1)}$. Notice that this construction is "uniform in $n$". So we could have replaced $\mathbb{C}^n$ by an infinite-dimensional vector space $\mathbb{C}^\infty$. This space no longer contains an invariant vector, because the all-ones vector $(1,1,\ldots)$ is no longer a finite linear combination of basis vectors. Nevertheless, there is a mean-zero subrepresentation defined just as before, and it is irreducible. The map $\mathbb{C}^\infty \to \mathbb{C}$ given by $(x_1, x_2, \ldots) \mapsto \sum x_i$ is a homomorphism having this mean-zero subrepresetation as its kernel. Thus the trivial representation is a quotient, but not a subrepresentation of the permutation representation. Here we encounter a point of difference from the finite theory; these representations are no longer semisimple. Also instead of labelling this representation by the "partition" $(\infty -1, 1)$, we just remember all entries after the first; in this case it is just $(1)$. To construct more complicated Specht modules, we can instead look inside tensor powers $(\mathbb{C}^n)^{\otimes k}$; it turns out that the result will contain all Specht modules indexed by partitions with at most $k$ boxes below the top row. The construction again may be generalised to the case $n=\infty$, which according to our convention will give irreducible representations labelled by partitions of size at most $k$ (since we omit the first row which would contain the symbol $\infty$). I have been a bit vague about how to find a given Specht module inside $(\mathbb{C}^n)^{\otimes k}$, but this essentially relies on the Schur-Weyl duality that holds between the symmetric group and the partition algebra. The details of this construction, as well as some basic properties of the category $Rep(S_\infty)$ in which these Specht modules naturally live, are described in the work of Sam and Snowden. See section 6 of the following paper: Sam, Steven V.; Snowden, Andrew, Stability patterns in representation theory, Forum Math. Sigma 3, Paper No. e11, 108 p. (2015). ZBL1319.05146. The category $Rep(S_\infty)$ is closed under the usual tensor product of representations (although it is far from being all representations of the abstract group $S_\infty$). If we write $V_\lambda$ for the the object generalising the Specht module $S^{(n-|\lambda|, \lambda)}$, then the tensor product multiplicities are the reduced Kronecker coefficients. Now, notice that $S_\infty$ contains a subgroup $S_n \times H_n$, where the factor $S_n$ permutes the numbers $\{1,2, \ldots, n\} \subseteq n$, and the factor $H_n$ is abstractly isomorphic to $S_\infty$, but permutes the numbers $\{n+1, n+2, \ldots\}$. If $V$ is in $Rep(S_\infty)$, then the $H_n$ invariants $V^{H_n}$ retain an action of $S_n$ (because $S_n$ commutes with the $H_n$-action). Thus we get a "specialisation" functor $F_n$ from $Rep(S_\infty)$ to $Rep(S_n)$ for any $n$. This turns out to be a monoidal functor. A consequence of defining things in this way is that $F_n(V_\lambda) = S^{(n-|\lambda|, \lambda)}$, provided that $(n-|\lambda|, \lambda)$ is a partition (i.e. that $n-|\lambda| \geq \lambda_1$). If that is not the case ($n$ is too small), then instead $F_n(V_\lambda) = 0$. One may wish to compute the derived functors of $F_n$, and this can be done by constructing a suitable injective resolution in $Rep(S_\infty)$. One way to do this is using an analogue of the BGG resolution; see Theorem 2.3.1 of Sam, Steven V.; Snowden, Andrew, GL-equivariant modules over polynomial rings in infinitely many variables, Trans. Am. Math. Soc. 368, No. 2, 1097-1158 (2016). ZBL1436.13012. We may view the symmetric group as the subgroup of the general linear group consisting of permutation matrices. It turns out that objects obtained by restricting representations of $GL_\infty$ to $S_\infty$ are injective in $Rep(S_\infty)$. So a different strategy is to understand how representations of general linear groups restrict to symmetric groups. Although this is not very well understood (it is an open problem to give a combinatorial formula known for the restriction multiplicities), one can use this approach to construct such an injective resolution; see Ryba, Christopher, Littlewood complexes for symmetric groups, ZBL07383251. One upshot of this is as follows. There is a good notion of the "character" of irreducible representations of general linear groups - they are Schur functions, which form a basis of the ring of symmetric functions, $\Lambda$. This connection between general linear groups and symmetric groups lets us identify the Grothendieck ring of $GL_\infty$ (i.e. $\Lambda$) with the Grothendieck ring of $Rep(S_\infty)$. Thus we obtain "characters" for the $V_\lambda$, which are symmetric functions $\tilde{s}_\lambda$ introduced independently in the following two papers: Orellana, Rosa; Zabrocki, Mike, Symmetric group characters as symmetric functions, Adv. Math. 390, Article ID 107943, 34 p. (2021). ZBL1473.05309. Assaf, Sami H.; Speyer, David E., Specht modules decompose as alternating sums of restrictions of Schur modules, Proc. Am. Math. Soc. 148, No. 3, 1015-1029 (2020). ZBL1455.20007. The "character" of $V_\lambda$ allows us to compute the characters of the usual Specht modules $S^{(n-|\lambda|, \lambda)}$, strengthening the claim that $V_\lambda$ is the correct notion of Specht module for $S_\infty$. (Although this is a bit subtle - to get a character value, we must evaluate a symmetric function at the eigenvalues of a permutation matrix, just as one would evaluate a Schur function at the eigenvalues of an element of $GL_n$.) Analogue 2: Edrei-Thoma theorem In the previous discussion, I mentioned that $Rep(S_\infty)$ is quite far away from being all representations of $S_\infty$. For example, $S_\infty$ has a sign representation, which does not appear in that category. We will discuss a theory that includes the sign representations (and many others not incuded in $Rep(S_\infty)$). In this setting we will eschew Specht modules in favour of their characters. What is the right notion of a character of (possibly infinite-dimensional) representation of an infinite group? Well, suppose that $\chi : G \to \mathbb{C}$ is the character of a representation of a finite group. We normalise this character: let $\phi: G \to \mathbb{C}$ be defined by $\phi(g) = \chi(g)/\chi(1)$. This has the following properties: $\phi(\mathrm{Id}) = 1$ $\phi(g_1 g_2) = \phi(g_2 g_1)$ For any finite sequence of group elements $g_i$, the matrix $\phi(g_1g_2^{-1})$ is Hermitian and positive semidefinite Some comments are in order. The first condition is trivial, but hides the essential detail that characters in the usual sense are poorly behaved on infinite dimensional representations (e.g. the trace of the identity is infinite). So instead of looking for genuine characters, we will look for such normalised characters. The second condition says that our function is a class function, as any character would be. The third condition has two parts. Being Hermitian requires $\chi(g_1g_2^{-1}) = \overline{\chi(g_2g_1^{-1})}$ , which holds for finite groups because the eigenvalues of the representing matrices are roots of unity (more generally this works for compact groups). The definiteness condition for such sequences follows from the particular case where $g_1, g_2, \ldots, g_l$ enumerates all the elements in the group (by taking a submatrix of this matrix). However, in order to avoid dealing with infinite matricess when the group is infinite, we stick to finite submatrices. Finally let me add that the normalised character corresponding to $\chi_1 \oplus \chi_2$ is $\frac{\chi_1(g) + \chi_2(g)}{\chi_1(\mathrm{Id}) + \chi_2(\mathrm{Id})} = \frac{\chi_1(g)}{\chi_1(\mathrm{Id})} \cdot \frac{\chi_1(\mathrm{Id})}{\chi_1(\mathrm{Id}) + \chi_2(\mathrm{Id})} + \frac{\chi_1(g)}{\chi_1(\mathrm{Id})} \cdot \frac{\chi_2(\mathrm{Id})}{\chi_1(\mathrm{Id}) + \chi_2(\mathrm{Id})}$ which is in particular a convex combination of the normalised characters for $\chi_1$ and $\chi_2$. So if our representation was reducible, it is a non-trivial convex combination of other normalised characters. Thus we arrive at our definition of "irreducible character of $S_\infty$" - it is a function on the group obeying the three properties stated above, which cannot be written as a non-trivial convex-combination of other such functions. For $S_\infty$, such functions have been classified, and this is the Edrei-Thoma theorem. Since this post is already very long, I will just summarise the result. The functions in question are parametrised by pairs of infinite sequences of non-negative real numbers $(\alpha_i, \beta_i)$, where $\alpha_1 \geq \alpha_2 \geq \cdots$ and $\beta_1 \geq \beta_2 \geq \cdots$, and $\sum_i \alpha_i + \sum_j \beta_j \leq 1$. We let $\gamma = 1 - \sum_i \alpha_i - \sum_j \beta_j$ (it is a real number between 0 and 1). This odd-looking parametrisation can be interpreted in terms of partitions as follows. Let us consider a sequence of normalised characters, viewed as functions on successively larger symmetric groups. Provided some kind of suitable convergence holds, we will get a normalised character on $S_\infty$. We take these normalised characters to correspond to progressively larger partitions $\lambda^{(n)}$ (indexing Specht modules). Then $\lambda_i^{(n)} / |\lambda^{(n)}|$ describes the fraction of the partition that lies in the $i$-th row. Similarly $(\lambda^{(n)})_j^\prime / |\lambda^{(n)}|$ describes the fraction of the partition that lies in the $j$-th column. In the limit, these quantities converge to the parameters $\alpha_i$ and $\beta_j$, and the remainder, $\gamma$ describes the fraction that is left over and distributed in the "bulk" of the diagram. There are explicit formulae for the normalised character in terms of the $\alpha_i$ and $\beta_j$. For more detail, one can consult the following references (which don't seem to appear in the citation engine): Okounkov - On the representations of the infinite symmetric group Borodin and Olshanski - Representations of the Infinite Symmetric Group So for this definition of irreducible (normalised) character, we get uncountably many. While this approach may be broader than the first, what we get is in some ways quite different from the classical Specht modules.<|endoftext|> TITLE: Log determinant of quadratic form QUESTION [5 upvotes]: I am reading a paper Cook and Forzani - Likelihood-Based Sufficient Dimension Reduction where the author uses the following result from matrix analysis but does not explain why it is true nor provide any reference. More specifically, let $B \in \mathbb{R}^{p\times d}$ be a semi-orthogonal matrix, i.e $B^\top B = I_d$, and $d < p$. Let $\Sigma$ and $\Delta$ denote two symmetric positive definite matrices such that $\Sigma - \Delta$ is also positive definite. What they claim is that $$ \log \det \left\lvert B^\top \Sigma^{-1} B \right\rvert \leq \log \det \left\lvert B^\top \Delta^{-1} B \right\rvert.$$ Could someone point me in the direction of explaining why it is true? I am thinking of using the Poincaré separation theorem, which provides bounds on the eigenvalues of the matrices on both the left and right-hand sides; however, I did not make any progress with it. The claim is in the proof of Proposition 3, at the top of page 33. Here $\Sigma = \operatorname{Var}(X)$, and $\Delta = E (\operatorname{Var}(X\vert y))$, so $\Sigma-\Delta = \operatorname{Var}(E(X \vert y))$ is also positive definite. The $B$ in my question plays the role of $B_0$ in the paper. REPLY [2 votes]: Since $\Sigma \succ \Delta$, by operator monotonicity we have $\Sigma^{-1} \prec \Delta^{-1}$ and thus $B^{\top}\Sigma^{-1}B \prec B^{\top}\Delta^{-1}B$. Since log on positive real is increasing, the trace monotonicity (also Theorem 2.10 here) gives $\text{trace}\left(\log\left(B^{\top}\Sigma^{-1}B\right)\right) < \text{trace}\left(\log\left(B^{\top}\Delta^{-1}B\right)\right)$ where the log inside the last inequality is principal logarithm. Replacing $\text{trace}\left(\log\left(\cdot\right)\right)$ with $\log\det(\cdot)$, we get the desired inequality.<|endoftext|> TITLE: Inscribed $n$-polytope with $2^n$ vertices of maximal volume QUESTION [6 upvotes]: The question is in the title: Question: Which inscribed $n$-dimensional polytope (inscribed in the unit sphere) with $2^n$ vertices has the largest possible volume? Is it the $n$-dimensional cube? If not, how much larger can its volume be? REPLY [13 votes]: For $n=3$, the maximal volume polytope with 8 vertices is described in that paper. Berman, J. D.; Hanes, K., Volumes of polyhedra inscribed in the unit sphere in (E^3), Math. Ann. 188, 78-84 (1970). ZBL0187.19604. It is combinatorially very different from a cube: it is simplicial, and each vertex has degree 4 or 5. Edit. An inscribed polytope which maximizes the volume given the number of vertices is always simplicial (see Lemma 1 in Horváth, Ákos G.; Lángi, Zsolt, Maximum volume polytopes inscribed in the unit sphere, Monatsh. Math. 181, No. 2, 341-354 (2016). ZBL1354.52016.), so the answer is "no" as well in any dimension $>2$. The gap is even quantitative. Let $\mu_n$ be the largest volume of an $n$-dimensional polytope with $2^n$ vertices inscribed in the ball of radius $\sqrt{n}$ (this normalization is better since it counterbalances the fact that the ball of unit radius has a tiny volume for large $n$). The cube gives the lower bound $\mu_n \geq 2^n$. By taking direct products, we have $\mu_{m+n} \geq \mu_m\mu_n$, and therefore $\mu_{3n} \geq \mu_3^n \geq (2+\epsilon)^{3n}$, exponentially better that the cube. We cannot do much better since the upper bound $\mu_n \leq C^n$ for some constant $C$ holds trivially by comparing with the ball itself.<|endoftext|> TITLE: Cycling through $\{0,1\}^{(2^n)}$ such that all Hamming distance appear equally frequently QUESTION [5 upvotes]: Let $n\in\mathbb{N}$ be a positive integer. Let $\{0,1\}^{(2^n)}$ be the set of $0,1$-sequences of length $2^n$. For $a,b\in \{0,1\}^{(2^n)}$ let $d_h(a,b)$ be the Hamming distance between $a$ and $b$. Question. Is there a bijection $\varphi:\{0,\ldots,2^{(2^n)}-1\} \to \{0,1\}^{(2^n)}$ such that the multiset $$\Big\{d_h\Big(\varphi(k), \varphi(k+1)\Big): k< 2^{(2^n)}-1\Big\}\cup\Big\{d_h\Big(\varphi(2^{(2^n)}-1), \varphi(0)\Big)\Big\}$$ is composed of elements $1, 2, \dots, 2^n$ with the same multiplicity $2^{(2^n)} / 2^n = 2^{2^n - n}$? Example. For $n = 1$, the answer is yes; consider the "run" $00, 11, 10, 01$, that is, the possible Hamming distances $1$ and $2$ each appear $2$ times when going through all strings of length $2^n = 2^1$. REPLY [3 votes]: The answer is yes. The proof is as follows: If there is a Hamiltonian cycle $f: \{0,...,2^{2^n}-1\} \rightarrow \{0,1\}^{2^n}$ of the hypercube where the number of edges of a given direction in the Hamiltonian cycle are the same, then there is such a bijection $\varphi$, by setting $\varphi_m(x)=\sum_{k=1}^m f_k(x)$, $m=1,2, ... 2^n$ (i.e. taking the cumulative sum of $f(x)$). Such a Hamiltonian cycle will be called a balanced Hamiltonian cycle. My construction of balanced Hamiltonian cycles is inspired by this paper, which provides balanced Hamiltonian cycles for $n=1,2,3$. The construction for $n=2$ will serve as the base case of mathematical induction. Assume the existence of a balanced Hamiltonian cycle for $n$, and make it a directed cycle. Let $g$ be a function on $\{0,1\}^{2^n}$ where $g(x)$ is the next vertex of $x$. Let $h(x)$ be an involution on $\{0,1\}^{2^n}$ such that there are exactly $2^{2^n}/2^n$ values where $x$ and $h(x)$ differs at the $k$th bit, $k \in \{1 ... 2^n \}$. (The existence of $h$ will be proven later.) Then there is a balanced Hamiltonian cycle on the hypercube of dimension $2^{n+1}$. Let's identify the $2^{n+1}$-dimensional hypercube by the Cartesian product of the $2^n$-dimensional hypercube with itself. Let $(a_0, b_0)$ be a vertex of the $2^{n+1}$-dimensional hypercube. The sequence $(a_k, b_k)$ is defined as follows: if $k$ is odd, then $(a_{k+1}, b_{k+1})=(a_k, g(b_k))$; if $k=0 \text{ or } 2 (\text{mod } 2^{2^n+1})$, then $(a_{k+1}, b_{k+1})=(g(a_k), b_k)$; otherwise $(a_{k+1}, b_{k+1})=(h(a_k), b_k)$ One can see the function $k\rightarrow (a_k, b_k)$ plays the role of $f$ for $n+1$. The point is to analyze a cycle of length $2^{2^n+1}$, conclude that $(a_{2^{2^n+1}m}, b_{2^{2^n+1}m})=(g(g(a_{2^{2^n+1}(m-1)})),b_{2^{2^n+1}(m-1)})$ and it follows that $(a_k, b_k)$ traces out a Hamiltonian cycle (i.e. you need $m=2^{2^n-1}$ for the point to return to $(a_0,b_0)$). It's easy to check the Hamiltonian cycle is balanced. Now we only to prove the existence of $h$. If $n=2$, then such $h$ exists: 0000 - 0001 1110 - 1111 0010 - 0110 0011 - 0111 0100 - 1100 0101 - 1101 1000 - 1010 1001 - 1101 Assume the existence of $h$ for $n$. Choose a subset $K$ of $\{0,1\}^{2^n}$ with size $2^{2^n-1}$ that is closed under $h$ and there are exactly $2^{2^n-1}/2^n$ elements $x$ in $K$ where $x$ and $h(x)$ differs at the $k$th bit ($k \in \{1 ... 2^n \}$). Define a function $p$ on $\{0,1\}^{2^n} \times \{0,1\}^{2^n}$: $p(a,b)=(a,h(b))$ if $b \in K$, otherwise $p(a,b)=(h(a),b)$. Then $p$ plays the role of $h$ for $n+1$.<|endoftext|> TITLE: Density of restrictions of harmonic functions inside a ball QUESTION [9 upvotes]: Let $B$ be the closed unit ball in $\mathbb R^3$ centered at the origin and let $U= \{x\in \mathbb R^3\,:\, \frac{1}{2}\leq |x| \leq 1\}.$ Let $$ S_U= \{u \in C^{\infty}(U)\,:\, \Delta u =0 \quad\text{on $U^{\textrm{int}}$}\},$$ and $$ S_B= \{u \in C^{\infty}(B)\,:\, \Delta u =0 \quad\text{on $B^{\textrm{int}}$}\}.$$ Is the following statement true? Given any $\epsilon>0$ and any $u \in S_U$, there exists an element $v \in S_B$ such that $\|v-u\|_{L^2(U)} \leq \epsilon$. REPLY [7 votes]: Suppose that $u\in S_{U}$. Then I claim that $\inf\{\|u-v\|_{L^{2}(U)}:v\in S_{B}\}$ is bounded below by the standard deviation of the spherical symmetrization $u^{\sharp}$ of $u$. For this post, the $L^{2}(U)$ norm will be with respect to the normalized area probability measure on $U$. Let $\mu$ be the Haar probability measure on the group of all $3\times 3$ orthogonal matrices. Let $\nu$ be the normalized area probability measure on $S^{2}$. Define the spherical symmetrization $w^{\sharp}$ of a function $w$ by letting $$w^{\sharp}(x)=\int_{A\in O(3)}(w\circ A)(x)d\mu(A).$$ Observe that $$w^{\sharp}(x)=\int_{y\in S^{2}}w(\|x\|\cdot y)d\nu(y).$$ Observe that if $w$ is harmonic, then $w^{\sharp}$ is also harmonic, and there are constants $\alpha,\beta$ such that $w^{\sharp}(x)=\frac{\alpha}{\|x\|}+\beta$ (this fact generalizes to all dimensions $n\geq 2$). By Jensen's inequality, if $r\in[0,1)$ and $x\in S^{2}$, then $$(f^{\sharp}(rx)-g^{\sharp}(rx))^{2}=(\int_{y\in S^{2}}f(ry)-g(ry)d\nu(y))^{2}\leq\int_{y\in S^{2}}(f(ry)-g(ry))^{2}d\nu(y).$$ Therefore, by integrating, we obtain $$\int_{x\in S^{2}}(f^{\sharp}(rx)-g^{\sharp}(rx))^{2}dx\leq\int_{y\in S^{2}}(f(ry)-g(ry))^{2}d\nu(y).$$ Therefore, if $f,g:U\rightarrow\mathbb{R}$ are continuous, then $\|f^{\sharp}-g^{\sharp}\|_{L^{2}(U)}\leq\|f-g\|_{L^{2}(U)}$ Suppose $u\in S_{U},v\in S_{B}$. Since $v$ is harmonic on $B$, the function $v$ satisfies the mean-value property, so the function $v^{\sharp}$ is constant. Therefore, $$\text{Var}(u^{\sharp})\leq\|u^{\sharp}-v^{\sharp}\|_{L_{2}(U)}^{2}\leq\|u-v\|_{L^{2}(U)}^{2}.$$ There are plenty of functions $u$ that are harmonic on $U$ but where $u^{\sharp}$ is non-constant on $U$ (such as the Newtonian potential), and for each such function, we have $$\text{Var}(u^{\sharp})>0.$$ This proof generalizes to any dimension $n\geq 2$ where the balls $B,B\setminus U$ have any radii but are still centered at $0$.<|endoftext|> TITLE: Defining the abstract tensor product of W*-algebras via a universal property QUESTION [9 upvotes]: I am playing around a bit with $W^*$-algebras, and I'm trying to come up with a definition for the $W^*$-algebraic tensor product. Here is my first attempt: It is easy to show that such an object exists. Simply represent the $W^*$-algebras as concrete von Neumann algebras on Hilbert spaces and consider the usual von Neumann algebra tensor product (i.e. the weak closure of the algebraic tensor product $M \odot N$). However, I fail to show that the above definition determines the $W^*$-algebra $M \overline{\otimes} N$ uniquely (up to (normal) $*$-isomorphism). Is this true? If not, can we still modify the above universal property to get a well-working definition? Basically, I want my universal property to convey the intuition that the $W^*$-tensor product is simply the usual von Neumann tensor product when we appropriately represent the spaces involved. Thanks in advance for any help! REPLY [5 votes]: Something similar was studied by Wiersma in arXiv:1506.01671 [math.OA]. However, that paper takes a different definition of the universal property: it wants the individual maps $\sigma$ and $\tau$ to be weak$^*$-homeomorphisms onto their ranges, and the map $\iota$ needs to have this property as well. However, even here you do not recover the usual von Neumann tensor product, but rather something larger.<|endoftext|> TITLE: Unpublished result of Rosser in Sieve Methods book QUESTION [14 upvotes]: Erdős and Selfridge (1971) state that the following is "implied by an unpublished result of Rosser" which they claim appears in a forthcoming book on sieve methods by Halberstam and Richert. I guess I could try to find something implying it in that 400-page book (called "Sieve Methods", available online in unsearchable format) but I thought I'd ask if maybe somebody knows a precise reference [or a proof!] for this result? For every $\epsilon>0$, there is a constant $c=c(\epsilon)$ such that at least $c\frac{k}{\log k}$ integers between $n$ and $n+k$ have all their prime factors greater than $k^{1/2-\epsilon}$. REPLY [10 votes]: This is part of the theory of the "linear sieve." Chapter 8 of the book of Halberstam and Richert deals with this topic. Alternatively you could look at Iwaniec's paper On the error term in the linear sieve. The result you want can be extracted from the lower bound (1.4) on page 2 of Iwaniec's paper, together with the formula for $f(s)$ at the top of that page.<|endoftext|> TITLE: How to find the asymptotics of a linear two-dimensional recurrence relation QUESTION [8 upvotes]: Let $d$ be a positive number. There is a two dimensional recurrence relation as follow: $$R(n,m) = R(n-1,m-1) + R(n,m-d)$$ where $R(0,m) = 1$ and $R(n,0) = R(n,1) = \cdots = R(n, d-1) = 1$ for all $n,m>0$. How to analyze the asymptotics of $R(n, kn)$ for fixed $k$? It is easy to see that $$R(n, kn) = O\left( c_{k,d}^{n} \cdot (n+k+d)^{O(1)} \right)$$ Is there a way (or an algorithm) to find $c_{k,d}$ given $k$ and $d$? PS: I have calcuated the bivariate generating function of $R(\cdot, \cdot)$: \begin{align} f(x,y) &= \frac{1 - xy - y^{d} + xy^{d}}{(1 - x)(1 - y)(1 - xy - y^{d})} \\ &= \frac{1}{(1 - x)(1 - y)} + \frac{xy^{d}}{(1 - x)(1 - y)(1 - xy - y^{d})} \\ \end{align} REPLY [2 votes]: The book Analytic Combinatorics in Several Variables by Pemantle and Wilson covers problems like this extensively.<|endoftext|> TITLE: Hensel's proof that $e$ is transcendental QUESTION [19 upvotes]: When he introduced $p$-adic numbers, Kurt Hensel produced an incorrect local/global proof of the fact that $e$ is transcendental. Apparently, the intended proof goes along the following lines: studying the well-defined $p$-adic number \begin{equation} e^p=\sum_{n=0}^{\infty}\frac{p^n}{n!}, \end{equation} one shows that $e$, if algebraic, should be of degree at least $p$ over $\mathbb Q$. As this holds for all $p$, $e$ cannot be algebraic. The fundamental mistake in the proof, I have read (e.g Éléments d'histoire des mathématiques), is that it incorrectly assumes that a rational power-series whose evaluation at a rational $r$ converges for the $p$-adic topology to an algebraic number and also converges for the real topology to an algebraic number necessarily converge to the same algebraic number for both topologies, and that is certainly not true. Does anyone know the details of the correct part of the proof, and in particular how Hensel's supposedly bounded below the purported degree of $e$ ? REPLY [23 votes]: You can read Hensel's original article here. The argument is really simple: from this expansion, we see that $e$ satisfies an equation of the form $y^p=1+pu$ where $u$ is a $p$-adic unit. But the polynomial $y^p-(1+pu)$ is irreducible over $\mathbb Q_p$ - writing $y=1+z$ and expanding, we get $$z^p+pz^{p-1}+\dots+pz-pu$$ which is irreducible by Eisenstein criterion. Therefore $e$ must have degree at least $p$ over $\mathbb Q_p$ (and, Hensel's argument goes, hence also over $\mathbb Q$).<|endoftext|> TITLE: Diffeomorphism group of the projective plane QUESTION [15 upvotes]: First of all, I am interested in the general case of a non-orientable manifold but let's for now consider the projective plane $\mathbb{R}P^2.$ In short, I am curious if there is any relation between the diffeomorphism group $\text{Diff}(\mathbb{R}P^2)$ of the projective plane and the diffeomorphism group $\text{Diff}(S^2)$ of its orientation double cover. As I understand, any diffeomorphism $\mathbb{R}P^2\to \mathbb{R}P^2$ can be lifted to a diffeomorphism $S^2\to S^2$ of the orientation bundle (same in the general case). That means that we can consider $\text{Diff}(\mathbb{R}P^2)$ as a subgroup in $\text{Diff}(S^2).$ I know that there are a lot of results on geometry and topology of this group and I wonder if some results remain true for $\text{Diff}(\mathbb{R}P^2).$ I tried to search for some literature on this topic but didn't find anything useful for me (partially because I am not sure what I am looking for). My final goal for now is to compute (or find results on) the curvature for a right-invariant metric on the group of volume-preserving diffeomorphisms $\text{SDiff}(\mathbb{R}P^2).$ I would be very much interested to learn anything on this topic. Thanks. REPLY [11 votes]: It is a theorem of A. Gramain from 1973 (Annales Sci. E.N.S.) that the diffeomorphism group of the projective plane has the homotopy type of $SO(3)$, the subgroup of isometries of the standard constant curvature metric. This is proved by projecting the diffeomorphism group onto the space of embeddings of a disk into the projective place by restricting diffeomorphisms to embeddings of a disk. This projection is a fibration whose fiber is the diffeomorphism group of the Moebius band fixing the boundary. Gramain showed this fiber is contractible. The base space of the fibration is easily seen to have the homotopy type of $SO(3)$.<|endoftext|> TITLE: Is every Schwartz function the product of two Schwartz functions? QUESTION [16 upvotes]: A Schwartz function on $\mathbb R^d$ is a $C^\infty$ function, such that all differentials of order $k \ge 0$ decay faster than any polynomial. They include the class $C^\infty_c(\mathbb R^d)$ of compactly supported, smooth functions. I would like two know, if for every Schwartz function $f$, there are Schwartz functions $g,h$ such that $f(x)=g(x)h(x)$ for all $x \in \mathbb R^d$. If $f \in C^\infty_c$, we can choose $g=f$ and $h$ as a cutoff function, such that $f(x) \neq 0 \implies h(x)=1$ to get such a (trivial) representation. For a general Schwartz function, I was unable to find a construction or a counter example. REPLY [10 votes]: (I do intend to read @AbdelmalekAbdesselam's literature review on this subject!) This sort of assertion has been known since A. Weil, at latest, and some forms are not very complicated. So, for example, yes, every Schwartz function is the product of two such. A stand-alone proof is given in my little essay https://www-users.cse.umn.edu/~garrett/m/fun/weil_schwartz_envelope.pdf<|endoftext|> TITLE: Example of an uncountable scattered space with some properties QUESTION [9 upvotes]: This might be an easy question, maybe the example I'm looking for is common knowledge. As always, recall that a topological space $X$ is scattered if and only if every non-empty subset $Y$ of $X$ contains at least one point which is isolated in $Y$. It is known that any first countable, $T_3$, Lindelöf and scattered space is countable. Is there an example of an uncountable, first countable, Hausdorff, Lindelöf and scattered space? REPLY [11 votes]: The answer to the question is yes: $\mathsf{ZFC}$ alone proves there is an example of an uncountable, first countable, Hausdorff, Lindelöf, scattered space. The example I will describe has scattered height $\omega$. By the second part of Taras' answer, having infinite scattered height is a necessary feature of the example, because we are not assuming $\mathfrak{b} = \aleph_1$. Recall that a Bernstein set is a subset $X$ of $\mathbb R$ such that both $X$ and its complement have nonempty intersection with every uncountable closed subset of $\mathbb R$. Such subsets of $\mathbb R$ can be proved to exist using the Axiom of Choice. In fact, the standard proof of the existence of Bernstein sets can be modified slightly to get a partition of $\mathbb R$ into infinitely many Bernstein sets: Lemma: There is a partition of $\mathbb R$ into a countably infinite number of sets, such that each of these sets has nonempty intersection with every uncountable closed subset of $\mathbb R$. (Actually, it is possible to partition $\mathbb R$ into $\mathfrak{c}$ Bernstein sets. But to construct our example, we want a partition into just $\aleph_0$ Bernstein sets.) Let $X_0, X_1, X_2, X_3, \dots$ be a countable collection of pairwise disjoint Bernstein sets, as described in the lemma. The underlying set of our example is $\mathbb R$, endowed with the topology generated by the following basic open sets. Given $x \in \mathbb R$, there is a unique $n$ such that $x \in X_n$; we take the basic open neighborhoods of $x$ to be sets of the form $\{x\} \cup \left( U \cap \bigcup_{i < n}X_i \right)$, where $U$ is a rational open interval in $\mathbb R$ containing $x$ (i.e., $U$ is a basic open neighborhood of $x$ in the normal Euclidean topology on $\mathbb R$). This space is clearly uncountable, because its underlying set is $\mathbb R$. It is first countable, because the neighborhood bases described above are all countable. This space is Hausdorff because rational Euclidean-open intervals are still open in this topology (i.e., it refines the usual topology on $\mathbb R$). It is scattered because for any $Y \subseteq \mathbb R$, there is some minimal $n$ such that $Y \cap X_n \neq \emptyset$, and any point of $Y \cap X_n$ is isolated in $Y$. Finally, it remains to show this topology is Lindelöf. Roughly, the idea is that for each $n$, we can use countably many of the neighborhoods of the points on level $n$ of our space (i.e., points in $X_n$) to reach down and cover all but countably many of the points on lower levels. Suppose $\mathcal U$ is an open cover for this space. By shrinking the elements of $\mathcal U$ if necessary, we may (and do) assume that each member of $\mathcal U$ is a basic open set as described above. For each $n$, let $\mathcal U_n$ denote the set of all those $U \in \mathcal U$ such that $|U \cap X_n| = 1$; and for each $U \in \mathcal U_n$, let $I_U$ denote the (unique) rational Euclidean-open interval such that $U \cap \bigcup_{i < n}X_i = I_U \cap \bigcup_{i < n}X_i$. Fix $n \in \omega$ with $n > 0$. For each Euclidean-open interval $I \in \{ I_U :\, U \in \mathcal U_n \}$, choose some particular $U(I) \in \mathcal U_n$ such that $I_{U(I)} = I$. The set of all rational Euclidean-open $I$'s is countable, so the set of all such $U(I)$'s is a countable subset of $\mathcal U_n$. Let us denote this countable subset of $\mathcal U_n$ by $\mathcal V_n$. Now observe that $W_n = \bigcup \{ I(U) :\, U \in \mathcal V_n \}$ is a Euclidean-open subset of $\mathbb R$ containing $X_n$. Because $X_n$ is a Bernstein set, this means $\mathbb R \setminus W_n$ is countable. In particular, this implies that $\bigcup \mathcal V_n$ contains all but countably many points of $\bigcup_{i < n}X_i$. Now unfix $n$. By the previous paragraph, $\bigcup_{n \in \omega \setminus \{0\}} \mathcal V_n$ is a countable subset of $\mathcal U$ that covers all but countably many points of our space. Adding in one more open set for each of the points not covered already by these sets, we obtain a countable subcover of $\mathcal U$.<|endoftext|> TITLE: Is this function monotonically increasing? QUESTION [6 upvotes]: Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and \begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots& m_{1N}\\ m_{21}&1&\cdots& m_{2N}\\ \vdots&\vdots&&\vdots\\ m_{N1}&m_{N2}&\cdots& 1 \end{matrix} \right] \end{equation} where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$; Let \begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) \,d\boldsymbol{u} \end{equation} where $\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma_1\right)\times\cdots\times \left(-\infty,\gamma_N\right)$ Now, if let the $(i,j)\text{th}$ element of $\boldsymbol{M}$ be an independent variable $a,$ $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$. Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u}) \, d\boldsymbol{u}$. My question is: Is $F(a)$ a monotonically increasing function of $a$? REPLY [5 votes]: Yes, this is a special case of Slepian's inequality.<|endoftext|> TITLE: Unnecessary uses of the axiom of choice QUESTION [55 upvotes]: What examples are there of habitual but unnecessary uses of the axiom of choice, in any area of mathematics except topology? I'm interested in standard proofs that use the axiom of choice, but where choice can be eliminated via some judicious and maybe not quite obvious rephrasing. I'm less interested in proofs that were originally proved using choice and where it took some significant new idea to remove the dependence on choice. I exclude topology because I already know lots of topological examples. For instance, Andrej Bauer's Five stages of accepting constructive mathematics gives choicey and choice-free proofs of a standard result (Theorem 1.4): every open cover of a compact metric space has a Lebesgue number. Todd Trimble told me about some other topological examples, e.g. a compact subspace of a Hausdorff space is closed, or the product of two compact spaces is compact. There are more besides. One example per answer, please. And please sketch both the habitual proof using choice and the alternative proof that doesn't use choice. To show what I'm looking for, here's an example taken from that paper of Andrej Bauer. It would qualify as an answer except that it comes from topology. Statement Every open cover $\mathcal{U}$ of a compact metric space $X$ has a Lebesgue number $\varepsilon$ (meaning that for all $x \in X$, the ball $B(x, \varepsilon)$ is contained in some member of $\mathcal{U}$). Habitual proof using choice For each $x \in X$, choose some $\varepsilon_x > 0$ such that $B(x, 2\varepsilon_x)$ is contained in some member of $\mathcal{U}$. Then $\{B(x, \varepsilon_x): x \in X\}$ is a cover of $X$, so it has a finite subcover $\{B(x_1, \varepsilon_{x_1}), \ldots, B(x_n, \varepsilon_{x_n})\}$. Put $\varepsilon = \min_i \varepsilon_{x_i}$ and check that $\varepsilon$ is a Lebesgue number. Proof without choice Consider the set of balls $B(x, \varepsilon)$ such that $x \in X$, $\varepsilon > 0$ and $B(x, 2\varepsilon)$ is contained in some member of $\mathcal{U}$. This set covers $X$, so it has a finite subcover $\{B(x_1, \varepsilon_1), \ldots, B(x_n, \varepsilon_n)\}$. Put $\varepsilon = \min_i \varepsilon_i$ and check that $\varepsilon$ is a Lebesgue number. REPLY [3 votes]: My favourite example is from Reverse Mathematics, namely Pincherle's theorem stating that a locally bounded function on Cantor space is bounded there. The obvious proof proceeds by contradiction and uses AC: Suppose $F:2^\mathbb{N}\rightarrow \mathbb{N}$ is unbounded, i.e. $(\forall n\in \mathbb{N})(\exists f \in 2^{\mathbb{N}})(F(f)>n)$. Apply (countable) choice to obtain a sequence $(f_n)_{n\in \mathbb{N}}$ such that $F(f_n)>n$ for all $n \in \mathbb{N}$. Use the sequential compactness of Cantor space to show that this sequence has a subsequence $(g_n)_{n\in \mathbb{N}}$ which converges to $g\in 2^\mathbb{N}$. Since $F$ is locally bounded, $F$ is bounded in a neighbourhood of $g$. However, as $n$ increases, $g_n$ approaches $g$ and $F(g_n)$ becomes arbitrary large. Contradiction. There is a proof in ZF (and weaker systems) that is more delicate: in step 2., one considers: $(\forall n\in \mathbb{N})(\exists \sigma\in 2^{<\mathbb{N}})[(\exists f \in 2^{\mathbb{N}})(F(f)>n) \wedge \sigma = (f(0),..., f(|\sigma|) ]$. One can apply `numerical choice' to obtain a sequence $(\sigma_n)_{n\in \mathbb{N}}$ such that: $(\forall n\in \mathbb{N})[(\exists f \in 2^{\mathbb{N}})(F(f)>n) \wedge \sigma_n = (f(0),..., f(|\sigma_n|) ]$. This `numerical' choice principle is provable in ZF. Now use the sequence $(\sigma_n)_{n\in \mathbb{N}}$ instead of the sequence $(f_n)_{n\in \mathbb{N}}$; the rest of the proof then can be modified to obtain a contradiction on the same way.<|endoftext|> TITLE: Existence of a real structure on the tangent bundle of a complex manifold QUESTION [6 upvotes]: Let $X$ be a compact complex manifold. What are obstructions to the existence of a real subbundle $V$ of $TX$ such that $TX = V \otimes \mathbb{C}$? For example, does $\mathbb{CP}^n$ have such a structure? REPLY [7 votes]: For a vector bundle to have real structure, it must be isomorphic to its complex conjugate, hence it's $i$th Chern class must be equal to its own negation (i.e. $2$-torsion) for all odd $i$. So $2c_1, 2c_3, 2 c_5, \dots$ are all obstructions. Since $2c_1$ is nonzero for the tangent bundle of projective space, $\mathbb C\mathbb P^n$ has no such structure.<|endoftext|> TITLE: A singular stochastic differential equation QUESTION [7 upvotes]: We consider the following SDE: $$dX_t = 1(X_t = 0) \, dt + 1(X_t >0) \, dB_t, \quad X_0= x > 0,$$ where $(B_t, \, t \ge 0)$ is linear Brownian motion. Let $\tau: = \inf\{t >0: X_t = 0\}$ be the first time at which $X$ hits $0$. It is obvious that $(X_t, \, 0 \le t \le \tau)$ is Brownian motion up to the first hitting to $0$. Question: Can we say something for $X$ after $\tau$? Is it well-defined? It is clear that $X$ is not reflected Brownian motion but is also supported on $[0,\infty)$. It seems that there is accumulation of zeros after $\tau$ which would lead to local times. REPLY [8 votes]: This is sticky reflecting Brownian motion, see for example this relatively recent paper. You can alternatively construct it by taking a reflected Brownian motion and then "stretching out" the local time accumulated at the origin, turning it into real time. Different constants in front of the $dt$ term yield different stretching factors.<|endoftext|> TITLE: Interview of Connes, Caramello, and L. Lafforgue about topos theory QUESTION [14 upvotes]: In a recent blog post, Lieven le Bruyn, discussing an interview with Connes, Caramello, and Lafforgue on France Culture, wrote Towards the end of the programme Connes, Caramello and Laforgue [sic] lament that topos theory is still not taken seriously by the mathematical community at large So, what have they said about topos theory not being taken seriously? Edit (YCor): here's a transcription of the involved part of the interview. I don't include an English translation but automatic translators should give quite accurate one. J= journalist (Nicolas Martin, France Culture) OC= Olivia Caramello, AC= Alain Connes, LL= Laurent Lafforgue J: On entendait parler dans ce reportage des topos, c'est très important, j'aimerais qu'on l'évoque maintenant, c'est la grande création, ce que Grothendieck disait être le plus fier d'avoir créé, une volonté unificatrice des mathématiques, et pourtant, paradoxalement, pour des raisons que vous allez peut-être pouvoir m'expliquer, les uns, les unes et les autres, ces mathématiques-là ont très mauvaise presse, aujourd'hui dans la communauté mathématique, il est difficile d'effectuer des recherches et de travailler sur les topos. Alain Connes, vous avez donné des leçons au Collège de France sur les topos mais ça n'a pas duré très longtemps; vous-même Olivia Caramello, ça a été difficile et vous aviez des pressions pour ne pas continuer dans cette voie. Comment est-ce que ça s'explique tout ça... euh, Alain Connes me fait signe que non. AC: Non non! en fait, si vous voulez, non. Ça c'est une version complètement externe de la réalité. La réalité, c'est que la notion conceptuelle de topos... J: ... que vous pouvez nous rappeler, brièvement, s'il vous plait AC: ah! Je peux vous rappeler ce que c'est... en gros, on avait, avant Grothendieck, l'habitude, pour regarder un espace — tout le monde sait que le rôle de l'espace est assez essentiel dans la géométrie et dans les mathématiques - avant Grothendieck, on quand on voulait connaître un espace, on le regardait directement, et on essayait de le comprendre. Ce que fait l'idée du topos, qui est une idée merveilleuse, c'est... elle met l'espace dans les coulisses, et ce que l'on fait, c'est des mathématiques ordinaires avec un paramètre, ce paramètre est dans l'espace en question: il est dans les coulisses. Pour vous donner l'exemple le plus simple possible, supposez que l'espace en question, ce soit simplement deux points, eh bien les mathématiques que vous faites: vous faites deux fois les mathématiques qui sont la théorie des ensemble ordinaire. Eh bien ce qui est merveilleux, dans la théorie des topos qui est là, c'est que, elle a deux caractéristiques. La première, c'est qu'en analysant ce qui se passe dans le contexte ordinaire de la théorie des ensembles, mais fait avec paramètre dans le topos, on arrive à une connaissance de cet espace (qui est l'espace des paramètres, qui est le topos) bien plus fine que si on l'avait regardé directement. Je ne peux pas m'empêcher de dire qu'en ce moment je suis en train d'écrire un livre avec un psychanalyste, Patrick Gauthier-Lafaye, dans lequel on utilise cette métaphore, mais par rapport à la psychanalyse. La deuxième chose, qui est absolument extraordinaire, dans cette idée du topos, c'est qu'elle revient à regarder le mathématicien au travail de manière structuraliste, c'est-à-dire: le mathématicien au travail va manipuler des ensembles, mais le structuraliste va se fiche du fait que ce sont des ensembles, il va regarder le mathématicien qui manipule des objets et des flèches et il va dire: ce mathématicien travaille sur ce qu'on appelle, en mathématiques, une catégorie, et, ce structuraliste va dire, mais quelles sont les propriétés de cette catégorie, qui font que le mathématicien peut travailler. Eh bien là on est au coeur des topos. J: Comment expliquer, alors, Olivia Caramello que les topos — alors corrigez-moi, évidemment, si je le décris de manière trop caricaturale — aient si mauvaise presse, ou soient un champ de travail thématique qui finalement a été très vite rejeté ou repoussé par les institutions? OC: En fait, j'ai réfléchi beaucoup par moi-même, en lisant Récoltes et Semailles, et aussi sur la base de celle ce qui a été ma propre expérience de vie. Effectivement j'ai reçu beaucoup d'oppositions, en fait, depuis le début de ma carrière, tout simplement parce que je voulais développer d'une façon globale et systématique cette théorie, justement dans le but de réaliser cette inspiration d'unification qui avait été déjà exprimée par Grothendieck, notamment dans Récoltes et Semailles. Donc tout mon travail de recherche a été dirigé vers le but d'élaborer des techniques, des méthodes, pour transférer des connaissances entre des parties complètement différentes des mathématiques par le biais des topos. Donc en fait, les topos peuvent être utilisés de façon incroyablement efficaces comme des objets-ponts pour relier des textes mathématiques les plus divers les uns et les autres. On peut penser métaphoriquement un topos comme un lieu dans lequel des points de vue différents se rencontrent en se reflétant les uns dans les autres. Donc je donne cette métaphore pour souligner cet aspect d'unification, car je pense que c'est celui qui a vraiment engendré le plus d'hostilité. Je pense que ce n'est pas la technicalité des topos en tant qu'objet mathématique comme d'autres objets — bien sûr il y a toute une technicalité, la théorie est quand même très sophistiquée, très profonde, sur le plan purement technique — mais ce n'est pas l'aspect technique qui a été à l'origine de l'ostracisme. Je pense que c'est vraiment cette dimension globale et interdisciplinaire qui dérange les gens. Aujourd'hui les mathématiques sont devenues hyper-spécialisées donc chaque spécialiste travaille dans son coin avec ses propres méthodes, il s'habitue à penser d'une certaine façon. Or avec ces ponts, qu'on arrive à engendrer avec les topos, on peut arriver notamment à démontrer un résultat dans un secteur des mathématiques en utilisant des mathématiques complètement étrangères à ce secteur-là. On peut arriver à établir des ponts entre des secteurs complètement éloignés en apparence et donc on peut arriver chez un spécialiste d'un certain domaine avec un résultat qui le surprend beaucoup, qu'on arrive à démontrer avec des méthodes qui ne sont pas les siennes. Donc vous pouvez déjà comprendre que ça peut être inquiétant pour certains si on n'a pas assez d'ouverture d'esprit pour accepter cette pluralité de points de vue. Donc je pense qu'il y a un certain dogmatisme, dans un certain cercle mathématique, qui fait qu'on s'habitue à un certain langage, et après, on se renferme, en quelque sorte, après des années d'hyper-spécialisation. Il faut quand même comprendre que travailler dans n'importe quel secteur des mathématiques aujourd'hui demande un investissement technique colossal, donc c'est quand même humainement compréhensible qu'on s'affectionne beaucoup à certaines méthodes et après on dit "je ne veux pas voir autre chose". Moi, ça m'est arrivé plusieurs fois que je donne des exposés ou par exemple je présente dans un domaine qui n'est pas le mien... J: par exemple? OC: la théorie des modèles, par exemple, ça m'est arrivé de démontrer un résultat parmi mes premiers, où je faisais une ample généralisation du théorème de Fraïssé, en théorie des modèles, qui est un résultat très important, et en fait, dans l'auditoire, je me rappelle bien, un théoricien des modèles important, qui ne pouvait pas croire que mon résultat était correct, parce que c'était trop général. Et en fait il a passé tout l'après-midi à essayer de trouver un contre-exemple, bien sûr sans y parvenir, parce que ma démonstration était tout-à-fait correcte... Sauf qu'il était formulé dans un langage que lui, il m'a dit "Je ne me mets même pas à essayer de comprendre, parce que de toute façon, je ne vais pas y arriver", il me l'a dit comme ça. Alors il a préféré passer quatre heures de son temps, il me tourmentait aussi parce que moi j'étais là, il essayait de me fabriquer tous ces contre-exemples, ça a été assez pénible (rire)... c'est juste pour vous donner une idée. Il y a vraiment ce côté interdisciplinaire qui dérange... en fait il y a beaucoup d'autres aspects évidemment. J: Laurent Lafforgue, un mot, parce que vous, vous avez quitté le milieu académique pour passer dans le privé, et parce que dans le privé, en l'occurrence, votre employeur vous demande de travailler sur les topos, et utilise cet outil, qui a l'air d'être particulièrement efficace, pour des travaux appliqués, qui ont l'air d'être des travaux passionnants et effectifs. LL: oui, c'est une histoire qui est totalement stupéfiante pour moi, que je n'aurais jamais imaginé il y a encore quelques années. Depuis une dizaine d'années, en fait depuis que je connais Olivia Caramello et ses travaux, je suis devenu, dans le monde académique, un fervent supporteur du développement de la théorie des topos, et comme toutes les personnes qui ont voulu développer les topos ou contribuer à leur développement, je me suis heurté pour cela à une très grande hostilité, et à ma totale surprise, j'ai trouvé, dans un milieu d'ingénieurs, en l'occurrence de la firme Huawei, en France, des oreilles beaucoup plus favorables. C'est une chose que je n'aurais jamais attendu, qui me stupéfie aujourd'hui encore. Et donc, depuis quelques mois, j'ai quitté le monde académique, je suis chez Huawei, mon environnement est constitué d'ingénieurs et de responsables de la hiérarchie de la recherche de Huawei qui sont totalement favorables au développement des topos, qui pensent dès aujourd'hui, c'est-à-dire seulement quelques années après avoir appris l'existence de cette théorie, qu'elle est extrêmement importante, et, certains parmi eux pensent que les topos de Grothendieck vont devenir, ou peuvent devenir, les mathématiques de l'intelligence artificielle. Donc, c'est-à-dire, quelque chose d'une importance absolument colossale. Pour moi c'est inimaginable, parce que ça fait 60 ans que la théorie des topos a été introduite par Grothendieck, qu'elle a été développée par lui, déjà à longueur de centaines et de centaines de pages, que lui-même, dont tout le monde sait qu'il est l'un des plus grands génies scientifiques de l'histoire a énormément insisté sur la puissance des topos, sur l'importance des topos, même au-delà des mathématiques. En fait dans Récoltes et Semailles, un certain nombre de pages sont consacrées à ça: Grothendieck dit pourquoi les topos sont tellement importants à ses yeux; il le dit dans des termes que les mathématiciens peuvent comprendre, mais aussi, que même un lecteur qui ne connaît pas les mathématiques peut être sensible à la beauté et à la profondeur de ce que dit Grothendieck quand il parle des topos. Donc il a écrit ses pages-là, et ça n'a eu aucun effet dans le monde académique. Donc là il y a un mystère, que Grothendieck lui-même ne s'explique pas: il constate cette hostilité, il ne la comprend pas. Olivia vient de proposer des éléments d'explication, mais pour moi ça reste un mystère. En fait les topos sont un sujet sensible et c'est bizarre, parce que, habituellement quand on dit qu'un sujet sensible, on comprend un sujet sensible c'est par exemple un sujet politique, sur lequel les gens ne s'accordent pas. On ne comprend pas qu'un sujet scientifique, qu'une définition théorique puisse être un sujet sensible, or en fait elle l'est. C'est un fait, que personnellement je ne m'explique pas, ou en tout cas, pas de manière satisfaisante. REPLY [9 votes]: I think that part of the answer lies in the folk tale Nail Soup. I can provide you a nourishing mathematical supper, says the toposopher to the hungry research student, but first I need you to provide me with a handful of carrots, and maybe a pinch of this and a pinch of that, ... . I have much sympathy with the first answer, because excessive hype, even of a beautiful subject, opens the door to bad mathematics. Category theory in its early days was also received badly in some quarters. Please excuse these superficial remarks from one who retired from academia twenty three years ago.<|endoftext|> TITLE: Every Grothendieck topos can be built from localic topoi QUESTION [6 upvotes]: Theorem 2 in these notes[1] states that, roughly, that each Grothendieck topos can be built (using limits and colimits) from localic topoi. To what extent is that related to the theorem of Joyal and Tierney which states that each Grothendieck topos is equivalent to the topos of equivariant sheaves on a groupoid in the category of locales? Jacob Lurie, 2018, lecture notes from Math 278X Categorical Logic, https://www.math.ias.edu/~lurie/278x.html, Lecture 16 Enumerations REPLY [9 votes]: They are (it is?) the same theorem, but emphasising different aspects. We can exploit the object classifier to get from the formulation in terms of (pseudo)colimits to the "elementary" formulation in terms of equivariant sheaves. The point is that, for every Grothendieck topos $\mathcal{E}$, the functor $\textbf{Geom} (\mathcal{E}, [\textbf{Set}_\textrm{fin}, \textbf{Set}]) \to \mathcal{E}$ sending a geometric morphism $f : \mathcal{E} \to [\textbf{Set}_\textrm{fin}, \textbf{Set}]$ to the object $f^* A$ in $\mathcal{E}$, where $A : \textbf{Set}_\textrm{fin} \to \textbf{Set}$ is the inclusion, is fully faithful and essentially surjective on objects. That is, $[\textbf{Set}_\textrm{fin}, \textbf{Set}]$ represents the contravariant forgetful 2-functor that sends a Grothendieck topos to its underlying category. It immediately follows that (bi)colimits in the 2-category of Grothendieck toposes are (bi)limits of the underlying categories. (It is awkward to keep using the 2- / bi- prefixes, so I will omit it from here onwards.) So, for example, the initial Grothendieck topos is the terminal category, the coproduct of Grothendieck toposes is the product of the underlying categories, the pushout of Grothendieck toposes is the pullback of the underlying categories, etc. What about the colimit of the diagram appearing in Lurie's formulation of the theorem? Well, the underlying category is the limit of the diagram of the underlying categories, and because the forgetful functor is contravariant, the shape of the diagram is also dualised. If you write out explicitly the universal property of the limit, you will find that it is an extremely terse way of writing something that looks like a groupoid action. Indeed, suppose given categories $\mathcal{C}^0$, $\mathcal{C}^1$, and $\mathcal{C}^2$ and functors $d^0, d^1 : \mathcal{C}^0 \to \mathcal{C}^1$, $d^0, d^1, d^2 : \mathcal{C}^1 \to \mathcal{C}^2$, satisfying the cosimplicial identities (strictly, to preserve our sanity). So we have a truncated cosimplicial object in $\textbf{Cat}$, and its limit is (up to equivalence!) the following category: An object consists of the following data: An object $X^i$ in each $\mathcal{C}^i$. An isomorphism $d^j X^i \cong X^{i+1}$ for each $i$ and $j$, such that for all $j$ and $k$, the composite $$\require{AMScd} \begin{CD} d^k d^j X^0 @>{d^k (\cong)}>> d^k X^1 @>{\cong}>> X^2 \end{CD}$$ depends only on the composite $d^k d^j$ in the simplex category $\mathbf{\Delta}$. So, for example, because $d^2 d^0 = d^0 d^1$ in $\mathbf{\Delta}$, the following diagram in $\mathcal{C}^2$ commutes: $$\begin{CD} d^0 d^1 X^0 @>{d^0 (\cong)}>> d^0 X^1 @>{\cong}>> X^2 \\ @| && @| \\ d^2 d^0 X^0 @>>{d^2 (\cong)}> d^2 X^1 @>>{\cong}> X^2 \end{CD}$$ A morphism consists of a morphism in each $\mathcal{C}^i$, compatible with the structural isomorphisms in the obvious way. Composition and identities are inherited from each $\mathcal{C}^i$. Manipulating the structural isomorphisms and eliminating the objects $X^1$ and $X^2$ gives us an isomorphism $d^1 X^0 \cong d^0 X^0$ in $\mathcal{C}^1$ and the following commutative diagram in $\mathcal{C}^2$: $$\begin{CD} d^2 d^1 X^0 @= d^1 d^1 X^0 @>{d^1 (\cong)}>> d^1 d^0 X^0 \\ @V{d^2 (\cong)}VV && @| \\ d^2 d^0 X^0 @= d^0 d^1 X^0 @>>{d^0 (\cong)}> d^0 d^0 X^0 \end{CD}$$ The isomorphism $d^1 X^0 \cong d^0 X^0$ in $\mathcal{C}^1$ can be thought of as the groupoid action on $X^0$, and the commutative diagram in $\mathcal{C}^2$ can be thought of as the compatibility between the groupoid action on $X^0$ and composition in the groupoid itself. If this still seems vague, then take the case where we have a groupoid $\mathcal{G}$ and $\mathcal{C}^i = \textbf{Set}^{G_i}$, where $G_0$ is the set of objects of $\mathcal{G}$, $G_1$ is the set of morphisms of $\mathcal{G}$, and $G_2$ is the set of composable pairs of morphisms of $\mathcal{G}$, and each of the functors $d^j : \mathcal{C}^i \to \mathcal{C}^{i+1}$ are defined by reindexing families of sets in the following way: $d^0 : \mathcal{C}^0 \to \mathcal{C}^1$ sends each $G_0$-indexed family of sets to the $G_1$-indexed family where the $g$-th component is the component of the original family at the codomain of $g$. $d^1 : \mathcal{C}^0 \to \mathcal{C}^1$ is defined similarly, except for using the domain of $g$ instead of the codomain of $g$. $d^0 : \mathcal{C}^1 \to \mathcal{C}^2$ sends each $G_1$-indexed family of sets to the $G_2$-indexed family where the $(h, g)$-th component is the $h$-th component of the original family. $d^1 : \mathcal{C}^1 \to \mathcal{C}^2$ sends each $G_1$-indexed family of sets to the $G_2$-indexed family where the $(h, g)$-th component is the $h \circ g$-th component of the original family. $d^2 : \mathcal{C}^1 \to \mathcal{C}^2$ sends each $G_1$-indexed family of sets to the $G_2$-indexed family where the $(h, g)$-th component is the $g$-th component of the original family. It is straightforward, if extremely tedious, to verify that the limit category is equivalent to $[\mathcal{G}, \textbf{Set}]$. The definition of an action of a localic/topological groupoid is based on this observation, replacing $\textbf{Set}^{G_i}$ with $\textbf{Sh} (G_i)$. I do not think there is an abstract nonsense argument to get back from the "elementary" formulation to the colimit formulation. If the colimit of the diagram of Grothendieck toposes exists then its underlying category must be the limit, but a priori we do not even know that the limit category is a Grothendieck topos, let alone one that has the required universal property. So the colimit formulation is stronger in some sense.<|endoftext|> TITLE: Holonomic = annihilated by some differential operator QUESTION [5 upvotes]: Let $X$ be a smooth variety over a characteristic zero field $k$. It seems to be well-known that "A coherent $\mathcal{D}_X$-module is holonomic if and only if 'every element is annihilated by a nonzero differential operator'." I know that, when $X=\mathbb{A}^1$, this is true in the sense that "a finitely generated $k[x]\langle\partial\rangle$-module $M$ is holonomic if and only if, for every $m\in M$, there exists $P\in k[x]\langle\partial\rangle\setminus 0$ such that $Pm=0$." Is a similar statement true in more generality? (For all $X$? For affine $X$?) If so, why? REPLY [8 votes]: "A coherent $D_X$-module is holonomic if and only if 'every element is annihilated by a nonzero differential operator'." This is very much not true in dim>1. For example, let $D=k[x_1,x_2]\langle \partial_1,\partial_2\rangle$ be the ring of differential operators on $\mathbb A^2$, and consider the left $D$-module $$M= D/D\partial_1.$$ Then every element is annihilated by a suitably high power of $\partial_1$, but $M$ is not holonomic (its characteristic variety is the hyperplane in $T^\ast\mathbb A^2= \operatorname{Spec} k[x_1,x_2,\xi_1,\xi_2]$ cut out by the equation $\xi_1=0$). EDIT: In fact, we can make a stronger statement. Let $M$ be a $D$-module, and suppose we can find $m\in M$ which is not annihilated by any non-zero differential operator. Then the homomorphism $D\to M$ given by $P \mapsto Pm$ is injective. It follows that $\mathit{SS}(D) \subseteq \mathit{SS}(M)$ (the characteristic variety operation takes short exact sequences to unions). Thus $\mathit{SS}(M)$ is the entire cotangent bundle in this case. Put another way, any $D$-module whose characteristic variety is a proper subset of the cotangent bundle will satisfy the property that each element is annihilated by a non-zero differential operator. In dimension 1, every proper closed coisotropic subset of the cotangent bundle is lagrangian, so every such $D$-module is holonomic. But this is definitely not the case in higher dimensions.<|endoftext|> TITLE: Degree-2 étale covers of curves in characteristic 2 vs torsion points on the Jacobian QUESTION [6 upvotes]: It can be found, at Hartshorne exercise 4.2.7 for example, that in the case where $\operatorname{char}(k) \neq 2$ we have a nice correspondence between etale degree 2 covers of a curve $C$ and 2-torsion in $JC$. This correspondence, as constructed in Hartshorne, breaks down in the case where $\operatorname{char}(k) = 2$. However it has been suggested in Pries and Stevenson - A survey of Galois theory of curves in characteristic $p$, at the bottom of section 4.7, that "remarks following proposition 4.13 in chapter 3 of Milne's Étale cohomology" show that some correspondence between $JC[2](k)$ and etale degree 2 covers of $C$ still exists. Does anyone know how to construct such a correspondence explicitly? I believe it is erroneous (there is no proposition 4.13, it is a lemma!). Otherwise, do étale degree 2 covers of $C$ correspond to anything else? Artin–Schreier theory suggests that it should be $H^1(C,\mathcal{O}_C)^F$. However from such a nontrivial cover $\pi:X \to C$ we also obtain a nontrivial extension of $\mathcal{O}_C$-algebras $$0 \to \mathcal{O}_C \to \pi_*\mathcal{O}_X \to \mathcal{O}_C \to 0$$ by looking at affines of $C$ and writing explicit Artin–Schreier extensions, and observing that the cocycles on $\pi_*\mathcal{O}_X$ are unipotent $2\times2$ matrices. Is there any way to relate this to $JC$? REPLY [6 votes]: There is a duality between degree 2 coverings and two-torsion points on the Jacobian — i.e. both form elementary abelian 2-groups, and these groups are naturally dual. This is the Artin–Milne Poincaré duality pairing in fppf cohomology (Corollary 4.9 of Duality in the Flat Cohomology of Curves) $$ H^1(C_{\overline{k}}, \mathbb Z/2) \times H^1 (C_{\overline{k}},\mu_2) \to H^2(C_{\overline{k}}, \mu_2) \to \mathbb Z/2 $$ with $H^1(C_{\overline{k}}, \mathbb Z/2)$ classifying degree 2 coverings and $H^1 (C_{\overline{k}},\mu_2)$ classifying 2-torsion points of $J$. Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$. More generally, for $C$ in characteristic $p$ and a $\mathbb Z/p$-covering of $C$ (and thus an isogeny $0 \to \mathbb Z/p \to A \to J\to 0$), we define the pairing with a $p$-torsion point by taking any lift of that point, multiplying by $p$, and obtaining an element of the kernel $\mathbb Z/p$. This doesn't depend on our choice of preimage because two lifts differ by an element of $\mathbb Z/p$ which is killed by multiplication by $p$. This is linear in the $A[p]$ variable since every step is compatible with multiplication by $p$. This is linear in the covering variable since the addition map on coverings is fiber product followed by adding the kernels, and the product of two lifts is a lift to the fiber product. To check this pairing has no kernel in the covering variable, the main point is to check that $A[p]$ and $J[p]$ have the same number of elements (since the rank is the multiplicity of the slope 0 of the Newton polygon, which is an isogeny invariant). Since $A \to J$ is a degree $p$ finite étale covering, its kernel contains a single $p$-torsion point, so the image of $A[p] \to J[p]$ has codimension $1$, forcing the pairing to be nontrivial. To check it has no kernel in the other variable, factor the multiplication by $p$ map $A \to A$ (which kills all the $p$-torsion) into a totally inseparable morphism (which preserves points and thus $p$-torsion and an étale morphism (which therefore must kill all the $p$-torsion). The étale morphism is a fiber product of $\mathbb Z/p$-coverings, so if a point was in the image of $p$-torsion on each $\mathbb Z/p$-covering it would have to be in the image of $p$-torsion on this covering, forcing it to be trivial. You're correct on your interpretation of Artin–Schreier theory. (I think you meant unipotent instead of unitary.) I don't know how to connect that perspective to the duality with the Jacobian — in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin–Schreier map $\mathbb Z/2 \to \mathcal O_C$.<|endoftext|> TITLE: How to numerically compute $x \ln x$ and related functions near $0$? QUESTION [6 upvotes]: I was recently trying to find a numerical solution to a thermodynamics problem and the expression $x\ln x$ appeared in one of the computations. I did not have to find its value very near $0$, so the computer managed fine, but it got me thinking - can one make a stable numerical algorithm to compute $x\ln x$ for values near 0? It is easy to prove that $\lim_{x\to 0^-} x\ln x=0$. However, simply multiplying $x$ by $\ln x$ is not a good solution for small $x$. The problem seems to be that we are multiplying a small number ($x$) by a large number ($\ln x$). My first thought would be to approximate it somehow. But I quickly saw that Taylor series wouldn't work, as the derivative is $\ln x + 1$, which blows up (or rather down :-)) to $-\infty$. Some kind of iterative method like Newton's method does not seem to be the solution either, because the operations needed seem to be even more messy than what we are trying to compute. So my question is - is there some numerically stable method to compute $x\ln x$ for small values of $x$? And preferably one that is more general, so that it could be used on functions like $x^n \ln x$ - but these at least have a finite first derivative at $0$ for $n>1$. REPLY [7 votes]: Modern mathematics libraries should be able to find $\log x$ precisely for all floating-point numbers, as the algorithms for doing that have long been known and adopted. My experiments on a fairly recent Intel chip with gnu mathematics library and gcc 10 compiler confirm that. Multiplication is even more definite: correct rounding of the last bit is guaranteed (though there can be different options available for what "correct rounding" means). It might appear from the above that precise computation of $x\log x$ is guaranteed for any small $x>0$. However there is a reason why that doesn't happen for really tiny $x$ and there is no way to avoid it. Floating-point numbers are usually stored with the mantissa normalized (leading bit 1, sometimes implicit). However, when the number is too small it may be impossible to normalize the number without the exponent underflowing. In this case (usually) the number is kept in unnormalized form and the number of bits of precision is reduced. This situation is called partial underflow and such numbers are subnormal or denormalized. So, if you try to compute $x\log x$ when $x\log x$ is in the partial underflow range, $\log x$ will be computed precisely but the multiplication by $x$ cannot produce more bits of precision than numbers of the size of $x\log x$ have. Short of using different floating-point numbers, there is no solution. If $x\log x$ is in the partial underflow range, then $x$ will be too, or maybe it will be so small as to underflow to 0. In practice $x$ will come from some earlier computation and the partial underflow means it may not be so precise as thought, which is another source of error. It isn't the fault of the function $x\log x$. Incidentally I tested this explanation empirically and behaviour was exactly as predicted.<|endoftext|> TITLE: Computation of $\pi_1$ for a Mazur manifold and its boundary QUESTION [5 upvotes]: If we attach a $4$-dimensional $1$-handle $D^1 \times D^3$ to a $4$-dimensional $0$-handle $B^4$, we obtain $S^1 \times D^3$. The null homologous knot in $S^1 \times S^2$ indicated in the picture lives in a solid torus/attaching region $S^1 \times D^2$ and $$ S^1 \times D^2 \hookrightarrow S^1 \times S^2 \hookrightarrow S^1 \times D^3.$$ In pg. 356 of his book Knots and Links, Dale Rolfsen notes that $\pi_1(W) = 1$, $\pi_1(\partial W) = \langle x,y \ \vert \ y^5 = x^7, y^4 = x^2 y x^2 \rangle \neq 1$. How we compute the relevant fundamental groups using the following diagram? What is the strategy? REPLY [6 votes]: Thanks to the dotted circle notation of Professor Akbulut, we can easily (compare with your picture) draw the handle diagram of a typical Mazur manifold as follows: Since $1$-handles and $2$-handles respectively give the generators and relations of the fundamental group of a $4$-manifold, we have (for my $W$): $$\pi_1(W) = \langle \alpha \ \vert \ \alpha^2 \alpha^{-1} \rangle = 1.$$ The boundary $3$-manifold $\partial W$ is already in the picture. Further, we can compute $\pi_1(\partial W)$ from the diagram by using Wirtinger's presentation. Because today is Sunday (which makes me so lazy) and Eylem already did such a computation (another purpose but for a Mazur manifold) in Example 1 in [Yıldız, 2018], the rest is for you, mimic the argument by keeping Ryan Budney's comments in your mind: Bonus: In particular, $W$ is a contractible $4$-manifold: By using Mayer-Vietoris sequences, observe that $W$ is a homology $4$-ball, i.e., we have $H_*(W, \mathbb{Z}) = H_*(B^4, \mathbb{Z})$. Then apply the classical theorems of Hurewicz and Whitehead. Bonus 2: Since $W$ is contractible, $\partial W$ must be a homology $3$-sphere, i.e., $H_*(\partial W, \mathbb{Z}) = H_*(S^3, \mathbb{Z})$. Using Kirby calculus, show that $\partial W \approx \Sigma(2,5,7)$ where $\Sigma(p,q,r)$ denotes the Brieskorn sphere given coprime positive integers $p,q$ and $r$.<|endoftext|> TITLE: Why is there an unexpected increase in the density of certain types of Goldbach primes? QUESTION [8 upvotes]: Note: Posted in MO since it was unanswered in MSE. I was checking how quickly we can verify Goldbach's conjecture for a given even number $n$ and it was clear that searching backward starting from the largest prime below $n$ was much faster than a forward search starting from the small odd prime $3$. I observed the following. Consider the set of even numbers $n$ which can be written as the sum of two primes and $p$ be the largest prime less than $n$ such that $n-p$ is also a prime. Let $p_k$ be the $k$-th prime. Then, $p_{\pi(n)-2}$, $p_{\pi(n)-1}$ and $p_{\pi(n)}$ are the three largest primes preceding $n$. Now $p$ can either be the largest prime below $n$, or the prime just before the largest prime or the prime before that and so on. Accordingly, we define: $d_1$ is the density of $n$ such that $p = p_{\pi(n)}$ $d_2$ is the density of $n$ such that $p = p_{\pi(n)-1}$ $d_3$ is the density of $n$ such that $p = p_{\pi(n)-2}$ Data for $n \le 3 \times 10^{10}$ shows that $d_1$ and $d_2$ decreased as $n$ increased. This was intuitively expected as explained in the answer for the related question "If $p$ is the largest prime less than $2n$, what is the probability that $2n-p$ is a prime?". However, quite counter intuitively, $d_3$ increased with $n$ as shown in the graphs below. Question 1: Why does $d_3$ increase with $n$ while $d_1$ and $d_2$ decreased? Question 2: What is the limiting value of $d_3$? Since $d_3 < 1$ and is increasing, it must converge to a positive limiting value. Note: I have not computed $d_k$ for $k \ge 4$ so I do not know if this increasing trend is observed other values of $k$. Graphs in normal scale REPLY [4 votes]: Here are a few thoughts. First, $d_i$ is not the density of a certain type of prime, it is the density (as you say) of a certain type of even integer. So let me abuse notation and say that an even integer $n$ is of type $c_i$ if $q=n-p_{\pi(n)+1-i}$ is prime and type $d_i$ if, in addition, its farthest spaced Golbach pair is $n=q+p_{\pi(n)+1-i}.$ That the apparent increase you see in $d_3$ eventually goes away has been explained convincingly by others. I wanted to suggest why it might take a very very long time. To oversimplify dramatically, $0\frac{16}{28}$ which retards that (tiny) effect ever so slightly. Of course there are many other considerations. Two other comments: I find the question interesting but the stated motivation unconvincing. Given an even $n$ you want to find one representation of it as a sum of two primes. So you suggest checking $n-3$ and, if that is not prime working down through the odd integers $q$ (not congruent to $n \bmod 3$) and , when one is prime, checking if $n-q$ is. It might be faster, for $n \equiv 2 \bmod 6,$ to check $n-3$ and then $n-7,n-13,n-19,n-31,n-37,n-43,n-61$ (though one would need to know when to stop) You might find it helpful to study $n \equiv 0 \bmod 6$ vs $n \equiv \pm 2 \bmod 6.$ What does the trend of the densities $d_1,d_2,d_3$ look like in each type? I'd expect $n \equiv 0 \bmod 6$ to be perhaps only $1/3$ as likely to be counted by $d_3$ as other even $n:$ Suppose $p\ \equiv 1 \bmod 3$ is a prime. Given only this information, of the even $n$ up to the next prime, those of the form $6m$ could be counted by $d_1,d_2$ or $d_3$ but those of the form $6m + 2$ could possibly be counted by $d_2$ or $d_3$ but not $d_1$. It might be that these kinds of effects balance out.<|endoftext|> TITLE: Existence of a generalized stable ordinal QUESTION [5 upvotes]: An ordinal $\alpha$ is called +1 stable, if $L_\alpha <_1 L_{\alpha+1}$ By considering $Σ_n$ elementary submodel we can generalize it. I'm curious about its further generalizations. Does there exist an ordinal $\alpha$, such that $L_{\alpha+1}$ is elementary equivalent to $L_{\beta+1}$ for some larger $\beta$ where every element of $L_\alpha$ is added in the language? Can we change "+1" to "+n", or "+α"? Is it possible that $L_{\alpha2} \equiv L_{\beta 2}$ for the language $\{\in\}\cup L_\alpha$? REPLY [2 votes]: Yes, all of this configurations do exist provably in $\mathsf{ZFC}$. Let us construct $\alpha<\beta$ such that $(L_{\alpha+n};\in,\alpha,\langle c\mid c\in L_\alpha\rangle)\cong (L_{\beta+n};\in,\beta,\langle c\mid c\in L_\alpha\rangle)$. Consider the structure $(L_{\omega_1+n};\in,\omega_1)$, where the last $\omega_1$ is a constant. Next we construct a sequence of countable elementary substructures of this structure $$\mathfrak{M}_0\prec \mathfrak{M}_1\prec \ldots,$$ such that $\mathfrak{M}_{i+1}\supseteq L_{\delta +1}$, for all $\delta<\omega_1$ for which $\mathfrak{M}_i\cap (L_{\delta+1}\setminus L_\delta)\ne \emptyset$. Let $\mathfrak{M}_{\omega}=\bigcup_{i<\omega} \mathfrak{M}_i$. By Condensation Lemma, $\mathfrak{M}_\omega$ is isomorphic to a structure of the form $(L_\gamma;\in,\alpha)$, where $\gamma$ is countable and $\alpha<\gamma$. Furthermore by construction of $\mathfrak{M}_\omega$, $L_\alpha\supseteq \mathfrak{M}_\omega$ and the isomorphism keeps all the elements of $L_\alpha$ in place. Also using the elementary equivalence $(L_\gamma;\in,\alpha)\cong (L_{\omega_1+n};\in,\omega_1)$ it is easy to see that $\gamma=\alpha+n$. In the same manner we construct countable $\gamma$ s.t. $(L_{\gamma};\in,\alpha,\langle c\mid c\in L_\alpha\rangle)\cong (L_{\omega_12};\in,\omega_1,\langle c\mid c\in L_\alpha\rangle)$ and again using the elementary equivalence we show that $\gamma=\alpha2$. Given any ordinal $\delta$, we consider $(L_{\omega_{\delta+1}+\delta};\in,\omega_{\delta +1},\delta)$ and construct its elementary submodel of the cardinality $\le \max(|\delta|,\aleph_0)$ that will contain all ordinals $\le \delta$ and has the property that for any $\eta<\omega_{\delta+1}$, whenever some element of $L_{\eta+1}\setminus L_\eta$ is in the submodel, then whole $L_{\eta+1}$ is contained in the submodel. This gives us $\delta<\alpha<\gamma< \omega_{\delta+1}$ s.t. $(L_{\gamma};\in,\alpha,\langle c\mid c\in L_\alpha\rangle)\cong (L_{\omega_{\delta+1}+\delta};\in,\omega_{\delta +1},\langle c\mid c\in L_\alpha\rangle)$. Again using the elementary equivalence we see that $\gamma=\alpha+\delta$.<|endoftext|> TITLE: Identity of J. L. Rabinowitsch (of Rabinowitsch Trick) QUESTION [62 upvotes]: For some time, it seemed widely accepted that G. Y. Rainich was the author of the note Rabinowitsch, J. L., Zum Hilbertschen Nullstellensatz., Math. Ann. 102, 520 (1929). JFM 55.0103.04., which describes a short proof of Hilbert's Nullstellensatz by what became known as Rabinowitsch Trick. This claim appears to be solely based on Bruce Palka's ``Editor's Endnotes", p. 460 of Am. Math. Monthly 111, No. 5 (May, 2004), p. 460, who reports: Peter May shares the following correspondence that he received from Richard Swan. [..] The following anecdote may explay why you couldn't find him [Rabinowitsch]. Unfortunately I can't remember who told me this. It seems that Rainich was giving a lecture in which he made use of a clever trick which he had discovered. Someone in the audience indignantly interrupted him pointing out that this was the famous Rabinowitsch trick and berating Rainich for claiming to have discovered it. Without a word Rainich turned to the blackboard, picked up the chalk, and wrote RABINOWITSCH He then put down the chalk, picked up an eraser and began erasing letters. When he was done what remained was RA IN I CH He then went on with his lecture. Surprisingly, this anecdote without proper source has become universally accepted knowledge, making it, e.g., into Wikipedia, or the MO thread on mathematician's pseudonyms. In several comments, KConrad (some as recent as Feb 11, 2022 - many thanks to Manfred Lehn for notifying me about this) questions this attribution, observing that Rainich's work was mainly in general relativity, rather unrelated to algebra. Rainich changed his name from Rabinovich upon arrival in the U.S. in 1922, and is unlikely to publish under his old name with affiliation in Moscow (where he never lived or worked). The initials J.L. don't fit to Rainich/Rabinovich's G. Y./Y. G. initials. There are several versions of this story before 1929 (naturally, without mentioning of the trick). For the attribution in zbMATH, I looked further into this issue, and I think it is worth a separate thread. The topic argument is likely less relevant - his main field until 1922 was algebraic number theory, including G. Rabinowitsch, Eindeutigkeit der Zerlegung in Primzahlfaktoren in quadratischen Zahlkörpern., J. für Math. 142, 153-164 (1913). JFM 44.0243.03. which earned him a talk of 1912 ICM G. Rabinowitsch, Eindeutigkeit der Zerlegung in Primzahlfaktoren in quadratischen Zahlkörpern., Proc. 5. Intern. Math. Congr. 1, 418-421 (1913). JFM 44.0244.01. (more about this later). Clearly, he had sufficient algebraic expertise. However, all other objections are valid. I checked the biographies at the Faculty history of U Mich, Inna Rikun's Russian Biography and Notes on Jewish History and "Orbiting in Eccentric Circles: A Family History" by Rainich's daughter Alice Rainich Nichols. First, one notes that none of them mentions the Rabinowitsch Trick, least of all its attribution to Rainich. All make it clear that Rainich didn't use his old name after fleeing to the U.S. in early 1923. The reasons for his escape from Odessa (he was arrested in 1922 - some sources say for unknown reasons, while his daughter claims that his work in General Relativity was incompatible with then Communist Doctrine, so he was declared enemy of the state - and nearly died in prison) make it highly unlikely to claim an affiliation in Moscow (actually, he published just in the preceding volume of the Annalen his G. Y. Rainich, Über die analytische Funktion auf einer Minimalfläche., Math. Ann. 101, 386-393 (1929). JFM 55.0400.05. as ``G. Y. Rainich in Ann Arbor, Mich. U. S. A.) Though his change in first names was somewhat unusual (from Yuri Germanovich to George Yuri, which is basically a duplication since Yuri is the Slavic George, to the effect of switched initials, and dropping his father's name Germanovich), the initial L. appears nowhere else in his work (some caution is required here since the paper is often quoted with wrong initials, e.g., in the Am. Math. Monthly anecdote). Perhaps the strongest indication comes from the versions of the anecdote. Mordell's 1971 Reminiscences give the original: In 1923, I attended a meeting of the American Mathematical Society held at Vassar College in New York State. Some one called George Yuri Rainich from the University of Michigan at Ann Arbor, gave a talk upon the class number of quadratic fields, a subject in which I was then very much interested. I noticed that he made no reference to a rather pretty paper written by one Rabinowitz from Odessa and published in Crelle's journal. I commented upon this. He blushed and stammered and said, "I am Rabinowitz." He had moved to the U.S.A. and changed his name. This story is known all over the U.S.A. Occasionally some one from Ann Arbor dines at John's and I ask them if they know Rainich. Yes, they say, there is a funny story about him. "Stop," I say, "let me tell you the story." This clearly refers to his 1913 work cited above, not the Nullstellensatz. Apparently, this anecdote enjoyed wide circulation (Mordell himself points this out). In his daughter version from her book (pp. 30-31): Yuri Germanovich Rabinovich and George Yuri Rainich In the American world of mathematics, Yuri Germanovich Rabinovich and George Yuri Rainich came together at Vassar University in 1925, when Professor L J. Mordell challenged the originality of a paper given by my father. Upon immigrating Yuri, in order to simplify, changed his surname to Rainich by omitting the third, sixth and seventh letters of Rabinovich (he had hoped to make the change by omitting the third, fifth and seventh, as consecutive primes, but it didn't work out), and had also reversed the initials of his first and his middle name (patronymic), thus becoming G. Y. Rainich, the name by which he was known in America, Yuri had just presented a paper on 'Expansion of Simple Factors in Quadratic Fields', when Mordell commented: "Perhaps you are unaware, Professor Rainich that this work was already done by Rabinovich, in Odessa." "But I am Rabinovich!" Yuri replied. This was the origin of the story that has followed him, as well as me, throughout our lives. This incident is part of American mathematical history. Yuri had still been Rabinovich when he presented the paper in 1912 at the International Congress of Mathematicians at Cambridge. The paper was immediately published in the very prestigious German publication: 'Zeitschrift fur Angewandte Mathematik and Physik' (Journal of Applied Mathematics and Physics), better known as 'Crelle's Journal'. Though Yuri was already a well-known scholar at the time having presented papers at Twelfth and Thirteenth meetings of 'Russian Mathematicians and Physicists,' this publication proved to play a very important role in the young scholar's life. Eighty-seven years after this scene played out I read for the first time a different conclusion to the incident. In the newer (to me) version, Yuri, upon being challenged, walked over to the blackboard, and picking up the chalk wrote 'RABINOVICH' in large letters. He then erased the 'B', the 'O' and the 'V'. We shall never know which version is the true one. Despite small differences (1925 instead of 1923, and the misnomer of Crelle's journal), this agrees with Mordell, and the referred second variant from 2012 has been obviously triggered by the Am. Math. Monthly variant of the anecdote. Alice Nichols doesn't notice that in the meantime, also the discussed work changed to the Trick. A good indication how many embroidered variants circulated is the anecdote accompanying the Rainich Lecures at U Mich Professor Rainich is a Russian by birth and an interesting story is told of an experience he had while lecturing at Columbia University. Speaking on Relativity before the faculty he quoted a number of the foremost authorities known upon the subject. One of the members of the Columbia faculty spoke up saying, "That is all very well but why don't you quote what Rabinovich has said upon this subject? "Professor Rainich was somewhat embarrassed as he replied, "Well, you see, I am Rabinovich." Here, already, the subject has been made up to fit better into Rainich's later main field (and Vassar College has been changed to Columbia). Since Rainich's only work on General Relativity under the name of Rabinovich was a 1923 Russian translation of Eddington's "Space, time and gravitation", this version is obviously fabricated. Given the reported wide circulation, it was perhaps inevitable that a version including the Rabinowitsch trick would eventually come up. For all these reasons, I think we can safely conclude that there is no proof of attributing the Rabinowitsch Trick to Rainich (the paper is now excluded from his zbMATH author profile), and it is indeed highly likely it was discovered by another person. So the main question is, by whom? Nineteen years later, there is a Yu. L. Rabinovich in the records, working in analysis and, according to Rabinovich, Yu. L.; Nesterov, S. V., General form of linear differential equations whose order is lowered by means of a generalized differentiation operator (D^ \alpha), Sov. Math., Dokl. 2, 476-479 (1961); translation from Dokl. Akad. Nauk SSSR 137, 1309-1311 (1961). ZBL 0102.07302., working at Moscow State University in 1961. Though such publication gaps during the time of terror and war are not uncommon, there is, however, no clear link to the J. L. Rabinowitsch from 1929 Nullstellensatz. Question 1. Are there any sources about the biography of Yu. L. Rabinovich, who published 1948-1961 on analysis (at least partially based in Moscow), that may confirm or rule out a link to the author of Rabinowitsch's Trick? (He did also three reviews for Math Reviews 1948-1963). Perhaps the closest link, however, is to the author of "Sur les courbes planes du quatrième ordre possédant deux points doubles", Mathesis 45, 286-290 (1931). JFM 57.0827.02, a M. Rabinowitch from Liège. It's not just the next chronological appearance of this name in the math literature, it deals also with a topic closely related to the note in Mathematische Annalen two years earlier - indeed, one may easily imagine how studying hyperplane sections of singular varieties in the singular points may lead to the Rabinowitsch Trick. So, as a working hypothesis, the author of the Mathesis paper appears to be a natural candidate to identify with the inventor of the Rabinowitsch Trick, which leads to Question 2. Is there any information available on a mathematician named Rabinowitch who worked in Liège around 1931, and can it be confirmed (or ruled out) that he moved there from Moscow shortly before? And, of course, more general: Question 3. Are there any other sources which may contribute biographical information about the author of the Rabinowitsch Trick? REPLY [20 votes]: The book "Математика в СССР за 40 лет 1917-1957" (Mathematics in USSR in 40 years 1917-1957) mentions what seems to be both relevant Rabinivitsches. This is not definite, though, since the bibliography only has papers published in USSR. One is Ю. Г. Рабинович (Yu. G. Rabinovitsch), the book only mentions two of his papers, in 1918 and 1921. This is probably Rainich. Another is Юлий Лазаревич Рабинович. That can be transliterated either as Juli L. Rabinivitsch, or as Yuli L. Rabinivich, so it is consistent with the initials in Math. Ann. paper. The book only mentions his papers from 1948 to 1957, and those are mostly in Functional analysis, PDE and Special functions. He was born in 1894 in Bobruisk and graduated from Moscow State University in 1924, where he worked since 1932 and was awarded a Ph.D. and a title of Docent in in 1935. So, the bio is also consistent with him being research-active in Moscow in 1929. Update: his birth and death dates are 1.09.1894 - 18.03.1968 (the former is from the above-mentioned source, and the latter from the preface to the 2nd edition of Courant-Hilbert that he translated with Libin.) I found mentions of J. L. Rabinovich in several memoirs (including one mentioned by Carlo Beenakker below) as a docent in the Physics department of the Moscow State University. He was remembered as a brilliant lecturer.<|endoftext|> TITLE: Spheres with the same homotopy groups QUESTION [24 upvotes]: What is known about the existence of other pairs of spheres (such as $S^2$ and $S^3$) whose homotopy groups coincide starting from some index. A sufficient condition for this is the existence of a fiber bundle $S^m \to S^n$ with fiber having a finite number of nonzero homotopy groups (as in the case of the Hopf fibration) P.S. I don't know if my question has a research level. If it is not, then feel free to close it. REPLY [50 votes]: It is a result from Serre's thesis that for $n\geq 3$ and a prime $p$, the first $p$-torsion in $\pi_*S^n$ occurs precisely for $* = n+2p-3$. This shows that $(m,n) = (2,3)$ is the only pair of (edit: positive integers) $m TITLE: Is the square root of the Kullback-Leibler divergence a convex map? QUESTION [8 upvotes]: $\newcommand{\KL}{\operatorname{KL}}$Let $X$ be a Polish metric space and $P(X)$ the space of probability measures on $X$. Given $\mu, \nu\in P(X)$, recall that $$\KL(\mu\parallel\nu) = \begin{cases}\mathbb E_\mu[\log\tfrac{d\mu}{d\nu}]&\text{if $\mu\ll\nu$;}\\+\infty&\text{otherwise.}\end{cases}$$ I know that both $\mu\mapsto \KL(\mu\parallel\nu)$ and $\nu\mapsto \KL(\mu\parallel\nu)$ are convex maps $P(X)\to\mathbb [0,+\infty]$. Are $$\mu\mapsto \sqrt{\KL(\mu\parallel\nu)}$$ and $$\nu\mapsto \sqrt{\KL(\mu\parallel\nu)}$$ convex as well? If this is not true in general, does it exists a measure $\mu\in P(X)$ such that $\nu\mapsto \sqrt{\KL(\mu\parallel\nu)}$ is a convex map $P(X)\to \mathbb [0,+\infty]$? Or a measure $\nu$ such that $\mu\mapsto \sqrt{\KL(\mu\parallel\nu)}$ is convex? REPLY [10 votes]: $\newcommand\de\delta\newcommand{\KL}{\operatorname{KL}}\newcommand{\p}{\,\|\,}$The maps $$\mu\mapsto\sqrt{\KL(\mu\p\nu)}$$ and $$\nu\mapsto\sqrt{\KL(\mu\p\nu)}$$ are not convex in general. Indeed, let $\mu_p:=p\de_0+(1-p)\de_1$, where $p\in(0,1)$ and $\de_a$ is the Dirac measure supported on $\{a\}$. Then the second partial derivative with respect to $p$ of $\sqrt{\mathrm{KL}(\mu_p\p\mu_r)}$ at $(p,r)=(1/10,1/11)$ is $-7.17\ldots<0$. So, $\sqrt{\mathrm{KL}(\mu\p\mu_r)}$ is not convex in $\mu$. Also, the second partial derivative with respect to $r$ of $\sqrt{\mathrm{KL}(\mu_p\p\mu_r)}$ at $(p,r)=(1/10,1/9)$ is $-11.50\ldots<0$. So, $\sqrt{\mathrm{KL}(\mu_p\p\nu)}$ is not convex in $\nu$. You also asked: "If this is not true in general, does it exists a measure $\mu\in P(X)$ such that $\nu\mapsto \sqrt{\mathrm{KL}(\mu\p\nu)}$ is a convex map $P(X)\to \mathbb [0,+\infty]$? Or a measure $\nu$ such that $\mu\mapsto \sqrt{\mathrm{KL}(\mu\p\nu)}$ is convex?" The answer to each of these two questions is yes, at least when $X=\{0,1\}$, say. For $p\in(0,1)$, let \begin{equation} F(p):=\sqrt{\mathrm{KL}(\mu_p\p\mu_{1/2})}, \end{equation} \begin{equation} f(p):=F''(p)4 \mathrm{KL}(\mu_p\p\mu_{1/2})^{3/2}, \end{equation} \begin{equation} f_1(p):=f'(p)(1-p)^2 p^2. \end{equation} Then $f_1(1/2)=f'_1(1/2)=f''_1(1/2)=0$ and \begin{equation} f'''_1(p)=\frac{2+4 p(1-p)}{(1-p)^2 p^2}>0. \end{equation} It follows that $F''(p)\ge0$, so that $\sqrt{\mathrm{KL}(\mu\p\mu_{1/2})}$ is convex in $\mu$. For $r\in(0,1)$, let \begin{equation} G(r):=\sqrt{\mathrm{KL}(\mu_{1/2}\p\mu_r)}, \end{equation} and \begin{equation} g(r):=G''(r)4\mathrm{KL}(\mu_{1/2}\p\mu_r)^{3/2}. \end{equation} Then $g(1/2)=g'(1/2)=g''(1/2)=g'''(1/2)=0$ and \begin{equation} g''''(1/2+h)\frac{(1 - 4 h^2)^4}{16}=9- 16 h^4 + 156 h^2 + 64 h^6 \\ >9- 1 + 156 h^2 + 64 h^6>0 \end{equation} if $|h|<1/2$. It follows that $G''(r)\ge0$, so that $\sqrt{\KL(\mu_{1/2}\p\nu)}$ is convex in $\nu$. Remark 1: The existence of a probability measure $\nu\in P(X)$ such that $\sqrt{\KL(\mu\p\nu)}$ is convex in $\nu$ holds for any Polish space $X$. Indeed, the case when $X$ is a singleton set is trivial. Otherwise, take any distinct points $x$ and $y$ in $X$ and let $\nu:=\frac12\,\de_x+\frac12\,\de_y$. Then the condition $\mu\ll\nu$ implies $\mu=p\,\de_x+(1-p)\,\de_y$ for some $p\in[0,1]$, so that here $\sqrt{\KL(\mu\p\nu)}=F(p)$. Remark 2: Let us say that a probability measure $\mu\in P(X)$ is good if $\sqrt{\KL(\mu\p\nu)}$ is convex in $\nu$. Then it is easy to see that the Dirac measure $\de_x$ is good for any $x\in X$. It also follows from above that $\mu:=\frac12\,\de_x+\frac12\,\de_y$ is good if $X=\{x,y\}$ for some $x,y$. Finally, using reasoning similar to that above, one can show that, in the case when the cardinality of $X$ is $\ge3$, the only good probability measures $\mu\in P(X)$ are the Dirac measures. Since this answer has already grown rather long, I will leave the latter assertion as an exercise to interested readers. Now the answer is quite complete.<|endoftext|> TITLE: Torus bundles and compact solvmanifolds QUESTION [6 upvotes]: I asked this question on MSE 9 days ago and it got a very helpful comment from Eric Towers providing the Palais Stewart reference, but no answers. So I'm crossposting it here. Let $$ T^n \to M \to T^m $$ be a principal torus bundle over a torus. Then $ M $ is a solvmanifold, even a nilmanifold (in fact $ M $ is the total space of a principal torus bundle over a torus if and only if it is a compact nilmanifold for a 2 step nilpotent Lie group, this is theorem 3 of [Palais, Stewart, TORUS BUNDLES OVER A TORUS]). What if the bundle is not necessarily principal? Is every torus bundle over a torus a solvmanifold? In other words, if we have a fiber bundle $$ T^n \to M \to T^m $$ then can we conclude that the total space $ M $ is a solvmanifold? EDIT: The answer, comments and references from Igor Belegradek prove that something much stronger is true: A manifold $ M $ is the total space of a bundle $$ N \to M \to T^n $$ where $ N $ is a compact nilmanifold and $ T^n $ is a torus if and only if $ M $ is homeomorphic to a compact solvmanifold. The smooth case is also addressed. Corollary 2.3 of https://arxiv.org/pdf/1307.3223.pdf states that "If $ h: T^n → T^n $ , $ n ≥ 6 $, is a diffeomorphism given by Proposition 2.1 then the mapping torus $ M_h $ is a fake torus." By fake torus the authors mean $ M_h $ is homeomorphic to a torus but not PL homeomorphic. Since $ M_h $ is not PL homeomorphic to a torus it is also not diffeomorphic to a torus. Thus already in dimension 7 we have an exotic homotopy torus $ M_h $ which is the total space of a torus bundle over the torus $$ T^6 \to M_h \to S^1 $$ but which is not diffeomorphic to any solvmanifold. So in every dimension $ d \geq 7 $ there are smooth torus bundles over a torus (both base and fiber with the standard smooth structure) that are not diffeomorphic to any solvmanifold. For $ d \leq 3 $ we have by Moise's theorem that there are no exotic smooth structures. So all torus bundles over a torus with total space of dimension $ d \leq 3 $ are diffeomorphic to a solvmanifold. For $ d=4 $ the existence of exotic 4 tori (or any exotic closed aspherical manifold) seems to be an open problem see Do there exist exotic 4-tori? In that same question it is pointed out that there are exotic tori in $ d=5,6 $ but it is not clear that these can be constructed as smooth fiber bundles with base and fiber standard tori. That said, for the $ d=4 $ case a torus bundle over a torus has base and fiber with dimension $ \leq 3 $ so there shouldn't be any exotic gluing maps and it seems like the total space of such a bundle should just always have the standard smooth structure. For $ d=5,6 $ I have no idea what kind of smooth structures might or might not be possible on torus bundles over the torus. REPLY [7 votes]: A group $G$ is isomorphic to the fundamental groups of a compact solvmanifold if and only if it fits into the short exact sequence $1\to N\to G\to\mathbb Z^n\to 1$ where $N$ is a finitely generated torsion-free nilpotent group. This is stated on p.253 and explained is chapter III of Auslander's An exposition of the structure of solvmanifolds. Part I: Algebraic theory. In particular, every torus bundle over a torus is homotopy equivalent to a solvmanifold. You may also be interested in Theorem 3 of Wilking's paper Rigidity of group actions on solvable Lie groups which gives an analogous result for infrasolvmanifolds.<|endoftext|> TITLE: Possible sign of scalar curvature for Einstein warped product manifold with Ricci-flat QUESTION [5 upvotes]: Let $(M, g_M)$ where $M= B \times_f F$ and $g_M=g_B + f^2g_F$, an Einstein warped product manifold (i.e., $Ric_M= \lambda g_M$), with Ricci flat fiber-manifold $F$, i.e., $Ric_F=0$. Then $M$ can admit only constant negative Ricci curvature or zero Ricci curvature (i e., $\lambda \le 0$) or $M$ could also have positive constant Ricci curvature (i.e., $\lambda >0$)? In other words, $Ric_F = 0$ necessarily implies $\lambda \le 0$, or can solutions be obtained with $\lambda > 0$? REPLY [4 votes]: It can not have constant positive Ricci curvature. By Bonnet-Myers constant positive Ricci curvature implies that $M$ is compact. If $V$ is a vertical vector then by the formula for Ricci curvature of warped product (page 266 in Besse's book) $$ Ric(V,V)=Ric_F(V,V) -|V|^2(\frac{\Delta f}{f}+(p-1)\frac{|\nabla f|^2}{f^2}) $$ where $p=\dim F$. If the fiber is Ricci flat then at the point on the base where $f$ achieves minimum (which exists by compactness) it holds that $Ric(V,V)\le 0$<|endoftext|> TITLE: A set with positive upper density whose difference set does not contain an infinite arithmetic progression QUESTION [9 upvotes]: For $S \subset \mathbb{N}$ define $S-S=\{x-y:x \in S, y \in S\}$. As noted below there is a simple example showing that a set $S \subset \mathbb{N}$ with positive upper density has a sumset $S+S=\{x+y:x \in S, y \in S\}$ with $S+S$ containing only finite length arithmetic progressions. However the case for the difference set seems not so obvious to me hence the question: What is an example of a set S with positive upper density in $\mathbb{N}$ such that $S-S$ does not contain an infinite arithmetic progression? Here is the example for the sumset $S+S$, in fact for any $hS=S+\dots+S$, taken from Erdos, Nathason and Sarkozy's paper "Sumsets Containing Infinite Arithmetic Progressions": "Let $(t_n)$ be a strictly increasing sequence of positive integers such that $t_{n+1}/t_n$ tends to infinity, and let the set $A$ be the union of the intervals $[t_{2n}+1, t_{2n+1}]$. Then $A$ has upper asymptotic density $d_U(A) = 1$ and lower asymptotic density $d_L(A)=0$. For fixed $h$ and all sufficiently large $n$, the sumset $hA$ is disjoint from the interval $[h t_{2n-1} + 1, t_{2n}]$. Thus, $hA$ contains arbitrarily long gaps, and so cannot contain an infinite arithmetic progression." REPLY [6 votes]: If the complement of $S-S$ contains two integers that are relatively prime, the density of $S$ is strictly smaller than 1/2. So a density of 1/2 is not possible. It is possible to get the density of $S$ arbitrarily close to 1/2. To show this, note that there are countably many infinite arithmetic sequences and number them $L_1,L_2,L_3,\ldots$. For each natural number $k$ we will recursively define $a_k\in L_k$ and an $a_k+1$-periodic set $S_k$ such that $a_i\not\in S_k-S_k$ for $i\leq k$. The sequence $(a_k)$ will be increasing. We can assume that $a_1=1$ and choose $S_1$ the odd numbers. For the recursion step, pick $a_{k+1} \in L_{k+1}$ large enough. Let $S_{k+1}$ be the $a_{k+1}$-periodic extension of the set $S_k\cap \{1,2,\ldots,a_{k+1}-a_k\}$. The difference between the different 'parts' of $S_{k+1}$ is larger than $a_k$, so $S_{k+1}-S_{k+1}$ still does not contain $a_i$ for $i\leq k$. Moreover, since $1\not\in S_{k+1}-S_{k+1}$ and $S_{k+1}$ is $a_{k+1}+1$-periodic, also $a_{k+1}\not\in S_{k+1}-S_{k+1}$. The density of $S_{k+1}$ will be at most $\frac{a_k}{a_{k+1}}$ less than the density of $S_k$. Since the first $a_k$ numbers do not change after $S_k$ is fixed, we can define the limit set $S$, which will have density arbitrarily close to 1/2. It will contain all $a_k$, so it will not contain an infinite arithmetic sequence.<|endoftext|> TITLE: How hard is it to say "not exactly $p$" with a Horn sentence? QUESTION [17 upvotes]: EDIT: immediately after bountying the question (whoops ...) I found, while looking for something else entirely, that Sauro Tulipani gave an explicit algorithm for producing a Horn sentence $\varphi_p$ saying "There are not exactly $p$ objects" and $length(\varphi_p)=O(p^5\log p).$ So my question should be rephrased as: Is Tulipani's solution optimal? (From Tulipani's article I also learned that the construction mentioned below was originally due to Appel, and was rediscovered later by Blass independently.) This question is motivated by Blass' paper There are not exactly five objects. Briefly, Wilkie observed that a general semantic result implies that "There are not exactly $p$ objects" must be expressible by a Horn sentence$^*$ in the equality-only language for each prime $p$, but actually finding such sentences is nontrivial. Blass' example is a bit long (with length exceeding $10^{10^{22}}$ in the case $p=5$); he goes on to say that a somewhat shorter example exists for $p=5$ (due to Morley) but that it doesn't obviously generalize to arbitrary primes. I'm curious what the optimal length is, as a function of the index of the prime involved: For $i\in\mathbb{N}$ let $H(i)$ be the length of the shortest Horn sentence equivalent to "There are not exactly $p_i$ objects," where $p_i$ is the $i$th prime. What can we say about the growth rate of $H$? In particular, it is consistent with my current understanding that Blass' construction is asymptotically optimal. I suspect that's not the case, but I don't have any actual evidence for that. (The tag "universal-algebra" is a bit tentative; my understanding is that, although not strictly equational, Horn logic is still of interest to UA, but I could easily be overestimating its relevance.) $^*$That is: "a first-order sentence in prenex normal form whose matrix is a conjunction of conditionals in each of which the consequent is atomic or "false" and the antecedent is a conjunction of atomic formulas." The obvious first-order formulation $$\forall x_1,...,x_p[(\bigwedge_{1\le i TITLE: Find the determinant of a matrix given the determinant of all $p\times p$ sub-matrices? QUESTION [16 upvotes]: Is it possible to find the determinant of an $n\times n$- matrix, only given the determinant of all $p\times p$ sub-matrices in it? Here $p\leq n$ is fixed. This is obviously true if $p=1,n$. But what happens in other cases? REPLY [16 votes]: Here is a way to see that, for all $p$ (even if $p \nmid n$), you can reconstruct the absolute value of the matrix's determinant, which was suggested by Will Sawin's comment. The condition $p \mid n$ is only needed to get the sign (or complex phase, if the ground field is $\mathbb{C}$). If you know the determinants of all $p \times p$ submatrices of $A$, that means exactly that you know $C_p(A)$: the $p$-th compound matrix of $A$. Since $\det(C_p(A)) = (\det(A))^\binom{n-1}{p-1}$, you can compute $$ \lvert\det(A)\rvert = \lvert\det(C_p(A))\rvert^{1/\binom{n-1}{p-1}}. $$<|endoftext|> TITLE: Relation between Ricci curvature and sectional curvature for 3-manifolds QUESTION [6 upvotes]: Let $(M^n,g)$ be a smooth Riemannian manifold. It is well known that if $sec(M)\geq \kappa$ then $Ric(M)\geq (n-1)\kappa$. If I understand correctly in dimensions $n\geq 4$ a lower bound on $Ric(M)$ does not imply a lower bound on $sec(M)$. However in $n=2$ this implication is true for trivial reason. Is it true that in $n=3$ a lower bound on $Ric(M)$ implies a lower bound on $sec(M)$? REPLY [11 votes]: This is definitely false. In dimension 3 if $\lambda_1,\lambda_2,\lambda_3$ are eigenvalues of the curvature operator then Ricci curvatures of eigenvectors are $\lambda_1+\lambda_2, \lambda_1+\lambda_3, \lambda_2+\lambda_3$. If one of $\lambda_i$'s is very negative but the other two are very positive then all these sums can be bounded below. Every algebraic curvature tensor can be realized at a point so locally this can definitely happen. Complete examples exist too. I can't seem to find a reference to an explicit example but it's well known that there are examples of metrics on $\mathbb R^3$ with $Ric\ge 0$ which have some negative sectional curvature. If you rescale such metric by a small number Ricci curvature remains nonnegative but sectional curvature can become arbitrary negative in some 2-planes.<|endoftext|> TITLE: Multiplicative structures on truncated Moore spectra QUESTION [10 upvotes]: As discussed for example in this paper (and this MO thread), there are obstructions for the existence of $A_n$-structures on Moore spectra $M(p^r):=\mathbb{S}/p^r$, which in general don't vanish. In particular, for odd primes, $M(p)$ does not admit an $A_p$-strucutre, and if I understand correctly, it is expected that for no value of $r$ the spectrum $M(p^r)$ admits an $A_\infty$-structure. I am interested in the analogous question for the $d$-truncated Moore spectra $M(p^r,d) := \tau_{\le d}(\mathbb{S}/p^r)$: What is known about the existence of $A_n$ (or perhaps even $E_n$) structures on $M(p^r,d)$? For $d=0$ one gets $M(p^r,0) \simeq H\mathbb{Z}/p^r$, which is an $E_\infty$-ring. However, I am not sure what is the situation even for $d=1$. Specifically, I am interested in the case where $p$ and $d$ are fixed and $r \gg 0$. It would be best if one could get an $E_\infty$-strucutre (the goal being to approximate $\tau_{\le d}(\mathbb{S}_p)$ by $\pi$-finite $E_\infty$-rings). REPLY [6 votes]: EDIT: In case you missed it, there's been a huge breakthrough in this area by Robert Burklund. I believe that Prasit Bhattacharya's methods, as linked to in the question, can be used to show that for fixed $p,d$ and $r$ sufficiently large, the truncated Moore spectrum $M(p^r,d)$ is $A_\infty$, and probably even $E_\infty$. Prasit shows that $M(p^r)$ is the Thom spectrum of any map $f_{p,r,u} : S^1 \to BGL_1(\mathbb S^\wedge_p)$ which picks out $1 + p^r u \in \mathbb Z_p \subset \mathbb Z/(p-1) \times \mathbb Z_p = \pi_1(BGL_1(\mathbb S^\wedge_p)$ for any unit $u \in \mathbb Z^\times_p$. His bounds on $A_n$-ness come from finding a choice of unit $u$ for which the map $f_{p,r,u}$ is $A_n$. Now, if we're interested in $M(p^r,d)$, then if I'm not mistaken, from facts like $\tau_{\leq d} \Sigma^\infty_+ X = \tau_{\leq d} \Sigma^\infty_+(\tau_{\leq d} X)$ we can deduce that $M(p^r, d)$ is the $\tau_{\leq d}$-truncation of the the Thom spectrum of the map $S^1 \xrightarrow{f_{p,r,u}} BGL_1(\mathbb S^\wedge_p) \to\tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$. [1] Since passage to Thom spectra takes $O$-maps to $O$-algebras for any operad $O$, it will suffice to show that $u$ can be chosen so that $S^1 \xrightarrow{f_{p,r,u}} BGL_1(\mathbb S^\wedge_p) \to \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$ is an $A_n$ map. As Prasit shows, this means we have to lift the composite map $$S^2 \to \Sigma \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p) \to B\tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$$ through the map $S^2 \to \mathbb C\mathbb P^n$ (the significance of $\mathbb C \mathbb P^n$ is that it's the $n$-truncated bar construction on $S^1$). Prasit shows that this is possible before taking $\tau_{\leq d+1}$ for some $n = n(r)$ if $r$ is sufficiently large. But then if we're truncating, because the maps $\mathbb C \mathbb P^n \to \mathbb C \mathbb P^{n+1}$ are increasing in connectivity, we can just automatically extend, as long as $2n(r) \geq d$. Since $n(r) \to \infty$ as $r \to \infty$, this is possible. I believe the obstruction theory to get an $E_\infty$ map works similarly in that the obstructions lie in higher and higher homotopy groups, so as long as you get some lift up to level $d+1$, you can extend to an $E_\infty$ structure when you're $d$-truncating. But perhaps somebody who is actually familiar with the relevant $E_\infty$ obstruction theory could say something more definitive. [1] The argument I have in mind constructs the Thom spectrum as a bar construction $M(f) = |\Sigma^\infty_+ F \wedge \Sigma^\infty_+ BGL_1(\mathbb S^\wedge_p)^\bullet|$ where $F$ is the fiber of $f$. So $\tau_{\leq d} M(f) = \tau_{\leq d} |\tau_{\leq d} \Sigma^\infty_+ \tau_{\leq d} F \wedge \Sigma^\infty_+ \tau_{\leq e} BGL_1(\mathbb S^\wedge_p)^\bullet|$ so long as $e \geq d$. When we take $e = d+1$, we observe that because the homotopy groups of $S^1$ (the domain of $f$) are easy, we have a fiber sequence $\tau_{\leq d} F \to S^1 \to \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$. So the bar construction we're taking is now, up to $\tau_{\leq d}$-truncation, exactly the same bar construction as for the Thom spectrum of the map $S^1 \to \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$. There's probably a nicer way to say this, though. REPLY [5 votes]: The $1$-type of $M\mathbb{Z}/(2^r)$ does have an $E_\infty$-ring structure for $r> 1$. I'm going to show it by using the algebraic models for $1$-truncated connective commutative ring spectra from: MR2405894 Reviewed Baues, Hans-Joachim; Jibladze, Mamuka; Pirashvili, Teimuraz Third Mac Lane cohomology. Math. Proc. Cambridge Philos. Soc. 144 (2008), no. 2, 337–367. MR2793446 Reviewed Baues, Hans-Joachim; Muro, Fernando The algebra of secondary homotopy operations in ring spectra. Proc. Lond. Math. Soc. (3) 102 (2011), no. 4, 637–696. These algebraic models, called $E_\infty$-quadratic pair algebras, give rise to bipermutative categories (see Remarks 5.9, 6.10, and 9.14 in the second paper) which, in turn, can be used to construct 1-truncated connective commutative ring spectra as per: A. D. Elmendorf and M. A. Mandell, Rings, modules, and algebras in infinite loop space theory, Adv. Math. 205 (2006) 163–228. I'm sure you can find other proofs, e.g. via an explicit topological construction or lifting the first $k$-invariant to topological André-Quillen cohomology. Consider the following $E_\infty$-quadratic pair algebra $$\begin{array}{ccc} \mathbb{Z}/2\times \mathbb{Z}&\stackrel{\partial}\longrightarrow& \mathbb{Z}\\ \nwarrow^P&&\swarrow_H \\ &\mathbb{Z}& \end{array}$$ Here, $$ \begin{array}{rcl} \partial([a],n)&=&2^rn,\\ P(a)&=&([a],0),\\ H(n)&=&\frac{n(n-1)}{2}. \end{array} $$ Moreover, $\mathbb{Z}$ is endowed with the usual product (in a quadratic pair algebra it could have a product which is only right distributive), but it doesn't in this case) and $\mathbb{Z}/2\times \mathbb{Z}$ is endowed with the usual left and right product by elements of $\mathbb{Z}$ (again, the product from the left need not be distributive in general). There could be an additional $\smile_1$ operation that in this case is trivial because all previous products commute. The $k$-invariant of such a structure is the homomorphism $$\operatorname{coker}\partial\otimes\mathbb{Z}/2\longrightarrow \ker\partial$$ defined by $$[a]\otimes[1]\mapsto P(H(2a)-2H(a)).$$ In the previous example, this morphism is $$\mathbb{Z}/2^r\otimes\mathbb{Z}/2\cong \mathbb{Z}/2$$ which coincides with the map $$\pi_0M\mathbb{Z}/2^r\otimes\pi_1S\longrightarrow \pi_1M\mathbb{Z}/2^r\colon [f]\otimes[g]\mapsto [fg].$$ This is the first $k$-invariant of $M\mathbb{Z}/2^r$. You may wonder what fails for $r=1$. For the second equation of the second set in Definition 6.1 (in the second of the aforementioned papers) we need $PH\partial=0$. This holds iff $r>1$ since $$PH\partial([a],n)=\left(\left[\frac{2^rn(2^rn-1)}{2}\right],0\right)=([2^{r-1}n],0).$$<|endoftext|> TITLE: When is a basis of a topological space a Grothendieck pretopology? QUESTION [7 upvotes]: Bases of a topological space in point set topology will in general form a coverage on its category of inclusion on open subsets and on its category of inclusion on basic opens, but it takes a bit more work to check whether either forms a Grothendieck pretopology. Is there a useful or natural criterion for when a (point-set) basis does give a (Grothendieck) basis? The criteria may apply either to the bases themselves, or to any particularly nice property of a topological space that forces some class of bases to have that property. REPLY [11 votes]: This is a matter of expanding the definition, in this case Definition II.1.3 in SGA 4, which defines pretopologies. By a “base” in this answer I mean what appears to be the most common definition: a collection of subsets of a fixed set $A$ such that any finite intersection of elements in the base is a union of elements in the base. There are also multiple constructions of a site (i.e., a category with a coverage) from a base of a topological space $A$. One can either (A) take the category of all open subsets of $A$, or (B) the category whose objects are open subsets of $A$ that belong to the base. For covering families of some object $V$, one can either take (a) those open covers of $V$ whose elements belong to the base, or (b) open covers of $V$ whose elements are given by the intersection of $V$ and some element of the base. Altogether, there are three different options: A-a, A-b, B-a, and only option A-b produces a pretopology in the sense of SGA 4. Axiom PT0 for pretopologies says that any morphism in a covering family admits base changes. Such base changes are always given by the corresponding intersection, provided that the intersection belongs to the category. Thus, PT0 is satisfied for options A-a, A-b and not satisfied for option B-a. If the base is closed under intersections, then PT0 is also satisfied for option B-a. Axiom PT1 says that base changes of covering families are covering families. In our case, the base change is given by the intersection with some open subset $V$. Thus, PT1 is satisfied for option A-b and not satisfied for options A-a, B-a. If the base is closed under intersections, then PT1 is also satisfied for options A-a, B-a. Axiom PT2 says that covering families can be composed. This is trivially true for the case under consideration. Axiom PT3 says that the singleton family consisting of the identity map is a covering family, which is tautologically true in our case. Thus, option A-b always gives a pretopology, whereas options A-a, B-a give a pretopology if and only if the base is closed under intersections of pairs. In particular, open balls in a metric space as a coverage on the category of open subsets trivially form a pretopology using option A-b, and do not for a pretopology in options A-a, B-a.<|endoftext|> TITLE: Hamilton cycles in $\{0,1\}^n$ with fixed Hamming distance QUESTION [5 upvotes]: Let $n>1$ be an integer. For $a,b\in \{0,1\}^n$ let $d_h(a, b)$ denote the Hamming distance of $a$ and $b$. For $k\in \{1,\ldots,n-1\}$ let $H(n,k)$ be the graph on $\{0,1\}^n$ given by the edge set $$E(n,k) = \{(a, b)\in (\{0,1\}^n)^2 : d_h(a, b) = k\}.$$ Question. For what values of $k\in \{2,\ldots n-1\}$ does $H(n,k)$ have a Hamilton cycle? Note 1. Hamilton cycles in $H(n,1)$ are called Gray codes. Note 2. A necessary (but maybe not sufficient) condition for $H(n,k)$ to have a Hamilton cycle is that $\text{gcd}(k,2^n) = 1$, otherwise $H(n,k)$ is not a connected graph. REPLY [5 votes]: Here is a different perspective on the problem (which also settles the case left open by Max Alekseyev's answer). $H(n,k)$ is a Cayley graph of the group $G = \{0,1\}^n$ with respect to the set $M_k$ of elements with exactly $k$ non-zero entries. It is well known (and easy to prove by induction on the number of generators) that any connected Cayley graph of an abelian group contains a Hamilton cycle. So we only need to check when $H(n,k)$ is connected, or equivalently, if $M_k$ is a generating set of $G$. Note that any element of $M_2$ is the sum of two elements in $M_k$, hence $M_{k-2}$ is contained in the subgroup generated by $M_k$. If $k$ is odd, then this implies that $M_k$ generates $M_1$ and thus all of $G$. If $k$ is even, then (as noted by Max Alekseyev) $M_k$ generates the same subgroup of index $2$ as $M_2$.<|endoftext|> TITLE: Regular epimorphisms in the category of topological rings QUESTION [7 upvotes]: I am cross-posting this question from math stack exchange (link below) as it has not received any comments or answers over the past month. A regular epimorphism is a morphism $f: X \to Y$ that is the coequalizer of some parallel pair $a,b: Z \to X$. I believe regular epimorphisms in the category of topological groups are precisely the surjective open maps. Is this also true for the category of topological rings? If not, is there some other characterization of the regular epimorphisms in the category $\mathbf{TopRing}$? A reference would be very much appreciated. https://math.stackexchange.com/questions/4357004/regular-epimorphisms-in-the-category-of-topological-rings REPLY [9 votes]: It is true and hods more generlaly for models of any algebraic theory with a Mal'tsev operation, i.e. a trinary operation $f$ such that $f(x,x,y)=y=f(y,z,z)$. For groups, you can take $f(x,y,z)=xy^{-1}z$. This is Theorem 10 in A.I. Mal'tsev, On the general theory of algebraic systems, Mat. Sb. (N.S.), 1954, Volume (35)(77), Number 1, 3-20. A proof in English can be found on page 396 of V.V. Uspesnki$\check{\text i}$. The Ma'tsev operation on countably compact spaces. Comment. mat. Univ. Carolinae30 (1989) 395-402. available at https://dml.cz/bitstream/handle/10338.dmlcz/106758/CommentatMathUnivCarol_030-1989-2_22.pdf The idea is to first simplify the situation of a regular epimorphism $X\to Y$ by using the fact it is the coequalizer of its kernel pair, hence the quotient of an equivalence relation on $X$, whence we want to show that the set of equivalence classes intersecting an open subset $U$ of $X$ is open, i.e. that any $y\in X$ equivalent to $x\in U$ has an open neighborhood of points $z$ equivalent to points in $U$. But if this relation is kernel pair in the category of models of a theory, then whatever operations the theory has, the equivalence relation has to respect. In particular, $x\sim y$ implies $f(x,y,z)\sim f(x,y,z)\sim f(x,x,z)=z$. Then because $f$ is continuous, there is an open that contains $z$ if and only if $f(x,y,z)\in U$, in which case $z\sim f(x,y,z)\in U$ for any $z$ in that open. Moreover, $x=f(x,y,y)$ implies the open contains $y$ and so is the desired neighborhood.<|endoftext|> TITLE: Annihilate a simple Lie algebra using two commutators QUESTION [14 upvotes]: Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra over an arbitrary field $K$. For any nonzero $x\in\mathfrak{g}$ we must have $[\mathfrak{g},x]\neq\{0\}$, or else we violate simplicity. Can it happen that there are two nonzero elements $x,y\in\mathfrak{g}$ such that $[[\mathfrak{g},x],y]=\{0\}$? If so, do $x,y$ have to satisfy some conditions? Are the previous questions easier to answer if we restrict to some $K$ (say algebraically closed, or of characteristic $0$) or to some $\mathfrak{g}$? REPLY [3 votes]: I will sketch the proof that over the complex numbers, the answer is no. The set $$\{x\in \mathbb{P}(\mathfrak{g}) \mid \exists y\neq 0, [[\mathfrak{g},x],y]=0\}$$ is closed and $G$-invariant. Therefore it suffices to assume that $x$ lies in a closed $G$-orbit in $\mathbb{P}(\mathfrak{g})$. So we can assume that $x\in \mathfrak{g}_\alpha$ for some root $\alpha$, where we have also fixed a Cartan subalgebra $\mathfrak{h}$ to talk about root spaces. [I'll give a proof of this fact at the end.] Now write $y=h+\sum_\beta c_\beta X_\beta$, where $h\in \mathfrak{h}$ and $X_\beta\in \mathfrak{g}_\beta$. Then $[y,X_\gamma]=0$ for all $\gamma$ with $\gamma-\alpha$ a root or zero. The set of possible $\gamma$ does not lie in a hyperplane, which forces $h=0$. For every root $\beta$ there exists such a $\gamma$ with $\gamma+\beta$ a root or zero, which forces $c_\beta=0$, QED. Proof of the classification of closed $G$-orbits on $\mathbb{P}(\mathfrak{g})$: If $x$ is in a closed $G$-orbit, then its stabiliser contains a Borel $B$. To be stable under the torus $T$ implies that $x$ lies in a single weight-space. To be stable under the unipotent radical implies that that weight space must be the highest weight. (An alternative approach to this result is the argument in the proof of Theorem 4.3.3 in Collingwood and McGovern's "Nilpotent Orbits in Semisimple Lie Algebras".)<|endoftext|> TITLE: Lie groups generated by finitely many Lie algebra elements QUESTION [22 upvotes]: Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. A standard fact is that $G$ is generated by $\exp(\mathfrak{g})$, i.e. every $g \in G$ can be written as $g=\exp(x_1)\cdots\exp(x_n)$ for some $x_i\in\mathfrak{g}$. A natural follow-up question is whether there is a bound on the number $n$ of Lie algebra elements. I suppose this depends on the group, so let me formulate my question as follows. Question. What are natural conditions on a Lie group $G$ which imply that there is a finite number $n$ such that $$G = \underbrace{\exp(\mathfrak{g}) \cdots \exp(\mathfrak{g})}_{n \text{ times}}?$$ For instance, is this true for complex semisimple groups? Of course, there are many cases in which $\exp$ is surjective (e.g. $G$ compact or nilpotent) so that $n = 1$. I'm looking for a larger class of groups, preferably in the linear algebraic category, or just interesting specific examples where $n > 1$. REPLY [25 votes]: $\def\fg{\mathfrak{g}}$The following answer is a paraphrase of material from Philips, Christopher N., How many exponentials?, Am. J. Math. 116, No. 6, 1513-1543 (1994). ZBL0839.46054. We'll say that a Lie group $G$ has exponential rank $\leq k$ if every element of $G$ is a product of $\leq k$ exponentials. We'll say that $G$ has exponential rank $\leq k+\epsilon$ if the set of products of elements $\leq k$ exponents is dense in $G$. In other words, in a group of exponential rank $\leq k+\epsilon$, for any open neighborhood $\Omega$ of the identity, we have $G = \exp(g_1) \exp(g_2) \cdots \exp(g_k) u$ for $g_1$, $g_2$, ..., $g_k\in\fg$ and $u \in \Omega$. Since it is known that $\exp(\fg)$ contains an open neighborhood of the identity, any group of exponential rank $\leq k+\epsilon$ also has exponential rank $\leq k+1$. We define the exponential rank of $G$ to be the minimum $r$ in $\{ 1 < 1+\epsilon < 2 < 2+\epsilon < 3 < \cdots \}$ such that $G$ has exponential rank $\leq r$. Note that, if $G = AB$ (for Lie subgroups $A$ and $B$) and the exponential ranks of $A$ and $B$ are $(a,b)$, $(a+\epsilon, b)$, $(a, b+\epsilon)$ or $(a+\epsilon, b+\epsilon)$, then the exponential rank of $G$ is bounded above by $a+b$, $a+b+\epsilon$, $a+b+\epsilon$, $a+b+\epsilon$ respectively. In this vocabulary, Phillips proves (Section 5) that Connected compact groups have exponential rank $\leq 1$. Unipotent groups have exponential rank $\leq 1$. Connected complex groups have exponential rank $\leq 1+\epsilon$. This bound cannot be improved, because $SL_2(\mathbb{C})$ has exponential rank $1+\epsilon$ (the matrix $\left[ \begin{smallmatrix} -1&1 \\ 0&-1 \end{smallmatrix} \right]$ is not an exponential)). Connected solvable groups have exponential rank $\leq 1+\epsilon$. This bound cannot be improved: Consider the Borel $\left[ \begin{smallmatrix} z & u \\ 0 & z^{-1} \\ \end{smallmatrix} \right]$ inside $SL_2(\mathbb{C})$ and the same matrix as before. All connected Lie groups have rank $\leq 2+\epsilon$. $SL_2(\mathbb{R})$ has exponential rank $2$. The proof that all connected Lie groups have rank $\leq 2+\epsilon$ is credited to Djokovic: The point is that any connected $G$ has an Iwasawa decomposition $K(ANR)$ where $K$ is compact connected (so exponential rank $\leq 1$) and $ANR$ is connected solvable (so exponential rank $\leq 1+\epsilon$). Phillips states it as an open problem whether there is a connected Lie group of exponential rank $2+\epsilon$, or whether, in fact, exponential rank $2$ is always enough.<|endoftext|> TITLE: Is irreducibility of polynomials $\in \mathbb{Z} [X]$ over $\mathbb{Q}$ an undecidable problem? QUESTION [13 upvotes]: There are a number of criteria for determining whether a polynomial $\in \mathbb{Z} [X]$ is irreducible over $\mathbb{Q}$ (the traditional ones being Eisenstein criterion and irreducibility over a prime finite field). I was wondering if the decision problem of "Given an arbitrary polynomial $\in \mathbb{Z} [X]$, is it irreducible over $\mathbb{Q}$ or not?" is decidable or undecidable. REPLY [22 votes]: There is a quick way to see that this is decidable (with terrible complexity). Let $h(x) \in \mathbb{Z}[x]$ have degree $d$. Evaluate $h$ at $d+1$ points $u_1$, $u_2$, ..., $u_{d+1}$. If any of the $h(u_i)$ are $0$, we have found a factor and we can reduce to a problem of lower degree, so suppose that $z_i:=h(u_i)$ is nonzero for $1 \leq i \leq d+1$. There are finitely many ways to split each $z_i$ as $x_i y_i$ for integers $x_i$ and $y_i$, which we can find using a prime factorization of $z_i$. For each splitting $(x_1, x_2, \dots, x_{d+1}, y_1, y_2, \dots, y_{d+1})$ with $x_i y_i = h(u_i)$, we can use Lagrange interpolation to find the unique polynomials of degree $\leq d$ with $f(u_i) = x_i$ and $g(v_i) = y_i$. If $\deg f + \deg g \leq d$, then we have $f(x) g(x) = h(x)$ at $d+1$ values of $x$, so we must have $f(x) g(x) = h(x)$ as a polynomial identity and we have found a factorization. Conversely, if a factorization $h(x) = f(x) g(x)$ exists, then Lagrange interpolation will find it for the splitting where $x_i = f(u_i)$, $y_i = g(u_i)$. Of course, this is a terrible algorithm, since it involves taking the prime factorization of $d+1$ integers and then doing exponentially many cases of Lagrange interpolation. But it is the only algorithm which I know how to explain in ten minutes, so it is useful when a student asks this question. See GH from MO's answer for a good algorithm.<|endoftext|> TITLE: Extended binomial coefficients and the gamma function QUESTION [5 upvotes]: For which $(a,b,n) \in \mathbb{Z}^3$ satisfying $a+b=n$ does $\frac{\Gamma(z+1)}{\Gamma(x+1)\Gamma(y+1)}$ approach a limit as $(x,y,z) \rightarrow (a,b,n)$ in $\mathbb{C}^3$, and what is that limit? (The case where $a$, $b$, and $n$ are all negative is the problematic one.) What if we restrict to $\{(x,y,z) \in \mathbb{C}^3 \ | \ x+y=z\}$? I'm guessing that where the limit exists it agrees with the "windmill" shown in https://i.stack.imgur.com/osFsj.png (taken from page 197 of Hilton, Holton, and Pedersen's "Mathematical Reflections: In a Room with Many Mirrors"). REPLY [6 votes]: There's nothing special about the gamma function; the failure of the limit to exist when $a$, $b$, and $n$ are negative is exactly the same as the failure of $$\lim_{(x,y,z)\to(0,0,0)} \frac{xy}{z}$$ to exist. I will renormalize in a way that I think makes clearer what's going on. First let's look at $$\lim_{(x,y,z)\to(0,0,0)} \frac{\Gamma(z)}{\Gamma(x)\Gamma(y)}.$$ We have $$\frac{\Gamma(z)}{\Gamma(x)\Gamma(y)} = \frac{xy}{z}\frac{\Gamma(z+1)}{\Gamma(x+1)\Gamma(y+1)}.$$ Since $\Gamma(z+1)/\Gamma(x+1)\Gamma(y+1)$ is analytic and nonzero near $(0,0,0)$, $\Gamma(z)/\Gamma(x)\Gamma(y)$ behaves just like $xy/z$ near $(0,0,0)$. In particular, if you restrict to $x+y=z$, this is $xy/(x+y)$, for which the limit as $(x,y)\to(0,0)$ does not exist. Now let's consider $$\lim_{(x,y,z)\to(0,0,0)} \frac{\Gamma(z-a)}{\Gamma(x-b)\Gamma(y-c)}$$ where $a$, $b$, and $c$ are nonnegative integers. We have $$\frac{\Gamma(z-a)}{\Gamma(x-b)\Gamma(y-c)} = \frac{\Gamma(z)}{\Gamma(x)\Gamma(y)}\cdot \frac{(x-b)\cdots (x-1)(y-c)\cdots (y-1)}{(z-a)\cdots(z-1)}.$$ The second factor on the right is analytic and nonzero near $(0,0,0)$, so $\Gamma(z-a)/\Gamma(x-b)\Gamma(y-c)$ behaves just like $\Gamma(z)/\Gamma(x)\Gamma(y)$ (and thus like $xy/z$) near $(0,0,0)$.<|endoftext|> TITLE: Are isomorphic subgroups of the symmetric group conjugate in some larger group? QUESTION [5 upvotes]: $\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$Let $S_\infty$ be the group of all permutations of a countable infinite set (enriched with the product Polish topology). Does there exist a (continuous) group embedding $\phi: S_\infty\to S_\infty$ such that $\phi(G)$ is conjugate to $\phi(H)$ for every pair of (closed) isomorphic groups $G, H\leq S_\infty$? Note that two subgroups $A, B$ of a group $G$ are conjugate in $G$ if there is $g\in G$ for which $A=gBg^{-1}$. Additional note: It would be stil of interest for me, if there is an embedding $\phi: S_\infty \to T$ for some Polish group $T$, such that $\phi(G)$ is conjugate to $\phi(H)$ in $T$ for every pair of (closed) isomorphic groups $G, H\leq S_\infty$? My first guess was to consider $T$ to be the semidirect product of $S_\infty$ with $\Aut(S_\infty)=\Inn(S_\infty)=S_\infty$. REPLY [5 votes]: No, there's no such endomorphism. More generally: Proposition. for every infinite set $X$, there exists no set $Y$ and nontrivial continuous homomorphism $f:S_X\to S_Y$ for which the following assertion holds: for any two isomorphic discrete infinite cyclic subgroups of $S_X$, the images $f(S_X)$ and $f(S_Y)$ are conjugate. Note: every group endomorphism of $S_\omega$ is continuous (automatic continuity, here essentially due to Dixon-Neumann-Thomas: see Prop 4.3 and Thm 4.4 in Rosendal's survey). [The notation $S_\infty$, which implicitly assumes that every infinite set is countable (Cantor disproved this) is inconvenient.] Let $c$ be any nontrivial permutation of $X$ consisting only of infinite cycles (at least one) and fixed points. Let $c'$ be any permutation of $X$ with both at least one infinite cycle and finite cycles of all sizes. Since both $c$ and $c'$ have an infinite cycle, the cyclic subgroups $\langle c\rangle$ and $\langle c'\rangle$ are infinite and discrete (hence closed). Proposition. For every set $Y$ and homomorphism $f:S_X\to S_Y$, the subgroups $\langle f(c)\rangle$ and $\langle f(c')\rangle$ are not conjugate. More precisely, $f(c)$ has no finite cycle of size $\ge 2$, while (unless $f$ is the trivial homomorphism) $f(c')$ has finite cycles of unbounded size. (This immediately implies the first proposition, in a strong sense since the embedding property fails for a given pair of isomorphic closed subgroups.) Lemma Let $H$ be an open subgroup of $S_X$. Then there exists a finite subset $F$ of $X$ such that $H$ is trapped between the pointwise and global stabilizer of $F$. $\Box$ [This is classical. The main case is when $H$ is transitive, in which case the conclusion is $H=S_X$. The general case easily follows.] Through $f$, we can view $Y$ as an $S_X$-set. For both statements it enough to assume that $Y$ is a transitive $S_X$-set (through $f$). Hence $Y=S_X/H$ for some open subgroup $H$. For the statement about $c$ (no nonsingleton finite cycle), we can use the lemma and find an open finite index subgroup $H'$ of $H$, such that $H'$ is the pointwise stabilizer of a finite subset. We can identify $S_X/H'$ with the set of pairwise distinct $n$-tuples in $X^n$ for the product action, for some $n\ge 0$. Indeed we see that $c$ has no nonsingleton finite cycle on $\omega^n$. Since the equivariant surjection $S_X/H'\to S_X/H$ is finite-to-one, we deduce that $c$ has also no nonsingleton finite cycle on $S_X/H$. For the statement about $c'$ (existence of unbounded finite cycles), we can use the lemma again, now defining $H'$ as finite index overgroup of $H$, such that $H'$ is the global stabilizer of a finite subset $F$. If $F$ is empty, we have $H=H'=S_X$ and the statement is void. Assume now $F$ nonempty. We can identify $S_X/H'$ with the set of $n$-elements subsets of $X$ for some $n\ge 1$. Since $c'$ admits a $kn$-cycle for every $k\ge 2$, we see that $c'$ has a $k$-cycle for its action on $S_X/H'$. Hence $c'$ has a $kk'$-cycle for some $k'\ge 1$ on $S_X/H$. Using automatic continuity, let me mention, out of curiosity: Corollary. For $X=Y=\omega$ and $c,c'$ as above, the HNN extension of $S_\omega$ over the isomorphism $\langle c\rangle \to \langle c'\rangle$ mapping $c$ to $c'$, has no faithful action on any countable set. (Indeed, finding groups of at most continuum cardinal with no faithful action on any countable set has been an open problem long ago, and while various examples are now known, such a construction seems to be a new one. Note that no group topology is involved in the above statement.) This argument doesn't address general Polish groups as targets. However it suggests that it already makes sense to attempt to make discrete subgroups conjugate, even infinite cyclic ones.<|endoftext|> TITLE: Unnecessary uses of the Continuum Hypothesis QUESTION [19 upvotes]: This question was inspired by the MathOverflow question "Unnecessary uses of the axiom of choice". I want to know of statements in ZFC that can be proven by assuming the Continuum Hypothesis, but can also be proven by a more elaborate (perhaps even significantly more elaborate) proof without assuming the Continuum Hypothesis. REPLY [7 votes]: A Dowker space is a normal Hausdorff space whose product with the closed unit interval $I$ is not normal. In 1971, Mary Ellen Rudin constructed the first ZFC Dowker space, which had cardinality $\aleph_\omega^\omega$. This space was considered "large" and for a long time it was an open problem to construct a "small" Dowker space. Various people constructed small Dowker spaces by assuming extra hypotheses. Most relevant to the current question is the construction, by I. Juhász, K. Kunen and M. E. Rudin (Two more hereditarily separable non-Lindelöf spaces), of a small Dowker space assuming CH. Finally, in 1996, Zoltán Balogh constructed A small Dowker space in ZFC. There might be other examples stemming from the work of M. E. Rudin. Initially, I thought that the normality of ${\mathbb R}^\omega$ in the box topology might be an answer to this question, but apparently it is still open whether it can be proved without CH.<|endoftext|> TITLE: Two identities involving Ext functors in the context of D-modules QUESTION [5 upvotes]: I have several questions regarding proposition 2.3 in "Cherednik and Hecke algebras of varieties with a finite group action", by Pavel Etingof. Let $X$ be a complex affine algebraic variety and $g$ an automorphism of finite order. We define the $D(X)$-bimodule $D(X)g$ by $a(bg)c=abg(c)g$. In the aforementioned proposition, it is stated that the standard equivalence of left and right $D$-modules on the second factor induces two left $D$-modules $\tilde{D}(X)$ and $\tilde{D}(X)g$ on $X\times X$ coming from $D(X)$ and $D(X)g$ respectively. I would like to understand how this equivalence can be used to induce $D$-modules on $X\times X$. Let $i:X\rightarrow X\times X$ be the morphism given by $x \mapsto (x, g^{-1})$ and $\Delta$ be the diagonal. The proposition states that $\tilde{D}(X)=\Delta_{*}O_{X}$ and $\tilde{D}(X)g=i_{*}O_{X}$. Then, the following chain of equalities is stated: $$Ext_{D(X)\otimes D(X)^{op}}^{m}(D(X),D(X)g)=Ext_{D(X\times X)}^{m}(\tilde{D}(X),\tilde{D}(X)g)= Ext_{D(X\times X)}^{m}(\Delta_{*}O_{X},i_{*}O_{X})=Ext_{D(X)}^{m}(O_{X},\Delta^{!}i_{*}O_{X})$$ I would like to understand the first and third equalities. REPLY [2 votes]: I asked Prof. Etingof and he answered the following: A $D$-module on $X\times X$ is a module over $D(X)\otimes D(X)$, while initially, these are bimodules over $D(X)$, which are modules over $D(X)\otimes D(X)^{op}$. The equivalence I mentioned is a Morita equivalence between $D(X)$ and $D(X)^{op}$ which defines a Morita equivalence between $D(X)\otimes D(X)$ and $D(X)\otimes D(X)^{op}$. The first equality follows from my answer to (1). The third equality follows because the functors $\Delta_{ * }$ and $\Delta^{ * }$ are adjoint.<|endoftext|> TITLE: The origin of the natural base in statistical mechanics QUESTION [10 upvotes]: In modern treatments of statistical mechanics, the natural base is conventionally used for the Gibbs and Boltzmann entropy without careful justification. While I am aware that the properties of the Shannon entropy are invariant to the choice of base of the logarithm, I suspect that physicists might have careful theoretical justifications for this particular choice. Question: Might there be a sound theory behind this convention? So far a couple reasons occurred to me: Stirling's log-factorial approximation: \begin{equation} \ln N! \sim N \cdot \ln N - N \tag{1} \end{equation} which is an essential tool in statistical mechanics. The exponential function diagonalises the derivative operator: \begin{equation} \forall \lambda \in (0,1), \frac{\partial}{\partial H} \exp(\lambda \cdot H) = \lambda \cdot \exp(\lambda \cdot H) \tag{2} \end{equation} which may be useful whenever one wants to analyse variations in the exponential of entropy. The advantage of the exponential of entropy is that it is parameterisation invariant as pointed out by Tom Leinster on a related question. As there might be subtle reasons I have ignored, any useful references on this question are more than welcome. References: von Neumann, John (1932). Mathematische Grundlagen der Quantenmechanik (Mathematical foundations of quantum mechanics) Princeton University Press., . ISBN 978-0-691-02893-4. Landau and Lifshitz. Statistical Physics. Butterworth-Heinemann. 1980. David J.C. MacKay. Information Theory, Inference and Learning Algorithms. Cambridge University Press 2003. REPLY [4 votes]: Something Gian-Carlo Rota made a big deal about emphasizing was that there is a real difference between the entropy of discrete and continuum probability distributions. For a discrete distribution, the Shannon entropy with $\sum p_{n}\log_{2}p_{n}$ is clearly dimensionless. However, the Boltzmann entropy, with $\int dx\, f(x)\ln f(x)$, is a horrible mess in terms of units. Fortunately, in quantum mechanics, Planck's constant $\hbar$ swoops in to fix the unit problem. (If you've never seen it before, it can seem like magic how such a fundamental-looking problem just evaporates.) So for real gasses (and other statistical physics systems), the problem goes away, but Rota emphasized that if we want to use entropy to describe other kinds of continuous distributions, the problem may still be there. However, the quantum mechanical entropy only really cleans itself nicely if the natural logarithm is used. You can define an entropy using any basis $a$ for the logarithm of the available phase space $S=k\log_{a}\Omega$; that just means changing the definition of the Boltzmann constant $k$. For a lot of expressions, changing $k$ is all that's needed, but that logarithm defining $S$ is not actually the only place where the constant $e$ enters in the statistical mechanics of ideal gasses. It also shows up in the Sacker-Tetrode equation, $$S=Nk\ln\left[\frac{Ve^{5/2}}{N\hbar^{3}}\left(\frac{mkT}{2\pi}\right)^{3/2}\right],$$ and there is no way to make the $e$ in the argument go away entirely. Moreover, there is no way to get that weird factor of $e^{5/2}(2\pi)^{-3/2}\hbar^{-3}$ without the quantum mechanical treatment I alluded to above. (This is in contrast to the dependences on $N$, $k$, $V$, $T$, and $m$, which can be determined from purely classical statistical mechanics.) Now, maybe one can say that $e$ is just showing up in that argument because, like the $\pi$ that is also present, $e$ is simply a ubiquitous constant that pops up all sorts of places that don't seem appropriate. However, it is clear that this is one instance where the natural logarithm gives an expression that is manifestly nicer than what you would get using a different base.<|endoftext|> TITLE: When is a symmetric tensor equal to gradient times gradient? QUESTION [5 upvotes]: On a ball in $\mathbb R^n$, a vector field $v_j$ is a gradient of a function when its exterior derivative vanishes. In other words, if $$\partial_i v_j-\partial_j v_i=0$$ then there exists a function $u$ such that $v_j=\partial_j u$. I am looking for something similar for a symmetric tensor. In particular: do you know under which conditions a symmetric tensor $A_{ij}$ can be written as $$A_{ij}=\partial_i u \partial_j u$$ where $u$ is any (smooth) function? Written in a different way, under which conditions is $A=\nabla u \otimes \nabla u$? Thanks. REPLY [4 votes]: Your question is answered in Props. 1 and 2 of Krongos, D. S.; Torre, C. G., Geometrization conditions for perfect fluids, scalar fields, and electromagnetic fields, J. Math. Phys. 56, No. 7, 072503, 17 p. (2015). ZBL1322.83005. arXiv:1503.06311. The precise conditions are \begin{gather*} A_{i[j} A_{k]l} = 0 , \\ A_{ij} A_{k[l,m]} + A_{ik} A_{j[l,m]} + A_{jk,[l} A_{m]i} = 0 , \end{gather*} where $(-)_{,k} = \partial_k (-)$ and $[ij]$ denotes antisymmetrization in the indices $i$ and $j$. The first condition implies that $A_{ij} = \pm v_i v_j$ for some vector $v_i$. Plugging that result into the left-hand side of the second condition gives $2 v_i v_j v_k v_{[l,m]}$, which obviously vanishes iff $v_{[l,m]} = 0$. There may be a more classical source for this result or other results of this kind, but I'm not aware of it.<|endoftext|> TITLE: Undecidability of irreducibility of infinite families of integer polynomials? QUESTION [11 upvotes]: A recent question, Is irreducibility of polynomials $\in\mathbb{Z}[X]$ over $\mathbb{Q}$ an undecidable problem? was quickly answered in the negative. I am wondering if there is a simple example of a family of families of integer polynomials whose irreducibility is undecidable. For example, consider the following computational problem: Instance: A positive integer $n$. Question: Does the family of polynomials $\{x^d + x + n : d \in \mathbb{N}\}$ contain infinitely many members that are irreducible over $\mathbb{Q}$? I don't know off the top of my head whether the above computational problem is undecidable. If it is, then that would answer my question affirmatively. If not, or if its undecidability is unknown, then is there some other problem of comparable simplicity that we can prove is undecidable? EDIT: Upon further reflection, I suspect that the most promising route for getting an interesting answer to this question is to define some kind of "dynamical system" that generates a sequence of polynomials, and ask if (for example) the process eventually produces an irreducible polynomial. Interesting prior results with a dynamical-systems flavor include The undecidability of the generalized Collatz problem by Kurtz and Simon, and Turing-completeness of various families of PDEs as shown by Tao and others. Such results seem to say something about the complexity of the systems in question, in a way that "artificially" encoding an uncomputable set directly in the parameters of a problem (intuitively) does not. Unfortunately, I do not have a concrete proposal for how to define a suitable dynamical system. REPLY [10 votes]: I don't know about the family in question but there are families of families for which this is undecidable. Given a total computable function $f:\mathbb{N}\times\mathbb{N}\to\{0,1\}$ consider the family of families $\mathcal{F}_n = (x^2+(-1)^{f(n,k)})_{k=0}^\infty$. Then asking whether $\mathcal{F}_n$ has infinitely many irreducible elements over $\mathbb{Q}$ is equivalent to asking whether the set $F_n = \{k \in \mathbb{N}\mid f(n,k)=1\}$ is infinite. So every $\Pi^0_2$ set can be encoded as a problem of this form. This is optimal since it is decidable whether a given polynomial is irreducible over $\mathbb{Q}$ according to the question mentioned by the OP.<|endoftext|> TITLE: Structure theorems for compact sets of rationals QUESTION [8 upvotes]: Everyone knows the Heine-Borel theorem characterizing compact subsets of Euclidean space. For any $n \in \mathbb N$ a set $A \subseteq \mathbb R^n$ is compact just in case it is closed and bounded (in the sense that it is contained in some large enough finite diameter ball). There is a similar theorem for the irrationals, due to Rothberger. The irrationals are homeomorphic to the space $\mathbb N^\mathbb N$ of functions $f:\mathbb N \to \mathbb N$ topologized by the product topology of the discrete topology on $\mathbb N$. With this presentation a set $A \subseteq\mathbb N^\mathbb N$ is compact just in case it is closed and bounded where bounded means there is an $f:\mathbb N \to \mathbb N$ so that for all $g \in A$ we have $\forall n \, g(n) \leq f(n)$. My question is whether there is a similar (or at least relatively neat) characterization of compact sets of rationals (with the subspace topology). Does some relatively simple (topological, geometric, combinatorial) property characterize when a closed $A \subseteq \mathbb Q$ is compact? REPLY [5 votes]: If we consider the ordering on $\mathbb{Q}$, then we obtain characterizations of the compact subsets of $\mathbb{Q}$ which give more information about these compact subsets than the homeomorphism type in Ville Salo's answer. A linearly ordered set $X$ is said to be scattered if $X$ does not contain any order isomorphic copy of the rational numbers. Let $\mathcal{A}_{\alpha}$ be the class of all totally ordered sets of the form $\bigcup_{a\in A}X_{a}$ where $A$ is either well-ordered or dual-well ordered, if $ax\}\notin\mathcal{A}_{\infty}$. Therefore, by recursion, one can construct a set $X_{\gamma}$ along with an element $x_{\gamma}$ for each binary string $\gamma$ where we set $X_{\epsilon}=X$ and where for each binary string $X_{\gamma}$, there is an element $x_{\gamma}$ where if $X_{\gamma 0}=\{y\in X_{\gamma}\mid yx\}$, then $X_{\gamma 0}\notin\mathcal{A}_{\infty}$ and $X_{\gamma 1}\notin\mathcal{A}_{\infty}$. In this case, the set $\{x_{\gamma}\mid \gamma\}$ is order isomorphic to $\mathbb{Q}$. Q.E.D. Proposition: A linearly ordered set $X$ is compact in the order topology if and only if $X$ is complete as a lattice. Suppose that $A$ is a subset of a linearly ordered set $X$. Then an element $a\in A$ is said to be a gap point if there exists some $x\in X$ where either $aa\}$ but where the set $\{b\in A\mid b>a\}$ has no minimal element or $\{b\in A\mid b TITLE: Is ku reflexive as a spectrum? QUESTION [5 upvotes]: Let $S$ and $\rm ku$ denote the sphere spectrum and the connective K-theory spectrum, respectively. Then is the morphism ${\rm ku} \to F(F({\rm ku},S),S)$ an equivalence? Here $F$ denotes the mapping spectrum. REPLY [8 votes]: Writing $D(-) = F(-, S)$, I believe that there is an equivalence $$D ku \simeq \prod_{n=0}^\infty \Sigma^{-2n-1} H(\widehat{\mathbb Z} / \mathbb Z),$$ where $\widehat{\mathbb Z}$ is the profinite completion of the integers. This would split as a similar product after dualizing a second time, so can't be equivalent to $ku$. The claimed equivalence is a combination of the following facts: $D H \mathbb Q \simeq \Sigma^{-1} H \widehat{\mathbb Z} / \mathbb Z$: Writing $H \mathbb Q = \operatorname{colim}( \cdots \to S \xrightarrow n S \to \cdots)$, we find $D H \mathbb Q = \lim( \cdots \leftarrow S \xleftarrow n S \leftarrow \cdots)$, and the conclusion follows from the Milnor sequence and the finite torsion exponents of the higher homotopy groups of spheres. $D H A \simeq 0$ for $A$ torsion abelian: Since $S$ is harmonic, $D H \mathbb F_p \simeq 0$, and $A$ torsion abelian has a filtration whose quotients are prime torsion. $D H \mathbb Q \simeq D H \mathbb Z$: There's a fiber sequence $$D H \mathbb Z \leftarrow D H \mathbb Q \leftarrow D H(\mathbb Q / \mathbb Z),$$ and the last spectrum is null by the above. For $X$ coconnective, $D X \simeq D (\mathbb Q \otimes X)$: $X$ can be written as the colimit of a filtration whose quotients are Eilenberg–Mac Lanes, to which one applies the previous facts. $L_{H \mathbb Z} KU \simeq \mathbb Q \otimes KU$: Using Snaith's theorem $KU \simeq \Sigma^\infty_+ \mathbb CP^\infty[\beta^{-1}]$, one calculates the integral homology of $KU$ to be rational. $D KU \simeq \prod_{n=-\infty}^\infty \Sigma^{-2n-1} H(\widehat{\mathbb Z} / \mathbb Z)$: Since $S$ is $H\mathbb Z$–local, $DKU \simeq D(L_{H\mathbb Z} KU) \simeq D(\mathbb Q \otimes KU)$. Rational spectra are Eilenberg–Mac Lane, hence determined by their homotopy groups, so one combines Bott periodicity and the first fact. The main claim then follows from the fiber sequence $$D ku \leftarrow D KU \leftarrow D(\tau_{< 0} KU).$$ We've analyzed the middle term, and the last term starts off coconnective so (after applying the above facts) turns out to be the connective part of $D KU$, which finally determines the first term to be the remaining coconnective part.<|endoftext|> TITLE: Equivalent definitions of Kodaira dimension QUESTION [5 upvotes]: The Kodaira($-$Iitaka) dimension of a line bundle $L$ on a complex manifold $X$ can be defined either in three ways: The maximal dimension of the image of the rational maps $φ_{|mL|} : X \dashrightarrow \mathbb{P}(H^0(X, m L)^*)$ The unique integer $k$ such that $h^0(m L) = O(m^k)$. $\operatorname{trdeg} \operatorname{Frac} \left( \bigoplus_{m ≥ 0} H^0(X, mL) \right) - 1$ I know why the first two definitions are equivalent but I'm struggling to find a reference treating the third. Several people cite Ueno's book Classification Theory of Algebraic Varieties and Compact Complex Spaces for this, but as far as I can see he only proves that first two are equivalent. Can anyone provide a proof of $(1) \iff (3)$ or $(2) \iff (3)$ or point me to a reference? REPLY [10 votes]: For $(1) \iff (3)$, we have the following chain of identities: $\operatorname{trdeg} \operatorname{Frac} \left( \bigoplus_{m ≥ 0} H^0(X, mL) \right)$ is equal to $\max \operatorname{trdeg}(F)$ where $F$ is a finitely generated subfield of $ \operatorname{Frac} \left( \bigoplus_{m ≥ 0} H^0(X, mL) \right)$, which is equal to $\max \operatorname{trdeg}( \operatorname{Frac}(R))$ where $R$ is a finitely generated subring of $\bigoplus_{m ≥ 0} H^0(X, mL) $ (take $R$ generated by the numerators and denominators of $F$ ), which is equal to $\max \operatorname{trdeg}( \operatorname{Frac}(R))$, where $R$ is the subring generated by $\bigoplus_{m \in S} H^0(X, mL) $ for $S$ a finite set of natural numbers (take the supports of the generators), which is equal to $\max \operatorname{trdeg}( \operatorname{Frac}(R))$, where $R$ is the subring generated by $ H^0(X, nL) $ (take the least common multiple of $S$, and note the old generators are finite over this ring), which is equal to the max of the dimension of the spectrum of $R$, where $R$ is the subring generated by $ H^0(X, nL) $, which is equal to the max of the dimension of the affine cone on the closure of the image of the rational map $φ_{|nL|} : X \dashrightarrow \mathbb{P}(H^0(X, n L)^*)$ (since the spectrum is equal to that cone) the max of the dimension of the image of the rational map $φ_{|nL|} : X \dashrightarrow \mathbb{P}(H^0(X, n L)^*)$, plus 1 (since the affine cone has dimension one higher).<|endoftext|> TITLE: Dual norm of a subspace of $\ell_\infty^3$ QUESTION [5 upvotes]: We define a norm on $\mathbb C^2$ as $\|(\alpha,\beta)\|:=\max\left\{|\alpha|,|\beta|,\big|\frac{\alpha+\beta}{\sqrt{2}}\big|\right\}.$ Can the dual norm be calculated explicitly? REPLY [4 votes]: $\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$This is to detail and correct the answer by Onur Oktay, which is based on a nice idea, leading to simplified and speedier calculations of the dual norm. (That Onur Oktay's answer contains at least two mistakes was pointed out in comments by Nathaniel Johnston and me.) As in Onur Oktay's answer, let $X:=\ell^3_\infty$ and $V:=\{av_1+bv_2:a,b\in\C\}\subset X$, where \begin{equation*} v_1:=(1,0,1/\sqrt2),\quad v_2:=(0,1,1/\sqrt2),\quad v_3:=(1,1,-\sqrt2). \end{equation*} Note that $v_3\cdot v=0$ for all $v\in V$, where $\cdot$ denotes the dot product. Consider \begin{equation*} V^\perp:=\{x^*\in X^*\colon x^*(v)=0\ \forall v\in V\}, \end{equation*} the so-called annihilator of $V$. Note that, if, as usual, $X^*=(\ell^3_\infty)^*$ is identified with $\ell^3_1$, then $V^\perp$ will be identified with the span $\C v_3=\{tv_3\colon t\in\C\}$ of $\{v_3\}$. It is then said in Onur Oktay's answer that $V^*=X^*/V^\perp$, which is isometrically isomorphic to $\C^2$ equipped with the norm \begin{equation*} \|(a,b)\|=\inf\{\|av_1+bv_2-tu\|_{\ell^3_1}: t\in\C\}, \end{equation*} where $u$ is our $v_3$. This statement is incorrect. First here is the minor point that $V^*$ is, not equal, but isometrically isomorphic to $X^*/V^\perp$ (Theorem 4.9), with the isometric isomorphism $\si$ given by the formulas \begin{equation*} V^*\ni v^*\mapsto\si(v^*):=x^*+V^\perp \end{equation*} and \begin{equation*} \|x^*+V^\perp\|:=\inf\{\|x^*+y^*\|\colon y^*\in V^\perp\}, \tag{0}\label{0} \end{equation*} where $x^*\in X^*$ is any extension of the continuous linear functional $v^*$. More importantly, the expression for $\|(a,b)\|$ in Onur Oktay's answer is incorrect. Indeed, consider the isometric isomorphism \begin{equation*} \C^2\ni(\al,\be)\mapsto\iota((\al,\be)):=\al v_1+\be v_2\in V\subset X=\ell^3_\infty \end{equation*} from $\C^2$ onto $V$, with the norm on $\C^2$ induced by $\iota$: \begin{equation*} \|(\al,\be)\|=\|\al v_1+\be v_2\|_\infty=\max(|\al|,|\be|,|\al+\be|/\sqrt2). \tag{1}\label{1} \end{equation*} For any $a$ and $b$ in $\C$, define linear functionals $l_{a,b}\in(\C^2)^*$ and $L_{a,b}\in V^*$ by the conditions \begin{equation*} l_{a,b}((1,0))=a=L_{a,b}(v_1)\quad\text{and}\quad l_{a,b}((0,1))=b=L_{a,b}(v_2); \end{equation*} here, of course, the norm $\|\cdot\|_{(\C^2)^*}$ on $(\C^2)^*$ is dual to the norm on $\C^2$ given by \eqref{1}. Also for $a$ and $b$ in $\C$, define the vector $x_{a,b}\in\R^3$ by the conditions \begin{equation*} x_{a,b}\cdot v_1=a,\quad x_{a,b}\cdot v_2=b,\quad x_{a,b}\cdot v_3=0, \end{equation*} so that \begin{equation*} x_{a,b}=\tfrac14\,(3 a-b,3 b-a,\sqrt{2}(a+b)). \end{equation*} So, the vector $x_{a,b}\in\R^3$ is the canonical representation of the extension -- say $\tilde L_{a,b}$ -- of the linear functional $L_{a,b}\in V^*$ such that $\tilde L_{a,b}\in X^*=(\ell^3_\infty)^*\simeq\ell^3_1$ and $\tilde L_{a,b}(v_3)=0$. The canonical representation of $\tilde L_{a,b}$ by $x_{a,b}$ here is of course the one induced by the canonical isometric isomorphism $(\ell^3_\infty)^*\simeq\ell^3_1$, so that $\tilde L_{a,b}(x)=x\cdot x_{a,b}$ for all $x\in\ell^3_\infty$. Then $l_{a,b}=L_{a,b}\circ\iota$ and, in view of \eqref{0}, \begin{equation*} \|l_{a,b}\|_{(\C^2)^*}=\|L_{a,b}\|_{V^*} =\inf_{t\in\C}\|x_{a,b}+tv_3/4\|_1. \end{equation*} Thus, \begin{equation*} \|l_{a,b}\|_{(\C^2)^*} =\tfrac14\,\min_{t\in\C}(|3 a - b + t|+|3 b-a + t|+\sqrt2\, |a + b - t|). \tag{2}\label{2} \end{equation*} Finding the minimum in \eqref{2} is a problem of real algebraic geometry. So, in principle, the dual norm can be computed purely algorithmically; however, this calculation can take too much time -- as it actually is the case with the minimum in \eqref{2}. However, finding the minimum in \eqref{2} numerically for any particular $(a,b)\in\C^2$ is not a problem. The numerical results obtained according to \eqref{2} completely agree with the numerical results obtained according to my other answer on this page, but the execution time using \eqref{2} is about 6 times as small. As Matthew Daws noted in a comment, one could use the vector $y_{a,b}:=(a,b,0)$ in place of $x_{a,b}=\tfrac14\,(3 a-b,3 b-a,\sqrt{2}(a+b))$ -- in the sense that the linear functional $X\ni x\mapsto x\cdot y_{a,b}$ is, just as the linear functional $\tilde L_{a,b}$, an extension of the linear functional $L_{a,b}$. This results in an expression for $\|l_{a,b}\|_{(\C^2)^*}$ looking slightly simpler than the one in \eqref{2}: \begin{equation*} \|l_{a,b}\|_{(\C^2)^*} =\inf_{s\in\C}\|y_{a,b}+sv_3\|_1 =\min_{s\in\C}(|a + s|+|b + s|+\sqrt2\, |s|), \tag{2a}\label{2a} \end{equation*} which seems to match the final expression in Nathaniel Johnston's answer. In fact, the ultimate expression in \eqref{2a} can be obtained from that in \eqref{2} by substitution $t=a+b+4s$.<|endoftext|> TITLE: Product of complex numbers on the unit circle with largest real part QUESTION [7 upvotes]: Let $T = \{z_1, \ldots z_n\}$ be a finite set of complex numbers on the unit circle. I would like an algorithm which can quickly compute the nonempty subset $S \subset T$ which maximizes $$\left| \operatorname{Re}\left(\prod_{z \in S} z\right) \right|.$$ I suspect this problem may be NP-Hard, but I'm not sure. I'm fine with an algorithm which computes an approximately maximal product. REPLY [5 votes]: If you write $z = \exp(2\pi i\theta_z)$ for arguments $\theta_z\in [0,1)$, then as $$\Re(\prod_{z\in S}z) = \Re(\exp(2\pi \sum_{z\in S}\theta_z)) = \cos(2\pi\sum_{z\in S}\theta_z).$$ the problem is equivalent to finding $S$ such that $\sum_{z\in S}\theta_z\bmod 1$ is closest to either of $0$ (when the absolute value is positive) or $1/2$ (when it is negative). If one assumes that all $\theta_z$ are rational, and moreover that for $N\in\mathbb{N}$ they are all of the form $\theta_z = a_z / n$ for $a_z\in[0,1,\dots,n-1]$, one can rewrite this as finding $S$ such that $\sum_{z\in S}a_z \bmod n$ is closest to $0$ or $n/2$. This seems to be a modular version of the 0-1 knapsack (optimization) problem. This is a simultaneous generalization of the standard 0-1 knapsack problem in 2 ways there is a modular constraint --- the objective function is $\sum_{z\in S}a_z \bmod n$ rather than $\sum_{z\in S}a_z$, and it is an optimization problem rather than a decision problem. If you only had one of these generalizations, casual searching would lead to known results. For example, if you knew that $\sum_{z\in S}a_z\bmod n = 0$ (or $n/2$) exactly, then there is a standard reduction to lattice problems, which is even efficient if the instance is "sparse" in a certain sense. I expect one can similarly attack this problem by reducing it to SVP/CVP on a suitable lattice. As a brief sketch, if one defines the matrix $$B =\begin{pmatrix} N &0&0 &\dots & 0\\ a_{z_1} & 1 & 0 & &0\\ a_{z_2} & 0 & 1 & \dots &0\\ \vdots &&&\ddots&\vdots\\ a_{z_k} & 0 & 0 & \dots & 1 \end{pmatrix},$$ Then one can check that any (integer) linear combination of the rows of the above takes the form $$ \begin{pmatrix} Nx_0 + a_{z_1}x_1 + \dots + a_{z_k}x_k\\ x_1\\ \vdots\\ x_k \end{pmatrix} $$ where $\vec x = (x_0,\dots, x_k)\in\mathbb{Z}^{k+1}$. Clearly, the norm of this is minimized for $\vec x = 0$. One can argue that the shortest non-zero vector will lead to $Nx_0 + \sum_i a_{z_i}x_i$ being small. There is still of course work to do (for example, the shortest vector in the aforementioned lattice may lead to a "solution" to a knapsack problem without 0/1 weights. This could plausibly be fixed by using a lattice basis with $M$ on the diagonal instead of $1$, to penalize choosing $x_1,\dots, x_k > 1$). Still, I expect a solution to your problem will proceed in a route similar to this, as this is fairly typical for "modular knapsack" problems.<|endoftext|> TITLE: Multiplicativity of the Euler characteristic for fibrations QUESTION [16 upvotes]: For a Serre fibration $$ F\to E \to B , $$ with $F,E,B$ having the homotopy type of finite complexes, it is known that the Euler characteristic is multiplicative: $$ \chi(E) = \chi(F)\chi(B) . $$ However, if we more generally assume that $B$ and $F$ are finitely dominated spaces, then does multiplicativity hold as well? (Recall that a finitely dominated space is a retract of a homotopy finite one.) If true I am looking for a reference. If false, please explain. Added later: it's true if the base is homotopy finite since we can take the fiberwise double suspension to obtain a fibration with homotopy finite fibers having the same Euler characteristic. So we only need to consider the case when the base is a finitely dominated and the fibers are homotopy finite. Second Addition: I can solve the problem in general if I can solve it in the following case: Let $\tilde B \to B$ be a finite, regular, $n$-sheeted covering space, where $B$ is finitely dominated. Then $\chi(\tilde B) = n\chi(B)$. (Note that $\tilde B$ is again finitely dominated, since there is a finite covering $\tilde B\times S^1 \to B\times S^1$ and by a theorem of Mather, a space $X$ is finitely dominated if and only if $X\times S^1$ is homotopy finite. But since the base $B\times S^1$ is homotopy finite, we can put a finiteness structure on the total space as well.) Third Addition: The case alluded to in my "Second Addition" holds, by the second answer I gave below. REPLY [5 votes]: I have already written a lot, but Oscar Randal-Williams's comment below John Klein's second answer seems to really simplify the matter, so I decided to add an answer to that effect. I can remove any of the answers I have written if somehow they're "obfuscating the view" or anything. This argument is just a simplification, that bypasses Pedersen-Taylor's result, and also the need for Swan's other theorem (suggested by Oscar) that $\tilde K_0(\mathbb Z[\pi])$ is finite for a finite group $\pi$ (I think the proof is harder, although I might be mistaken here). The argument is therefore due to Oscar and John in this respect, this is just my write-up of it. Namely, as in all other answers reduce to the case of a finite covering space $p: E\to B$. Letting $F$ denote the fiber, and $Q$ the (finite) image of $\pi_1(B)\to Aut(F)$, let $f: B\to BQ$ denote the corresponding classifying map. Define the following functor from perfect $\mathbb Z[\Omega B]$-modules to $\mathbb Q$-modules: $P\mapsto B_!( (P\otimes_\mathbb Z \mathbb Q) \otimes_\mathbb Q C_*(F;\mathbb Q))$. The observation is that $C_*(F;\mathbb Q)$, as a $\mathbb Z[\Omega B]$-module, is of the form $f^* C_*(F;\mathbb Q)$, so this is $\mathrm{colim}_{BQ} f_!((P\otimes_\mathbb Z\mathbb Q)\otimes_\mathbb Q f^*C_*(F;\mathbb Q))\simeq \mathrm{colim}_{BQ} f_!(P\otimes_\mathbb Z\mathbb Q) \otimes C_*(F;\mathbb Q)$ (where $\mathrm{colim}_{BQ}$ is derived) In other words, this functor factors as $Perf(\mathbb Z[\Omega B])\to Perf(\mathbb Z[Q])\to Perf(\mathbb Q[Q])\to Perf(\mathbb Q)$, but the most important part is the first part. Indeed, $Q$ is finite, and so on $K$-theory, $K_0(\mathbb ZQ)\to K_0(\mathbb QQ)$ lands in the image of $K_0(\mathbb Q)$ (under the "induction" morphism) : this is (as far as I know) a theorem of Swan's (specifically, theorem 4.2 from his book "K-theory of finite groups and orders") Now if $P= \mathbb Q[Q]^n$ is a free $\mathbb Q[Q]$-module, $\mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F;\mathbb Q)) = \mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F^{triv}; \mathbb Q)) = P_{hQ}\otimes_\mathbb Q C_*(F; \mathbb Q)$. In particular, on $K$-theory, this composite has the same effect as $\mathrm{colim}_B (-)\otimes_\mathbb Q C_*(F;\mathbb Q)$, and so it sends the trivial coefficient system $\mathbb Z$ to $C_*(B;\mathbb Q)\otimes_\mathbb Q C_*(F;\mathbb Q)$, which of course is another name for $\chi(B)\chi(F) \in \mathbb Z\cong K_0(\mathbb Q)$. But this map also sends this trivial coefficient system to $C_*(E;\mathbb Q)$, which is also another name for $\chi(E)\in \mathbb Z\cong K_0(\mathbb Q)$. This proves the claim.<|endoftext|> TITLE: Steenrod squares in terms of chain maps QUESTION [6 upvotes]: $\DeclareMathOperator\Sq{Sq}$The Steenrod squares $\Sq^i: H^n({-};\mathbb{F}_2) \to H^{n+i}({-};\mathbb{F}_2)$ are fundamental cohomological operations. By the Yoneda lemma, they induce a map between the Eilenberg–MacLane spaces $K(\mathbb{F}_2; n) \to K(\mathbb{F}_2; n +i)$. By the Dold–Kan correspondence, this map should be expressible as a chain map (if I'm not mistaken): $$\widehat{\Sq^i}: \mathbb{F}_2[-n] \to \mathbb{F}_2[-(n+i)].$$ Question: How do we explicitly describe $\widehat{\Sq^i}$? Is it much harder to do this for other finite fields $\mathbb{F}_q$? REPLY [9 votes]: You do have a map of spaces $Sq^i : K(\mathbb F_2 , n) \to K(\mathbb F_2, n+i)$, but it is not a map of topological abelian groups unless $i=0$ or $i=1$. The Dold-Kan correspondence says that maps of chain complexes correspond to maps of topological abelian groups. So there is no map of chain complexes realizing $Sq^i$ for $i\neq 0, 1$. For $i=0$, there is -- it's the identity map. For $i=1$, $Sq^1$ is the Bockstein, and is again represented by a map of chain complexes, but it's not $\mathbb F_2$-linear, only $\mathbb Z$-linear. In fact, if $A$ is a free resolution of $\mathbb F_2$ as a chain complex, you can see that there are no $\mathbb Z$-linear maps $A \to A$ of degree $>1$ because $\mathbb Z$ has homological dimension 1. You can see that there are no $\mathbb F_2$-linear maps of degree $>0$ because $\mathbb F_2$ has homological dimension $0$. On the spectrum side, these observations correspond to the fact that $Sq^i : H \mathbb F_2 \to H\mathbb F_2 [i]$ is not an $H\mathbb F_2$-linear map for $i > 0$, and not $H\mathbb Z$-linear for $i > 1$.<|endoftext|> TITLE: Is St. Petersburg a good place for the 2022 Int. Congress of Mathematicians QUESTION [113 upvotes]: There might be just enough time to pick another location, but I am curious what mathematicians think. Will Ukrainian mathematicians be able to attend a conference in Russia if Russia no longer recognizes their passports? To be clear: I love Russia, and I am not trying to hurt the feelings of Russian mathematicians or people. The International Mathematical Union (IMU) have made a decision on moving the ICM to a virtual event, but there is still the (less exciting) decision to make concerning the location of IMU General Assembly. REPLY [50 votes]: The IMU reached their decision, the ICM will take place entirely virtually except for the IMU general assembly and awards ceremony which will take place in-person outside of Russia. No word on whether Russian government involvement in organizing will continue. Further details to come. Terry Tao, who is the chair of the ICM Structure Committee, is inviting discussion of how to best use the virtual format on his blog. The IMU executive committee has released some further clarifications, including announcing that there will be no Russian government involvement in the ICM.<|endoftext|> TITLE: Infimum of Fourier transform of singular measure QUESTION [5 upvotes]: Let $\mu$ be a finite non negative singular measure on $\mathbf{R}^d$. I would like to know if there exists some result on the infimum of the absolute value of its Fourier Transform $$\hat{\mu}(t)=\displaystyle\int \mathrm{e}^{2i\pi t\cdot x} ~\mu(dx).$$ It is known that we don't necessarily have $\hat{\mu}(t)\rightarrow0$ for singular measure. But does one have like Lebesgue measure $$\mathrm{inf}~|\hat{\mu}(t)|=0~?$$ Edit : To be more precise, $\mu$ is continuous singular, so does not have a discret part (https://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem) REPLY [4 votes]: Yes, that's true. By Wiener's lemma we have $\sum_{x\in \mathbb{R}^d} |\mu(\{x\})|^2 = \lim_{R\to \infty} \frac{1}{(2R)^d} \int_{[-R,R]^{d}} |\widehat{\mu}(t)|^2 dt$. If $\mu$ is continuous, then the left-hand side is $0$, so the infimum of $|\widehat{\mu}|$ has to be zero as well, since the right-hand side is bounded below by this infimum (squared).<|endoftext|> TITLE: Understanding Balmer spectra QUESTION [7 upvotes]: $\DeclareMathOperator\Spec{Spec}\newcommand{\perf}{\mathrm{perf}}\DeclareMathOperator\SHC{SHC}$I have just finished reading the paper "The spectrum of prime ideals in tensor triangulated categories" in which Balmer proposes his notion of spectrum which nowadays is considered central in the understanding and classification of the homotopy categories which we want to study in the concrete mathematical practice (to name a few examples: the $G$-equivariant stable homotopy category for $G$ a compact Lie group, or the derived category of quasi-coherent sheaves on a scheme). Since I am not familiar with this notion I wanted to ask here various questions about the underlying ideas of such concept. (1) I noticed that all the examples proposed by Balmer in his paper deal with compact objects, in the sense that the proposed tensor triangulated categories (t.t. categories from now on) can be identified with the full-subcategories of compact objects in a larger t.t. category. And from what I remember every other example which I read in different sources does the same thing: we study the Balmer spectrum of compact objects in a larger t.t. category. Balmer does not explicitly state that this must be the case, indeed his definition does not require the involved objects to be compact a priori. For this abstract machinery to work we only need the t.t. category to be essentially small. I could think that this is the problem: in general we cannot guarantee that the t.t. category we are interested in is essentially small so we restrict to the subcategory of its compact objects for this property to be more likely. But I have other reasons to believe that this justification is not completely correct: if we indulge in the intuition suggested by the choice of words, we should think of the support of an object in our t.t. category as an higher categorical analogue of the usual support of a function. Fixing the domain of our functions to be compact spaces ensures that the support will also be compact. So if we consider also non-compact objects the support could be non "topologically small". Thus I am inclined to believe for the complete t.t. categories either the Balmer spectrum is too big to be computed or its is not the correct notion we want to use to classify their tensor subcategories. (2) Related to the previous question: if the proposed notion of Balmer spectrum should be applied only to categories of compact objects, what can we deduce about the whole category of possibly non-compact objects? Suppose we consider an essentially small t.t. category $\mathcal{T}$ and we manage to compute the Balmer spectrum of $\mathcal{T}^c$, can we deduce any information regarding the thick tensor ideals or localizing tensor ideals of $\mathcal{T}$? Two classical examples of this are $D(R)$, the derived category of a commutative ring $R$, and $\SHC$, the stable homotopy category. For $D^{\perf}(R)$ this is homeomorphic to the usual Zariski spectrum $\Spec(R)$, while for $\SHC^\mathrm{c}$ we have the classification provided by the thick subcategory theorem from chromatic homotopy theory. But I have never seen a classification (even partial) of their thick tensor subcategories or thick localizing subcategories. (3) What information does the Balmer spectrum encode? Balmer proves that there is a bijection between the Thomason subsets of this spectrum and the radical thick tensor ideals of the t.t. category. But other than this? At first I expected that if two t.t. categories had isomorphic spectrum then they would have a sufficiently compatible t.t. structure. Then I found the following interesting example: we have that the Balmer spectrum of the category of compact rational $S^1$-equivariant spectra is homeomorphic to $\Spec(\mathbb{Z})$. If $H \leq S^1$ is a closed subgroup then the kernel of $\phi^H$, the non-equivariant geometric $H$-fixed points, provides a Balmer prime. Then $\ker \phi^{S^1}$ corresponds to the generic point $(0)$, while $\ker \phi^{C_n}$ can be mapped to $(p_n)$ where we order the prime numbers $\{p_n : n \geq 1 \}$. Therefore $S^1\text{-}\SHC^\mathrm{c}_{\mathbb{Q}}$ and $D^{\perf}(\mathbb{Z})$ have the same Balmer spectrum, but they are very different t.t. categories: for one, the latter has a compact generator given by the tensor unit, while this is not the case in the former category. I would have thought that the t.t. structure would have been more rigid with respect to the Balmer spectrum, but this seems not to be the case. If you wanted a more precise question: if two t.t. categories have homeomorphic Balmer spectra, can we translate this to any information on the two categories? What if the homeomorphism is induced by a monoidal exact functor? Can we deduce it is fully faithful, essentially surjective or any other property? I hope that my questions are not too vague or naïve. REPLY [2 votes]: To help with (3), let me point out that in 'nice' situations (e.g., in the derived category of a noetherian commutative ring), the Balmer spectrum classifies all localizing tensor ideals of the category, in terms of arbitrary subsets of the spectrum (so the topology plays no role). This uses a theory of support developed by Balmer--Favi and Stevenson. Along with Beren Sanders and Tobias Barthel, we investigated when this occurs in some detail in a recent preprint.<|endoftext|> TITLE: How far away can we get by multiple rounding and unit change? QUESTION [9 upvotes]: This question is inspired by xkcd #2585 (Rounding): Let $u_0,\ldots,u_n$ be positive real numbers (we can assume w.l.o.g. that $u_0=1$) or “units”. Consider the following directed graph: its vertices are pairs $(i,k)$ where $0\leq i\leq n$ designates a unit, and $k\in\mathbb{Z}$ is a “measurement” performed in that unit. We place a directed edge from $(i,k)$ to $(j,\ell)$ whenever $$ \ell = \lfloor k \cdot u_i/u_j \rceil $$ where $\lfloor—\rceil$ means “closest integer to” (i.e., $\lfloor x\rceil = \lfloor x+\frac{1}{2}\rfloor$). Maybe assume that all $u_i/u_j$ are irrational so there is never any ambiguity as to what the closest integer means. In other words, we can get from $(i,k)$ to $(j,\ell)$ by converting unit $u_i$ into unit $u_j$ and rounding the result to the nearest integer. Question: does there exist $u_0,\ldots,u_n$ such that the graph just defined has an infinite strongly connected component? (In other words, can we find units such that infinitely many different values are reachable from one another by converting between these units and rounding to the nearest integer?) REPLY [7 votes]: For a vertex $(i,k)$, it is reasonable to define its value as $ku_i$. We show that the value cannot grow too large, which establishes a negative answer to the question. Let $v$ be the current value, and let $V>v$ be a real number such that $\{V/u_i\}<1/2$ for all $i$. Such a $V$ exists: we can start by choosing a very small $w>0$ such that $0<\{w/u_i\}<\frac12$, and then approximate those values by $\{V/u_i\}$ due to Kronecker. We claim that the value cannot exceed $V$. Indeed, if the current value $ku_i$ does not exceed $V$, then $ku_i/u_j$ rounds to at most $\lfloor V/u_j\rfloor$ (check the brackets’ corners!), so the new value does not exceed $V$.<|endoftext|> TITLE: Examples and properties of spaces with only trivial vector bundles QUESTION [7 upvotes]: Let $B$ be a paracompact space with the property that any (topological) vector bundle $E \to B$ is trivial. What are some non-trivial examples of such spaces, and are there any interesting properties that characterize them? For simple known examples we of course have contractible spaces, as well as the 3-sphere $S^3$. This one follows from the fact that its rank $n$ vector bundles are classified by $\pi_3 (BO(n)) = \pi_2 (O(n)) = 0$. I'm primarily interested in the case where $B$ is a closed manifold. Do we know any other such examples? There is this nice answer to a MSE question which talks about using the Whitehead tower of the appropriate classifying space to determine whether a bundle is trivial or not. This seems like a nice tool (of which I am not familiar with) to approaching this problem. As a secondary question, could I ask for some insight/references to this approach? EDIT Now that we know from the answers all the examples for closed $3$-manifolds, I guess I can now update the question to the case of higher odd dimensions. Does there exist a higher dimensional example? REPLY [4 votes]: Here is another obstruction. Suppose $M^n$ is a closed simply connected manifold which admits only trivial vector bundles. Then $M$ cannot be a $\mathbb{Z}/2\mathbb{Z}$-homology sphere, unless $n=3$. I'm not sure if the hypothesis that $M$ is simply connected is necessary, but it's certainly necessary in the proof. Together with the end of Michael Albanese's answer, this implies that every simply connected $5$-manifold admits a non-trivial vector bundle. Proof Sketch: By the work in the other answers, we already know that if $M$ admits only trivial vector bundles, then $M$ is odd dimensional, orientable, and it has the rational homology of a sphere. Thus, we may assume $n\geq 5$. Now, assume for a contradiction that all the torsion in $H^\ast(M)$ is of odd order. Let $k$ denote the least common multiple of the orders of the torsion and note that $k$ is odd. Because $M$ is orientable, it has a degree $1$ map $f:M\rightarrow S^n$. We will construct a non-trivial vector bundle on $M$ by pulling back a non-trivial bundle $E$ over $S^n$ along $f$. When $n\neq 7$, we may use the tangent bundle $E=TS^n$. When $n=7$, we let $E$ denote the rank $3$ vector bundle corresponding to a generator of $\pi_7(BO(3))\cong \pi_6(O(3)) = \pi_6(S^3) = \mathbb{Z}/12\mathbb{Z}$. Let $\phi:S^n\rightarrow BO(s)$ denote the classifying map of the bundle $E$ (where $s = n$ if $n\neq 7$, and $s= 3$ when $n = 7$.) We claim that $f^\ast E$ is a non-trivial vector bundle over $M$. To see this, note that since $M$ is simply connected and all torsion is annihilated by $k$, there is a map $g:S^n\rightarrow E$ of degree $k^r$ for some integer $r\geq 1$. (See the answer here for a proof.) It is enough to show that $g^\ast(f^\ast E)$ is a non-trivial vector bundle over $S^n$. Of course, it is enough to show that the map $\phi\circ g\circ f:S^n\rightarrow BO(s)$ is homotopically non-trivial. We'll show this by showing the induced map on $\pi_n$ is non-trivial. So, consider the induced map on $\pi_n$. For $n\geq 5$ odd (except $n=7$), Kervaire has shown that $\pi_n(BO(n))\cong \pi_{n-1}(O(n))$ is either $\mathbb{Z}/2\mathbb{Z}$ or $(\mathbb{Z}/2\mathbb{Z})^2$. When $n\neq 7$, the fact that $E$ is non-trivial implies that $\phi_\ast$ is non-trivial, so, for $n\neq 7$, the kernel of the map $\phi_\ast:\pi_n(S^n)\rightarrow \pi_n(BO(n))$ is the even integers. On the other hand, by our choice of $E$ when $n=7$, we have $\ker \phi_\ast = 12\mathbb{Z}$. In either case, the image of $(g\circ f)_\ast:\pi_n(S^n)\rightarrow \pi_n(S^n)$ is multiplication by the odd number $k^r$, so not contained in $\ker \phi_\ast$. The completes the sketch. $\square$.<|endoftext|> TITLE: Is there a Cantor set $C$ in $\mathbb{R}^{2}$ so the graph of every continuous function $[0,1]\rightarrow [0,1]$ intersects $C$? QUESTION [19 upvotes]: Consider the Cantor ternary set on the real line with the usual topology and define a Cantor set to be any topological space $C$ homeomorphic to the Cantor ternary set. The idea is to construct a Cantor set $C$ in $\mathbb{R}^{2}$ such that for every continuous function $f:[0,1]\rightarrow [0,1]$ we have $C\cap\operatorname{Graph}(f)\neq\emptyset$, where $\operatorname{Graph}(f) = \{(x,f(x)):x\in[0,1]\}$. Does the result generalize to $\mathbb{R}^{n}$, $n\ge 3$? That is, for every positive integer $k$, let $I^{k}$ denote the product $[0,1]\overbrace{\times\cdots\times}^{k\rm\ times}[0,1]$. We seek to find a Cantor set $C$ in $\mathbb{R}^{n}$ such that for every continuous function $f:I^{n-1}\rightarrow I^{n-1}$ we have $C\cap\operatorname{Graph}(f)\neq\emptyset$. A related question can be found at Is there a cantor set in $\mathbb{R}^{n}-\{0\}$ which intersects every ray from the origin?. REPLY [8 votes]: Here's one simple other example of such a Cantor subset. First, the theoretical side. In the closed unit square $[0,1]^2$, define a decreasing sequence of "simple" closed subsets $K_n$ (with complement $U_n$), and define $K=\bigcap_n K_n$. The conditions we can to ensure is that (a) $K$ is totally disconnected (b) $K$ meets the graph of every continuous map (written "graph" below). A trivial but essential remark is that (b) holds if and only if no $U_n$ contains a graph. (Indeed, by compactness, if $U=\bigcup U_n$ contains a graph, then this graph is contained in $U_n$ for some $n$.) As regards (a), the condition to ensure is that the sup $r_n$ of diameters of connected components of $K_n$ tends to zero. (I don't insist on $K$ being Cantor because it's a trivial issue, which can always be ensured by enlarging it, e.g., replacing isolated points with small Cantor subsets.) The interest is that while $K$ is a bit mysterious, the individual $U_n$ are meant to be easy to understand. Now for the example. Each $U_n$ will be blue and $K_n$ white. Each $K_n$ will be a finite union of horizontal squares tilted by $45$ degrees, then intersected with $[0,1]$. The basic construction is as follows: start from a square, cut it into $5\times 5$ squares and fill the 9 ones as follows: Next, fill the 16 remaining $5$-times-smaller white squares in the same fashion, but in the orthogonal direction. Next, do the same with the $25$-times-smaller white squares, again in the original direction. Eventually tilt everything by 45 degrees. At each step, this defines $U_1\subset U_2\subset U_3$... The pictures are as follows, showing, for $i=1,2,3$, $U_i$ (blue), $K_i$ (white): (at the bounding square, the white part should be understood as the closure of the inner white part — it was written as a blue line only for readibility) It can easily be checked that the sup of diameter of components of $K_n$ tends exponentially to zero. Next, $U_n$ contains no graph for any $n$. The point is that in $U_n$, the smallest $\Pi$-shaped paths (those in $U_n-U_{n-1}$) can't be used for travelling "to the right". Hence showing that $U_n$ contains no graph can be checked by induction. Of course details would be a little cumbersome to be written in Bourbaki-style, but this is quite visible from the picture, even for a non-mathematician. Edit: here are pictures of subsets $U_n$ (blue color) for Nik Weaver's example, after 4 steps and after 11 steps: (the increasing slopes are $(4/3)^n$, while the apparently decreasing slopes are actually vertical segments) More precisely, define functions $f_i$ as in Nik's example, where $f_i$ has slope $(4/3)^i$ at each continuity point: $f_0(x)=x$, $f_1(x)=(4/3)x$ for $x<1/2$ and $=(4/3)x+(2/3)$ for $x>1/2$, etc. Note that $f_i$ is defined on the set $C$ of dyadic expansions, rather than on $[0,1]$ where it bi-valued at some dyadic numbers. Thus the $f_n$, viewed as continuous function on $C$, converge uniformly to a function $f$ and $f(C)$ is the desired Cantor set. We have $\|f_{n+1}-f_n\|_{\infty}\le \frac16(2/3)^n$, so $\|f_n-f\|_\infty\le \frac12(2/3)^n$. Thus I defined $U_n$ as the set of $(x,y)$ such that $|y-f_n(x)|>\frac12(2/3)^n$ (properly speaking this is ill-defined for for the few points at which $f_n$ is discontinous, in which case I mean that $\max(|y-f_n(x^+)|,|y-f_n(x^-)|)>\frac12(2/3)^n$). PS pictures written with SageMath.<|endoftext|> TITLE: Who introduced the discrete Fourier transform? QUESTION [8 upvotes]: I am trying to find the original reference which introduced the definition of discrete Fourier transform as used today. When did this modern formulation (which includes the indexing from n to N-1) of DFT appear in the literature? $$ \begin{aligned} X_{k} &=\sum_{n=0}^{N-1} x_{n} \ e^{-\frac{i 2 \pi}{N} k n} \\ &=\sum_{n=0}^{N-1} x_{n} \left[\cos \left(\frac{2 \pi}{N} k n\right)-i \ \sin \left(\frac{2 \pi}{N} k n\right)\right] \end{aligned} $$ Fourier original on Analytical Theory of Heat does not deal with discrete versions. History related articles credit Gauss well before Fourier. For example here, Gauss and the History of the Fast Fourier Transform, Archive for History of Exact Sciences , 1985, Vol. 34, No. 3 (1985), pp. 265-277 ([Link])1, shows a table, but the article just defines the DFT in "modern notation" for Gauss's Latin work and credits Gauss rather. REPLY [10 votes]: You can go earlier than Gauss if you allow for a DFT involving only sines or only cosines: I quote from Gauss and the history of the fast Fourier transform Alexis-Claude Clairaut (1713-1765) published in 1754 what we currently believe to be the earliest explicit formula for the DFT (the computation for series coefficients from equally spaced samples of the function), but it was restricted to a cosine Fourier series. Joseph Louis Lagrange (1736-1813 published a DFT-like formula for finite Fourier series containing only sines, in 1759 and in 1762. The earliest explicit DFT formula containing both sines and cosines is due to Carl Friedrich Gauss (1777-1855) in "Theoria Interpolationis Methodo Nova Tractata". It was published only posthumously in 1866 [10], but was originally written, most likely, in 1805. Here is Gauss's formula ( source)<|endoftext|> TITLE: Is this relation about elementary embedding transitive? QUESTION [5 upvotes]: For ordinals $\alpha<\beta$, we say $\alpha<_{el}\beta$, if there is an elementary embedding with domain $L_\beta$ and critical point $\alpha$. Is $<_{el}$ transitive? REPLY [11 votes]: EDIT: If the codomain $N$ is allowed to be illfounded, then the answer is yes. (Here if $\kappa=\mathrm{crit}(j)$ then this will mean that $\kappa\subseteq N$, but it might be that $N$ is illfounded and $\kappa$ is exactly the wellfounded part of the codomain, in which case $\kappa\notin N$.) For let $\kappa_0<\beta_0=\kappa_1<\beta_1$ be ordinals and $j_i:L_{\beta_i}\to N_i$ be elementary (for $i=0,1$) with $\mathrm{crit}(j_i)=\kappa_i$. Let $U_0$ be some non-principal ultrafilter derived from $j_0$ (let $x\in N_0$ such that $N_0\models$"$\alpha\in x\in j_0(\kappa_0)$" for each $\alpha<\kappa_0$, and let $U_0$ be the filter over $\kappa_0$ derived from $x$). Then $U_0$ is an $L_{\beta_0}$-ultrafilter over $\kappa_0$ which is $L_{\beta_0}$-$\kappa_0$-complete (i.e. closed under length ${<\kappa_0}$-intersections of sequences $\left_{\alpha<\gamma}\in L_{\beta_0}$). And because $\kappa_1=\mathrm{crit}(j_1)$, $\kappa_1$ is a (regular) cardinal in $L_{\beta_1}$. Therefore $\mathcal{P}(\kappa_0)\cap L_{\beta_1}\subseteq L_{\beta_0}=L_{\kappa_1}$. But then $U_0$ is also an $L_{\beta_1}$-ultrafilter over $\kappa_0$ which is $L_{\beta_1}$-$\kappa_0$-complete. So letting $N=\mathrm{Ult}(L_{\beta_1},U_0)$ and $j:L_{\beta_1}\to N$ the ultrapower map, then $j$ is elementary and $\mathrm{crit}(j)=\kappa_0$. EDIT 2: On the other hand, if the codomain is required to be wellfounded, and $0^\sharp$ exists, the answer is no. For let $\kappa_0$ be the least $L$-indiscernible and let $\pi:L_{\omega_1}\to L_{\omega_1}$ be elementary with $\mathrm{crit}(\pi)=\kappa_0$. (Here $\omega_1$ is as computed in $V$.) There is an elementary $\sigma:L_{\kappa_0}\to L_{\kappa_0}$. For there is an elementary $\sigma':L_{\kappa_\omega}\to L_{\kappa_\omega}$ where $\kappa_\omega$ is the $\omega$th $L$-indiscernible, for example the one with $\sigma'(\kappa_n)=\kappa_{n+1}$. But the existence of some such $\sigma'$ is forced over $L$ by $\mathrm{Coll}(\omega,\kappa_\omega)$, and so by indiscernibility, the existence of a $\sigma$ as stated is forced over $L$ by $\mathrm{Coll}(\omega,\kappa_0)$, and therefore there is actually some such embedding $\sigma$ in $V$. So letting $\kappa=\mathrm{crit}(\sigma)$, then $\kappa<_{\mathrm{el}}\kappa_0<_{\mathrm{el}}\omega_1$. But I claim $\kappa\not<_{\mathrm{el}}\omega_1$. For since $\kappa<\kappa_0$, $\kappa$ is not an $L$-indiscernible. But if $j:L_{\omega_1}\to L_{\lambda}$ is elementary (where $\lambda$ is some ordinal) then $\mathrm{crit}(j)$ is an $L$-indiscernible. (This is a standard fact. The key is that if $U$ is the normal ultrafilter derived from $j$, then $\mathrm{Ult}(L,U)$ is wellfounded.) These observations leave open the version where the codomain is required to be wellfounded, but $0^\sharp$ does not exist.<|endoftext|> TITLE: Is Hurwitz's theorem true in constructive mathematics? QUESTION [17 upvotes]: Hurwitz's theorem says that the only division composition algebras over the real numbers $\mathbb{R}$ are the real numbers themselves $\mathbb{R}$, the complex numbers $\mathbb{C}$, the quaternions $\mathbb{H}$, and the octonions $\mathbb{O}$. However, in pure constructive mathematics without any weak axiom of choice, the notion of the set of real numbers bifurcates into multiple incompatible notions, such as the Cauchy real numbers $\mathbb{R}_C$, the Dedekind real numbers $\mathbb{R}_D$, the Escardó-Simpson real numbers $\mathbb{R}_E$, and the MacNeille real numbers $\mathbb{R}_M$, and I'd imagine the same for $\mathbb{C}$, $\mathbb{H}$, and $\mathbb{O}$ (i.e. Cauchy complex numbers, Dedekind quaternions, etc). For which of these sets of real numbers, complex numbers, quaternions, and octanions, if any at all, does Hurwitz's theorem still hold true? Edit: Swapped out "normed division algebra" for "division composition algebra" for the following reason: Classically, Hurwitz's theorem is also expressed in terms of finite-dimensional normed division algebras over that set of real numbers. However, finite-dimensional normed divison algebras and division composition algebras over the real numbers do not coincide in constructive mathematics because there are multiple different types of real numbers in constructive mathematics. In a division composition algebra, the norm $\lvert a \rvert := \langle a, a\rangle$ has a codomain of the ground field. However, that is not necessarily true of finite-dimensional normed division algebras over some field of real numbers $\mathbb{R}_X$ in constructive mathematics, because the notion of "multiplicative norm" bifurcates into multiple definitions based upon which set of real numbers $\mathbb{R}_Y$ is used as the codomain of the norm. It is perhaps more appropriate to call them finite-dimensional $\mathbb{R}_Y$-normed division $\mathbb{R}_X$-algebras. One can have a finite-dimensional normed division $\mathbb{R}_C$-algebra with a norm valued in $\mathbb{R}_D$, where $\mathbb{R}_C$ are the Cauchy real numbers and $\mathbb{R}_D$ are the Dedekind real numbers, but such a finite-dimensional normed division algebra is not a composition algebra, and is not covered under Hurwitz's theorem. Thus, for sets of real numbers $\mathbb{R}_X$ and $\mathbb{R}_Y$, every division composition $\mathbb{R}_X$-algebra for some set of real numbers $\mathbb{R}_X$ is a finite-dimensional $\mathbb{R}_X$-normed division $\mathbb{R}_X$-algebra, but a finite-dimensional $\mathbb{R}_Y$-normed division $\mathbb{R}_X$-algebra is only a division composition $\mathbb{R}_X$-algebra if $\mathbb{R}_Y$ is isomorphic to $\mathbb{R}_X$. REPLY [3 votes]: There is a weakening of Hurwitz's theorem that is true constructively, with essentially the same proof: Let $A$ be a division composition algebra. Then any chain of proper subalgebras $\mathbb{R} = A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n = A$ has length $n \leq 3$ (where "proper" means "contains an element with positive distance from the previous algebra). We can also show that in general, any inclusion of subalgebras generated by adding one element must come from (a quotient of) the General Cayley-Dickson construction with parameter $\cdot \gamma$, $\gamma \leq 0$.<|endoftext|> TITLE: Why is choice needed in Ellis' Lemma? QUESTION [9 upvotes]: Ellis Lemma on idempotent elements asserts that: Lemma (Ellis). Every compact semigroup has an idempotent. The proof below is excerpted from Todorcevic's Introduction to Ramsey Spaces, Lemma 2.1. Let $S$ be a compact semigroup. Pick by Zorn’s Lemma a minimal compact subsemigroup $R \subseteq S$ and an arbitrary $s \in R$. Then $Rs$ is also a compact subsemigroup and $Rs \subseteq R$. Hence $Rs = R$. Let $P = \{x \in R : xs = s\}$. Then $P \neq \emptyset$, since $s \in Rs$. Note that $P$ is also a compact subsemigroup of $S$. Hence $P = R$ and therefore $s^2 = s$. It appears that Zorn's Lemma is required to obtain a minimal compact subsemigroup $R \subseteq S$. I do not understand why, as it appears to me that we can simply let: $$ R := \bigcap \{Q \subseteq S : Q \text{ is a compact subsemigroup of $S$}\} $$ What is the error in reasoning here? REPLY [9 votes]: The issue is that $R$ could be empty. When you apply Zorn's lemma, a nested intersection of non-empty compact sets is non-empty, guaranteeing minimal non-empty elements.<|endoftext|> TITLE: Is Solèr’s theorem true in constructive mathematics? QUESTION [6 upvotes]: Solèr’s theorem says that for every star division ring $R$ and every $R$-module $H$ with an orthomodular Hermitian form $\langle (-),(-) \rangle:H \times H \to R$ such that there exists an infinite orthonormal sequence $e:\mathbb{N} \to H$, $R$ is either the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, or the quaternions $\mathbb{H}$, and $H$ is a Hilbert space over $R$. Assuming that the star division rings used are Heyting division rings (or else Solèr’s theorem is most likely false), is Solèr’s theorem true in constructive mathematics? REPLY [8 votes]: Suppose you have a classical classification theorem saying Each structure (of a certain kind) is either an $A$ or a $B$. Then you cannot exhibit constructively a $C$ which is neither $A$ nor $B$ because every constructive proof is also classical, and so that would contradict the classical classification. What happens instead is that the classical meaning of “$p$ or $q$” corresponds to the constructive “not ($\neg p$ and $\neg q$)“ so the constructive reading of the classical classification theorem is Apart from $A$ and $B$, there is no other structure (of a certain kind). To give you an idea on how to play tricks with excluded middle in constructive mathematics, consider any proposition $p$ and define $$K_p = \{ z \in \mathbb{C} \mid p \Rightarrow z \in \mathbb{R} \}.$$ It is easy to check, regardless of what $p$ is, that $K_p$ is a subfield of $\mathbb{C}$. Moreover, if $p$ holds then $K_p = \mathbb{R}$, and if $\neg p$ holds then $K_p = \mathbb{C}$. But stating that $K_p$ is either $\mathbb{R}$ or $\mathbb{C}$ implies $\neg \neg p \lor \neg p$, which lets us decide $\neg p$, which is not generally possible in constructive mathematics. It is still the case that $K_p$ cannot be different from both $\mathbb{R}$ and $\mathbb{C}$, because that amounts to $\neg (\neg p \land \neg\neg p)$, which is constructively true (obviously). Consequently, if Solèr’s theorem were true constructively (in the version that says that every structure is either this or that), we could decide $\neg p$: just ask the theorem to classify $K_p$ for you. The usual solution to the conundrum is to strengthen the assumptions to something that does not matter classically. For example, you might want to assume that the vector space $E$ featuring in the definition of a Hermitian form has a given basis, and that the basis has size which is either a natural number or is countably infinite (this cannot be shown to hold for $K_p$ above seen as a real vector space). But that is only the first step, there will be further complications, and one would have to dig into the details of the classification theorem. Unfortunately, I do not know whether anyone has done so.<|endoftext|> TITLE: Is it possible to start a PhD in mathematics at the age of 29? QUESTION [21 upvotes]: I graduated with a bachelor’s degree in mathematics. I was initially focused on branches in analysis like operator algebra. At the third year of my undergraduate study, I experienced a financial loss in my family. It was only a slight loss and would not influence the life and regular plans of my family. But at that time I was not mentally strong enough and I could not concentrate on study. I postponed two years to graduate, in 2020. These days I am trying to apply for a master program in mathematics. My GPA is not top, but fair enough, and I also did my graduation thesis carefully. I applied for several programs in Europe and received the conditional admission of Uni of Göttingen, but my Toefl grade did not meet the requirements. This year I have prepared all the things and I am going to apply for several master’s programs in Germany. I am currently interested in low-dimensional topology and want to select this area as my direction. But when I apply for a PhD, I am 29 years old, is it a huge disadvantage? I also referred to several persons working on geometric topology, and the time cost seems to be high. But I am still enthusiastic about mathematics and want to get a bread. Anyone could give me some suggestions? REPLY [2 votes]: I started my PhD at age 30, and don't feel my age was a significant obstacle. However, I believe almost no one should do a PhD, regardless of age.<|endoftext|> TITLE: Aphantasiac mathematicians? QUESTION [9 upvotes]: Over the past few years there's been a fair amount of publicity given to the phenomenon of aphantasia, the condition of being unable to form visual images in one's mind or remember what things look like. Such a person after closing their eyes may be unable to remember what colors look like a second or two after seeing them, or unable to imagine what green looks like when nothing green is where they can see it. In thinking about simple geometric problems and graphs I rely heavily on being able to imagine their visual appearance. (For that matter I can name the fifty states in the U.S.A. and the ten provinces and three territories of Canada in seconds by remembering the visual appearance of the maps, but if I tried to do it alphabetically I would hesitate for long periods between list items and almost certainly miss some.) It seems that a mathematician who is an aphantasiac would be unable to do that, and their cognition would function differently. Is anything known about this? Has anything been published about it? REPLY [12 votes]: Some personal experiences are described at Mathematicians with aphantasia (inability to visualize things in one's mind) For the more specific question "How does aphantasia affect doing mathematics?" see Brailey Sims talk on Aphantasia and Mathematical Thinking. It seems to me likely that the freedom from the precision and detail of a visual image is a boon to my ability to think and work abstractly. Where I do feel a lack of being able to conceive visual images on which to work is a handicap is in anticipating and seeing through a symbolic/algebraic argument, or carrying out any lengthy algebraic computations/manipulations in my head. A study from 2020 by the University of Exeter, based on interviews with 2000 individuals, found that more than 20% of people with aphantasia worked in science, computing or mathematics. This is attributed to an enhanced ability to "assimilate complex information into new ideas and approaches".<|endoftext|> TITLE: Mañé's example of an attractor with no natural measure QUESTION [6 upvotes]: I'm reading Milnor's notes on dynamical systems and in Lecture 3 he gives an example of an attractor with no natural measure, which he attributes to Mañé. I can find no other reference in which this example is discussed; no paper by Ricardo Mañé, no books or papers in which this example is mentioned. Milnor also states that an earlier paper by Zakharevich has a similar example, but again no reference is given, and I cannot find this paper anywhere. Has this or any other counterexample been published anywhere else? Any references would be highly appreciated. REPLY [3 votes]: Q: Has this or any other counterexample been published anywhere else? A: In A continuous Bowen-Mañé type phenomenon examples of vector fields without physical measure for certain parameters are discussed under the name "Bowen-Mañé type phenomena". No primary reference to either scientist is given. The Bowen-Mañé example is also called "Bowen's eye" in the literature, see arXiv: 2010.08945 and arXiv:1609.05356. The attribution to Bowen is due to Takens, who writes in a 1994 paper: "I attribute this example to Bowen: although he never published it, I learned this example through a paper by Ruelle who referred to Bowen (we have no reference: even Ruelle could not localize that paper)." [source]<|endoftext|> TITLE: Mapping class group and pure mapping class group QUESTION [5 upvotes]: "A Primer on Mapping Class Groups" wrote Let $\mathrm{Homeo}_+(S, \partial S)$ denote the group of orientation-preserving homeomorphisms of $S$ that restrict to the identity on $\partial S$. $\mathrm{Mod}(S)$ is the group of isotopy classes of elements of $\mathrm{Homeo}_+(S, \partial S)$, where isotopies are required to fix the boundary pointwise. and Let $\mathrm{PMod}(S_{g,n})$ denote the pure mapping class group of $S_{g,n}$, which is defined to be the subgroup of $\mathrm{Mod}(S_{g,n})$ consisting of elements that fix each puncture individually. What is the difference between the two definitions? Doesn't "restrict to the identity on $\partial S$" mean for for all $x\in \partial S$ we have $\phi(x)=\mathrm{id}(x)=x$, which implies $\phi$ fixes each puncture individually? REPLY [12 votes]: Just to give an explicit description of the difference: if one takes a loop "around a boundary component," the Dehn twist around this loop is not isotopic to the identity. On the other hand, if one takes a loop around a puncture, it is isotopic to the identity, by "sliding the loop" to the puncture. In fact this is the only difference between the two groups if I remember correctly. If $(S,\partial S)$ is the surface with boundary obtained by deleting open discs around each puncture of $S_{g,n}$, the inclusion $(S, \partial S)\to S_{g,n}$ induces a map $\text{Mod}(S,\partial S)\to \text{PMod}(S_{g,n})$. This map is surjective with kernel the free abelian group generated by the Dehn twists I describe above.<|endoftext|> TITLE: Dimension reduction for non-negativity of elementary symmetric polynomials QUESTION [5 upvotes]: Fix integers $1 \leq k \leq n$ and suppose $\mathbf{x} \in \mathbb{R}^n$ is such that $e_j(x_1,x_2,\ldots,x_n) \geq 0$ for all $1 \leq j \leq k$, where $e_j$ is the $j$-th elementary symmetric polynomial. Question 1: Is it true that $x_1 + x_2 + \cdots + x_{n-k+1} \geq 0$? The answer is trivially "yes" in the edge cases $k = 1$ and $k = n$, but the intermediate cases seem much less obvious. While Question 1 is the one that I'm really interested in, there is a natural generalization of it that is perhaps known, so I'll ask it now: Question 2: Is it true that $e_j(x_1,x_2,\ldots,x_{n-1}) \geq 0$ for all $1 \leq j \leq k-1$? In other words, can we use non-negativity of elementary symmetric polynomials in $n$ variables to infer non-negativity of elementary symmetric polynomials in $n-1$ variables? If the answer to Question 2 is "yes" then we can use it $k-1$ times to see that the answer to Question 1 is "yes" too. REPLY [4 votes]: The answer to question 2, and therefore question 1, is "yes". We abbreviate $e_k(x_1, x_2, \ldots, x_{n-1})$ to $b_k$ and $e_k(x_1, x_2, \ldots, x_{n-1}, x_n)$ to $a_k$, so $a_k = x_n b_{k-1} + b_k$. We are trying to show that, if $a_1$, $a_2$, ..., $a_k \geq 0$ then $b_1$, $b_2$, ..., $b_{k-1} \geq 0$. We prefer to prove the contrapositive: If $b_j<0$ then one of $a_1$, $a_2$, ..., $a_j$, $a_{j+1} <0$. We may assume that $j$ is minimal with $b_j<0$, so $b_1$, $b_2$, ..., $b_{j-1} \geq 0$. Case 1: $x_n<0$. Then $a_j = x_n b_{j-1} + b_j < 0$. Case 2: $x_n \geq 0$. First of all, if $b_{j-1} \leq 0$, then $a_j<0$ and, if $b_{j+1} \leq 0$, then $a_{j+1} < 0$. So we may assume that $b_{j-1}$ and $b_{j+1}>0$. Now, Newton's inequalities give $$\frac{b_{j-1} b_{j+1}}{\binom{n-1}{j-1} \binom{n-1}{j+1}} \leq \frac{b_j^2}{\binom{n-1}{j}^2}$$ or $$b_{j-1} b_{j+1} \leq \frac{j(n-j-1)}{(j+1)(n-j)} b_j^2 < b_j^2.$$ So $$0<\frac{b_{j+1}}{- b_j} < \frac{-b_j}{b_{j-1}}.$$ We must either have $x_n > \tfrac{b_{j+1}}{- b_j}$ or $x_n < \tfrac{-b_j}{b_{j-1}}$. In the first case, $0 > x_n b_j + b_{j+1} = a_{j+1}$; in the second case, $a_j = x_n b_{j-1} + b_j < 0$. Either way, we have found a negative $a_j$.<|endoftext|> TITLE: Refining a triangulation QUESTION [8 upvotes]: I'm reading Thurston's article "Shapes of polyhedra and triangulations of the sphere." In the introduction he claims the following: "${}^{(1)}$There are procedures to refine and modify any triangulation of a surface until every vertex has either 5, 6 or 7 triangles around it, or with more effort, ${}^{(2)}$so that there are only 5 or 6 triangles if the surface has positive Euler characteristic, only 6 triangles if the surface has zero Euler characteristic, or only 6 or 7 triangles if the surface has negative Euler characteristic..." I divide the quote into two claims. I proved the first claim, but I'm stuck with the second one. There must be a global argument I'm not seeing. Does anyone know how to do it? REPLY [7 votes]: Suppose that $S$ is a closed, connected surface with negative Euler characteristic. Suppose that $T$ is a triangulation of $S$. Define "refine" to mean "replace each triangle by four triangles" (so that edge midpoints become vertices of valence six). This does not improve any of the vertices of "concentrated positive or negative curvature", but it does isolate them. Suppose that there are no vertices of degree five or lower. Refine as above to isolate all vertices of degree greater than six. Suppose that the vertex $v$ has degree eight or higher. We "split" at $v$ - we choose two edges $e$ and $e'$ at $v$, cut $T$ along $e \cup e'$ and insert a pair of triangles. This increases the number of triangles by two, the number of vertices by one, and the number of edges by three. The new vertices $u$ and $u'$ have degree less than that of $v$. Also, two old vertices of degree six will now have degree seven. Repeat. As a bit of motivation: the original triangulation gives a metric on $S$. Refinement has the effect of scaling the metric up, which "brings the curvature down". Cutting and inserting triangles disperses concentrated negative curvature (at a vertex) into adjacent (almost) flat regions. In general, if there are vertices of degree less than six, we refine and then pair them with vertices of higher degree, and (carefully!) cancel curvature. In the case of Euler characteristic zero, after pairing in this way, only vertices of degree six remain. In the case of positive Euler characteristic, we have some vertices of degree five (either six or twelve) left at the end.<|endoftext|> TITLE: Thom spectra, tmf, and Weierstrass curve Hopf Algebroid QUESTION [5 upvotes]: Let $X(4)$ be the Thom spectrum associated to $\Omega SU(4) \to \Omega SU \simeq BU$. Since $X(4)$ is a homotopy commutative ring spectrum, for any spectrum Y we can construct a resolution $$ Y \wedge X(4) \to Y \wedge X(4) \wedge X(4) \to Y \wedge X(4) \wedge X(4) \wedge X(4) \to \dots $$ of Y. (Sorry but I don't know how to type the cosimplicial diagram. This is just construction of the ANSS with respect to $X(4)$.) In Rezk's note on tmf, he claims that the resolution for $Y = tmf$ is the same as the cobar complex obtained from the Weierstrass Hopf algebroid $(\mathbb{Z}[a_1, a_2, a_3, a_4, a_6], \mathbb{Z}[a_1, a_2, a_3, a_4, a_6][r, s, t])$ (Thm 14.5). I want to know a proof of this theorem because this construction is used in Bauer's paper, but I haven't find a written proof of it. I think Hopkins-Mahowald report also states similar result (Cor 2.4), but I don't see how to conclude this from the theorem above. Does anybody know how to compute this, at least? Following is my scatterer thoughts: Since $Y \wedge X(4)$ has a complex orientation of degree 4 (green book, 6.5.3), we know that $\pi_*(tmf \wedge X(4) \wedge X(4)) = \pi_* (tmf \wedge X(4))[r, s, t]$. So we have to compute $\pi_* (tmf \wedge X(4))$ first. Can we compute this from $\pi_* (tmf \wedge MU)$? Mathew's paper calculate the latter so I am wondering if I can use this. As Hopkins-Mahowald says, homotopy classes $a_1 \dots a_6$ come from $\pi_*(X(4))$. However I don't know the computation of $\pi_*(X(4))$ or image of Hurewicz homomorphsim. Also, what are so special about $tmf$ or $eo_2$ so that these elements forms a homotopy group? Do we have a explanation of this computation? Why $X(4)$ is used here? REPLY [2 votes]: $\newcommand{\MU}{\mathrm{MU}} \newcommand{\SU}{\mathrm{SU}} \newcommand{\tmf}{\mathrm{tmf}} \newcommand{\ko}{\mathrm{ko}} \newcommand{\BGL}{\mathrm{BGL}} \newcommand{\ku}{\mathrm{ku}} \newcommand{\GL}{\mathrm{GL}} \newcommand{\RP}{\mathbf{R}P}$ Suppose $R$ is a homotopy commutative ring such that $X(n)_\ast(R)$ is concentrated in even degrees for some $n\geq 0$. Then we can form the graded stack $M_R$ associated to the graded Hopf algebroid $(A, \Gamma):= (X(n)_\ast(R), X(n)_\ast(X(n) \otimes R))$. This will be isomorphic to the graded stack associated to the graded Hopf algebroid $(A', \Gamma'):= (\MU_\ast(R), \MU_\ast(\MU \otimes R))$. To see this, note that there is an isomorphism of algebras $$\MU_\ast(R) \cong X(n)_\ast(R)[x_{n+1}, x_{n+2}, \cdots],$$ where $|x_i| = 2i$. Similarly, $$\MU_\ast(\MU \otimes R) \cong X(n)_\ast(X(n) \otimes R)[x_{n+1}, x_{n+2}, \cdots][t_{n+1}, t_{n+2}, \cdots].$$ where $|t_i| = 2i$. (In fact, these can be lifted to equivalences at the level of spectra $$\MU\otimes R \simeq \MU\otimes_{X(n)} (X(n) \otimes R) \simeq X(n) \otimes R \otimes \Omega(\SU/\SU(n))_+;$$ but this is at the expense of multiplicativity: $\Omega(\SU/\SU(n))$ isn't an $\mathbf{E}_2$-space.) So $(A', \Gamma')$ is isomorphic to $(A[x_{n+1}, \cdots], \Gamma[x_{n+1}, \cdots, t_{n+1}, \cdots])$. One can now check from the Hopf algebroid structure that they must present the same stack, namely $M_R$. Now, let us try to understand $X(4) \otimes \tmf$. As a warmup, let us first understand $X(2) \otimes \ko$ (this is in Section 3.2 of "From spectra to stacks", see the comments). Observe that $X(2)$ is the Thom spectrum of the map $\Omega S^3 \to \BGL_1(S)$ which detects $\eta\in \pi_1(S)$ on the bottom cell of the source. There is an EHP sequence $S^2 \to \Omega S^3 \to \Omega S^5$, using which one can show that $X(2) \simeq C\eta \otimes S/\!/\nu$, where $S/\!/\nu$ is the Thom spectrum of the map $\Omega S^5 \to \BGL_1(S)$ which detects $\nu\in \pi_3(S)$ on the bottom cell of the source. Therefore, $\ko \otimes X(2) \simeq \ko \otimes C\eta \otimes S/\!/\nu$. We also know that $\ko \otimes C\eta\simeq \ku$ (the Wood equivalence), so that $\ko \otimes X(2) \simeq \ku \otimes S/\!/\nu$. Since $\nu = 0$ in $\ku$, we have $\ku \otimes S/\!/\nu \simeq \ku[\Omega S^5]$, so we conclude that $\ko \otimes X(2) \simeq \ku[\Omega S^5]$. At the level of homotopy, this is $\mathbf{Z}[\beta, x_4]$ with $|\beta|=2$ and $|x_4|=4$. (Remark: the above argument only shows that the equivalence $\ko \otimes X(2) \simeq \ku \otimes S/\!/\nu$ is true additively. Indeed, we used the Wood equivalence $\ko \otimes C\eta\simeq \ku$, which is a priori just an equivalence of spectra. To show that $\ko \otimes X(2) \simeq \ku \otimes S/\!/\nu$ as ring spectra, we need to equip $\ko/\eta$ with an $\mathbf{E}_\infty$-ring structure and show that the Wood equivalence is one of $\mathbf{E}_\infty$-rings. There are a few ways of doing this. For example, it is a consequence of the fact that the map $\eta: S^1 \to \GL_1(\ko)$ detecting $\eta$ is a map of infinite loop spaces; in fact, this map factors through an infinite loop map $\RP^\infty \to \GL_1(\ko)$. It corresponds to a map $\Sigma \mathbf{Z}/2 \to \mathrm{gl}_1(\ko)$, which is the inclusion of a summand. But this is getting to be a bit of a digression.) Let us now return to $X(4) \otimes \tmf$. (I don't know the details of the original argument due to Hopkins and Mahowald, so let me explain a possibly different approach. Most of the details of this story, such as the spectra $T(2)$ and $B$ below, are in my paper https://sanathdevalapurkar.github.io/files/thom.pdf. ) For simplicity, let me localize at $2$. Then $X(4)$ splits as a wedge of copies of an $\mathbf{E}_2$-ring called $T(2)$. In fact, $T(2)$ is the Thom spectrum of the bundle over $\Omega \mathrm{Sp}(2)$ given by the $\mathbf{E}_2$-map $$\Omega \mathrm{Sp}(2) \to \Omega \SU(4) \to \Omega\SU \simeq \mathrm{BU}.$$ Since $\SU(4)/\mathrm{Sp}(2) \simeq S^5$, we have that $X(4)$ is the Thom spectrum of a map $\Omega S^5 \to \BGL_1(T(2))$. This map is $2$-locally null, so $X(4) \simeq T(2)[\Omega S^5]$, i.e., $T(2)[a_2]$ with $|a_2|=4$. So we only need to understand $\tmf \otimes T(2)$. There is a map $T(2) \to \mathbf{F}_2$ which is injective on mod $2$ homology, and its image is $\mathrm{H}_\ast(T(2); \mathbf{F}_2) \cong \mathbf{F}_2[\zeta_1^2, \zeta_2^2]$. There is an $\mathbf{E}_1$-ring $B$ such that $T(2) \simeq B \otimes DA_1$, and which has a map $B \to T(2)$. The composite $B \to T(2) \to \mathbf{F}_2$ on mod $2$ homology is injective, and its image is $\mathrm{H}_\ast(B; \mathbf{F}_2) \cong \mathbf{F}_2[\zeta_1^8, \zeta_2^4]$. Then, $$\tmf \otimes T(2) \simeq \tmf \otimes DA_1 \otimes B \simeq \mathrm{BP}\langle 2\rangle \otimes B,$$ whose homotopy groups are $\pi_\ast(\mathrm{BP}\langle 2\rangle)[x_8, y_{12}] \cong \mathbf{Z}_{(2)}[v_1, v_2, x_8, y_{12}]$. Here, $|v_1| = 2$, $|v_2| = 6$, $|x_8|=8$, and $|y_{12}|=12$. (Note that we used the Wood-type equivalence $\tmf \otimes DA_1 \simeq \mathrm{BP}\langle 2\rangle$; just as with the Wood equivalence $\ku \otimes C\eta \simeq \ku$, this is a priori just an equivalence of spectra. But I think one can actually equip $\tmf \otimes DA_1$ with at least an $\mathbf{E}_2$-ring structure such that the Wood-type equivalence is one of $\mathbf{E}_2$-rings.) Therefore, $$\pi_\ast(\tmf_{(2)} \otimes X(4)) \simeq \pi_\ast(\tmf_{(2)} \otimes T(2)[\Omega S^5]) \cong \pi_\ast(\mathrm{BP}\langle 2\rangle)[a_2, x_8, y_{12}] \cong \mathbf{Z}_{(2)}[v_1, a_2, v_2, x_8, y_{12}].$$ This is exactly the desired calculation (recall that $|a_2| = 4$), at least $2$-locally. The same sort of argument works $3$-locally.<|endoftext|> TITLE: Homotopy properties of Lie groups QUESTION [10 upvotes]: Let $G$ be a real connected Lie group. I am interested in its special homotopy properties, which distinguish it from other smooth manifolds For example $G$ is homotopy equivalent to a smooth compact orientable manifold. In particular, Poincaré duality holds for $G$. $\pi_1(G)$ is abelian, $\pi_2(G) = 0$ The impossibly perfect answer to my question is a list of properties that make up a complete homotopy characterization of Lie groups (that is, in every homotopy type (of smooth manifolds) with such properties, there exists a smooth manifold admitting the structure of Lie groups). P.S. In this question, I am not interested in the homotopy properties of manifolds that distinguish them from other CW-complexes, for this see resp. question on MO REPLY [8 votes]: A very useful fact is that every connected Lie group is rationally homotopy equivalent to the product of several odd dimensional spheres where the number of spherical factors is equal to the rank of the group. For example, $SU(n)$ is rationally equivalent to $S^3\times S^5 \ldots \times S^{2n-1}$ and $Sp(n)$ is rationally equivalent to $S^3\times S^7\times \ldots \times S^{4n-1}$. For simple Lie groups the first factor is always $S^3$ and there is only one $S^3$ in each simple group. Also, every simple $G$ has $\pi_3\cong\mathbb Z$. For reference see for example the book "Topology of transitive transformation groups" by Onishchik.<|endoftext|> TITLE: Abelian groups such that $A \cong \mathrm{End}(A)$ and "complete rings" QUESTION [6 upvotes]: Motivation: for any ring $R$ there is the natural monomorphism $\mathrm{in} \colon R \to \mathrm{End}(R_{add}): r \mapsto (x \mapsto rx)$, where $R_{add}$ is an additive abelian group ( rings are assumed to be associative with identity, but not necessarily commutative). So a ring is exactly an abelian group with a distinguished subgring of its endomorphisms (and a fixed bijection between elements and distinguished endomorphisms). Some rings are "complete" in the sense that they "contain" all endomorphisms of the underlying abelian group. For example, $\mathrm{in}$ is an isomorphism for $R = \mathbb{Z}, \mathbb{Q}, \mathbb{Z}_n$. What is known about the classification of abelian groups $A$ such that there is an isomorphism between the abelian groups $\mathrm{End}(A)$ and $A$? Each such isomorphism gives some "complete ring" structure on $A$. What is known about the uniqueness (up to isomorphism) of the "complete ring" structure on an abelian group? I'm interested in the answers to these questions, with any additional assumptions that seem natural to you. I am especially interested in the answers for commutative rings. REPLY [6 votes]: The rings, you call ``complete'' are known as $E$-rings (as Ulrich Pennig mentioned in the comments). Some comments on your questions There are too many results on the $E$-rings to list them here and I'd rather direct you to the book by Göbel and Trlifaj Approximations and endomorphism algebras of modules. However to give you some sense that we have no hope to obtain any reasonable classification - every Abelian cotorsion-free group embeds in an $E$-ring. Formally a group $A$ is cotorsion-free if there are only null homomorphisms $\mathbb{Z}^\wedge_p\to A$. This is equivalent to the claim that if another Abelian group $C$ admits a compact topology then every homomorphism $C\to A$ is null. To make long story short - in the absence of compactness you may add to $A$ new elements to get more endomorphisms and then add yet more elements to kill the unwanted endomorphisms. Given an Abelian group $A$ the $E$-ring structure on $A$ is unique up to the choice ot the identity element. Every invertible element of an $E$-ring $A$ can be chosen as the identity element for another ring structure on $A$.<|endoftext|> TITLE: What is the connection between Lurie's definition of shape and Čech homotopy? QUESTION [10 upvotes]: It seems there are many subtly different notions of the shape of a topological space (and, more generally, toposes). For instance, Lurie [Higher topos theory] defines this one: Definition 1. The shape of a topos $\mathcal{E}$ is the pro-object in $\mathcal{S}$ representing the endofunctor $p_* p^* : \mathcal{S} \to \mathcal{S}$, where $p$ is the unique geometric morphism from the $\infty$-sheaf topos $\mathcal{E}$ to the $(\infty, 1)$-category $\mathcal{S}$ of $\infty$-groupoids. Here is another, perhaps closer to what is studied in classical shape theory: Definition 2. The shape of a topological space $X$ is the pro-object $\varprojlim_{\mathcal{U} : \textrm{Cov} (X)} \mathrm{B} \mathcal{U}$ where $\mathcal{U}$ ranges over the poset $\textrm{Cov} (X)$ of covering sieves $\mathcal{U}$ of non-empty open subspaces of $X$ and $\mathrm{B} \mathcal{U}$ is the geometric realisation of the nerve of $\mathcal{U}$. There is a canonical comparison from definition 1 to definition 2. Roughly speaking, it comes from the Grothendieck plus construction for sheafification. The sheaf $p^* K$ is the sheafification of the constant presheaf with value $K$, so there is a comparison morphism $\varprojlim_{U : \mathcal{U}} K \to p_* p^* K$ for each covering sieve $\mathcal{U}$. We have $\varprojlim_{U : \mathcal{U}} K \cong \mathcal{S} (\mathrm{B} \mathcal{U}, K)$, and taking the colimit over all covering sieves $\mathcal{U}$ yields a morphism $${\textstyle \varinjlim}_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} \mathcal{S} (\mathrm{B} \mathcal{U}, K) \to {\textstyle \varinjlim}_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} p_* p^* K \cong p_* p^* K$$ which yields a morphism from the pro-object representing $p_* p^*$ to $\varprojlim_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} \mathrm{B} \mathcal{U}$. Question. Is this an equivalence? When $K$ is discrete, the constant presheaf with value $K$ is almost separated – there is a problem over $\varnothing$, but it can be ignored – so $\varinjlim_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} \mathcal{S} (\mathrm{B} \mathcal{U}, K) \to p_* p^* K$ is in fact a bijection in this case. I am less clear about the situation for non-discrete $K$ – do we need to iterate the plus construction even for constant presheaves? Can we fix the problem by replacing $\textrm{Cov} (X)$ with some suitable category of hypercovers? REPLY [3 votes]: The plus construction has to be iterated, yes. The topological space from this answer provides a simple counterexample. Let $X=\{a,b,c,d\}$ with opens $\{a\},\{b\},\{a,b\},\{a,b,c\},\{a,b,d\}$. Then the plus construction does not change the global sections of the constant presheaf on $X$ with fiber $K$. However, the global sections of the associated sheaf is the free loop space $\mathcal LK=K\times_{K\times K}K$. So the shape of $X$ is a circle, and in this case Definition 2 "does not work". The problem can more or less be fixed by using hypercovers, because the analogue of the plus construction with hypercovers always produces the hypersheafification (this is essentially Verdier's hypercovering theorem). So the analogue of Definition 2 gives the shape of the $\infty$-topos of hypersheaves on $X$.<|endoftext|> TITLE: Finding non-inner derivations of simple $\mathbb Q$-algebras QUESTION [5 upvotes]: What's a good example of a simple algebra over a field of characteristic $0$ which has a non-inner derivation but also has the invariant basis number property (IBN)? I'm under the impression that when an algebra is simple Artinian, all derivations are inner. If I'm missing something subtle about what can happen please let me know, because that would be a lot easier to use than what follows. Here is what I've been pursuing. In this article, K. Goodearl. Simple self-injective rings need not be Artinian. (1974) Two examples are given of self-injective, simple, non-Artinian rings that have the IBN property. Roughly recapitulating them, they are The coordinatizing ring for an irreducible continuous geometry $L$ "in case $\infty$" (i.e. fails the DCC) as constructed in von Neumann's Examples of Continuous Geometries. The maximal right ring of quotients of an $AW^\ast$-factor type $\mathrm{II}_f$ I'm new to non-inner derivations and AW* algebras and what the coordinatizing ring for a continuous lattice looks like, so it's tough to make progress. If there is a way to demonstrate such a non-inner derivation, or alternatively a reason that neither of these algebras has such derivations, either one would settle this question. REPLY [4 votes]: Let $K$ be any field. I will give a simple $K$-algebra with IBN and a noninner derivation. My example will be a contracted monoid algebra. These have IBN by @PaceNielsen's nice answer here. Let $X$ be an infinite set and let $M$ be the monoid with zero given by the presentation $$\langle X\cup X^*\mid x^*y=\delta_{x,y}, x,y\in X\rangle$$ where $X^*$ is a bijective copy of $X$. Let $R=K_0M$ be the contracted monoid algebra. So it has $K$-basis $M\setminus \{0\}$ and the product extends that of $M$, where we identify the zero of $M$ with the zero of $K$. This is the Leavitt path algebra of an infinite bouquet of circles and is well known to be simple. This was first proved by Douglas Munn (W. D. Munn. Simple contracted semigroup algebras. In Proceedings of the Conference on Semigroups in Honor of Alfred H. Clifford (Tulane Univ., New Orleans, La., 1978), pages 35–43. Tulane Univ., New Orleans, La., 1979) to the best of my knowledge. I claim that it has a noninner derivation. This paper studies outer derivations of Leavitt path algebras and the derivation I am using is from there, although technically that paper does not allow digraphs with infinite out-degree. It is easy to check that every nonzero element of $M$ can be written uniquely in the form $pq^*$ where $p,q$ are words in $X$ (possibly empty) and you extend $\ast$ to words in the obvious way, so that $*$ is an involution on $M$. Now define $d\colon R\to R$ on the basis $M\setminus \{0\}$ as follows. Fix $x\in X$ and put $$d_x(pq^*) = (|p|_x-|q|_x)pq^*.$$ Here $|w|_x$ is the number of occurrences of the letter $x$ in $w$. It is easy to check that $d_x$ is a derivation. It is also not too hard to check that it is not inner using that $d_x(x)=x$ and you cannot find $m\in R$ with $xm-mx=x$. Details added (update). I found a little time to add in some details. First of all the presentation of $M$ is a complete rewriting system (it is length reducing and there are no overlaps of rules) and the nonzero reduced elements are the elements $pq^*$ with $p,q$ words in $X$. To see that $R=K_0M$ is simple, let $I$ be a nonzero ideal and let $$0\neq a=\sum_{m\in M}c_mm\in I.$$ We shall use the following observation. Use $|w|$ for the length of a word $w$ over $X$. If $p,q,w,a,b$ are words in $X$ and $0\neq pq^*w=sb^*w$ with $|b|,|q|\geq |w|$, then $pq^*=sb^*$. Indeed, we must have $q=wz$ and $b=wy$ for the product to be nonzero and then $pq^*w = pz^*$ and $sb^*w = sy^*$, whence $p=s$ and $z=y$, i.e., $q=b$. First suppose that $supp(a)$ contains $1$. Since $X$ is infinite, we can find a letter $x\in X$ such that $x$ appears in no $p,q$ with $pq^*\in supp(a)$. Then $x^*ax=x^*1x=1$ as $x^*pq^*x=0$ whenever $x$ does not appear in $p, q$ and at least one of $p$ or $q $ is nonempty. Next we show that we can modify $a$ to contain $1$ in its support. Suppose that some word $p$ over $X$ is in the support of $a$ and assume $p$ has minimal length with this property. Then $p^*a$ has $1$ in its support and it is nonzero since $p^*wz^*=1$ in $M$ implies $z=1$ and $w=p$, and so the coefficient of $1$ in $p^*a$ is the coefficient of $p$ in $a$. Thus we are in the previous case, and hence done. So suppose that all elements of the support of $a$ are of the from $pq^*$ with $q$ nonempty. Choose $pq^*$ in the support of $a$ with $|q|$ minimal. By our observation above, the coefficient of $p$ in $aq$ is the same as the coefficient of $pq^*$ in $a$ because $|q|\leq |b|$ for any $sb^*$ in the support of $a$ and so $sb^*q\neq pq^*q=p$ by the observation. This puts us in the previous case, and hence we are done. It now follows that $R$ is simple. Next lets check that $d$ is a derivation. Call a mapping $\theta\colon M\setminus \{0\}\to G$ with $G$ a group a partial homomorphism if $\theta(mn)=\theta(m)\theta(n)$ whenever $mn\neq 0$. For example, there is a partial homomorphism $\theta\colon M\to F_X$ (where $F_X$ is the free group on $X$) with $\theta(pq^*) = pq^{-1}$, and this is in fact the universal partial homomorphism from $M$ to a group, although we don't need this. Now there is a homomorphism $\eta\colon F_X\to \mathbb Z$ sending our fixed $x$ to $1$ and $X\setminus \{x\}$ to $0$. Then $\nu:=\eta\theta\colon M\to \mathbb Z$ is a partial homomorphism with $\nu(pq^*) = |p|_x-|q|_x$. Therefore, $d(m) = \nu(m)m$ for $m\in M$, where we consider $\nu(0)0=0$ (even though technically $\nu(0)$ is not defined). Then we easily check that, for $m,n\in M$, we have $$d(mn) = \nu(mn)mn = (\nu(m)+\nu(n))mn = m\nu(n)n+\nu(m)mn = md(n)+d(m)n$$ where this works out even if $mn=0$. Finally, we observe that $d$ is not inner. Indeed, if $d$ is inner then $x=d(x)=xr-rx$ for some $r\in R$. Note that since $1$ commutes with $x$, we may assume without loss of generality that $1\notin supp(r)$. Since left multiplication by $x$ is injective on $M$ (because $x^*x=1$), we cannot have $x\in x\cdot supp(r)$ because $1\notin supp(r)$. Thus $x\in supp(r)x$. But the only way that can happen is if $xx^*\in supp(r)$ since if $pq^*x=x$ and $p\neq 1$, then $q=p=x$ (use the observation). Let $n\geq 1$ be maximal with $(x^n)(x^n)^*\in supp(r)$. Then $xr$ has $(x^{n+1})(x^n)^*$ in its support (since left multiplication by $x$ is injective) and hence $(x^{n+1})(x^n)^*\in supp(r)x$ since $xr-rx=x$. But if $pq^*\in supp(r)$ with $pq^*x = (x^{n+1})(x^n)^*$ with $n\geq 1$, then $|q|\geq 2$ and the only possibility is $p=x^{n+1}$, $q=x^{n+1}$ by the observation. Thus $x^{n+1}(x^{n+1})^*\in supp(r)$, contradicting the choice of $n$. We deduce that $d$ is not inner.<|endoftext|> TITLE: Non-commutative knot invariants QUESTION [5 upvotes]: $\newcommand{\ab}{\mathrm{ab}}$Let $L=K_1\cup \dots \cup K_r$ be a link embedded in a 3-sphere. Here, $K_1,\dots, K_r$ are the component knots of $L$. A prototypical invariant associated with $L$ is its Alexander polynomial. This is a polynomial in $\Lambda_r=\mathbb{Z}[T_1^{\pm 1}, \dots, T_r^{\pm 1}]$, the Laurent series ring in $r$-variables. It is defined as follows. Let $X_L$ be the complement of $L$ in $S^3$ and $\widetilde{X}_L^{\ab}\rightarrow X_L$ be its universal abelian cover. The Deck group of this cover is the abelianization of $\pi_1(X_L)$, which by the Hurewicz theorem is isomorphic to $H_1(X_L;\mathbb{Z})$. This homology group is generated by $x_1,\dots, x_r$, where $x_i$ is the class associated with the meridian looping around that $i$-th knot $K_i$. The Alexander module is $H_1(\widetilde{X}_L^{\ab}, \mathcal{F}_{x_0};\mathbb{Z})$, where $x_0$ is a chosen base-point of $X_L$ (where all meridians begin and end at), and $\mathcal{F}_{x_0}$ its fibre in $\widetilde{X}_L^{\ab}$. This becomes a module over $\pi_1(X_L)^{\ab}$ and hence, is a module over $\Lambda_r$ as well. The multivariable Alexander polynomial is then defined to be the 1-st elementary ideal of the Alexander module. An account of this can be found in any basic text in knot theory. Fix a prime $p$. There is a p-adic version of this construction due to Hillman, Matei, and Morishita, an account of which can be found in the book "Knots and primes" by Morishita. The p-adic analog of the Alexander polynomial is defined in the power series ring $\widehat{\Lambda}_r:=\mathbb{Z}_p[\![X_1,\dots, X_r]\!]$ and is the topological equivalent of the Iwasawa polynomial defined for number fields. In fact, there is an asymptotic formula for the growth of $p$-parts of homology classes in various $\mathbb{Z}/p^n\mathbb{Z}$-covers that can be determined in terms of Iwasawa invariants $\mu,\lambda$ and $\nu$, once a $\mathbb{Z}_p$-tower is fixed. For more information, see the last few chapters in Morishita's book, or the paper of Hillman-Matei-Morishita. Given these developments, it is natural to extend the analogy between number theory and topology further. In Iwasawa theory, it is natural to work with non-commutative towers and extensions and study non-commutative versions of Iwasawa invariants and polynomials. This has been done for number fields. There is a nice book on the subject called "Noncommutative Main conjectures over totally real fields" for instance. Perhaps one can study non-commutative analogs of the Alexander polynomial which will have significant knot theoretic applications as well, and perhaps one can compute these in some examples. After all Alexander polynomials can be computed using a variety of techniques, like using Seifert surfaces. I wonder how difficult it would be to tackle such questions since, after all, I'm not a knot theorist. But I don't see anything in the literature that suggests this problem has been studied. REPLY [2 votes]: Perhaps one can study non-commutative analogs of the Alexander polynomial which will have significant knot theoretic applications as well, and perhaps one can compute these in some examples. After all Alexander polynomials can be computed using a variety of techniques, like using Seifert surfaces. I think one of the major papers on the topic is Cochran's paper "non-commutative knot theory": https://arxiv.org/abs/math/0206258. Other papers on the topic includes work of Harvey (e.g. https://arxiv.org/abs/math/0207014), Friedl-Harvey (https://arxiv.org/abs/math/0608409). This is not an exhaustive list. There are also many papers on non-commutative/higher order signatures and Blanchfield forms.<|endoftext|> TITLE: Potential counterexamples to Bass' trace conjecture QUESTION [5 upvotes]: Motivation: The following is a theorem of Berrick-Hesselholt (essentially also due to Linnell, though not in this form): Let $G$ be a group. Suppose that for every subgroup of $G$ isomorphic to $\mathbb Q$, $G$ has a quotient in which the image of this subgroup is central and nontrivial. In this case the Bass trace conjecture holds for $G$. I can add it for context, but for my question it is not important to know what this conjecture is - I just state this as motivation. Question : What are some examples of finitely presented groups which do not have this property ? That is, they have a subgroup isomorphic to $\mathbb Q$ such that for each quotient of $G$, its image is central only if it is trivial. REPLY [9 votes]: a) There are old results which directly imply the existence of such groups: (1) Boone Higman 1972: every f.g. group with solvable word problem embeds into a simple subgroup of a finitely presented group. (2) Every countable group with solvable word problem embeds into a f.g. group with solvable word problem (reference? the original HNN construction directly works since it consists of explicit amalgams — alternatively here Ph. Hall produced in the 50s an explicit 3-generated metabelian group with solvable word problem, with copies of $\mathbf{Q}$). b) A more explicit example is the group $\tilde{T}$ obtained as the set of self-homeomorphisms of $\mathbf{R}/\mathbf{Z}$ commuting with $\sigma:n\mapsto n+1$, that are piecewise affine with dyadic slopes and breakpoints. The center of $\tilde{T}$ is the infinite cyclic group $\langle\sigma\rangle$ and the quotient is naturally identified with Thompson's group $T$, which is a finitely presented simple group. The normal proper subgroups of $\tilde{T}$ are precisely the subgroups of $\langle\sigma\rangle$. If $Q$ is any copy of $\mathbf{Q}$ in $\tilde{T}$, it follows that the image of $Q$ in $T$ is non-central (since $\mathbf{Q}$ has no nontrivial cyclic quotient). Hence the same holds in every nontrivial quotient of $\tilde{T}$. Finally, that $\tilde{T}$ contains a copy of $\mathbf{Q}$ (and even continuum many such copies) is an original observation of Belk, Matucci, Hyde (arXiv). (The other answer is closely related as it refers to a more complicated finitely presented simple group containing $\tilde{T}$. The group $\tilde{T}$ itself is not finitely presented but has few enough normal subgroups for the condition to hold.) REPLY [8 votes]: As explained in the comments, Jim Belk's answer to this question answers mine as well : he points ou that the group $T\mathcal A$ he constructed with his coauthors is finitely presented, simple and contains a copy of $\mathbb Q$. In particular, it has no quotient where $\mathbb Q$ can be central and non trivial, so this answers the question.<|endoftext|> TITLE: When does the diagonal cohomology class of a non-compact oriented manifold vanish? QUESTION [9 upvotes]: Let $M$ be a non-compact, connected and oriented topological $d$-manifold without boundary. My understanding is that there are two (equivalent) ways of defining the diagonal class $\delta_M \in H^{d}(M \times M; \mathbb{Z})$: Consider the fundamental class $[M] \in H^{lf}_d(M;\mathbb{Z})$ in locally finite (=Borel-Moore) homology. The diagonal map $M \rightarrow M \times M$ is proper, hence induces a map in locally finite homology. From here we get a class $$[M] \mapsto \Delta_M \in H^{lf}_d(M \times M;\mathbb{Z})\,.$$ By Poincaré duality in the connected oriented 2d-manifold $M \times M$ we get a dual class $\delta_M \in H^d(M \times M;\mathbb{Z})$, that is, $\delta_M$ is the unique class with $\delta_M \cap [M \times M] = \Delta_M$. Take $\delta_M$ as the image of 1 under the Gysin map $\mathbb{Z} \cong H^0(M;\mathbb{Z}) \rightarrow H^d(M \times M;\mathbb{Z})$ induced by the diagonal map. Question: Is there a characterization of the class of manifolds $\mathcal{C} := \{M: \delta_M = 0\}$? One obvious sufficient condition for $M \in \mathcal{C}$ is that $M$ has no cohomology to begin with, e.g. is contractible. Second, $\delta_{M \times N} = \delta_M \times \delta_N$ under the cohomological cross product so if $M \in \mathcal{C}$ then $M \times N \in \mathcal{C}$ for any $N$. Third, for any open subset $U \subseteq M \in \mathcal{C}$ we have $U \in \mathcal{C}$ because in general $\delta_M$ pulls back to $\delta_U$. Are there other easy to check sufficient or necessary conditions? REPLY [4 votes]: There are some closely related notions appearing the first part of "Espaces de configuration généralisés - Espaces topologiques i-acycliques - Suites spectrales basiques" by Arabia. Note first that an equivalent condition is that the Gysin map vanishes in all degrees, i.e. $H^\ast(M) \to H^{\ast+d}(M \times M)$ is zero. This is because the Gysin image of a cohomology class $\alpha$ is the cup-product of $\delta_M$ with the pullback of $\alpha$ to one of the factors. Arabia considers the linear dual condition: he defines a topological space to be $\cup$-acyclic if the cup product in compact support cohomology is identically zero. On an oriented manifold this just means then that the intersection product on homology $H_{\ast+d}(M \times M) \to H_\ast(M)$ is zero. With field coefficients this is equivalent to your condition. Proposition 1.2.4 of his paper gives some more characterizations of this property, in particular that on an oriented manifold it is equivalent to asking that $H^\ast_c(X)\to H^\ast(X)$ vanishes. He also writes down exactly the same examples as you do.<|endoftext|> TITLE: Existence of a strongly continuous topologically irreducible representation of a compact group on an infinite dimensional Banach space? QUESTION [10 upvotes]: Does there exists a triple $(G, X, \pi)$, where $G$ is a compact group, $X$ an infinite dimensional Banach space over $\mathbf{C}$, and $\pi : G \to B(X)$ a strongly continuous representation of $G$, such that if $Y$ is a non-zero closed invariant subspace of $X$ in the sense that $\pi(g)Y \subseteq Y$ for all $g \in G$, then $Y = X$ (we call nonzero strongly representations of $G$ satisfying this property topologically irreducible)? I am not sure that if this is settled already, as the question seems natural, yet I am ignorant of the relevant literature, nor could I come up with a proof or disproof. Let me be (perhaps overly) precise about the question asked in the title. By compact group, I mean Hausdorff plus quasi-compact, and topologically irreducible means that the only invariant closed subspaces are either $0$ or the whole space. More generally, can we find such an example by replacing infinite dimensional Banach spaces by complete infinite dimensional locally convex Hausdorff spaces instead? While we are on the subject, I'd also appreciate some recommendation of books, papers/surveys on representations of topological groups on Banach spaces that is not focused on unitary representations on Hilbert spaces? REPLY [5 votes]: Another reference is Hofmann, Morris, "The structure of compact groups", Zbl 1277.22001 (There is a 4th edition, but my library has access to this 3rd edition which I'll give references to). Chapter 3 considers this question, and the "Big Peter and Weyl Theorem", Theorem 3.51, is exactly the result which Mikael de la Salle mentions, but is here stated and proved for any "$G$-complete locally convex" space. In particular, this applies to any complete locally convex topological space, see page 72. (There are some very brief references given, but no bibliographic notes, so I don't know where this result was first proved.)<|endoftext|> TITLE: Area-differences for lattice triangles in a checkerboard QUESTION [27 upvotes]: For positive integers $m$ and $n$, what is the integral of the function $(-1)^{\lfloor x \rfloor + \lfloor y \rfloor}$ on the triangle with vertices $(0,0)$, $(m,0)$, and $(0,n)$? Pictorially, we are putting a red/black checkerboard coloring on the plane and finding the signed difference between the red region enclosed by the triangle and the black region enclosed by the triangle. Call this integral $I(m,n)$. Here are some values of $I(m,n)$ (if my calculations of signed sums of areas of little triangles and trapezoids are correct): $\begin{array}{c|ccccc} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1/2 & 1/2 & 1/2 & 1/2 & 1/2 \\ 2 & 1/2 & 0 & -1/6 & 0 & 1/10 \\ 3 & 1/2 & -1/6 & 1/2 & 5/6 & 1/2 \\ 4 & 1/2 & 0 & 5/6 & 0 & -1/2 \\ 5 & 1/2 & 1/10 & 1/2 & -1/2 & 1/2 \end{array}$ It would be good to tabulate more values. Can any computer algebra systems handle such computations for specific $m,n$? Obviously $I(m,n) = I(n,m)$, and an easy telescoping-sum argument shows that $I(1,n)=1/2$. A symmetry argument shows that $I(m,n)=1/2$ when $m$ and $n$ are both odd, and from this it can be proved that $I(m,n)=0$ when $m$ and $n$ are both even. The values of $I(m,n)$ where $m$ and $n$ have opposite parity seem more subtle. REPLY [2 votes]: Some people have noticed that $mnI(m,n)$ is an integer when $m, n$ opposite parity. It is actually proved in jcdornano's reply with a nice argument I think. As it might not be very clearly expressed, I took the liberty to rewrite it here. Let $m$ be even number and let $T$ be the triangle of vertices $(0,0)$, $(m,0)$, and $(0,n)$ with a black and white coloring $C$ as explained in the initial post. Let us consider $L$ the linear application that multiplies the absiceas by $n$ and the ordinates by $m$ and let us consider $T' $ the image of $T$ by $L$. $L$ naturally induces a coloring $C'$ on $T'$: the images of the black subset of $T$ is a black subset of $T'$ and the images of the white subset of $T$ is a white subset of $T'$. Language convention: Let $X$ be any black and white coloring of the plane. we can associates to $X$ a function $f $ whose value is $1$ on the white parts and $-1$ on the black parts. By abuse, we will speak of the integral of $X$ to mean the itegral of $f$. Let $I' $ be the integral of $C'$ on $T'$. It is immediate that $I(n,m)=I'/mn$. To prove that $I(n,m).nm$ is an integer, we just have to prove that $ I' $ is an integer. Let us prove this fact. Since $T'$ is isocele, the points of its hypothenus with integer ordinate also have an integer absicea. Let us denote these points by $A_0, A_1, A_2, ..., A_{nm}$. For any $0\leq i\leq nm-1$, let $T_i$ be the right triangle whose hypothenuse is $[A_i A_{i+1}]$. $T'$ is the union of : $\cup_{i=1}^{nm} T_i$ $S = T'\setminus \cup_{i=1}^{nm}T_i$. Let $I_1$ be the integal of $C'$ on $S$ and let $I_2$ be the integal of $C'$ on $ \cup_{i=1}^{nm} T_i$. $I' = I_1+I_2$. Let us prove that $I_1=0 $ et that $ I_2$ is an integer. Because the points $A_i$ have integers coordinates, $S$ is the union of squares of the form $[j,j+1]*[k,k+1]$ where $j,k\in \mathbb N$. Since such squares are either entirerly white or entirerly black, the value of the integral of $C'$ on them is $1 $ or $-1$ and the integral of C' on S in an integer. Let us now prove that $I_2 $ is equal to $0$. Because $T' $ is an isocele triangle , all the triangles $T_i $ are isometrics one to the other. There are an even number of them with half of them being black while the other half being white. Hence they "compensate" each other and $I_2=0$.<|endoftext|> TITLE: Is every finite subgroup the integer points of a linear algebraic group? QUESTION [6 upvotes]: Cross Posting this from MSE since it's been there for almost a month and it got a couple upvotes but no answers. MSE link Is every finite subgroup the integer points of a linear algebraic group? Let $ K $ be a compact connected Lie group. For every finite subgroup $ \Gamma $ of $ K $ does there exist a linear algebraic group $ G $ such that the integer points are $$ G_\mathbb{Z} \cong \Gamma $$ and the real points are $$ G_\mathbb{R} \cong K. $$ I'm interested in this because sometimes the integer points are cool like $$ \operatorname{SO}_3(\mathbb{Z}) \cong S_4. $$ EDIT: Here is an attempt to clarify what I am looking for. Consider 3 by 3 matrices with complex entries. For a $ 3 \times 3 $ complex matrix the conditions $$ I=MM^T $$ and $$ det(M)=1 $$ are polynomial in the entries of $ M $. The polynomials defining these conditions all have integer coefficients. The subset of matrices that satisfy these two constraints is the Lie group $ SO_3(\mathbb{C}) $. Now if we restrict the entries to be real then we get exactly the group $ SO_3(\mathbb{R}) $ which is a compact connected Lie group. Finally, if we restrict the entries to be integers we get $ SO_3(\mathbb{Z}) $ which is a finite group with 24 elements isomorphic to the symmetric group $ S_4 $. So what I was really interested in was the idea that for any compact connected lie group $ K $ and finite subgroup $ \Gamma $ we can find a (finite collection of integer coefficient) polynomial constraints on the entries of a square matrix such that the complex matrices satisfying those constraints form a Lie group, the matrices satisfying those constraints and having real entries form a compact connected Lie group, and finally the matrices satisfying those constraints and having integer entries form a finite group isomorphic to $ \Gamma $. REPLY [6 votes]: The answer to this question is no. There exist finite groups $\Gamma \subset K$ of a compact connected Lie group $K$, such that for any algebraic $\mathbb Q$-group $G$ with $G(\mathbb R)=K$, the finite group $\Gamma $ cannot be a subgroup of $G(\mathbb Z)$: We take $K=SU(2)$ and $\Gamma $ to be a Dihedral group of the form $(\mathbb Z/l\mathbb Z)\rtimes {\mathbb Z}/2{\mathbb Z}$ where the nontrivial element of the group ${\mathbb Z}/2{\mathbb Z}$ operates by $x\mapsto -x$ on ${\mathbb Z}/l{\mathbb Z}$. Here $l$ is a large prime. Suppose $G$ is an algebraic group defined over $\mathbb Q$ such that $G({\mathbb R})=K$, and $\Gamma \subset G(\mathbb Z)$. Then for almost all primes $p$, $\Gamma $ injects into $G({\mathbb F}_p)=G(\mathbb Z/p\mathbb Z)$. Moreover, for almost all primes $p$, the order of $G({\mathbb F}_p)$ is $(p^2-1)(p^2-p)/(p-1)=p(p^2-1)$. Further, $l$ divides this order since $\Gamma $ is a subgroup of $G({\mathbb F}_p)$. Therefore, for almost all primes $p$, we have $p(p^2-1)\equiv 0 \quad (mod \quad l)$. But by Dirichlet's theorem on primes in arithmetic progressions, the residue class of the generator of the unit group of ${\mathbb Z}/l{\mathbb Z}$ is represented by infinitely many primes $p$. Hence the order of such a $p$ (modulo $l$) is $l-1$. On the other hand $p(p^2-1)$ is divisible by $l$ which means that $l-1\leq 2$, and $l$ cannot be a large prime. I am pretty sure that a much simpler proof can be found, but this is "a proof". ADDED later: The proof shows that the "large prime" $l$ need only satisfy $l\geq 5$. Moreover, the gcd of the numbers $p(p^2-1)$ as $p$ varies over primes large enough, is just $24$. Hence the order of $\Gamma $ is $\leq 24$.<|endoftext|> TITLE: Is a $*$-automorphism $M(A) \to M(A)$ automatically strictly continuous? QUESTION [9 upvotes]: Let $A$ be a (non-unital) $C^*$-algebra with multiplier $C^*$-algebra $M(A)$. Let $\phi: M(A) \to M(A)$ be a $*$-automorphism. Is it true that $\phi$ is automatically strictly continuous (on bounded subsets)? Some remarks/observations: (1) If $A = B_0(H)$, then this is true because $*$-automorphisms of $B(H) = M(B_0(H))$ are automatically strict (as they are given by conjugation with a unitary). (2) If $A$ is separable, this is true due to a result by Woronowicz which says that $$A= \{x \in M(A): xM(A) \mathrm{\ is \ separable}\}$$ so that we can reconstruct $A$ from its multiplier $C^*$-algebra $A$. (3) I tried to see what happens in the commutative case, so $A=C_0(X)$. Then a $*$-automorphism of $M(A) = C_b(X)= C(\beta X)$ corresponds to a homeomorphism $\beta X \to \beta X$. I have hope that if a counterexample exists, then a smart example of such a homeomorphism can lead to a counterexample. (4) The following question seems to be related: Does a strict $*$-automorphism $\phi: M(A) \to M(A)$ preserve the subalgebra $A$, i.e. do we have $\phi(A)\subseteq A?$ REPLY [5 votes]: I think that the answer is no. Let $\mu$ be a non-trivial homeomorphism of $\beta \bf N$ with distinct points $y,z\in \beta\bf N\setminus \bf N$ such that $\mu(y)=z$ and $\mu(z)=y$. Set $A=\{f\in C(\beta{\bf N} ): f(y)=0\}$. Then $M(A)=C(\beta {\bf N})$. Let $\phi: M(A)\to M(A)$ be given by $\phi(f)(x)=f(\mu(x))$ $(f\in C(\beta {\bf N}), x\in \beta {\bf N})$. Let $(f_{\alpha})$ be a bounded approximate identity for $A$. Then $(f_{\alpha})$ converges strictly to $1\in M(A)$ but $(\phi(f_{\alpha}))$ does not converge strictly to $\phi(1)=1$. To see this, take $g\in A$ such that $g(z)=1$. Then $\phi(f_{\alpha})(z)g(z)=\phi(f_{\alpha})(z)=f_{\alpha}(\mu(z))=f_{\alpha}(y)=0$ but $\phi(1)(z)g(z)=1$. Hence $\phi(f_{\alpha})g$ does not converge in norm to $g$.<|endoftext|> TITLE: Condensed math and cofiltered limits QUESTION [6 upvotes]: I have a question involving preservation of cofiltered limits. Ordinarily this would be a very boring question, but it comes up in condensed math in its analogue of the completeness concept. The concept is that of "solid", which says that for each profinite set $S = \text{lim} S_i$, maps of condensed abelian groups $\mathbb{Z}[S] \rightarrow M$ lift against the canonical map $\mathbb{Z}[S] \rightarrow \text{lim} \mathbb{Z}[S_i]$. One broad goal of condensed math is to extend commutative algebra to subsume some areas of functional analysis. The concept of solid plays a role similar to completion. But I don't as of yet understand the condition very well. I don't understand why the free condensed abelian group $\mathbb{Z}[S]$ is not identically $\text{lim} \mathbb{Z}[S_i]$. I am trying to come up with a broad class of examples as follows: Let $\mathcal{S}$ be a site. Consider the category of sheaves of sets on the site $\mathcal{S}$, $\text{Sh}(\mathcal{S}, \text{Set})$, and the category of sheaves of abelian groups $\text{Sh}(\mathcal{S}, \text{Ab})$ on the site $\mathcal{S}$. I am looking for a general class of examples showing that the free functor $\text{Sh}(\mathcal{S}, \text{Set}) \rightarrow \text{Sh}(\mathcal{S}, \text{Ab})$ does not preserve cofiltered limits. REPLY [10 votes]: It should be noted that already in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)} \colon \mathbf{Set} \to \mathbf{Ab}$ does not preserve cofiltered limits. For a cofiltered diagram $D \colon \mathcal I \to \mathbf{Set}$, write $S_i$ for its value at $i \in \mathcal I$, write $S$ for its limit, and write $\pi_i \colon S \to S_i$ for the canonical projection. There is always a map \begin{align*} \phi \colon \mathbf Z^{(S)} &\to \lim_\leftarrow \mathbf Z^{(S_i)}\\ \sum_{k=1}^n n_k s_k &\mapsto \left( \sum_{k=1}^n n_k\pi_i(s_k) \right)_i. \end{align*} Denote its $i^{\operatorname{th}}$ component by $\phi_i$. For a set $X$ and an element $z=\sum_{k=1}^n n_k x_k \in \mathbf Z^{(X)}$ with $x_k \neq x_{k'}$ for $k \neq k'$, write $\operatorname{Supp}(z)$ for $\{x_1,\ldots,x_n\}$, and denote by $|z|$ its cardinality. If $f \colon X \to Y$ is a map, we denote the induced map $\mathbf Z^{(X)} \to \mathbf Z^{(Y)}$ by $f$ as well (by abuse of notation). For $z \in \mathbf Z^{(X)}$, we have $\operatorname{Supp}(f(z)) \subseteq f(\operatorname{Supp}(z))$, so $|f(z)| \leq |z|$ with equality if and only if $f_* \colon \operatorname{Supp}(z) \to \operatorname{Supp}(f(z))$ is a bijection. Lemma. The map $\phi$ is injective, and its image consists of those $(x_i)_{i \in \mathcal I}$ for which there exists $n \in \mathbf Z$ with $|x_i| \leq n$ for all $i \in \mathcal I$. In other words, the image consists of the sequences $(x_i)_i$ of bounded support. Proof. For injectivity, if $x = \sum_{k=1}^n n_ks_k \in \ker(\phi)$ is such that $s_k \neq s_{k'}$ for $k \neq k'$, then there exists $i \in \mathcal I$ such that $\pi_i(s_k) \neq \pi_i(s_{k'})$ for $k \neq k'$ (here we use that the sum is finite and that $\mathcal I$ is cofiltered). Then $\phi_i(x) = \sum_{k=1}^n n_k\pi_i(s_k)$ is zero by assumption, so all $n_k$ are zero. For the image, it is clear that $|\phi_i(x)| \leq n$ for all $i \in \mathcal I$ if $x \in \mathbf Z^{(S)}$ has $|x| = n$. Conversely, if $|x_i| \leq n$ for all $i \in \mathcal I$, then decreasing $n$ if necessary, we may assume $|x_{i_0}| = n$ for some $i_0 \in \mathcal I$. Then $|x_i| = n$ for all $f \colon i \to i_0$ since $n = |x_{i_0}| = |D(f)(x_i)| \leq |x_i| \leq n$. Replacing $\mathcal I$ by the coinitial segment $\mathcal I/i_0$ we may therefore assume $|x_i| = n$ for all $i \in \mathcal I$. Constancy of $|x_i|$ means that for every morphism $f \colon i \to j$ in $\mathcal I$, the map $f_* \colon \operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$ is a bijection. Setting $T = \lim\limits_\leftarrow \operatorname{Supp}(x_i)$, we see that each projection $\pi_i \colon T \to \operatorname{Supp}(x_i)$ is a bijection. Functoriality of the limit gives an injection $T \hookrightarrow S$, giving elements $s_1,\ldots,s_n \in S$ such that $\pi_i(\{s_1,\ldots,s_n\}) = \operatorname{Supp}(x_i)$ for all $i \in \mathcal I$. The coefficients must also be constant under the bijections $\operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$, so we get an element $x = \sum_{k=1}^n n_ks_k \in \mathbf Z^{(S)}$ with $\phi(x) = (x_i)_i$. $\square$ Example. An example where $\phi$ is not surjective: let $S = \mathbf Z_3$ with $S_i = \mathbf Z/3^i$. Define the element $(x_i)_i \in \prod_i \mathbf Z^{(S_i)}$ where $x_i \in \mathbf Z^{(S_i)} = \mathbf Z^{S_i}$ has coordinates (for $k \in \{0,\ldots,3^i-1\}$) given by $$x_{i,k} = \begin{cases} 1, & \text{the first $3$-adic digit of } k \text{ is } 1, \\ -1, & \text{the first $3$-adic digit of } k \text{ is } 2, \\ 0, & k=0.\end{cases}$$ These form an element of $\lim\limits_\leftarrow \mathbf Z^{(S_i)}$: any fibre of $\mathbf Z/3^{i+1} \to \mathbf Z/3^i$ above $k \in \{0,\ldots,3^i-1\}$ consists of $\{k,3^i+k,2 \cdot 3^i+k\}$, of which $x_{i+1,k} = x_{i,k}$ and the others are $1$ and $-1$ since $3^i+k$ starts on $1$ and $2 \cdot 3^i+k$ starts on $2$. Since $\operatorname{Supp}(x_i) = S_i \setminus \{0\}$, we see that $(x_i)_i$ does not have bounded support, hence is not in the image of $\phi$. Corollary. For any presheaf topos $\mathbf T = \mathbf{PSh}(\mathscr C) = [\mathscr C^{\operatorname{op}},\mathbf{Set}]$ on a small nonempty category $\mathscr C$, the free functor $\mathbf Z^{(-)} \colon \mathbf T \to \mathbf{Ab}(\mathbf T)$ does not preserve cofiltered limits. Proof. Let $f \colon \mathscr C \to *$ be the map to the terminal category $*$, on which $\mathbf{PSh}(*) = \mathbf{Set}$. Note that $f$ has a section $g$ since $\mathscr C$ is nonempty. We saw above that in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)}$ does not preserve cofiltered limits. The pullback $f^* \colon \mathbf{Set} \to \mathbf{PSh}(\mathscr C)$ takes a set $S$ to the constant sheaf $\underline{S}$ on $\mathscr C$. Since limits, colimits, and free abelian group objects in presheaf categories are pointwise, $f^*$ commutes with formation of limits, colimits, and free abelian group objects. Thus pulling back everything along $f^*$ gives the natural map $\underline{\mathbf Z}^{(\underline S)} \to \lim\limits_\leftarrow \underline{\mathbf Z}^{(\underline S_i)}$, which is not an isomorphism since it isn't after applying the section $g^*$ (this is just evaluation at an object $g(*) \in \mathscr C$). $\square$ The class of topoi where the free functor does not commute with cofiltered limits is probably much larger still. For instance, my example does not include condensed sets, which is close to a presheaf topos but not quite. I'm not even sure what a topos would look like where these do commute!<|endoftext|> TITLE: How much induction does a p-adic valuation need? QUESTION [11 upvotes]: Recently I learned a nice constructive proof of the irrationality of $\sqrt{2}$, which uses the 2-adic valuation of an integer: the count of how many times a number is divisible by 2. The valuation requires some induction to construct, and this nice answer by François Dorais talks about how Robinson's Arithmetic $Q$ isn't strong enough to prove $\sqrt{2}$ irrational. By the question "how much induction...?" I mean what is the complexity of the statement that is used in the application of induction to prove the existence of the valuation (I think the particular case $p=2$ is not special here). Further, I think the only property really needed in this irrationality proof is that the parity of the valuation is well-defined, so in principle it is this specific property that I need to know the strength of: there is a well-defined multiplicative function $p_2\colon \mathbb{N}\to \{\pm 1\}$ encoding the parity of the 2-adic valuation. I can easily think of a recursion (say in some dependent type theory, or a proof assistent) that defines this function, but I don't know how to classify the precise strength of the induction principle needed, in the usual arithmetic hierarchy. [As an aside, I really like this proof, not just because it gives a constructive lower bound on how far a rational is from $\sqrt{2}$, but also because it doesn't rely on more extensive factorisation properties of integers, like one of the most common proofs relying on fractions in 'lowest terms', or on the beautiful, but more subtle, use of infinite descent] REPLY [9 votes]: If you want to stick to theories in the basic language of arithmetic $\langle0,1,+,\cdot,<\rangle$, the irrationality of $\sqrt2$ can be easily proved in the theory $IE_1$ (i.e., using induction for bounded existential formulas), since it proves the $\gcd$ property; or even more directly, you can just prove $$\forall a,b TITLE: Are there functions $\mathbb{F}_2^n \to \mathbb{F}_2$ satisfying these special relations? QUESTION [9 upvotes]: Let $\mathbb{F}_2=\{0,1\}$ be the field with two elements, and let $u:\mathbb{F}_2^n\rightarrow \mathbb{F}_2$. Suppose that $n$ is odd. Is it possible that $$ \sum_{x \in \mathbb{F}_2^n}(-1)^{u(x)+u(x+a)}= 0, $$ for every $a \neq 0$ in $\mathbb{F}_2^n$? I treat the sum here as natural number, not as an element of $\mathbb{F}_2$. This question arose in the context of this question. (Trying to derive a lower bound on the approximate multiplicativity of a Boolean function). When $n$ is even there are such functions; this is related to Bent functions. REPLY [12 votes]: I think the answer is no. To see this, observe that \begin{equation*} \left(\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2=\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\mathbb{E}_{\mathbf{y}}[(-1)^{u(\mathbf{y})}]=\mathbb{E}_{\mathbf{x},\mathbf{y}}[(-1)^{u(\mathbf{x})+u(\mathbf{y})}]=\mathbb{E}_{\mathbf{x},\mathbf{a}}[(-1)^{u(\mathbf{x})+u(\mathbf{x}+\mathbf{a})}], \end{equation*} where we use independence and then rewriting $\mathbf{y}=\mathbf{x}+\mathbf{a}$, and where all expectations are uniform over $\mathbb{F}_2^n$. The assumption then is that for $\mathbf{a}=\mathbf{0}$, the expectation over $\mathbf{x}$ is clearly $1$, while for all other values of $\mathbf{a}$, the expectation over $\mathbf{x}$ is $0$. It follows that \begin{equation*} \left(\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2 = \frac{1}{2^n}\iff \mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}] = \frac{\pm 1}{2^{n/2}} \end{equation*} However, it's not hard to see that this expectation must also be an integer multiple of $1/2^n$. But then this integer must be $\pm 2^{n/2}$, but this is integral if and only if $n$ is even.<|endoftext|> TITLE: Proving short consistency: can we do better than brute force search? QUESTION [6 upvotes]: This is a minor variation of a question originally asked on MSE by user779130 and bountied by me, without success. Throughout, "length" refers to the number of symbols, not lines, in a proof. For $T$ an "appropriate" theory, let $p_T(n)$ be the length of the shortest $T$-proof of "There is no proof of $0=1$ in $\mathsf{PA}$ of length $\epsilon>0$ such that $$n^\epsilonn^\epsilon,$$ but that’s just about all we know for a fact. We have every reason to expect that if $T$ is sufficiently stronger than $S$, then the trivial upper bound $\eqref{ub}$ cannot be significantly improved, i.e., $p_{S,T}$ is exponential. It seems likely that this should hold already when $T\supseteq S+\con_S$. However, we cannot even prove the much weaker statement Conjecture 2: For every $S$, there is $T$ such that $p_{S,T}$ is not polynomially bounded. In fact, Krajíček and Pudlák [3] proved that Conjecture 2 is equivalent to Conjecture 3: There is no optimal propositional proof system. Here, a propositional proof system (pps) is a polynomial-time predicate $P(\pi,\phi)$ (meaning “$\pi$ is a $P$-proof of $\phi$”) such that $$\phi\in\def\taut{\mathrm{TAUT}}\taut\iff\exists\pi\:P(\pi,\phi),$$ where $\taut$ is the set of all classical propositional tautologies. A pps $P$ simulates a pps $Q$ if there exists a polynomial $p$ such that $$Q(\pi_Q,\phi)\implies\exists\pi_P\:\bigl(|\pi_P|\le p(|\pi_Q|+|\phi|)\land P(\pi_P,\phi)\bigr),$$ and a pps $P$ is optimal if it simulates any other pps. A pps $P$ is polynomially bounded if there is a polynomial $p$ such that $$\phi\in\taut\implies\exists\pi\:\bigl(|\pi|\le p(|\phi|)\land P(\pi,\phi)\bigr).$$ Any polynomially bounded pps is clearly optimal, but the converse is not necessarily true. It is generally assumed that every pps requires proofs of exponential size to prove some tautologies. However, we are far from proving this; the weaker statement that there are no polynomially bounded pps is equivalent to the famous $\mathrm{NP\ne coNP}$ problem. We can only prove unconditional superpolynomial (or even exponential) lower bounds on weak proof systems such as Resolution. No nontrivial lower bounds are known even for the Frege proof system (called Hilbert outside proof complexity), which is the textbook proof system using a finite set of axiom schemata and schematic rules (modus ponens). The proof of the equivalence of Conjecture 2 and Conjecture 3 works roughly as follows. On the one hand, for any theory $S$, we can define a proof system $Q_S$ by $$Q_S(\pi,\phi)\iff \pi\text{ is an $S$-proof of “$\phi\in\taut$”}$$ (this is called the strong proof system of $S$ in Pudlák [4]) and we show that if $S$ violates Conjecture 2, then $Q_S$ is optimal. On the other hand, if $P$ is an optimal pps, then the theory axiomatized by the reflection principle $$S=\pv+\forall\pi,\phi\:(P(\pi,\phi)\to\phi\in\taut)$$ violates Conjecture 2. As already mentioned, it is consistent with current knowledge that Frege is polynomially bounded, and therefore optimal. Since the reflection principle for Frege is provable already in the base theory $\pv$, it follows that we cannot at present even disprove the statement $p_{S,T}$ is polynomially bounded for all $S$ and $T$. Thus, for example, we cannot prove anything about $p_{\pra,\pa}$ besides the $n^\epsilon$ lower bound. See also [5] for a discussion of statements related to Conjectures 2 and 3. References: [1] Pavel Pudlák: On the length of proofs of finitistic consistency statements in first order theories, in: Logic Colloquium '84 (J. B. Paris, A. J. Wilkie, G. M. Wilmers, eds.), Studies in Logic and the Foundations of Mathematics vol. 120, 1986, pp. 165–196, doi 10.1016/S0049-237X(08)70462-2. [2] Pavel Pudlák: Improved bounds to the length of proofs of finitistic consistency statements, in: Logic and Combinatorics (S. G. Simpson, ed.), Contemporary Mathematics vol. 65, AMS, 1987, pp. 309–331. [3] Jan Krajíček, Pavel Pudlák: Propositional proof systems, the consistency of first order theories and the complexity of computations, Journal of Symbolic Logic 54 (1989), no. 3, pp. 1063–1079, doi 10.2307/2274765, jstor 2274765. [4] Pavel Pudlák: Reflection principles, propositional proof systems, and theories, arXiv:2007.14835 [math.LO]. [5] Pavel Pudlák: Incompleteness in the finite domain, Bulletin of Symbolic Logic 23 (2017), no. 4, pp. 405–441, doi 10.1017/bsl.2017.32.<|endoftext|> TITLE: Topological K-theory of Riemann surface QUESTION [7 upvotes]: Let $X$ be a compact Riemann surface of genus $g$, then $K^1_{\mathrm{top}}(X)\cong\mathbb{Z}^{2g}$. Is there a explicit description of a set of basis of $K^1_{\mathrm{top}}$? (e.g., For cohomology $H^1(X,\mathbb{Z})\cong\mathbb{Z}^{2g}$ we may take the 1-cochains ``around the holes'') Furthermore, we define the Mukai vector of $\kappa\in K^1_{top}(X)$ to be $v(\kappa)=\mathrm{ch}(\kappa)\sqrt{\mathrm{td}(X)}$, and the Euler pairing on $K^*_{top}(X)$ by $\langle a,b\rangle=(v(a^\vee),v(b))$ where $(-,-)$ is the pairing in $H^*(X,\mathbb{Q})$. Do we know the pairing with respect to the basis? (The definition of $\mathrm{ch}\colon K^*(X)\to \oplus H^*(X)$ is given in https://www.maths.ed.ac.uk/~v1ranick/papers/ahvbh.pdf) REPLY [5 votes]: Following @Kiran's suggestion in the comments, I'll outline why the map $U(1)\to U$ induces an isomorphism between cohomology and K-theory in this setting. At the end I'll also explain a different perspective that might be helpful. The inclusion maps $U(n)\to U(n+1)$ are $(2n-1)$-connected, so the map $U(1)\to U$ is 1-connected. This means that for any 2-dimensional CW complex X, the induced map Map$(X, U(1))\to$ Map$(X, U)$ is $-1$-connected, and in particular a surjection on $\pi_0$ (I believe this fact about connectivity appears in May's Concise Course; it's proven by induction on skeleta of X). So $[X, U(1)]\to [X, U]$ is surjective. (This is a a borderline case of the result I'm quoting and I would recommend checking carefully that it does work... Note that there's no difference between based and unbased homotopy classes of maps in this setting, because the action of $\pi_1 U(1) = \pi_1 U$ is trivial on $\pi_* U(1)$ and on $\pi_* (U)$, as these are groups.) Now, say $X = M^g$, a closed Riemann surface of genus $g$. Knowing in advance that $[M^g, U(1)] = H^1 (M^g; \mathbb Z) = \mathbb{Z}^{2g} = K^1 (M^g) = [M^g, U]$, this surjection must be an isomorphism (note that the group structures are induced by the group structures on $U(1)$ and $U$, so the function between homotopy sets is a group homomorphism.) Another way of thinking about $K^1 (M^g)$ is to consider vector bundles over the suspension $\Sigma M^g$. Since the attaching map of the 2-cell in $M^g$ is a commutator in $\pi_1 (\bigvee_{2g} S^1)$, the attaching map of the 3-cell in $\Sigma M^g$ is a commutator in $\pi_2 (\bigvee_{2g} S^2)$, and hence is nullhomotopic. This means $\Sigma M^g \simeq (\bigvee_{2g} S^2) \vee S^3$, which gives $[\Sigma M^g, BU] \cong \pi_2 (BU)^{2g} \oplus \pi_3 (BU) = \mathbb{Z}^{2g}$ (and by Bott periodicity $[\Sigma M^g, BU] \cong [M^g, \Omega BU] \cong [M^g, U]$). See Tyrone's comment below for a way to make this explicit.<|endoftext|> TITLE: Hopf algebra with a non-invertible antipode QUESTION [9 upvotes]: What is an example of a Hopf algebra with a non-invertible antipode? REPLY [11 votes]: Theorem of Takeuchi (in Free Hopf algebras generated by coalgebras, 1971) asserts that free Hopf algebra $H(C)$ over a coalgebra $C$ has injective antipode, and it is bijective precisely (at least over alg. closed field) when $C$ is pointed. On the other hand, in a paper Faithful flatness over Hopf subalgebras - counterexamples, 2000 P. Schauenburg constructs a Hopf algebra with surjective, but non-injective antipode. Example is constructed as follows. There's a "free Hopf-with-bijective-antipode" functor from coalgebras to Hopf algebras; one can find a biideal in such free-with-bijective-antipode Hopf algebra over matrix coalgebra $M_4(k)$ that is stable under antipode, but not under its inverse, and quotient will be the example. Paper can be found on author's homepage http://schauenburg.perso.math.cnrs.fr/personnelle.html<|endoftext|> TITLE: In the not necessarily abelian cat setting, is there a Grothendieck spectral sequence for computing the homotopy of a composition of derived functors? QUESTION [8 upvotes]: Recall the Grothendieck spectral sequence which computes the homology groups of a composition of left derived functors $F$ and $G$ on abelian categories: \begin{align*} E_{p,q}^{2}(A)=L_{p}G\circ L_{q}F(A)\Rightarrow L_{p+q}(G\circ F)(A) \end{align*} (note the nLab link is written cohomologically instead). To consider derived functors on a not necessarily abelian category $\mathcal{C}$, we will adopt the language of animation as described in this paper due to Česnavičius and Scholze - in particular section 5. Namely, the animation of a category $\mathcal{C}$ is an $\infty$-category $\mathrm{Ani}(\mathcal{C})$ freely generated by $\mathcal{C}^{sfp}$, the strongly of finite presentation objects of $\mathcal{C}$, and a functor $F:\mathcal{C}\to\mathcal{D}$ which preserves sifted colimits has a unique animation $\mathrm{Ani}(F):\mathrm{Ani}(\mathcal{C})\to\mathrm{Ani}(\mathcal{D})$ which is computed on $\mathcal{C}^{sfp}$ and defined via Kan extension. For example, for a fixed ring $k$, the cotangent complex $L_{-/k}$ is the animation of the functor of Kahler differentials $\Omega_{-/k}$ on the nonabelian category of $k$-algebras, and $L_{A/k}$ is explicitly computed by taking a polynomial simplicial resolution (or more generally a cofibrant replacement) $P\to A$ and then $L_{A/k}:=\Omega_{P/k}\otimes_{P}A$. And in the abelian setting, animation is the derived functor; namely, given $F:\mathcal{C}\to\mathcal{D}$ which is right exact, a cofibrant replacement is a resolution by projective modules and $\mathrm{Ani}(F)$ is the functor $LF:D(\mathcal{C})\to D(\mathcal{D})$. For instance, $\mathrm{Ani}(-\otimes_{k}M)=-\otimes_{k}^{L}M$. The Grothendieck spectral sequence operates under the following hypotheses (Theorems 2.1 and 2.2 in the above nLab link): first, under certain "nice" behavior of $F$ and $G$, the composition of derived functors $LF\circ LG$ is the derived functor of the composition $L(F\circ G)$, and second, if $F$ takes projective objects (or more generally $F$-acyclic objects) to $G$-acyclic objects, then the Grothendieck spectral sequence exists, page two is defined as above, and abuts as above. Now let $F$ and $G$ be functors on not necessarily abelian categories. In the Česnavičius-Scholze paper linked above, at Proposition 5.1.5, one has similar "nice" behavior of $F$ and $G$ which will imply that the composition of derived functors $\mathrm{Ani}(F)\circ\mathrm{Ani}(G)$ is the derived functor of the composition $\mathrm{Ani}(F\circ G)$. Here is my question: suppose we have the aforementioned "niceness" and also $F:\mathcal{C}\to\mathcal{D}$ takes $\mathcal{C}^{sfp}$ to $\mathcal{D}^{sfp}$ (my best understanding for an analog of the second hypothesis). Is it the case that we get a Grothendieck-style spectral sequence where \begin{align*} E_{p,q}^{2}(A)=\pi_{p}\mathrm{Ani}(G)\circ\pi_{q}\mathrm{Ani}(F)(A)\Rightarrow\pi_{p+q}\mathrm{Ani}(G\circ F)(A)? \end{align*} It seems to me that the proof of the spectral sequence in the abelian setting goes through. If not, what is the obstruction? REPLY [10 votes]: Not in general, no. The problem is that animated functors play well with colimits, and homotopy groups play better with limits. However, if your functors $\mathcal{A}\xrightarrow{F}\mathcal{B}\xrightarrow{G}\mathcal{C}$ are such that $G$ is a right-exact functor of abelian categories, you do get a Grothendieck spectral sequence. In that case, $\operatorname{Ani}(\mathcal{B})\subseteq \mathcal{D}(\mathcal{B})$ is the full subcategory on connective objects, and $\mathcal{D}(\mathcal{B})$ is stable, same for $\mathcal{C}$. The animated functor $\operatorname{Ani}(G\circ F)$ can be recovered from the composite $LG \circ \operatorname{Ani}(F): \operatorname{Ani}(\mathcal{A})\to \mathcal{D}(\mathcal{C})$. For any $X\in \operatorname{Ani}(\mathcal{A})$, the Whitehead tower $\tau_{\geq *} \operatorname{Ani}(F)(X)$ gives a filtration on $\operatorname{Ani}(F)(X)$, which is taken by the exact $LG$ to a filtration on $\operatorname{Ani}(G\circ F)(X)$ with associated graded $LG(\pi_* \operatorname{Ani}(F)(X))[*]$. This gives a spectral sequence $$ \pi_* LG(\pi_* \operatorname{Ani}(F)(X)) \Rightarrow \pi_*(G\circ F)(X) $$<|endoftext|> TITLE: Galois groups of specific classes of polynomials with one coefficient fixed QUESTION [5 upvotes]: Let $f = x^n + a_{n-1}x^n + \cdots + a_0$ be a monic polynomial of degree $n \geq 2$ with integer coefficients. By $\text{Gal}(f)$ we mean the Galois group over $\mathbb{Q}$ of the Galois closure of $f$. Define $H(f) = \max\{|a_i|\}$ denote the naive or box height of $f$. Hilbert's irreducibility theorem asserts that for most integer $n$-tuples $(a_0, \cdots, a_{n-1})$ with $H(f) \leq B$ say, the corresponding polynomial $f$ has symmetric group $S_n$. If we assume that one of the coefficients $a_i$ of the polynomials are fixed, then is it still true that the number of such polynomials whose galois group is not symmetric group is $o(B^{n-1})?$ Thanks in advance for any assistance. REPLY [9 votes]: This is equivalent (by Hilbert) to asking whether the partially specialized polynomial still has symmetric Galois group (over the respective function field). This holds unless you specialized the constant coefficient to $0$. According to Cohen, S. D., The Galois group of a polynomial with two indeterminate coefficients, Pac. J. Math. 90, 63-76 (1980), ZBL0408.12011, you may even prescribe all but two of the coefficients and still get the full symmetric group, unless you specialized such that the resulting polynomial is a polynomial in $x^r$ for some $r>1$.<|endoftext|> TITLE: $S_3 \times 2^2 \cdot PSU_6 (2) \overset ? \subset 2^2 \cdot {}^2 E_6 (2)$ QUESTION [5 upvotes]: Does $2^2 \cdot {}^2 E_6 (2)$ have a conjugacy class of maximal subgroups isomorphic to $S_3 \times 2^2 \cdot PSU_6 (2)$? REPLY [7 votes]: Yes. Here is a sketch why. The Monster $F_1$ contains a four-group $V$ such that $C=C_{F_1}(V)\cong [2^2]\,{^2}\!E_6(2)$ and $N=N_{F_1}(V)$ satisfies $N/C\cong S_3$. $C$ contains a subgroup $T\cong Z_3$ such that $N_{C/V}(TV/V)\cong S_3\times PSU_6(2)$. The issue that you raise is the structure of the preimage $L$ in $C$ of the $PSU_6(2)$ direct factor. Using a $3$-element of $N-C$ that centralizes $T$, one sees that either $L$ splits over $V$ or $L$ is quasisimple, i.e., isomorphic to $[2^2]PSU_6(2)$. It therefore suffices to show that $L$ does not split over $V$. This can be seen in $C_T:=C_{F_1}(T)$. Since $|C_T|_2\ge |L|_2=2^{17}$, $C_T$ is isomorphic to the $3$-fold cover of $Fi_{24}'$. If $L$ did split over $V$, then $Fi_{24}'$ would have a four-subgroup $V_1$ such that $C_{Fi_{24}'}(V_1)\cong Z_2\times Z_2\times PSU_6(2)$. Write $V_1=\langle v,w\rangle$. Then $C_{Fi_{24}'}(v)\cong 2Fi_{22}2$. But $Fi_{22}2\cong Aut(Fi_{22})$ does not contain $Z_2\times PSU_6(2)$. (It does, of course, contain $2U_6(2)$.) The relevant information about local subgroups of $F_1$, $Fi_{24}'$, and $Fi_{22}$, as well as the desired maximality, can be found in the Atlas of Finite Groups by Conway, Curtis, Norton, Parker, and Wilson.<|endoftext|> TITLE: Relationship between doubling constant of a metric space and of a metric measure space QUESTION [6 upvotes]: Let $(X,d,m)$ be a metric measure space. We say that it is doubling in the sense of metric spaces if for every: $x\in X$ and every $r>0$ there exists some (metric) doubling constant $C_d\geq 0$ such that $$ Ball(x,r) \mbox{ can be covered by at-most $C_d$ balls of radius $r/2$}. $$ There is a different, related and in some spirit "equivalent", notion of doubling in the sense of metric measure spaces, which states that there is a constant $C_m\geq 0$ such that: for every $x\in X$ and each $r>0$ $$ m(Ball(x,r)) \leq C_m m(Ball(x,r/2)). $$ If $(X,d,m)$ is doubling in the sense of metric measure spaces, with constant $C_m$, then is it doubling in the sense of metric spaces? And if so, can we $C_m$ to deduce an upper-bound for $C_d$? Note, I'm most interested in the case where $m$ is an $s$-dimensional Hausdorff measure. REPLY [10 votes]: Apart from the obvious counterexample of the measure being $0$, if $(X,d,m)$ is doubling in the sense of metric measure spaces it will be doubling in the sense of metric spaces. Consider a ball $B(x,r)$. If for some $n$, $B(x,r)$ cannot be covered by $n$ balls of radius $\frac{r}{2}$, then we can obtain by recursion a sequence of points $x_1,\dots,x_n$ in $B(x,r)$ which are pairwise at distance $\geq\frac{r}{2}$. Thus the balls $B(x_i,\frac{r}{4})$ are disjoint, and they are all contained in $B(x,2r)$. Suppose $m(B(x_1,\frac{r}{4}))$ is the smallest of all the $m(B(x_i,\frac{r}{4}))$. Then $m(B(x_1,4r))\geq m(B(x,2r))\geq\sum_{i=1}^nm(B(x_i,\frac{r}{4}))\geq nB(x_1,\frac{r}{4})$, so either $n\leq C_m^4$ or $m(B(x_1,4r))=0$. If you only consider finite distances the second option implies that $m(X)=0$. So an upper bound would be $C_d\leq C_m^4$.<|endoftext|> TITLE: Vanishing linear combinations of minors QUESTION [6 upvotes]: Let $V$ be the set of $k$ by $n$ matrices ($k TITLE: Gordon's approach: slice knots and contractible $4$-manifolds QUESTION [8 upvotes]: Let $K \subset S^3$ be a slice knot. Then it bounds a smooth embedded disk $D \subset B^4$. Let $S^3_{p/q}(K)$ denote a $3$-manifold obtained by $p/q$-surgery on $K \subset S^3$. The following theorem is due to Gordon: Gordon, C. M. (1975). Knots, homology spheres, and contractible 4-manifolds. Topology, 14(2), 151-172. Theorem: For a slice knot $K \subset S^3$, $S^3_{\pm 1}(K)$ bounds a contractible $4$-manifold. I wonder that Gordon's theorem can be generalized to $1/n$ surgeries on slice knots for all $n$? REPLY [4 votes]: There is a more general result that this follows from. This is in Gordon's paper cited above; see Theorem 4. In any event, you don't really need the handle arguments in Marco's answer. Here is a sketch of Gordon's construction, in slightly different language. Of $K$ and $K'$ are concordant, then for any $p/q$ there is a homology cobordism between $S^3_{p/q}(K)$ and $S^3_{p/q}(K')$ with fundamental group normally generated by the meridian of $K$ or the meridian of $K'$. To see this, consider a tubular neighborhood $S^1 \times D^2 \times I$ of the concordance, and do $p/q$ surgery at each level. The meridian of the concordance is conjugate to the meridian of either end, and normally generates the fundamental group of the complement, and hence the fundamental group of the homology cobordism. When $K$ is slice, apply this to $K' = $ the unknot to get a homology cobordism to a lens space $L(p,q)$ with fundamental group $\mathbb{Z}/p$. In particular when $p=1$, you get a simply connected homology cobordism to $S^3$, which gives a contractible $4$-manifold with boundary $S^3_{1/q}(K)$ as desired.<|endoftext|> TITLE: The maximal subset of a finite field where the sum of any subset is non-zero QUESTION [10 upvotes]: Given a finite field $\mathbb{F}_q$ with $q=p^m$ where $p$ is the characteristic. For any subset $S=\{a_1,\dots,a_n\}$ of $\mathbb{F}_q$, if any partial sum (i.e. the sum of elements in a non-empty subset of $S$) is non-zero, then we may call $S$ a good subset. My question is what's the maximal cardinality $f(q)$ of a good subset $S$? Or are there any (lower) bounds for $f(q)$? REPLY [9 votes]: To complete Seva's answer, the article (“Subset sums modulo a prime”) of Nguyen, Szemerédi and Vu gives an asymptotic result for $p$ large enough, and this was also independently proved by Deshouillers and Prakash in “Large zero-free subsets of $\mathbb{Z}/p\mathbb{Z}$”, Integers, Procedings of Integers conference 2009 11A (2011), article 8. The result in $\mathbb{F}_p$ for any prime $p$ is established in “An addition theorem and maximal zero-sum free sets in $\mathbb{Z}/p\mathbb{Z}$”, E. Balandraud, Israel Journal of Mathematics, 188 (2012), 405-429.<|endoftext|> TITLE: Infinite groups that admit a discrete, co-compact, bilipschitz action on $\mathbb{R}^3$ QUESTION [5 upvotes]: Apart from the abstract types of the crystallographic groups, are there any other abstract groups that admit a proper, co-compact, uniformly bilipschitz action on $\mathbb{R}^3$? REPLY [5 votes]: Fix $k\ge 0$. Let $\Gamma$ be a discrete group. Then $\Gamma$ (a) has a proper cocompact, uniformly bilipschitz action on the Euclidean space $\mathbf{R}^k$ if and only if (b) it has an isometric one, if and only if (c) it has a finite index subgroup isomorphic to $\mathbf{Z}^k$. Proof: That (c) implies (b) is very standard and (b) trivially implies (a). Suppose (a): consider such an action $(g,x)\mapsto gx$, assume each map $(1/C,C)$-bilipschitz. Fix $x$ in $\mathbf{R}^k$ and a symmetric generating subset $S$ of $\Gamma$. Write $M=\max_{s\in S}d(x,sx)$ and $C'=CM$. Then $g\mapsto gx$ is a quasi-isometry. For $g,h\in\Gamma$, let $n=d_S(g,h)$ be their word distance: consider elements $g_0,\dots,g_n$ with $g_0=g$, $g_n=h$, $d(g_i,g_{i+1})\le 1$. Then $$d(gx,hx)\le \sum_{i=0}^{n-1} d(g_ix,g_{i+1}x)\le C\sum_{i=0}^{n-1} d(x,g_i^{-1}g_{i+1}x)\le nCM=C'd_S(g,h),$$ and $d(gx,hx)\ge C^{-1}d(x,g^{-1}h)$ which tends to infinity when $d_S(g,h)$ tends to infinity. So $x\mapsto gx$ Lipschitz, uniformly proper, and has cobounded image and is a map between geodesic spaces. Hence it is a quasi-isometry. We conclude by using that a group that is quasi-isometric to $\mathbf{R}^k$ satisfies (c). (This is a standard consequence of Gromov's theorem on groups with polynomial growth, e.g. using Pansu's theorem, and also admits proofs not appealing to Gromov's theorem.)<|endoftext|> TITLE: Cardinality of the maximum points of the determinant on matrices with entries in [-1, 1] QUESTION [8 upvotes]: By multilinearity, the maximum of the determinant on matrices with entries in the interval [-1, 1] is attained at a {-1, 1}-matrix. By the following example, the maximum is attained at infinitely many points on matrices of order $3$ \begin{bmatrix} 1 & 1 & 1\\ -1 & 1 & 1\\ t & 1 & -1 \end{bmatrix} $-1 \leq t \leq 1$. However, I verified that there are only finitely many maximum points for orders $4, 5, 6$. So, I ask, is the maximum attained only at finitely many points for order above $3$, or perhaps for orders not equal to $3$ mod 4? I suspect this is known, as there has been a lot of work on Hadamard Maximum determinant problem, but I can't find such statement so far. If this is not known, can we at least prove that the dimension of the space of maximum points is less than the order? REPLY [9 votes]: This response was too long for a comment, but is far from a complete answer. Much of the research on the maximal determinant problem has focused on the Gram matrix, and used the theory of positive definite matrices to obtain bounds. For orders $n \equiv 0 \bmod 4$ the optimal Gram matrix is $nI_{n}$ and the maximal determinant matrices are Hadamard matrices. Since the entries must be $\pm 1$ and rows must be orthogonal, the set of solutions will be finite, and there will be no free parameters of the type in the question since these would appear in the Gram matrix, and you could apply Fischer's Inequality to push the determinant away from the Hadamard bound. For orders $1, 2 \bmod 4$ there are Gram matrices which are known to be optimal, but these cannot always be attained for number theoretic reasons. When the optimal Gram matrices are known and attained, the number of solutions will be finite. For orders $3 \bmod 4$ there are not even conjectures for what the optimal Gram matrices should be. Forgive the self-promotion, but with some collaborators, I wrote a survey on the maximal determinant problem which might be of interest: https://arxiv.org/abs/2104.06756 The free parameter you observe above occurs because the matrix has an $(n-1) \times (n-1)$ minor equal to $0$. This cannot occur in the Hadamard case, because the inverse matrix is proportional to the transpose. Outside of the Hadamard case, the inverse matrix doesn't figure directly in the maximal determinant problem, so I don't know that this problem has been considered. Hadamard matrices do contain many $(n-2)\times (n-2)$ minors equal to $0$, complementary to any vanishing $2\times 2$ minor, so the question may well be quite difficult in general. Physicists have explored the existence of free parameters in complex Hadamard matrices, though they normally allow multiplication of a subset of entries in the matrix by a complex number of norm 1. Many examples are here: https://chaos.if.uj.edu.pl/~karol/hadamard/<|endoftext|> TITLE: Who is M. Meyniel? QUESTION [10 upvotes]: Does anyone know anything about M. Meyniel? According to zbMath, he published precisely one mathematics paper, in which he gave a sufficient condition for hamiltonicity of digraphs: "Une condition suffisante d'existence d'un circuit hamiltonien dans un graphe oriente" (JCTB 14 (1973), 137–147). In that paper, he was listed with an address but no affiliation. The address was 13, rue Poirier de Narçay, Paris 14ᵉ, which appears to be an apartment above a game store for what it is worth. I assume that M. Meyniel is distinct from Henri Meyniel of Meyniel graphs and Meyniel's conjecture. That said, the paper "Sufficient conditions for a digraph to be Hamiltonian" (J. Graph Th. 22 (1996) 181–187) is dedicated to the memory of Henri Meyniel and lists Henri Meyniel as the author of M. Meyniel's paper (item [15] in the bibliography). REPLY [22 votes]: You can be quite sure that M. Meyniel means "Monsieur Meyniel" (a common usage in French). Here is what I think is definite proof that M. Meyniel is H. Meyniel: The acknowledgement of the 1973 paper by M. Meyniel thanks J.C. Bermond, so evidently Bermond knew the author. In the article Cycles in digraphs - a survey Bermond and Thomassen cite the 1973 paper as follows: Meyniel, H., Une condition suffisante d'existance d'un circuit hamiltonien dans un graphe orienté, J. Combinatorial Theory B 14(1937), 137–147<|endoftext|> TITLE: Conceptual reason why the sign of a permutation is well-defined? QUESTION [133 upvotes]: Teaching group theory this semester, I found myself laboring through a proof that the sign of a permutation is a well-defined homomorphism $\operatorname{sgn} : \Sigma_n \to \Sigma_2$. An insightful student has pressed me for a more illuminating proof, and I'm realizing that this is a great question, and I don't know a satisfying answer. There are many ways of phrasing this question: Question: Is there a conceptually illuminating reason explaining any of the following essentially equivalent statements? The symmetric group $\Sigma_n$ has a subgroup $A_n$ of index 2. The symmetric group $\Sigma_n$ is not simple. There exists a nontrivial group homomorphism $\Sigma_n \to \Sigma_2$. The identity permutation $(1) \in \Sigma_n$ is not the product of an odd number of transpositions. The function $\operatorname{sgn} : \Sigma_n \to \Sigma_2$ which counts the number of transpositions "in" a permutation mod 2, is well-defined. There is a nontrivial "determinant" homomorphism $\det : \operatorname{GL}_n(k) \to \operatorname{GL}_1(k)$. …. Of course, there are many proofs of these facts available, and the most pedagogically efficient will vary by background. In this question, I'm not primarily interested in the pedagogical merits of different proofs, but rather in finding an argument where the existence of the sign homomorphism looks inevitable, rather than a contingency which boils down to some sort of auxiliary computation. The closest thing I've found to a survey article on this question is a 1972 note "An Historical Note on the Parity of Permutations" by TL Bartlow in the American Mathematical Monthly. However, although Bartlow gives references to several different proofs of these facts, he doesn't comprehensively review and compare all the arguments himself. Here are a few possible avenues: $\Sigma_n$ is a Coxeter group, and as such it has a presentation by generators (the adjacent transpositions) and relations where each relation respects the number of words mod 2. But just from the definition of $\Sigma_n$ as the group of automorphisms of a finite set, it's not obvious that it should admit such a presentation, so this is not fully satisfying. Using a decomposition into disjoint cycles, one can simply compute what happens when multiplying by a transposition. This is not bad, but here the sign still feels like an ex machina sort of formula. Defining the sign homomorphism in terms of the number of pairs whose order is swapped likewise boils down to a not-terrible computation to see that the sign function is a homomorphism. But it still feels like magic. Proofs involving polynomials again feel like magic to me. Some sort of topological proof might be illuminating to me. REPLY [3 votes]: There is a comment by Benjamin Steinberg, saying that "Usually the sign is used to show that the top exterior power is nonzero" I think it is possible to turn things around and first prove the existence of non-zero top-forms on an arbitrary $n$-dimensional $K$-vector-space (by induction on $n$, essentially by using Laplace expansion) and extract the sign of permutations from such non-zero top forms by a straightforward group-action: For $n=1$, the existence of non-zero top forms is vacuously true. So let $n>1$, $V$ an arbitrary $n$-dimensional $K$-vector-space, $\varphi\in V^*\setminus\{0\}$, $\psi\in \varphi^{-1}(\{1\})$ and define the following linear "projection on $\ker \varphi$": $$P:V \to \ker \varphi : v \mapsto v-\frac{\varphi(v)}{\varphi(\psi)}\psi = v-\varphi(v)\psi.$$ $$\forall j \in \{1,...,n\}:\,P_j:V^n \to (\ker \varphi)^{n-1}: v \mapsto (P(v_1),...,P(v_{j-1}),P(v_{j+1}),...,P(v_n))$$ Since $\ker \varphi$ is an $(n-1)-$dimensional $K$-vector-space, the induction hypothesis implies that there exists a non-zero multi-linear alternating map $\epsilon_{n-1}: (\ker \varphi)^{n-1} \to K$. Then define $$\epsilon_n: V^n \to K: v \mapsto \sum_{j=1}^n (-1)^{j+1} \varphi(v_j)\epsilon_{n-1}(P_j(v)).$$ Checking that $\epsilon$ is non-zero and multilinear is easy. The "alternation" becomes easy to check after noting that it suffices to prove that $\forall v\in V^n:\,\forall j\in \{1,...,n-1\}:\,v_{j}=v_{j+1}\Rightarrow \epsilon_n(v)=0$. To proceed to get the sign of permutations, one still has to show that all permutations are compositions of a number of transpositions (maybe of neirest-neighbour-elements $(j,j+1)$). Strictly speaking, for the goal of getting the sign quickly we could have omitted here the proof that $\epsilon_n$ is multilinear, although in that case we have to prove the version of alternation/antisymmetry which says that $$\forall v\in V^n:\,\forall j\in \{1,...,n-1\}:\, \epsilon_n(v_1,...,v_n)=-\epsilon_n(v_{\pi_{j,j+1}(1)},...,v_{\pi_{j,j+1}(n)}).$$<|endoftext|> TITLE: Topos with enough points but not coherent QUESTION [8 upvotes]: By Deligne's theorem, each coherent topos has enough points. What would be an example of a Grothendieck topos with enough points which is not coherent? REPLY [16 votes]: Here are some examples : For any topological space $X$, the topos of sheaf $\operatorname{Sh}(X)$ has enough points. In most cases this is not a coherent topos. If I remember correctly (for $X$ sober), this is only coherent if $X$ is a spectral space. In any case, spaces like $\mathbb{R}$ are definitely not coherent toposes. For any topos $\mathcal{T}$, and any set of points $X$ of $\mathcal{T}$, you get geometric morphism $X \to \mathcal{T}$ (where by $X$ I mean the topos $\operatorname{Set}/X$), you can take its image factorisation $X \twoheadrightarrow I \hookrightarrow \mathcal{T}$ and $I$ is a subtopos of $\mathcal{T}$ with enough points. Generally, non coherent toposes don't have a lot of coherent subtoposes (though of course this can happen), so this will often gives example of non-coherent topos with enough topos. There is also a way to do this for $X$ the class of all points of $\mathcal{T}$ despite the size issues. There is another completeness theorem like Deligne's which says that any "separable" Grothendieck topos has enough points. Separable essentially means that the topos can be defined by a site whose underlying category is countable and whose topology is generated by a countable family of basic covering Sieve. This is done in Makkai and Reyes's book "First order categorical logic" (theorem 6.2.4). There are many separable toposes that are not coherent. In terms of classifying toposes, coherent means you only use finitary logic, while separable means you can use infinitary logic but the theory should have a countable signature and a countable set of axioms. So none of the two class is included in the other, but I would personally consider that in term of which classical topos they contains, separable toposes form a much larger class.<|endoftext|> TITLE: Show that these matrices are invertible for all $p>3$ QUESTION [25 upvotes]: I am working on a paper which will extend a result in my thesis and have boiled one problem down to the following: show that the symmetric matrix $M_p$, whose definition follows, is invertible for all odd primes $p$. Letting $p>3$ be prime and $\ell = \frac{p-1}{2}$, we define $$M_p = \begin{pmatrix} 2ij - p - 2p\left\lfloor\frac{ij}{p}\right\rfloor\end{pmatrix}_{1\leq i,j\leq \ell}$$ Examples: For $p=5$ we have $M_5 = \begin{pmatrix} -3 & -1 \\ -1 & 3 \end{pmatrix}$ and $\det(M_5) = -1\cdot 2\cdot 5$. For $p=7$ we have $M_7 = \begin{pmatrix} -5 & -3 & -1 \\ -3 & 1 & 5 \\ -1 & 5 & -3 \end{pmatrix}$ and $\det(M_7) = 2^2 \cdot 7^2$. For $p=11$ we have $M_{11} = \begin{pmatrix} -9 & - 7 & -5 & -3 & -1 \\ -7 & -3 & 1 & 5 & 9 \\ -5 & 1 & 7 & -9 & -3 \\ -3 & 5 & -9 & -1 & 7 \\ -1 & 9 & -3 & 7 & -5 \end{pmatrix}$ and $\det(M_{11}) = -1\cdot 2^4\cdot 11^4$. Though this (seemingly) nice formula that we see above fails for primes greater than 19, though the determinant has been checked to be non-zero for primes less than 1100. (My apologies if this question is not as motivated or as well discussed as is desired. If there are any questions or if further clarification is needed just let me know!) REPLY [7 votes]: Here are some elements to solve the question. 1st step. Extend $M_p$ as a $(2\ell)\times(2\ell)$-matrix $N_p$, with the same definition of entries. By the way the entries may be writen as $a_{j,k}=p(2\left\{\frac{jk}p\right\}-1)$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$. Remark that $a_{j,p-k}=-a_{j,k}$ and likewise $a_{p-j,k}=-a_{j,k}$. Remark that $a_{j,k}$ depends only upon the product $jk\mod p$, thus can be written $a_{jk}$ instead. Therefore $$N_p=\begin{pmatrix} M_p & -M_pF \\ -FM_p & FM_pF \end{pmatrix}=\binom{I}{-F}M_p\begin{pmatrix} I & -F \end{pmatrix}$$ where $F$ is the anti-unit matrix, $f_{j,k}=\delta_j^{\ell+1-k}$. 2nd step. From the formula above, $N_p$ and $M_p$ have the same rank. Thus $M_p$ is invertible if and only if $N_p$ has rank $\ell$. This rank is unchanged by a permutation of rows. Thus we examine instead the matrix $N_p'$ obtained by the involution of rows given by the inversion $j\mapsto j^{-1}$ $\mod p$. That way, the entries of $N_p'$ are the numbers $a_{j^{-1}k}$. 3rd step. The matrix $N_p'$ is diagonalizable, its eigenpairs being explicit, because this is a group matrix (entries of the form $a_{g^{-1}h}$) for the multiplicative group ${\mathbb F}_p^\times$ (the non-zero elements of ${\mathbb Z}/p$). Recall that because ${\mathbb Z}/p$ is a field, this group is isomorphic to the additive group ${\mathbb Z}/2\ell$ : there exists an element $\theta\in{\mathbb Z}/p$ of order exactly $2\ell$, and an isomorphism is $\psi(k)=\theta^k$ ($k\mod2\ell$, $\theta^k\mod p$). The eigenvectors $v^\omega$ have coordinates $$v^\omega_k=\omega^{\psi^{-1}(k)},\qquad 1\le k\le2\ell,$$ where $\omega$ is any complex solution of $\omega^{2\ell}=1$. The corresponding eigenvalue is $$\lambda_\omega=\sum_{r=0}^{2\ell-1}a_{\psi(r)}\omega^r=:Q_p(\omega).$$ 4th step. There remains to see that among these $2\ell$ eigenvalues, $\ell$ of them only vanish, so that $N_p$ has rank $2\ell-\ell=\ell$. I leave this as an open question (though I am rather optimistic). At least, the fact that $\ell$ of them vanish is clear, because we have $\theta^\ell=-1$, hence $\psi(k+\ell)=-\psi(k)$. There follows that $a_{\psi(k+\ell)}=-a_{\psi(k)}$, so that $Q_p$ factorizes: $$Q_p(X)=(X^\ell-1)R_p(X),\qquad R_p(X)=\sum_{r=0}^{\ell-1}a_{\psi(r)}X^r.$$ Thus $\lambda_\omega=0$ for every root of $\omega^\ell=1$. Thus there remains only to prove that if instead $\omega^\ell=-1$ (the other roots of unity), one has $R_p(\omega)\ne0$. I have done a few calculations for small prime numbers $p$, which make me optimistic. Edit. One can conclude at least when $\ell$ is a power of $2$ (that is when $p$ is a Fermat prime). Because then $X^\ell+1$ is irreducible over $\mathbb Q$, while $R_p\in{\mathbb Z}[X]$ has degree $\ell-1$, hence they cannot share a root. In the general case, the question reduces to whether some cyclotomic polynomial $\Phi_m$ with $m|\ell$ can be such that $\Phi_m(-X)|R_p(X)$. Notice that $\Phi_m$ is reciprocal, while $R_p$ is not, thus this divisibility will imply some other ones, ...<|endoftext|> TITLE: Can $y^2-4$ be a divisor of $x^3-x^2-2 x+1$? QUESTION [8 upvotes]: Do there exist integers $x$ and $y$ such that $\frac{x^3-x^2-2 x+1}{y^2-4}$ is an integer? In other words, can any integer representable as $x^3-x^2-2 x+1$ have any divisor representable as $y^2-4$? This is the simplest non-trivial example of my earlier question Integer points of rational function in 2 variables . REPLY [28 votes]: No. The roots of $x^3 - x^2 - 2x + 1$ are $-(\zeta + \zeta^{-1})$ where $\zeta$ is a 7th root of unity; this soon implies [see below] that any prime factor is either $7$ or $\pm 1 \bmod 7$, and thus that all factors of $x^3 - x^2 - 2x + 1$ are congruent to $0$ or $\pm 1 \bmod 7$. In particular it is not possible for two factors to differ by $4$, so no number of the form $y^2 - 4 = (y-2) (y+2)$ can divide $x^3 - x^2 - 2x + 1$. added later: To show that any prime factor $p$ of $x^3 - x^2 - 2x + 1$ is either $7$ or $\pm 1 \bmod 7$, let $k$ be the finite field of order $p^2$, and $\zeta \in k$ a root of the quadratic equation $\zeta^2 + x\zeta + 1 = 0$ (any quadratic equation with coefficients in the $p$-element field has a root in $k$). Then $\zeta^7 = 1$, so either $\zeta = 1$ or the multiplicative group $k^\times$ of $k$ has a subgroup of size $7$. In the former case, $x = -2$, and then $x^3 - x^2 - 2x + 1 = -7$ so $p=7$. In the latter case, Lagrange's theorem gives $7 \mid \#k^\times = p^2-1$, so $p \equiv \pm 1 \bmod 7$. QED<|endoftext|> TITLE: Irreducibility of the $n$th symetric power of the reduction of the Galois representation of a non-CM newform QUESTION [5 upvotes]: In "On $\ell$-adic representations attached to modular forms II", Ribet proved that the $\ell$-adic representation $\rho_{f,\ell}$ attached to a non-CM newform form $f$ satisfies $${\rm SL}_2(\mathbb{F}_\ell)\subset \overline{\rho}_{f,\ell}(G_\mathbb{Q}).\qquad\qquad(*)$$ I am wondring : (i)- Why $(*)$ implies that $\overline{\rho}_{f,\ell|G_{\mathbb{Q}(\zeta_\ell)}}$ is absolutely irreducible? (ii)- Let $n$ be a positive integer. Dose $(*)$ implies that ${\rm Sym}^n\overline{\rho}_{f,\ell|G_{\mathbb{Q}(\zeta_\ell)}}$ is irreducible? If the answer is negative, under what condition does this hold? REPLY [5 votes]: Here's an answer that's more work than Kevin Ventullo's, but works for $\ell=2$ and $\ell=3$, and gives you a method that will work in various more general cases. There is a surjection $\pi_1\colon G_\mathbb Q \to \overline{\rho}_{f,l} (G_{\mathbb Q})$ and a surjection $\pi_2 \colon G_\mathbb Q \to \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q)$. The group $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the image under $\pi_1$ of the kernel of $\pi_2$. In other words, it is the intersection of the image of the product map $(\pi_1,\pi_2) \colon G_\mathbb Q \to \overline{\rho}_{f,l} (G_{\mathbb Q}) \times \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q)$ with $G_{\mathbb Q}) \times 1$. By Goursat's lemma, the image of the product map is $ \{ (a,b) \mid f_1(a) =f_2 (b) \}$ where $f_1 \colon \overline{\rho}_{f,l} (G_{\mathbb Q}) \to Q$ and $f_2 \colon \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q) \to Q$ are surjections for some finite group $Q$. Expressed in this language, $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the kernel of $f_1$. Now because $f_2$ is surjective, $Q$ is a quotient of a cyclic group of order $\ell-1$, and thus is a cyclic group of order dividing $\ell-1$. So $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the kernel of a map from $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ to a cyclic group of order dividing $\ell-1$. Thus it contains the kernel of a map from $SL_2(\mathbb F_\ell)$ to the kernel of a cyclic group of order dividing $\ell-1$. Since the abelianization of $SL_2(\mathbb F_\ell)$ is trivial (for $\ell>3$) or has order relatively prime to $\ell-1$ (for $\ell=2,3$, since $2$ is relatively prime to $1$ and $3$ is relatively prime to $2$), $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ contains $SL_2(\mathbb F_\ell)$. Since $SL_2(\mathbb F_\ell$ acts absolutely irreducibly on the standard representation for all $\ell$, it follows that $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ acts absolutely irreducibly on the standard representation, as desired. It immediately follows that the image of $\operatorname{Sym}^n \overline{\rho}_{f,\ell} \mid_{ G_{\mathbb Q(\zeta_\ell)}}$ containsthe image of $SL_2(\mathbb F_\ell)$ under the $n$th symmetric power representation. Since the $n$th symmetric power representation of $SL_2(\mathbb F_\ell)$ is absolutely irreducible for $n \leq \ell-1$, the representation $\operatorname{Sym}^n \overline{\rho}_{f,\ell} \mid_{ G_{\mathbb Q(\zeta_\ell)}}$ is absolutely irreducible for $n\leq \ell-1$.<|endoftext|> TITLE: Augmentation ideal of a free group QUESTION [8 upvotes]: If $F$ is a free group then it has cohomological dimension one, which implies that the augmentation ideal $IF=\operatorname{ker}(\epsilon:\mathbb{Z}G\to \mathbb{Z})$ of its group ring is a projective $\mathbb{Z}F$-module. Hence $IF$ is a direct summand of a free $\mathbb{Z}F$-module $M$. Question: Is it possible to give an explicit construction of such a free module $M$? Perhaps it has a basis related to a free basis of $F$? REPLY [10 votes]: Let $X$ be a free basis for $F$. The Cayley graph of $F$ is a tree $T$ on which $F$ acts freely. The augmented chain complex gives a resolution $$0\to \mathbb ZF^{(X)}\to \mathbb ZF\to \mathbb Z\to 0$$ (since $T$ is a tree) where we identify $\mathbb ZF^{(X)}$ as the free abelian group on the edges of $T$ and the image of the boundary map sends the edge $(g,x)$ to $gx-x$ and is hence the augmentation ideal (which in any event is clearly the kernel of the augmentation map). So the augmentation ideal is a free module on $X$ with generators $x-1$ with $x\in X$.<|endoftext|> TITLE: Does every monoidal category admit a braiding? QUESTION [9 upvotes]: The question is in the title. To make the statement more precise, is is true that for any given monoidal category $(\mathcal C, I, \otimes)$ there exists at least one braiding $\beta$? In other words, does the forgetful functor from braided monoidal categories admit a section (not a left adjoint!)? I strongly suspect the answer to be 'no', or at least 'yes but not functorially'. REPLY [25 votes]: No, sometimes there are even $x,y$'s with no abstract isomorphism $x\otimes y \cong y\otimes x$. Here are two families of examples: Monoids, viewed as discrete categories. The tensor product is just the multiplication, and the existence of a braiding would simply mean that the monoid is commutative, which it typically isn't. Functor categories : if $C$ is a category, then $Fun(C,C)$ is monoidal with respect to $\circ$, but it's quite rare for endofunctors to commute.<|endoftext|> TITLE: Is there a purely constructive presentation of the HK integral? QUESTION [7 upvotes]: Treating the Riemann integral in a constructive setting is easy and straightforward. Treating the closely related but much more powerful Henstock-Kurzweil integral constructively is almost easy, except for the dependence on Cousin's lemma where every proof seems to involve the excluded middle. In a constructive setting, is there a way to: define the HK integral prove that the HK integral is unique define another notion of integrability which constructively implies HK-integrability and is classically equivalent to Lebesgue integrability and prove that every derivative, even if discontinuous, is HK-integrable? One approach would be restricting the choice of gauge to Baire functions, since under fairly general assumptions the gauge can be chosen to be Baire 2. This would require proving Cousin's lemma only for Baire functions, which looks more promising than proving the full Cousin’s lemma constructively. I have seen a paper on the reverse mathematics of Cousin’s lemma, but that is not in a constructive setting. REPLY [2 votes]: Firstly, the following paper deals with the HK-integral and constructive math, see [2] below for details. Taschner R., The swap of integral and limit in constructive mathematics. Secondly, the general HK integral may be (too) difficult to salvage, as the associated versions of Cousin's lemma (even the restrictions you mention) seem fundamentally non-constructive. Thirdly, the HK integral restricted to absolutely integrable functions is the Lebesgue integral (on suitable domains). Hence, there should be a constructive treatment of this restriction. In particular, in the case of bounded functions, one can replace the use of Cousin's lemma by Vitali's covering theorem (see Section 3.5 in [1]), where the latter is 'more constructive' than the former. Also, since constructive measure theory exists, there is a 'constructive way around' Vitali's covering theorem, central as the latter is to Lebesgue measure theory (see again [1]). References [1] Dag Normann and Sam Sanders, the logical and computational properties of the Vitali covering theorem, arxiv: https://arxiv.org/abs/1902.02756 [2] Taschner R., The swap of integral and limit in constructive mathematics, Math. Log. Quart. 56, No. 5, 533 – 540 (2010) / DOI 10.1002/malq.200910107<|endoftext|> TITLE: Given a quasi-convex subgroup $H$ of hyperbolic $G$, can we decide if two elements $x,y \in G$ lie in the same double coset of $H$? QUESTION [5 upvotes]: I've come across the following question in my research, which seems elusive but is almost surely decidable. Let $H$ be a quasi-convex subgroup of the hyperbolic group $G$. Given $x, y \in G$, we wish to decide whether $HxH = HyH$. This is equivalent to asking whether there exists $h, h' \in H$ such that $hx = yh'$. It is easy to see that the question $x \in yH$ is easily decidable, since $H$ has a solvable membership problem and this reduces to checking whether $y^{-1} x \in H$. The double coset problem seems harder, but the instinct is that this might solved by bounding the lengths of $h, h'$, akin to the solution to the conjugacy problem. Does this problem appear in the literature anywhere? Any references or thoughts are appreciated. Thanks! REPLY [5 votes]: This is answered for free groups in Membership to double cosets in free groups and the same method basically works for hyperbolic groups. You might as well do double cosets $HgK$ with $H,K$ both quasiconvex. Then $w\in HgK$ if and only if $Hw\cap gK$ intersect nontrivially. Now the set of geodesic words belonging to $Hw$ is a regular language and there are known algorithms for constructing an automaton recognizing these languages (HJRW mentioned this in the comments for subgroups, but it is more or less straightforward to generalize for cosets) and similarly for $gK$. Therefore the geodesic words in $Hw\cap gK$ are recognized by a finite automaton that you can construct. Since emptiness is decidable for finite automata, you are done. You need the quasi-convexity constants of course to build these automata. I should mention the algorithms for these things I have seen do not appear super-effecitve. The point is if $H$ is $L$-quasiconvex, then to get the automaton for $Hw$ you construct the $(L+|w|)$-neighborhood of the coset $H$ in the Schreier graph associated to $G/H$ and make $H$ the start state and $Hw$ the final state and intersect with the automaton computing the language of geodesic words for $G$. Any geodesic for an element of $Hw$ does not leave this $(L+|w|)$-neighborhood by quasi-convexity. One can build this fragment of the Schreier graph via a Todd-Coxeter style method: see section 4 of https://arxiv.org/pdf/1408.1917.pdf. However, there are some classes of groups (like surface groups) where more efficient means to build this core are known.<|endoftext|> TITLE: Can we make distances in a finite subset of a manifold whatever we want? QUESTION [22 upvotes]: Given a connected smooth manifold $M$ of dimension $m>1$, points $p_1,\dots,p_n\in M$ and positive values $\{d_{i,j};1\leq i TITLE: PBW basis for the quantized enveloping Lie algebra of $\mathfrak{g}_2$ QUESTION [5 upvotes]: I would like to know if you have any reference where I can find the canonical PBW basis for $U_q(\mathfrak{g}_2),$ computed using the action of the braid group as defined by Luzstig. Alternatively I would like to know if I should expect Luztig's root vectors to be given by some sort of $q$-commutators or linear combinations of $q$-commutators. REPLY [6 votes]: The canonical basis and the PBW basis are two different things. I don't expect a simple computation of the canonical basis (in G2 the appropriate cluster algebra is not of finite type) while the PBW basis is not hard to compute for G2 as there are only six positive roots. Each of Lusztig's root vectors for the non-simple roots are scalar multiples of q-commutators of appropriately chosen other root vectors. This dates back to Levendorskĭ-Soibelman (Proposition 5.5.2), but the only place I know of where the scalar multiples are determined is in my paper with Brundan and Kleshchev (Theorem 4.2). The results mentioned in this paragraph are true in all finite types, not just G2.<|endoftext|> TITLE: Prove or disprove that the power of positive term polynomial will be eventually single peak QUESTION [7 upvotes]: This is a question that a classmate asked me three years ago. Let $P(x)=\sum_{i=0}^n a_ix^i$ be a polynomial such that each $a_i>0$. Prove or disprove that there exists a positive integer $r$ such that $P(x)^r=\sum_{i=0}^{nr} b_ix^i$ and there exists $0\le j\le nr$ such that $b_0\le b_1\le \dots\le b_{j}$ and $b_{j+1}\ge\dots\ge b_{nr}$. This problem may have a probability problem as its original problem, since the classmate asked this while taking a probability class. What I could do is to try to apply CLT to claim that the "central terms" has only one peak and try to select some $r$ to make the first few terms and last few terms increasing/decreasing (that is, I proved that I can choose an $r$ to make $b_0\le b_1\le\dots \le b_k$ for any $k$, and same at the other side). But I failed to prove the problem... I also tried to factorize it into smaller quadratic polynomials but also failed... Also, I can't come up with a counterexample, too... Note: if polynomials $p,q$ are single peak, this DOES NOT imply that $pq$ is single peak. For example: $p(x)=1+x+100x^2$ and $q(x)=10000+10000x+10100x^2+9000x^3+9000x^4$. But $p(x)q(x)=900000x^6+909000x^5+1028000x^4+1019100x^3+1020100x^2+20000x+10000$ P.S. This is exactly same as the problem in MSE. As @ChrisSanders in the comment pointed out, I just ask the question here... REPLY [8 votes]: This is answered affirmatively by Odlyzko and Richmond, On the unimodality of high convolutions of discrete distributions, Annals of probability (1985) 299--306: all sufficiently large powers of the polynomial (with positive coefficients and no gaps) are strongly unimodal, that is, the coefficients form a log concave sequence. The proof uses estimates of contour integrals over circles of just the right radius.<|endoftext|> TITLE: Any name for this special function? QUESTION [5 upvotes]: We know $$ \sum_{m=0}^\infty \frac{x^m}{(a-m)!m!} = \frac{1}{a!}(1+x)^m $$ where we understand the factorial as Gamma function $\Gamma(x)$ such that it is divergent if the argument is negative integer. We also know $$ \sum_{m=0}^\infty \frac{x^m}{(b+m)!m!} \sim \,_0F_1(b,x) $$ as hypergeometric function while this can be generalized to any $p,q$ for $_pF_q(\cdots,x)$. Now I want to study $$ f_{abc}(x)=\sum_{m=0}^\infty \frac{x^m}{m!(a-m)!(b+m)!(c-m)!}, \qquad a,b,c \in \mathbb{Z}^+ $$ Although the function is defined to sum over infinite numbers of integers, but it is effectively truncated at min$(a,c)$. I want to ask whether I can find any reference studying the polynomial $f_{abc}(x)$? Is there any special name for it? I can imagine one possible manipulation is to use the reflection formula of Gamma function $$ \Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin \pi z} $$ to change the sum over $(c-m)!$ to $(m-c)! \sin(\pi(m-c))$. And you might then call this function some kind of $_2F_1$ type hypergeometric function. However, it is not that good. First of all, the relection formula is better used if arguments $z$ are not integer, which is different from my purpose. Second, even after doing that, the function is not fully hypergeometric as there are extra sine functions as coefficients. I was expecting whether any literature has ever studied these kinds of functions and name them? REPLY [17 votes]: This is a standard hypergeometric function. Note that $$ \frac{1}{(a-m)!} = (-1)^m \frac{(-a)_m}{a!}\quad\text{and}\quad \frac{1}{(b+m)!} = \frac{1}{b!\,(b+1)_m}$$ in terms of the rising Pochhammer symbol $(q)_m = q(q+1)\cdots(q+m-1)$. Hence, $$f_{abc} = \frac{1}{a!b!c!} \sum_{m=0}^\infty \frac{(-a)_m(-c)_m}{(b+1)_m}\frac{x^m}{m!} = \frac{1}{a!b!c!}\,{_2F_1}(-a,-c;b+1;x).$$<|endoftext|> TITLE: Conjugated subgroups in $\mathsf{GL}(m+n,\mathbb{Z})$ implies conjugated subgroups in $\mathsf{GL}(n,\mathbb{Z})$? QUESTION [15 upvotes]: In my research I came up with the following question: Question: Let $H_1$ and $H_2$ be finite abelian subgroups of $\mathsf{GL}(n,\mathbb{Z})$. Define $$ H_1'=\left\{\begin{pmatrix} I_m &0\\0&h_1\end{pmatrix}\mid h_1\in H_1\right\},\quad\text{and}\quad H_2'=\left\{\begin{pmatrix} I_m &0\\0&h_2\end{pmatrix}\mid h_2\in H_2\right\}.$$ Suppose $H_1'$ is conjugated to $H_2'$ as subgroups of $\mathsf{GL}(m+n,\mathbb{Z})$ (i.e, there exists $\alpha\in \mathsf{GL}(m+n,\mathbb{Z})$ s.t. $H_2'=\alpha H_1' \alpha^{-1}$). Does this imply that $H_1$ is conjugated to $H_2$ as subgroups of $\mathsf{GL}(m,\mathbb{Z})$? Some thoughts: $H_1$ is isomorphic to $H_1'$ and $H_2$ is isomorphic to $H_2'$, so $H_1$ and $H_2$ are at least isomorphic, but of course this does not imply that are conjugated. On the other hand, writing $\alpha=\begin{pmatrix} X&Y \\ Z&W \end{pmatrix}$, we have that for every $h\in H_1$ there exists $h'\in H_2$ such that \begin{align*} \alpha \begin{pmatrix} I_m&0\\0&h\end{pmatrix}=\begin{pmatrix} I_m&0\\0&h' \end{pmatrix} \alpha&\iff \begin{pmatrix} X& Yh \\ Z& Wh\end{pmatrix}=\begin{pmatrix}X&Y\\h'Z&h'W\end{pmatrix}\\&\iff Yh=Y, Z=h'Z,Wh=h'W. \end{align*} These conditions do not say too much, because $W$ a priori does not have to be invertible. I've checked carefully (and certainly not in the more efficient way) using GAP and/or MAGMA the case when $n=2$ and for cyclic subgroups of $\mathsf{GL}(3,\mathbb{Z})$ and the answer to my question is affirmative, but I'm afraid there could be some example for much larger dimensions. Some update (14/3): $\bullet$ I've checked for subgroups of $\mathsf{GL}(3,\mathbb{Z})$ and $\mathsf{GL}(4,\mathbb{Z})$ (enlarging them with $I_1$ or $I_2$) and the answer is also affirmative. $\bullet$ In the comments the case when $H_1$ and $H_2$ have order 2 is settled. I suspect now that it could be true. If it helpful, in this page one can download the finite subgroups of $\mathsf{GL}(n,\mathbb{Z})$ up to $n=6$. Any idea would be greatly appreciated! Thanks! REPLY [10 votes]: $\def\ZZ{\mathbb{Z}}\def\GL{\text{GL}}$We can make partial progress using: Warfield, R. B. jun., Cancellation of modules and groups and stable range of endomorphism rings, Pac. J. Math. 91, 457-485 (1980). ZBL0484.16017. In particular, we can show that if $H_1$ and $H_2$ are conjugate in $GL(n+m)$ then they are conjugate in $GL(n+1)$. We first rephrase in the language of modules. Let $H$ be the abstract abelian group which is isomorphic to both $H_1$ and $H_2$. Let $R$ be the group ring $\ZZ[H]$. Embedding $H$ into $\GL_n(\ZZ)$ is equivalent to equipping $\ZZ^n$ with the structure of an $R$-module; call our $R$-modules $M_1$ and $M_2$. Conjugating $H_1$ to $H_2$ is equivalent to giving an isomorphism of $R$-modules. I'll write $A$ for $\ZZ$ considered as an $R$-module with the trivial action of $R$. The hypothesis that $H_1$ and $H_2$ are conjugate in $\GL_{n+m}(\ZZ)$ means that $A^{\oplus m} \oplus M_1 \cong A^{\oplus m} \oplus M_2$. So your question is: Given $M_1$ and $M_2$ are $R$-modules, both of which are free as rank $n$ as $\ZZ$-modules, and $A \oplus M_1 \cong A \oplus M_2$, do we have $M_1 \cong M_2$? Warfield tells us that we should consider the endomorphism ring $E:=\text{Hom}_R(A,A)$. In this case, $E = \ZZ$. We should then compute the stable rank of $E$: The stable rank of $\ZZ$ is known to be $2$. (This also follows from Theorem 4.1 in Warfield, taking $R=S$ as here and $A$ as here.) Warfield then tells us that, if $E$ has stable range $2$ then $A$ obeys $2$-substitution. In Theorem 1.3, Warfield tells us that $2$-substitution has the following consequence (his $X'$ is my $M_1$ and his $Y$ is my $A \oplus M_2$): If $A^2 \oplus M_1 \cong A^2 \oplus M_2$ then there is some module $L$ such that $A \oplus M_1 \cong L \oplus M_2$ and $A^2 \cong A \oplus L$. In our special case, I claim that $L$ must be $A$. To see this, use the hypothesis that $A^2 \cong A \oplus L$. Looking at ranks as $\ZZ$-modules, we must have $L \cong \ZZ$ as a $\ZZ$-module. But $L$ is supposed to be a submodule of $A^2$, which is just $\ZZ^2$ with the trivial $H$-action. So the $H$-action on $L$ is likewise trivial and we have $A \cong L$. So we have the simpler statement If $A^2 \oplus M_1 \cong A^2 \oplus M_2$ then $A \oplus M_1 \cong A \oplus M_2$. This means that conjugacy in $\GL_{n+2}(\ZZ)$ implies conjugacy in $\GL_{n+1}(\ZZ)$.<|endoftext|> TITLE: Groups with unusual cohomological dimension of direct product QUESTION [6 upvotes]: $\DeclareMathOperator\cd{cd}$Are there any known examples of non-free groups with a property that $\cd(G)+1 = \cd(G \times G)$, or, less restrictive, $G, H$ with $\cd \neq 1, \infty$ such that $\cd(H)+1 = \cd(G \times H)$? REPLY [6 votes]: Let $G=(\mathbb{Q},+)$. Then ${\rm cd}(G)=2$ and ${\rm cd}(G\times G)=3$.<|endoftext|> TITLE: Measurable cardinals from saturated ideals QUESTION [7 upvotes]: Assume that $\omega<\kappa_1<\dotsb< \kappa_n$ are infinite cardinals such that for each $1\le i\le n$ there is a $\kappa_i$-complete, $\kappa_i^+$-saturated ideal $\mathcal I_i\subset \mathcal P(\kappa_i)$. Can you obtain a ZFC model which contains $n$-many measurable cardinals? The natural candidate is $L[\mathcal I_1,\dotsc, \mathcal I_n]$. It is well known that the answer is yes for $n=1$ (see Kunen: Some applications of iterated ultrapowers in set theory. Ann. Math. Logic 1 (1970), 179–227.) REPLY [4 votes]: Yes. And if the $\mathscr{I}_k$s are normal, then the suggested candidate works (I haven't really thought about whether the candidate still works without normality). Here is the argument, which is a typical one: First, we may assume that each $\mathscr{I}_k$ is normal, by Jech Lemma 22.28. Case 1. There is an proper class transitive inner model of ZFC with $n+1$ measurable cardinals (not just $n$). Let $M$ be the minimal proper class transitive inner model satisfying ZFC with $n$ (not $n+1$) measurable cardinals $\mu_1<\ldots<\mu_n$, as witnessed by (unique) normal measures $U_1,\ldots,U_n$. Then (this uses the case hypothesis) $\mu_1,\ldots,\mu_n$ are countable. So we can iterate $M$ out at its measurables (using the $U_k$s and their images), eventually sending each $\mu_k$ to $\kappa_k$. Let $\mathcal{T}$ be this iteration (consisting of the sequence of iterates $M^{\mathcal{T}}_\alpha$, iteration maps $i_{\alpha\beta}^{\mathcal{T}}$, etc), and $M'=M^{\mathcal{T}}_{\kappa_n}$ be the final iterate and $i:M\to M'$ the iteration map. Then I claim that $i(U_k)\subseteq\mathscr{F}_k$, the filter dual to $\mathscr{I}_k$. For let $j:V\to M$ be a generic embedding given by $\mathscr{I}_k$. So $j(\mathcal{T})$ is an iteration with last model $j(M')$, and $j(\mathcal{T})\upharpoonright(\kappa_k+1)=\mathcal{T}\upharpoonright(\kappa_k+1)$, but the $\kappa_k$th measure used in $j(\mathcal{T})$ is $i(U_k)$, and $j(\mathcal{T})$ eventually sends $\kappa_k$ further out to $j(\kappa_k)$. Now standard calculations show that the iteration map $i^{\mathcal{j(\mathcal{T})}}_{\kappa_kj(\kappa_k)}$ agrees with $j$ over $\mathcal{P}(\kappa_k)\cap M^{\mathcal{T}}_{\kappa_k}$, which implies that $j(U_k)\subseteq G$, the generic filter, and since this is independent of $G$, therefore $j(U_k)\subseteq\mathscr{F}_k$, as desired. Since $i(U_k)$ is an ultrafilter in $M'$ and $i(U_k)\subseteq\mathscr{F}_k$, it follows that $$L[\mathscr{F}_1,\ldots,\mathscr{F}_n]=L[i(U_1),\ldots,i(U_n)]\subseteq M',$$ and that $i(U_k)=\mathscr{F}_k\cap L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$; let $\bar{\mathscr{F}}_k$ denote this measure. Therefore $L[\mathscr{F}_1,\ldots,\mathscr{F}_k]\models$"$\bar{\mathscr{F}_k}$ is a $\kappa_k$-complete normal measure on $\kappa_k$". Of course $L[\mathscr{I}_1,\ldots,\mathscr{I}_n]=L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$, so we are done. (Using the minimality of $M$, can also show that $M'=L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$.) Case 2: Otherwise. Then the core model $K$ exists. Let $j:V\to M$ be a generic embedding given by $\mathscr{I}_k$. Then (using core model theory for this level) $j(K)$ is an iterate of $K$ and $j\upharpoonright K$ is the iteration map. But $\mathrm{crit}(j)=\kappa_k$. Therefore $\kappa_k$ is measurable in $K$ (as witnessed by its extender sequence), and because $j\upharpoonright K$ is the iteration map, letting $D_k$ be the unique normal measure on $\kappa_k$ in $K$ (uniqueness because otherwise we get a measure of Mitchell order 1), we have $D_k\subseteq\mathscr{F}_k$. It follows that $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]=L[D_1,\ldots,D_n]\subseteq K$, and that letting $\bar{\mathscr{F}}_k=\mathscr{F}_k\cap L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$, then $\bar{\mathscr{F}}_k$ is a normal, $\kappa_k$-complete measure on $\kappa_k$ in $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$, as desired. (With the case hypotheses as they are, it might be that $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]\subsetneq K$. I could have made the case hypothesis to case 1 be that the sharp for an inner model with $n$ measurables exists, and case 2 its negation. Then we would get that $K$ and $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$ have the same universe.)<|endoftext|> TITLE: From topological actions on $\mathbb{R}^3$ to isometric actions QUESTION [6 upvotes]: It is known that if a finite group $G$ admits a faithful topological action on the 3-sphere $S^3$, then $G$ admits a faithful action on $S^3$ by isometries. (Pardon proved that a topological action implies a smooth action, and Dinkelbach & Leeb proved that a smooth action implies an isometric one.) I wonder if this extends to infinite groups acting on $R^3$: Question: Let $G$ be a finitely generated group that admits a faithful, co-compact, topological action on $R^3$, such that no orbit has an accumulation point. Must $G$ admit an action by isometries on one of Thurston’s geometries, preserving the above properties (i.e. faithful, co-compact, accumulation-free)? Update: The comments below suggest that the answer is negative in this generality (an "official" answer with references and explanation would be welcome). What if G is assumed to be Gromov-hyperbolic? I'm most interested in the 1-ended case, anticipating an isometric action on $\mathbb{H}^3$. (1-endedness excludes $\mathbb{S}^2 \times \mathbb{R}$.) This is partly motivated by Cannon's conjecture. By topological action I mean an action by homeomorphisms. Update: instead of just assuming that no orbit has an accumulation point, I'm happy with stronger discreteness conditions such as proper discontinuity REPLY [2 votes]: I believe (though have not checked carefully) that the argument in my paper proves: If $\Gamma$ (discrete) acts continuously and properly discontinuously on a smooth three-manifold $M$, then that action can be uniformly approximated by a smooth action. The point is simply that each step in the argument is local on the quotient space $M/\Gamma$ (which is a reasonable topological space given proper discontinuity). Here is a (sketched) better argument, which proves the indented statement above as a consequence of my paper. Fix $x\in M$, and consider the stabilizer $\Gamma_x\leq\Gamma$, which is finite. Choose coset representatives $g_i\in\Gamma/\Gamma_x$, so $\Gamma x=\{g_ix\}_i$. Fix a $\Gamma_x$-invariant open neighborhood $U$ of $x$ whose translates $g_iU$ are all disjoint (should exist by proper discontinuity). Now smooth the action of $\Gamma_x$ on $U$ using my paper, and smooth the homeomorphisms $g_i:U\to g_iU$ using Bing--Moise. This determines a smoothing of the action of $\Gamma$ on $\Gamma U\subseteq M$. By making the approximations sufficiently $C^0$-close, we ensure that this smoothed action of $\Gamma$ on $\Gamma U\subseteq M$ splices together with the original action of $\Gamma$ on $M\setminus\Gamma U$ to define a new action of $\Gamma$ on $M$, which is now smooth over $\Gamma U$. Now iterate a (locally) finite number of times to cover all of $M$.<|endoftext|> TITLE: Stronger (?) form of Vopenka's principle QUESTION [5 upvotes]: A category $\mathcal{C}$ is called $\textbf{discrete}$ if the only morphisms are identity morphisms. Consider the following weaker notion: a category $\mathcal{C}$ is called $\textbf{totally disconnected}$ if $\text{Hom}_\mathcal{C}(C,D)=\varnothing$ for all $C\neq D$. Vopenka's principle ($\textbf{VP}$) states that a full large subcategory $\mathcal{D}$ of a locally presentable category $\mathcal{C}$ cannot be discrete. I want to consider an anaologus statement: ($\textbf{VP2}$) Full large subcategory $\mathcal{D}$ of a locally presentable category $\mathcal{C}$ cannot be totally disconnected. Since discrete categories are totally disconnected, $\textbf{VP2}$ implies $\textbf{VP}$. Hence $\textbf{VP2}$ can't be proven in $\textbf{ZFC}$, because $\textbf{VP}$ can't. Now I would like to know: (1) Is there a counter-example to $\textbf{VP2}$ in $\textbf{ZFC}$? As far as I know, no counter-example to $\textbf{VP}$ has been found but since $\textbf{VP2}$ is in principle a stronger statement, perhaps we can construct one here? (2) If not, what is the relation between $\textbf{VP}$ and $\textbf{VP2}$ inside $\textbf{ZFC}$? Are they equivalent? Or does at least consistency of $\textbf{VP}$ implies that of $\textbf{VP2}$? (I wrote $\textbf{ZFC}$ as my set theory of choice but if there are any problems in using such a weak theory, feel free to consider some other). Thanks! REPLY [5 votes]: VP2 is equivalent to VP because every set carries a rigid binary relation. This is similar as Lemma 6.3 in my book with Adámek. In fact, VP2 is an original formulation of VP (see Jech, Set Theory).<|endoftext|> TITLE: Questions about coherent topology QUESTION [8 upvotes]: For each coherent category $C$, let $J_C$ be the topology on $C$ in which a sieve $\{f_i\colon U_i\to X\}_{i\in I}$ is covering if and only if there exists a finite set $I_0\subseteq I$ such that $\bigcup_{i\in I_0} \operatorname{im}(f_i)=X$ as subobjects of $X$. (This is a Grothendieck topology by Proposition 12 of Lecture 8 (Grothendieck topologies) of Lurie's "Categorical logic" notes.) Is $J_C$ equivalent to what other people call "the coherent topology on $C$"? If $C$ is a coherent category which is Boolean, is the topos $\operatorname{Sh}(C, J_C)$ Boolean too? Can one find for each Boolean coherent topos $\mathcal E$ a Boolean coherent category $C$ such that both $\mathcal E\simeq \operatorname{Sh}(C, J_C)$ and there exists an object $X\in C$ such that every object of $C$ is a subobject of $X^n$? Is the étale topos of the spectrum of any field coherent? REPLY [10 votes]: Edit : I should clarify that I've interpreted "Etale topos" to mean the petit/small étale topos everywhere. What I've said about Grothendieck-Galois duality only apply to the petit étale topos. If you are talking about the Gros topos, then these part no longer holds. I actually don't know if the Gros étale topos of a fields has a boolean category of coherent object or not. Yes. I can't give you a proof because as far as I'm concerned this is the definition of the coherent topology. If you see a different definition, maybe edit your question! Essentially no. Take for example a Boolean algebra $B$. It can be seen as a coherent category (I see $B$ as a poset, and every poset as a category in the usual way). Then the associated topos is the topos of sheaves over the Stone spectrum of $B$, and unless $B$ is finite it has plenty of open that are not also closed (in fact the open that are complemented corresponds exactly to the element of $B$). The general case looks like this though: a Boolean coherent category will gives a topos that "looks like" a Stone spectrum. The answer is yes for the first half, no for the second, but only because there are very few coherent boolean topos. I would say, a coherent topos is essentially never Boolean (the only exception being the framework of Galois theory): A coherent topos has always enough points, and it can be proved that a boolean topos with enough points is "atomic", that is a disjoint union of topos of the form $BG_i$ where the $G_i$ are localic group. (Here $BG$ is the topos of sets endowed with a continuous action of the localic group $G$.) Adding back the fact that we want this topos to be coherent, we get that the Boolean coherent toposes are exactly the toposes that are finite coproducts of $BG_i$ where the $G_i$ are profinite groups. The second condition you ask for doesn't hold if some of the $G_i$ are non-discrete though. If $G$ is a profinite group (take $G = \mathbb{Z}_p$ for example), then a coherent $G$-set $X$ is a finite $G$-set, so there is going to be an open normal subgroup of $G$ that stabilise all the points of $X$, and the things you get as subobjects of $X^n$ will all be stabilised by the same subgroup. Yes. The étale topos of any affine scheme is coherent. (For a general scheme it is locally coherent. I'll let an algebraic geometer give you the precise condition under which we get coherence.) In fact, thanks to Grothendieck Galois duality, the étale topos of a field is one of the rare examples of Boolean coherent toposes: it is $BG$ where $G$ is the absolute Galois group of the field, with its profinite topology. Answering some of the follow up question in the comments. I would recomand to double checking what I'm going to say here if you plan on using it - I haven't looked at it in enough to details, but I think toposes associated to boolean coherent categories can be characterized as the coherent toposes in which coherent subobject of coherent object have complement. A topos satisfying these condition is clearly the topos of coherent sheaf on a booleancoherent category (by taking all coherent objects) but the converse also seems true. The other condition you have (every object of $C$ is a subobject a power of some fixed object $X$) should corresponds to have a coherent "pre-bound". The class descriebd you in (1) (including both conditions) are the topos that classyfies single sorted "boolean" first order theory. This additional condition of having a coherent prebound is not automatic at all as the exemple of the topos $BG$ for $G$ a non discrete profinite group shows. Yes. étale topos of fields are Boolean topos (from their explicit descrition given by Galois duality) so in particular coherent subobject have complements.<|endoftext|> TITLE: Do second-order theories always have irredundant axiomatizations? QUESTION [12 upvotes]: It's a standard exercise to show that every countable first-order theory has an irredundant axiomatization. For uncountable first-order theories, the result is much more difficult and was proved by Reznikoff. For $\mathcal{L}_{\omega_1,\omega}$, I believe the state-of-the-art for the analogous question is due to Hjorth/Souldatos following X. Caicedo, Independent sets of axioms in $L_{\kappa,\alpha}$, Canadian Mathematical Bulletin 22 (1981), 219–223. I don't recall seeing anything, however, about second-order logic: Suppose $T$ is a second-order theory - of arbitrary cardinality, in an arbitrary language. Must there be a second-order theory $S$ in the same language such that $(i)$ $S$ and $T$ are semantically equivalent (= have the same classes of models) but $(ii)$ no proper subset of $S$ is semantically equivalent to $S$? Again the countable case is easy, but the uncountable case is unclear to me (at a glance I don't think Reznikoff's argument generalizes). More generally, I'd be interested in any sources treating the "irredundance property" in a broader class of logics than just the $\mathcal{L}_{\kappa\lambda}$s. REPLY [9 votes]: Here is a proof of Reznikoff’s theorem I found among my notes, not quite following Reznikoff’s proof. I stared at it for a while, and it seems to apply to second-order logic just the same; in fact, the only property of first-order logic it uses is that any sentence contains only finitely many non-logical symbols, and there are only countably many sentences in any given finite language. Let me know if I missed something. We say that a theory $T$ in a language $L$ depends on a predicate or function symbol $s\in L$ if there are models $A\models T$, $B\let\nmodels\nvDash\nmodels T$ such that $A\let\res\restriction\res(L\let\bez\smallsetminus\bez\{s\})=B\res(L\bez\{s\})$. Let $L_T$ denote the set of symbols $s\in L$ on which $T$ depends. If $L'\subseteq L$, let $T\res L'$ be the set of $L'$-sentences valid in $T$. Lemma 1: If $A\models T$ and $A\res L_T=B\res L_T$, then $B\models T$. Proof: Fix $\phi\in T$, we will show $B\models\phi$. Let $\{s_i:i TITLE: Smallest relation in complement of partial order that prohibits its extension QUESTION [7 upvotes]: Let $P$ be a partial order on a finite set $S$ (assume that every element is related to at least one other element besides itself…this raises a few quick questions: is this implied by the definition of partial order and if not, what are the "isolated" points called and what is a partial order with no such points called?) and $R$ be the smallest subset of $(S\times S)\setminus P$ such that for all partial orders $Q\supseteq P$, $$R\cap Q=\varnothing\implies P=Q.$$ Is the relation $R$ unique? Does it have a name? Is it equivalent to something else that does? I asked a similar (perhaps the same) question in an unwieldy way on MSE at Minimal generating sub-relation of complement of partial order and got no response. ======= edit added by mathematrucker 16 Mar 2022 ======= While searching the literature unsuccessfully for @JosephVanName's nice "one pair extension property of finite partially ordered sets" below it occurred to me that it follows immediately from the fact that the poset under set containment of all unlabeled posets on $n$ points is graded by cardinality. This appears as Lemma 2.1 in New results from an algorithm for counting posets by Culberson and Rawlins (1990), which cites note (3.1) in Aigner, Producing posets (1981). Neither reference supplies a proof, so if anyone knows of one that does, please add it in the comments. REPLY [7 votes]: I claim that the relation $R$ is in fact unique. The uniqueness follows from the one pair extension property of finite partially ordered sets. Proposition: Suppose that $X$ is a finite set with partial ordering $P$. Then whenever $Q$ is a partial ordering on $X$ with $P\subseteq Q,P\neq Q$, there exists an ordered pair $(x,y)\in Q\setminus P$ such that $P\cup\{(x,y)\}$ is a partial ordering. Proof: Suppose that $(x,y)\in Q\setminus P$. Then $x\not\leq_{P}y$ and $y\not\leq_{P}x$. Let $x',y'$ be a pair such that $x'\leq_{P}x,y\leq_{P}y'$, $x'\not\leq_{P}y'$, and $$(x''\leq_{P}x',y'\leq_{P}y'')\implies(x''\leq_{P}y''\mbox{ or }x''=x',y''=y').$$ Let $T=P\cup\{(x',y')\}$. Note that since $y\leq y'$, $x'\leq x$, and $y\not\leq_{P}x$ we have $y'\not\leq_{P}x'$ and thus $(y',x')\not\in T$. I claim that $T$ is a partial ordering. The ordering $T$ is clearly reflexive. Suppose that $r\leq_{T}s,s\leq_{T}t$. If $(r,s),(s,t)\in P$, then $r\leq_{P}t$ by the transitivity of $P$. In the case that $(r,s)=(x',y')=(s,t)$, we have $r=s=t$, so $(r,t)\in T$ by reflexivity. Now, assume that $(r,s)=(x',y')$ and $s\leq_{P}t$. Then either $r\leq_{P}t$ or $s=t$. In either case, $r\leq_{T}t$. If we assume $(s,t)=(x',y'),r\leq_{P}s$, then either $r=s$ which would imply that $r\leq_{T}t$ or we would have $r\leq_{P}t$ which would imply that $r\leq_{T}t$ as well. Therefore, $T$ is transitive. Now, assume that $r\leq_{T}s\leq_{T}r$. If $(r,s)\neq(x',y')$ and $(s,r)\neq(x',y')$, then we know that $r\leq_{P}s\leq_{P}r$ which implies that $r=s$. On the other hand, we can assume without loss of generality that $(r,s)=(x',y')$. Since it was shown above that $y'\not\leq_{T}x'$ we conclude that $T$ is antisymmetric. Hence $T$ is a partial ordering. Observe that $x'\leq_{Q}x\leq_{Q}y\leq_{Q}y'$, so $(x',y')\in Q$. Q.E.D. Therefore, if we let $R$ be the collection of all ordered pairs $(x,y)\in X^{2}\setminus P$ where $P\cup\{(x,y)\}$ is a partial ordering, then if $Q$ is a partial ordering on $X$ with $P\subseteq Q$ and $Q\cap R=\varnothing$, then $Q=P$ (otherwise a new pair could be added to $R$, contradicting its definition).<|endoftext|> TITLE: Kirby diagrams: sliding 1-handles over 1-handles and ribbon disks QUESTION [5 upvotes]: Consider the Kirby diagram $ D$ given by a 2-component unlink, both dotted circles. In general, when performing a 1-handle slide over another 1-handle, the band chosen must not link any dotted circle, therefore the handle slide $D\to D'$ resulting in the diagram $D'$ below is illegal. This is because the collection of dotted circles is required to bound a collection of boundary-parallel disks (so that their removal is equivalent to the addition of 1-handles). On the other hand, the blue dotted circle in $D'$ can be given a meaning because it bounds a ribbon disk. This is shown in Gompf-Stipsicz, 4-manifolds and Kirby calculus pg. 213 where they also extend the Kirby diagram language by allowing such kind of dotted circles in diagrams (that now will denote a ribbon disk deletion). They explicitely say that this has also the advantage of performing "illegal" 1-handle slides. What I am looking for is a clear statement and an explanation. 1)Is it correct that we can slide dotted-circles over a dotted-circle, even if the band goes inside some dotted-circles, just by interpreting the result as deleting a ribbon disk (instead of the usual boundary-parallel disk)? If this was true the example below would be a diagram for $(\mathbb{S}^1\times \mathbb{D}^3)\#_b(\mathbb{S}^1\times \mathbb{D}^3)$ (boundary connected sum) . If the answer is positive, then how does this affect cancellation of 1h/2handles pairs? For example, add to $D'$ a 2-handle along a meridian to the red dotted circle. That is a pair of complementary handles, but it is not clear if we can cancel them since the blue circle enters inside the red one (and we do not slide 1-handles over 2-handles). The diagram $D'$ REPLY [2 votes]: 1) Yes, it's correct. In sec. 6.2 of their book Gompf-Stipsicz extend their notation so that now a dotted circle denotes the removal of a ribbon disk from $\mathbb{D}^4$, notice that this extension of notation also contemplates the use of some bands in the diagram to reconstruct the ribbon moves (and hence the disk) without ambiguities. Also, I should say that in this extended notation a dotted circle will not correspond anymore to a 1-handle so the title of the post is a bit misleading. Anyway, consider two dotted circles $A$, and $B$ in a diagram, then given any band a slide of $A$ following the band makes sense and does not change the topology of the 4-manifold. This is because you are actually performing an isotopy of the ribbon disk bounded by $A$, lets call it $D_A$. Imagine to push $D_A$ with your finger tip along the band and finally spreading it over a parallel copy of $D_B$ (i.e. consider a tubular neighbouhood of $D_B$, of the form $D_B\times \mathbb{D}^2$ and use a copy $D_B\times \{pt\}$, $pt\neq 0$ as a guide to finalize the move) the result of this "spreading" introduces the push-off of $B$ that one sees after the slide is performed. Observe that this is similar (see comment section below) to the slide of a 2-handle over a 2-handle, except that now all it's happening in $\mathrm{int}(\mathbb{D}^4)$, so inside the 4-ball instead of above it. You can check that at each time $T\in [0,1]$ of the isotopy, the deformed disk $D_A$ does not meet $D_B$. For a time $T$ before reaching the parallel copy of $D_B$ you can see it with a picture like the one below which shows an embedding of the deformed $D_A$ and $D_B$ at time $T$ inside $\mathbb{S}^3\times [0,1]$ (where $\mathbb{S}^3\times \{0\} = \partial \mathbb{D}^4$) so that the two disks do not meet. In the picture at the levels $\mathbb{S}^3\times\{r\}$ for $r TITLE: Is orientability a miracle? QUESTION [57 upvotes]: $\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}$This question is prompted by a recent highly-upvoted question, Conceptual reason why the sign of a permutation is well-defined? The responses made me realize that my intuition differs from that of many other mathematicians, in a way that I had previously been unaware of. As a bit of personal background, I recall being taught in middle school that (in not so many words) that there are "24 symmetries of a cube," since you can rotate any face to the bottom (a factor of 6) and then rotate that face in place (a factor of 4). Similarly, for each of the Platonic solids, we can count $4\times 3 = 12$ "symmetries" of a tetrahedron, $8\times 3 = 24$ "symmetries" of an octahedron, and so forth. One of the various equivalent formulations of the aforementioned MO question is, "What is the conceptual explanation for the existence of the alternating group?" In part because of my background, the answer that came to my mind was, "Because it's the group of symmetries of a simplex." However, people rightly pointed out that to conform to standard usage, this answer should really be phrased, "Because it's the group of orientation-preserving symmetries of a simplex." But once the statement is phrased this way, it suggests that maybe the existence of the alternating group isn't such a "basic" fact after all; maybe it's only the symmetric group whose existence can be taken as basic, and we have to "explain" the existence of a subgroup of index 2 using some kind of algebraic argument. There's no question that Poonen's version of Cartier's argument is slick and beautiful. Nevertheless, something doesn't quite sit right with me if we call this an "explanation" of the existence of the alternating group. It still seems to me that orientation-preserving rigid motions in Euclidean space are mathematically fundamental, because they are rooted in our physical intuition. Admittedly, in modern mathematics, formally capturing our physical intuition in a direct manner is a cumbersome process; we have to construct the real numbers, and then talk about continuous transformations that preserve the metric. But for me, the complexity of this formal definition does not imply that the underlying concept is complex; rather, it says more about the difficulties involved in formalizing our geometric intuition. It is not hard for me to imagine an alternative universe in which we begin with $\SO(n)$, and think of $\O(n)$ as an extension of $\SO(n)$, much as Spin came before Pin. But let me now finally come to my question. For those who take the symmetric group as given, and feel the need for an algebraic explanation of a subgroup of index 2, do you feel the same way about other finite subgroups of $\SO(n)$? For example, do you think the existence of the hyperoctahedral group of order $2^n n!$ is "obvious," but the subgroup of index 2 is in need of an algebraic "explanation"? If so, is there an algebraic proof that you feel explains all these "miraculous" subgroups of index 2 in a uniform way? Or is the existence of orientability itself a miraculous fact? EDIT: Let me make a few additional comments to nudge the discussion in a more technical and less opinion-based direction. Suppose I don't buy the "argument from geometric intuition" and I seek a more algebraic (or combinatorial) explanation of the existence of the alternating group. It still seems to me that an argument based on determinants, or the distinction between $\SO(n)$ and $\O(n)$, would be a more "conceptual" route than an argument based on actions on the complete graph. For example, suppose I ask the corresponding question for the hyperoctahedral group ("why is there this subgroup of index 2?"). If we've already decided that the relationship between $\SO(n)$ and $\O(n)$ is the key point, then the same explanation works for both questions. Whereas, the graph-theoretic approach doesn't seem to have the same ability to give a unified explanation for both cases. Am I mistaken about this? Does Cartier's proof generalize to cover all situations that I might try to explain by appealing to the distinction between $\SO(n)$ and $\O(n)$? Does it perhaps generalize even further to explain index-2 subgroups that don't seem to be explicable in terms of $\SO(n)$? REPLY [6 votes]: In the spirit of the answer by David Speyer, one could also point out that perhaps the important role of 2 (or bilateral symmetry) in our study of symmetry can be seen as an artefact of living in a world where the "special valuation at $\infty$" dominates our intuition. (Since $\mathbb{Z}/\langle 2\rangle$ occurs as the maximal finite subgroup of $\mathbb{R}^{*}$.) If we lived in a $p$-adic universe for a specific $p\neq\infty$, then the maximal finite subgroup of the group $\mathbb{Z}_p^{*}$ (the units in $p$-adic integers) may have played a bigger role in our intuition. The above does not quite work the way I imagined! The group generated by rotations (defined as the product of two reflections) always (over a field of characteristic different from 2) generates a subgroup of index 2 in the group generated by reflections. My answer to the earlier question mentioned above gives an elementary proof. So, in this context, the homomorphism to $\mathbb{Z}/\langle 2\rangle$ does appear to play a special role. Another way to see is by appealing to the Cartan-Dieudonné theorem and determinants.<|endoftext|> TITLE: What is the liminf of a sum of i.i.d. random variables with heavy tails? QUESTION [7 upvotes]: Let $\{X_i\}_{i=1}^{\infty}$ be i.i.d. random variables such that: i) $X_i > 0$; and ii) $\textrm{Pr}[X_i > x] \sim x^{-\alpha}$ at large $x$ for some $\alpha \in (0, 1)$. Define the quantities \begin{equation} S_n \equiv \frac{X_1 + \cdots + X_n}{n^{1/\alpha}}. \end{equation} By Kolmogorov's zero-one law (right?), there is some $\mu \in [0, \infty]$ such that $\liminf_{n \rightarrow \infty} S_n = \mu$ almost surely. I just want to know whether $\mu$ is zero, finite, or infinite (and see a proof). I haven't been able to find an answer to this question anywhere (note that I'm specifically interested in $\alpha < 1$ and that $S_n$ is normalized differently than the usual mean). For a bit of context, I can easily show/find in textbooks that $\limsup_{n \rightarrow \infty} S_n = \infty$. But those textbooks never seem to mention the liminf. The closest result I could find is that when $\alpha = 1$, $\liminf_{n \rightarrow \infty} S_n = \infty$. But this is shown via truncation + SLLN, and I don't see a way to generalize it to $\alpha < 1$. More generally, one could define \begin{equation} S_n^{(\beta)} \equiv \frac{X_1 + \cdots + X_n}{n^{\beta}}. \end{equation} Then all the following are not too difficult to prove: i) for any $\beta < 1/\alpha$, $\lim_{n \rightarrow \infty} S_n^{(\beta)} = \infty$; ii) for any $\beta > 1/\alpha$, $\lim_{n \rightarrow \infty} S_n^{(\beta)} = 0$; and iii) for $\beta = 1/\alpha$, $\limsup_{n \rightarrow \infty} S_n^{(\beta)} = \infty$ (all of these holding almost surely). Thus the question of the liminf when $\beta = 1/\alpha$ is the one missing (to me) piece to understanding the scaling of the partial sums, and I'm really surprised that I haven't been able to find it anywhere. REPLY [3 votes]: $\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$Here we implement the idea proposed in mike's answer. The proof below works for all $\alpha\in(0,2]$. Let $a:=1/\alpha$. Let \begin{equation*} Z_n:=X_1+\cdots+X_n, \end{equation*} so that \begin{equation*} S_n=Z_n/n^a. \end{equation*} Take any real $\ep>0$. We have \begin{equation*} S_n\to S \tag{1}\label{1} \end{equation*} in distribution as $n\to\infty$ for some nonnegative random variable (r.v.) $S$ such that $P(S\ge\ep)<1$, so that \begin{equation*} q:=\frac{1+P(S\ge\ep)}2<1; \tag{2}\label{2} \end{equation*} see e.g. Attraction domain of a stable distribution and Stable distribution. For each natural $k$, let $C_k\in(0,\infty)$ be such that $P(S\ge C_k)\le1/k^2$. It follows from \eqref{1} that for some strictly increasing sequence $(n_k)$ of natural numbers and all natural $k$ we have \begin{equation*} n_{k+1}-n_k\to\infty \tag{3}\label{3} \end{equation*} as $k\to\infty$, \begin{equation*} P(S_{n_k}>C_k)\le P(S\ge C_k)+1/k^2\le2/k^2, \end{equation*} and \begin{equation*} \ep_k:=\frac{2\ep n_{k+1}^a-C_k n_k^a}{(n_{k+1}-n_k)^a}>\ep. \tag{4}\label{4} \end{equation*} Introducing now the event \begin{equation*} A_m:=\{\forall k\ge m\ S_{n_k}\le C_k\}, \end{equation*} we have \begin{equation*} 1-P(A_m)\le\sum_{k\ge m}P(S_{n_k}>C_k)\le\sum_{k\ge m}2/k^2\to0 \tag{5}\label{5} \end{equation*} as $m\to\infty$. Moreover, letting \begin{equation*} T_k:=\frac{Z_{n_{k+1}}-Z_{n_k}}{(n_{k+1}-n_k)^a}, \end{equation*} we have \begin{equation*} \begin{aligned} & P(A_m,\forall k\ge m\ S_{n_k}>2\ep) \\ &=P(A_m,\forall k\ge m\ Z_{n_k}>2\ep n_k^a) \\ &\le P(\forall k\ge m\ Z_{n_{k+1}}-Z_{n_k}>2\ep n_{k+1}^a-C_k n_k^a) \\ &=P(\forall k\ge m\ T_k>\ep_k) \\ &\le P(\forall k\ge m\ T_k>\ep) \\ &=\prod_{k=m}^\infty P(T_k>\ep), \end{aligned} \tag{6}\label{6} \end{equation*} by \eqref{4} and the independence of the $X_i$'s. By \eqref{3}, $T_k\to S$ as $k\to\infty$ (cf. \eqref{1}), so that, by \eqref{2}, $P(T_k>\ep)\le q<1$ for all large enough $k$. So, by \eqref{6}, for all natural $m$ we have $P(A_m,\forall k\ge m\ S_{n_k}>2\ep)=0$ and hence for \begin{equation} B_m:=\{\forall k\ge m\ S_{n_k}>2\ep\} \end{equation} we have \begin{equation*} P(B_m)\le1-P(A_m)\to0 \end{equation*} as $m\to\infty$, by \eqref{5}. But $B_1\subseteq B_2\subseteq\cdots$. So, $P(B_m)=0$ for all natural $m$ and hence \begin{equation*} P(\exists m\ \forall k\ge m\ S_{n_k}>2\ep)=P(B_1\cup B_2\cup\cdots)=0. \end{equation*} So, \begin{equation*} P(\liminf_{n\to\infty}S_n\le2\ep)\ge P(\forall m\ \exists k\ge m\ S_{n_k}\le2\ep)=1, \end{equation*} for any real $\ep>0$. Thus, \begin{equation*} P(\liminf_{n\to\infty}S_n=0)=1. \end{equation*}<|endoftext|> TITLE: Coordinates on $N_+ \backslash \overline{B_+ w B_+} / N_+$ QUESTION [8 upvotes]: Let $G = \text{GL}_n(\mathbb{C})$ and let $N_+$ be the subgroup of upper triangular matrices with $1$'s on the diagonal. Let $w$ be a permutation, let $B_+ w B_+$ be the Bruhat cell and let $\overline{B_+ w B_+}$ be its closure in $G$. I want to describe the ring of functions on $\overline{B_+ w B_+}$ which are invariant for $N_+ \times N_+$ acting on the left and right. I believe I know what the answer is, and I would like to know if I am right and get references to previous work. Additional notation: Let $[n] := \{ 1,2,\ldots, n \}$. Let $T$ be the torus of diagonal matrices in $G$, and let $(t_1, t_2, \dots, t_n)$ be the entries of the diagonal matrix. Recall that the permutation matrix $w$ has $1$'s in position $(w(j), j)$. Here is what I believe the answer to be. For any $I$, $J \subset [n]$ with $|I| = |J|$, and $g \in \text{GL}_n$, let $\Delta_{I,J}(g)$ be the minor with rows $I$ and columns $J$. Define a partial order $\prec$ on $[n]$ by $j_1 \prec j_2$ iff $j_1 < j_2$ and $w(j_1) > w(j_2)$. So, for $w= w_0$, this is the total order $1 \prec 2 \prec \cdots \prec n$ and, for $w = e$, all the elements of $[n]$ are incomparable. Let $J$ be any lower order ideal for this partial order. Then one can verify that the minor $\Delta_{w(J), J}(g)$ is invariant for the $N_+ \times N_+$ action on $\overline{B_+ w B_+}$. We'll call this $\Delta(J)$. In addition, we have $\Delta([n]) = \Delta_{[n], [n]}(g)=\det(g)$, so $\Delta([n])$ is a unit and we have to include $\Delta([n])^{-1}$ in our ring of invariants. So, question: Is the ring of $N_+ \times N_+$ invariant functions genearted by the $\Delta(J)$'s, and by $\Delta([n])^{-1}$? What is a reference for this? I'll close by noting there is a good, explicit description of the ring generated by the $\Delta(J)$'s. If we restrict $\Delta(J)$ to $wT \subset B_+ w B_+$, we see that $\Delta(J) = \pm \prod_{j \in J} t_j$, so we can think of this as the monomial $\prod_{j \in J} t_j$ on $wT \cong N_+ \backslash B_+ w B_+ / N_+$. Since each orbit of $N_+ \times N_+$ on $B_+ w B_+$ contains a unique point in $wT$, the relations between the $\Delta(J)$'s are exactly the same as the relations between these monomials on $T$. Any product of the $\Delta(J)$'s gives a monomial of the form $\prod t_j^{a_j}$ where $j_1 \prec j_2$ implies $a_{j_1} \geq a_{j_2}$. In other words, the ring generated by the $\Delta(J)$'s is the semigroup ring corresponding to the semigroup of linear extensions of $([n], \prec)$, and is closely related to Stanley's order polyope $\mathcal{O}(\prec)$. REPLY [2 votes]: I have proved that the invariant ring is as I expected. As discussed in the question, $N_+ \backslash B_+ w B_+ / N_+ \cong T$, so the ring of $N_+ \times N_+$ invariants on $B_+ w B_+$ is the coordinate ring of $T$, which is the Laurent polynomial ring in the $t_i$. Note, let $R$ be the ring of $N_+ \times N_+$ invariants on $\overline{B_+ w B_+}$, so $R$ is a subring of $\mathbb{C}[t_1^{\pm}, t_2^{\pm}, \ldots, t_n^{\pm}]$. Moreover, the $T$ action on $\overline{B_+ w B_+}$ normalizes $N_+ \times N_+$, so we get a $T$-action on $R$, so $R$ must be a $T$-invariant subspace of $\mathbb{C}[t_1^{\pm}, t_2^{\pm}, \ldots, t_n^{\pm}]$. So $R$ is a monomial subring of $\mathbb{C}[t_1^{\pm}, t_2^{\pm}, \ldots, t_n^{\pm}]$. We need to show that a monomial $t_1^{a_1} t_2^{a_2} \cdots t_n^{a_n}$ extends to a function on $\overline{B_+ w B_+}$ if and only if $j_1 \prec j_2$ implies $a_{j_1} \geq a_{j_2}$. The question shows that, if $a$ obeys these inequalities, then $t_1^{a_1} t_2^{a_2} \cdots t_n^{a_n}$ does extend to a function on $\overline{B_+ w B_+}$ , so what we need is the reverse implication. So, suppose that there are $j_1$ and $j_2$ with $j_1 \prec j_2$ and $a_{j_1} < a_{j_2}$. We may assume that $(j_1, j_2)$ is a cover of $\prec$, meaning that there is no $j$ with $j_1 \prec j \prec j_2$. In this case, $w$ covers $w (j_1 j_2)$ in the strong Bruhat order, so $B_+ w (j_1 j_2) B_+$ is a divisor in $\overline{B_+ w B_+}$. We will show that the function $\prod t_j^{a_j}$ does not extend to $B_+ w (j_1 j_2) B_+$. Put $i_1 = w(j_1)$, $i_2 = w(j_2)$. Consider matrices $X$ whose only nonzero entries are in positions $(w(j), j)$, together with $(i_1, j_2)$ and $(i_2, j_1)$. As long as $X_{i_1 j_1}$ is nonzero, this is in $B_+ w B_+$; when $X_{i_1 j_1}$ becomes $0$, we pass into $B_+ w (j_1 j_2) B_+$. We evaluate the functions $t_j$ on such a matrix: We have $t_j = X_{w(j) j}$ for $j \neq j_1$, $j_2$, we have $t_{j_1} = X_{i_1 j_1}$, and we have $t_{j_2} = \frac{X_{i_2 j_1} X_{i_1 j_2} - X_{i_1 j_1} X_{i_2 j_2}}{X_{i_1 j_1}}$. We assumed that $a_{j_1} < a_{j_2}$. Plugging the above formulas into the product $\prod t_j^{a_j}$, we see that $X_{i_1 j_1}$ appears with exponent $a_{j_2} - a_{j_1}$ in the denominator. So, as $X_{i_1 j_1} \to 0$, the monomial $\prod t_j^{a_j}$ blows up, and thus doesn't extend to $B_+ w (j_1 j_2) w B_+$. Still glad to hear references!<|endoftext|> TITLE: How to generate all triangulations of an orientable surface? QUESTION [12 upvotes]: $\newcommand{\comb}{\mathrm{comb}}$Consider an orientable surface $S$ with punctures and boundaries (each boundary having at least a marked point). A triangulation, up to orientation preserving homeomorphisms, of $S$ is described by a collection $T_{\comb}$ of triangles $t$ which themselves are cyclically ordered triples of labels associated to each edge, $t=(e_1,e_2,e_3)$. Let me call these data a combinatorial triangulation. A combinatorial triangulation $T_{\comb}$ does not label unically a triangulation up to homeomorphism, because one can further permute the edge labels $e_i \to e_{\sigma(i)}$. What are efficient ways to generate all combinatorial triangulations $T_{\comb}$ up to permutation of the edges? In other words, how to generate one combinatorial triangulation for each triangulation of the surface up to homeomorphism? One way which I know is to use the Whitehead theorem, which says that any two triangulations are related by a sequence of flips (remove one edge from the triangulation and replace it with the other diagonal in the square thus formed). Then one can start from a triangulation, and sequentially flip all edges and then repeat with the triangulations thus obtained. Are there better options? Clarification: The triangulations I consider are such that the edges are at the marked points/punctures. For instance for a disk with 4 marked points on the boundaries we have two triangulations: Let a=(12), b=(23),c=(34),d=(41),e=(13),f=(24), then the two triangulations are ((a,b,e),(c,d,e)) and ((d,a,f),(b,c,f)). REPLY [10 votes]: Thom Sulanke has a program and a paper about this. Check out https://tsulanke.pages.iu.edu/graphs/surftri/ . Incidentally, Sulanke's method does not require the triangulations to be stored, so memory is not a limitation. An implementation for the planar case is plantri which we have used to make more than $10^{13}$ triangulations .<|endoftext|> TITLE: Short time existence for fully nonlinear parabolic equations QUESTION [6 upvotes]: I am trying to assert short time existence for a fully nonlinear equation of the general form \begin{equation} \begin{cases} u_t = F(x,u,Du,D^2u) & \text{in }(0,T)\times\Omega\\ u(\cdot,0) = u_0(\cdot) & \text{in }\Omega \\ u(x,t) = 1 & \text{on }(0,T)\times\partial\Omega \end{cases} \end{equation} where $u_0$ is smooth, $F=F(x,z,p,A)$ is smooth in all entries and concave in $A$, and $\Omega\subset\mathbb{R}^n$ is a smooth bounded domain. I know that the initial data is uniformly elliptic, in the sense that there are constants $0 TITLE: If $L_\alpha \vDash ZFC$, then do we have $L_{\alpha+1} \vDash \alpha\text{ is inaccessible}$? QUESTION [6 upvotes]: Here we choose the definition of "is a cardinal" as there is no surjective map from a smaller ordinal to it. It's easy to prove that, if $L_{\alpha+1} \vDash\ \alpha\text{ is inaccessible}$, then $L_\alpha \vDash ZFC$. Also it's easy to prove that if $L_\alpha \vDash ZFC$, then α is a cardinal in $L_{\alpha+1}$. However, I don't know how to prove confinally many cardinals won't collapse at the next step of L. If the answer is false, then is it true at the least such α? REPLY [12 votes]: Yes. The elements of $L_{\alpha+1}$ are exactly those subsets of $L_\alpha$ which are definable from parameters over $L_\alpha$. But $L_\alpha\models\mathrm{ZFC}$, so from here we can just use the usual proof that second order ZFC implies inaccessibility. That is: (i) every bounded subset of $L_\alpha$ which is in $L_{\alpha+1}$, is already in $L_\alpha$, and so $L_{\alpha+1}$ agrees with $L_\alpha$ regarding power sets and cardinalities of elements of $L_\alpha$; that is, for each $x\in L_\alpha$, we have $\mathcal{P}(x)^{L_{\alpha+1}}=\mathcal{P}(x)^{L_\alpha}$, and $\mathrm{card}(x)^{L_{\alpha+1}}=\mathrm{card}(x)^{L_\alpha}$; in particular, $L_{\alpha+1}\models$"$\alpha$ is a strong limit cardinal", and (ii) for every $\beta<\alpha$ and every $f:\beta\to\alpha$ which is in $L_{\alpha+1}$, $f$ is bounded in $\alpha$; that is, $L_{\alpha+1}\models$"$\alpha$ is regular".<|endoftext|> TITLE: Infinite series for $1/\pi$. Is it known? QUESTION [7 upvotes]: Indirect method (associated with a certain problem of electrostatics) indicates that $$\sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!}=\frac{2}{3\pi}.$$ Is this result known? REPLY [19 votes]: Using the standard power series for the complete elliptic integral of the second kind $$E(k) = \frac{\pi}{2} \sum_{j=0}^\infty \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j},$$ we find \begin{align*} \sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} k^{2j}&=\sum_{j=1}^\infty\frac{-j}{j+1} \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j} \\ &= -\frac{1}{k^2} \int_0^k\mathrm{d}k\,k^2 \frac{\mathrm{d}}{\mathrm{d}k}\left(\frac{2}{\pi}E(k)\right)\\ &= \frac{2}{3}\frac{k^2-1}{k^2}\frac{2}{\pi}K(k) - \frac{k^2-2}{3k^2}\frac{2}{\pi}E(k). \end{align*} In the limit $k\to 1$ only the second term survives with $E(1)=1$ and therefore \begin{align*} \sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} &=\frac{2}{3\pi}. \end{align*}<|endoftext|> TITLE: A knot in the solid torus and a Mazur manifold QUESTION [9 upvotes]: Part 1: The following picture is from Saveliev's book Lectures on Topology of 3-manifolds, page 130: He indicates that the knot drawn in the solid torus $S^1 \times D^2$ is homologous to $S^1 \times \{ 0\} \subset S^1 \times D^2$. How can we prove such a claim? Do we know the classification of null-homologous knots in $S^1 \times D^2$? Part 2: To obtain a Mazur manifold $W$, he says that we attach a $2$-handle to $S^1 \times D^3$ (the dotted circle) with framing $3$. Diagrammatically, we have: If the knot is not null-homologous, where does the framing $3$ come from? REPLY [4 votes]: As @MarcoGolla mentioned, the framing can be controlled due to Akbulut's carving technology: a dotted circle notation. It was introduced in the following article: Akbulut, Selman. "On 2-dimensional homology classes of 4-manifolds." Mathematical Proceedings of the Cambridge Philosophical Society. Vol. 82. No. 1. Cambridge University Press, 1977. Let $U$ be the unknot in $S^3$ and $D_U$ be the ribbon disk in $B^4$ with $\partial D_U = U$. Observe that $S^1 \times B^3$ is the ribbon disk exterior of $D_U$, i.e., it is diffeomorphic to $B^4 \setminus \nu(D_U)$ where $\nu(D_U) \approx D_U \times B^2$. Consider a ribbon knot with a ribbon disk $(K,D) \subset (S^3,B^4)$. Similarly, one can try to understand the $4$-manifold $B^4 \setminus \nu(D)$. The procedure of the construction of the Kirby diagram was answered, for instance here. Once it is understood, one can also put a dot on the ribbon disk exterior. The reference is again Akbulut's book Section 1.1 and 1.4 about carving ribbon disks: Akbulut, Selman. 4-manifolds. Vol. 25. Oxford University Press, 2016. It equivalently represents the ribbon disk exterior. See again Exercise 1.10 and Figure 1.22 in Akbulut's book.<|endoftext|> TITLE: Discrete cocompact group of isometries of Nil QUESTION [7 upvotes]: Is it true that every group quasi-isometric to the Heisenberg group admits a proper cocompact action by isometries on Nil? REPLY [7 votes]: This is true. The first step is that the group $\Gamma$ then has a subgroup of finite index that embeds as a lattice in the Heisenberg group $H$. This step is sketched in this answer (I repeated the argument here below "original answer"). To deduce the general case, let me relax assumptions.$\DeclareMathOperator\Aut{Aut}$ Proposition. Let $\Gamma$ be a discrete group with a finite index subgroup $\Lambda$ that embeds as a lattice into a simply connected nilpotent Lie group $G$. Let $K$ be a maximal compact subgroup of $\Aut(G)$. Let $W$ be the finite radical of $\Gamma$ (= the maximal finite normal subgroup, which exists here). Then $\Gamma/W$ embeds as a lattice into $K\ltimes G$. Proof: first step: there exists a Lie group $H$ of the form $H^0\rtimes L$ with $L$ finite, $H^0$ simply connected nilpotent, in which $\Gamma$ maps as a lattice with finite kernel. Fix an embedding of $G$ into the upper unipotent subgroup of $\mathrm{GL}_n(\mathbf{R}$ for some $n$. This defines a representation of $\Lambda$. Induce it to $\Gamma$. We obtain a representation $f$ of $\Gamma$, for which $\Lambda$ acts unipotently. Then the Zariski closure of $f(\Lambda)$ is a simply connected nilpotent Lie group in which $f(\Lambda)$ is a lattice. Let $H$ be the Zariski closure of $f(\Gamma)$. Then $H^0$ is the Zariski closure of $f(\Lambda)$, and has finite index in $H$. Let $L$ be a maximal compact subgroup of $H$. Then $LH^0=H$ (this is true for every virtually connected Lie group, result of Mostow). Since $L\cap H^0=1$, we deduce $H=H^0\rtimes L$ and $L$ is finite. So the first step is proved. Now conclude. Consider $H=H^0\rtimes L$ as in the first step. Without loss of generality, we can suppose that $L$ acts faithfully on $H$ (otherwise, mod out by the kernel). So $L\subset\Aut(H)$. Hence $L$ is contained in a maximal compact subgroup of $\Aut(H)$. We are now done (the assumption then now force the kernel to be exactly the finite radical of $\Gamma$). Corollary. There exists a left-invariant Riemannian metric on $G$ and a geometric (=proper isometric cocompact) action of $\Gamma$ on $G$. Proof: choose a left-$G$-invariant right-$K$-invariant Riemannian metric on $G$. Then the action of $G\ltimes K$ on $G$ is geometric (=continuous proper isometric cocompact), and hence so is that of $\Gamma$. Original answer (which passes to a finite index subgroup) It (essentially) directly follows from Gromov's theorem that a f.g. group quasi-isometric to $\mathsf{Nil}$ is virtually isomorphic to the integral Heisenberg group. (No need to assume the existence of an action.) Indeed, such a group has polynomial growth, so is virtually nilpotent and hence virtually isomorphic to a lattice in some simply connected nilpotent Lie group $G$. Once we know that $G$ is the 3-dimensional Heisenberg group, one is done since every lattice therein contains with finite index a copy of the integral Heisenberg group. One can directly invoke Pansu's 1988 theorem to deduce that the associated Carnot Lie algebra of $G$ is isomorphic to the Heisenberg group, and hence $G$ itself as well (since $G$ is 2-step-nilpotent it is Carnot). Alternatively, one can avoid Pansu's theorem, saying that $G$ has growth $n^4$ which leaves only 2 candidates: $\mathbf{R}^4$ and the Heisenberg group. So one just has to know that they are not quasi-isometric and this follows form various arguments, e.g., using asymptotic cones (which is homeomorphic to $\mathbf{R}^4$ for $\mathbf{R}^4$ and to $\mathbf{R}^3$ for the other).<|endoftext|> TITLE: Function that produces primes QUESTION [65 upvotes]: For any $n\geq 2$ consider the recursion \begin{align*} a(0,n)&=n;\\ a(m,n)&=a(m-1,n)+\operatorname{gcd}(a(m-1,n),n-m),\qquad m\geq 1. \end{align*} I conjecture that $a(n-1,n)$ is always prime. To verify it one may use this simple PARI prog: a(n)=my(A=n, B); for(i=1, n-1, B=n-i; A+=gcd(A,B)); A; The sequence begins $$3, 5, 7, 11, 11, 13, 17, 17, 19, 23, 23, 29, 29, 29, 31, 41, 53, 37, 41, 41$$ The sequence is not in the OEIS. Is there a way to prove it? REPLY [23 votes]: Extended comment, generalizing @IlyaBogdanov's comment about $2n-1$. Fix $n$ and let $$x_m = a(m, n) + n - m - 1.$$ Then $(x_m)$ obeys the similar recurrence $$ x_m = x_{m-1} + \gcd(x_{m-1}, n-m) - 1. $$ Also $x_0 = 2n-1$ and $x_{n-1} = a(n-1, n)$. Now from the recurrence it is clear that if $x_m$ is prime for any value of $m < n$ then $x_{n-1} = x_m$. While $x_m$ is not prime it is increasing slowly and erratically, so it has plenty of chances to hit a prime.<|endoftext|> TITLE: Geometric definition of divergence using curvature mentioned in Tristan Needham QUESTION [8 upvotes]: In page-479 of Visual Complex Analysis, Tristan Needham derives the flux of a vector field in Geometric form: $$ \nabla \cdot X = \partial_s |X| + \kappa_p |X|$$ The $\partial_s$ is a derivative along streamlines of the vector field $X$ an $\kappa_p$ is the curvature of the orthogonal streamline at the point we are taking divergence at. Where did this formula originate from, and, what other sources discuss it? There is nothing of it on wikipedia. The book itself claims it couldn't find any reference to the formula in Modern literature, but is it still true in 2022 ? REPLY [10 votes]: Given any hypersurface of a Riemannian manifold with a unit normal vector field $\nu$, extend $\nu$ to be unit length. Then $$\operatorname{div}(u\nu)=u\operatorname{div}\nu+\text{d}u(\nu).$$ This proves that if $X$ is any nonvanishing vector field on a neighborhood of the hypersurface, which is orthogonal to the hypersurface, then $$\operatorname{div}X=|X|H+\text{d}|X|(\nu)$$ where $H$ is the mean curvature. And so, when only given a Riemannian manifold, this holds also when $X$ is a hypersurface-orthogonal vector field (also called "complex lamellar vector field"); $H$ is the mean curvature of the orthogonal hypersurface and $\nu$ is $\frac{X}{|X|}$. Every vector field on a two-dimensional space is hypersurface-orthogonal, so this fully contains Needham's formula as a special case.<|endoftext|> TITLE: When are bundles of odd and even differential forms isomorphic? QUESTION [9 upvotes]: Let $M$ be a compact oriented $n$-manifold. Denote $\Omega^k := {\bigwedge}^k T^*M$ the vector bundle of differential $k$-forms, and let $\Omega^{\text{odd}} := \bigoplus_{\text{$k$ odd}} \Omega^k$ and $\Omega^{\text{even}} := \bigoplus_{\text{$k$ even}} \Omega^k$ be the bundles of odd and even differential forms, respectively. When are $\Omega^{\text{odd}}$ and $\Omega^{\text{even}}$ isomorphic as vector bundles over $M$? A few observations: If $n$ is odd they are isomorphic, $\Omega^{\text{odd}} \simeq \Omega^{\text{even}}$, as can be seen e.g. by introducing a Riemannian metric $g$ on $M$, and observing that the Hodge star $\star: \Omega^k \to \Omega^{n - k}$ is an isomorphism for eack $k$ and hence also between $\Omega^{\text{odd}}$ and $\Omega^{\text{even}}$. More generally, if $\chi(M) = 0$, then $\Omega^{\text{odd}} \simeq \Omega^{\text{even}}$. Proof: there is a non-vanishing section $s$ of $\Omega^1 \simeq TM$. Write $\Omega^1 = \mathbb{R}s \oplus V$ for some vector bundle $V$. Then $$\Omega^{\text{odd}} = s \wedge \Omega^{\text{even}}(V) \oplus \Omega^{\text{odd}}(V) \simeq \Omega(V) \simeq s \wedge \Omega^{\text{odd}}(V) \oplus \Omega^{\text{even}}(V) = \Omega^{\text{even}},$$ where $\Omega(V)$ is the full exterior bundle of $V$, and $\Omega^{\text{odd/even}}(V)$ are the odd/even parts of $\Omega(V)$. However, if $n = 2$ they are not isomorphic unless $M$ is the torus: the Euler class of $\Omega^{\text{odd}} = \Omega^1$ is then non-zero, while $\Omega^{\text{even}} = \Omega^0 \oplus \Omega^2 \simeq \mathbb{R}^2$ has trivial Euler class. If $n = 2d$, $\Omega^{\text{even}}$ and $\Omega^{\text{odd}}$ can be seen to have identical Chern classes $c_i$, except possibly the top one $c_d$. (I define the Chern class of a real vector bundle to be the Chern class of its complexification.) Namely, if $SM \subset TM$ denotes the unit sphere bundle, and $\pi: SM \to M$ is the footpoint projection, then the pullbacks $\pi^*\Omega^{\text{odd}} \simeq \pi^*\Omega^{\text{even}}$ are isomorphic. (Proof: since there is a non-vanishing tautological section $\tau(x, v) := g_x(v, \bullet)$ of $\Omega^1$, we may repeat the argument in the second point.) By Gysin sequence, $\pi^*$ is an isomorphism on $H^\bullet(M)$ for $\bullet \leq n - 1$, so $c_i(\Omega^{\text{odd}}) = c_i(\Omega^{\text{even}})$ for $i \leq d - 1$, and $c_d(\Omega^{\text{odd}}) - c_d(\Omega^{\text{even}})$ is a multiple of $\chi(M)$ in $H^{2d}(M, \mathbb{Z})$. If $n = 4$, my computations suggest that $\Omega^{\text{even}}$ and $\Omega^{\text{odd}}$ have equal all Chern, and Stiefel–Whitney characteristic classes. One needs to use that $\Omega^2 = \Omega^1 \wedge \Omega^1$ and the formula for $c_2$ of a wedge product. In particular, $c_2(\Omega^{\text{odd}}) - c_2(\Omega^{\text{even}}) = 0$ is the zero multiple of the Euler characteristic. My guess after all of this is that $\Omega^{\text{even}} \not \simeq \Omega^{\text{odd}}$ if $n$ is even and $\chi(M) \neq 0$, but I was not able to prove this. REPLY [4 votes]: Strengthening your fourth bullet point, one can see all the Chern classes are equal using the complex splitting principle. So one cannot use Chern classes to distinguish them. Indeed, I will prove the stronger statement: Let $V$ be a complex vector bundle of even rank $n$. Then the differences between the Chern classes of $\bigoplus_{i=0}^{\lfloor n/2\rfloor } \wedge^{2i} V$ and $\bigoplus_{i=0}^{\lfloor (n-1)/2 \rfloor} \wedge^{2i+1} V$ is divisible by $c_n(V)$ and thus, in particular, vanishes in degrees $<2n$. This implies the Chern classes are equal in this setting by taking $V$ to be the complexified cotangent bundle. To prove this, it suffices by the splitting principle to handle the case when $V$ is a direct sum of the $n$ universal complex line bundles on $(\mathbb C\mathbb P^{\infty})^n$. The Cherm classes of $\bigoplus_{i=0}^{\lfloor n/2\rfloor } \wedge^{2i} V$ and $\bigoplus_{i=0}^{\lfloor (n-1)/2 \rfloor} \wedge^{2i+1} V$ are then polynomials in the Chern classes of these line bundles. If we restrict to $(\mathbb C\mathbb P^{\infty})^{n-1}$ by trivializing one of the line bundles, then the pullbacks of these polynomials must be equal, as the pullbacks of the vector bundles are isomorphic. This pullback is equivalent to quotienting the cohomology ring by the first Chern class of that line bundle. It follows that the difference between the two polynomials is divisible by the first Chern class of each universal line bundle, hence (since the polynomial ring is a UFD), by their product, which is the $n$th Chern class, as desired.<|endoftext|> TITLE: Monotonicity of eigenvalues QUESTION [6 upvotes]: We consider block matrices $$\mathcal A = \begin{pmatrix} 0 & A\\A^* & 0 \end{pmatrix}$$ and $$\mathcal B = \begin{pmatrix} 0 & B\\C & 0 \end{pmatrix}.$$ Then we define the new matrix $$T(t) = \begin{pmatrix} \mathcal A+t & \mathcal B \\ \mathcal B^* & \mathcal A-t\end{pmatrix}.$$ Numerical experiments seem to show that the eigenvalues of $[0,\infty) \ni t\mapsto T(t)$ have the property that their absolute values are monotonically increasing in $t \ge 0.$ However, I do not have a proof of this, does anybody know how this follows? (The eigenvalues of $T(t)$ seem to come in pairs $\pm \lambda$ with $\lambda = \lambda(t) \ge 0$, i.e. $+\lambda(t)$ is increasing, while $-\lambda(t)$ is decreasing. To illustrate the effect, consider $$T(t)=\begin{pmatrix} t & 1& 0& 2\\ 1 & t & 0& 0\\ 0 & 0& -t & 1\\ 2& 0 & 1 & -t \end{pmatrix},$$ then the eigenvalues of $T(t)$ are $$ \pm 1 \mp \sqrt{2+t^2}.$$ Please let me know if anything is unclear. REPLY [6 votes]: The idea is to apply a unitary congruence $$U=\dfrac{1}{\sqrt{2}}\begin{pmatrix}I&-I\\I&I\end{pmatrix}.$$ I consider here $\mathcal{B}$ to be hermitian, $\mathcal{B}=\mathcal{B}^*$, whereas the general case may be 'different'. So $$R=UT(t)U^*=\begin{pmatrix}\mathcal{A}-\mathcal{B}&tI\\tI&\mathcal{A}+\mathcal{B}\end{pmatrix}.$$ Similarly $$ R-\lambda I=\begin{pmatrix}-\lambda I&(A-B)&tI&0\\(A^*-C)&-\lambda I&0&tI\\tI&0&-\lambda I &(A+B)\\0&tI&(A^*+C)&-\lambda I\end{pmatrix}. $$ Using the well known determinant formula for block matrices with a commuting off-diagonal block, you obtain that the eigenvalues $\lambda$ satisfy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$ Equivalently, $$\det\begin{pmatrix}(\lambda^2-t^2)I +F(2A^*-G)&-2\lambda A\\-2\lambda A^*&(\lambda^2-t^2)I+G(2A-F)\end{pmatrix}=0.$$ The monotonicity of the eigenvalues follows and $-\lambda$ is also an eigenvalue since $$\begin{pmatrix}Q&X\\Y&Z\end{pmatrix}$$ is unitarily congruent to $$\begin{pmatrix}Q&-X\\-Y&Z\end{pmatrix}.$$ If $\mathcal{B}$ is not hermitian the block matrix $$T(t)=\begin{pmatrix}tI&A&0&B\\A^*&tI&C&0\\0&C^*&-tI&A\\B^*&0&A^*&-tI\end{pmatrix}$$ may not be similar to $-T(t)$ for $2\times 2$ blocks, $t=2$, $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}, B=\begin{pmatrix}0&0\\0&5\end{pmatrix}$ and $C= \begin{pmatrix}0&0\\3&0\end{pmatrix}.$ The property seems to hold for $T(t)\in \mathbb{M}_4(\mathbb{C})$, consider the characteristic polynomial of $T(t)$ and $-T(t)$.<|endoftext|> TITLE: Matrices of combinatorial sequences that are inverse in two ways QUESTION [17 upvotes]: I'm interested in pairs $A=(a_{i,j})_{i,j=0,1,\ldots}$ and $B=(b_{i,j})_{i,j=0,1,\ldots}$ of infinite matrices for which: They are uni-lower-triangular, i.e., $a_{i,i}=1$ for all $i$ and $a_{i,j}=0$ for all $j>i$. They are inverses, i.e. their product $AB$ is the identity. The generating function of the $k$th column of $A$ is the reciprocal of the generating function of the $(k+1)$st row of $B$, that is, $$ \sum_{j=k}^{\infty}a_{j,k} \; x^{j-k}= \big(\sum_{i=0}^{k+1}b_{k+1,k+1-i} \; x^{i}\big)^{-1}.$$ Notice how for the generating function of columns we start at the $1$ on the diagonal and go down, while for the generating function of rows we start at the $1$ on the diagonal and go left (and all g.f.'s are normalized to start with the constant term). I have a handful of examples of this occurring with matrices of combinatorial significance: Example 1. We let $a_{i,j}=\binom{i}{j}$ and $b_{i,j}=(-1)^{i-j}\binom{i}{j}$. Note the generating function of the $k$th row of $B$ is $(1-x)^{k}$, and the generating function of the $(k-1)$th column of $A$ is $1/(1-x)^{k}$. Example 2. We let $a_{i,j}=S(i+1,j+1)$ and $b_{i,j} = s(i+1,j+1)$, where $S(n,k)$ and $s(n,k)$ are the Stirling numbers of the 2nd and 1st kind, respectively (the shift by one is just to match my convention that indexing of rows/columns starts at $0$). This looks like $$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 1 & 1 & 0 & 0 \\ 1 & 3 & 1 & 0 \\ 1 & 7 & 6 & 1 \\ \vdots & & & & \ddots \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ -1 & 1 & 0 & 0 \\ 2 & -3 & 1 & 0 \\ -6 & 11 & -6 & 1 \\ \vdots & & & & \ddots \end{pmatrix}$$ Note that the generating function of the $k$th row of $B$ (and the reciprocal of the generating function of the $(k-1)$th column of $A$) is $(1-x)(1-2x)\cdots(1-kx)$. Example 3. The previous examples were over $\mathbb{Z}$, this example is over $\mathbb{Z}[q]$; actually it is a $q$-analog of Example 1. We let $a_{i,j} = \binom{i}{j}_q$ be the usual $q$-binomial $\binom{i}{j}_q = \frac{(1-q^i)(1-q^{i-1})\ldots(1-q^{i-j+1})}{(1-q^j)(1-q^{j-1})\ldots(1-q)}$, and we let $b_{i,j} = (-1)^{i-j} \; q^{\binom{i-j}{2}} \; \binom{i}{j}_q$. This looks like: $$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 1 & 1 & 0 & 0 \\ 1 & q+1 & 1 & 0 \\ 1 & q^{2}+q+1 & q^{2}+q+1 & 1 \\ \vdots & & & & \ddots \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ -1 & 1 & 0 & 0 \\ q & -(q+1) & 1 & 0 \\ -q^3 & q^3+q^2+q & -(q^2+q+1) & 1 \\ \vdots & & & & \ddots \end{pmatrix}$$ Note that the generating function of the $k$th row of $B$ (and the reciprocal of the generating function of the $(k-1)$th column of $A$) is $(1-x)(1-qx)\cdots(1-q^{k-1}x)$. Question 1. What is going on here? Why are these pairs of matrices of combinatorial sequences inverse "in two ways"? How is being inverse in one way related to being inverse in the other way? Note that all of these examples come from sequences of uniform posets: they are the Whitney numbers of the 2nd and 1st kind of these posets. Hence, this question is related to my previous question. (But it's easy to come up with sequences of uniform posets whose matrices are not "inverse in the 2nd way.") Also note that in all the examples, the generating function of the $(k+1)$st column of $A$ is obtained from the generating function of the $k$th column by multiplying by a simple factor; but (except for Example 1), it is not exactly the same factor every time, so these matrices are not quite Riordan arrays. Question 2. Can you find some more examples of matrices of combinatorially significant sequences with these properties? EDIT: One further observation is that you can take any such pair $A=(a_{i,j})$, $B=(b_{i,j})$ of matrices and get another one $A'$, $B'$ by choosing some constant $\kappa$ and setting $a'_{i,j}=\kappa^{i-j} \; a_{i,j}$, $b'_{i,j}=\kappa^{i-j} \; b_{i,j}$. So for example from the Pascal's triangle example you can (basically) get the $f$-vectors of the cross polytope this way. REPLY [9 votes]: Gjergji Zaimi's answer appears to completely address question 1, but that leaves question 2. On the assumption that OEIS is a good place to look for known matrices of combinatorial significance, I wrote a program to scan an offline copy of the OEIS sequence data for candidate $A$ or $B$ matrices. Unsurprisingly, most of them fall into a small number of families related to binomial coefficients or Stirling numbers. There are relatively few for which both the $A$ and $B$ matrices are present. First, for completeness, OEIS sequence numbers for those directly related to the ones mentioned in the question, and moderately direct generalisations thereof. Coefficient sequence $(s, s, s, \ldots)$: when $s = 1$ we get $A = \textrm{A007318}$, Pascal's triangle, and $B = \textrm{A130595}$, signed Pascal's triangle. When $s = 0$ we get $A=B=\textrm{A023531}$, which can be interpreted as the identity matrix. For $s = 2, 3, \ldots, 15$ the $A$ matrix is in OEIS, and it's also the $B$ matrix for the negated value of $s$. These sequences are A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467. We can generalise further to $(s, t, t, t, \ldots)$ and find a lot of triangles in OEIS: s t A matrix B matrix Comments -2 -4 A193735 -2 -3 A193723 Matrix product A200139 (B matrix for s=-1, t=-2) * A007318 (A matrix for s=t=1) -1 -3 A136158 -3 -2 A209149 -1 -2 A200139 1 -2 A251634 A251636 -10 -1 A093645 (10,1) Pascal triangle -9 -1 A093644 (9,1) Pascal triangle -8 -1 A093565 etc. -7 -1 A093564 -6 -1 A093563 -5 -1 A093562 -4 -1 A093561 -3 -1 A093560 -2 -1 A103316 A029653 -1 -1 A130595 A007318 0 -1 A097805 1 -1 A112468 A112467 10 -1 A164881 -2 0 A156319 t=0 gives subdiagonal -s and all diagonals below it are zero -1 0 A097807 A097806 1 0 A000012 A167374 2 0 A130321 A251635 3 0 A140303 4 0 A262616 -1 1 A112465 A112466 0 1 A097805 2 1 A055248 3 1 A106516 100 1 A164847 1 2 A112857 3 2 A112626 There are a handful of other sequences which are $A$ or $B$ matrices for coefficient sequences which become constant after a small number of exceptional terms: Coefficients A-matrix B-matrix Comments -1,-1,-5,-5,-5,... A107310 Not many values given and I haven't tracked down the reference, so may be a false positive -2,-1,0,0,... A167684 0,-1,0,0,... A167371 0,1,0,0,... A123110 1,-1,0,0,... A135839 A071022 B-matrix relates to cellular automata 1,0,-1,0,0,... A071023 Relates to cellular automata 1,1,0,0,... A193592 1,0,1,1,... A135225 Pascal's triangle augmented with left column of 1s 1,2,3,3,3,... A250118 Set partitions avoiding 12343 1,2,3,4,4,... A250119 Set partitions avoiding 123454 Coefficient sequence $(1, q, q^2, q^3, \ldots)$: these are the $q$-binomial coefficients. The $A$ matrices for $q=2, 3, \ldots, 24$ are A022166 to A022188, and for $q=25$ it's A173583. The $A$ matrix for $q=-1$ is A051159, although this property is hidden in the comments. The $A$ matrices for $q=-2, -3, \ldots, -24$ are in the range A015109 to A015151, with a few other sequences interleaved. The $B$ matrix for $q=2$ is A135950. There are also $B$ matrices for coefficients which are scaled by $\kappa = -q$ for $q=2,3,4$, respectively A108084, A173007, A173008. And there are some sequences which skip the $q$ term: i.e. the coefficient sequence is $(1, q^2, q^3, \ldots)$. For $q=2,3,4$ the $A$ matrix is A176242, A176243, A176244. Coefficient sequence $(s, s+k, s+2k, s+3k, \ldots)$. With $s=k=1$ this is the Stirling pair $A = \textrm{A008277}$, $B = \textrm{A008275}$. A lot of members of this family are present: those which don't have other names are usually called something like "A generalised Stirling triangle". k s A matrix B matrix Comments 1 -1 A105794 A105793 1 0 A048993 A048994 Stirling numbers with different offset 1 1 A008277 A008275 Stirling numbers 1 2 A143494 A049444 2-Stirling numbers 1 3 A143495 A049458 3-Stirling numbers 1 4 A143496 A049459 etc. 1 5 A193685 A049460 1 6 A051338 1 7 A051339 1 8 A051379 1 9 A051380 1 10 A051523 -1 -4 A143493 Unsigned 4-Stirling numbers of the first kind -1 -3 A143492 Unsigned 3-Stirling numbers of the first kind -1 -2 A136124 = A143491 -1 -1 A130534 Unsigned Stirling numbers of the first kind -1 0 A132393 Unsigned Stirling numbers of the first kind -1 1 A094645 -1 2 A094646 2 1 A039755 A039757 B-analogs of Stirling numbers; A matrix is Stirling-Frobenius subset numbers of order 2 2 2 A075497 A039683 -2 -1 A028338 Stirling-Frobenius cycle numbers of order 2 3 1 A111577 Galton triangle 3 2 A225468 Stirling-Frobenius subset numbers of order 3 3 3 A075498 A051141 -3 -1 A286718 -3 -2 A225470 Stirling-Frobenius cycle numbers of order 3 4 1 A111578 4 3 A225469 Stirling-Frobenius subset numbers of order 4 4 4 A075499 A051142 -4 -3 A225471 Stirling-Frobenius cycle numbers of order 4 5 1 A166973 5 5 A075500 A051150 6 6 A075501 A051151 7 7 A075502 A051186 8 8 A075503 A051187 9 1 A166979 9 9 A075504 A051231 10 10 A075505 A048176 Making some minor changes to the first term yields $A$-matrix Stirling differences A269952 for $(0,2,3,4,\ldots)$, and $A$-matrix A137596 (matrix product of all-ones lower-triangular matrix and triangle of Stirling numbers of the second kind) for $(1,1,2,3,4,\ldots)$. Other families which are less directly related (although by no means unrelated in all cases) to the ones explicitly mentioned in the question are: Coefficient sequences based on squares, cubes, etc. Squares $(1, 4, 9, \ldots)$ has $A$-matrix A036969, "central factorial numbers $T(2n,2k)$"; and $B$-matrix A204579. Squares $(0, 1, 4, 9, \ldots)$ has $A$-matrix A269945, central factorial numbers $T(2n,2k)$ with a different offset. Negated squares $(0, -1, -4, -9, \ldots)$ has $B$-matrix A269944, unsigned central factorial numbers $t(2k,2k)$. Odd squares $(1, 9, 25, \ldots)$ has $A$-matrix A160562, "scaled central factorial numbers". Negated odd squares $(-1, -9, -25, \ldots)$ has $B$-matrix A160563, "number of (n,k)-Riordan complexes". Cubes $(1, 8, 27, \ldots)$ has $A$-matrix A098436, 3rd central factorial numbers or Stirling set numbers of order 3. With zero, $(0, 1, 8, 27, \ldots)$ has $A$-matrix A269948 of the same 3rd central factorial numbers offset by one. Negative cubes $(0, -1, -8, -27, \ldots)$ has $B$-matrix A269947, Stirling cycle numbers of order 3. Alternating sequences. Coefficients A-matrix B-matrix Comments 0,-1,0,-1,... A103633 Repeated stepped binomial coefficients 0,1,0,1,... A103631 abs(q-Stirling2) for q=-1 1,0,1,0,... A065941 q-Stirling2 for q=-1; repeated stepped signed binomial coeffs 1,2,1,2,... A140068 2,1,2,1,... A140069 Binomial transform of A135225 (A-matrix of 1,0,1,1,...) 1,-2,1,-2,... A140166 2,-1,2,-1,... A140168 1,3,1,3,... A140070 3,1,3,1,... A140071 Also near-alternating sequences with early exceptions: 1,1,-1,1,-1,... A207974 Has interpretation in terms of binary strings 1,1,0,1,0,... A194005 Likewise; also "companion to A103631" (A-matrix of 0,1,0,1,...). The sequence $(1,3,7,\ldots,2^k-1,\ldots)$ gives $A$-matrix A139382, $q$-Stirling2 with $q=2$. Sequences $\binom{k+\alpha}{\alpha}$ or $\alpha! \binom{k+\alpha}{\alpha}$ give a few hits: Coefficients A-matrix B-matrix Comments 1,3,6,10,... A080248 2,6,12,20,... A071951 A130559 Legendre-Stirling numbers 0,2,6,12,20,... A129467 ... with different offset 1,4,10,20,... A080249 6,24,60,120,... A089504 24,120,360,840,... A090215 120,720,2520,6720,... A090217 Repeating coefficients gives other Stirling-related sequences: 0,1,1,2,2,3,3,... A246118 Set partitions with certain constraints 1,1,2,2,3,3,... A256161 Same with different offset 0,-1,-1,-2,-2,... A246117 Parity-preserving permutations by number of cycles Other sequences based on combinatorial coefficient sequences: Coefficients A-matrix B-matrix 1,1,2,3,5,... (Fibonacci) A111669 1,1,3,5,11,... (Jacobsthal) A114163 1,1,2,6,24,... (factorials) A136457 -1,2,-3,4,-5,6,... A140956 Rows are coefficients of alternating factorial polynomial And there's a handful based on number-theoretic sequences: Coefficients A-matrix B-matrix 2,3,5,7,... (primes) A124960 A070918 -1,-2,-4,-6 (1-p) A096294 Fraction of positive integers with exactly k of the first n primes as divisors 1,-1,-1,0,-1,1,... (mu) A124961 -1,0,0,0,-9,0,0,-15,... A109409 Coeffs of product of (x+i) for odd non-prime i<|endoftext|> TITLE: For each $k$, is there a vector bundle $E$ such that $E\oplus\varepsilon^k$ is trivial but $E\oplus\varepsilon^{k-1}$ is not? QUESTION [8 upvotes]: A vector bundle $E$ is stably trivial if $E\oplus\varepsilon^k$ is trivial for some $k \geq 0$; here $\varepsilon^k$ denotes the trivial bundle of rank $k$. For such a bundle, let $s(E)$ denote the smallest value of $k$ such that $E\oplus\varepsilon^k$ is trivial; note that $s(E) = 0$ if and only if $E$ is trivial. Question: Does every value of $k$ arise as $s(E)$ for some stably trivial bundle $E$? For $k = 1$, the tangent bundle of a non-parallelisable sphere provides an example, e.g. $s(TS^2) = 1$. In fact, the tangent bundle of any stably parallelisable manifold $M$ which is not parallelisable satisfies $s(TM) = 1$, see this answer. Some other values of $k$ can be obtained using tangent bundles of spheres as follows. Consider $TS^n$ where $n$ is odd, $n \neq 1, 3, 7$ - note that $TS^n$ is stably trivial but not trivial. As $n$ is odd, it admits a nowhere-zero section. Adams showed that the maximum number of linearly independent sections is $\rho(n+1) - 1$ where $\rho(n+1)$ is the $(n+1)^{\text{st}}$ Radon-Hurwitz number: if $n + 1 = 2^{4a+b}c$ where $a \geq 0$, $0 \leq b \leq 3$, and $c$ is odd, then $\rho(n+1) = 8a + 2^b$. Therefore there is a non-trivial vector bundle $E$ such that $E\oplus\varepsilon^{\rho(n+1)-1} \cong TS^n$ and hence $E\oplus\varepsilon^{\rho(n+1)} \cong TS^n\oplus\varepsilon^1$ is trivial, so $s(E) = \rho(n+1)$. Every $k \equiv 0, 1, 2, 4 \bmod 8$ arises as $\rho(n+1)$ for some odd $n$ with $n \neq 1, 3, 7$. For the remaining values of $k$ (i.e. $k \equiv 3, 5, 6, 7 \bmod 8$), one can find a bundle $E'$ with $s(E') = k$ in the following way. First choose $a$ such that $k < 8a$. As $\rho(16^a) = \rho(2^{4a}) = 8a$, by the above there is a bundle $E \to S^N$ with $N = 16^a-1$ such that $s(E) = \rho(16^a) = 8a$. As $E\oplus\varepsilon^{8a} = (E\oplus\varepsilon^{8a-k})\oplus\varepsilon^k$, we see that $s(E\oplus\varepsilon^{8a-k}) = k$. Answer: Yes. For $k \equiv 0, 1, 2, 4 \bmod 8$, the examples constructed above do not admit a nowhere-zero section, while the examples for $k \equiv 3, 5, 6, 7 \bmod 8$ certainly do. This leads to the following refinement of the original question. Refined Question: Does every value of $k$ arise as $s(E)$ for some stably trivial bundle $E$ which does not admit a nowhere-zero section? By the above, we only have the cases $k \equiv 3, 5, 6, 7 \bmod 8$ left to deal with. In particular, the first unknown case is $k = 3$. One might hope we can modify the construction above by replacing $TS^n$ with $TM$ for some stably parallelisable manifold $M$. However, a theorem of Bredon and Kosinski states that if $M$ is a $n$-dimensional stably parallelisable manifold, then either $M$ is parallelisable or the maximum number of linearly independent vector fields on $M$ is the same as that of $S^n$, namely $\rho(n+1)-1$. REPLY [8 votes]: This is more of an extended comment on ways to approach this problem than an answer. If we have a vector bundle $E$ on $X$ and have a trivialization $E ⊕ \epsilon^k \to \epsilon^{n+k}$, then this determines a map from $X$ to the Stiefel manifold $V_{k,n+k}$ of $k$-dimensional frames in $(n+k)$-dimensional space: at each point $x$ of $X$, we have an isometric embedding $\Bbb R^k \to \Bbb R^{n+k}$ whose orthogonal complement is identified with the fiber $E_x$. As a result, the Stiefel manifold $V_{k,n+k}$ is a universal example. The vector bundle $E$ is pulled back from the $n$-dimensional vector bundle $E_{univ}$ on $V_{k,n+k}$. Therefore, $E_{univ} \oplus \epsilon^{k-1}$ is trivial if and only if every $n$-dimensional vector bundle $E$ such that $E \oplus \epsilon^k$ is trivial also has that $E \oplus \epsilon^{k-1}$ is trivial. We can also recast this in terms of homotopy theory. The space $V_{k,n+k}$ is the homotopy fiber of the map $BO(n) \to BO(n+k)$ of classifying spaces for vector bundles. We've reduced this to asking if the map $V_{k,n+k} \to BO(n) \to BO(n+k-1)$ is homotopic to the constant map. Moreover, that map factors through the composite $V_{k,n+k} \to S^{n+k-1} \to BO(n+k-1)$, the first map sending a $k$-frame $(v_1,\dots,v_k)$ of ortogonal vectors in $\Bbb R^{n+k}$ to the first element $v_1 \in S^{n+k-1}$, and the second map classifying the tangent bundle of $S^{n+k-1}$. In terms of the refined question, we also need to assure that the map $V_{k,n+k} \to BO(n)$ does not lift to a map $V_{k,n+k} \to BO(n-1)$. For the same reason as above, that's equivalent to a cross-section of the map $V_{k+1,n+k} \to V_{k,n+k}$ of Stiefel manifolds. Cross-sections like that almost never exist (see this question). Now that this is re-expressed in terms of asking if the map $V_{k,n+k} \to BO(n+k-1)$ is homotopic to a trivial map, we could test it by asking more basic questions about homotopy groups. So I'd like to explore that a little. (I apologize, this may not be particularly satisfying: I'm going to start with this geometric problem, and said that we could potentially solve it with hard spectral sequence calculation. Such is the danger with homotopy theorists!) We can ask: is the map $\pi_*(V_{k,n+k}) \to \pi_*(BO(n+k-1))$ trivial? If this map is nontrivial, then $E_{univ} \oplus \epsilon^{k-1}$ is not trivial. The fibration sequence $V_{k,n+k} \to BO(n) \to BO(n+k)$ gives a long exact sequence on homotopy groups, $$ \dots \to \pi_* V_{k,n+k} \to \pi_* BO(n) \to \pi_* BO(n+k) \to \pi_{*-1} V_{k,n+1}. $$ This identifies the image of $\pi_*(V_{k,n+k})$ in $\pi_* BO(n)$ with the collection of elements that map to zero in $\pi_* BO(n+k)$. This allows us to rephrase the questions about homotopy groups without needing to reference the Stiefel manifold at all, and ask the equivalent questions: Does there exist an element in $\pi_* BO(n)$ that maps to zero in $\pi_* BO(n+k)$, and which does not map to zero in $\pi_* BO(n+k-1)$? Next, let's take a look at what's actually going on with the homotopy groups. In the sequence of spaces $BO(1) \to BO(2) \to BO(3) \to \dots$, each participates in a fibration sequence $S^{m-1} \to BO(m-1) \to BO(m)$, inducing a long exact sequence on homotopy groups: $$ \dots \to \pi_* S^{m-1} \to \pi_* BO(m-1) \to \pi_* BO(m) \to \pi_{*-1} S^{m-1} \to \dots $$ Putting all of these long exact sequences together gives a big spectral sequence that starts with the homotopy groups of spheres and converges to the homotopy groups of $\bigcup BO(m) = BO$: $$ E^1_{p,q} = \pi_{p+q-1} S^p \Rightarrow \pi_{p+q} BO $$ This spectral sequence is gigantic and extraordinarily messy, even though the thing it is converging to is known by Bott periodicity. It has been heavily studied previously (especially by Mahowald, who related it to the Atiyah-Hirzebruch spectral sequence for the stable homotopy groups of $\Bbb{RP}^\infty$). But our question about homotopy groups is equivalently formulated in terms of this spectral sequence. A class in $\pi_* BO(n)$ whose image is nonzero in $\pi_* BO(n+k-1)$ but zero in $\pi_* BO(n+k)$ will give rise to a differential in this spectral sequence that is a $d_k$ or higher, from the line $p=n+k-1$. The upshot is this. Take this spectral sequence, starting with all of the homotopy groups of spheres and converging to the homotopy groups of $BO$. If we have arbitrarily long differentials in this spectral sequence, then for any $k$ there is a vector bundle of the type you are asking for.<|endoftext|> TITLE: Is the bundle map of the Eguchi-Hanson metric a Riemannian submersion? QUESTION [7 upvotes]: Background. (Can be skipped if you already know what is the Eguchi-Hanson metric.) The Eguchi-Hanson metric $g$ is a complete Ricci-flat Riemannian metric on the cotangent bundle of the 2-sphere, $T^*S^2$. By removing the zero-section, we can identify it with $S^3/\mathbb{Z}_2 \times (0, \infty) = (\mathbb{R}^4\setminus\{0\}) / \mathbb{Z}_2$, where $\mathbb{Z}_2$ acts by antipodal reflections. It then has the explicit description $$g = \frac{r^2}{\sqrt{1+r^4}}(dr^2 + r^2 \alpha_1^2) + \sqrt{1 + r^4}(\alpha_2^2+\alpha_3^2),$$ where $r^2 = x_0^2 + x_1^2 + x_2^2 + x_3^2$, $$\alpha_1 = \frac{1}{r^2} (x^0 dx^1 - x^1 dx^0 + x^2 dx^3 - x^3 dx^2)$$ and $\alpha_2, \alpha_3$ are defined with the same formula by cyclic permutations of $(1, 2, 3)$. It has also been described by Calabi as a Kähler metric on $T^*\mathbb{CP}^1$, where the Kähler form is $\pi^*\omega_{FS} + i\partial\bar{\partial} (u \circ t)$, where $\omega_{FS}$ is the Fubini-Study Kähler form on $\mathbb{CP}^1$, $t : T^*\mathbb{CP}^1 \to \mathbb{R}$ is the squared-norm function with respect to the Fubini-Study metric, and $u(t) = 4\sqrt{1+t} - 4\log(1+\sqrt{1+t})$. Question. Is the bundle map $\pi : T^*S^2 \to S^2$ a Riemannian submersion? In other words, does $\pi$ restrict to isometries $d\pi : (T_\xi(T^*S^2))^{\mathrm{horizontal}} \to T_xS^2$? I tried to work this out explicitly using the above coordinate expressions, but it is very messy, and I couldn't do it. I was wondering if there is another argument, or perhaps a way to see that it's not a submersion using asymptotic properties of the metric. REPLY [9 votes]: No. The reason is basically the same why if you take flat $\mathbb R^4\setminus 0$ and quotient by the standard Hopf $\mathbb S^1$ action you get a punctured cone over $\mathbb S^2$ but the projection to that $\mathbb S^2$ is not a Riemannian submersion because the horizontal spaces scale with $r$. The Euguchi-Hanson metric is asymptotic to this one and the same exact thing happens. What IS true for the Euguchi-Hansen metric is that if you fix $r$ then the projection from $\mathbb S^3/\mathbb Z_2$ to $\mathbb S^2$ is a Riemannian submersion up to scaling. But the scaling changes with $r$. This can be seen as follows. The forms $\alpha_1,\alpha_2,\alpha_3$ form the standard left invariant orthornormal basis in the round binivariant metric on $\mathbb S^3$ (and its factor $SO(3)=\mathbb S^3/\mathbb Z_2$). Here $\alpha_1$ can be thought of as the inner product (with respect to the binivariant metric) with the Hopf vector field $X$ generating a free isometric $\mathbb S^1$ action on $SO(3)$. For any fixed $r$ the difference between the round metric on $SO(3)$ and the Euguchi Hanson one is that this circle is scaled by a function depending on $r$. But this does not affect the quotient metric at all and the projections $SO(3)\to \mathbb S^2$ are Riemannian submersions (up to scaling) with respect to both metrics on $SO(3)$. However, as I said you get a scaling depending on $r$ so the global map $TS^2\to S^2$ is not a Riemannian submersion.<|endoftext|> TITLE: A cohomological variant of the second Riemann's extension theorem QUESTION [5 upvotes]: Let $X$ be a connected compact complex manifold, $U$ an open subset of $X$ such that the complement of $U$ in $X$ is an analytic subset of codimension at least 2 in $X$. Let $O_X$ (resp. $O_U$) be the sheaf of holomorphic functions on $X$ (resp. on $U$). If $n$ is a nonnegative integer then there is a natural homomorphism of complex vector spaces $$r_n: H^n(X,O_X) \to H^n(U,O_U).$$ The second Riemann extension theorem actually asserts that $r_n$ is an isomorphism for $n=0$. I am looking for a reference where it is proven that $r_n$ is an isomorphism, say, for $n=1$ or $2$ (may be, under some additional assumptions). Thanks! REPLY [3 votes]: For the first cohomology statement, you need the codimension to be at least three and for the first and second cohomology, codimension four. This theorem was proved by G Scheja in [1]. You can also find a proof in the book by Banica and Stanasila ([2] Chapter II, §II.3 pages 66-67). References [1] Constantin Banica, Octavian Stanasila, Algebraic methods in the global theory of complex spaces. Rev. English ed. (English) Bucuresti: Editura Academiei; London-New York-Sydney: John Wiley&Sons, pp. 296 (1976), MR0463470, Zbl 0334.32001. [2] Günter Scheja, "Riemannsche Hebbarkeitssätze für Cohomologieklassen" (German) Mathematische Annalen 144, 345-360 (1961), MR0148941, Zbl 0112.38001.<|endoftext|> TITLE: Interpreting Conway's remark about using the surreals for non-standard analysis QUESTION [16 upvotes]: In Conway's "On Numbers And Games," page 44, he writes: NON-STANDARD ANALYSIS We can of course use the Field of all numbers, or rather various small subfields of it, as a vehicle for the techniques of non-standard analysis developed by Abraham Robinson. Thus for instance for any reasonable function $f$, we can define the derivative of $f$ at the real number $x$ to be the closest real number to the quotient $$\frac{f[x + (1/\omega)] - f(x)}{1/\omega}$$ The reason is that any totally ordered real-closed field is a model for the elementary states about the real numbers. But for precisely this reason, there is little point in using subfields of $\mathbf{No}$ when so many more visible fields will do. So we can say in fact the field $\mathbf{No}$ is really irrelevant to non-standard analysis. Conway here makes clear that you could, if you wanted, use the surreal numbers for non-standard analysis, because they are a real-closed field and thus a model for the theory of all elementary (first-order) statements about the reals. Conway does also make the point, in the last two sentences, that he doesn't view nonstandard analysis as the ultimate application of the surreals. But, for the sake of curiosity, I'm quite interested in understanding how you could do what Conway is hinting at above. However, the statement that you can use the surreals for nonstandard analysis is really quite strong, and much stronger than just the field being real-closed. The real meat of the claim being made is the expression $f(x + 1/\omega)$ even exists at all. This would demand some kind of "transfer principle" for $f(x)$ to the surreals. Just being real-closed wouldn't be enough for this: the real algebraic numbers are real closed, but that doesn't mean we can use the real algebraic numbers for nonstandard analysis. But Conway says this is "of course" possible with the surreals. So the main question is: how would such a transfer principle work? Or an even stronger question: do we need anything like an ultrafilter lemma for Conway's claim to be true? Unlike the hyperreals, the surreals don't require any ultrafilter at all -- which is a pretty significant achievement, really, since we are at least certain something like $1/\omega$ exists even in ZF, and that the resulting field is real-closed. It would seem to be possible to just go through the motions with taking the nonstandard derivative of, for instance, $\frac{\exp(x + 1/\omega) - \exp(x)}{1/\omega} = \exp{x}\left(\frac{\exp^{1/\omega} - 1}{1/\omega}\right) \approx \exp(x)$, treating $1/\omega$ as a formal expression satisfying the first-order properties of the reals and deriving $\exp(x)$ as the closest real number to the result, none of which seems to have required any kind of ultrafilter. (Or has it, implicitly?) I've tried to keep this short but there is an enormous amount of subtlety to this question, so I will go into some of that below. Some later results have clarified the relationship between the surreals and hyperreals, so some additional detail regarding what is being asked is probably necessary. There has been a little bit of prior discussion about this, for instance in this post, where it is talked about the much more modern result that the surreals are isomorphic to the proper-class sized ultrapower of the reals. These isomorphisms can be thought of as various ways to transfer real functions to the surreal numbers. So in one sense, the answer is yes, a transfer principle exists in theory. But the pitfall with this approach is that everything requires ultrafilters, and is non-constructive, and there is no canonical choice of isomorphism. This is very different from the way that the surreals are built, which do not require ultrafilters. On the other hand, Conway's book was written before any of the above results were published (with possibly an exception regarding one paper of Keisler). So partly the question is informal - what did Conway have in mind? But the other part of it is to formally ask if there is some other way to do this that doesn't involve this very particular method of using these isomorphisms, or even to use ultrafilters at all. For instance, what if we don't have the ultrafilter lemma? Then the hyperreals don't necessarily exist at all, but we can still build the surreals, which don't even require choice. Even if we don't have the ultrafilter lemma, can we still just go ahead anyway and say that $\frac{f(x + 1/\omega) - f(x)}{1/\omega}$ is a well-defined expression, and look for the closest real number to it, using some other way to derive a transfer principle? The other part of the question is admittedly a soft question, but still well worth answering. The ultrafilter construction makes it very easy to see how such a transfer principle would work. Every hyperreal is a set of reals (or an equivalence class thereof), and to transfer any first-order predicate to some hyperreal, you simply ask the predicate of every real in the set and see if it's true of "most" of them (where "most" means "in the ultrafilter"). Thus you have a real-closed field, a "transfer principle," and all of that. Conway, on the other hand, has a very interesting way of building up the surreals in his book which is somewhat agnostic to the choice of set theory, using "birthdays," "left and right sets," etc. I am curious if there is some way to interpret Conway's assertion regarding the existence of $f(x+1/\omega)$ using his own machinery for the surreals, perhaps doing something clever and inductive with the left and right sets, rather than using these later developments involving isomorphisms with the ultrapower. The last subtlety involves a philosophical point that has sometimes been raised with the topic of surreals vs hyperreals, but it is also worth addressing. There is, for instance, some debate regarding how functions like $\sin$ and $\cos$ should be transferred to the surreals. In theory, you could say that since we have these isomorphisms to the hyperreals, which have a transfer principle, these guarantee the existence of some kind of function on the surreals with the required first-order properties. But the surreals are very tangible in a very constructive sort of way, whereas these isomorphisms are typically totally non-constructive, so there is no way to use them to see what $\sin(\omega)$ should be, if it's positive or negative, etc. On the other hand, you could raise the same philosophical issue with the hyperreals, because there also is no real answer regarding what $\sin((1,2,3,4,...))$ should be, where $(1,2,3,4...)$ is a particular hyperreal number. The answer depends on the ultrafilter, which determines what $(1,2,3,4...)$ even means to begin with, or what properties it has, or if you like, which hyperreal it's referring to. But what you can do with the hyperreals, which is part of the appeal, is you can kind of get "part of the way there" in a totally constructive manner. You know, for instance, that whatever $\sin((1,2,3,4,...))$ is, in some sense it's $(\sin(1), \sin(2), \sin(3), \sin(4), ...)$. Since we don't know what the ultrafilter is we don't know exactly what properties that has, but we do know something: we know that for any ultrafilter this value will not be an integer, for instance, or any rational number. So, we have some idea of what the transferred sin function would have to look like on the surreals as a result, at least given that $\omega$ is some hypernatural. So even though the ultrafilter is non-constructive, you can at least get "part of the way there" in an entirely constructive manner, which is part of what makes the entire thing interesting. And of course you don't really need to know much more than these few constructive things to actually do nonstandard analysis, just kind of happily plodding along formally doing nonstandard derivatives, with the understanding that the ultrafilter handles all of the various pathological, undefinable sets of indices in some logically consistent way or another. So the last question is if there is some way for us to do something similar with the surreals, to get "part of the way there," in this sense. That is, to at least have enough constructive "transfer" for us to play around with all of this stuff, but using the framework of the surreals rather than the hyperreals, so that we can see that $f(1 + 1/\omega)$ even makes sense to begin with and play around with it. Something like Terry Tao's "cheap nonstandard analysis", perhaps. REPLY [7 votes]: Conway was of course correct in saying that NSA is irrelevant to the surreals, but like Mike I found Conway’s further remarks about NSA puzzling and I am not sure what he had in mind. What I think Conway might have said is: "$\mathbf{No}$ is really irrelevant to nonstandard analysis", and, vice versa. After all, whereas the transfer property of hyperreal number systems, a property not possessed by $\mathbf{No}$, is central to the development of nonstandard analysis, the s-hierarchical (i.e the algebraico-tree-theoretic) structure of $\mathbf{No}$, which is absent from hyperreal number systems, is central to the theory of surreal numbers and is responsible for its canonical nature. On the other hand, as I have noted on occasion, I do not rule out the possibility that down the road there might be cross-fertilization between the two theories. However, even if not, it seems to me that to not appreciate that both theories—which are used for quite different purposes--are remarkably powerful and beautiful is a demonstration of ignorance. While surrealists have thus far shown little interest in applying surreal numbers to nonstandard analysis or in providing an infinitesimalist approach to classical analysis based on surreal numbers more generally, a number of surrealists beginning with Norton, Kruskal, and Conway have fostered the idea of extending analysis to the entire surreal domain. While this has been a comparatively small focal point of the theory thus far and progress has been slow, I am happy to say that, contrary to the assertions of Sam and Emanuele, there is now a reasonably strong theory of surreal integration. After laying dormant for quite some time and in need of much revision, work on surreal integration has recently made substantial progress and I expect that a paper on the subject by Ovidiu Costin and myself will be posted before too long. The theory extends integration from the reals to the surreals for a large portion of Écalle’s class of resurgent functions, in particular, a large subclass of the resurgent functions that are found in applied analysis. On the other hand, there are considerations in the foundation of mathematics that preclude the theory from being extended very much further. So, for example, the theory cannot be applied in general to the class of smooth functions. Thus far, a sizable portion of the literature on the surreals has dealt with the theory of ordered algebraic systems and (as Emanuele suggests) model-theoretic issues thereof—issues involving ordered abelian groups, ordered domains, ordered fields, ordered exponential fields and ordered differential fields. In the latter two cases, there has been considerable work on Hardy fields and Écalle’s ordered differential field of transseries and generalizations thereof. Some of the work on ordered exponential fields and ordered differential fields, however, go well beyond ordered algebraic and model-theoretic considerations and are concerned with developing asymptotic differential algebra--the subject that aims at understanding the asymptotics of solutions to differential equations from an algebraic point of view—for the surreals. In their 2017 ICM talk (On Numbers, Germs, and Transseries), Aschenbrenner, van den Dries and van der Hoeven outline the program they (along with Mantova, Berarducci, Bagayoko and Kaplan) are engaged in for developing an ambitious theory of asymptotic differential algebra for all of the surreals, though one that would require a derivation on $\mathbf{No}$ having compositional properties not enjoyed by the derivation introduced by Mantova and Berarducci in the paper cited by Emanuele. Such a program, if successful, would provide the most dramatic advance towards interpreting growth rates as numbers since the pioneering work of Paul du Bois-Reymond, G. H. Hardy and Felix Hausdorff on "orders of infinity" in the decades bracketing the turn of the 20th century. Unlike the surreals, the framework on NSA does not appear to be particularly well suited to this end, despite the fact that Robinson and Lightstone did make modest contributions to the theory of asymptotics in their nice monograph that applies NSA to the subject. In his response, Sam noted that (as far as he knew) the surreals do not have a visible copy of surnatural numbers. Sam is, of course, correct; but given the recursive nature of the construction of the surreals no one should expect to find one, and for the purposes for which it has been used it has not proven to be a limitation. On the flip side, of course, unlike the surreals, the number systems employed in NSA do not have a canonical copy of the ordinals or significant initial segments thereof. However, for the purpose of developing an infinitesimalist approach to classical analysis, I don’t see this to be a problem either. Given all the areas and questions to which mathematicians apply finite, infinite and infinitesimal numbers, it is extremely unlikely there will ever be one theory ideally suited for all applications. An assertion like "For all its weaknesses" made by Sam about the surreals, fails to appreciate this. What would be a weakness in one context need not necessarily be a weakness in another. All of the people I know who are working in the theory of surreal numbers are familiar with, and have great respect for, NSA. Sadly, however, there appears to be a small segment of the contemporary NSA community who, while repeatedly demonstrating their lack of knowledge of the subject and its applications, attack it time and again, one person (as is evident from the comments) even describing those who work on it as members of a cult. During the 19th century, Cantor repeatedly attacked the works of du Bois-Reymond, Stolz, and Veronese on their non-Cantorian theories of the infinite (and infinitesimal), theories designed to deal with issues not addressed by Cantor’s theory—non-Archimedean geometry, the rates of growth of real functions, and non-Archimedean ordered algebraic systems. Abraham Robinson, who was as gracious a person as he was a great and knowledgable mathematician, attempted to soften his well-deserved implicit critique of Cantor’s misguided and narrow-minded attacks by noting: "It may be recalled that, at that time, Cantor was fighting hard in order to obtain recognition for his own theory" (The Metaphysics of the calculus, p. 39). I wonder to what extent some of the aforementioned attacks are motivated by similar considerations, despite the fact that NSA is already widely, albeit not universally, regarded as a major contribution. However, whatever the motivation may be, I believe losing sight of the lesson of Cantor or the humanity of Robinson would be an unfortunate mistake indeed. P.S. One of the longstanding bugaboos in the aforementioned attacks on the surreals has been that there is no natural sine and cosine functions for the surreals. For a proof that this contention is mistaken, see Section 11 of Kaplan and the author’s recent Surreal Ordered Exponential Fields, The Journal of Symbolic Logic, 86 (2021) pp.1066-1115.<|endoftext|> TITLE: A "surnatural numbers" as a largest model of the natural numbers QUESTION [7 upvotes]: One characteristic of the surreal numbers is that they are a monster model of the first-order theory of real numbers, according to Joel David Hamkins in this post. Thus they are real-closed, and every other real-closed field embeds into them with very nice properties. So we may ask if one can similarly create a monster model of the first-order theory of natural numbers as a kind of "surnaturals." One natural question to ask is if the non-negative elements in Conway's ring of "omnific integers" would fit the bill, given that they are often promoted as the "surreal version of the integers" or something like that. However, it is rather easy to see that this is not true: the omnific integers have the interesting property that their field of fractions is the entire surreal number field, thus there are two omnific integers whose quotient is the square root of 2, whereas no such integers exist in any model of the naturals. Thus in some sense, the omnific integers are not quite the most direct correspondant of the natural numbers within the surreal numbers. So one question is if the monster model can be built constructively, similarly to the surreals. I would suspect that such a construction exists, given that Joel David Hamkins was able to explicitly construct a monster model of all groups (!) in the above post, which would seem to suggest a monster model of the integers also exists, and thus the naturals as those non-negative integers. How can one build such a model? EDIT: I initially asked if the Grothendieck ring of the ordinals with commutative addition/multiplication could be a monster model for the integers, but it is apparently too small, since for $\omega$ we should have that $\omega$ is either even or odd. This means there should be some element $x$ such that $x + x = \omega$, or $x + x = \omega + 1$, so either $\omega/2$ or $(\omega + 1)/2$ should be in our set. Also, there will need to be some element in the monster model which is divisible by every standard finite number, and if that were $\omega$, we'd thus need to have $\omega \cdot q$ for every rational q. Thanks to Noah Schweber and Emil Jeřábek for pointing this out, and also to Emil for clearing up some confusion I had about whether the non-unique factorization of omnific integers necessarily implies they are not a model of the naturals (apparently it does not, but there are other reasons why, such as there are two omnifics whose ratio is sqrt(2), which is provably not true in any model of PA). REPLY [5 votes]: I asked (and also answered) a more general version of this question a while ago. To summarize the answer, some results of Kanovei and Shelah have the following corollary: Fact. In $\mathsf{ZFC}$ there is a uniform procedure for building 'set-saturated,' class-sized elementary extensions of arbitrary structures. That is to say there are formulas $S(M,L,x)$ and $F(M,L,f,x)$ in the language of set theory such that in any model $V \models \mathsf{ZFC}$ if $L \in V$ is a language and $M \in V$ is an $L$-structure, then the following hold (where $M^\ast = \{x \in V : V \models S(M,L,x)\}$): $M \subseteq M^\ast$, if $\varphi \in V$ is an $L$-formula with free variables $x_0,\dots,x_n$ and $\bar{a} \in M^\ast$ is an $n$-tuple, then $V \models F(M,L,\exists x_n\varphi,\bar{a})$ if and only if $V \models (\exists x \in M^\ast) F(M,L,\varphi,\bar{a}x)$ (where we are using some fixed coding of tuples in $\mathsf{ZFC}$), furthermore, if $\bar{c} \in M$ is an $(n+1)$-tuple, then $V \models F(M,L,\varphi,\bar{c})$ if and only if $V \models “M \models \varphi(\bar{c})”$ (in particular, if $\varphi$ is a sentence, then $V \models F(M,L,\varphi,\varnothing)$ if and only if $V \models “M \models \varphi”$), $F$ is compatible with Boolean combinations (i.e., $V\models F(M,L,\varphi\wedge \psi,\bar{a})$ if and only if $V\models F(M,L,\varphi,\bar{a})\wedge F(M,L,\psi,\bar{a})$, etc.), and if $A \subseteq M^\ast$ is a set and $p(x)$ is a finitely satisfiable set of $L_A$-formulas with free variable $x$, then there is $b \in M^\ast$ such that for any $\varphi(x,\bar{a}) \in p(x)$, $V \models F(M,L,\varphi,b\bar{a})$. So to state it informally, $S(M,L,x)$ defines the universe of a class-sized elementary extension of $M$ and $F(M,L,f,x)$ is its truth predicate. Applying this to the naturals tells us that there is a formula that defines a proper class monster model of $\mathrm{Th}(\mathbb{N})$ in any model of $\mathsf{ZFC}$. One thing to note, though, is that without global choice (which makes my original question trivial), it's unclear whether there's always a definable isomorphism between different set-saturated class-sized models of a given theory. I believe this is related to an unanswered MathOverflow question of Hamkins. That said, if $M$ and $N$ are $L$-structures and $M \equiv N$, then there will be an isomorphism between $M^\ast$ and $N^\ast$ that is definable with certain parameters. Another thing to note is that some constructions that model theorists commonly use with the monster model are unclear in the context of these class monster models. There isn't necessarily a good way to talk about arbitrary global types, for instance. You do, however, get a good homogeneity property: There is a subgroup $G$ of $\mathrm{Aut}(M^\ast)$ that can be represented as a class in a definable way which has the property that if $\bar{a}$ and $\bar{b}$ are set-sized tuples that realize the same type, then there is a $\sigma \in G$ such that $\sigma \bar{a} = \bar{b}$. REPLY [4 votes]: I think it should be said that the first order theory of the semi-ring $\mathbb{N}$ of non-negative integers is much more difficult to work with than that of the real ordered field. The existence of certain proper elementary extensions of $\mathbb{N}$ usually relies on non-constructive methods, so it is unlikely that one could give a nice description of such an object within the class $\mathbf{No}$ of surreal numbers. An easier task would be to try to identify models of Peano Arithmetic (PA) within $\mathbf{No}$ (the class of non-negative omnific integers only satisfies the fragment of PA called Open Induction, which only contains weak instances of the induction axioms). This seems already difficult enough, but for this one at least there is some literature. Specifically about integer parts of real-closed fields which are models of (the full integers version of) PA. I don't know this literature, but I know Salma Kuhlmann and Paola D'Aquino have look into related questions.<|endoftext|> TITLE: Flatness of schemes QUESTION [6 upvotes]: I am learning about flatness for the first time and I cannot wrap my head around why the definition with tensor products of a flat module implies geometrically that 1-parameter families of schemes have limits. I came across these lecture notes by Alexander Ritter which give a very explicit example: However, I don't understand why $\pi$ flat over $0 \iff$ $X_0$ is $\lim_b X_b$. Any help to understand this particular example or a different intuition of this geometric implication would be very much appreciated. REPLY [4 votes]: Note that $\overline{X^*}$ is the scheme-theoretic image of $X^* \to \mathbf A^n_B$: if $X^*$ is reduced, this agrees with the reduced induced structure (see for instance exercise II.3.11 in Hartshorne), and otherwise this is the only sensible definition of the closure of a locally closed subscheme. In particular, if $X^* \subseteq \mathbf A^n_{B^*}$ is cut out by some ideal $I^*$, then by definition $\overline{X^*} \subseteq \mathbf A^n_B$ is cut out by the contracted ideal $\overline{I^*} = \iota^{-1}(I^*)$, where $\iota \colon B[x_1,\ldots,x_n] \to B^*[x_1,\ldots,x_n]$ is the inclusion. In other words, the factorisation $X^* \to \overline{X^*} \to \mathbf A^n_B$ corresponds to the image factorisation $$B[x_1,\ldots,x_n] \twoheadrightarrow B[x_1,\ldots,x_n]\big/\ \overline{I^*} \hookrightarrow B^*[x_1,\ldots,x_n]/I^*$$ of rings. Any module $M$ over the principal ideal domain $B$ is flat over $0$ if and only if it is $t$-torsion-free. Since $B^*[x_1,\ldots,x_n]/I^*$ is a $B^*$-algebra, the element $t$ is invertible there, so $B^*[x_1,\ldots,x_n]/I^*$ is always flat over $0$. Thus, $B[x_1,\ldots,x_n]/\overline{I^*}$ is also flat, since a submodule of a $t$-torsion-free module is $t$-torsion free. So if $I \subseteq B[x_1,\ldots,x_n]$ is an ideal and $I^*$ its extension to $B^*[x_1,\ldots,x_n]$, then we certainly see that $I = \overline{I^*}$ implies that $B[x_1,\ldots,x_n]/I$ is flat. Conversely, we always have $I \subseteq \overline{I^*}$, and the kernel of $$B[x_1,\ldots,x_n]/I \twoheadrightarrow B^*[x_1,\ldots,x_n]\big/\ \overline{I^*}$$ is $t$-torsion since this map becomes an isomorphism after inverting $t$. Thus, we see that $B[x_1,\ldots,x_n]/I$ is $t$-torsion-free if and only if $I = \overline{I^*}$. Of course there is nothing special about $B[x_1,\ldots,x_n]$, and the same argument works for any $B$-algebra $R$ (or $B$-scheme $Y \to \operatorname{Spec} B$).<|endoftext|> TITLE: Abelianization of $\mathrm{GL}_2(R)$ QUESTION [8 upvotes]: $\DeclareMathOperator\GL{GL}$Let $R$ be a number ring. Are there known lower bounds for $H_1(\GL_2(R);\mathbb Q)$ or $H_1(\GL_2(R),\GL_1(R);\mathbb Q)$ in terms of properties of $R$ (class number, number of real and complex embeddings, etc.)? REPLY [9 votes]: There is an exact sequence $\operatorname{SL}(2,R)_\text{ab}\rightarrow \operatorname{GL}(2,R)_\text{ab} \xrightarrow{\ \det\ }R^*\rightarrow 1 $. Now, if $R$ is not a ring of imaginary quadratic integers, $\operatorname{SL}(2,R)_\text{ab}\otimes \mathbb{Q} $ is zero: this follows from the Corollary to Theorem 3 in Le problème des groupes de congruence pour $\operatorname{SL}_2 $ by J.-P. Serre, Ann. Math. 92, no. 3, 489–527 (1970). Therefore $\det$ induces an isomorphism $\operatorname{GL}(2,R)_\text{ab}\otimes \mathbb{Q}\rightarrow R^*\otimes \mathbb{Q}$ in that case. You'll find how to treat the imaginary quadratic case in the same paper.<|endoftext|> TITLE: Group scheme with an isotrivial maximal torus QUESTION [5 upvotes]: Let $G$ be a reductive group scheme over a normal ring $A$. Then, we know that Zariski locally it admits a maximal torus. Let us assume that it admits a maximal torus after a finite surjective (resp. finite flat) cover, is it possible to replace it by a finite étale cover? REPLY [6 votes]: Edit. I realized after the original post that there are even easier examples. These examples also show that the Quillen–Suslin Theorem fails already for smooth affine quadric hypersurfaces (in some sense, the "simplest" smooth affine varieties after affine spaces, where Quillen–Suslin does hold). Let $\overline{B}$ be a Grassmannian parameterizing $2$-dimensional quotient vector spaces of a fixed vector space $V$ of dimension $n\geq 4$, $$\overline{B}=\operatorname{Grass}_k(V,2),$$ let the rank $2$, locally free $\mathcal{O}_{\overline{B}}$-module be the tautological quotient, $$V\otimes_k \mathcal{O}_{\overline{B}} \twoheadrightarrow \mathcal{E},$$ and let $D\subset \overline{B}$ be a general hyperplane section for the Plücker embedding. For the open complement $B=\overline{B}\setminus D$, all of the arguments in the original post still apply. Moreover, by van Kampen's theorem, the (tame) fundamental group of $B$ is trivial. So that is definitely a better example than in the original post. For completeness, I include below the original post. Original post. Sorry for the delay. I had an overly elaborate example starting with an Enriques surface in characteristic $2$. In fact, sufficiently general K3 surfaces in characteristic $0$ give even simpler examples. Let $\overline{B}$ be a K3 surface over $\mathbb{C}$. Recall by Mumford's Theorem that the group $\operatorname{CH}_0(\overline{B})$ is uncountably generated, in fact, it cannot even be parameterized by any (finite dimensional) algebraic variety, in an appropriate and precise sense. MR0249428 Mumford, D. Rational equivalence of 0-cycles on surfaces. J. Math. Kyoto Univ. 9 (1968), 195–204. https://projecteuclid.org/journals/journal-of-mathematics-of-kyoto-university/volume-9/issue-2/Rational-equivalence-of-0-cycles-on-surfaces/10.1215/kjm/1250523940.full Moreover, in every ample divisor class on $\overline{B}$, there exists a divisor $D$ whose irreducible components have normalization isomorphic to $\mathbb{P}^1$, cf. Beauville's exposition of the Yau–Zaslow theorem. MR1682284 Beauville, Arnaud Counting rational curves on K3 surfaces. Duke Math. J. 97 (1999), no. 1, 99–108. https://arxiv.org/abs/alg-geom/9701019 In fact, by a theorem of Beauville–Voisin, the subgroup of $\operatorname{CH}_0(\overline{B})$ generated by all pushforwards of all $0$-cycles on (possibly singular) rational curves in $\overline{B}$ is a cyclic subgroup commensurate with the cyclic subgroup generated by the cycle class $c_2(T_{\overline{B}})$ and containing all cup products of first Chern classes of divisors. MR2047674 Beauville, Arnaud; Voisin, Claire On the Chow ring of a K3 surface. J. Algebraic Geom. 13 (2004), no. 3, 417–426. https://arxiv.org/abs/math/0111146 Now assume that $\overline{B}$ is a very general K3 surface, so that the Picard group is generated by an ample $\mathcal{O}_{\overline{\mathcal{B}}}$-module $\mathcal{L}$. For definiteness, assume that the degree of $c_1(\mathcal{L})\cap c_1(\mathcal{L})$ equals $2$. Let $D$ be an irreducible, nodal curve in the linear system of $\mathcal{L}$ such that the normalization of $D$ is a rational curve, i.e., an irreducible, nodal curve of arithmetic genus $2$ and geometric genus $0$. For the affine surface $B=\overline{B}\setminus D$, there exists a closed point $p\in B$ such that the cycle class of $\{p\}$ in $\operatorname{CH}_0(\overline{B})$ is not contained in the cyclic subgroup coming from pushforwards of $0$-cycles from $D$. In particular, the cycle class of $\{p\}$ in $\operatorname{CH}_0(B)$ does not equal a cup product of two divisor classes. Apply Serre's construction to the ideal sheaf $\mathcal{I}_p\subset \mathcal{O}_{\overline{B}}$ with determinant invertible sheaf $\mathcal{L}=\mathcal{O}_{\overline{B}}(D)$, i.e., consider the local-to-global spectral sequence computing $\operatorname{Ext}^*_{\mathcal{O}_{\overline{B}}}(\mathcal{I}_p,\mathcal{L})$. The long exact sequence of low degree terms is $$ 0 \to H^1(\overline{B},\mathcal{L}) \to \operatorname{Ext}^1_{\mathcal{O}_{\overline{B}}}(\mathcal{I}_p,\mathcal{L}) \to L\rvert_p \to H^2(\overline{B},\mathcal{L}) .$$ Here $L\rvert_p$ is the $1$-dimensional vector space that is, essentially, the fiber at $p$ of $\mathcal{L}$. Since the dualizing sheaf on $\overline{B}$ is isomorphic to the structure sheaf, Kodaira vanishing gives that $H^q(\overline{B},\mathcal{L})$ is the zero vector space for $q=1,2$. Thus, there is a unique isomorphism class of a nontrivial extension $\mathcal{E}$ of $\mathcal{I}_p$ by $\mathcal{L}$. The sheaf $\mathcal{E}$ is a locally free $\mathcal{O}_{\overline{B}}$-module of rank $2$ whose first Chern class equals $c_1(\mathcal{L})$ and whose second Chern class equals the cycle class of $\{p\}$. In particular, the restriction $\mathcal{E}\rvert_B$ is a locally free $\mathcal{O}_B$-module of rank $2$ whose first Chern class is zero and whose second Chern class is the nonzero class of $\{p\}$. Consider the group scheme $G$ over $\overline{B}$ parameterizing $\mathcal{O}_B$-automorphisms of $\mathcal{E}$ such that the induced automorphism of $\det(\mathcal{E}) \cong \mathcal{L}$ is the identity automorphism. This is an inner form of $\mathbf{SL}_2$ over $\overline{B}$. Since $\mathcal{E}$ is locally free, this inner form is Zariski locally isomorphic to $\mathbf{SL}_2$, and thus has split maximal tori Zariski locally. However, the restriction over $B$ does not have a maximal torus. If it did, the corresponding representation $\mathcal{E}\rvert_B$ would split as a direct sum of locally free sheaves of rank $1$. Then, by the Whitney sum formula, the second Chern class of $\mathcal{E}\rvert_B$ would be in the image of the cyclic subgroup of Beauville–Voisin, i.e., the second Chern class would be a torsion class in $\operatorname{CH}_0(B)$, contrary to the choice of $p$. Now consider the projective space bundle over $\overline{B}$ that parameterizes invertible quotients of the pullback of $\mathcal{E}$. This is a projective scheme. By Bertini theorems, an intersection of this projective scheme with a sufficiently general hyperplane is a smooth closed subscheme $\overline{A}$ whose projection to $\overline{B}$ is finite surjective, and thus finite flat. In particular, the restriction $A$ over $B$ is a finite, flat cover of $B$. The pullback of $\mathcal{E}\rvert_B$ to $A$ has an invertible quotient. Since $A$ is affine, thus surjection automatically splits, so that the pullback of $\mathcal{E}\rvert_B$ to $A$ is isomorphic to a direct sum of two invertible sheaves. Thus, the pullback of $G$ to $A$ does have a maximal torus.<|endoftext|> TITLE: Coherence theorem in braided monoidal categories QUESTION [8 upvotes]: In MacLane's Categories for the working mathematician, the author shows that the evaluation at 1 gives an equivalence of categories $\mathrm{hom}_{\mathrm{BMC}}(B,M)\simeq M_0$ where $B$ is the braid category, $M$ is a braided monoidal category and $M_0$ is the underlying (ordinary) category to $M$. As a result of that, he states a second theorem (coherence theorem) claiming that each composite of canonical maps in $M$ induces a braiding (element of the braid group), and that two such composites are equal for all $M$ if and only they induce the same element braiding. I understand how to construct the braiding from a given composition of canonical morphisms, but I fail to write it properly using the previous theorem and I don't know how to prove that two such maps are equal if and only if they induce the same braiding. I also read the original preprint of Joyal and Street which appears in the references given by MacLane (http://maths.mq.edu.au/~street/JS1.pdf) but I still don't see how to write a proper proof. REPLY [2 votes]: Theorem 1 says precisely that for every braided monoidal category $M$ and every object $V \in M$, any isotopy class of braid on $n$ strands induces an isomorphism $$ V^{\otimes n}\longrightarrow V^{\otimes n}. $$ In particular this theorem also says this isomorphism is well defined, i.e. it does not depend one the particular braid one choose in the equivalence class. In other words, it does indeed says that if two braids are representatives of the same element in the braid group $B_n$, then they define the same morphism for all pair $(M,V)$. Conversely, applying this to the case $M=B$ and $V=\bullet$, i.e. to the identity functor $B\rightarrow B$, one sees that this is an "only if": if two braids are not isotopic, then there exists at least one braided monoidal category (e.g. $B$ itself) for which these two braids induce non-equal morphisms. So to me theorem 2 is really just a slightly weaker formulation of theorem 1 but they're still pretty much saying the same thing.. Also as Donald says, theorem 1 stats precisely that $B$ is the free braided monoidal category on one object. As a useful analogy, if $A$ is an abelian group and $A_0$ its underlying set, the statement that there is a natural bijection $$ Hom_{Ab}(\mathbb{Z},A)\cong A_0$$ states precisely that $\mathbb{Z}$ is the free abelian group on one generator.<|endoftext|> TITLE: Geometric explanation for coincidence in lengths of 16-dimensional even unimodular lattices? QUESTION [13 upvotes]: Question Up to equivalence, there are two positive-definite even unimodular lattices in $16$ dimensions: $D_{8}^{+}\oplus D_{8}^{+}$ and $D_{16}^{+}$. As observed by Witt in 1941, the theory of modular forms implies the "strange and interesting" fact that these inequivalent lattices have identical theta functions. He comments that while modular forms make this fact obvious, geometrically it remains opaque, seemingly a pure coincidence. Is there any known geometric explanation for this fact? For example, is it possible to construct a (geometrically meaningful) explicit length-preserving bijection between $D_{8}^{+}\oplus D_{8}^{+}$ and $D_{16}^{+}$? Reference: Witt, Ernst. "Eine Identität zwischen Modulformen zweiten Grades." Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg. Vol. 14. No. 1. Springer-Verlag, 1941. Background Definitions By lattice I mean a free abelian group $L$ of finite rank $n$ equipped with a positive-definite inner product. Two lattices are equivalent if they are isomorphic as groups by an isomorphism which preserves the inner product. The theta function of a lattice $L$ is the generating function $\theta_{L}(q)$ counting the number of lattice points of a given squared-length in $L$: $$\theta_{L}(q):=\sum_{x\in L}q^{x\cdot x}=\sum_{\alpha\in\left\{ x\cdot x\mid x\in L\right\} }N_{\alpha}q^{\alpha},\quad N_{\alpha}:=\#\left\{ x\in L\bigm|x\cdot x=\alpha\right\}.$$ (Commonly there is a factor of $\tfrac{1}{2}$ in the exponent.) The $D_{n}$ root lattice is the sublattice of $\mathbb{Z}^{n}$ consisting of all vectors whose components sum to an even number. Let $c_{n}\in\mathbb{Z}^{n}$ denote the characteristic vector $(1,1,\ldots,1)$. Then $D_{n} =\left\{ x\in\mathbb{Z}^{n}\bigm|c_{n}\cdot x\equiv0\pmod{2}\right\}$, and we define $$ D_{n}^{+}:=D_{n}\cup\left(D_{n}+\tfrac{1}{2}c_{n}\right). $$ Note that $D_{8}^{+}$ is the usual $E_{8}$ lattice. Collected results The lattices are even. $D_{n}^{+}$ is a lattice when $n\equiv0\pmod{2}$. If additionally $n\equiv0\pmod{4}$, then $D_{n}^{+}$ is integral: $\left\{ x\cdot x\mid x\in L\right\} \subset\mathbb{Z}$. If moreover $n\equiv0\pmod{8}$, then $D_{n}^{+}$ is even: $\left\{ x\cdot x\mid x\in L\right\} \subset2\mathbb{Z}$. The lattices are unimodular. $D_{n}$ has index $2$ in both $\mathbb{Z}^{n}$ and $D_{n}^{+}$. Since $\det\mathbb{Z}^{n}=1$ and $\det(L')/\det(L)=[L:L']^2$, it follows that $\det D_{n}=4$ and $\det D_{n}^{+}=1$. The lattices have the same theta functions. As a consequence of Poisson summation, if $L$ is even and unimodular with dimension $n$, then $\theta_{L}$ is a modular form of level $1$ and weight $n/2$. The unital ring of modular forms of level $1$ is graded by weight and is freely generated by the Eisenstein series $$ \begin{align} E_{4}(q) &=1+240q^{2}+2160q^{4}+\cdots, \\ E_{6}(q) &=1-504q^{2}-16632q^{4}+\cdots. \end{align} $$ Thus the modular forms of weight $8$ are spanned by $E_{4}(q)^{2}=1+480q^{2}+61920q^{4}+\cdots$, and any positive-definite even unimodular lattice of dimension $16$ must have $E_{4}(q)^{2}$ as its theta series. The lattices are inequivalent. The 480 vectors of length $\sqrt{2}$ in $D_{16}^{+}$ generate a proper sublattice, while those in $D_{8}^{+}\oplus D_{8}^{+}$ generate the full lattice. Concretely, the 480 vectors of length $\sqrt{2}$ in $D_{16}^{+}$ are the $2^{2}\cdot\binom{16}{2}$ vectors of the form $\left\{ \pm e_{i}\pm e_{j}\mid1\leq i TITLE: Is there a good mathematical explanation for why orbital lengths in the periodic table are perfect squares doubled? QUESTION [48 upvotes]: $\DeclareMathOperator\SO{SO}\newcommand{\R}{\mathbb{R}}\newcommand{\S}{\mathbb{S}}$The periodic table of elements has row lengths $2, 8, 8, 18, 18, 32, \ldots $, i.e., perfect squares doubled. The group theoretic explanation for this that I know (forgive me if it is an oversimplification) is that the state space of the hydrogen atom is made of functions on $\R^3$, which we can decompose as functions on $\S^2$ times functions on $\R^+$. Then $\SO(3)$ acts on the functions on $\S^2$ and commutes with the action of the Hamiltonian, so we can find pure states inside irreducible representations of $\SO(3)$. The orbital lengths depend on how these representations line up by energy, which is a function on $\R^+$. It happens that the spaces with the same energy have the form $(V_0 \oplus V_2 \oplus V_4 \oplus \cdots ) \otimes W$, where $V_i$ is the irreducible representation of $\SO(3)$ of dimension $i+1$, and $W$ is a $2$-dimensional space that represents the spin. So the orbital length is the dimension of $V_0 \oplus V_2 \oplus V_4 \oplus \cdots$ is the sum of the first $k$ odd numbers, which is a perfect square, and you double it because of the spin. So, in short, the perfect squares arise as the sums of the first $k$ odd numbers, and the invariant subspaces arrange themselves into energy levels that way because... well, here I get stuck. Factoring out the spin, which explains the doubling, can anyone suggest a more conceptual (symmetry-based?) explanation for why perfect squares arise here? REPLY [9 votes]: In this answer, I'm going to crib from this presentation by @JohnBaez and the paper On the Regularization of the Kepler Problem. Milnor's paper includes a lot of the same information. First, I'm going to state a few facts without proof. One can compose the stereographic projection of $\mathbb{R}^3$ to $S^3$ with the symplectomorphism swapping $p$s and $q$s on $T^*(\mathbb{R}^3)$ to get a symplectomorphism from the punctured $T^*(S^3)$ to $T^*(\mathbb{R}^3)$. Furthermore, one can show that the Hamiltonian flow of $p^2$ on a a constant energy surface in $T^*(S^3)$ maps to the Hamiltonian flow of the Kepler potential on $T^*(\mathbb{R}^3)$. This maps the constant energy classical mechanics of a negative energy state in the Kepler potential to a free particle on $S^3$ with fixed energy. This also exhibits the $SO(4)$ symmetry as rotations on $S^3$. Thus (and I'm still undecided if there's some handwaving here), the energy eigenstates in the quantum theory should be irreps of $SO(4)$. You can also exhibit the $SO(4)$ symmetry directly in the quantum theory, so any handwaving isn't a problem. To see what the representations are, the $SO(4)$ action on $S^3$ can be exhibited by the two $SU(2)$ factors in $\operatorname{Spin}(4)$ acting on both sides of $S^3 \cong SU(2)$. The element $(-I,-I) \in SU(2) \times SU(2)$ acts trivially, so you get an $SO(4)$ action. With this, we can decompose a la the Peter–Weyl theorem: $$ L^2(S^3) \cong \bigoplus_i \rho_i \otimes \rho_i^\star $$ Each $\rho_i$ is an irrep of $SU(2)$, and those irreps can be labelled by an integer $n$. Thus, the problem decomposes into $n^2$-dimensional irreps of $SO(4)$, which explains the question asked. [N.B. -- I'd be interested in understanding if this can be done "all at once" as opposed to working with constant energy surfaces and arguing by scaling as I see in the references. If I have time, I'd also want to show that the different irreps of SO(4) have different energies, or maybe I'm missing something obvious. This can all be done by looking at the symmetry explicitly in the quantum theory, I'm sure, but it would be nice to see it geometrically.]<|endoftext|> TITLE: Tangent bundle of a tensor product bundle QUESTION [6 upvotes]: This question was also asked here on math-stackexchange. Let $E\to M$ and $F\to M$ be vector bundles. The structure of their tangents $TE$ and $TF$ is well known. In particular, connectors map $K_E: TE \to E\times_M E$ and $K_F: TF \to F\times_M F$ induce isomorphisms $TE \simeq E \times_M (E\oplus TM)$ and $TF \simeq F \times_M (F\oplus TM)$ (I am using here a fibered product notation, rather than the equivalent pullbacks). Consider now the vector bundle $E\otimes F\to M$. Its tangent can be characterized in the same way as the tangent bundle of every vector bundle. My question is whether there exists a canonical isomorphism of $T(E\otimes F)$ involving the tangent bundles $TE$ and $TF$. I would suspect a positive answer, which in particular looks like a "Leibniz rule". Also, what is the connector $K_{E\otimes F}$ induced by $K_E$ and $K_F$? After all, covariant derivative of tensor products satisfy a Leibniz rule. Here is yet an alternative formulation. How does the choice of horizontal bundles for $TE$ and $TF$ determine a horizontal bundle for $T(E\otimes F)$? The same question can be asked regarding the vector bundle $\operatorname{Hom}(E,F)$. In fact, the question seems reducible to the case of $M$ being a point, so that $E$ and $F$ are just vector spaces. Then, $TE \simeq E\times E$ and $TF \simeq F\times F$. If $E$ has dimension $m$ and $F$ has dimension $k$, then the tensor product $TE\otimes TF$ has dimension $4mk$, which is twice the dimension of $T(E\otimes F)$. How to proceed from here? REPLY [3 votes]: Over a point: $$ T(E\otimes F) = (E\otimes F)\oplus (E\otimes F) = E\otimes (F\oplus F) $$ which is naturally isomorphic to $(E\oplus E)\otimes F$ using the canonical flip $E\otimes F = F\otimes E$. Likewise $$ T\operatorname{Hom}(E,F) = \operatorname{Hom}(E,F)\oplus \operatorname{Hom}(E,F)=\operatorname{Hom}(E,TF) $$ The Leibniz rule mixes the two representations: Consider first curve $\sum_i e_i(t)\otimes f_j(t)$; its velocity at $t=0$ is then $$\Big(\sum_i e_i(0)\otimes f_j(0), \sum_i e_i'(0)\otimes f_j(0) + \sum_i e_i(0)\otimes f_j'(0)\big).$$ Counting entries you have $2mk$. It is more clear to consider a curve in terms of bases $$ \sum_{i,j} c_{ij}(t)\; e_i\otimes f_j = \sum_j\Big(\sum_{i} c_{ij}(t)\; e_i\Big)\otimes f_j = \sum_{i} e_i\otimes \Big(\sum_jc_{ij}(t)\;f_j \Big), $$ then its derivate via (footpoint, speed vector) is $\Big(\sum_{i,j} c_{ij}(0)\; e_i\otimes f_j, \sum_{i,j} c_{ij}'(0)\; e_i\otimes f_j\Big)$. You see that you can move the function part from left to right which explains the isomorphism above. For vector bundles it is similar: the $TM$-part should be there only once. Added: Now let $p_E:E\to M$ and $p_F:F\to M$ be vector bundles. Then $$E\otimes F = \operatorname{Hom}(E^*, F) = \operatorname{Hom}(F^*,E)$$ where the last natural isomorphism is via transpose using $E^{**}=E$. Then $$ T\operatorname{Hom}(E^*,F) = \operatorname{Hom}(E^*,TF) \xrightarrow{\operatorname{Hom}(E^*,\pi_F)} \operatorname{Hom}(E^*,F), $$ where the middle ${\operatorname{Hom}}$ abuses notation and uses unsaid conventions. Note the second vector bundle structure $$ \operatorname{Hom}(E^*,TF)\xrightarrow{\operatorname{Hom}(E^*,T(p_F))} \operatorname{Hom}(E^*,TM), $$ see 8.12 ff of this book or 6.11 in that book. Your next question is essentially, how to write the induced connector $K_{E\otimes F}: T(E\otimes F) \to (E\otimes F)\times_M (E\otimes F)$ whose kernel would identify the pullback of $TM$ to $E\otimes F$ with the horizontal bundle. See 19.12 ff of this book for background. Here we need a name for the canonical isomorphism $\rho:E\otimes TF = F\otimes TE$ (abuse of notation here). Then $K_{E\otimes F} = Id_E \otimes K_F + \rho \circ Id_F\otimes K_E \circ \rho$. Note that the horizontal bundle is not natural. A remark to the formulation at end of your question: $TE$ is NOT a vector bundle over $M$, it has two vector bundle structures $$ TM \xleftarrow{Tp} TE \xrightarrow{\pi_E} E, $$ and the chart changes over $M$ are quadratic (like for the Christoffel symbols). So $TE\otimes TF$ does make sense only with a lot of abuse of notation and unsaid conventions.<|endoftext|> TITLE: Does "$X \not\to (\omega)^\omega_2$ for every infinite $X$" imply ${\sf AC}$? QUESTION [9 upvotes]: For any set $X$ and cardinal $\mu \neq \emptyset$, we denote by $[X]^\mu$ the collection of subsets of cardinality $\mu$. If $\kappa, \mu \neq \emptyset$ are cardinals and $f: [X]^\mu\to \kappa$ is a map, we say that $H\subseteq X$ is homogeneous with respect to $f$ if the restriction $f|_{[H]^\mu}: [H]^\mu \to \kappa$ is constant. For cardinals $\lambda, \mu, \kappa\neq \emptyset$ and any set $X\neq \emptyset$ we write $$X \to (\lambda)^\mu_\kappa$$ if for every map $f: [X]^\mu\to\kappa$ there is $H\subseteq X$ such that $H$ is homogeneous with respect to $f$ and $|H|=\lambda$. With the help of the Axiom of Choice ${\sf (AC)}$ one can prove that $X \not\to (\omega)^\omega_2$ for every infinite $X$ (see Theorem 7, p. 5 of this recommended introduction to infinite combinatorics, thank you to Burak for writing it!). Question. Does the statement "$X \not\to (\omega)^\omega_2$ for every infinite set $X$" imply ${\sf (AC)}$? REPLY [11 votes]: The answer is no, the statement that for every set $X$ we have $$X\not\to(\omega)^\omega_2$$ does not imply the axiom of choice. This was shown by Kleinberg and Seiferas in 1973, see MR0340025 (49 #4782) Kleinberg, E. M.; Seiferas, J. I. Infinite exponent partition relations and well-ordered choice. J. Symbolic Logic 38 (1973), 299–308. https://doi.org/10.2307/2272066 For $\kappa$ a (well-ordered) infinite cardinal, $\kappa$-well-ordered choice, $\mathsf{AC}_\kappa$, is the statement that every $\kappa$-sequence of nonempty sets admits a choice function. The axiom of well-ordered choice $\mathsf{WOC}$ is the statement that $\mathsf{AC}_\kappa$ holds for all infinite well-ordered $\kappa$. This statement is strictly weaker than the axiom of choice: it does not imply that $\mathbb R$ is well-orderable, and even if we add this assumption, the result is still weaker than choice. See for instance theorem 5.1 in MR1351415 (96h:03087) Higasikawa, Masasi Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434. https://doi.org/10.1305/ndjfl/1040149358 However, as shown in the paper by Kleinberg and Seiferas, $\mathsf{WOC}$ plus the existence of a well-ordering of $[\omega]^\omega$ rules out all infinite exponent partition relations. It is still open (as far as I know) whether $\mathsf{WOC}$ suffices for this result. What Kleinberg and Seiferas show is that, under $\mathsf{WOC}$, either all infinite exponent partition relations fail, or else $\omega\to(\omega)^\omega_2$. (And the latter fails if $[\omega]^\omega$ is well-orderable.)<|endoftext|> TITLE: Number of points on schemes modulo $p^k$ QUESTION [5 upvotes]: Let $X$ be a finite type scheme over $\mathbb{Z}_p$ for some prime $p$. Assume that $X_{\mathbb{Q}_p}$ is smooth of dimension $n$, but not necessarily irreducible. Then is $$X(\mathbb{Z}/p^k\mathbb{Z}) = O(p^{kn})$$ as $k \to \infty?$ REPLY [5 votes]: Yes, this is true, for elementary reasons (i.e. not to do with the Igusa zeta function or something). By passing to an open cover, we may assume $X$ is affine, say $X = \operatorname{Spec} \mathbb Z_p[x_1,\dots, x_N]/ (f_1,\dots, f_m)$. The Jacobian of this system of equations is an $N \times m$ matrix. Consider for each $x \in X(\mathbb Z/p^k)$ the minimum $p$-adic valuation of the determinant of an $N-n \times N-n$ submatrix of this Jacobian, considering the $p$-adic valuation of $0$ in $\mathbb Z/p^k$ to be $k$. I claim this is bounded by some constant $c$. If not, we can choose a sequence where this increases to $\infty$ (necessarily with increasing $k$) and then a subsequence which converges $p$-adically, which must converge to a singular point of $X$. So we may assume that one of these determinants has $p$-adic valuation at most $c$. Summing over the finitely many possible choices, it suffices to bound the number of points assuming a particular determinant divides $p^c$. We may as well throw away all the equations not involved in our fixed submatrix. We will show that, setting all variables not involved in this submatrix to specific values, the number of solutions is $O(1)$. This suffices as the number of ways to assign the other variables is $p^{kn}$. To do this, consider two solutions $x_1,\dots, x_{N-n}$ and $y_1,\dots, y_{N-n}$. Let $\ell$ be the minimum $p$-adic valuation of $x_j-y_j$. We have $f_i (x_1,\dots, x_{N-n}) = 0= f_i (y_1,\dots, y_{N-n}) = f_{i} (x_1,\dots, x_{N-n}) + \sum_j \frac{\partial f_i}{\partial x_j} (y_j-x_j) + O \max_j ( y_j -x_j )^2$. Since the determinant divides $p^c$, for some $i$ the middle term divides $p^{\ell+c}$. So we must either have $\ell +c \geq k$ or we have the middle term cancelled by the following term, meaning $\ell +c \geq 2\ell$, i.e. $\ell \leq c$. So $\ell$ is either in $[0,c]$ or in $[k-c k]$. This means that, if $x_1,\dots, x_{N-n}$ and $y_1, \dots, y_{N-n}$ are congruent modulo $p^{c+1}$, they are congruent modulo $p^{k-c}$. So the number of solutions in each congruence class modulo $p^{c+1}$ is at most $p^{(N-n)c}$, and thus the total number of solutions is at most $p^{2 (N-n)c}$. This is $O(1)$, and we're done.<|endoftext|> TITLE: Surprising applications of the theory of games? QUESTION [8 upvotes]: I am currently studying the applications of games in quantum information theory and related fields and I am aware of its uses in places like model theory and set theory. So I was curious, what are some other somewhat surprising places(somewhat because I am aware of the subjectivity of the term) where games were/are useful tools? REPLY [2 votes]: Games are used to solve the problem of "full abstraction for the lambda-calculus", which should count as fairly surprising. Spelling it out a bit: a game can define what a computer program means, satisfying some stringent requirements. The problem is stated such that we need a "meaning" that is not syntactic, i.e. it is quite unlike the source program, and is also expected to be unlike the state of a concrete machine that implements it. And furthermore, it is expected that the distinct objects in the semantics (the "meanings") should exactly match the behaviorally distinct programs. It has turned out that over time, games have been the only kind of solution to that puzzle (I believe). Wikipedia has a brief overview: https://en.wikipedia.org/wiki/Game_semantics<|endoftext|> TITLE: Algebraically-free monadicity theorem QUESTION [8 upvotes]: The monadicity theorem characterises when a functor $u : \mathbf B \to \mathbf E$ is the forgetful functor from the category of algebras for some monad on $\mathbf E$ (up to an equivalence over $\mathbf E$). We can also define the category of algebras for an endofunctor on $\mathbf E$. Is there a characterisation of those functors $u : \mathbf B \to \mathbf E$ which are the forgetful functors from the category of algebras for some endofunctor on $\mathbf E$ (up to an equivalence over $\mathbf E$)? Such forgetful functors do not always admit left adjoints. However, when they do, they are in particular monadic (the conditions of the monadicity theorem being easy to verify). So we can expect such a characterisation to imply that $u \colon \mathbf B \to \mathbf E$ creates $u$-split coequalisers. Furthermore, when $u$ is monadic, the induced monad is algebraically-free (by definition). So (in the presence of a left adjoint) my question can really be seen as asking when a monadic functor induces an algebraically-free monad. This motivates my second question. Is there an intrinsic characterisation of those monads on $\mathbf E$ that are algebraically-free? By "intrinsic", I mean in terms of the data $(T, \mu, \eta)$, without reference to the forgetful functor from the category of algebras. These seem natural questions, and I expect there are answers in the literature somewhere, but so far I have not been successful in finding them. REPLY [2 votes]: I don't have a complete answer, but here's a couple of related observations. If $T$ is the algebraically free monad on an endofunctor $P$ in a category with binary coproducts, then for all objects $X$ we have natural isomorphisms $T X \cong P(T(X)) + X$ such that the unit map is the coproduct inclusion $X \hookrightarrow P(T(X)) + X$. This can be used to show there are examples of monads that are not algebraically free on any endofunctor. The second observation is a negative result, suggesting that the question is difficult. For monads we have the nice result that you can recover the monad up to isomorphism from the category of algebras by composing the left and right adjoints. This is no longer possible for endofunctors. One example is to define endofunctors $P_n : \mathsf{Set} \to \mathsf{Set}$ by $$P_n(X) := \begin{cases} n & X = \emptyset \\ 1 & X \neq \emptyset \end{cases}$$ For $f : X \to Y$ we take $P_n(f)$ to be the quotient map when $X = \emptyset$ and otherwise the identity on $1$. The algebras are the same for all $n > 0$: an algebra structure on $X$ is a point of $X$. Hence they generate the same algebraically free monad, so we cannot recover the endofunctor from the algebraically free monad.<|endoftext|> TITLE: First Pontrjagin class and generator of $\pi_3(\mathrm{SO}(d))$ QUESTION [7 upvotes]: It is well-known that $H^4(B\mathrm{SO}(d), \mathbb{Z}) \cong \mathbb{Z}$, with a canonical generator given by $p_1$, the first universal Pontrjagin class. Let's assume $d\geq 5$ so that everything is stable. We have the following diagram of group homomorphisms: $$H^4(B\mathrm{SO}(d), \mathbb{Z}) \stackrel{\alpha}{\longrightarrow} \mathrm{Hom}(\pi_4(B\mathrm{SO}(d)), \mathbb{Z}) \stackrel{\beta}{\longleftarrow} \mathrm{Hom}(\pi_3(\mathrm{SO}(d)), \mathbb{Z})$$ The left map is the usual map coming from the universal coefficient theorem and the Hurewicz homomorphism $\pi_4(B\mathrm{SO}(d) \to H_4(B\mathrm{SO}(d))$. The right map is induced by the connecting homomorphism for the long exact sequence of homotopy groups for the classifying space fibration $\mathrm{SO}(d) \to E\mathrm{SO}(d) \to B\mathrm{SO}(d)$. Now, there exists a generator $x$ of $\pi_3(\mathrm{SO}(d))$ such that the homomorphism $\varphi_x$ such that $\varphi_x(x) = 1$ satisfies $\alpha(p_1) = \beta(\varphi_x)$. Q: What is an explicit description of this generator? In particular, how does $x$ compare to the canonical map $y: \mathrm{SU}(2) \to \mathrm{SO}(4) \to \mathrm{SO}(d)$ for $d \geq 5$ (which, I think, is twice a generator)? Is $y = 2x$ or $y=-2x$? (Edit: As pointed out by Achim Krause, one has to fix an isomorphism $S^3 \cong \mathrm{SU}(2)$ to make this really well-defined). Pontrjagin classes are commonly defined through Chern classes, which in turn are determined by their values on complex projective spaces. My problem is that it seems quite involved to figure out what this really means in this concrete situation. So what is an efficient way to determine the sign ambiguity? REPLY [4 votes]: Let me try to summarize various observations from the comments and put it all together. Let $d\ge 5$ and let $f: S^4\to BSO(d)$ be a generator of $\pi_4(BSO(d))\cong \mathbb Z$. Let $\gamma^d$ be the universal orientable $\mathbb R^d$ vector bundle over $BSO(d)$. Then we have the following diagram extending the one by the OP. It is easily seen to commute by naturality. Note that as indicated on the diagram the bottom $\alpha$ and the right $f^*$ are isomormphisms. Observe that $f^*(p_1(\gamma^d))=p_1(f^*(\gamma^d))$. It is well known that $p_1$ of any vector bundle over $S^4$ is divisible by 2 (if we identify $H^4(\mathbb S^4)$ with $\mathbb Z$) and in fact every even number is realized. In particular $p_1$ of the Hopf $\mathbb R^4$ bundle is $\pm 2$. The above is done in detail for example in Milnor's classical paper on exotic $7$-spheres. Since the two arrows ending in $Hom(\pi_4(\mathbb S^4), \mathbb Z)$ are isomorphisms this implies that the image of the top row $\alpha$ is exactly $2\mathbb Z$ (we again identify $Hom(\pi_4(BSO(d)),\mathbb Z)$ with $\mathbb Z$). Moreover this also shows that $\alpha(p_1)=\pm 2\beta(x)$ (recall that $x\in Hom(\pi_3(SO(d),\mathbb Z)$ is a generator). Furthermore this shows that the clutching map of the stabilized Hopf bundle is exactly the generator of $\pi_3(SO(d))$ since if it wasn't then $p_1$ of the Hopf bundle would be a nontrivial multiple of $2$. The clutching map of the Hopf bundle is just the identity map $\mathbb S^3=Sp(1)\to Sp(1)\cong SU(2)\subset SO(4)$ and therefore the generator of $\pi_3(SO(d))$ comes from the inclusion $SU(2)\to SO(4)\to SO(d)$. This can also be seen in other ways without using Pontryagin classes. For example, recall the following well known picture of $SO(4)$. Think of $\mathbb R^4$ as quaternions and $\mathbb S^3$ as $Sp(1)$ then consider the ineffective isometric linear $\mathbb S^3\times \mathbb S^3$ action on $\mathbb R^4$ given by $(q_1,q_2)(v)=q_1vq_2^{-1}$. The kernel is $\mathbb Z_2=\{\pm (1,1)\}$ and we get $SO(4)=(\mathbb S^3\times \mathbb S^3)/\mathbb Z_2$. The first $\mathbb S^3$ commutes with multiplication by $i$ on the right and the second with multiplication by $i$ on the left. This identifies the two $Sp(1)$'s with two different embeddings of $SU(2)$ in $SO(4)$. The right $Sp(1)$ is easily seen to correspond to the standard embedding. The two $SU(2)$' give generators of $\pi_3(SO(4))\cong\mathbb Z\oplus \mathbb Z$. Now look at the inclusion $SO(4)\to SO(5)$ and the induced map on $\pi_3$. From the homotopy sequence of the bundle $SO(4)\to SO(5)\to S^4$ we get that the map $\pi_3(SO(4))\to \pi_3(SO(5))\cong \mathbb Z$ is onto. Since there is a unique irreducible real representation of $SU(2)$ of dimension 4 the two $SU(2)$'s are conjugate in $O(4)$ (but of course not in $SO(4)$). This conjugation can be made explicit by observing that $\overline{q(\bar v)}=v\bar q=vq^{-1}$. Hence quaternionic conjugation written as an element of $O(4)$ conjugates the two $SU(2)$'s. Under the standard identification of $\mathbb R^4$ with $\mathbb H$ given by $(a,b,c,d)\mapsto a+bi+cj+dk$ the conjugating matrix is diagonal $diag(1,-1,-1,-1)$. The two $SU(2)$'s become conjugate in $SO(5)$ by $diag(1,-1,-1,-1, -1)$. Hence the $SU(2)$'s have the same image in $\pi_3(SO(5))$ which must be the generator else $\pi_3(SO(4))\to \pi_3(SO(5))$ would not be onto. Lastly, this could be seen using general theory by computing the Dynkin index of $SU(2)$ in $SO(5)$. Given a simple compact Lie group $G$ and its simple subgroup $H$ normalize their Killing forms $(\cdot, \cdot)_G$ and $(\cdot, \cdot)_H$ so that the longest roots have length $\sqrt 2$. Then the restriction of $(\cdot, \cdot)_G$ to the Lie algebra of $H$ is proportional to $(\cdot, \cdot)_H$. The coefficient is called the Dynkin index of $H$ in $G$. It is always an integer and up to sign is equal to the coefficient in the induced map $\pi_3(H)\cong \mathbb Z\to \pi_3(G)\cong \mathbb Z$. For the proof see the book "Topology of transitive transformation groups" by Onishchik. It is not hard to compute that the Dynkin index of $SU(2)$ in $SO(5)$ is 1 but won't do it as we have two other proofs already.<|endoftext|> TITLE: Asymptotics of a quadratic recursion QUESTION [9 upvotes]: Consider the sequence defined by \begin{align} c_0 &{}= 1 \\ c_n &{}= 2\,n\,c_{n-1}-\frac{1}{2}\sum_{m=1}^{n-1}c_m\,c_{n-m}. \end{align} How can you prove that it has the following asymptotics (strongly suggested by numerics) $$ c_n\sim \frac{2}{\pi}\Gamma(n)\,2^n\ \ \ ? $$ To make the question more motivated: the associated divergent series is the asymptotic expansion of a simple ratio of modified Bessel functions $$ R(z) = \frac{K_0(-\frac{1}{4z})}{K_1(-\frac{1}{4z})}\sim \sum_{n\ge 1}c_n z^n = 1+2z+6z^2+24z^3+\dotsb. $$ Indeed, as pointed out in the comment by Richard Stanley, one has $$ 4z^2 R'+4z R+1-R^2 = 0. $$ So the question boils down to what can be said about the large order behaviour of the terms of the asymptotic expansion of a particular known function. If $R$ were analytic one could have used Darboux theorems to relate the answer to the type of nearest singularity. For a divergent non-Borel summable function I don't know whether there are general results. By the way, having the Borel transform in closed form could help because then one could deform the integration contour off the positive real axis to get information, in the spirit of Nevanlinna theorems. REPLY [7 votes]: TL;DR: I have a proof of your conjectured asymptotic formula, modulo the correctness of a certain alternative description of your $c_n$ sequence. I tried to complete Iosif Pinelis's elegant analysis by finding a way to derive the value of the constant $2/\pi$ (denoted $a$ in Iosif's answer) from the quadratic recurrence relation. The problem with that relation is that the behavior it implies for $c_n$ for large $n$ seems to depend in a very sensitive way on the initial values of the sequence, so I concluded that this approach has little chance of working. Fortunately, I've now discovered another, linear recurrence relation for the same sequence $c_n$ that has better behavior and gives your claimed asymptotics without much effort. The relation is: $$ c_0=1, \qquad c_n = g_n + \sum_{k=1}^n h_k c_{n-k} \quad (n\ge 1), \qquad (*) $$ where I define \begin{align} g_n &= \frac{((2n)!)^2}{2^{3n}(n!)^3}, \\ h_n &= \frac{((2n)!)^2}{2^{3n}(n!)^3}\cdot \frac{2n-1}{2n+1}. \end{align} (Edit: this corrects a small typo from the earlier version. As you pointed out in a comment, $g_n$ can also be expressed as $\frac{2^n \Gamma(n+1/2)^2}{\pi \Gamma(n)}$.) I haven't verified rigorously that this relation is equivalent to your quadratic relation, but numerically it gives the correct sequence 1, 2, 6, 24, 126, 864, 7596, ..., and I believe this should be straightforward to prove. The reasoning that led me to it involves your description of the sequence as coming from an asymptotic expansion for a ratio of two Bessel functions. I started with the relation $$ K_0(-1/4z) = K_1(-1/4z) \times \sum_{n} c_n z^n, $$ and, expanding both Bessel functions in a power series (actually not quite a traditional power series because of some nasty-looking transcendental terms, but those can be factored out), massaged this into a linear system of equations satisfied by the $c_n$'s, which gave me the recurrence after a bit of additional guesswork. To rigorously prove the relation, one can work with this Bessel function picture and do the analysis more carefully, or one can try to prove directly that the linear recurrence is equivalent to the quadratic recurrence without any reference to Bessel functions. I suspect this is doable through an inductive argument, probably involving formulating and proving some auxiliary hypergeometric summation identities. Finally, if we assume that $(*)$ is correct, we can prove your claim that $$ c_n \sim \frac{2}{\pi} 2^n (n-1)!. $$ Observe that $$ \frac{c_n}{2^n (n-1)!} = \frac{g_n}{2^n (n-1)!} + \frac{h_n}{2^n (n-1)!} + \frac{1}{2^n (n-1)!} \sum_{k=1}^{n-1} h_k c_{n-k} $$ Using Stirling's formula, you can check that each of the first two terms in this expression converges to $1/\pi$. The third term (the normalized sum) can be easily shown to be $O(1/n)$ (the first summand is $O(1/n)$, and the remaining summands are $O(1/n^2)$ and there are $n-2$ of them). Here I am using the fact that the sequence $c_n/2^n (n-1)!$ is bounded, as shown in Iosif Pinelis's answer (and as can probably also be shown from the linear recurrence without much effort).<|endoftext|> TITLE: Reporting inconclusive experimental searches QUESTION [6 upvotes]: In many areas of mathematics it is informative to conduct numerical experiments. But, it not uncommon that the searches do not lead to the examples or data one was hoping for. Since the numerical searches can be quite time consuming, it seems useful to share these negative results, so that others avoid spending time attempting the same searches. What would be best practices in this regard? And would setting up and maintaining an open-access database be a realistic prospect? As an example, consider the question of the (non-)existence of $n$-body choreographies when allowing the bodies to have different masses and different time-lags (see "$n$-body choreographies" by Montgomery for background as of 2013, and Minton's fine applet for the type of numerical searches that one could try to adapt). I'd be interested to know what has been already attempted by others. I could ask experts, but it may well overlook work by other experts and by Ph.D. students. REPLY [5 votes]: An easy and reliable way to share code is via Zenodo --- works much like arXiv, you get a DOI, can update your files, and it's free. We use it regularly to document computer simulations in physics, I imagine computational mathematics is not that different. Note that Zenodo explicitly welcomes both positive and negative results.<|endoftext|> TITLE: Derived functor of functor tensor product QUESTION [7 upvotes]: Suppose $\mathcal{A}$ is a Grothendieck abelian category with enough projectives, then $\mathcal{A}$ is tensored and cotensored over $\mathrm{Ab}$ with $\mathbb{Z}^{\oplus S}\otimes X\cong \bigoplus_S X$ for any set $S$, and for any abelian group $A$, with presentation $\mathbb{Z}^{\oplus R}\xrightarrow{f} \mathbb{Z}^{\oplus G}\to A \to 0$ we define $$A\otimes X := \mathrm{coker}(\mathbb{Z}^{\oplus G}\otimes X \xrightarrow{f\otimes A} \mathbb{Z}^{\oplus G}\otimes X).$$ Now if we have some $\mathrm{Ab}$-enriched category $Q$ and functors $P:Q^{op}\to \mathrm{Ab}$ and $X:Q\to \mathcal{A}$ we can define the functor tensor product of $P$ and $X$ to be $$ P\otimes_Q X:= \int^{p \in Q}P(p)\otimes X(p) \in \mathcal{A}. $$ This is clearly functorial in both variables. I am wondering does anyone know a balancing result, similarly to balancing Tor. That is is $\mathbb{L}_i(P\otimes_Q -)(X)\cong \mathbb{L}_i(-\otimes_Q X)(P)$. In general I would be very interested in any result on these derived functors or their relation to $\mathrm{Ext}$ in $\mathrm{Fun}(Q,\mathcal{A})$. I am interested in these, as they relate to certain "homology" functors related to $\mathrm{Fun}(Q,\mathcal{A})$. REPLY [5 votes]: The answer is yes if you assume enough things. In particular, the notion of a left flat object of $\mathcal A$ comes up : Definition: An object $L\in\mathcal A$ is left flat if $-\otimes L$ is exact. My assumption will be that $\mathcal A$ has enough left flat objects. This allows you to even define $\mathbb L(P\otimes_Q -)$ in terms of left flat resolutions as I will explain below. (Note : Here I'm using a definition of $\mathbb L$ which is slightly more general than "take projective resolutions", namely I'm using the one using "left defomations", see Riehl's book on homotopical algebra. If you want to use projective resolutions, the assumption is that $\mathcal A$ has enough left flat projectives) Warning: Note that if $\mathcal A$ has projectives that are not left flat, then the answer is no even for $Q=*$, as is easy to convince yourself of. In particular, if $\mathcal A$ has enough projectives (to define $\mathbb L$ in the classical homological algebra sense), the assumption is that enough (equivalently all) projectives are left flat. In this case, the answer is yes, and basically the proof is the same as in the ordinary case. Let me work in the bounded below case because there are the same subtleties as in the ordinary case for the unbounded case. In particular, up to shifting, I will work in the connective case. Let me sketch a proof below (convention : my functor categories are categories of $Ab$-enriched functors, I'm assuming this is what you meant - otherwise, replace $\hom_Q$ with $\mathbb Z[\hom_Q]$): Definition: $P$ is right flat if $P\otimes_Q -$ is exact, and same for $X$ being left flat. Lemma: Projectives in $Fun(Q^{op},Ab)$ are right flat, and there are enough left flats in $Fun(Q,\mathcal A)$. Proof: By the enriched Yoneda lemma and smallness of $Q$, every projective is a summand of $\bigoplus_i \hom_Q(-,q_i)$ for some family of $q_i$'s. Therefore it suffices to prove it for those ones, because flatness is stable under retracts. Flatness is also stable under direct sums because they are exact ($\mathcal A$ is Grothendieck, in particular AB5), so it suffices to prove it for $\hom_Q(-,q)$. But now by the enriched Yoneda lemma again (the "canonical colimit of representable presheaves" version), we have $\hom_Q(-,q)\otimes_Q X \cong X(q)$, which is manifestly an exact functor. The dual case is dual, you simply need to allow $\hom_Q(q,-)\otimes L$ for enough flat objects $L$ of $\mathcal A$. In particular this lemma tells you that the two things you want to compare are well-defined. Corollary: If $C$ is a chain complex of right flat functors $Q^{op}\to Ab$, $C\otimes_Q -$ preserves quasi-isomorphisms (where you define $\otimes_Q$ on chain complexes in the obvious way); and dually. Proof : Let $n\in \mathbb N$, and let $C_{\leq n}$ denote the so-called stupid truncation of $C$, i.e. $0\to C_n\to ... \to C_0$. Because $-\otimes_Q -$ manifestly preserves filtered colimits in each variable, because filtered colimits are exact ($\mathcal A$ is Grothendieck, in particular AB5), and finally because $C= \mathrm{colim}_n C_{\leq n}$, it suffices to prove that each $C_{\leq n}\otimes_Q -$ preserves quasi-isomorphisms. Now we have a short exact sequence $0\to C_{\leq n-1}\to C_{\leq n}\to C_n\to 0$ which is split as a sequence of graded objects. Note that the underlying graded object of $A\otimes_Q B$ only depends on the underlying graded objects of $A,B$ respectively. It follows that this exact sequence remains exact after tensoring (over $Q$) with anything. In particular, by the long exact sequence in homology, and by induction, we reduce to proving that $C_n\otimes_Q-$ preserves quasi-isomorphisms for each $n$. But this is by the assumption that each $C_n$ is right flat. The dual case is completely dual. Note : I am implicitly using the fact that $-\otimes -: Ab\times\mathcal A\to\mathcal A$ preserves colimits in each variable. This is an exercise I'll leave to you :) Note : This lemma is essentially what allows you to define $\mathbb L$ using left/right flat resolutions. Note: This lemma is often packaged as a spectral sequence argument, but it's really about the filtration. Corollary: $\mathbb L(P\otimes_Q -)(X)\simeq \mathbb L(-\otimes_Q X)(P)$. Proof : Pick a right flat resolution $\tilde P$ of $P$, and a left flat resolution $\tilde X$ of $X$ (those exist by the first lemma). You have a zigzag $$\mathbb L(P\otimes_Q -)(X) =P\otimes_Q \tilde X \to \tilde P\otimes_Q \tilde X \leftarrow \tilde P\otimes_Q X = \mathbb L(-\otimes_Q X)(\tilde P)$$ The extreme equalities are by definition, and the two middle arrows are quasi-isomorphisms by the previous corollary. Note : this quasi-isomorphism can be made as natural as left flat resolutions in $\mathcal A$.<|endoftext|> TITLE: How often does the omega theorem hold? QUESTION [6 upvotes]: Write $\psi(x) = \sum_{n\le x} \Lambda(n)$. The classical omega theorem says that $\psi(x) - x = \Omega_{\pm}(x^{1/2})$. Question: How often does this hold? For example, what do we know about the size of the set $\{ n\le x: \psi(x) - x > c x^{1/2} \}$ for some $c$? Ditto for $\psi(x) - x < c x^{1/2}$. What about replacing e.g. $x^{1/2}$ by $x^{\alpha}$ for some fixed $0 < \alpha < 1/2$? What's a good reference for such results? Thanks! REPLY [7 votes]: For any $\varepsilon>0$, there exist $c(\varepsilon)>0$ and $X_0(\varepsilon)>0$ such that for any $X>X_0(\varepsilon)$ we have $$\sup_{X\leq x\leq X^{1+\varepsilon}}\frac{\psi(x)-x}{\sqrt{x}\log\log x}>c(\varepsilon)\qquad\text{and}\qquad \inf_{X\leq x\leq X^{1+\varepsilon}}\frac{\psi(x)-x}{\sqrt{x}\log\log x}<-c(\varepsilon).$$ More precisely, Ingham (1935) proved a stronger result for the case when the real parts of the zeta zeros have a maximum, while Pintz (1980) proved a stronger result for the case when the real parts of the zeta zeros do not have a maximum. There might be even stronger results in the literature, please check.<|endoftext|> TITLE: What is the least dimension of the Euclidean space into which every Riemannian flat n-torus embeds isometrically? QUESTION [8 upvotes]: Given an integer $n > 0$, let $f(n)$ denote the least dimension of the Euclidean space into which there exists an isometric embedding of every Riemannian flat $n$-dimensional torus $\Bbb T^n =\Bbb R^n / L$, where $L$ is an $n$-dimensional lattice in $\Bbb R^n$. What is known about $f(n)$ in terms of exact values, upper and lower bounds, or asymptotics? REPLY [6 votes]: By the Nash Embedding Theorem, $f(n)\le \tfrac12n(3n+11)$. Here's an idea of a way to show that $f(n)\le n(n{+}1)$ by constructing an explicit isometric embedding: Let $\mathbb{T}^n = \mathbb{R}^n/L$ where $L\subset\mathbb{R}^n$ is a lattice and $\mathbb{R}^n$ is endowed with the usual inner product and the standard Riemannian metric $g=\mathrm{d}x\cdot\mathrm{d} x$. Let $L^*\subset\mathbb{R}^n$ be the dual lattice, i.e., the set of vectors $\xi\in \mathbb{R}^n$ such that $\xi\cdot x$ is an integer for all $x\in L$. Now choose $N = \tfrac12n(n{+}1)$ elements $\xi_1,\ldots,\xi_N$ that generate $L^*$ and $N$ nonnegative constants $r_1,\ldots, r_n$ and define a map $\Phi: \mathbb{T}^n\to \mathbb{C}^N\simeq \mathbb{R}^{n(n+1)}$ $$ \Phi(x) = \frac1{2\pi}\bigl(\sqrt{r_1}\,\mathrm{e}^{2\pi i\,\xi_1\cdot x},\ldots, \sqrt{r_N}\,\mathrm{e}^{2\pi i\,\xi_N\cdot x}\bigr). $$ Because of the hypothesis about the set $\xi_1,\ldots,\xi_N$, it follows that $\Phi$ is a well-defined injection, provided that the $r_i$ are all positive, and moreover it follows that, giving $\mathbb{R}^{n(n{+}1)} = \mathbb{C}^N$ the standard product metric with each copy of $\mathbb{C}$ given its standard metric, the metric pulled back by $\Phi$ is of the form $\mathrm{d}x\cdot G\,\mathrm{d}x$ where $G$ is the symmetric positive definite constant $n$-by-$n$ matrix $$ G = {r_1}\,\xi_1\,\xi_1^T + \cdots + {r_N}\,\xi_N\,\xi_N^T. $$ In particular, the induced metric on the image of $\Phi$ is flat. Now, if it's possible to choose the $\xi_1,\ldots,\xi_N$ in $L^*$ so that a multiple of the identity matrix $I_n$ is in the convex hull of the $N$ rank 1 positive semi-definite symmetric matrices $$ \xi_1\,\xi_1^T,\ \ldots,\ \xi_N\,\xi_N^T, $$ then one can choose the $r_i\ge0$ so that $\Phi$ will be an isometric embedding. This is trivial when $n=1$ (of course, $f(1) = 2$), and not difficult when $n=2$: After a rotation in $\mathbb{R}^2$, we can assume that $L^*$ is generated by $\xi_1 = (a,0)$ and $\xi_2 = (b,c)$ where $a>0$ and $c>0$ and $b^2+c^2>a^2$ and $0\le b\le \tfrac12 a$. Then, taking $\xi_3 = (b-a,c) = \xi_2-\xi_1$ works. Thus, $4\le f(2)\le 6$. I'm pretty sure that I have a method for choosing $\xi_1,\ldots ,\xi_6$ when $n=3$ that works, but the details are messy. Probably, there's a cleaner way to do it.<|endoftext|> TITLE: Intuition/meaning behind/physical content of the concept of a smooth structure QUESTION [7 upvotes]: Some mathematical structures are visualized very well. I imagine how a shapeless bunch of points (a set; the only property of which is quantity) is collected in one or another soft form (topological space), depending on what topology is set on it. I imagine how this soft (floating) shape becomes hard when I set the metric. The first examples of deformation retractions give a vivid geometric intuition of the homotopy type as a "(homotopy) framework". Thus, the concepts of homotopy type, topology, and metrics have a clear physical content. In contrast to these three examples, for the concept of a smooth structure on a manifold, everything is not so clear due to the fact that manifolds up to dimension 3 have a unique smooth structure Do you have any intuition for a smooth structure on a manifold? Do you see any meaning/physical content in it? Do visually homeomorphic non-diffeomorphic spaces differ for you? After a smooth manifold is provided with additional structures (Riemannian, symplectic, etc.), the meaning of these objects becomes clear. But a smooth manifold devoid of additional structures is still very mysterious for me. P.S. I'm not sure if this question is for a forum (on the other hand, the system showed me many questions with similar titles). If so, then feel free to close it. REPLY [2 votes]: $\newcommand{\R}{\mathbb{R}}$The purpose of a manifold structure is to be able to differentiate functions. And initially we know how to differentiate functions only if the domain is an open set in $\R$. Moreover, once you know how to differentiate functions, this opens the door to doing all the other stuff you do with manifolds. Here's how I think of it: Initially, think of a manifold as just a set $M$ and nothing else. Think of a coordinate map as a bijection $\phi: O \rightarrow \R^n$, where $O\subset M$ and $\phi(O)$ is an open subset of $\R^n$. An atlas is a collection of coordinate maps such that the domains of the coordinate maps cover $M$. No assumptions on topology or differentiability yet. A topological manifold is $M$ with an atlas $\mathcal{A}$ such that for any two coordinate maps $\phi_1: O_1 \rightarrow \R^n$ and $\phi_2: O_2 \rightarrow \R^n$ such that $O_1\cap O_2\ne \emptyset$, then the map $\phi_2\circ\phi_1^{-1}: \phi_1(O_1\cap O_2) \rightarrow \phi_2(O_1\cap O_2)$ is a homeomorphism. Notice that such an atlas immediately defines a topology on $M$ where the domains of the coordinate maps in $\mathcal{A}$ form a base of open sets. Assume that this topology is Hausdorff, and you have a topological manifold. A smooth manifold is defined in exactly the same way, except you assume that the change of coordinate maps, $\phi_2\circ\phi_1^{-1}: \phi_1(O_1\cap O_2) \rightarrow \phi_2(O_1\cap O_2)$, are smooth. Clearly, a smooth manifold is a topological manifold. If you start with a topological manifold, extend its atlas to a maximal atlas, then you can ask whether there is a subatlas (i.e., a subset of coordinate maps) that satisfies the definition of an atlas of a smooth manifold. If so, you say that the topological manifold is smoothable. There is no reason why there couldn't be two different smooth subatlases of a topological manifold. And, if there are two such subatlases, there is no reason why they should be compatible with each other. In other words, if you have a coordinate map $\phi_1: O_1 \rightarrow \R^n$ in the first smooth atlas and a coordinate map $\phi_2: O_2 \rightarrow \R^n$ in the second smooth atlas, it does not necessarily follow that $\phi_2\circ\phi_1^{-1}: \phi_1(O_1\cap O_2) \rightarrow \phi_2(O_1\cap O_2)$ is smooth. It is homeomorphic, since both lie in the topological atlas. Now, to make things even more complicated, it is possible that there is a global homeomorphism $\Phi: (M,\mathcal{A}_1) \rightarrow (M,\mathcal{A}_2)$ that is a smooth diffeomorphism of the two apparently different smooth manifolds. So, even though the two subatlases are incompatible, they actually define two smooth structures on $M$ that are diffeomorphic. Finally, we say that a topological manifold has more than one smooth structure on it if there are two smooth subatlases such that no map $\Phi$ as described in the previous paragraph exists.<|endoftext|> TITLE: Forcing out of L[U] when we have a precipitous ideal in V QUESTION [7 upvotes]: The following theorem of Jech, Magidor, Mitchell and Prikry is well-known. Theorem. (1) If $\kappa$ is a regular cardinal that carries a precipitous ideal, then $\kappa$ is a measurable cardinal in an inner model of ZFC. (2) If $\kappa$ is a measurable cardinal, then there is some $\mathbb{P}$ such that $\mathbb{P}$ forces that $\kappa$ carries a precipitous ideal. Problem. Suppose there is a precipitous ideal on $\omega_{1}$. Then by the theorem, we know there is some $U$ such that in $L[U]$, $U$ witnesses that $\omega_{1}$ is measurable. Fixing such a $U$, is there a poset $\mathbb{P}$ and $\mathbb{P}$-generic $G\in V$ (this is the rub) over $L[U]$ such that $L[U][G]$ satisfies that $\omega_{1}$ has a precipitous ideal? REPLY [8 votes]: If more large cardinals exist, then there is such a filter. That is, suppose that the least $\kappa_0$ such that there is a proper class transitive model of "$V=L[U_0]$", is countable, (correction) and so is $\kappa_0^{+L[U_0]}$. (This holds if there is a proper class transitive model with two measurables, for example.) Then there is a such a filter. (EDIT: I had ignored the possibility that $\kappa_0^{+L[U_0]}=\omega_1$ in the previous version. But if there is a precipitous filter on $\omega_1$ and $\kappa_0<\omega_1$ then $\kappa_0^{+L[U_0]}<\omega_1$, because otherwise if $j:V\to M$ is a generic embedding with $\mathrm{crit}(j)=\omega_1$ then $j(\kappa_0^{+L[U_0]})>\kappa_0^{+L[U_0]}$, although $j(\kappa_0)=\kappa_0$, which gives us two distinct $L[U]$-type models with the same critical point in $V[G]$, which is impossible.) The construction is along the lines suggested by @AsafKaragila in his comment above. There are also similar such constructions in the literature; I think Sy Friedman used such constructions, for example. It suffices to find an $(L[U],\mathrm{Coll}(\omega,{<\omega_1}))$-generic filter. For this, first observe that because $\kappa_0^{+L[U_0]}<\omega_1$, the linear iteration of $L[U_0]$ of length $\omega_1$ does not send the measurable past $\omega_1$ at any stage ${<\omega_1}$, so the eventual $\omega_1$st iterate has measurable $\omega_1$, so it is just $L[U]$. Let $C\subseteq\omega_1$ is the club of critical points resulting from iterating $L[U_0]$ out to $\omega_1$, let $L[U_\alpha]$ be the $\alpha$th iterate of $L[U_0]$, with measurable $\kappa_\alpha\in C$. Choose a sequence $\left_{\kappa\in C}$ such that $G_\alpha$ is $\mathrm{Coll}(\omega,{<\kappa_\alpha})$-generic over $L[U_\alpha]$, and such that $G_\alpha\subseteq G_\beta$ when $\alpha<\beta$. We can proceed from $G_\alpha$ to $G_{\alpha+1}$ because (i) $\mathrm{Coll}(\omega,{<\kappa_\alpha})^{L[U_\alpha]}=\mathrm{Coll}(\omega,{<\kappa_\alpha})^{L[U_{\alpha+1}]}$ (ii) $\mathcal{P}(\kappa_\alpha)\cap L[U_\alpha]=\mathcal{P}(\kappa_\alpha)\cap L[U_{\alpha+1}]$ is countable, (iii) $\mathrm{Coll}(\omega,{<\kappa_{\alpha+1}})$ factors $\mathrm{Coll}(\omega,{<\kappa_\alpha})\times\mathrm{Coll}(\omega,[\kappa_\alpha,\kappa_{\alpha+1}))$ (iv) for limit $\lambda\leq\omega_1$, $\mathrm{Coll}(\omega,{<\kappa_\lambda})$ is just the direct limit of the $\mathrm{Coll}(\omega,{<\kappa_\alpha})$ under the iteration maps, which is the same thing as their finite support product, and $L[U_\lambda]$ is the direct limit of the $L[U_\alpha]$'s, so all dense subsets of $\mathrm{Coll}(\omega,{<\kappa_\lambda})$ in $L[U_\lambda]$ are met by some $G_\alpha$ where $\alpha<\lambda$. In particular, $G_{\omega_1}$ is the desired $(L[U],\mathrm{Coll}(\omega,{<\omega_1}))$-generic filter. But things seem more subtle if $\kappa_0=\omega_1$.<|endoftext|> TITLE: Solving functional equation $f(xy)=f(x+y)$ and Diophantine equations QUESTION [7 upvotes]: It is well known that the only solution is $f$ a constant function. However, by putting some restrictions on the functional equation, we might get other solutions, with potential implications to solving Diophantine equations or factoring an integer. The restrictions are as follows: $f(xy)=f(x+y)$ if $x,y\geq 3$ are positive integers and $\gcd(x,y)=1$. Now my question is whether or not there is a non-constant function satisfying these requirements. I consider that $f(u)$ is defined for integers $u\geq 7$. Let $f$ be a solution. I use the following notation: $u\sim v\bmod{f}$ if and only if $f(u)=f(v)$. This defines equivalence classes, just like modulo operators define residue classes. Various functional equations similar to the one discussed here, produce various sets of equivalence classes, in the same way that various moduli produce various sets of residue classes. Link to Diophantine equations and factoring This is just a trivial example to illustrate the concept. Using various non-constant $f$ solutions of various related functional equations, we can replace the equation $x^2 + y^2 = z^2$ by $(xy)^2 \sim z^2 \bmod{f}$ if we could come up with a large class of $f$'s that make the two equations equivalent. In short, we reduce an additive equation with $k$ variables (here $k=3$) to one having $k-1$ variables. Likewise, factoring $z$ consists of finding $x,y$ such that $xy=z$. It is equivalent to solving the additive equation $x+y \sim z$ for various $f$'s, hopefully easier to handle once turned from a multiplicative into an additive problem. But do non-constant solutions $f$ really exist? That is my question. A corollary, assuming such functions exist, is how difficult they are to handle, and even whether finding such a function could be an unsolved problem for years to come. If such non-constant functions exist and are rather simple, then Fermat's theorem would have been proved long ago. This makes me think that either they don't exist, or if they exist, they are a tough nut to crack. Possible approach to finding a non-constant $f$ I define $f(u)$ as the smallest integer $v\geq 7$ such that $u\sim v \bmod{f}$. Again, see the analogy with residue classes. Now let's start with $u=7$. We have $7=3+4$, thus $7\sim 12$. We have $12=5+7$, thus $7\sim 12\sim 35$. Apply the principle recursively, and you will get an infinite equivalence class for $7$. Does it cover all integers $\geq 7$? If yes the only solution for $f$ is $f(u)=7$ (a constant function). Now let's start with $u=11$. We also end up with an infinite class, seemingly larger than the previous one, and apparently containing all prime numbers $>7$ and many non-primes. In particular $$11 \sim 13 \sim 17\sim 19 \sim 23 \sim 24\sim 25 \sim 28 \sim 29 \sim 30\sim 31 \sim 36 \sim 37 \sim 40\sim 41$$ How many equivalence classes do we end up with? One? (then $f$ is constant.) Two? Three? More than three? Note that finding the equivalence classes consists of finding all the connected components of an undirected (infinite) graph. If you prove that $7\sim 11$, that won't answer my question, but it will be a big blow, big enough that I may stop doing research on this topic. And I would accept the answer. Likewise, if you prove that $7$ and $11$ are not equivalent ($\bmod{f}$) it will show that a non-constant $f$ exists, and this would be a big encouragement. Update See solutions provided by readers. The possibility of the existence of a non-constant function $f$ was "too good to be true". That said, the functional equation discussed here is a particular case of $a f(xy)= b f(x+y) +c$ with $a,b,c$ integers. Or something more general like (say) $f(xy)=f(g(x+y))$ for some function $g$. Maybe a non-constant solution can be found in this setting. In the end, the goal is to transform an additive problem into a multiplicative one, or the other way around. And to do it in such a way that it leads to simplifications. REPLY [9 votes]: Too long for a comment. My guess is that for $x\geq 7$ we that that $f(x)=c$ by strong induction. We need to check the cases $7\leq x\leq 20$ by hand . Let us now suppose we have $y\geq 20$ and integer. If it can be factored as $a\cdot b$ with $a,b\geq 3$ coprime then obviously $6\leq a+b TITLE: Permanent invertible elements QUESTION [5 upvotes]: Let $A$ be a unital complex algebra with the unit $\bf1$. Let $\mathcal{N}$ be the family of all norms on $A$ making it a unital normed algebra with the same unit $\bf1$. Let us put $B_{\|\cdot\|}=\{x\in A : \|x-{\bf1}\|<1\}$ where $\|\cdot\|\in \mathcal{N}$. Clearly, the intersection $\bigcap_{\mathcal{N}}B_{\|\cdot\|}$ is contained in $A^{-1}_{\|\cdot\|}$ where $A_{\|\cdot\|}$ is the completion of $A$ with respect to norm $\|\cdot\|$ in $\mathcal{N}$. Q. Does there exist any non-scalar element in the intersection $\bigcap_{\mathcal{N}}B_{\|\cdot\|}$? REPLY [8 votes]: Here is a non-trivial condition: If $x$ belongs to this intersection, then $x$ commutes with every nilpotent element. Proof. For every invertible $a\in A$, the function $$N(z):=\lVert a^{-1}za\rVert$$ is a norm of unital algebra. Let $n$ be a nilpotent element, of order $k$, and choose $a_t=\mathbf1-tn$ in the construction above, with $t\in\mathbb R$ a parameter. Then $$a_t^{-1}xa_t=(\mathbf 1+tn+\dotsb+t^{k-1}n^{k-1})x(\mathbf1-n)=x+t(nx-xn)+\cdots-t^kn^{k-1}xn$$ is a polynomial function of $t$. By assumption, $t\mapsto\lVert a_t^{-1}xa_t\rVert$ is a bounded (by $1$) function, hence the polynomial above needs to be constant. In particular $nx-xn=0$. As a corollary, the answer for the case of $A=M_r({\mathbb C})$ with $r\ge2$ is as you expected. Write $X$ instead of $x$ (it is a matrix). Let $u\in{\mathbb C}^r$ be a non-zero vector. Choose $v$ such that $v^Tu=0$ (it exists). Then $uv^T$ is nilpotent, hence $Xuv^T=uv^TX$, which implies that $Xu$ is parallel to $u$. Thus every vector is an eigenvector, which implies that $X$ is scalar: $X=\alpha I_r$. Addition. Suppose now that $(A,\|\cdot\|)$ is a unital Banach algebra. The spectral radius $$r(u)=\lim\inf\|u^k\|^{1/k}$$ is well-defined. As above, $\cal N$ contains all norms $N_a=\|a^{-1}\cdot a\|$ for $a\in A^\times$. If $x$ belongs to the OP's intersection, then the set $$\{a^{-1}xa\;|a\in A^\times\}$$ is bounded. Consider an element $u\in A$ for which $r(u)=0$ ($u$ can be be nilpotent, but this is not necessary if $A$ is infinite dimensional). Then ${\bf1}-zu$ is invertible for every $z\in\mathbb C$. Thus $$z\mapsto({\bf1}-zu)^{-1}x({\bf1}-zu)$$ is a bounded entire function, hence a constant function, $\equiv x$. In other words $x({\bf1}-zu)\equiv ({\bf1}-zu)x$, that is $xu=ux$. Thus the property mentionned above extends to: If $x$ belongs to this intersection, then $x$ commutes with every element of spectral radius $0$.<|endoftext|> TITLE: What is known about exceptional slopes of hyperbolic knots? QUESTION [7 upvotes]: For $K$ a hyperbolic knot in $S^3$, a rational number $p/q$ is an exceptional slope if the manifold $M_{p/q}$ obtained from $(p,q)$-Dehn surgery on $K$ does not admit a hyperbolic structure. Thurston showed that there are finitely many exceptional slopes for any knot complement, and it's now known that there are at most $10$. I'm interested in understanding what sorts of numbers $p/q$ can occur. Is anything known in general? For example, when $K$ is the figure-eight knot, the exceptional slopes are [T, Theorem 4.7] $1/0, 0/1, 1/1, 2/1, 3/1$, and $4/1$, along with their negatives. All of these except $(1,0)$ have $1$ in the denominator. Does this pattern hold more generally? The exception to this pattern is $(1,0)$ which is always an exceptional slope because it gives $S^3$ for any $K$, unless I have the conventions backwards and $(0,1)$ does. (Usually we consider $1/0 = \infty$ to correspond to no surgery for this reason, but that doesn't seem to be Thurston's convention.) [T] W. Thurston, Geometry and topology of three-manifolds, http://library.msri.org/books/gt3m/ REPLY [5 votes]: For an overview, see the survey of Cameron Gordon and older survey. Any integer can occur as the slope of a Seifert fibered Dehn filling on a knot. Conjecture 3.3 of Gordon’s survey states that Seifert fillings must have integral slope. Thus one is left to consider reducible and toroidal fillings. Reducible fillings have integer slope. As indicated in Sam Nead’s answer, toroidal slopes are either integeral or half integral and one of Eudave-Muñoz’s examples. Hence the conjecture would give a complete characterization of exceptional Dehn filling slopes of hyperbolic knots. I’m not sure how much progress has been made on this conjecture in general. Exceptional surgeries have been classified for arborescent knots. Originally Gordon had conjectured that any Seifert surgery on a knot came from Dean’s construction, but other examples were found by Mattman-Katura-Kimihiko . Other specific classes of Seifert spaces which can arise as Dehn fillings and classes of knots have been considered, as well as collections of exceptional slopes that can occur. But that seems to be the state of the art that is known about possible exceptional slopes.<|endoftext|> TITLE: Interesting properties of "coadjoint" orbits inside $V\in \operatorname{Rep}G$ QUESTION [5 upvotes]: Let $G$ be a reductive group over $\mathbf{C}$. It acts on the dual of its Lie algebra $\mathfrak{g}^*$ by conjugation. One can describe the orbits of $\mathfrak{g}^*$ explicitly (e.g. using Jordan blocks for $\operatorname{GL}_n$), There is an especially interesting class, of nilpotent orbits. There are finitely many, labelled by combinatorial data attached to $G$ (e.g. partitions of $n$), and are related to a lot of interesting maths. My question is: what when you replace $\mathfrak{g}^*$ by a general finite dimensional representation $V$ (with some conditions, if you like)? What can be said about their orbits: are they interesting, and can they be classified? Possibly the answer might just be no, and that $\mathfrak{g}^*$ is special because it's Poisson/has other special properties, but hopefully at least some other $V$'s are interesting. REPLY [5 votes]: Generally they can be classified when the action of $G$ on $V$ is visible, which by definition means there are finitely many orbits in the nullcone. Irreducible visible pairs $(G, V)$ were classified in Kac - Some remarks on nilpotent orbits (with some corrections in Dadok–Kac - Polar representations). Most of them come from Vinberg's theta-groups, which are graded Lie algebras/groups (graded by the integers or integers mod $n$). The visible action in those cases is then to consider the adjoint action of the Lie group on its Lie algebra, restricted to the zero-graded piece $G_0$ of the group acting on the 1-graded piece $\mathfrak{g}_1$ of the Lie algebra. In those cases being in the nullcone and being nilpotent in the usual sense coincide, and the Jordan decomposition respects the grading.<|endoftext|> TITLE: Chebotarev density theorem and pure weight local systems QUESTION [5 upvotes]: How do we deduce the following statement from the Chebotarev density theorem? The statement is from Ngo's Fundamental Lemma paper. Let $U$ be a scheme of finite type over $\mathbb{F}_q$. Let $\mathcal{L}_1$, $\mathcal{L}_2$ be two local systems of pure weight $i$. Suppose for any $u \in U$ and $k \in \mathbb{N}$, we have $Tr(\sigma_u^k,\mathcal{L}_1)=Tr(\sigma_u^k,\mathcal{L}_2)$, where $\sigma_u$ is the Frobenius conjugacy class of $u$ in $\pi_1(U)$. Then $\mathcal{L}_1$ and $\mathcal{L}_2$ are isomorphic up to semisimplification. REPLY [11 votes]: As Piotr says, we must assume $U$ normal. The purity assumption is not needed. There are two steps to this proof (1) Suppose for any $u \in U$ and $k \in \mathbb{N}$, we have $Tr(\sigma_u^k,\mathcal{L}_1)=Tr(\sigma_u^k,\mathcal{L}_2)$, where $\sigma_u$ is the Frobenius conjugacy class of $u$ in $\pi_1(U)$. Then for any $\sigma \in \pi_1(U)$, we have $Tr(\sigma^k,\mathcal{L}_1)=Tr(\sigma^k,\mathcal{L}_2)$ (2) Suppose that for any $\sigma \in \pi_1(U)$, we have $Tr(\sigma^k,\mathcal{L}_1)=Tr(\sigma^k,\mathcal{L}_2)$ Then $\mathcal{L}_1$ and $\mathcal{L}_2$ are isomorphic up to semisimplification. The first one follows from the Chebotarev density theorem. Because the traces are $\ell$-adic numbers, it suffices to prove the identity mod $\ell^n$ for each $n$. Whether this is true or not for a given $\sigma$ depends only on the image of $\sigma$ in some finite group (a subgroup of $GL_m (\mathbb Z/\ell^n\mathbb Z)^2$ where $m$ is the rank of these representations). But in a finite quotient of a Galois group (here is where we use normalcy, to get that the fundamental group is a quotient of a Galois group), every element arises from a Frobenius conjugacy class. Since we have the identity on every Frobenius element, we have it on every element. The second one is a general fact about representations of groups over fields of characteristic zero. In particular, we do not use the $\ell$-adic topology, so we may assume the base field is $\mathbb C$. Let $G$ be the Zariski closure of the image of $\pi_1$ in $GL_m \times GL_m$ acting on the two (semisimplified) representations. Because $G$ has a faithful semisimple representation, it is reductive. Take a maximal compact subgroup. If two representations are isomorphic on restriction to this compact subgroup they are isomorphic (Weyl). Now use character theory on this compact subgroup.<|endoftext|> TITLE: Limit of a sum with binomial coefficients QUESTION [5 upvotes]: Let $$A_k = \frac{\sum_{i=1}^ki{2k-i-1 \choose i-1}{i-1 \choose k-i}}{k{2k-1\choose k}}$$ $$B_k = \frac{\sum_{i=1}^ki{2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}$$ $$C_k = \frac{\sum_{i=1}^k(2k-2i-1){2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}$$ for $k\in\mathbb{N}$, where the binomial coefficients are to be taken as zero if any of the parameters are negative. I want to prove that $S_k:=A_k+B_k+C_k$ is decreasing from $k=3$ and $S_k\to2/3$ as $k\to\infty$. I have been struggling with a formal mathematical proof for a few days, and I hope that somebody can point me to the right direction. Note that based on the first 10000 values, the above statements seem to hold, and $A_k,B_k$ and $C_k$ seem to tend to $2/9$ as $k\to\infty$, furthermore, $A_k$ and $B_k$ are decreasing whereas $C_k$ is increasing from $k=3$. Also note that $B_k+C_k$ is simply $$\frac{\sum_{i=1}^k(2k-i-1){2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}.$$ The reason for not making this simplification is that I found it interesting that each of $A_k,B_k$ and $C_k$ tends to $2/9$. It may be better to handle $B_k+C_k$ as a unite. Motivation: This question is related to a preceding question. In the setting explained in the other question, $S_k$ is the probability of the marked red ball staying red in a random permutation of $(2k-1)$ balls. The above statements are already proven in an excellent answer to the preceding question. The aim of the present question is to give a new proof using $A_k,B_k,C_k$. REPLY [3 votes]: Jorge Zuniga showed the identity \begin{equation*} S_{n+2}=p_n\,S_{n+1}+q_n\,S_n, \tag{1}\label{1} \end{equation*} where \begin{equation*} p_n=\frac{21 n^2 + 44 n + 16}{24 n^2 + 52 n + 24},\quad q_n=\frac{3 n^2 + 8 n + 5}{24 n^2 + 52 n + 24}. \end{equation*} Based on \eqref{1}, Jorge Zuniga showed that \begin{equation*} S_{n+1} TITLE: A vertical line with many intersections with $n$ non-parallel lines QUESTION [8 upvotes]: Pick $n\ge 3$ non-vertical lines $\mathscr{L}:=\{\ell_1,\ldots,\ell_n\}$ in the plane which are pairwise non-parallel, and they are not all concurrent in a single point. Question. Does there exist a vertical line $L$ which has $n-1$ or $n-2$ intersections with $\mathscr{L}$? Note: More explicitly, "$L$ has $k$ intersections with $\mathscr{L}$" means that the set $\{L\cap \ell_i: i=1,\ldots,n\}$ has $k$ distinct points. Partial note 1: By using a dual result due to Ungar here one can see that there are at least $n-1$ vertical lines $L$ which have at most $n-1$ intersections with $\mathscr{L}$ Partial note 2: The Sylvester-Gallai problem shows the existence of an ordinary line, that is, given $n$ points in the plane, the exists a connecting line passing through exactly two of those points (this has been improved "optimally" here by Green and Tao). In the setting above, dually, there exists a point which is the intersection of exactly two lines of $\mathscr{L}$. Partial note 3: This seems to be somehow related to (dual of) Problem 1 at the end of this article, as suggested by domotorp in his answer at this previous question. However, the motivation of the latter problem was a conjecture of Erdos and Purdy, namely, $n$ points in the plane determine at least $\lfloor (n-1)/2\rfloor$ triangles with distinct positive area; finally, this has been proved by Pinchasi in 2008 here. REPLY [11 votes]: For the solution see the attached figure here. It is the counterexample with $n=8$. Vertical lines represent all possible vertical lines which have less that $n$ intersection points with given lines. How I found this counterexample? By duality of the real projective plane, the problem is equivalent to the following: Given $n$ distinct points on the plane such that they are not collinear and no two of them lie on the vertical line, is it true that there exists a direction $\mathcal{D}$ such that when passing a line from $\mathcal{D}$ through every given point we obtain $n-1$ or $n-2$ distinct lines? And the answer is no. The counterexample consists of the vertices of the regular octagon rotated so that no two vertices lie on the vertical line. The figure corresponds to such octagon. More notes. After the discussion with my colleagues from Katowice we concluded a little more about the problem. Concerning the equivalent (dual) problem, given a set of $n$ points on the plane, we may consider the set $I$ of all natural numbers $j$ such that there exists a direction $\mathcal{D}$ yielding exactly $j$ distinct lines (when passing a line from $\mathcal{D}$ through every given point). For a regular polygon with $n=2k$ vertices we have $I=\{k,k+1,2k\}$ and for a regular polygon with $n=2k+1$ vertices we have $I=\{k+1,2k+1\}$. Therefore $n=7$ is sufficient for a counterexample to the original problem. One more observation is that transforming a regular polygon through an affine bijection of $\mathbb{R}^2$ preserves parallel lines, so the image of $n$ points under such transformation has the same set $I$ as the original points.<|endoftext|> TITLE: Estimates about prime numbers: a lemma in Bourgain's article QUESTION [7 upvotes]: For $n\in \mathbb{N}$ with prime decomposition $n=p_1^{r_1}\cdots p_k^{r_k},p_i\neq p_j$, let $A=\{p_1,\cdots,p_k\}$; then the following holds: \begin{equation} |\{q\in \mathbb{N},q q' \ge \prod_{p \in A'}p,$$ we have $|A'| := k' \le (1+o(1))\log Q/\log\log Q$ by PNT. We next observe that $|f^{-1}(q')|$ is upper bounded by the number of $q TITLE: Multiple root of resultant QUESTION [5 upvotes]: Let us suppose that we have two polynomials $F_1(x,y)$ and $F_2(x,y)$. Generally speaking, each of them defines a curve on the plane and the system of polynomial equations defined by them computes the intersection points of these curves. If, additionally, their tangents vector are collinear at some common point $(x_1, y_1)$ then, according to my experiments with Groebner Bases, their resultant $\operatorname{Res}(F_1, \, F_2)(y)$ should have a multiple root at the point $y = y_1$. Does this fact has simple explanation? REPLY [8 votes]: This is one of the most important steps in the proof of Bézout theorem on plane curves by elimination theory. Without loss of generality, let me work with resultants in $x$. Assume that the point $p=(0, \,0)$ belongs to both the affine curves $C_1$ and $C_2$ of equations $F_1=0$ and $F_2=0$, respectively. Then, calling $r$ the multiplicity of $C_1$ at $p$ and $s$ the multiplicity of $C_2$ at $p$, we have $$R(x)=x^{r+s} \det \Delta(x),$$ were $\Delta(x)$ is obtained from the resultant matrix by applying suitable elementary operations on its rows and columns. This shows that $x^{r+s}$ divides $R(x)$. Moreover, a simple analysis of the matrix $\Delta(x)$ shows that $\Delta(0)$ is invertible (namely, $x$ does not divide $\det \Delta(x)$) if and only if the tangent cones of $C_1$ and $C_2$ at $p$ have no common component. So we obtain the following Proposition. Let $C_1$, $C_2$ be two plane curves and let $p \in C_1 \cap C_2$. Then, denoting by $\nu_p(C_1, \, C_2)$ the intersection multiplicity of $C_1$, $C_2$ at $p$, we have $$\nu_p(C_1, \, C_2) \geq \operatorname{mult}_p(C_1) \operatorname{mult}_p(C_2),$$ and equality holds if and only if the tangent cones of $C_1$, $C_2$ at $p$ have no common component. Your situation is the case where $C_1$ and $C_2$ are both smooth at $p$, namely, $r=s=1$. Then $x$ divides $R(x)$ and, additionally, $x^2$ divides $R(x)$ if and only if $C_1$ and $C_2$ have the same tangent line at $p$. You can find all the details in Marco Manetti's lecture notes in Algebraic Geometry (in Italian, but they are very clear), see in particular Subsection 5.3.<|endoftext|> TITLE: Zero trace elements in finite fields QUESTION [8 upvotes]: There is so much literature on the relation between the multiplicative structure of a finite field and elements having zero trace, that I am hoping that the following is known. Let $q$ be a prime power, let $n$ be an odd prime number, let $\mathbb{F}_{q^{2n}}$ be the field of cardinality $q^{2n}$ and let $\mathrm{Tr}:\mathbb{F}_{q^{2n}}\to \mathbb{F}_{q^2}$ be the trace map. Let $\mathcal{A}$ be the subgroup of $\mathbb{F}_{q^{2n}}^\times$ having order $(q^n+1)/(q+1)$. Is there an element $a\in \mathcal{A}$ and an element $y$ in the multiplicative group $\mathbb{F}_{q^n}^\ast$ of the field $\mathbb{F}_{q^n}$ with $\mathrm{Tr}(y)\ne 0$ and $\mathrm{Tr}(ay)=0$? It seems to me that $(n,q)=(3,2)$ is the only exception. REPLY [4 votes]: I can show that exceptions occur at most for $n=3$. (Primality of $n$ is never used.) Since $n$ is odd, $\mathbb F_{q^{2n}} = \mathbb F_{q^2} \otimes_{\mathbb F_q} \mathbb F_{q^n}$. The trace map $\operatorname{Tr}:\mathbb F_{q^{2n}} \to \mathbb F_{q^2}$ is obtained by tensoring the identity map $\mathbb F_{q^2} \to \mathbb F_{q^2}$ with the trace map $\operatorname{tr} : \mathbb F_{q^n} \to \mathbb F_q$. Thus, choosing an arbitrary basis of $\mathbb F_{q^2}$, we can write any $a$ as a pair of elements $a_1,a_2 \in \mathbb F_{q^n}$, and your condition that $y \in \mathbb F_{q^n}$ satisfies $\operatorname{Tr}(y)\neq 0$ but $\operatorname{Tr}(ay)=0$ is equivalent to the condition that $\operatorname{tr} (y) \neq 0$ but $\operatorname{tr} (a_1 y ) =\operatorname{tr}(a_2 y)=0$. (We can ignore the condition that $y\neq 0$ as it is implied by the condition that $y$ has trace zero.) Since the trace map of a product is a perfect $\mathbb F_q$-linear pairing on $\mathbb F_q^n$, such a $y$ exists unless $1$ is an $\mathbb F_q$-linear combination of $a_1$ and $a_2$. I will show there must exist a member of $\mathcal A$ that has this unusual property by bounding the number of members of $\mathcal A$ that do have this unusual property. Note that every member of $\mathcal A$ is in the subgroup of order $q^n+1$, thus has norm to $\mathbb F_{q^n}$ equal to $1$. This is a nonsingular quadratic equation in $a_1,a_2$. For each $\lambda_1,\lambda_2$ in $\mathbb F_q$, not both zero, $\lambda_1 a_1 + \lambda_2 a_2 =1$ is a linear equation. There can be at most two solutions to a linear equation together with an nonsingular quadratic equation in two variables, since it gives a nontrivial quadratic equation in one variable. Summing over possible choices of $\lambda_1,\lambda_2$, the number of members of $\mathcal A $ with this unusual property is at most $2 (q^2-1)$. So we can only have all members of $\mathcal A$ with this property if $$ \frac{q^n+1}{q+1} \geq 2 (q^2-1)$$ i.e. $$q^n+1 \geq 2 (q^2-1) (q+1).$$ For $n\geq 5$, the left side dominates the right side for any $q$.<|endoftext|> TITLE: Can countable ordinals start gaps of every order in the constructible universe? QUESTION [5 upvotes]: Define "$\alpha$ starts a gap of order $n+1$ and length $\beta$" iff $\mathcal P^n(\omega)\cap (L_{\alpha+\beta}\setminus L_\alpha)=\emptyset\land\forall\gamma\in\alpha: L_\alpha\setminus L_\gamma\neq\emptyset$ where $\mathcal P^n$ is the powerset operation iterated $n$ times. Define "$\alpha$ is in a gap of order $n$" as $\forall m0$); let $\alpha$ be that sup. Then the sequence $\left<\alpha_n\right>_{n<\omega}$, where $\alpha_n$ is the first start of a gap of order $n$, is $\Sigma_1$-definable over $L_\alpha$ without parameters, and in $L_\alpha$, every set is countable. But then it follows that $L_\alpha$ is the $\Sigma_1$-hull of the empty set in $L_\alpha$ (that is, every element of $L_\alpha$ is $\Sigma_1$-definable over $L_\alpha$ from no parameters), which implies that $\alpha$ does not start a gap. No (ignoring the fact that "starts a gap of order $0$" was not defined); $\alpha$ can not simultaneously start a gap of two distinct orders (this is alluded to in the case of orders 2 and 3 on p. 371 of the paper, near the bottom of the page). E.g. for orders 2 and 3, if cofinally many $\gamma<\alpha$ project to $\omega$, then $\omega$ is the largest cardinal in $L_\alpha$. (Note Monroe's comment was written when the definition of "starts a gap" was different.)<|endoftext|> TITLE: Kernel of the Laplacian + a function QUESTION [7 upvotes]: It is known that the kernel of the (non-negative) Laplacian operator on a closed manifold consists of constant functions. I would like to ask if some similar phenomena happens for the modified operator: $$ Lu=\Delta u+ fu,$$ where $f$ is a smooth function. More specifically: If $f$ equals to minus an eigenvalue of $\Delta$, then $Lu=0$ has non-trivial solutions. Are these the only $f$ with non-trivial solutions? Can we conclude that $Lu=0$ has only zero (or constant solutions) if we assume $f$ non-constant? Otherwise, can you parametrize its kernel (as you parametrize constant functions by their integrals, or by their value in one point)? Thank you very much. REPLY [11 votes]: Q: Can we conclude that $Lu=\Delta u+ fu=0$ has only zero (or constant solutions) if we assume $f$ non-constant? A: No, a counter example in one dimension is the Mathieu equation, which has non-constant $\pi$-periodic or $2\pi$-periodic solutions $u(x)$ when $f(x)=a-2q\cos 2x$ for any given real $q$ at an infinite sequence of values of $a_n(q)$, $n=1,2,3,\ldots$. More generally, $-L$ is the Hamiltonian of a particle in the potential $-f$, and we can readily adjust the potential so that it has a bound state at zero energy – simply by adding a constant to the potential to shift the bound state up or down. Do note that the answer to the question would be affirmative for any generic $f$. To have a nonzero $u$ with $Lu=0$ requires fine tuning of the function $f$, for a generic $f$ such a solution will not exist.<|endoftext|> TITLE: What is Euler's method in linear algebra? QUESTION [6 upvotes]: I have been interested in the following paper ("On systems of linear indeterminate equations and congruences" by Henry J. Stephen Smith, Philosophical Transactions of the Royal Society of London, Vol. 151 (1861), pp. 293-326) JSTOR open link. On page 300, the author states "if we apply Euler's method for the resolution of indeterminate equations to the system (16.)" The latter is a system of $n$ linear equations in $n+m$ variables, with no constants. I am not following what he means by Euler's method in this context. Does someone know what it means? REPLY [9 votes]: Euler's method is described, for example, by James Fogo in Linear indeterminate problems. This applies to systems of equations where there are more unknowns than there are equations, and a solution can be found by restricting the solutions to integers. The method was published by Euler in his book Elements of Algebra (1770). For the historical context, see The historical background of a famous indeterminate problem and some teaching perspectives. Here is an example described by Euler, for the case of one equation with two unknowns: express the two unknowns in terms of a single auxiliary variable and then use the integer condition to restrict that single variable. page 312 of Euler's Algebra, describing the Regula Caeci ("blind man's rule") also known as the The Rule of False Position. I'm not sure why this name is appropriate here; also note that the English translation from 1822 reproduced above is corrupted, for "Position, or The Rule of False" read "or The Rule of False Position"<|endoftext|> TITLE: Reverse Mathematics strength of fixed radius covering theorem QUESTION [8 upvotes]: I am curious about the reverse math status of the below statement. Note that we work in second-order RM, i.e. 'closed set' is interpreted as in Simpson's excellent SOSOA. For any closed $E\subset [0,1]$ and $\epsilon >0$, there are $x_0, \dots, x_k \in E$ such that $\cup_{i\leq k} B(x_i, \epsilon)$ covers $E$. It seems that enough induction suffices to prove this theorem, but I cannot seem to formulate a nice minimal/small fragment. Note that the above statement can be interpreted as asking for a finite sub-covering of the uncountable covering $\cup_{x\in E} B(x, \epsilon)$ of $E$. REPLY [8 votes]: Let me strengthen Emil's excellent answer to a finer reverse mathematical result. Recall that $\mathsf{RCA}_0^\star=\mathsf{EA}+\mathsf{I}\Delta^0_1+\Delta^0_1\textsf{-CA}$ and $\mathsf{WKL}_0^\star=\mathsf{RCA}_0^\star+\mathsf{WKL}$. Over $\mathsf{WKL}_0^{\star}$ the principle is equivalent to $\Sigma^0_1$-induction and over $\mathsf{RCA}_0^{\star}$ the principle is equivalent to $\forall X\Sigma^0_1$-induction. Since over $\mathsf{WKL}_0^{\star}$ every $\forall X\Sigma^0_1$-predicate could be transformed to an equivalent $\Sigma^0_1$-predicate, I'll simply prove that over $\mathsf{RCA}_0^{\star}$ the principle is equivalent to $\forall X\Sigma^0_1$-induction. First let me prove the principle using $\forall X\Sigma^0_1$-induction. We fix $\varepsilon>0$ and a closed set $E\subseteq [0,1]$. Let $N=\lceil\frac{2}{\varepsilon}\rceil$. Clearly it will be enough to construct the finite set $A=\{0\le i TITLE: Simpson's motivicity conjecture QUESTION [11 upvotes]: My question is about Simpson's motivicity conjecture, that is the conjecture that for any (cohomogically) rigid irreducible connection $(M,\nabla)$ on a smooth complex scheme $X$ is of geometric origin in the sense that there exists $Y\overset{f}{\to}X$ such that $(M,\nabla)$ is a subquotient of $R^nf_*\mathcal{O}_Y$ with the Gauss-Manin connection. My main question can be put over the top as "why should we believe it?" I'm aware that predictions of this conjecture have been proved, for instance by H.Ésnault and her colaborators. I'd be more interested in a "plausability" sort of criterion. For instance, there are some deformation theoretic arguments explaining why the Fontaine-Mazur conjecture should hold (I'm choosing F-M because it is of a very similar "shape"). Considering this came from Simpson's work on non-abelian Hodge theory, presumably there is a non-abelian Hodge theoretic reason to expect the validity of this conjecture... Also, how "sharp" do we expect this conjecture to be? Do we know that subquotients of Gauss-Manin connections are rigid? REPLY [3 votes]: I'm not sure if this is the kind of evidence you are looking for, but since you mention the Fontaine-Mazur conjecture, let me remark that the relative version of the Fontaine-Mazur conjecture implies Simpson's motivicity conjecture. A rigid irreducible local system, in particular, is arithmetic: it extends to an etale local system on the descent of your complex variety to some finitely generated field. This is Theorem 4 in Simpson's http://www.numdam.org/article/PMIHES_1992__75__5_0.pdf (note that every local system of geometric origin is arithmetic, by a spreading out argument). Now, one can show that for any irreducible arithmetic local system, the extension to an etale local system over a finitely generated field can be chosen to satisfy the assumptions of the relative Fontaine-Mazur conjecture (e.g. as formulated by Liu and Zhu here https://arxiv.org/abs/1602.06282) -- this follows from the main result of https://arxiv.org/abs/2012.13372 See Lemma 6.2 there for this particular statement. This deduction of Simpson's conjecture from the Fontaine-Mazur conjecture is also outlined on the last page of the paper https://arxiv.org/abs/2101.00487 by Esnault and Kerz.<|endoftext|> TITLE: Solid tensor product of pro-discrete space with Laurent series QUESTION [5 upvotes]: Consider the category $\operatorname{Solid}_{\mathbf{Z}}$ of solid abelian groups in the sense of Clausen-Scholze. This category is a full subcategory of condensed abelian groups, $\operatorname{Cond}_{\mathbf{Z}}$. These are, modulo set theoretical technicalities, abelian sheaves on the site of profinite sets, with finite families of jointly surjective maps as coverings. There is a left adjoint to the inclusion, and thus one can define a tensor product by tensoring in $\operatorname{Cond}_{\mathbf{Z}}$ and then solidifying. Question. Let $V$ be a pro-discrete topological abelian group. Note that $\mathbf{Z}((T))$ is solid, as it is the limit along open inclusions of pro-discrete spaces, and any map from a compact space must factor through some pro-discrete subspace. What is the solid tensor product of $V$ with $\mathbf{Z}((T))$? I presume it is $V\{T\}$, the module of two way infinite Laurent series over $V$, whose Laurent tail coefficients tend to $0$ in $V$? To prove this it would seem I need to prove that the solid tensor product commutes with certain cofiltered limits, which I have not been able to do. REPLY [4 votes]: This is not true in general, the most important observation being that it fails already when $V$ is discrete. In that case $V\otimes^{\blacksquare} \mathbb Z((T))$ is just the usual algebraic tensor product. This agrees with $V((T))$ only if $V$ is finitely generated. In a different direction, for those pro-discrete abelian groups $V$ that are limits of finitely generated abelian groups, a variant of the claim is true; namely, one gets $V\otimes^\blacksquare \mathbb Z((T)) = V((T))$ (but the Laurent tail is finite). Indeed, in that case $V$ is a compact object of the category of solid abelian groups, so has a finite resolution by product of copies of $\mathbb Z$, reducing on to that case; and then it follows from $\prod_I \mathbb Z\otimes^\blacksquare \prod_J \mathbb Z=\prod_{I\times J} \mathbb Z$. As Z. M observes in the comment, a related true statement is that $V\otimes_{\mathbb Z_\blacksquare} \mathbb Z[T]_\blacksquare$ is given by $V\langle T\rangle$ (those series $\sum_{i\geq 0} v_i T^i$ for which $v_i\to 0$ as $i\to \infty$). The tensor product here is not a solid tensor product, but a base change from solid $\mathbb Z$-modules to solid $\mathbb Z[T]$-modules (where being $\mathbb Z[T]$-solid is stronger than being solid over $\mathbb Z$).<|endoftext|> TITLE: Looking for some interesting complex integration contours QUESTION [35 upvotes]: I am currently working on some tools to make contour integration in a proof assistant less painful and I'm looking for interesting examples of contours in the complex plane used in the literature. I am most interested in contours that add little epsilon detours in order to avoid singularities/branch cuts/etc. One example of what I am looking for is the following contour that is used in the study of modular functions. [An interesting integration contour from Apostol's "Modular Functions and Dirichlet Series in Number Theory"] REPLY [2 votes]: See page 382 in [1], a paper I have found very enlightening. Reference [1] HARDY, G. H. (1905), "A method for determining the behavior of certain classes of power series near a singular point on the circle of convergence", Proceedings of the London Mathematical Society 2, pp. 381-389, JFM 36.0474.01.<|endoftext|> TITLE: If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, must the fixed points of $1_\mathbb{Z}$ be a finitely generated group? QUESTION [5 upvotes]: I've copied over this question from what I asked on StackExchange, in the hope that an expert here can readily answer the question. Is there an example of a group $G=K\rtimes \mathbb{Z}$ satisfying the following three conditions? $G$ is finitely generated; $K$ is not finitely generated; the fixed points of $\phi(1)$, which is the automorphism on $K$ corresponding to $1_\mathbb{Z}$, are not a finitely generated group. I suspect there is an example, but I don't have enough experience with infinite groups to come up with one right away. $\\\\$ One idea, which may or may not work: finding matrices $M_1,\ldots,M_k\in\text{GL}_n(\mathbb{C})$ and $g\in\text{GL}_n(\mathbb{C})$ such that $S=\langle g^{-n}\,M_j\,g^n\rangle_{1\leq j\leq k,\;n\in\mathbb{Z}} $ is not finitely generated but such that $g$ commutes with surprisingly many matrices in $S$. REPLY [7 votes]: Here is another example, which I am fond of, because it played a role in a PhD thesis of a student that I supervised a long time ago. It is constructed as a central extension of $C_p \wr {\mathbb Z}$ for a prime $p$. The subgroup $K$ is generated by elements $x_i,y_k$ with $i,k \in {\mathbb Z}$ and $k > 0$, and it has defining relations \begin{eqnarray*} x_i^p &=& y_j^p= 1\ \mbox{for all}\ i,j,\\ [x_j,x_i] &=& y_{j-i}\ \mbox{for}\ j>i,\\ [y_k,x_i] &=& 1\ \mbox{for all}\ i,k, \end{eqnarray*} so it is nilpotent of class $2$ with $K$ and $K/Z(K)$ both infinite elementary abelian, where $Z(K)$ is generated by the elements $y_k$. Now we define the action of $({\mathbb Z},+)$ on $K$ to make the element $1 \in {\mathbb Z}$ map $x_i$ to $x_{i+1}$ for all $i \in {\mathbb Z}$. So the induced action on $Z(K)$ is trivial. Now let $G$ be the semidirect product $K \rtimes {\mathbb Z}$. Then $G$ is generated by $x_0$ and $1 \in {\mathbb Z}$, but $K$ and $Z(G) = Z(K)$ are not finitely generated.<|endoftext|> TITLE: Combinatorics related plane geometry QUESTION [5 upvotes]: There are $n$ men, standing one at each vertex of a convex $n$-gon. If they are allowed to move together along sides or diagonals of the polygon to reach another vertex, how many different ways are there to do so without meeting another one? See OEIS A350599 for the first few numerical values. REPLY [6 votes]: Assume that the paths may not cross and each man must move. Label the vertices $1,2,\dots,n$ in clockwise order. Let the man at vertex $i$ move to vertex $\pi(i)$, so $\pi$ is a permutation of $1,2,\dots,n$. If we draw an arrow from vertex $i$ to $\pi(i)$, then we get a disjoint union of noncrossing cycles of length $\geq 3$. We can obtain such cycles by choosing a noncrossing partition of the vertices with no blocks of size 1 and 2, and then orienting the boundary of the convex hull of each block in two ways. Thus in Exercise 5.35(b) of Enumerative Combinatorics, vol. 2, we should set $f(i)=2$ for $i\geq 3$ and $f(1)=f(2)=0$. If the desired answer is $h(n)$, then by this exercise we have $$ x+\sum_{n\geq 1}h(n)x^{n+1} = \left( \frac{x}{1+2\sum_{n\geq 3} x^n}\right)^{\langle -1\rangle} $$ $$ = \left( \frac{x(1-x)}{(1+x)(1-2x+2x^2)}\right)^{\langle -1\rangle}, $$ where $\langle -1\rangle$ denotes compositional inverse. If a man is allowed to stand still, replace $1+2\sum_{n\geq 3}x^n$ by $1+x+2\sum_{n\geq 3}x^n$. Possibly you can get some kind of explicit formula for $f(n)$ out of this, but it will be messy.<|endoftext|> TITLE: Cartesian-closed full subcategory of locally ringed spaces containing smooth manifolds QUESTION [8 upvotes]: This coming fall, I will be teaching a course on differential topology to a small group of strong students. In preparation for it, I'm trying to find a category $\mathrm{GDiff}$ with the following properties: $\mathrm{GDiff}$ is the category of topological spaces with additional structure $\mathrm{Diff}$ (the category of smooth manifolds) is a full subcategory of $\mathrm{GDiff}$. $\mathrm{GDiff}$ is Cartesian-closed. The natural functor $\mathrm{GDiff} \to \mathrm{LocallyRingedSpace}$ is a full embedding. Question: can such a category exist, or are these conditions inconsistent? If the conditions are inconsistent, then you can deviate somewhat from property 2 (in fact, for the most part I just want to be able to talk about infinite-dimensional manifolds such as spaces of smooth mappings). On the other hand, any additional nice categorical properties are welcome as long as you keep properties 1 and 3. I think that thanks to the 3rd property, $\mathrm{GDiff}$ is as close as possible to the classical differential topology, since the most important concepts of differential topology are defined in terms of locally ringed spaces. In particular, there is a natural canonical functor $\mathrm{Tangent} \colon \mathrm{GDiff}_{*} \to \mathrm{Vect}$ extending resp. functor for $\mathrm{GDiff}$ (here $\mathrm{Vect}$ is the category of vector spaces over different fields, morphisms in it are pairs from a morphism of scalar fields and a morphism of vector spaces). When, as far as I understand, for diffeological spaces there is no standard definition of a tangent space at the moment and different definitions are being explored now (this is related to my other question about the existence of a full embedding of diffological spaces into locally ringed spaces) REPLY [4 votes]: Since you said that you are interested in infinite-dimensional manifolds such as mapping spaces, I wouldn't give up to use diffeological spaces. Diffeological spaces satisfy your requirements 0,1,2, but probably not 3. The reason why I write "probably" is that 3 is not satisfied for the naive way of mapping diffeological spaces to locally ringed spaces (see the comment by Simon Henry at Is the category of diffeological spaces a full subcategory of locally ringed spaces?); however, there might be another way which we currently don't know. The naive way is equip a diffeological space $X$ with the D-topology and with the sheaf that associated to a D-open subset the set of smooth functions on $X$. This defines a functor that is faithful but in general not full. My point is that the naive functor is full when restricted to a very general class of manifolds such as mapping spaces. This is because the D-topology coincides with the usual topology on such mapping spaces (often called the Fréchet topology or Whitney $C^{\infty}$-topology). This follows from Lemma 4.13 and Theorem 5.14 in these Lecture Notes of Christoph Wockel: https://www.math.uni-hamburg.de/home/wockel/teaching/data/HigherStructures2013/hs.pdf the smooth functions (in the diffeological sense, when the mapping space is equipped with the functional diffeology) are precisely the smooth functions (in the manifold sense); this is Theorem 7.6 (c) in above notes of Wockel. Thus, viewing a mapping space as a diffeological space, and then using the naive construction of a locally ringed space, gives the same result as viewing the mapping space (the manifold) as a locally ringed space in the usual way - no information will be lost.<|endoftext|> TITLE: How can I detect the homology image of a unipotent group in the general linear group? QUESTION [7 upvotes]: Suppose $n$ is a positive integer greater than 2, and $F$ is an arbitrary field with at least 4 elements. Denote $\text{GL}_n(F)$ the general linear group in the usual sense and $U_n(F)$ the unipotent group in the sense that it consists of $n\times n$ upper-triangular matrices with 1's on the diagonal. The inclusion $U_n(F)\hookrightarrow\text{GL}_n(F)$ induces a (unstable) homology homomorphism $f_k: H_k(U_n(F);\mathbb{Z})\to H_k(\text{GL}_n(F);\mathbb{Z})$ for any positive integer $k$. And there is another map, a stable homology map: $g_k: H_k(U_n(F);\mathbb{Z})\to H_k(\text{GL}(F);\mathbb{Z})$. How can I detect the image of $f_k$ (or $g_k$)? Is $\text{im}(f_k)$ or $\text{im}(g_k)$ always zero? Background: In Suslin's paper [1] (Sublemma 4.4.2), he asserts that when $k=n=\text{char}(F)=3$ then $\text{im}(f_k)=\text{im}(g_k)=0$ with a reference directed to [2]. But from [2] it's quite unclear to me how to deduce the assertion made in [1]. [1] A. Suslin, K3 of a field, and the Bloch group (Russian), Trudy Mat. Inst. Steklov 183 (1990), 180-199. English transl. in Proc. Steklov Inst. Math. (1991), 217-239. [2] A. Suslin, Stability in algebraic K-theory, pp. 304-333 in Lecture Notes in Math. 966, Springer-Verlag, 1982. REPLY [10 votes]: Suppose first that $F$ is a finite field of characteristic $p$. Then $U_n(F)$ is a Sylow $p$-subgroup of $GL_n(F)$, and so using the transfer in group homology one sees that the image of $f_k$ (for $k>0$) is precisely the $p$-torsion in $H_k(GL_n(F);\mathbb{Z})$. Unstably, this is not well understood, but it cannot be trivial by general principles in group cohomology (see the top answer to this MO question). But it follows that the image of $g_k$ is contain in the $p$-torsion of $H_k(GL(F);\mathbb{Z})$, and this is completely understood by Quillen's calculations: it is trivial. Suppose now that the field $F$ is infinite; it is then a "ring with many units" in the sense of Nesterenko--Suslin (Homology of the full linear group over a local ring, and Milnor's K-theory), and we may apply the results of that paper. There is a group $UT_n(F)$ of all upper triangular matrices (not necessarily having 1's on the diagonal), and a split extension $$1 \to U_n(F) \overset{i}\to UT_n(F) \overset{q}\to D_n(F) \to 1$$ where $D_n(F)$ are the diagonal matrices. If follows by iteratedly applying Theorem 1.11 of Nesterenko--Suslin that the map $q$ induces an isomorphism in integral homology, but then $i$ induces the trivial map in integral homology. As the inclusion of $U_n(F)$ into $GL_n(F)$ factors over $UT_n(F)$, it follows that $f_k$ (and hence $g_k$) is trivial.<|endoftext|> TITLE: Geometric series in algebraic number fields QUESTION [12 upvotes]: For which algebraic numbers $\alpha$ is there a valuation on the number field ${\mathbb {Q}}(\alpha)$ for which the infinite series $\sum_{n=0}^\infty \alpha^n$ converges to $1/(1-\alpha)$? REPLY [18 votes]: This holds precisely for elements which aren't roots of unity. Indeed, by the product formula we have $\prod_v|\alpha|_v=1$, where the product runs over all (finite and infinite) primes. This shows that either all $|\alpha|_v$ are equal to $1$, or at least one of them is smaller than $1$. If all $|\alpha|_v$ are equal to $1$, then $\alpha^n$ doesn't tend to zero in any valuation, so $\sum\alpha^n$ cannot converge. This happens precisely when $\alpha$ is an algebraic integer (that property follows from having $|\alpha|_v\leq 1$ for all the finite places $v$) with all conjugates of absolute value $\leq 1$ (infinite places), which by a theorem of Kronecker characterizes roots of unity. If $|\alpha|_v<1$ for some place $v$, then $\sum\alpha^n=\frac{1}{1-\alpha}$ in the valuation induced by $v$, regardless of whether it is a finite or infinite place. The above answer is assuming you consider valuations/absolute values coming from all places in your question. If you only consider finite places, then there isn't such a complete characterization. Rather, you just want $\alpha$ to have positive valuation at some finite place, which you can characterize by saying $1/\alpha$ is not an algebraic integer.<|endoftext|> TITLE: Finite domination and compact ENRs QUESTION [8 upvotes]: Edit: In the comments, Tyrone points out that West's positive answer to Borsuk's conjecture implies that every compact ENR is homotopy equivalent to a finite CW complex. It follows that the only finitely dominated spaces which are homotopy equivalent to compact ENRs are those finitely dominated spaces whose Wall finiteness obstruction is trivial. This completely settles the question. A space $X$ is said to be finitely dominated if there is a finite CW complex $K$ and maps $r: X \to K$, $s: K \to X$ such that $r \circ s: K \to K$ is homotopic to the identity map, i.e., $X$ is a homotopy retract of $K$. Equivalently, $X$ is finitely dominated if there is a space $K'$ homotopy equivalent to a finite CW complex and maps $r: X \to K'$, $s: K' \to X$ such that $r' \circ s': K' \to K'$ is the identity, i.e., $X$ is a (strict) retract of $K'$. A space $X$ is said to be an ENR (Euclidean neighborhood retract) if there is an embedding $i: X\to \Bbb R^n$ such that $i(X)$ is a retract of some open neighborhood $U \subset \Bbb R^n$. It is known that $X$ is a compact ENR if and only if $X$ is a (strict) retract of a finite CW complex (cf. Hatcher's book App. A). The above leads to the following question: Question Is every finitely dominated space homotopy equivalent to a compact ENR? If not, what are the obstructions? Notes: (1) The question asks whether or not the property of being a homotopy retract of a finite complex is the same as that of being homotopy equivalent to a strict retract of a finite complex. (2) If $X$ is simply connected and finitely dominated, then Wall shows that X is homotopy equivalent to a finite CW complex. It follows that X is homotopy equivalent to a compact ENR. So if there a counterexample, if it exists, is necessarily not 1-connected. REPLY [11 votes]: It was originally conjectured by Borsuk that every compact ANR should be homotopy equivalent to a finite CW complex. While it was known that every separable ANR has the homotopy type of a countable CW complex, the finer statement was an open problem for some time in the '60s and '70s. It was J. West Mapping Hilbert Cube Manifolds to ANR's: A Solution of a Conjecture of Borsuk, Ann. Math., 106, (1977), 1-18. who finally arrived at a positive solution to Borsuk's conjecture (see Corollary 5.2). T. Chapman had previously shown that every compact Hilbert Cube manifold has the homotopy type of a finite CW complex. In the paper linked above, West showed that every compact ANR has the homotopy type of compact Hilbert cube manifold, thus concluding; Theorem: [West] Every compact ANR has the homotopy type of a finite CW complex. Since Wall has shown that there are finitely dominated spaces not of the homotopy type of a finite complex, it must be that there are fintiely dominated spaces not of the homotopy type of any compact ANR.<|endoftext|> TITLE: Is there a simple explanation for why rational plane curves of degree $>2$ are singular? QUESTION [8 upvotes]: Its a well-known result that smooth projective plane curves of degree $d$ have genus $(d-1)(d-2)/2$, so in particular, smooth curves of degree $1$ and $2$ are genus 0, and those of higher degree have positive genus. It seems like there should be a simple explanation for this in terms of the behavior of dimension 3 subspaces of degree $>2$ polynomials (something about them having too many zeros), using the standard criterion that the map determined by a linear system is an isomorphism if and only if contains functions that distinguish points, and with non-zero derivative at each point, but I have not managed to find it in any reference, or figure out what it is myself. Do you have a simple explanation of this fact? REPLY [3 votes]: After reading the reference suggested by Abdelmalek, I think I have a proof I’m happy with (which is basically the one from the book in more modern language); I suspect this is also equivalent to the observation in terms of the secant variety from Kapil’s answer. This is a little more advanced than I was looking for, but I’ve already given my lecture, so this is more for my own satisfaction at this point. Given three homogeneous polynomials $f_1,f_2,f_3$ of degree $d$ on $\mathbb{P}^1$, we can look for point where this map fails to be injective (we have a node in the image) by looking at whether the section $g_{ij}([X_0:X_1],[Y_0:Y_1])=\frac{f_i(X_0,X_1) f_j(Y_0,Y_1)-f_j(X_0,X_1)f_i(Y_0,Y_1)}{X_0Y_1-X_1Y_0}$ of $\mathcal{O}(d-1)\boxtimes\mathcal{O}(d-1)$ on $\mathbb{P}^1\times \mathbb{P}^1$ vanishes for all $i,j$. Of course, enough to check this for $i=1$ and $j=2,3$. The multi-Bézout theorem (equivalently, cup product) tells me that there should be $2(d-1)^2$ simultaneous zeros of $g_{12}$ and $g_{13}$ (up to multiplicity). A simultaneous zero of these corresponds to a cusp if $[X_0:X_1]=[Y_0:Y_1]$ and to a node if $[X_0:X_1]\neq [Y_0:Y_1]$, except when $f_1(X_0,X_1)=f_1(Y_0,Y_1)=0$ and the intersection multiplicity is 1 (in which case, it’s a smooth point). This final exceptional case accounts for $d^2-d$ points (up to multiplicity). So, there are $(2(d-1)^2)-d(d-1)=(d-1)(d-2)$ intersection points left that give singularities. The symmetry under swapping X’s and Y’s tells us that generically we have $(d-1)(d-2)/2$ nodes, as expected from the fact that a smooth curve of this degree has that genus.<|endoftext|> TITLE: Are standard Koszul algebras with the same Kazhdan-Lusztig polynomials Morita equivalent? QUESTION [6 upvotes]: Let $A,B$ be a pair of quasi-hereditary algebras and assume that $A$ and $B$ are both standard Koszul. Further assume that the graded decomposition matrices of $A$ and $B$ coincide (that is, the multiplicities of standard modules in projective modules coincide, as do their grading shifts). Under what circumstances could I conclude that these algebras $A$ and $B$ are graded Morita equivalent? It feels to me that one should be able to make a general statement. If not, are there any counterexamples to the following claim? If $A$ and $B$ are standard Koszul and have the same decomposition matrices, then they are Morita equivalent. REPLY [3 votes]: Let $k$ be an infinite field. I will descibe an infinite family of standard Koszul algebras with the same graded decomposition matrix. They will be algebras given by a quiver with relations. The quiver will be the same in all cases: nine vertices $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$, with an arrow from $a_{i}$ to $b_{j}$ and an arrow from $b_{i}$ to $c_{j}$ for all choices of $(i,j)$, so $18$ arrows in total. For the relations, for each $(i,j)$ pick one of the three paths from $a_{i}$ to $c_{j}$ and for each of the other two paths, make it a scalar multiple of the first path. All $18$ scalars may be chosen independently. So this gives an $18$ parameter family of sets of relations. Different choices may give isomorphic algebras. But since there is at most one arrow between each pair of vertices, the only ambiguity is that we can permute vertices (but only in finitely many ways) and multiply arrows by nonzero scalars. However, some choices of these scalars (for example, choosing them all to be equal) do not affect the relations, so we have at most $17$ parameter families of choices of relations giving the same algebra (in fact, less than $17$ parameters). So we must have infinitely many nonisomorphic algebras. Since the quiver is acyclic, we can order the vertices so that the standard modules are just the simple modules. If we do this, it is straightforward to check that the algebras are all standard Koszul and have the same graded decomposition matrices.<|endoftext|> TITLE: About closed points in symmetric product schemes over a finite field QUESTION [5 upvotes]: Let $k=\mathbb{F}_q$ be a finite field with $q$ elements and let $X$ be a quasi-projective $k$-scheme. I saw somewhere claims the following results (without explanation): Let $N$ be a positive integer and let $i: Z\hookrightarrow X$ be the closed immersion of the (finite) disjoint union of ${\rm Spec}(\kappa(x))$ for all the closed points $x\in X$ of degree $[\kappa(x):k] TITLE: Two vague questions about TFT QUESTION [9 upvotes]: Question 1. Take a smooth projective Calabi-Yau $X$. Then $D^b(X)$ is a fully-dualizable category and there's an associated 2d TFT. This the usual 2d B-model with target $X$. But $D^b(X)$ is actually a monoidal category, so there's a 2-category $D^b(X)$-mod of categories with an action of $D^b(X)$. Apparently this 2-category is also fully-dualizable, and it defines a 3d TFT. This is the Rozansky-Witten theory with target $T^*X$. (Here I'm just repeating words I've been told.) But $D^b(X)$ is actually symmetric monoidal, so $D^b(X)$-mod is monoidal, so there's a 3-category ($D^b(X)$-mod)-mod. Does this give us a 4d TFT? And if so does this ever stop? $\mathcal{O}_X$ is genuinely commutative (or $E_\infty$) so for every $n$ there should be an $n$-category of modules over modules over $\;$ ...$\;$ over modules over $\mathcal{O}_X$. So potentially an $n$-d TFT. Does this (a) work, and (b) produce anything interesting? And a related: Question 2. Evaluating an $n$-d TFT on the $n-1$ sphere apparently gives us an $E_n$-algebra. If the $n$-d theory came from modules over ... an $E_{n-1}$ algebra then this is some version of taking the centre. For example the 2d B-model on $X$, evaluated on $S^1$, produces the $E_2$ algebra $HH^*(X)$. What happens if I take modules over modules over ... this $E_n$ algebra? If I'm lucky I might get a dualizable $n$-category and then an $(n+1)$-TFT. Does this work in general? (Presumably not). Does it ever work? For example is there a 3d TFT arising from modules over modules over the $E_2$-algebra $HH^*(X)$, and if so does it have a name? REPLY [11 votes]: I will only attempt to answer the first question. Given a symplectic manifold $M$, there is a TQFT in any odd dimension. In dimension 1 this is the topological quantum mechanics and in dimension 3 this is the Rozansky--Witten theory. In higher dimensions these have no name and do not arise from topological twists of supersymmetric sigma-models. These are all examples of AKSZ theories. If you squint your eyes, you can think of $M=B G$ as a symplectic manifold. The corresponding TQFTs are the Chern--Simons theories: in 1d it is the theory considered in https://arxiv.org/abs/1005.2111, in 3d is is the usual Chern--Simons theory, in 5d it is the theory considered in https://arxiv.org/abs/1710.02841 (a topological twist of 5-dimensional superysmmetric Yang--Mills). In higher dimensions it again does not arise from topological twisting. Given any manifold $X$ you can consider $M=T^* X$. So, you get TQFTs associated to any manifold $X$. These will not be fully defined in the top dimension as $M$ is noncompact. For $M = T^* BG$ these are called the BF theories. Similarly, if you have an odd symplectic manifold $M$, you get the AKSZ TQFT in any even dimension. E.g. for $M = \Pi T^* X$ for a manifold $X$ you get the $B$-model. To conclude: the TQFTs you mention (obtained from $D^b(X)$ as a symmetric monoidal category) exist in any dimension. The partition functions will not be well-defined in dimensions $\dim\geq 3$ (the corresponding object is not fully dualizable), but they have interesting spaces of states, categories etc.<|endoftext|> TITLE: Asymptotics of error function integral with square root QUESTION [5 upvotes]: I am interested in the asymptotics of the integral $$I(a):=\int_0^\infty \sqrt{x}\operatorname{Erfc}(x+a)\,\mathrm{d}x$$ for $a>0$. I think that $I(a)$ should be decaying exponentially as $I(a)\lesssim e^{-a^2}$ for large $a$. Numeric integration indicates that $$ \lim_{a\to\infty} I(a) e^{a^2} a^{5/2} =C \in(0,1)$$ for some constant $C$. Are there integral tables where I could obtain explicit expressions for $I(a)$ for finite $a$? If not, are the asymptotics above correct, and what is $C$? So far I could only find results when the error function is integrated against integer powers of $x$ e.g. in Section 2.14 of https://intra.ece.ucr.edu/~korotkov/papers/Korotkov-book-integrals.pdf REPLY [7 votes]: $$\int_0^\infty \sqrt{x}\,\mathrm{Erfc}(x+a)\,\mathrm{d}x\rightarrow \frac{e^{-a^2}}{(2a)^{5/2}},\;\;\text{for}\;\;a\rightarrow \infty,$$ so $C=2^{-5/2}.$ Corrections are smaller by a factor $1/a$. I obtained this asymptotics by integrating the large-$a$ expansion of the error-function, $$\text{Erfc}\,(x+a)\rightarrow \frac{e^{-(a+x)^2}}{\sqrt{\pi } a},\;\;\text{for}\;\;a\rightarrow \infty.$$<|endoftext|> TITLE: Deductive system involving only geometric sequents QUESTION [10 upvotes]: A geometric theory is made up of sequents of restricted form: It may only be of the form $$\phi \vdash \psi$$ possibly with free variables (which are implicitly taken universal closure). $\phi, \psi$ are geometric formulae which can only involve $\top, \bot, =, \exists, \wedge$ (binary conjunction) and $\bigvee$ (arbitrary disjunction). Geometric theories enjoy some good properties. In particular, it plays well with geometric morphisms, hence the name. My question is: is there a (complete, sound) deductive system that involves only geometric sequents? I think we can do this just by writing down the obvious rules. But I'm not familiar with infinitary disjunction. I'm also not quite sure once it gets to the predicative logic part. Is there any existing work on this? An answer or a pointer to relevant material is welcome. REPLY [12 votes]: For geometric theories in a countable fragment, (axiomatized by countably many sequents where the disjunctions are also countable), there exists a sound and complete system that appears already in Makkai and Reyes "First-order categorical logic". In page 159, it is described in the section "Completeness of a one-sided system for coherent logic". One should be aware that what Makkai and Reyes called there "coherent" is what we call now "geometric". Also, although they use a Gentzen style sequent calculus, their system is easily seen to be equivalent to the system of geometric logic through sequents as described in the question, which appears, e.g. in Johnstone's "Sketches of an Elephant" section D1.3 If one removes the countability condition, then the system is no longer complete for usual set models but there is still a completeness theorem in terms of models in toposes (and, in fact, in terms of Boolean-valued models). In the same book Makkai and Reyes also treat these cases. If one insists in having a complete system in terms of set models even when the theory has $\kappa$ sequents for some uncountable $\kappa$, the workaround is to relax the notion of geometric formula to that of a $\kappa$-geometric formula, in which conjunctions are not necessarily finitary, but of size less than $\kappa$. These are preserved by the $\kappa$-geometric morphisms, which are those geometric morphisms whose inverse image additionally preserves all $\kappa$-small limits. The deductive system in this case features a new infinitary rule and is described in my paper "Infinitary generalizations of Deligne's completeness theorem", The Journal of Symbolic Logic, volume 85, Issue 3, pp. 1147 - 1162.<|endoftext|> TITLE: What are the rational functions on a noetherian affine scheme? QUESTION [8 upvotes]: Let $A$ be a noetherian ring and $X=\operatorname {Spec}A$ the corresponding affine scheme. There are three rings which might reasonably be called the ring of rational functions on $X$. a) The total ring of fractions $S^{-1}A$ obtained by inverting the monoid $S$ of regular elements of $A$ (=non zero divisors): $$\mathcal K (X)=\operatorname {Tot} A$$ b) The product of the localizations of $A$ at the finitely many minimal prime ideals of $A$: $$\mathcal R(X)=\prod_{\mathfrak p \in \operatorname {Specmin}A} A_\mathfrak p$$ c) The product of the localizations of $A$ at the finitely many associated prime ideals of $A$: $$\mathcal A(X)=\prod_{\mathfrak P \in \operatorname {Ass}A} A_\mathfrak P$$ Questions What are the relations between these rings? Is there a name or use for these rings? REPLY [7 votes]: The ring $\mathcal K(X)$ is called the ring of meromorphic functions on $X$ in EGA, the Stacks project or by Kleiman and many others. The terminology is disastrous if one considers schemes over $\mathbb C$, since these schemes have a holomorphic structure for which "meromorphic" has a completely different meaning. Be that as it may, there is a canonical ring morphism $$u=(u_\mathfrak P):\mathcal K (X)=\operatorname {Tot} A \to \mathcal A(X)=\prod_{\mathfrak P \in \operatorname {Ass}A} A_\mathfrak P$$ Indeed $S=A\setminus \cup_{\mathfrak P \in \operatorname {Ass}(A)}\mathfrak P\subset S_\mathfrak P =A\setminus \mathfrak P$ so that we have ring morphisms $$u_\mathfrak P:\operatorname {Tot} A=S^{-1}A\to A_\mathfrak P=S^{-1}_\mathfrak P A$$ which taken together give the morphism $u$. We also have a natural projection $v:\mathcal A(X)\to \mathcal R(X)$, as well as the composition of those natural morphisms $w=v \circ u:\mathcal K (X)\to \mathcal R(X)$. The ring $\mathcal R(X)$ is called the ring of rational functions in EGA 1, the Stacks project and many other places; this is an excellent terminology generalizing the standard case of an integral scheme. On the other hand I have never seen any allusion to $\mathcal A(X) $ in the literature. An example Let $k$ be a field and consider the ring $A= k[X,Y]/\langle Y^2,XY\rangle=k[x,y]$ ($y^2=xy=0$), for which the associated prime ideals are $\mathfrak p=\langle y\rangle\subset \mathfrak P=\langle x,y\rangle$. The canonical morphism $w=v \circ u:\mathcal K (X)\to \mathcal R(X)$ is thus the morphism $w:k[x]_{\langle x\rangle}[y]\to k(x)$, in which $w(x)=x$ and $w(y)=0$. That morphism is not injective since the nonzero nilpotent meromorphic function $y$ is sent to zero, nor surjective since the rational function $\frac 1x$ is not the image of a meromorphic function on $X$. Some remarks As already mentioned the ring $\mathcal A(X)= k[x]_{\langle x\rangle}[y]\times k(x)$ doesn't seem very useful. To tell the truth I don't quite understand the reason for the introduction of $\mathcal K(X)$ either, but I hope that some of the extremely competent algebraic geometers on this site will tell me how useful meromorphic functions are in certain situations. Recall also that the definition of the sheaf $\mathcal K$ on a general non-affine scheme is not trivial at all. Kleiman showed in his celebrated article, mischievously called Misconceptions about $\mathcal K_X$, that Grothendieck, Hartshorne and Kleiman himself had given false definitions for that sheaf!<|endoftext|> TITLE: Induced homeomorphism from a quasi-isometry between hyperbolic spaces QUESTION [7 upvotes]: Theorem. Let $\phi:X\rightarrow Y$ be a quasi-isometry between two (Gromov) hyperbolic spaces $X$ and $Y$. If $X$ and $Y$ are proper, then ϕ induces a homeomorphism between their boundaries. The proof of the above statement is well-written in Bridson and Haefliger's book. My question is that `can we drop the condition that $X$ and $Y$ are proper?'. In some papers about boundaries of hyperbolic spaces, the authors usually say that the above theorem is true without mentioning that $X$ and $Y$ are proper. If you know the answer or any references, then let me know. REPLY [8 votes]: Properness is already needed to have a well-defined boundary at infinity, i.e., with a topology not depending on the chosen base point. This is Proposition III.3.7 in Bridson-Haefliger, which builds on some previous lemmata that are applications of the Arzela-Ascoli theorem. To apply the Arzela-Ascoli theorem one needs bounded sets to be compact.<|endoftext|> TITLE: Polynomials for which roots can be expressed as polynomials in a single root QUESTION [11 upvotes]: Classical Galois theory gives necessary and sufficient conditions for the roots of a polynomial in $k[x]$ to be expressible in terms of nested radicals of the coefficients. Suppose instead that a single root $\alpha$ of $p(x)\in \mathbb{Q}[x]$ is known. Are there known necessary and sufficient conditions on $p(x)$ such that all remaining roots can be expressed as polynomial (or rational) functions of $\alpha$ and the coefficients of $p(x)$? For example, the cyclotomic polynomials have this property, since every primitive $n^{\textrm{th}}$ root of unity can be written as a power of some fixed root. REPLY [14 votes]: From a computational point of view, one should not try to compute the Galois group. Assuming $p(x) \in \mathbb{Q}[x]$ is irreducible, and $\alpha$ is a root of $p(x)$, it is sufficient to factor $p(x)$ over the number field $\mathbb{Q}(\alpha) = \mathbb{Q}[x]/(p(x))$, and look whether all the irreducible factors have degree 1. In this way, you also get the expression of the roots in terms of $\alpha$. This is much less expensive than computing the Galois group, which is feasible only in relatively low degree.<|endoftext|> TITLE: What is the right notion of a functor from an internal topological category to a topologically enriched category? QUESTION [11 upvotes]: Let $\mathcal{C}$ be a category internal to (some convenient model for) topological spaces (which I will denote by $\mathsf{Top}$). In the question Greg Arone asks: What is the correct notion of a topological functor $\mathcal{C} \to \mathsf{Top}$? In particular, Greg wanted to know about the homotopy theory of such functors. My questions can be seen as somewhat of a follow-up question. Let $\mathcal{M}$ be a topologically enriched category (feel free to assume $\mathcal{M}$ is (co)tensored over $\mathsf{Top}$ if that aids in an answer). What is the correct notion of a topological functor $\mathcal{C} \to \mathcal{M}$? Particular examples I am interested in include (some convenient model for) based topological spaces and spectra. I am also interested in understanding the homotopy theory of such functors. If, in addition, $\mathcal{M}$ is a topological model category, what do we know about the homotopy theory of topological functors $\mathcal{C} \to \mathcal{M}$? I am aware of one definition of such functors in the literature. In "Derivatives of embedding functors I: the stable case" Arone makes the following definition (Definition 3.1), which I will state only for based spaces, but Arone also defines for spectra. A functor from a small topological category $\mathcal{C}$ to the category of based spaces consists of the following data: A ex-space $F$ over the space $ob(\mathcal{C})$ of objects of $\mathcal{C}$, the fiber over $c \in \mathcal{C}$ is what we would usually call $F(c)$. A fiberwise map of objects over the space $mor(\mathcal{C})$ of morphisms of $\mathcal{C}$ $$ \alpha: s^* (F) \to t^*(F), $$ where $s^*(F)$ and $t^*(F)$ are the pullbacks of $F$ from $ob(\mathcal{C})$ to $mor(\mathcal{C})$ along the source and the larget maps respectively. This data is subject to unicity and composition law conditions, which I leave to the reference. The idea is that when $\mathcal{C}$ is discrete, this precisely recovers the standard definitions. I'm hopeful that there is some 'slicker' way to define such functors analogous to that of a $\mathsf{Top}$-internal diagram as in Emily's answer to Greg's original question. REPLY [4 votes]: I don't believe it is possible to recover the "correct" notion of "functor $\mathcal{C}\to \rm Top$", as described at the other question you linked to, by viewing $\rm Top$ only as a topologically enriched category. But it is possible if you view $\rm Top$ as a more richly structured object called a locally internal category. The examples of based spaces and spectra that you mention can also be enhanced to locally internal categories over $\rm Top$, and we thereby obtain notions of functor from an internal category $\mathcal{C}$ to these locally internal categories. Intuitively, a locally internal category over $\rm Top$ consists of a $({\rm Top}/X)$-enriched category $\mathcal{M}_X$ for each $X\in \rm Top$, such that each map $f:X\to Y$ in $\rm Top$ induces a functor $f^*:\mathcal{M}/Y \to \mathcal{M}/X$, in a coherent way. (To be precise, this $f^*$ doesn't typecheck, since its domain and codomain are enriched in different categories, so we apply $f^* : {\rm Top}/Y \to {\rm Top}/X$ homwise to its domain; see the nLab link and the references cited therein.) For instance, $\rm Top$ itself underlies such a locally internal category, where ${\rm Top}_X = {\rm Top}/X$. The category $\rm Ex$ of based spaces also underlies a locally internal category, where ${\rm Ex}_X$ is the category of pointed objects in ${\rm Top}/X$, i.e. the category of sectioned spaces or "ex-spaces". There is also a locally internal category $\rm Sp$ of spectra, where ${\rm Sp}_X$ is the category of parametrized spectra over $X$ (e.g. in the May-Sigurdsson sense). Now if $\mathcal{M}$ is a locally internal category and $\mathcal{C} = (\mathcal{C}_1 \rightrightarrows \mathcal{C}_0)$ is an internal category, a functor $D:\mathcal{C} \to \mathcal{M}$ consists of: An object $D\in \mathcal{M}_{\mathcal{C}_0}$ A morphism $s^* D \to t^* D$ in $\mathcal{M}_{\mathcal{C}_1}$ An associativity condition in $\mathcal{M}_{\mathcal{C}_1 \times_{\mathcal{C}_0} \mathcal{C}_1}$ and a unit condition in $\mathcal{M}_{\mathcal{C}_0}$. If you interpret this in the locally internal category $\rm Top$, you get the notion of functor $\mathcal{C} \to \rm Top$ mentioned in the other answer. Specializing to $\rm Ex$ and $\rm Sp$ will then answer your question. Of course, nothing is special about $\rm Top$ here; any category with finite limits will work to define locally internal categories, although it must be locally cartesian closed in order for itself to be an example. Locally internal categories can also be identified with fibrations or indexed categories that satisfy an internalized "local smallness" condition. I can't resist also mentioning my paper Enriched indexed categories, which studies a generalization (independently discovered by a number of people) of the notion of locally internal category to a notion of category that is simultaneously indexed by (or internal to) a base category and enriched in another category (which itself is indexed over the same base). In this way one can define and study categories that are "internal to $\rm Top$", with a topology on their set of objects, but also enriched over a (monoidal) locally internal category like $\rm Ex$ or $\rm Sp$, as well as functors from such categories to their enriching categories, or other categories enriched over those.<|endoftext|> TITLE: A generalization of the law of tangents QUESTION [5 upvotes]: The law of tangents is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides. Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle, and $\alpha$, $\beta$ and $\gamma$ be the angles opposite those three respective sides. The law of tangents states that $$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{a-b}{a+b}.\tag{1}\label{1}$$ The law of tangents can be used in any case where two sides and the included angle, or two angles and a side, are known. Although Viète gave us the modern version of the law of tangents, it was Fincke who stated the law of tangents for the first time and also demonstrated its application by solving a triangle when two sides and the included angle are given (see Wu - The Story of Mollweide and Some Trigonometric Identities). A proof of the law of tangents is provided by Wikipedia (see here). Generalization. Let $a$, $b$, $c$ and $d$ be the sides of a cyclic convex quadrilateral. Let $\angle{DAB}=\alpha$ and $\angle{ABC}=\beta$, then the following identity holds $$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{(a-c)(b-d)}{(a+c)(b+d)}.\tag{2}\label{2}$$ Proof. Using the sum-to-product formulas we can rewrite the left-hand side of $(2)$ as follows $$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{\sin\frac12(\alpha-\beta)\cos\frac12(\alpha+\beta)}{\cos\frac12(\alpha-\beta)\sin\frac12(\alpha+\beta)}=\frac{\sin{\alpha}-\sin{\beta}}{\sin{\alpha}+\sin{\beta}}.$$ The area of a cyclic quadrilateral can be expressed as $\Delta=\frac12(ad+bc)\sin{\alpha}$ (see $(12)$ at Killing three birds with one stone) and similarly for the other angles. Then substituting, simplifying and factorizing we have $$\begin{align*}\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}&=\frac{\frac{2\Delta}{ad+bc}-\frac{2\Delta}{ab+cd}}{\frac{2\Delta}{ad+bc}+\frac{2\Delta}{ab+cd}}=\frac{ab-ad+cd-bc}{ab+ad+cd+bc}=\frac{(a-c)(b-d)}{(a+c)(b+d)}\end{align*}.$$ $\square$ The formula \eqref{2} reduces to the law of tangents for a triangle when $c=0$. A related result can be found at A generalization of Mollweide's Formula (rather Newton's). Crossposted at MathSE. Question: Is this generalization known? REPLY [7 votes]: To place the formula of the OP into context, it is helpful to note the identity (from Josefsson - More characterizations of cyclic quadrilaterals) $$\frac{q-p}{q+p}=\frac{(a-c)(b-d)}{(a+c)(b+d)}.\tag{$*$}\label{star}$$ Here $q$ and $p$ are the lengths of the blue and red diagonals, with opposite angles $\alpha$ and $\beta$. The quadrilateral tangent formula can thus be rewritten in a form that looks more like the triangle tangent formula, $$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{q-p}{q+p}.$$ Q: Is it obvious that the law of tangents applies to the blue and red diagonals? The formula \eqref{star} is equivalent (upon rearrangement of the terms) to what is known as Ptolemy’s second theorem $$\frac{p}{q}=\frac{ad+bc}{ab+cd}.$$<|endoftext|> TITLE: Mathematical explanation of orbital shell sizes: why is it sufficient to consider single-electron wave functions? QUESTION [12 upvotes]: Motivation The question "Is there a good mathematical explanation for why orbital lengths in the periodic table are perfect squares doubled?" asks for an explanation of the sequence 2, 8, 8, 18, 18, 32, … of row lengths in the periodic table. The question currently has two very interesting answers (Carlo Beenakker's and Aaron Bergman's) providing good insight on the question. However, both of those answers revolve around analyzing eigenstate degeneracies in the single-electron atom quantum system (aka the hydrogen atom), and then explaining how the sequence 2, 8, 8, … can be understood in terms of dimensions of irreducible representations of the underlying symmetry group of that quantum system. While this is very nice and somewhat satisfying, I feel that these explanations still leave an annoying gap in the understanding of where electron shells come from. The point is that the hydrogen atom is not the correct quantum system that describes multi-electron atoms. The correct system (in non-relativistic quantum mechanics) involves a Hamiltonian that acts on a much larger Hilbert space, with a much larger symmetry group. So the missing piece in the story seems to be giving a convincing mathematical explanation for why it is sufficient to analyze the single-electron quantum system to understand electron shells and their sizes. What I'm able to understand from doing some cursory reading on the subject is that this has to do with certain approximation schemes to the multi-electron atom quantum system (some relevant technical terms are configuration interaction, Hartree–Fock method, and Slater determinant). However, the mathematical details of why such approximations produce acceptable results and what is the relationship of the approximate solutions to the solutions of the original system remain murky. Questions What is the mathematical justification for analyzing the multi-electron atom quantum system in terms of solutions to the hydrogen atom system? Is there some rigorous analysis that justifies this step, or is it something that gives accurate enough results in sufficiently many situations that physicists are happy to use it regardless? (If it is not rigorous, is there at least a good clean heuristic that makes it seem plausible?) In the original multi-electron atom system without any approximations, is there a well-defined notion of "electron shells"? Or is the electron shell picture only meaningful to speak of after taking the step of replacing the solutions to the original system with various approximations based on single-electron wave functions? Related to question 2, has there been any attempt to explain electron shell sizes through a representation-theoretic analysis of the symmetry group of the original multi-electron atom quantum system, in the spirit of the answers given to the earlier linked question? This would skip the approximation step so in my eyes would give a more satisfying explanation than the hydrogen atom-based representation-theoretic analysis. REPLY [8 votes]: Let me expand a bit on my comment, focusing on points 1 and 2. (I have no meaningful response to 3.) It may be instructive to consider the simplest multi-electron atom, Helium, with two electrons. In the shell model one says these occupy two hydrogenic 1s states with opposite spin. This is a qualitative, approximate statement, the exact two-electron wave function has only about 90% weight on the antisymmetric product of two 1s states. So one might think that the missing 10% is from other hydrogenic bound states, but even if all higher hydrogenic shells are included one does not reach the exact ground state energy of Helium, –79 eV, one ends up about 1 eV too high. This is because the hydrogenic bound states are not a complete basis set (I mistakenly stated that in a comment), for a complete basis set also the continuum states are needed. They are needed for quantitative agreement, but also to satisfy rigorous sum rules. All of this is explained clearly in The Spectral Decomposition of the Helium atom two-electron configuration in terms of Hydrogenic orbitals (see also this comment). So the answer to 1 and 2 would be: No, there is no mathematical justification for a description of a multi-electron atom in terms of hydrogenic shells, these have no well-defined meaning. There are more accurate representations of the multi-electron wave function (the cited paper discusses these), but these are less intuitive.<|endoftext|> TITLE: Question on KAK decomposition QUESTION [7 upvotes]: Let $G$ be a semisimple Lie group and let $G = KAK$ be a Cartan decomposition. For $\mathrm{SL}_2(\mathbb{R})$ it holds for every $g \in G$ that $KgK = Kg^{-1}K$. Does the same hold for every semisimple Lie group? REPLY [8 votes]: No. For instance it fails in $\mathrm{SL}_3(\mathbf{R})$. Indeed, choose $g$ the diagonal matrix $(2,2,1/4)$. Let $V_2$ be its $2$-eigenspace. Suppose by contradiction that $g^{-1}\in KgK$ ($K=\mathrm{SO}(3)$). Equivalently, this means that for some $k\in K$ we have $gkg\in K$. Then $V_2\cap k^{-1}V_2$ has dimension $\ge 1$. For $x\in V_2\cap k^{-1}V_2$ we have $gkg(x)=2gkx=4kx$, so $\|gkg(x)\|=4\|x\|$. For $x\neq 0$, this contradicts $gkg\in K$. REPLY [8 votes]: No, this fails already in $\mathrm{SL}_3(\mathbb{R})$. If $Kg_1K=Kg_2K$, then $g_1^Tg_1$ is conjugate (by an element of $K$) to $g_2^Tg_2$. So $g_1^Tg_1$ and $g_2^Tg_2$ have the same (positive eigenvalues). In particular, if $g_1$ and $g_2$ are positive diagonal matrices, then they have the same diagonal entries up to permutation. This rarely happens when $g_1$ and $g_2$ are inverses of each other. For example, $\mathrm{diag(2,2,1/4)}$ and $\mathrm{diag(1/2,1/2,4)}$ have very different diagonal entries, and they are inverses of each other.<|endoftext|> TITLE: Is there a closed manifold whose universal cover is $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$? QUESTION [11 upvotes]: There are many closed manifolds with universal cover homotopy equivalent to $\mathbb{R}^n$, they are precisely the closed aspherical manifolds. There are also many closed smooth manifolds with universal cover diffeomorphic to $\mathbb{R}^n$, e.g. those which admit a metric of non-positive curvature. If one weakens diffeomorphic to homeomorphic, then the only additional examples one could possibly obtain would be four-dimensional, but no such examples are known to exist, see this question. If one considers $\mathbb{R}^n\setminus\{x\}$ with $n > 2$ instead as a universal cover, there are plenty of examples. Any quotient of $S^{n-1}$ will have universal cover $S^{n-1}$ which is homotopy equivalent to $\mathbb{R}^n\setminus\{x\}$. If one upgrades to diffeomorphism, then the product of a smooth quotient of $S^{n-1}$ with $S^1$ yields a suitable manifold. Unlike the case of $\mathbb{R}^n$, one can construct smooth manifolds with universal cover homeomorphic but not diffeomorphic to $\mathbb{R}^n\setminus\{x\}$. For example, for any exotic $(n-1)$-sphere $\Sigma$, the universal cover of $\Sigma\times S^1$ is diffeomorphic to $\Sigma\times\mathbb{R}$ which is homeomorphic to $S^{n-1}\times\mathbb{R}$, and hence $\mathbb{R}^n\setminus\{x\}$, but is not diffeomorphic to it, see these comments by Igor Belegradek. What if we remove more than one point from $\mathbb{R}^n$? Is there a closed manifold whose universal cover is homotopy equivalent/homeomorphic/diffeomorphic to $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$? There are no such manifolds in dimension one or two, but I don't even know if such examples can arise in dimension three. The space $\mathbb{R}^n\setminus\{x_1,\dots, x_k\}$ is homotopy equivalent to $\bigvee_{i=1}^kS^{n-1}$. If $M$ is a closed manifold with the given universal cover, one might hope that an analysis of the natural $\pi_1(M)$-action on $\pi_{n-1}(M) \cong \mathbb{Z}^k$ could provide some insight. REPLY [18 votes]: If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing point (together with the "infinity" of the one-point compactification of $\mathbb{R}^n$) is an end of the covering space, and these are all the ends. Therefore the universal cover has $k+1 \ge 3$ ends. Freudenthal and Hopf proved that a finitely generated group has $0$, $1$, $2$, or infinitely many ends. The ends of the fundamental group biject with the ends of the universal cover, so the theorem contradicts our assumption. I suspect (but am really not sure) that it may be possible to extend the argument to closed $n$-manifolds with universal cover only homotopy equivalent to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$. Perhaps it can be shown that the universal cover must have $k+1$ ends by thinking of the separation properties of representatives of $H_{n-1}(\bigvee_i S^{n-1})$.<|endoftext|> TITLE: Universal covers of non-prime 3-manifolds QUESTION [8 upvotes]: Let $M$ be a closed, connected, oriented 3-manifold. If $M$ is prime, then we know what the universal cover of $M$ looks like: it is either $S^3, \mathbb{R}^3$ or $S^2 \times \mathbb{R}$ depending on whether the 3-manifold is spherical, aspherical or $S^2 \times S^1$. If $M$ is non-prime, what does its universal cover look like? It must be a simply connected, non-compact 3-manifold (without boundary), but I do not know whether they are well-understood. Perhaps something more concrete: what is the universal cover of the connected sum of lens spaces? REPLY [3 votes]: Here is a version of Ryan's answer (in the comments). Suppose that $M$ and $N$ are non-trivial lens spaces. Note that $M \# N$ is covered by $D_g$, the connect sum of $g$ copies of $T = S^2 \times S^1$, for some $g > 0$. (Also, $g = 1$ if and only if $M = N = \mathbb{RP}^3$). Suppose that we are not in that special case. Then $g > 1$ and $D_g$ covers $D_2 = T \# T$. So it suffices to understand the universal cover of $D_2$. The manifold $D_2$ is obtained by doubling a genus two handlebody $U_2$. Thus the universal cover of $D_2$ is obtained by taking the universal cover of $U_2$, and doubling. However, the universal cover of $U_2$ is homeomorphic to a closed ball, minus a Cantor set from its boundary. Thus the universal cover of $D_2$ is, as claimed, a copy of $S^3$ minus a Cantor set from its equatorial two-sphere. This is somewhat similar to my answer here.<|endoftext|> TITLE: Math Review #'s — Include the initial zeros? QUESTION [8 upvotes]: I like to use MathSciNet's BibTeX feature. However, I recently found out that sometimes their BibTeX entry removes the initial zeros on a MR number, and other times it includes the initial zeros. Is there a preferred choice between these two options? Alternatively, is there a reason I'm missing that explains why MathSciNet isn't consistent? REPLY [16 votes]: In most instances, when displaying items with MR numbers that are less than 1000000, we pad the number by prepending with zeros to obtain a 7-digit number. The numbers, however, are stored just as numbers in the database. When we export the record to either BibTeX or AMSRefs, we use the number, no padding. So, "short" numbers should be the norm for items with MR numbers below 1000000.<|endoftext|> TITLE: Distinguishing topologically weak topologies of Banach spaces QUESTION [15 upvotes]: Are the weak topologies of $\ell_1$ and $L_1$ homeomorphic? Strangely may it sound, the question seeks contrasts between norm and weak topologies of Banach spaces from the non-linear point of view. And indeed, no linear homomorphism between weak topologies of non-isomorphic spaces exists as prevented by the Closed Graph Theorem. Note that the fact that $\ell_1$ is a Schur space yet $L_1$ is not is a priori not helpful, unless I am missing something trivial. However, all infinite-dimensional, separable Banach spaces are homeomorphic. For weak topologies this is not so, as the weak topology of a reflexive space is $\sigma$-compact whereas this does not hold for $L_1$, say. We could then ask: Is there a pair of non-isomorphic, non-reflexive Banach spaces whose weak topologies are homeomorphic? 31.03.2022: As noticed by Jerzy, the weak topologies of $\ell_1$ and $C[0,1]$ are not homeomorphic. REPLY [3 votes]: The weak topologies of the spaces $\ell_1$ and $L_1$ are not homeomorphic because of the following Theorem. Assume that $X,Y$ are two Banach spaces whose weak topologies are homeomorphic. If $X$ has Shur's property, then $Y$ has Shur's property, too. Proof. For a topological space $T$ let $T_s$ be the sequential coreflection of $T$, which is $T$ endowed with the topology consisting of sequentially open sets. A subset $U\subseteq T$ is sequentially open if for any convergent sequence $(x_n)_{n\in\omega}$ in $T$ with $\lim x_n\in U$ there exists $m\in\omega$ such that $x_n\in U$ for all $n\ge m$. Assume that $h:X_w\to Y_w$ is a homeomorphism between the Banach spaces $X,Y$ endowed with the weak topologies. The homeomorphism $h$ remains a homeomorphism betweem the sequential coreflections $X_s$ and $Y_s$ of the spaces $X_w$ and $Y_w$, respectively. Since $X$ has Shur's property, the sequential coreflection $X_s$ of $X_w$ coincides with $X$. Since $X=X_s$ is a Baire space, so is the space $Y_s$. The Baireness of the space $Y_s$ implies that the closed unit ball $B_Y$ of $Y$ has non-empty interior in $Y_s$. Let $y_*$ be an interior point of the set $B_Y$ in $Y$. Assuming that $Y$ fails to have Shur's property, we can find a null sequence $(y_n)_{n\in \omega}$ in $Y_w$ consisting of vectors of norm $\|y_n\|=3$. Then $(y_n+y_*)_{n\in\omega}$ is a sequence in $Y_s\setminus B$ that converges to $y_*$, which is not possible as $y_*$ is an interior point of $B$ in $Y_s$. This contradiction implies that the Banach space $Y$ has Shur's property. The proof of the above theorem suggests the following characterization of the Shur property. Characterization. A Banach space $X$ has Shur's property if and only if the sequential coreflection $X_s$ of the weak topology of $X$ is a Baire space if and only if $X_s=X$. Remark. Concerning the second question, it can be shown that any reflexive separable infinite-dimensional Banach spaces endowed with the weak topologies are sequentially homeomorhic (= the sequential coreflections of their weak topologies are homeomorphic), see this paper for more information in this direction.<|endoftext|> TITLE: Bijectively counting labeled trees by number of leaves QUESTION [5 upvotes]: A rooted, labeled tree on $n$ vertices is a tree with vertex set $[n] := \{1,2,\ldots,n\}$ in which one vertex has been designated the root. A leaf of a rooted tree is a vertex $v$ for which either: $v$ is not the root and has degree $1$; or $v$ is the root and has degree $0$. Using standard generating function techniques (i.e. Lagrange inversion) it is not hard to prove that the number $t_{n,m}$ of rooted, labeled trees on $n$ vertices with exactly $m$ leaves is $t_{n,m} = \frac{n!}{m!} S(n-1,n-m)$, where $S(n,k)$ is the Stirling number of the $2$nd kind. Together with Cayley's formula, this gives a proof of the identity $n^{n-1} = \sum_{m=1}^{n} \frac{n!}{m!} S(n-1,n-m)$. But actually there is a much simpler proof of that identity: $\frac{n!}{m!} S(n-1,n-m)$ clearly counts the number of functions $f\colon [n-1]\to [n]$ whose image has cardinality $n-m$. This makes me wonder: is there a simple bijection between functions $[n-1]\to [n]$ and rooted, labeled trees on $n$ vertices for which $n$ minus the cardinality of the image becomes the number of leaves of the tree? Unless I'm mistaken, the Prüfer code does not accomplish this. REPLY [5 votes]: If you identify a function $f: [n-1] \to [n]$ with the Prüfer code $(f(1), f(2), \ldots, f(n-1))$ then it corresponds to an unrooted labelled tree on $n+1$ vertices in which the label $n+1$ is a leaf. Designate the neighbour of that leaf as the root and delete the leaf and you have the desired bijection.<|endoftext|> TITLE: Splitting the integers from $1$ to $2n$ into two sets with products as close as possible QUESTION [25 upvotes]: For each positive integer $n$, split the integers $1$ to $2n$ into two sets of $n$ elements each, and such that the products of the elements in each of these sets are as close as possible, say they differ by $a(n)$. It can be checked that $a(1)=1$, $a(2)=2$, $a(3)=6$, $a(4)=18$, and $a(5)=30$. Is this sequence strictly increasing? What about if it is not required that the two sets contain the same number of elements? REPLY [9 votes]: $\newcommand{\sub}[1]{_{\substack{m\in[1,2n] \\ #1}}}$ $\newcommand{\td}{{\widetilde d}}$ $\renewcommand{\cP}{{\mathcal P}}$ For the growth rate, we have the upper bound $a_n\le 2^nn!$ and, conditionally to the abc conjecture, the nearly-matching lower bound $a_n>ce^{-2(1+\varepsilon)n}\sqrt{(2n)!}$, for any fixed $\varepsilon>0$, with a constant $c$ depending on $\varepsilon$. Unconditional lower bounds seem to be much subtler; in this direction, I will show that $a_n>e^{(1+o(1))(\ln n\ln\ln n)^c}$ with an absolute constant $c>0$ (one can take $c=1/5$). The upper bound $a_n\le 2^nn!$ can be obtained as follows. Let $X_1:=1$, $Y_1:=2$, and $$ X_n:=\min\{(2n-1)Y_{n-1},2nX_{n-1}\}, \ Y_n:=\max\{(2n-1)Y_{n-1},2nX_{n-1}\}, \quad n\ge 2. $$ Clearly, one can partition $[1,2n]$ into two equal-sized subsets so that the product of all elements of the first subset is $X_n$, while for the second subset, the product is $Y_n$. We notice that $X_n\le 2^nn!$ by a straightforward induction. Moreover, arguing inductively, we obtain $Y_n-X_n\le 2^nn!$: \begin{align*} Y_n-X_n &= |(2n-1)Y_{n-1}-2nX_{n-1}| \\ &\le (2n-1)(Y_{n-1}-X_{n-1})+X_{n-1} \\ &\le (2n-1)2^{n-1}(n-1)! + 2^{n-1}(n-1)! \\ &= 2^nn!. \end{align*} Therefore, $a_n\le Y_n-X_n\le 2^nn!$. For a lower bound conditional to the abc conjecture, consider a decomposition $(2n)!=P_1P_2$ with $P_2>P_1$, and let $D:=\gcd(P_1,P_2)$ and $P_i':=P_i/D$. From $P_1'+(P_2'-P_1')=P_2'$, assuming the abc conjecture, for any fixed $\varepsilon>0$ we have $P_2'\le Cr^{1+\varepsilon}$, where $r$ is the radical of the product $P_1'P_2'(P_2'-P_1')$, and $C$ is a constant depending on $\varepsilon$. The radical does not exceed the product of all primes up to $2n$, which is $e^{(2+o(1))n}$ by the prime number theorem. Furthermore, from $P_1'\sqrt{(2n)!}/D$. This yields $$ \sqrt{(2n)!}/D < P_2' < Ce^{(2+o(1))(1+\varepsilon)n} < Ce^{2(1+\varepsilon)n} $$ resulting in $D>C^{-1}e^{-2(1+\varepsilon)n}\sqrt{(2n)!}$. It remains to notice that $P_2-P_1\ge D$ since both $P_2$ and $P_1$ are divisible by $D$. For an unconditional bound, given a partition $[1,2n]=\cP_1\cup\cP_2$, let $P_i:=\prod_{m\in\cP_i}m$; thus, $P_1P_2=(2n)!$. For $i,j\in\{1,2\}$ with $i\ne j$, let $Q_i$ be the product of all primes dividing $P_i$ but not dividing $P_j$. Furthermore, let $Q_0$ be the product of all primes in $[1,2n]$ dividing both $P_1$ and $P_2$; thus, $Q_0Q_1Q_2$ is the product of all primes in $[1,2n]$. The idea behind our argument is quite simple. Since both $P_1$ and $P_2$ are divisible by $Q_0$, we have $|P_1-P_2|\ge Q_0$; thus, it suffices to show that $Q_0$ is large. Assuming for a contradiction that $Q_0$ is small, at least one of $Q_1$ and $Q_2$, say the former, is large. This means, there are many primes dividing $P_1$ but not dividing $P_2$. For any such prime $p$, the set $P_1$ contains all multiples of $p$ in the range $[1,2n]$. But this would make $P_1$ significantly larger than $\sqrt{(2n)!}$, and hence make $P_2$ significantly smaller than $\sqrt{(2n)!}$, resulting in $|P_1-P_2|$ large. We now work out the details. For integer $d\in[1,2n]$ let $\td$ denote the product of all multiples of $d$ in the range $[1,2n]$; therefore writing $L:=\lfloor\frac{2n}{d}\rfloor$, we have $\td=d^L\cdot L!$ and then, from Stirling's formula, $$ \ln\td = \left(\frac 1d+\frac{\theta}{2n}\right) \,\ln((2n)!),\quad |\theta|<1. \tag{$*$} $$ The key observation is that $\cP_1$ contains only those $m\in[1,2n]$ with $\gcd(m,Q_2)=1$. Therefore, $$ P_1 \le \prod\sub{\gcd(m,Q_2)=1} m \le \prod\sub{\gcd(m,D)=1} m $$ for any divisor $D\mid Q_2$. Using the approximation ($*$), we get \begin{align*} \ln P_1 &\le \sum\sub{} \ln m \sum_{d\mid\gcd(m,D)} \mu(d) \\ &= \sum_{d\mid D} \mu(d) \ln \td \\ &\le \left( \sum_{d\mid D} \frac{\mu(d)}d + \frac{1}{2n}(2^k-1) \right) \ln((2n)!) \\ &= \left( \prod_{p\mid D}\left(1-\frac1p\right) + \frac{2^{k-1}}{n} - \frac1{2n}\right) \ln((2n)!) \end{align*} where $\mu$ is the Möbius function, and $k$ is the number of prime divisors of $D$. Assuming that $\ln P_1\ge\frac12\left(1-\frac1{n}\right)\,\ln((2n)!)$ (as we certainly can), we conclude that $$ \prod_{p\mid D}\left(1-\frac1p\right) > \frac12 - \frac{2^{k-1}}{n} \ge \frac14 $$ provided $k\le \log_2 n-1$. To summarize, for any $D\mid Q_2$ with at most $K:=\lfloor \log_2n\rfloor-1$ prime divisors we have $\prod_{p\mid D}\left(1-p^{-1}\right)>1/4$. By symmetry, this is also true with $Q_2$ replaced with $Q_1$. Consequently, for any $D\mid(Q_1Q_2)$ with at most $K$ prime divisors we have $\prod_{p\mid D}\left(1-p^{-1}\right)>1/16$; as a result, $\sum_{p\mid D}p^{-1}<3$. We notice that if $Q_1Q_2$ has fewer than $K$ prime divisors, then $Q_1Q_2<(2n)^K=e^{o(n)}$ whence $|P_1-P_2|\ge Q_0\ge e^{(1+o(1))n}$. Suppose therefore that $Q_1Q_2$ has at least $K$ prime divisors. We choose $D$ to be the product of the $K$ smallest prime divisors of $Q_1Q_2$, and we denote by $M$ the largest of these $K$ smallest divisors; notice that $M>(1+o(1))|K|\ln(|K|)>\ln n\ln\ln n$ by the prime number theorem. Let $m:=M^c$ with $c\in(0,1)$. If $c$ is sufficiently small, then by Mertens' second theorem, and recalling that $\sum_{p\mid D}p^{-1}<3$, we obtain $$ \sum_{\substack{m\le p\le M \\ p\nmid D}} \frac1p > \sum_{m\le p\le M} \frac1p - 3 > \ln\frac{\ln M}{\ln m} - 3 + o(1) > 1. $$ Denoting by $T$ the number of primes $p\nmid D$ in the range $[m,M]$, we thus have $T>m=M^c>(\ln n\ln\ln n)^c$. The product of these $T$ primes is at least as large as the product of the first $T$ primes, which is $$ e^{(1+o(1))T} = e^{(1+o(1))(\ln n\ln\ln n)^c}. $$ The assertion follows in view of $|P_1-P_2|\ge Q_0$ and since any prime in $[1,M]$ not dividing $D$ is a divisor of $Q_0$.<|endoftext|> TITLE: Equivalent definitions of topological weak mixing QUESTION [8 upvotes]: A dynamical system $f:X\to X$ is said to be topologically transitive if for any two nonempty open sets $U,V$ there exists $n \in \mathbb{Z}$ such that $f^{\circ n}(U) \cap V \neq \emptyset$. The following two definitions of topological weak mixing can be found in the literature: $f$ is topologically weak mixing if $f\times f$ is topologically transitive on $X\times X$ (e.g here) $f$ is topologically weak mixing if it has no non-constant continuous (with respect to the topology) eigenfunctions of the shift operator. (e.g. here) How are these definitions related? Can they be shown to be equivalent? REPLY [4 votes]: Which are your regularity assumptions on $f$? In general there are some subtle differences between the two definitions. If $X$ is a Baire space, then 1. implies (a slightly modified version of) 2. See on this Theorem 2.3 in H.B. Keynes and J.B. Robertson, Eigenvalue Theorems in Topological Transformation Groups. In the same paper, Theorem 2.5 provides conditions making the two definitions basically equivalent. Theorem 2.8 shows instead a case of non-equivalence. Notice that: the mentioned paper concerns the more general setting of topological transformation groups; the authors mean "almost everywhere" in a topological sense (= everywhere except on a meagre set); they allow (unless differently specified) a meagre discontinuity set.<|endoftext|> TITLE: Cardinality of a special set of continuous functions QUESTION [6 upvotes]: Let $C$ be a set of continuous functions with a domain $[0,1]$ and for every input $x$ in a domain there is a set $S(x)$ that contains all values that functions in $C$ will output given that input. For example, if $f$, $g$ and $h$ are all functions in $C$, than, for example $f(0.5)$, $g(0.5)$ and $h(0.5)$ are all in $S(0.5)$. Of course, if these are the only functions in $C$ and all three outputs $f(0.5)$, $g(0.5)$ and $h(0.5)$ are different, than cardinality of $S(0.5)$ is 3. If they all are the same, than cardinality of $S(0.5)$ is 1. The question is, given the fact that $card(S(x))= \aleph_0$ for all $x$ in a domain, can there be such $C$ that $card(C)> \aleph_0$ ? What about the case when $C$ is a set of smooth functions? Analytic? REPLY [3 votes]: For your initial question $-$ yes, we can find such $C$ as Saúl Rodríguez Martin showed in their comment. For the analytic functions, I believe it is exactly Wetzel's problem: the solution is independent from ZFC. But with continuum hypothesis being true, we can find such $C$. For smooth functions I didn't find solution, but here is an observation. If there is $C$ satisfying your conditions and contains only functions that can be differentiated at most $N$ times, than there surely is satisfying $C$ that with functions differentiable $N-1$ times as it is less restrictive. So if there is some $N$ for which there is no $C$, than there are no solutions above $N$ (because if there is solution for $N+1$ than there is for $N$). REPLY [2 votes]: Original problem with continuous functions: A counterexample is given by the set $C$ of continuous functions $f$ obtained by setting $f(1)=0$, $f(1-2^{-n})\in\{0,2^{-n}\}$ and interpolating linearly. $C^\infty$ functions: Let $\phi:\mathbb{R}\to\mathbb{R}$ be a $C^\infty$ bump function with support $[-1,1]$, and $\phi(p,r)(x)=\phi\left(\frac{x-p}{r}\right)$ be the same bump function but centered in $p$ and with support of radius $r$. Now for a given positive sequence $\mathbf{x}=(x_n)_n$ let $F_\mathbf{x}=\sum_{n\in\mathbb{N}}x_n\phi(1-3\cdot2^{-n},2^{-n-2})$. Then for any $f\in C$, $f\cdot F_\mathbf{x}$ is $C^\infty$ in $[0,1)$. Moreover, if we make the $x_n$ decrease fast enough, we can ensure that $\forall f\in C$, $\lim_{x\to1}(f\cdot F_\mathbf{x})^{n)}(x)=0$ $\forall n$, so that $f\cdot F_\mathbf{x}$ is $C^\infty$ in $[0,1]$. So $\{f\cdot F_\mathbf{x};f\in C\}$ does the job. Analytic functions: In this case the existence of $\mathcal{C}$ is equivalent to the continuum hypothesis. If the continuous hypothesis is true, let $\omega_1$ be the set of countable ordinals and consider a numbering $(x_\alpha)_{\alpha\in\omega_1}$ of $[0,1]$. We can create recursively functions $\{f_\alpha\}_{\alpha\in\omega_1}$ which only take countably many values at any point: it is enough to, for each ordinal $\alpha$, let $f_\alpha$ be a function which: For every $\beta<\alpha$, $f_\alpha(x_\beta)$ is rational. $f_\alpha$ is distinct from $f_\beta$ $\forall\beta<\alpha$. It is possible to create such $f_\alpha$ due to the following lemma: Lemma: Given a countable set $(x_n)_{n\in\mathbb{N}}$ in $[0,1]$ and some $y_0\in\mathbb{R}$ we can create an analytic function $f:[0,1]\to\mathbb{R}$ with $f(x_0)=y_0$ and $f(x_n)\in\mathbb{Q}$ $\forall n$ Proof of the lemma: We can define $f=y_0+\sum_{n\geq1}P_n$, where $P_n$ are polynomials such that: $P_n(x_m)=0\;\forall m TITLE: Is the crossing number of the line graph of $K_5$ determined? QUESTION [5 upvotes]: The line graph of an undirected graph $G$ is another graph $L(G)$ that represents the adjacencies between edges of $G$. $L(G)$ is constructed in the following way: for each edge in $G$, make a vertex in $L(G)$; for every two edges in $G$ that have a vertex in common, make an edge between their corresponding vertices in $L(G)$. I would like to know the crossing number of a line graph of a complete graph $K_5$. Furthermore, what is crossing-minimal drawing of $L(K_5)$? Here's what I know now and that's about it. $L(K_5)$ has crossing numer 3 or more. We can see above result in following paper. Kulli V R, Akka D G, Beineke L W. On line graphs with crossing number 1[J]. Journal of Graph Theory, 1979, 3(1): 87-90.] Ps: We see easily that $L(K_5)$ is the complement of Petersen graph. REPLY [6 votes]: Thanks for advice from Timothy Chow. I have now received an email from CRWS. The graph has a crossing number of 12. Its crossing-minimal drawing is as follows.<|endoftext|> TITLE: Can you get any natural number from 4 by performing given operations? QUESTION [10 upvotes]: You can perform the following operations on numbers: divide the number by 2, add 0 or 4 at the end of the number. Can you get any natural number from 4 by performing only these operations? So far I have come up with this algorithm to get 4 from any natural number N by performing opposite operations (multiplying by 2, deleting 0 or 4 from the end of the number). Check the last digit of N, a. If it's 0 or 4, remove the last digit. b. If it's neither 0 nor 4, multiply N by 2. If N is not 4 go to Step 1. The problem is I can't prove that this algorithm works for any natural number. REPLY [7 votes]: The following program (in Asymptote, but the syntax is pretty much the same as in C and, I hope, the algorithm is self-explanatory) considers all remainders modulo $10^n$ and prints those for which the steps up to the moment when we remove all $n$ known digits do not allow to conclude that the number got smaller. The printouts for $n=1,2,3$ are as mentioned in the comment. For $n=4$ the program terminates without printing anything. int n=4, M=10^n; for(int k=0;k0 && s>=-0.0000000001) { if(a%10==4 || a%10==0) {a=quotient(a,10);--nn;s-=log(10); MM=quotient(MM,10);} else {a=(a*2)%MM; s+=log(2);} } if(s>0) write(k); } pause();<|endoftext|> TITLE: A notion of 2-dimensional tree QUESTION [14 upvotes]: Summary: This post has got rather long after the discussion. The main still open Questions are 5 & 6 below. There is work in progress, and I'll post an update at some point. A tree is a connected 1-complex $X$ every two points $x,y$ of which can be separated by a point $z\neq x,y$. (We could let $X$ be a more general topological space here.) Going one dimension up, call a connected 2-complex (or topological space) $X$ a 2-tree, if for every $x,y\in X$, there is a homeomorph of a forest in $X - \{x,y\}$ separating $x$ from $y$. (A forest is a disjoint union of trees.) Question 1: Has this notion been studied? Question 2: Is it true that a 2-complex $X$ is a 2-tree iff it contains no homeomorph of a closed surface? (The forward implication is obvious.) Some examples of 2-trees: standard trees, graphs, $\mathbb{R}^2$, and most interestingly, cartesian products $\mathbb{R} \times T$, where $T$ is any tree (to see this, embed $\mathbb{R} \times T$ in $\mathbb{R}^3$, and cut it with a plane separating $x$ from $y$). Examples of non-2-trees: closed surfaces, $\mathbb{R}^3$, standard Cayley complex of $\mathbb{Z}^3$ (to see the latter, notice that it contains copies of $\mathbb{S}^2$). Remark: The definition can be applied recursively, to define $n$-trees for every $n\in \mathbb{N}$. Update 1: The answer to Question 2 is negative, as Geva Yashfe points out in the comments. As Sam Nead points out below, the variant of the definition with “forest” replaced by “tree” seems more sensible. Now a graph is a 2-tree iff it is a tree. Defining a 0-tree to be a point, the recursive definition now starts with 1-trees being the usual trees. The main question now is: Question 3: Is every CW-complex (or other nice space, see e.g. Ian Agol’s answer) which is an $n$-tree contractible? Simply-connected? Update 2 The answer to Question 3 is negative: as Geva Yashfe points out, an annulus is a 2-tree. A remaining question is: Question 4: Is every $(n-1)$-connected CW-complex which is an $n$-tree contractible? But here is another variant of the recurersive definition of n-tree that perhaps makes Q3 interesting again: Definition 3: a point is a strong 0-tree; A space X is a strong n-tree, if for every two disjoint strong $(n-1)$-trees $T$, $T'$ in X, there is a strong $(n-1)$-tree $S$, disjoint from $T$, $T'$, that separates them. Question 5: Is every strong $n$-tree contractible? Going through the above examples and non-examples with 2-trees replaced by strong 2-trees is interesting. My motivation goes back to a theorem of Manning, saying that a graph is quasi-isometric to a tree if (and only if) every two points $x$, $y$ are separated by a ball of uniformly bounded radius around any midpoint from $x$ to $y$. This could be reformulated as saying that $X$ is a quasi-1-tree iff every two disjoint strong quasi-0-trees in $X$ can be separated by a third strong quasi-0-tree disjoint from both. I'm looking for the right definition that will generalise this to higher dimensions. Thus I propose: Question 6: Let $X$ be a geodesic metric space in which every two disjoint strong quasi-$(n-1)$-trees can be separated by a third strong quasi-$(n-1)$-tree disjoint from both. Then $X$ is a strong quasi-$n$-tree. By strong quasi-$n$-tree here I mean a space quasi-isometric to a strong $n$-tree, and the constants are some function of the dimension and the constants used in dimension $n-1$. REPLY [2 votes]: In this answer a weak $2$-tree is a simplicial complex in which any two points can be separated by a forest. It is clear that a strong $2$-tree is also a weak $2$-tree. By passing to subdivisions, we may assume that any (specific) given subpolyhedron of a simplicial complex is a subcomplex, and we will do this implicitly when discussing trees in $2$-trees. Define the face-edge graph $G$ of a $2$-dimensional simplicial complex $X$ to be the following graph: its vertices are the $2$- and $1$- dimensional faces of $X$ (so just triangles and edges). There is an edge in $G$ between a $2$-face $\sigma$ and a $1$-face $\tau$ if $\tau\subset\sigma$. A free face in a simplicial complex $X$ is a face $\tau$ which is properly contained in precisely one face $\sigma$ (which must then be of dimension exactly $1+\dim\tau$). If $\sigma$ has a free face $\tau$ of codimension $1$, an elementary collapse can be performed which deletes both $\tau$ and $\sigma$: it is convenient to think of this as a strong deformation retract of $X$ onto the complement of $\sigma$. A complex which, by a sequence of elementary collapses, can be turned into a vertex is called collapsible. If some subdivision is collapsible the complex is called geometrically collapsible. It makes sense to speak of collapsing a complex onto a subcomplex, or geometrically collapsing a complex $X$ onto a subpolyhedron $Y$, and we will denote the situation that $X$ (geometrically) collapses onto $Y$ by $X \searrow Y$. (These notions are more or less standard in PL topology / combinatorial topology.) We may subdivide our complexes as much as convenient, although we won't really need to subdivide much. I won't keep track of when a certain sequence of collapses requires / does not require subdivisions. Given a collapse $X\searrow Y$ we can realize each elementary collapse as a PL map in a more-or-less obvious way. Composing these we obtain a $\pi:X\rightarrow Y$ which I will call the ``collapse map'' below. This may not be unique -- there are probably several reasonable ways to define the map for each elementary collapse -- but it is a strong deformation retract with the property that if $\dim X = 2$ and the collapsed faces are all $2$-dimensional then the preimage of any point is a tree. Claim: Let $X$ be a weak $2$-tree. If $\sigma\in X$ is a $2$-face, its component in $G$ is either infinite or contains a free edge of $X$. Proof: Suppose the component of $\sigma$ in $G$ is finite and denote by $Y$ the union of all $1$- and $2$- faces in this component (including their vertices). Let $x,y$ be two points in the interior of $\sigma$. Let $F\subset X$ be a forest separating $x$ from $y$. Then $F^{\prime}=F\cap Y$ is a forest which separates $x$ from $y$ within $Y$. If a leaf of $F^{\prime}$ is not on a free edge, there is a path ``around'' this leaf, and it is redundant: The leaf and its unique edge can be deleted from $F^{\prime}$, and the resulting forest still separates $x$ from $y$ within $Y$. Deleting all redundant leaves iteratively in this way until none remain, we end up with a nontrivial forest, and each of its leaves is in a free edge of $Y$. ${\scriptstyle \blacksquare}$ Claim: A subcomplex of a weak $2$-tree is a weak $2$-tree. Proof: Just intersect any forest which separates two given points with the subcomplex. It still intersects all paths between the points, and hence separates. ${\scriptstyle \blacksquare}$ Claim: A finite weak $2$-tree collapses onto a $1$-dimensional subcomplex. Proof: Iteratively collapse all $2$-faces with a free edge. By the previous two claims, at each step we still have a finite weak $2$-tree, and there is still a free edge. ${\scriptstyle \blacksquare}$ Claim: A finite strong $2$-tree is collapsible. Proof: Let $X$ be a finite strong $2$-tree. Collapse it onto a $1$-dimensional subcomplex $Y$ (a graph), and let $\pi:X\rightarrow Y$ be a corresponding PL retract satisfying that the preimage of any point is a tree. Assume there is a nontrivial loop, and let $x,y$ be two points on this loop. Their $\pi$-preimages are disjoint trees $T_{1},T_{2}$, which can be separated by a tree $T\subset X$. But then $\pi\left(T\right)$ is a connected set which separates $x$ from $y$, a contradiction. ${\scriptstyle \blacksquare}$ For lack of time to think and write in detail, I decided to include some ideas without proof. I would have preferred not to put something so "unripe" here, hopefully it is useful. I think we can prove the following claim by induction on the number of faces (the induction step involves examining an elementary collapse performed in reverse). Claim(?): A collapsible $2$-dimensional complex is a strong $2$-tree. I will leave this for another time or for another person. I think it's easy if true (and probably true) but I did not check carefully, there could be some weird edge-case. It is interesting is to ask what happens in the case of an infinite complex such as a triangulation of $\mathbb{R}^{2}$. This is probably necessary if we want to work with quasi-isometry... I suggest calling an infinite complex $X$ collapsible if there is an infinite chain $U_{1}\supseteq U_{2}\supseteq U_{3}\supseteq\ldots$ of open neighborhoods of the set of all ends of $X$ such that $X\setminus U_{i}$ is compact and collapsible for each $i$ and $\bigcap_{i}U_{i}=\emptyset$. I think I have a proof that any (possibly infinite) strong $2$-tree is contractible using this notion. It should be possible to strengthen it to get "honest" collapsibility, if the definition is weakend slightly. At each step the complex may only be collapsible onto a graph; but these graphs should themselves collapse onto vertices in some later stage $X\setminus U_j$. (Perhaps I will update this answer with it on some other evening). I have not seen definitions for infinite collapsibility of this form before, perhaps they are new. There is a decent chance these ideas extend to the higher dimensional notions.<|endoftext|> TITLE: Is $\mathbb{CP}^3$ minus two points the universal cover of a compact manifold? QUESTION [21 upvotes]: After reading some recent questions on mathoverflow about universal coverings, I am curious about the following: Is it possible to construct a closed $6$-manifold $M$, with universal cover homeomorphic to $\mathbb{CP}^3 \setminus \{p_1,p_2\}$? REPLY [35 votes]: More is true. Let $M^n$ be a closed connected simply connected manifold of dimension $\ge 3$. Let $p,q\in M$ be two distinct points. Suppose $M\setminus\{p,q\}$ is the universal cover of a closed manifold. Then $M$ is homeomorphic to $\mathbb S^n$. Suppose it is and $M=\tilde N$ is the universal cover of a closed manifold $N^n$. As HJRW suggested in a comment $\pi_1(N)$ must have two ends and hence must be virtually $\mathbb Z$. By passing to a finite index subgroup we can assume it's $\mathbb Z$. Since the action of $\pi_1(N)$ on $\tilde N$ must permute the ends we can again assume by passing to finite index subgroup that it fixes the ends. Then it can be extended to a continuous action on $M$ fixing $p$ and $q$. Take a peripheral $X_1=S^{n-1}$ around $p$. Let $g$ be a generator of $\pi_1(N)\cong \mathbb Z$. Since $\tilde N$ is quasiisometric to $\mathbb Z$ by a quasiisometry preserving the group action and $X_1$ is compact it follows that $g^l(X_1)$ is disjoint from $X_1$ for all large $l$. By passing to a finite cover of $N$ we can assume that this holds for all nontrivial powers of $g$. Let $\bar D^n$ be the disk centered at $p$ with boundary $X_1$. By possibly changing $g$ to $g^{-1}$ we can assume that $X_2=g(X_1)$ is a sphere in $ D^n$. By the Annulus theorem the region $W_1$ between $X_1$ and $X_2$ is homeo to $S^{n-1}\times [0,1]$. Then $W_2=g(W_1)$ is a manifold with boundary glued to $W_1$ along $X_2$. Then $g^2(W_1)=g(W_2)$ will be another copy of $\mathbb S^{n-1}\times [0,1]$ glued to $W_2$ along the second piece of the boundary. Continuing we get a copy of $\mathbb S^{n-1}\times \mathbb R$ sitting in $\tilde N$ with the standard action of $\mathbb Z$. Passing to the quotient by $\mathbb Z$ this gives an embedding of $\mathbb S^{n-1}\times \mathbb S^1$ into $N$ which must be a homeomorphism since these are connected closed manifolds of the same dimension. Then $\tilde N$ is homeo to $ \mathbb S^{n-1}\times \mathbb R$ and $M$ is homeo to $\mathbb S^n$.<|endoftext|> TITLE: Beilinson-Drinfeld local geometric class field theory QUESTION [12 upvotes]: There is the following version of categorical local geometric class field theory: Let $\mathbb{D}=\operatorname{Spec} \mathbb{C}((t))$, $L\mathbb{G}_m$: the loop group of $\mathbb{G}_m$ over $\mathbb{C}$ (i.e. $L\mathbb{G}_m(A)=\mathbb{G}_m(A((t)))$ for any $\mathbb{C}$-algebra $A$), then there is an equivalence of triangulated categories: $\operatorname{Dmod}(L\mathbb{G}_m) \simeq \operatorname{QCoh}(\operatorname{Locsys}_{\mathbb{G}_m}\mathbb{D})$, where the first category is the D-modules on the loop group and the second category is the quasi-coherent sheaves on the stack of rank 1 local systems on $\mathbb{D}$. I have seen this "theorem" in least two places. One is the paper Tate's thesis in the de Rham setting of Hilburn and Raskin in which they claimed the equivalence is due to Beilinson–Drinfeld. Another place is a talk of Braverman in which he claimed the equivalence is due to Laumon. Both places however did not seem to give a detailed reference (e.g. [BD, Thm x.xx] or [Lau, Cor x.xx]). I believe a proof has to do with certain Fourier–Mukai transform, but I'd really like to see the details if they were written down. Can someone point me to a reference for the proof? REPLY [8 votes]: As pointed out in the comments, this is Theorem 6.3.1.2 of Hilburn-Raskin. (It certainly was known much earlier, but I'm not sure what to give as a reference.) Their proof is stated quite elegantly, in a way that makes the canonicity of the equivalence clear. Let me give a less canonical explanation, which you may or may not find more digestible. (But note that if you want to show good properties of this equivalence it really does help to have a canonical description.) The point is that $L\mathbb{G}_m$ and $\operatorname{LocSys}_{\mathbb{G}_m}(\mathbb{D})$ are relatively simple and can be described explicitly. Let's start with $L\mathbb{G}_m$. Actually, let's start with its reduced (ind-)scheme $(L\mathbb{G}_m)_{\operatorname{red}}.$ By Kashiwara's lemma, this has the same category of D-modules as does $L\mathbb{G}_m.$ I claim there is an isomorphism $$(L\mathbb{G}_m)_{\operatorname{red}}\cong\mathbb{Z}\times\mathbb{G}_m\times\mathbb{A}^{\infty}_{\operatorname{pro}}$$ where $\mathbb{A}^{\infty}_{\operatorname{pro}}$ is $\operatorname{Spec}$ of a polynomial ring in infinitely many generators. The -pro subscript reflects that this is a scheme which is an inverse limit of finite-dimensional affine spaces; later we will see its cousin $\mathbb{A}^{\infty}_{\operatorname{ind}}$ which is the ind-scheme that is a colimit of finite-dimensional affine spaces. Let me justify this isomorphism on $\mathbb{C}$-points and leave verifying it at the level of functors of points as an exercise. The $\mathbb{C}$-points of $L\mathbb{G}_m$ parametrize invertible elements of $\mathbb{C}((t))$. There are of the form $$ct^d(1+a_1t+a_2t^2+\cdots)$$ with $d\in\mathbb{Z}$, $c$ an invertible complex number, and $a_i$ complex numbers. The $d$ corresponds to the $\mathbb{Z}$-factor, the $c$ gives the $\mathbb{G}_m$, and the $a_i$ give the $\mathbb{A}^{\infty}_{\operatorname{pro}}.$ Now we turn to $\operatorname{LocSys}_{\mathbb{G}_m}(\mathbb{D}).$ I claim there is an isomorphism $$\operatorname{LocSys}_{\mathbb{G}_m}(\mathbb{D})\cong B\mathbb{G}_m\times(\mathbb{A}^1/\mathbb{Z})\times(\mathbb{A}^{\infty}_{\operatorname{ind}})_{\operatorname{dR}}.$$ Here $B\mathbb{G}_m$ is the classifying stack of $\mathbb{G}_m$ and $\mathbb{A}^1/\mathbb{Z}$ is the quotient of the affine line by the translation $\mathbb{Z}$-action. The last term is a little more subtle. As mentioned before, there is an ind-scheme $\mathbb{A}^{\infty}_{\operatorname{ind}}$ defined as a colimit of finite dimensional affine spaces. I take its de Rham stack, which has functor of points $$\operatorname{Hom}(X,(\mathbb{A}^{\infty}_{\operatorname{ind}})_{\operatorname{dR}})\cong\operatorname{Hom}(X_{\operatorname{red}},\mathbb{A}^{\infty}_{\operatorname{ind}}).$$ The main property I will use of de Rham stacks is that $\operatorname{QCoh}((\mathbb{A}^{\infty}_{\operatorname{ind}})_{\operatorname{dR}})\cong\operatorname{D-mod}(\mathbb{A}^{\infty}_{\operatorname{ind}})$. Once again, let me check this expression for $\operatorname{LocSys}_{\mathbb{G}_m}(\mathbb{D})$ at the level of $\mathbb{C}$-points. Of course, you should be fundamentally unsatisfied with this, given that the de Rham stack is invisible if you only look at $\mathbb{C}$-points, but I leave the functor of points check as an exercise. The moduli space of local systems is the quotient $$(dt+\mathbb{C}((t)))/\mathbb{C}((t))^{\times}$$ where $\mathbb{C}((t))^{\times}$ acts by gauge transformations. Because we are in an abelian setting, the gauge action is super simple; the action of $f$ is just translation by $\frac{df}{f}.$ Again, write $$f=ct^d(1+a_1t+a_2t^2+\cdots).$$ Then we have $$\frac{df}{f}=\frac{d}{t}+\operatorname{dlog}(1+a_1t+a_2t^2+\cdots).$$ The $c$ does not appear in the final expression, indicating that the $\mathbb{G}_m$ corresponding to $c$ acts trivial, which contributes a $B\mathbb{G}_m$. The $\operatorname{dlog}(1+a_1t+a_2t^2+\cdots)$ can be any formal power series, so all in all the $\mathbb{C}$-points of our quotient are elements of $$B\mathbb{G}_m\times\mathbb{C}((t))/(\mathbb{Z}\frac{1}{t}+\mathbb{C}[[t]]).$$ The coefficient of $\frac{1}{t}$ gives a $\mathbb{A}^1/\mathbb{Z}$, and the coefficients of $\frac{1}{t^n}$, $n\geq 2$, give a $(\mathbb{A}^{\infty}_{\operatorname{ind}})_{\operatorname{dR}}.$ To see that we get $\mathbb{A}^{\infty}_{\operatorname{ind}}$ and not $\mathbb{A}^{\infty}_{\operatorname{pro}},$ note that only finitely many coordinates are allowed to be nonzero. (If you want to see where the de Rham stack comes from, you'll have to do this calculation at the level of functors of points.) So that gives our expression for $\operatorname{LocSys}_{\mathbb{G}_m}(\mathbb{D}).$ Once we have these formulas, we can see that geometric local class field theory is the product of three simpler equivalences: $$\operatorname{D-mod}(\mathbb{Z})\cong\operatorname{QCoh}(B\mathbb{G}_m)$$ $$\operatorname{D-mod}(\mathbb{G}_m)\cong\operatorname{QCoh}(\mathbb{A}^1/\mathbb{Z})$$ $$\operatorname{D-mod}(\mathbb{A}^{\infty}_{\operatorname{pro}})\cong\operatorname{QCoh}((\mathbb{A}^{\infty}_{\operatorname{ind}})_{\operatorname{dR}})\cong\operatorname{D-mod}(\mathbb{A}^{\infty}_{\operatorname{ind}}).$$ The first equivalence follows because $\operatorname{QCoh}(B\mathbb{G}_m)$ is the representation category of $\mathbb{G}_m$, which can be described as the category of $\mathbb{Z}$-graded vector spaces. The second equivalence is the Fourier transform for D-modules on $\mathbb{G}_m$ (it sends the D-module $Dt^{\lambda}$ on $\mathbb{G}_m$ to $\lambda$.) The third equivalence is a limit of finite-dimensional additive Fourier transforms of D-modules. Combined they give the desired equivalence.<|endoftext|> TITLE: Finding maximal prefix of a simple curve QUESTION [7 upvotes]: Let $S$ be a piecewise-linear simple curve. For a point $p_1$ on $S$ I want to determine a maximal interval $I$ on $S$ starting in $p_1$ such that $I$ is contained in a unit disk. Is this possible, or has it perhaps already been solved in the past, and I am just unable to find an answer? I thought of something like the following: Place points uniformly spaced along $S$ and greedily include as many points as possible while the minimum enclosing disk is at most the unit disk. Assume that $p_{k+1}$ is a point on $S$ that cannot be included as the minimum enclosing disk containing $p_1, p_2,...,p_{k+1}$ would become larger than the unit disk. A unit disk would be able to cover points $p_1, p_2,...,p_{k}$ as well as some segment of the curve between $p_k$ and $p_{k+1}$, however actually determining this seems non-trivial. Edit: I have noted that the boundary of the unit disk containing $I$ need not intersect $p_1$. See the figure below. Here the black simple curve is $S$, the orange circle is a unit disk and the blue interval on $S$ is the maximal interval $I$. This seems useful, but I cannot quite tell why yet though. REPLY [3 votes]: I haven't worked this out carefully, but here's an approach. Use J.J.Green's nice idea to identify the last partially covered segment $p_k p_{k+1}$. Let $H_k$ be the convex hull of $p_1,\ldots,p_k$. Clearly covering $H_k$ covers the segments from $p_1$ to $p_k$. Any pair of vertices of $H_k$ determine a unit disk through those two points. Some of these enclose all of $H_k$, some do not. Among those that include $H_k$, record how much of $p_k p_{k+1}$ is captured by each unit disk. In the image below, the red circle is the winner.       But there remains the possibility that the unit disk touches only one vertex $p_i$ of $H_k$, and is free to rotate about $p_i$ until it bangs into $H_k$ (when it then touches two points). So in that interval of rotation, it encloses all of $H_k$ and covers a portion of $p_k p_{k+1}$. So now we have reduced the problem to a unit-circle through one point $p_i$, and as it rotates about $p_i$, computing where the circle intersects $p_k p_{k+1}$. This is the part I haven't worked out, but it is a calculation with one variable $\phi$, the rotation about $p_i$, and it should not be too complicated to find the maximum over a range of $\phi$. The overall idea is to "walk" or roll the unit circle around $H_k$, recording the maximum extent of that last edge captured.<|endoftext|> TITLE: A (possible) generalization of the unknot, inverses and the knot concordance QUESTION [6 upvotes]: The notion of a knot concordance is a rich subject in low-dimensional topology, see Livingston's survey. More precisely: For $i=0,1$, let $K_i$ be knots in $S^3$. A knot concordance from $K_0$ to $K_1$ is a smooth annulus $A=S^1 \times [0,1]$ in $S^3 \times [0,1]$ such that $\partial A= -(K_0) \cup K_1$ where $-$ denotes the reversed orientation. Using this relation, we can form a group structure on the set of oriented knots in $S^3$, denoted by $\mathcal{C}$. Let $K$ be a slice knot in $S^3$, that is, $K$ bounds a smooth disk $D$ embedded in $B^4$. We can also show that $K$ is slice if and only if $K$ is concordant to the unknot in $S^3$. Let $M$ be a closed oriented $3$-manifold. I wonder: Can we talk about a knot $M$ behaves like the unknot in $S^3$? Can we generalize the notion of knot concordance to the oriented knots in $M$? (Extra) Can we define inverses of knots in $M$ as in the case of $S^3$? REPLY [4 votes]: One can certainly define concordance by cylinders in $M \times I$ as you suggest. Making the set of concordance classes into a group is problematic, however. Most of the troubles come from the observation that for two knots to be concordant, they must be freely homotopic. This was explored in the Indiana U. PhD thesis of Prudence Heck, Knot concordance in non-simply connected manifolds. I don't think you get a group in any obvious way, for much the same reason that you don't get a group out of the set of free homotopy classes of loops. There is a notion of homology concordance of oriented pairs $(M,K)$ where $M$ is a homology sphere. The equivalence relation is then concordance in a homology cobordism between $M_0$ and $M_1$. This becomes a group under pairwise connected sum. The unknot in $S^3$ is the 0 element, and pairwise orientation reversal provides inverses.<|endoftext|> TITLE: Does the category of commutative and cocommutative Hopf algebras have enough injectives? QUESTION [7 upvotes]: It is well-known that the category of commutative and cocommutative Hopf algebras is abelian (see https://arxiv.org/abs/1502.04001v2 and its references). But does it have enough injectives? What about projectives? REPLY [3 votes]: Over a field $k$, the answer is yes for injectives; I'm not sure about projectives. Over $\mathbb Z$ or other commutative rings, I really don't know -- the use of the fundamental theorem of coalgebra below seems pretty essential (and the fundamental theorem of coalgebra fails over $\mathbb Z$). Over a field, in fact more is true: Claim: Let $k$ be a field. Then the following categories are locally finitely presentable: The category of coalgebras over $k$; The category of cocommutative coalgebras over $k$; The category of algebra objects in either (1) or (2); The category of commutative algebra objects in (1) or (2); As in (3) or (4), but with restricting to objects with antipodes. Moreover, the natural tensor product on each of these categories is a symmetric monoidal structure preserving filtered colimits, with compact unit. Corollary: The category of commutative, cocommutative Hopf algebras over a field $k$ is a Grothendieck abelian category (and in particular has enough injectives). Proof: Every locally finitely presentable abelian category is Grothendieck. Proof of Claim: The conclusion for (1) and (2) follows from the fundamental theorem of coalgebra. Then (3) and (4) follow: in general if you have a locally finitely-presentable category with a monoidal structure with compact unit and which preserves filtered colimits, its category of monoids will be locally finitely-presentable, and similarly for commutative monoids. Finally, (5) follows because Hopf algebras are closed among bialgebras under filtered colimits (when the monoidal product has compact unit and preserves filtered colimits).<|endoftext|> TITLE: Does p-adic etale cohomology know the variety has ordinary reduction or not? QUESTION [5 upvotes]: For a smooth proper variety $X$ over discrete valuation ring $\mathcal{O}$ of mixed characteristic $(0,p)$, let $X_K$ be the generic fibre over a generic point $\textbf{Spec} K$ and let $X_k$ be the special fibre over a special point $\textbf{Spec} k$. We say that the variety $X_k$ has ordinary reduction iff the cohomology group $H^i(X_k,d\Omega^j)=0$ for any $i,j$. My question is that p-adic etale cohomology $H^i_{et}(X,\mathbb{Z}_p)$ of $X_K$ knows $X_k$ has ordinary reduction or not. REPLY [6 votes]: I think you meant the etale cohomology of the geometric generic fiber, equipped with the action of the Galois group of $K$. Moreover, in $p$-adic Hodge theory it is usually necessary to assume that the residue field $k$ is perfect. In this case, the answer is yes under the additional (customary) assumption that the crystalline and Hodge cohomology groups of $X$ are torsion free (this assumption is needed for the equivalence of two "standard" definitions of ordinarity: the one you gave and the one in terms of Hodge and Newton polygons). This essentially follows from the results of Bernadette Perrin-Riou "Representations p-adiques ordinaires" in Asterisque no. 223, 1994 "Periodes p-adiques" MR 1293973. See Proposition 6.7 in https://arxiv.org/abs/2005.02246 (I claim no originality here, but the statement gives what you want).<|endoftext|> TITLE: Is there a finitely axiomatizable class of structures whose equality-free theory is not finitely axiomatizable? QUESTION [11 upvotes]: This was originally an MSE question, but I was told to ask it on MathOverflow. Does there exist a class $C$ of $L$-structures for a finite signature $L$, which is finitely axiomatizable in first-order logic with equality, but whose equality-free theory $Th(C)$ is not finitely axiomatizable? Non-OP edit: To clarify, the question is whether the finite $\mathsf{FOL}$-axiomatizability of $Th_\mathsf{FOL}(C)$ necessarily implies the finite $\mathsf{FOL_{w/o=}}$-axiomatizability of $Th_{\mathsf{FOL_{w/o=}}}(C)$, regardless of whether the $\mathsf{FOL}$-deductive closure of $Th_{\mathsf{FOL_{w/o=}}}(C)$ coincides with $Th_\mathsf{FOL}(C)$ or not. (That is, we don't care whether $Mod(Th_{\mathsf{FOL_{w/o=}}}(C))\not=C.$) REPLY [7 votes]: EDIT: As pointed out by Emil Jerabek in the comments, the argument for relational languages fails in the last step. However, the question is already answered by the example with function symbols. EDIT: If function symbols are allowed, then this can happen. Let $L=\left\{P, f, a\right\}$, where $P$ is a property symbol, $f$ is a unary function symbol and $a$ is a proper name. Let $C$ be the class of models axiomatized by $P(a)$ and $a=f(a)$. The equality-free theory of $C$ is axiomatized by $T =\left\{P(a), P(f(a)), P(f(f(a))), ...\right\}$. In fact, if $M$ is a model of $T$, then a model $N$ of $P(a)\wedge(a=f(a))$ can be obtained from $M$ by taking a quotient (identifying all elements corresponding to $a$, $f(a)$, $f(f(a))$, ... in $M$). The canonical mapping from $M$ to $N$ is elementary for equality-free formulas. Since $N$ is in $C$, $M$ satisfies any equality-free sentence in $C$. However, $T$ is not finitely axiomatizable. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ If $L$ is relational (and finite), then finite axiomatizability with equality implies finite axiomatizability without equality. Since $L$ is finite, let $\phi_1(x,\bar{z})$, ..., $\phi_n(x,\bar{z})$ be all atomic formulas (with the exception of equality formulas) in the variables $x$, $\bar{z}=(z_1,...,z_k)$, in which $k+1$ is the greatest relational arity in $L$. Consider the formula $I(x,y)$ saying that $x$ and $y$ satisfy the same atomic relations: $\forall\bar{z}((\phi_1(x,\bar{z})\leftrightarrow\phi_1(y,\bar{z}))\wedge...\wedge(\phi_n(x,\bar{z})\leftrightarrow\phi_n(y,\bar{z})))$. [edit: In the compact notation sugested in the comments, $I(x,y)$ is $\forall\bar{z}\bigwedge_{i=1}^{n}(\phi_i(x,\bar{z})\leftrightarrow\phi_i(y,\bar{z}))$.] Now, let $A$ be an axiom for $Th_{FOL}(C)$. We may assume that $\bar{z}$ does not occur in $A$. Suppose that $\psi$ is an equality-free sentence such that $\psi\in Th_{FOL}(C)$. Therefore, $A\vdash\psi$. If $A_I$ is obtained from $A$ by replacing all occurrences of the form $u=v$ by $I(u,v)$, then $A_I, \forall x,y(x=y\leftrightarrow I(x,y))\vdash\psi$, by a standard result on equivalence. Since the equality axioms are consequences of $\forall x,y(x=y\leftrightarrow I(x,y))$, we have that $A_I, \forall x,y(x=y\leftrightarrow I(x,y))\vdash\psi$ in first-order logic without equality also. Hence, $A_I\vdash\psi$ in first-order logic without equality by the standard conservativity result on definitional extensions. Since $A$ implies $A_I$, we conclude that $A_I$ is an axiom for $Th_{FOL_{w/o=}}(C)$.<|endoftext|> TITLE: Expected rank of linear combination of matrices QUESTION [8 upvotes]: Let $A_1,\dots, A_s \in M_n (\mathbb{R})$ be symmetric matrices and suppose they are linearly independent over $\mathbb{R}$. This means that $$ m = \min_{(c_1, ..., c_s) \in \mathbb{R}^s \backslash \{0\}} \mbox{rank} \left( \sum_{i=1}^s c_i A_i \right) > 0 $$ I am wondering about what do we expect $m$ to be greater than generically, meaning if we choose the coefficients of the matrices "randomly" what can we expect as a lower bound of $m$? Does someone see a relatively simple heuristic that gives a non-trivial expected lower bound? any input would be greatly appreciated. thank you. This question is related to my earlier question here, rank of a linear combination of matrices, in which I asked about the maximal size of $m$. REPLY [11 votes]: In other words, given the "Grassmannian" $$GS_{m-1,n}=\{S\in{\bf Sym}_n\,|\,{\rm rk}\,S\le m-1\},$$ you ask what is the largest dimension $s$ of a subspace $E\subset{\bf Sym}_n$ intersecting $GS_{m-1,n}$ at the origin only, transversally (in order that this subspace be somewhat generic). This is a notoriously difficult and unsolved problem. At least the solution when $m=2$ is very simple: $$s=\frac{n(n+1)}2-1=\frac{(n+2)(n-1)}2.$$ Just take the subspace of trace-less symmetric matrices<|endoftext|> TITLE: Defining Hochschild homology of non-commutative DG-algebras with animated rings with a circle action QUESTION [5 upvotes]: A cool construction of Hochschild homology (that I saw on B. Antieau's website here ) is the following: Let $k$ be a commutative ring, then denote by $\mathfrak{a}\text{CAlg}_k$ the category of animated commutative rings. For any $x\in BS^1$, we have a natural forgetful functor $$x^*:\mathfrak{a}\text{CAlg}_k^{BS^1}\to \mathfrak{a}\text{CAlg}_k.$$ Then Hochschild homology is the left adjoint $x_!:\mathfrak{a}\text{CAlg}_k\to \mathfrak{a}\text{CAlg}_k^{BS^1}$. I've not found a proof that this does agree with the "usual" Hochschild homology of commutive rings, is there a good reference for this? My main question is in the non-commutative setting. Hochschild homology has a natural definition for non-commutative DG-algebras (see for instance here). DG-algebras are, at least in characteristic zero, equivalent to simplicial commutative algebras, so we can (probably) use $x_!$ as a definition for Hochschild homology of commutative DG-algebras (even if I'm not sure if this is backwards-compatible). My question is if we can use $x_!$ to define Hochschild homology for non-commutative DG-algebras. I think non-commutative DG-algebras are equivalent to simplicial non-commutative rings, and so we can consider the category of animated non-commutative rings $\mathfrak{a}\text{Alg}_k$ and the forgetful functor $$x^*:\mathfrak{a}\text{Alg}_k^{BS^1}\to \mathfrak{a}\text{Alg}_k$$ and the left adjoint $x_!$ thereof. Edit: Maybe animated non-commutative rings is not the right way of going about it, but what I'm probably most interested in is having a universal property for non-commutative Hochschild homology, which is maybe a more amenable quesion. REPLY [6 votes]: No, this does not work in the non-commutative case. In general we have $HH(A)=A\otimes_{A\otimes A^{\mathrm{op}}} A$, and this is only a $k$-module, not an algebra. If $A$ is commutative, the tensor product happens to compute coproducts/pushouts of commutative $k$-algebras, and we have $HH(A)=\operatorname{colim}_{S^1}A=x_!(A)$ since $S^1=*\coprod_{*\sqcup *}*$, but in the non-commutative case the tensor product does not have such an interpretation. To see that the group $S^1$ acts on $HH(A)$ in general, one can use the formalism of factorization homology, $HH(A)=\int_{S^1} A$, which makes the functoriality on $BS^1$ apparent. A more classical approach is to use the fact that the usual "Hochschild complex" extends to a cyclic $k$-module (a functor on Connes' cyclic category $\Lambda$), whose geometric realization acquires an action of $S^1$ due to the $\infty$-groupoid completion of $\Lambda$ being $BS^1$. A reference for the latter approach is Appendix B of the article by Nikolaus and Scholze: https://arxiv.org/pdf/1707.01799.pdf<|endoftext|> TITLE: Positive harmonic functions on nilpotent groups & Random walk on groups with a finite number of generators QUESTION [6 upvotes]: I want to read the following papers in the English version which I could not find anywhere (the only papers I can get are the Russian versions). Kindly help me out. Gregory A. Margulis, Positive harmonic functions on nilpotent groups. Dokl. Akad. Nauk SSSR 166 (1966), 1054–1057 (Russian); English translation in: Soviet Math. Dokl. 7 (1966), Dynkin, E. B.; Maljutov, M. B. Random walk on groups with a finite number of generators. (Russian) Dokl. Akad. Nauk SSSR 137 1961 1042–1045. I do apologize for asking this here as this is not the right platform. I have checked at Mathscinet but these are not available there. So I could not think of any better way to find it. If I can get any one these two I will be really grateful. REPLY [2 votes]: About Margulis' paper, the contents of his article are very well explained in Section 25 of Woess' book Random walks on infinite graphs and groups (available online). One of the most important result of Margulis' paper is that positive harmonic functions on finitely generated nilpotent groups are constant on left cosets of the commutator subgroup. This is Corollary 25.9 in Woess' book. In the same book, the results of Dynkin and Malyutov are considered (for instance in the preface). Their results about the integral representation of harmonic functions is proved (with more generality) in Section 26.A.<|endoftext|> TITLE: Lipschitz property of the symmetric rearrangement QUESTION [6 upvotes]: I'm currently reading Talenti's paper "Best constant in Sobolev inequality" and am rather stuck on an argument on pg 363 (or pg 11 if you're reading the pdf). In this section of the paper, Talenti is proving the Polya-Szego inequality. More specifically, I don't see how (23a) implies the Lipschitzness of $u^*$. For some background, recall that for a non-negative function $u \in C^{\infty}_{c}(\mathbb{R}^n)$, the symmetric decreasing rearrangement of $u$, denoted by $u^*$ is defined as follows $$ u^*(x) := \sup\{t\geq0 : \mu(t) > C_{n}\ |x|^n \} ,$$ where here $C_{n}$ is the volume of the unit ball in $\mathbb{R}^n$ and $\mu(t) = |\{x: u(x)>t\}|$ is the Lebesgue measure of the set where $u$ exceeds t. Then, evidently, $u^*$ is radially symmetric, is a decreasing function of the radius, and satisfies $$ |\{x: u(x)>t\}| =|\{x: u^*(x)>t\}| .$$ Using the coarea formula, together with the isoperimetric inequality, one can show the inequality: $$Ch\mu(t)^{\frac{n-1}{n}} \leq \mu(t-h)-\mu(t),$$ where $C$ is a dimensional constant and $h>0.$ I'd like to show that this implies that $u^*$ is a Lipschitz function provided $u \in C^{\infty}_{c}(\mathbb{R}^n).$ So far I've tried plugging in $u^*(x)$ and $u^*(y)$ into the above inequality taking the roles of $t$ and $t-h$ to try and bound their difference but one gets a problematic factor of $|x|^{n-1}$ on the left hand side. Also note that if one can prove the Polya-Szego inequality for the $p = \infty$ case then one can bypass Talenti's argument entirely. Appreciate if anyone could take a look at this. REPLY [6 votes]: $\newcommand{\R}{\mathbb R}$By the continuity of measure, the nonincreasing function $\mu$ is right-continuous. The function $u^*$ is radial, that is, \begin{equation*} u^*(x)=U(|x|) \end{equation*} for some nonincreasing function $U\colon[0,\infty)\to[0,\infty)$ and all $x\in\R^n$. So, \begin{equation*} U(a)=\sup\{t\ge0\colon\mu(t)>C_n a\}=\sup E_a \tag{1}\label{1} \end{equation*} for all real $a\ge0$, where \begin{equation*} E_a:=\{t\ge0\colon\nu(t)>B_n a\}, \quad \nu:=\mu^{1/n},\quad B_n:=C_n^{1/n}. \tag{2}\label{2} \end{equation*} Note that the function $\nu$ is nondecreasing and right-continuous. We want to show that the function $U$ is Lipschitz. Take any real $s$ and $t$ such that $0\le sL(b-a) \end{equation*} for some real $a$ and $b$ such that $0\le aL(b-a)$ and hence, by \eqref{3}, \begin{equation} \nu(t)-\nu(s)<-\frac Cn\,(t-s)<-\frac Cn\,L(b-a)=\frac Cn\,L(a-b). \tag{5}\label{5} \end{equation} On the other hand, by \eqref{1}--\eqref{2}, the conditions $t\in[0,U(a))$ and $s=U(b)$ imply that $t\in E_a$ and $s'\notin E_b$ for any real $s'>s$, so that $\nu(t)>B_na$ and $\nu(s)=\nu(s+)\le B_n b$. So, \begin{equation*} B_n(a-b)=B_na-B_nb<\nu(t)-\nu(s)<\frac Cn\,L(a-b) \end{equation*} by \eqref{5}, which indeed contradicts \eqref{4}. Thus, \begin{equation*} U(a)-U(b)\le L(b-a) \end{equation*} for all real $a$ and $b$ such that $0\le a TITLE: Nondegeneracy of kernel of map on homology induced by covering of surfaces QUESTION [5 upvotes]: Let $f:X \rightarrow Y$ be a finite covering map between compact oriented surfaces and let $K$ be the kernel of the induced map $f_\ast: H_1(X) \rightarrow H_1(Y)$. Here homology has rational coefficients. Question: must $K$ be nondegenerate with respect to the symplectic algebraic intersection pairing $\omega$ on $H_1(X)$? In other words, is it true for all nonzero $k \in K$, there exists some $k’ \in K$ with $\omega(k,k’) \neq 0$? REPLY [7 votes]: The answer is yes. Let me work with $H^1$, which is canonically isomorphic to $H_1$ by Poincaré duality. I claim that $K=\operatorname{Ker} f_*$ is the orthogonal of $\ \operatorname{Im}f^* $: this is because $(\beta \cdot f^*\alpha )=(f_*\beta \cdot \alpha )$ for $\alpha $ in $H^1(Y)$ and $\beta $ in $H^1(X)$. Now for $\alpha ,\beta \in H^1(Y)$ we have $(f^*\alpha \cdot f^*\beta )=d(\alpha \cdot \beta )$, where $d$ is the degree of the covering. Thus $\ \operatorname{Im}f^* $ is a non-degenerate subspace of $H^1(X)$, hence its orthogonal $K$ is also non-degenerate.<|endoftext|> TITLE: In search of a combinatorial proof on particular set of partitions QUESTION [5 upvotes]: Given a partition $\lambda=(\lambda_1\geq\lambda_2\geq\dots)$, denote the conjugate partition by $\lambda'=(\lambda_1'\geq\lambda_2'\geq\dots)$. For example, if $\lambda=(4,2,2)$ then $\lambda'=(3,3,1,1)$. The hook length of a cell $\square=(i,j)$ in the Young diagram of $\lambda$ is given by $h(i,j)=\lambda_i+\lambda_j'-i-j+1$. Define the symplectic content of cell $(i,j)$ of $\lambda$ as $$c_{sp}(i,j)=\begin{cases} \lambda_i+\lambda_j-i-j+2 \qquad \text{if $i>j$} \\ i+j-\lambda_i'-\lambda_j' \qquad \qquad \text{if $i\leq j$}.\end{cases}$$ I had an MO question here without any solution. So, I decided to rephrase the problem in case it helps. Define $\mathcal{syP}_0(n)$ to be the set of all partitions $\lambda\vdash n$ which has no zero symplectic content for any $\square\in\lambda$. Here are a few examples for $n=2, 4, 6, 8, 10$ and $12$: \begin{align*} &11 \\ &211 \\ &222, 3111 \\ &3221, 41111 \\ &3322, 42211, 511111 \\ &3333, 43221, 522111, 6111111. \end{align*} On the other hand, if ${}_2\mathcal{P}_4(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are congruent to $2$ mod $4$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$: \begin{align*} &2 \\ &22 \\&222, 6 \\ &2222, 62 \\ &22222, 622, (10) \\ &222222, 6222, 66, (10,2). \end{align*} Also, if ${}_0\mathcal{P}_2(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are distinct and congruent to $0$ mod $2$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$: \begin{align*} &2 \\ &4 \\ &42, 6 \\ &62, 8 \\ &64, 82, (10) \\ &642, 84, (10,2), (12). \end{align*} We already know (due to Euler) that $\#{}_2\mathcal{P}_4(n)=\#{}_0\mathcal{P}_2(n)$. So, I would like to inquire that: QUESTION 1. Is there a bijection proving either $\#\mathcal{syP}_0(n)=\#{}_2\mathcal{P}_4(n)$ or $\#\mathcal{syP}_0(n)=\#{}_0\mathcal{P}_2(n)$? QUESTION 2. Is it true that $\lambda\in\mathcal{syP}_0(n)$ iff $\vert c_{sp}(\square)\vert=h(\square)$? REPLY [6 votes]: Proposition. A partition $\lambda \in \mathcal{syP}_0$ iff it is empty, or both of the following hold: $\lambda'_1 - \lambda_1 = 1$, the partition $\mu$ obtained by removing first row and column of $\lambda$ is also in $\mathcal{syP}_0$. Informally, all partitions in $\mathcal{syP}_0$ (and only those) are obtained by starting from the empty partition, and repeatedly adding "L-shapes" to the bottom left corner, each L-shape one cell wider than it is taller, given that each intermediate step yields a proper (column-monotonic) partition. For example, $\lambda = 43221$ looks like this: A AB ABBB AAAAA and $\lambda = 3333$ looks like this ABCC ABBB AAAA To see this, reverse the sequence $\lambda_0 = \lambda, \lambda_1 = \mu, \ldots, \lambda_k = \varnothing$, where each subsequent partition is obtained by "shedding" the first row and column of the one before it. As explained below in the proof, if at any point $\lambda'_{i, 1} - \lambda_{i, 1} \neq 1$, then we present a cell with $c_{sp}(\square) = 0$, otherwise the claim is established by induction. Proof. Observe that symplectic content of cells of $\mu$ is carried over to corresponding cells of $\lambda$, as such it has to be/stays non-zero. If $\lambda'_1 - \lambda_1 > 1$, then $c_{sp}(i, j) = 0$ for $(i, j) = (\lambda_1 + 2, 1)$. Indeed $i > j$ and $\lambda_i = 1$, and $c_{sp}(i, j) = 1 + \lambda_1 - (\lambda_1 + 2) - 1 + 2 = 0$. If $\lambda'_1 - \lambda_1 < 1$, then similarly $c_{sp}(1, \lambda'_1) = 1+\lambda'-\lambda'-1=0$. Finally, if $\lambda'_1 - \lambda_1 = 1$, then for any $x \in \{1, \ldots, \lambda_1\}$ we have $c_{sp}(1, x) = 1 + x - \lambda'_1 - \lambda'_x \leq 1 + \lambda_1 - \lambda'_1 - \lambda'_x = -\lambda'_x < 0$. Similarly one obtains that $c_{sp}(x + 1, 1)=\lambda_{x+1}+\lambda_1-1-(x+1)+2=\lambda_{x+1}+\lambda_1-x\geq \lambda_{x+1} > 0$. $\square$ The bijection to partitions into even distinct parts is now obvious: use the parts as descending even sizes (hook-lengths $h(i,i)$ of the diagonals) of L-shapes, which answers Q1. Q2 is now also easy to answer in the positive. Both $h(\square)$ and $c_{sp}(\square)$ are preserved after adding/removing the first row and column. Direct substitution yields $h(1, x) = -c_{sp}(1, x)$, $h(x + 1, 1) = c_{sp}(x + 1, 1)$ iff we are allowed to substitute $\lambda'_1 = \lambda_1 + 1$.<|endoftext|> TITLE: direct proof of an identity regarding certain symmetry of integer partitions? QUESTION [5 upvotes]: Suppose $(a_1,\ldots, a_k)$ is an integer partition of $n$, and $(b_1,\ldots,b_k)$ is a rearrangement of the $a$-sequence. Prove the following identity (preferably combinatorially): $$ \sum_{j_2+\cdots+j_k=l,\atop l\geq 0, \, a_t>j_t\geq 0} \quad\frac{(-1)^l l!}{(a_1+l+1)_{l+1}} {a_2\choose j_2}\cdots {a_k \choose j_k}=\sum_{j_2+\cdots+j_k=l,\atop l\geq 0, \, b_t>j_t\geq 0} \quad\frac{(-1)^l l!}{(b_1+l+1)_{l+1}} {b_2\choose j_2}\cdots {b_k \choose j_k}, $$ where $(x)_k$ is the falling factorial. REPLY [4 votes]: I assume you meant $j_t\leq a_t$ (not $j_t TITLE: Looking for an interesting result on the Navier-Stokes equations QUESTION [29 upvotes]: I am in my second year of master in Mathematics and one of my courses consists of a reading of Navier-Stokes Equations by Roger Temam. We have proven the existence for the weak Stokes and Navier-Stokes (defined on a bounded domain) in a stationary and non-stationary regime using a Galerkin method. Now I have to do a 2 hours presentation on the subject of my choice (still related to hydrodynamics) and I would like to know if one of you had an idea of something I could talk about. I looked on the internet for an article that could match my critera but all I found was way too advance or too long for what I am looking for. I am looking for a paper that could be presented in two hours, that is accessible to a master student (who has read Temam's book) and that is an important result in the modern understanding of the Navier-Stokes equations (a paper by Leray, Ladyzhenskaya, Fujita-Kato, Lions, ... for example). Could one of you, who has a better knowledge of what has been done in hydrodynamics, give me some advice on this subject? I first asked the question on Stackexchange Mathematics but I don't know which site is best for what I'm looking for. You may find here my question on Stackexchange Mathematics. REPLY [3 votes]: Maybe the following papers will be helpful: Some open questions in hydrodynamics and Between Hydrodynamics and Elasticity Theory: The First Five Births of the Navier-Stokes Equation. Another option would be to talk about Arnold's result that solutions to Euler's equations correspond to geodesics on the infinite-dimensional Riemannian manifold of volume preserving diffeomorphisms: Geometric Hydrodynamics: from Euler, to Poincare, to Arnold.<|endoftext|> TITLE: What is the motivation behind symplectic/orthogonal content? QUESTION [6 upvotes]: Here $\lambda'$ is the conjugate partition of $\lambda=(\lambda_1,\lambda_2,\dots)$ and cells are in the Young diagram. The symplectic content of cell $(i,j)$ of $\lambda$ is defined by $$c_{sp}(i,j)=\begin{cases} \lambda_i+\lambda_j-i-j+2 \qquad \text{if $i>j$} \\ i+j-\lambda_i'-\lambda_j' \qquad \qquad \text{if $i\leq j$}.\end{cases}$$ The orthogonal content of cell $(i,j)$ of $\lambda$ is defined by $$c_{O}(i,j)=\begin{cases} \lambda_i+\lambda_j-i-j \qquad \qquad \text{if $i\geq j$} \\ i+j-\lambda_i'-\lambda_j'-2 \qquad \text{if $i< j$}.\end{cases}$$ Although I have used these in my analysis, I still wonder: QUESTION. What is the motivation behind these definition choices for the "contents"? REPLY [12 votes]: A hook-content formula, using the contents $c_{sp}(i,j)$ and $c_O(i,j)$, for the dimensions of the irreducible polynomial representations of the symplectic and orthogonal groups, goes back to Ron King. I believe the relevant paper is https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0008414X00053086, but I did not check for sure. REPLY [5 votes]: As I guessed in a comment above, there is apparently a symplectic/orthogonal hook-content formula which uses these notions. See "Hook-content Formulae for Symplectic and Orthogonal Tableaux" by Campbell and Stokke https://doi.org/10.4153/CMB-2011-105-7. Warning: they seem to have withdrawn the arXiv version of their paper (https://arxiv.org/abs/0710.4155) for reasons that are not clear to me. EDIT: Regarding priority, I see Campbell and Stokke cite El Samra and King for the formula for the evaluation of the symplectic/orthogonal characters at $(1,1,\ldots,1)$ (i.e., pure counting of tableaux); what they do that is new is the principal evaluation $(1,q,\ldots,q^{n-1})$ (i.e., $q$-counting). Perhaps the arXiv withdrawal is about what Per said: copyright.<|endoftext|> TITLE: Heegaard Floer homology of a genus two Heegaard splitting of $S^3$ QUESTION [6 upvotes]: This is a duplicate of a question (https://math.stackexchange.com/questions/4416204/heegaard-floer-homology-of-a-genus-two-diagram-of-s3) on stackexchange, which did not get any answer. Feel free to flag it if it does not respect the standards. I am reading the introductory paper "Heegaard diagrams and holomorphic disks" by Ozsváth and Szabó (https://arxiv.org/abs/math/0403029v1, Section 2.2), and I do not understand one of the examples. They compute the homology of the following Heegaard diagram of $S^3$: In order to count holomorphic disks connecting $x_3 \times y_3$ to $x_2 \times y_3$, they consider the uniformization of $\Delta$ as a standard annulus with four marked points on its boundary, corresponding to $x_2, x_3, y_2, y_3$. They call $a$ the angle of the arc in the boundary connecting $x_3$ to $x_2$ which comes from $\alpha_1$, and $b$ the angle of the arc in the boundary connecting $y_3$ to $y_2$ which comes from $\alpha_2$. Then, they consider the one-parameter family of conformal annuli with four marked points obtained from $\Delta \cup \Gamma$ by cutting a slit along $\alpha_2$ starting from $y_3$, where the marked points correspond to $x_2,x_3$ and $y_3$ counted twice. At this point, they state: A four-times marked annulus which admits an involution (interchanging the two $\alpha$-arcs on the boundary) gives rise to a holomorphic disk connecting $x_3 \times y_3$ to $x_2 \times y_3$. By analyzing the conformal angles of the $\alpha$ arcs in this one-parameter family, one can prove that the mod $2$ count of the holomorphic is $1$ iff $a < b$. I have many questions. First, I do not really get what $a$ and $b$ are. I cannot fully understand why the involution gives rise to a holomorphic disk, although I am quite convinced. The thing I do not get at all is the analysis of the conformal angles in the end. The idea seems to be that if the arc of $\alpha_2$ from $y_2$ to $y_3$ is too long, there cannot be a holomorphic disk, but I do not really get the reason. Any help will be greatly appreciated! REPLY [8 votes]: There are several things going on here, explained rather elliptically in the paper. Let me expand. First, there's the question of which holomorphic annuli double-cover the disk. More precisely, suppose you are given a holomorphic annulus with each boundary colored in two pieces, red and blue. (Suppose the boundary is sufficiently nice, e.g. locally connected, so that various notions of "boundary" coincide.) When does that annulus have a double-cover to the disk, again with the boundary colored in two colors? The answer is as follows. Uniformize the annulus to a Euclidean annulus of circumference $2\pi$ with bi-colored boundary. A certain proportion of the upper boundary will be red; call that angle $a$ with $0 < a < 2\pi$. Similarly call the angle of the red portion of the lower boundary $b$. Then the annulus has a double-cover of the disk if and only if $a = b$. I won't go through the proof in detail, but in the case $a=b$ you can get the double cover as follows: Connect the left end of the upper red segment to the right end of the lower red segment, and conversely connect the right end of the upper red segment to the left end of the lower red segment. These intersect in a point $p$. Similarly connect the left/right ends of the upper blue segment to the right/left ends of the lower blue segment. These intersect in a point $q$. Both $p$ and $q$ are exactly halfway up the Euclidean annulus (since we assume $a=b$). The $180^\circ$ rotation around $p$ and $q$ gives an involution of the annulus; the quotient is the disk, and the quotient map is the desired holomorphic map to the disk. OK, so now we know what kind of annulus we are looking for (and the relevance of comparing $a$ to $b$). Now let's look at the specific annuli in the problem. There is a 1-parameter space of potential conformal annuli here, depending on how long of a slit to cut from $y_3$ to $y_2$ along the boundary between $\Gamma$ and $\Delta$. The question is whether any of the annuli in this family have the balance mentioned above of the amount of $\alpha$ on the two components. At one end of the parameter space, there is the annulus $A_0 = \Gamma \cup \Delta$ with no slit. This has no $\alpha$-arc appearing on the inner ($y_2$ and $y_3$ boundary), so for that annulus we have $b = 0$. Near the other end of the parameter space, the slit reaches almost all the way from $y_3$ to $y_2$. Then the conformal structure on the resulting annulus is very close to the conformal structure on $\Delta$ itself: the extra near-bigon $\Gamma$ has almost no effect on the conformal structure. Thus, if we look at the curve in the space of $a$ and $b$ values, we go from some point with $b=0$ (so necessarily $a>b$) to whatever values you get for $\Delta$. You will have an odd number of crossings iff we have $a < b$ for $\Delta$. A couple notes. In case it's not obvious, one point of this example is that the holomorphic disk counts are not combinatorial and do depend on the complex structure. You can make other examples where the two choices of which disk is represented are combinatorially equivalent, I think maybe there's an example later in this same paper. I generally find it much easier to think about everything in Lipshitz's cylindrical reformulation, and probably implicitly use that above. But the foundational papers are written in this symmetric product language. I'd love it if this argument appeared more explicitly somewhere in the literature; maybe it's somewhere I don't know about. You can generalize this criterion to cases where there are more than four total marked points on the boundary. Then the condition is that the total angle of $\alpha$-arcs on one component equals the total angle of $\alpha$-arcs on the other. Again I don't know if this appears in the literature. I omitted several things, for instance justifying the treatment of the boundary (using stronger versions of the Riemann mapping theorem/uniformization of annuli), why the extra bigon has no effect on the conformal structure of the annulus (related to the Gromov compactness used everywhere in the theory), and discussion of perturbing the almost-complex structure to make it generic enough to be sure the moduli spaces have the expected dimension doesn't ruin the argument.<|endoftext|> TITLE: Counterexamples to an analog of Cannon's Conjecture which do not arise from manifolds? QUESTION [11 upvotes]: Let $\Gamma$ be a finitely presented hyperbolic group with boundary homeomorphic to $S^{n-1}$. Are there any examples of such $\Gamma$ which are known to not be the fundamental group of any $n$-dimensional compact, negatively curved, manifold? When $n=2$, such groups are known not to exist, and when $n=3$ this is an open question, originally posed by J. Cannon. For $n\geq 4$ this MathOverflow answer gives a number of examples where $\Gamma$ is not the fundamental group of a hyperbolic manifold. However, these examples still arise as fundamental groups of negatively curved (albeit not hyperbolic) manifolds of the appropriate directions. Are there any known examples of hyperbolic groups with sphereical boundary which do not arise at all as the fundamental group of negatively curved compact manifolds? REPLY [10 votes]: The requirement that the manifolds "do not arise at all from negatively curved compact manifolds" is somewhat vague, but here is one known construction. If one applies Charney-Davis strict hyperbolization to a closed oriented PL manifold with a non-integral Pontryagin number, the result is an aspherical manifold with hyperbolic fundamental group whose boundary is a sphere (by work of Davis and Januszkiewicz), and which is not the fundamental group of a closed aspherical smooth manifold. Such manifolds exist in all dimensions $4k-1$ where $k\ge 2$ is an integer. This is explained in "Aspherical manifolds with hyperbolic fundamental group" by Barthel, Lueck and Weinberger, on the bottom of p.12 .<|endoftext|> TITLE: Question about information measurement for continuous random variable QUESTION [13 upvotes]: I have a question about the information measurement for continuous random variables. Maybe it is some basic question, but it really drives me crazy. Suppose we have a random variable $x \sim N(0, 1)$ and another random variable $Y = 2X$, so that $Y \sim N(0, 4)$. Then, let's compute some information measurements of $X$ and $Y$: First, we can compute the entropy. Since both $X$ and $Y$ are normal distribution, we can compute the entropy easily: $H(X) = \ln(\sqrt{2 \pi e})$ and $H(Y) = \ln(2 \sqrt{2 \pi e})$. It is already strange, since there is only a linear mapping between $X$ and $Y$. We create some more information by just multiplying a constant. Then, let's try to measure the mutual information between $X$ and $Y$. We know that $I(X, Y) = H(X) - H(X | Y) = H(Y) - H(Y | X)$. Since $Y = 2X$, when we know $X$, $Y$ is also known, vice versa. So we should have $H(X | Y) = H(Y | X) = 0$. Plug it into the above equation, we get $I(X, Y) = H(X) = H(Y)$. But it is in contrast with the entropy we compute in the first step. What I am missing here? REPLY [17 votes]: There is actually nothing weird about this at all. What you have discovered is that for continuous quantities, information is scale-relative. Or, to put it another way, the "true" entropy of any continuous quantity not known exactly (or, known down to a finite number of exact candidates) is $\infty$. Which, of course, should make intuitive sense to you - it takes an infinite number of digits to name an arbitrary real number, and there are an infinite number of potential candidates (uncountably infinitely many actually, $\beth_1$), whenever you have a whole continuous smear of a probability distribution that is a nontrivial continuous function. But, of course, just assigning "$\infty$" all the time would not be very useful. Hence, we have to introduce some form of relativity, by which we measure the information versus a moveable reference level. It's kind of like decibel measurements - $0\ \mathrm{dB}$ as an absolute measure must always be set arbitrarily for similar reasons. The simple definition of entropy you give, i.e. $$H[X] = -\int_S f_X(x) \lg f_X(x)\ dx$$ does this in a very simple manner, which can be "reverse engineered" from it by feeding it a suitably-wide uniform distribution: namely if you feed it $$f_X(x) = \begin{cases}\frac{1}{b - a},\ \text{if $x \in [a, b]$}\\0,\ \text{otherwise}\end{cases}$$ you will find that it gives you $\lg(b - a)$ and this is 1 bit exactly when $b - a$ is 2, in other words, when you have a situation in which the position is known to within two units. When it is two bits, it is unknown to four units and, in general, $n$ bits, unknown to $2^n$ units. That is, we have a uniform grid, formed by whatever measuring unit we are measuring $x$ in (e.g. meters, centimeters, millimeters, whatever), and the entropy is how much worse you are informed compared to knowing "to the nearest unit", as in a simple "digital" or "quantized" sense. Hence, it is no surprise you can have a negative entropy if you know the position better than the nearest unit! E.g. if your base unit is meters and you know it to $\frac{1}{128}$ meter, then you should have an entropy of -7 bits and, indeed, you do! But the entropy can keep getting more negative, which is what I was saying before, because there's still infinitely more information (at least in theory) you could still have yet to know about where exactly that thing is (or isn't) located! And if you want to make this more explicit, if you have a set reference scale you want that is different from your measuring unit, i.e. you want "entropy relative to 0.001 m" when your positions are measured in m, you just add that to $H[X]$: $$H_\mathrm{adjust}[X] = -\int_S f_X(x) \lg f_X(x)\ dx - H_0$$ where $H_0 = \lg(l_\mathrm{ref})$, and $l_\mathrm{ref}$ is your reference length measured in your unit of choice. So if your measured in meters that you know to 2 m, i.e. $H[X]$ is 1 bit, then relative to 0.001 m, where that $H_0 \approx -9.96$ then $H_\mathrm{adjust}[X] \approx 10.96\ \mathrm{bit}$.<|endoftext|> TITLE: Is $\frac{1}{L(1+it)}$ unbounded? QUESTION [9 upvotes]: Let $\chi$ be a Dirichlet character and $L(s, \chi)$ be the corresponding L-function. Is $$\frac{1}{L(1+it, \chi)}$$ unbounded for $t \in \mathbb{R}$? I'm aware that this is true if $L=\zeta$, but I'm not sure of general $L$. REPLY [6 votes]: Yes, this is known to hold, and for more general families of $L$-functions. See in particular Theorem 3.1 and the discussion following it in "Extremal values of Dirichlet $L$-functions in the half-plane of absolute convergence " by Jörn Steuding.<|endoftext|> TITLE: The knot $K\subset \Bbb S^3$ is smoothly slice, but the disc $D\subset \Bbb D^4$ is only locally flat. Can $D$ be smoothed? QUESTION [10 upvotes]: Suppose I am given a smoothly slice knot $K\subset\Bbb S^3$. But I am only given a locally flat disc $D\subset \Bbb D^4$ with boundary $K$. Question: Is there a smooth disc $D'\subset\Bbb D^4$ with boundary $K$ that is $\epsilon$-close to $D$ and approaches $D$ towards the boundary? More precisely, I want $D'$ to be in $D_\epsilon\cap X$, where $D_\epsilon$ is an $\epsilon$-thickening of $D$ and $$X:=\bigcup_{x\in D} B_x(d(x,K)),$$ $B_x(r)$ being a ball around $x$ of radius $r$, and $d(x,K)$ being the distance of $x$ from $K$. REPLY [6 votes]: Nice question! Here's an example to think about. Let J be a knot that is topologically but not smoothly slice, and let $D_J$ be a locally flat disc that is smooth near the boundary, with boundary J. Note that $-J$ (the usual concordance inverse) is also slice, with disk $-D_J$. Let $D = D_J\natural(-D_J)$, where the $\natural$ denotes boundary connected sum done along a band near the boundary. Let $K = J \# (-J)$; then it is smoothly slice by a standard argument. At first glance, it looks hard for any slice disk $D'$ for $K$ to be close to $D$ without splitting into smooth slice disks for $J$ and $-J$. But the proof is not obvious because you are approximating in the $C^0$ topology whereas one would presumably need a stronger sense of approximation near the boundary. As I said, something to think about.<|endoftext|> TITLE: How could we define "recursively greatly Mahlos"? QUESTION [6 upvotes]: A common action in set theory is making a large cardinal axiom "recursive", i.e. turning it from a large uncountable cardinal to a large countable ordinal. For example: Recursively regular = $\alpha$ is admissible, i.e. $L_\alpha \vDash \text{KP}$. Recursively inaccessible = $\alpha$ is admissible and the admissible ordinals are unbounded in $\alpha$. Recursively Mahlo = for every function $f: \alpha \to \alpha$ definable in $L_\alpha$ by a $\Delta_1$ formula, there is an admissible $\beta < \alpha$ closed under $f$. Recursively weakly compact = $\Pi_3$-reflecting, i.e. for any $\Pi_3$ formula $\phi(x)$ and any $a \in L_\alpha$, $(L_\alpha, \in) \vDash \phi(a)$ implies $\exists \beta (\beta \in \alpha \land a \in L_\beta \land (L_\beta, \in) \vDash \phi(a))$. And so on... I was wondering how we could define a "recursively greatly Mahlo" ordinal. The original greatly Mahlo cardinals are defined like so: A cardinal $\kappa$ is greatly Mahlo iff it is inaccessible and there is a normal $\kappa$-complete filter on $\mathcal{P}(\kappa)$ which is closed under the function $X \mapsto \{\alpha \in S: \text{cof}(\alpha) \geq \omega_1 \land X \cap \alpha \text{ is stationary in } \alpha\}$. The resulting large countable ordinal should likely be in between recursively Mahlo and $\Pi_3$-reflecting, as the original was between Mahlo and weakly compact. However, some of my friends say that recursively greatly Mahlo is $\Pi_3$-reflecting. I don't believe this, as Stegert's PhD says that the recursive analogue of $\Pi^1_n$-indescribability is $\Pi_{n+2}$-reflection, but they may be correct after all and that recursive weak compactness is after all stronger, maybe on the level of stable or nonprojectible ordinals. REPLY [2 votes]: Let me preface this by stating the obvious in that there is no injection from a given uncountable cardinal to any countable ordinal, thus any system of describing "analogous properties" of large countable ordinals relative to an uncountable set of ordinals will be flawed in infinitely many ways. The connection between large cardinal axioms and large unrecursive countable ordinal axioms is based on intuition, not that there is anything wrong with that. The bottom line is, "recursively greatly Mahlo" ordinals can all be stable, or nonprojectible or $\Pi_3$-reflecting if you define them as such, but they absolutely don't have to be. I suggest reading this paper by Baldwin, where he defines a very structured hierarchy of Mahloness in analogy to the Mitchel rank for measurability of cardinals. I myself have not read more than a few pages but we only need to use the outline from the paper, so the descriptions I will give here are very oversimplified and do not constitute definitions, so keep that in mind. $m(\kappa)$ is a rank of Mahloness that essentially tells the that a filter on the stationary subsets of $\kappa$, if applied less than $m(\kappa)$ times to an equivalence class on $\kappa$ based on stationarity, will result in a set that is neither 0 nor the set of all nonstationary subsets of $\kappa$. This ranking does not provide much new insight for the structure of Mahloness as $m(\kappa)\leq\kappa^+$ is always the case by definition and for every greatly Mahlo $\kappa$ we already have $m(\kappa)=\kappa^+$ and vice versa, thus this notation is useless for trying to understand cardinals that go beyond greatly Mahlos in stationary properties that are themselves not on the level of the hierarchy of weakly and supercompacts. Further down the paper in section 3, functions $\bar m(\kappa), m^*(\kappa)$ are defined with more complex definitions that provide further structure to the different types of Mahlo cardinals and the different levels of iterated stationarity. According to Baldwin, for these to work as desired, we must assume a particular type of failure of GCH in $2^\kappa=2^{\kappa^+}$ at least for greatly Mahlo $\kappa$, that is, $|\mathcal P(\kappa)|$, which is the size of the domain and codomain of $m(), \bar m(), m^*()$, must be greater or equal to a limit of regular cardinals above $\kappa$. There is no analogy of this that can be translated down to the constructible universe of countable sets $L_{\omega_1}$, which is where we want the least "recursively greatly Mahlo" to fall. Let $M[\beta](0)$ be the first recursively $\beta$-Mahlo ordinal and let $M[\beta](\lambda+\alpha)$ be the $\alpha$th recursively $\beta$-Mahlo ordinal above $M[\beta](\lambda)$ whenever $\alpha < M[\beta+1](0)$ and let $\beta$-Mahlo ordinals be exactly those ordinals $\kappa$ such that $L_\kappa\models$KP+"$\forall\gamma<\beta:$ the $\gamma$-Mahlo ordinals form a stationary subset of the ordinals". We could define a collapse of $M[\beta+1](\alpha)$ inside $M[\beta](\text{___})$ to form an evaluation for the hierarchy of $\beta$-Mahlos, limits of such, inaccessible limits of such, hyper-inaccessible limits of such, hyper-hyper-inaccessible limits of such and so on, and keep going from there. If we use, say, the least $\Pi_3$-reflecting ordinal $K$ to collapse inside of $M[\text{___}](0)$ such that $M[K+\gamma](\beta)$ is the $(1+\beta)$th ordinal $\alpha$ that is $\alpha$-Mahlo and for all $\delta_0\in\gamma\land\delta_1\in\alpha:$ the ordinals of the form $M[K+\delta_0](\delta_1)$ form a stationary subset of $\alpha$ with respect to clubs definable in $L_{\omega_1}$. We can have an ordinal representation system $M[K+K\alpha+\beta](\gamma)$ being the $(1+\gamma)$th $\beta$-hyper$^\alpha$-Mahlo ordinal and so on. Then finally, we can define the recursively greatly Mahlo ordinals as exactly those that are of the form $M[\omega^{\rm CK}_{K+1}](\alpha)$ for some $\alpha$ as $\omega^{\rm CK}_{K+1}$ is admissible and above $K$ and thus not reacheable from below by some recursive well-ordering on $K+1$ and thus the least recursively greatly Mahlo will be greater than the $M$-collapse of any ordinal in $[K,K^+)$ or even simply in $K^+$. If $\alpha^+$ is the least regular above $\alpha$, that means that there is no surjection from $\alpha$ to $\alpha^+$. Analogously, if $\alpha^+$ is the least admissible above $\alpha$ then that means there is no surjection from $\alpha$ to $\alpha^+$ that consitutes a recursive or hyperarithmetical reordering of $\alpha$. This is not a definition but it suffices for making the analogy. I believe this way of defining the recursively greatly Mahlos fits the bill for placing them in the hierarchy of recursively Mahlos in such a way that they always have the recursively $\beta$-hyper$^\alpha$-Mahlo ordinals and all kinds of limits of such be unbounded in them, yet without the greatly Mahlos necessarily having $\Pi_3$-reflecting or stronger properties.<|endoftext|> TITLE: Model of an elliptic curve with p-torsion QUESTION [5 upvotes]: Suppose I have an elliptic curve $E$ defined over a number field $K$. I know that if it has a $2$ $K$-torsion, it has a model of the form: $E: Y^2=X^3+aX^2+bX$ a $3$ $K$-torsion, it has a model of the form: $E: Y^2 +cXY +dY=X^3$ My question is, do we have a nice description for elliptic curves with a $p$ $K$-torsion where $p$ is any rational prime? Any help or reference would be very much appreciated! REPLY [5 votes]: A standard method to find such an equation (in principle) is the following, shown to me by Tate, but maybe due to Mordell(?). If $P_0$ is a point of order at least $4$ on $E$ (infinite order is allowed), then after a change of coordinates, we can find an equation for $E$ of the form $$ E : y^2 + u x y + v y = x^3 + v x^2 \quad\text{with}\quad P_0=(0,0). $$ (It's a nice exercise to check this assertion.) So now we start with this equation and point, and we want $P_0$ to have order $N$. If $N$ is odd, say $N=2n+1$, then use the group law on $E$ to compute $x(nP_0)$ and $x\bigl((n+1)P_0\bigr)$. These will each be rational functions of $u$ and $v$; i.e., they're in $\mathbb Q(u,v)$. Setting $$ x(nP_0)=x\bigl((n+1)P_0\bigr)$$ and clearing the denominators will give you a polynomial in $(u,v)$ whose vanishing is an affine plane model for $X_1(N)$ if $N$ is prime. However it tends to be rather singular at points where the discriminant of $E$ vanishes. And if $N$ is composite, then the polynomial will factor, i.e., the curve will have components for each divisor of $N$ greater than $3$. Finally, if $N=2n$ is even, you can do the same thing with the decomposition $N=(n+1)+(n-1)$, or you could set the numerator of $(2y+ux+v)(nP_0)$ equal to $0$, which is another way of forcing $2nP_0$ to be $O$.<|endoftext|> TITLE: Placing pins on a Galton board to approximate an arbitrary distribution QUESTION [5 upvotes]: Inspired by this reddit post: https://old.reddit.com/r/math/comments/tv3cbg/how_do_you_unbell_curve_a_galtonplinko_board/ The Nth Galton Board, G(N), is a triangular lattice of pegs of height N-1. When a ball is dropped, at each level it will drop left or right with p=1/2. The path of the ball is a bernoulli distribution with N steps, and when N is large, this distribution converges to the normal distribution. My question, which I'm not even sure how to approach answering, is twofold: Which distributions can be constructed through a galton board process? (Assuming you may place pegs anywhere you like). Can all distributions be constructed like this? Does there exist a distribution which cannot be constructed? REPLY [2 votes]: $\newcommand{\N}{\mathbb N}$The question can be apparently formalized as follows. We have a random walk $(X_1,X_2,\dots)$ on the state space $\N:=\{1,2,\dots\}$ starting at point $1$ at time moment $1$ -- that is, $X_1=1$. If the walker is at point $j\in\N$ at a time moment $t\in\N$, then it moves to the right by $1$ with a probability $b_{t,j}\in[0,1]$ or stays at the point $j$ with probability $1-b_{t,j}$: \begin{equation*} \begin{aligned} &P(X_{t+1}=j+1|X_t=j,X_{t-1},\dots,X_1) \\ &=b_{t,j} =1-P(X_{t+1}=j|X_t=j,X_{t-1},\dots,X_1). \end{aligned} \end{equation*} The question is then this: Given any $n\in\N$ and any probability distribution $\pi$ on the set $[n]:=\{1,\dots,n\}$, do there necessarily exist numbers $b_{t,j}\in[0,1]$ for all natural $t$ and $j$ such that the distribution of $X_n$ is $\pi$? The answer to this question is yes, which can be proved by induction on $n$. Indeed, for $n=1$ this claim is trivial. To make the induction step, we only have to prove the following: Take any natural $n\ge2$. Take any nonnegative real numbers $p_1,\dots,p_n$ (corresponding to the desired distribution of $X_n$) such that $p_1+\dots+p_n=1$. Then there exist (i) nonnegative real numbers $q_1,\dots,q_{n-1}$ (corresponding to an appropriate distribution of $X_{n-1}$) such that $q_1+\dots+q_{n-1}=1$ and (ii) numbers $b_1,\dots,b_{n-1}$ in $[0,1]$ such that \begin{equation*} p_j=q_{j-1}b_{j-1}+q_j(1-b_j)\text{ for all $j\in[n]$,} \tag{1}\label{1} \end{equation*} where $q_0:=0$, $q_n:=0$, $b_0:=0$, and $b_n:=0$. But such numbers $q_1,\dots,q_{n-1},b_1,\dots,b_{n-1}$ can be explicitly and simply constructed e.g. as follows: \begin{equation*} q_1:=p_1+p_2,\ q_2:=p_3,\ \dots,\ q_{n-1}:=p_n, \end{equation*} \begin{equation*} b_1:=\frac{p_2}{p_1+p_2},\ b_2:=1,\ \dots,\ b_{n-1}:=1 \end{equation*} (if $p_1+p_2=0$, let $b_1:=1$). Indeed, then all the equalities \eqref{1} will hold. $\quad\Box$ So, to get the desired distribution, we are just successively splitting the probability of the state $1$ while carrying the other probabilities to the next time step. That is, once the walker has left the state $1$, it will march deterministically to the right, one unit step at a time. If at a time moment $t\in\{1,\dots,n-1\}$ the walker still remains in the state $1$, then at the next time moment, $t+1$, it will still be in the state $1$ with the (conditional) probability $\dfrac{p_1+\dots+p_{n-t}}{p_1+\dots+p_{n-t}+p_{n-t+1}}$ (or will move to the state $2$ with the complementary probability $\dfrac{p_{n-t+1}}{p_1+\dots+p_{n-t}+p_{n-t+1}}$). As an illustration, here is the tree diagram of getting the probabilities $0.2,0.3,0.4,0.1$ for the respective states $1,2,3,4$ at time $t=4$: The root of this tree, at the top, corresponds to time moment $t=1$, followed by the layers corresponding to time moments $t=2,3,4$. The nodes of the tree corresponding to the time moments $t=1,2,3,4$ are labeled by the probabilities $P(X_t=j)$ for $j=1,\dots,t$.<|endoftext|> TITLE: Fiber product of algebraic curves QUESTION [5 upvotes]: Let $ \phi : Y \to X $ and $ \psi : Z \to X $ be finite morphisms of integral algebraic curves over a field $ k $. Let $ \phi^* : K( X ) \to K( Y ) $ and $ \psi^* : K( X ) \to K( Z ) $ be the pullbacks of $ \phi $ and $ \psi $. Is the following true? The fiber product $ Y \times_X Z $ is an integral curve over $ k $ if and only if the tensor product $ K( Y ) \otimes_{ K( X ) } K( Z ) $ is a field. Thank you Edit: As shown in the comments, this is not true without the additional assumption that $ X $ is smooth. Now, is it true with this assumption? REPLY [4 votes]: This is ok when $X$ is smooth and $Y$ and $Z$ are integral. The point is that over a smooth curve, any finite morphism from an integral curve is automatically flat: over a Dedekind scheme, flat is the same as torsion-free [Tag 0AUW], and a finite integral extension is indeed torsion-free. Then $Y \times_X Z \to X$ is also finite flat, so satisfies both going up [Tag 00GU] and going down [00HS]. This implies that all irreducible components of $Y \times_X Z$ have dimension $1$, and for any point $p \in Y \times_X Z$ the height of $p$ is the same as the height of its image in $X$. In particular, every generic point of an irreducible component maps to the generic point of $X$, so there is only one irreducible component if and only if $K(Y) \otimes_{K(X)} K(Z)$ has only one irreducible component. Finally we have to say something about reducedness. Recall that a scheme $S$ is reduced if and only if it is (R$_0$) and (S$_1$) [Tag 031R]. For $Y \times_X Z$, the condition (R$_0$) means exactly that $K(Y) \otimes_{K(X)} K(Z)$ is reduced. Thus, we see that $K(Y) \otimes_{K(X)} K(Z)$ is a field if and only if $Y \times_X Z$ is (R$_0$) and irreducible. Recall also that a $d$-dimensional Noetherian scheme $S$ is Cohen–Macaulay if and only if it is (S$_d$), hence a $1$-dimensional scheme $S$ is (S$_1$) if and only if it is Cohen–Macaulay. So it remains to show that $Y \times_X Z$ is always Cohen–Macaulay. But $\Delta_X \colon X \hookrightarrow X \times X$ is a regular closed immersion [Tag 0E9J], hence so is $Y \times_X Z \hookrightarrow Y \times Z$ since it is the pullback of $\Delta_X$ along $Y \times Z \to X \times X$ [Tag 067P]. Since $Y$ and $Z$ are reduced curves, they are (S$_1$), hence Cohen–Macaulay as they have dimension $1$. Then $Y \times Z$ is Cohen–Macaulay as well [Tag 045J], hence so is $Y \times_X Z$ since it is a regular closed subscheme of $Y \times Z$ [Tag 02JN].<|endoftext|> TITLE: Are homeomorphic representations isomorphic? QUESTION [29 upvotes]: Let $G$ be a finite group. Let $V_1, V_2$ be two finite-dimensional real representations. Suppose $f: V_1 \to V_2$ is a $G$-equivariant homeomorphism. Can one conclude that $V_1$ and $V_2$ are isomorphic representations? REPLY [42 votes]: This is a famous problem, originating in work of de Rham, and the answer turns out to be No. The lowest-dimensional examples of non-linear similarity, as it is called, are in dimension 6, and examples only exist if the group has order divisible by (but not equal to) 4. This article contains a summary of the subject: Sylvain Cappell, Julius Shaneson, Mark Steinberger, Shmuel Weinberger, and James West. The classification of nonlinear similarities over ${\text{Z}}_{2^r}$, Bulletin of the American Mathematical Society 22 (1990). Mark Steinberger wrote another short summary of the subject, available here: http://math.albany.edu/topics/steinberger/msteinbergerrsch.pdf.<|endoftext|> TITLE: Examples of continua that are contractible but are not locally connected at any point QUESTION [10 upvotes]: A continuum is a compact, connected, metrizable space. What are examples of continua that are contractible but nowhere locally connected, meaning that no point has a neighbourhood basis consisting of connected sets? The following is an example of a contractible nowhere locally connected metrizable space, but I'm not aware of any compact examples (every segment on the "main line" has segments sticking out on a dense set). This question was previously asked on MSE but didn't get an answer there. REPLY [8 votes]: An example of such a space was constructed by Edwards in "R. Edwards, A contractible, nowhere locally connected compactum, Abstracts A.M.S. 20 1999), 494.". The example is the following space: Start with the space $X$ obtained stacking a sequence of Cantor fans (cones over the Cantor set) end to end as depicted. Then consider the space $X\times\Bbb R$, which looks like the upper half-plane with a lot of "fins" attached to it. Let $X'$ be the one point compactification of $X\times\Bbb R$, so that it looks like a closed disk with a lot of circular fins attached. The space $X'$ is a contractible continuum and it is locally connected precisely on the boundary of the circle. To make it nowhere locally connected construct $Y$ from $X'$ by wrapping one more Cantor fan around the circle, the resulting space is still contractible as explained here and clearly nowhere locally connected (in fact it is also nowhere connected im kleinen). Now I'm wondering whether a planar contractible continuum must be locally connected somewhere, but that should be a separate question.<|endoftext|> TITLE: Examples of non-self-induced algebras QUESTION [9 upvotes]: Let $A$ be a (possibly non-unital) algebra over $\mathbb C$. We say that $A$ is self-induced if the product map $m:A \otimes_A A \rightarrow A$ is an isomorphism. Here $A \otimes_A A$ is the balanced tensor product, the quotient of $A\otimes A$ by the linear span of elements of the form $ab\otimes c - a\otimes bc$. This notion seems to have been introduced by Gronbaek in Morita equivalence for self-induced Banach algebras and was further studied by Meyer in Smooth and rough modules over self-induced algebras. If $A$ is unital, or more generally, has local, one-sided, units, then $A$ is self-induced. I am interested in non-trivial examples of $A$ which are not self-induced. What is an example of $A$ which is not self-induced, but such that the product map is surjective, and such that the product is non-degenerate (so for each non-zero $a\in A$ there are $b,c\in A$ with $ba\not=0, ac\not=0$). I have tagged this functional analysis, as the notion seems to have arisen in the context of topological algebras (and so people working in this area might know examples). But the question does not ask about the topological case. I would be interested to know if this idea is studied in non-topological contexts under a different name? [Asked on Math.Stackexchange a few days ago, with no answers.] Partial leads: Some "factorisation" results (relevant to $m$ being surjective in this algebraic situation) are considered by Dales, Feinstein, Pham in Zbl 1471.46051 (full-text available here) which leads me to an old paper of Willis, Zbl 0742.46032. These give very complicated examples of $A$ for which $m$ is surjective, but without there being local units, and so maybe they are not self-induced. Unfortunately, I see no obvious way to check if they are self-induced. That $m$ is onto means that $A$ is an idempotent ring. Such rings, together with the non-degeneracy condition, and considered by Parvathi and Rao Zbl 0682.16031 and García and Simón Zbl 0747.16007. Unfortunately, these leads do not seem to lead to examples nor consideration of the "self-induced" idea. REPLY [3 votes]: In this paper the authors consider the analogous question in the context of semigroups and I think basically the contracted semigroup algebra of their semigroup on page 5 works. This argument should be ok over any base commutative ring with unit but to keep things easier I'll work over a field $K$. If $S$ is a semigroup with zero, then the contracted semigroup algebra $K_0S$ has basis the nonzero elements of $S$ and we interpret a zero product in $S$ as zero in $K_0S$. Let $S=\{a,b,c,d,0\}$ be the semigroup with zero with the following multiplication table: $$ \begin{array}{c| c c c c c} &0 & a& b &c& d\\\hline 0 & 0& 0& 0& 0& 0\\ a & 0& 0 &0& 0& a\\ b& 0& 0& 0& 0& b\\ c& 0& 0& a& 0& 0\\ d& 0& a& 0& c& d\end{array}$$ Put $A=K_0S$. Note that $S^2=S$ since $d$ is a left identity for $a,c,d,0$ and a right identity for $a,b,d,0$. Therefore, $A^2=A$, i.e., $m\colon A\otimes_A A\to A$ is surjective. Next I claim that $A$ is non-degenerate. Indeed, if $x=k_aa+k_bb+k_cc+k_dd$ with $xA=0$, then $0=xd = k_aa+k_bb+k_dd$, and so $x=k_cc$. But then $0=xb=k_ca$, hence $x=0$. Similarly, if $Ax=0$, then $0=dx = k_aa+k_cc+k_dd$ and so $x=k_bb$. But then $0=cx = k_ba$ and so $x=0$. Observe that $A\otimes_{K} A$ has basis all pairs $(s,t)\in S\setminus \{0\}\times S\setminus \{0\}$ and that $A\otimes_A A$ is the quotient of $A\otimes_K A$ by the subspace $V$ spanned by all differences $st\otimes u-s\otimes tu$ with $s,t,u,st,tu\neq S\setminus \{0\}$. Note that $b\notin sS$ for $s\in S\setminus \{b\}$. So a term in the spanning set for $V$ involving an element of the from $b\otimes c$ would have to look like $bt\otimes u-b\otimes tu$ with either $u=c$ or $tu=c$. If $tu\neq c$, then we would need $u=c$ and $bt=b$ to get $b\otimes c$. But then $t=d$ and we get $b\otimes c-b\otimes c=0$. If $tu=c$, then $t=d$, $u=c$. But then we again get $b\otimes c-b\otimes c=0$. Thus $b\otimes c$ does not appear in any term in the spanning set of $V$. Since the $s\otimes t$ with $s,t\in S\setminus \{0\}$ form a basis for $A\otimes_K A$, we must have that $b\otimes c\notin V$. Hence in $A\otimes_A A$, we must have $b\otimes c\neq 0$, but $m(b\otimes c)=bc=0$. So $A$ is our example. Nb. It looks to me like $S$ is a $\ast$-semigroup with $b^*=c$, $c^*=b$ and all other elements self-adjoint, so you can choose $A$ to be a $\ast$-algebra over $\mathbb C$ if you like and the $\ell_1$-norm will make it a $\ast$-Banach algebra since it is finite dimensional.<|endoftext|> TITLE: Transfinitely iterating the Levi-Civita, Hahn or Puiseux constructions QUESTION [8 upvotes]: This question was originally asked at MSE but seems too advanced, so I'm reposting it here. In short, the idea is that many constructions for non-Archimedean fields can naturally be iterated, in some way, to arrive at a transfinite sequence of increasingly large non-Archimedean fields, each of which embeds into the last. For instance, if we start with the reals, we can take a transfinite sequence of iterated ultrapowers to arrive at a proper class sized field, which is the maximal real-closed field. The surreal numbers have something similar with their "birthday structure," or what Philip Ehrlich calls an "$s$-hierarchical" structure, so that we can keep adding new levels of infinitesimals for each new ordinal and again arrive at the maximal real-closed field. The question, in a sense, is if we can get the same results iterating simpler power series-type constructions such as the Levi-Civita, Hahn, or Puiseux series. They also can easily be iterated: given any real-closed field, we can generate the Puiseux series in that field, which is a new real-closed field for which we can generate another Puiseux series (with an even smaller infinitesimal), etc. We can basically do the same thing with iterating the Levi-Civita construction, which is just the Cauchy completion of the Puiseux series. We get something similar with Hahn series, which also let us iterate the exponents at each level, so that they also can take values in the previous field, rather than only taking values in the rationals. On some level these fields are all just minor variations of one another, differing only in what kinds of rational exponents and infinite series are permitted, so the question is: to what extent can we iterate these kinds of things and ultimately arrive at the same, maximal, proper class sized real-closed field as the surreals and hyperreals? Or, put another way: what is the simplest power series-style construction like this which can be iterated transfinitely many times to get that same field? I say this because there are a few quirks - such as for Hahn series, we can either iterate the field of coefficients, or the group of exponents, or both, at each step, and I'm not sure just how important these details are. For instance, the surreals have about the simplest iteration procedure possible, requiring no ultrapowers or anything else, yet they succeed in this task. So, I'm curious if iterating something very simple like the Levi-Civita or Puiseux fields lead to a similar result. LATER EDIT: one important note is that the Levi-Civita construction, if iterated, only seems to add increasingly smaller infinitesimals and larger infinities, but doesn't quite add larger infinitesimals and smaller infinities in the right way. For instance, the Levi-Civita field has $\omega^1, \omega^{1/2}, \omega^{1/3}, \omega^{1/4}, ...$ at the first stage of iteration, but doesn't contain anything smaller than all of those but larger than the natural numbers. These numbers are never created at any successive stages either, because we aren't adding any new exponents to get something like $\omega^{1/\omega}$. So the short answer is that we will at least need something like Hahn series, so that we can modify the exponents at each stage. Alling apparently built something isomorphic to the surreals via a modified Hahn series construction in these papers: https://doi.org/10.1090%2FS0002-9947-1962-0146089-X https://doi.org/10.1090%2Fs0002-9947-1985-0766225-7 REPLY [7 votes]: Let us work in NBG set theory with global choice. There is, up to non unique isomorphism, a unique real-closed field that is $\kappa$-saturated for all infinite cardinals $\kappa$. Let's denote it by $\mathbf{K}$. For real-closed fields, being $\kappa$-saturated is the same as having no cut of size $<\kappa$, by which I mean an ordered pair $(L,R)$ of subsets $L,R$ of size $<\kappa$ such that $L TITLE: Are there interesting examples of theorems proved using ‘height’ extensions? QUESTION [13 upvotes]: It's well known that forcing is more than a tool for proving independence: We can prove theorems and formulate axioms in theories like $\mathsf{ZFC}$ by moving to forcing extensions (e.g. $\mathfrak{p}=\mathfrak{t}$, remarkable cardinals, Todorčević and Farah's book "Some applications of the method of forcing"). Are there nice examples of theorems/axioms that use "height" extensions (i.e. where we use "ordinals" longer than $\mathit{Ord}$ or where the resources required are strictly greater than second-order etc.) to prove some result in $\mathsf{ZFC}$ (or an extension thereof) or about the ground model? (From here on, let "height extension of $M$" denote any model $M'$ such that $\mathit{Ord}^M \in M'$). Some uses that occur to me: Uses of $\mathsf{ETR}$ in class theory (e.g. using iterated truth predicates and the connection to determinacy for class games, cf. Gitman and Hamkins "Open Determinacy for Class Games"). #-generation. This is a technical axiom to state, so I won't do so here, the core point is that we capture reflection properties of some model $M$ by taking it to be an initial segment of a model $M'$ generated by an ultrapower construction (this ultrapower, in turn, may be longer than $\mathit{Ord}^M$). See Honzik and Friedman "On Strong Forms of Reflection in Set Theory" for details. There are useful definable well-order longer than $\mathit{Ord}$ (e.g. the ordering on mice). I am curious as to whether the use of such "height" extensions pops up a lot in set-theoretic practice, and whether its as useful/ubiquitous as the use of forcing in proving theorems and formulating axioms (beyond relative consistency). REPLY [6 votes]: Here is another instance, which appears in my recent paper with Bokai Yao on second-order reflection in the context of KMU with abundant urelements. Joel David Hamkins and Bokai Yao, Reflection in second-order set theory with abundant urelemets bi-interprets a supercompact cardinal, 2022, arXiv:2204.09766. The following theorem is an immediate consequence of the main theorem. Theorem. Assume Kelley-Morse set theory with urelements KMU, with the abundant atom axiom and second-order reflection. Then there is a stationary proper class of measurable cardinals, partially supercompact cardinals, and more. I mention the theorem because the main proof method is to undertake the unrolling construction, which produces sets at ranks higher than Ord. Starting at lower right in the model $\langle V(A),\in,\mathcal{V}\rangle$ in which the hypothesis is satisfied, we undertake unrolling to produce the models $W$ and $\bar V$, in which there is a supercompact cardinal. In fact, it is the cardinal $\kappa=\text{Ord}^V$ that becomes supercompact in these taller models. And the supercompactness of $\kappa$ in $\bar V$ and $W$ reflects to diverse consequences in the original universe $V(A)$ and its class of pure sets $V$, such as a stationary proper class of measurable and partially supercompact cardinals, as much supercompactness as desired.<|endoftext|> TITLE: Adelization of automorphic forms for higher class number QUESTION [8 upvotes]: Short version: is there a canonical way to adelize a classical Hecke eigenform automorphic form when the adelic quotient has many components? If not, what are the different "choices", how many, etc.? Some sketched details: Let $A$ be a central simple algebra over a number field $F$, e.g. $A$ is the matrix algebra over $F$. Let $\mathcal{O} \subset A$ be an order, e.g. the matrix ring over the ring of integers $\mathfrak{o}$ of $F$. The adelic quotient $A^\times \backslash \hat{A}^\times / \hat{\mathcal{O}}^\times$ is (assuming some Eichler condition) a disjoint union parametrised by the class group $F^\times_{>0} \backslash \hat{F}^\times / \operatorname{nr}(\hat{\mathcal{O}}^\times)$ (see Thm. 28.5.5 in Voight's Quaternion Algebras). This class group can be non-trivial if $F$ has non-trivial narrow class group or if the order $\mathcal{O}$ is small, more precisely if the image of the local norm $\operatorname{nr}(\mathcal{O}^\times_\mathfrak{p})$ fails to be the whole unit group $\mathfrak{o}_\mathfrak{p}^\times$, where $\mathfrak{p}$ is a prime of $F$. Each such component corresponds to a locally symmetric space (e.g. for quaternion algebras this would be an arithmetic quotient of the upper half plane). In this way, each adelic automorphic form is given by a tuple of classical automorphic forms on these components (see e.g. Shimura's 1978 Hilbert modular forms paper). If one has a classical automorphic form on one of these components and we assume that it is also a Hecke eigenform, then is there a canonical way to choose forms on the other components so that the corresponding adelic form is a Hecke eigenform. Of course, Hecke eigenform here has two meanings, referring to the classical and to the global Hecke algebra accordingly. Are there any references for this? REPLY [8 votes]: $\newcommand{\p}{\mathfrak{p}}$Let $C$ be the class group parametrising the components, say $X = \bigcup_{c\in C}X_c$. Then the Hecke operator $T_\p$ sends component $X_c$ to $X_{c\p}$. In particular, the Hecke operators preserving the components are the $T_\p$ where the class of $\p$ in $C$ is trivial. If $f$ is an eigenform on one component $X_c$, then you can extend it by $0$ on the other components, but that is usually not going to be an eigenform for the whole Hecke algebra. If you look at the automorphic representation generated by this extension, it will in general not be irreducible, but it is going to be a finite sum $\bigoplus_{\chi}\pi\otimes\chi$ where $\pi$ is an automorphic representation and $\chi$ ranges over some subset of the characters of $C$. Proof of the last statement: Let $\pi$ and $\pi'$ be two irreducible representations occurring in the decomposition of the representation generated by $f$. Then the $T_\p$-eigenvalues of $\bigoplus_{\chi \in C}\pi \otimes \chi$ and $\bigoplus_{\chi \in C}\pi' \otimes \chi$ agree for almost all $\p$ (determined by $f$ if the class of $\p$ is trivial in $C$, and $0$ otherwise), so these representations are isomorphic, and $\pi'$ is one of the $\pi\otimes\chi$. Together with the multiplicity one theorem, this proves the claim.<|endoftext|> TITLE: Isomorphism type of the knot concordance group QUESTION [8 upvotes]: Isotopy classes of oriented knots in $S^3$ form a commutative monoid with respect to connected sum. Smoothly slice knots, i.e. knots that are the boundary of a smooth properly embedded disk in $B^4$, form a submonoid. The quotient of the monoid of all knots by the submonoid of slice knots is the smooth knot concordance group $\mathcal{C}$. It is a group because every knot has in inverse: its mirror image with reversed orientation. My question is: what is known about the isomorphism type of the group $\mathcal{C}$? Of course, much research has gone into studying $\mathcal{C}$, but here I'm only interested in purely group theoretic properties of $\mathcal{C}$, which can be formulated without making reference to the geometric meaning of $\mathcal{C}$. For example, satellite operations promise to reveal much about $\mathcal{C}$, but (as far as I know) nothing about the isomorphism type of $\mathcal{C}$. Here are the properties of $\mathcal{C}$ that I can think of: $\mathcal{C}$ is countable and abelian. There is an epimorphism $\mathcal{C}\to \mathbb{Z}^{\infty}$ (the sum of countably many copies of $\mathbb{Z}$). There is a split epimorphism $p\colon\mathcal{C}\to (\mathbb{Z/2})^{\infty}$ (i.e. there exists $i\colon (\mathbb{Z/2})^{\infty}\to \mathcal{C}$ such that $p\circ i$ is the identity of $(\mathbb{Z/2})^{\infty}$). Of course, any epimorphism as in 2. is automatically split, since $\mathbb{Z}^{\infty}$ is free. Are these three properties all that is known? Here's how the three properties can be proven: Every knot has a diagram, and there are clearly only countably many isotopy types of diagrams, so only countably many isotopy types of knots. Connected sum may be seen to be abelian by waving your hands around. Such an epimorphism is given by (half of) Levine-Tristram signatures at $e^{2\pi i/m}$ for $m$ ranging over the prime numbers. Such an epimorphism can e.g. be constructed by choosing amphichiral knots $K_1, K_2, ...$ whose Alexander polynomials $\Delta(K_j)$ are irreducible and pairwise not related by multiplication by a unit. Take $i$ to be the inclusion of the subgroup of $\mathcal{C}$ generated by the classes $[K_j]$. Let $p$ send a knot concordance class represented by a knot $J$ to the sum of those $[K_j]$ for which the maximal $e$ such that $\Delta(K_j)^e$ divides $\Delta(J)$ is odd. Instead of smoothly slice knots, one may quotient by topologically slice knots (boundaries of locally flat embedded disks), which gives the topological knot concordance group $\mathcal{C}_{t}$. The same question can be asked about $\mathcal{C}_{t}$. REPLY [3 votes]: The followings are not purely algebraic results but contain crucial information about the smooth concordance group $\mathcal C$: Every knot is concordant to a prime knot. Kirby, R. C., Lickorish, W. R. (1979). Prime knots and concordance. In Mathematical Proceedings of the Cambridge Philosophical Society (Vol. 86, No. 3, pp. 437-441). Cambridge University Press. Every concordance class admits a hyperbolic representative. Myers, R. (1983). Homology cobordisms, link concordances, and hyperbolic 3-manifolds. Transactions of the American Mathematical Society, 278(1), 271-288. Montesinos knots cannot generate $\mathcal C$. Hendricks, K., Hom, J., Stoffregen, M., Zemke, I. (2020). Surgery exact triangles in involutive Heegaard Floer homology. arXiv preprint arXiv:2011.00113.<|endoftext|> TITLE: Universal cover with one end QUESTION [8 upvotes]: Suppose that $M$ is a non-compact manifold of finite topological type with one end which is the universal cover of some closed manifold $N$. Is $M $ necessarily homeomorphic to the total space of some vector bundle over a compact manifold? In fact the only examples I can think up are much more limited, just of the form $M = \Sigma \times \mathbb{R}^n$ where $\Sigma$ is a closed simply connected manifold. Cross posted on stack exchange https://math.stackexchange.com/questions/4417368/universal-covers-with-one-end. REPLY [11 votes]: I think not. In any dimension $n\geq 4$ there are examples, constructed by Mike Davis, of contractible manifolds $M^n$ that are not homeomorphic to $\mathbb{R}^n$, and yet are the universal cover of a compact manifold $N$. The only way that $M$ could be a vector bundle over a compact closed manifold $X$ is if $X$ were a point and the fiber Euclidean space, which is evidently not true. (I suppose that the base space of the vector bundle would have to be closed, or else $M$ would itself have boundary.) Davis constructs his manifolds so that they are not simply connected at infinity; this distinguishes them from $\mathbb{R}^n$. At least if $n=4$ this would imply that they are not vector bundles over a non-compact manifold either. M. Davis, Groups generated by reflections and aspherical manifolds not covered by Euclidean space. Ann. of Math. (2) 117 (1983), no. 2, 293–324<|endoftext|> TITLE: Representation theory of $\text{SL}(2,\mathbb{Z})$ QUESTION [5 upvotes]: The group $\text{SL}(2,\mathbb{Z})$ is the group of two-by-two matrices with integer entries and determinant one. This is a very simple definition. Yet its representation theory seems quite wild to me (a PhD physics student) from the limited information I can find online... In particular, it seems it has uncountably many irreducible representations. I am also aware that if we restrict ourselves to the class of representations with finite image and whose kernel can be described by congruences, the representation theory becomes way more well-behaved. These are the so-called congruence subgroups. I would like to learn more about the representation theory of $\text{SL}(2,\mathbb{Z})$ beyond the classical textbook discussions restricted on congruence subgroups. Any good general references (papers, book, ...) I could look into? REPLY [13 votes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\GL{GL}$The reason that there are uncountably many irreducible representations is not so bad, and gets at an important point: You shouldn't think of irreducible representations of a group like $\SL_2(\mathbb Z)$ individually, but rather as points in a space of representations, i.e. objects parameterized by a geometric space. This is simplest to see for representations sending $-I \in \SL_2(\mathbb Z)$ to the identity, i.e. representations of $\PSL_2(\mathbb Z)$. By the presentation YCor gave $\PSL_2(\mathbb Z) = \langle x,y \mid x^2 =y^3 =1 \rangle$, such a representation is given by a matrix $X$ satisfying $X^2=1$ and a matrix $Y$ satisfying $Y^3=1$, and representations up to isomorphism are given by pairs of matrices up to conjugation. For $n$-dimensional representations, say $n$ a multiple of $6$, it's not so hard to check that the space of matrices $X \in \GL_n(\mathbb C)$ with $X^2=1$ has dimension $n^2/2$ (over $\mathbb C$), and the space of matrices $Y \in \GL_n(\mathbb C)$ with $Y^3= 1$ has dimension $2n^2/3$, so the space of pairs has dimension $n^2/2 + 2n^2/3$, isomorphism classes of $n$-dimensional representations have dimension $n^2/2 +2n^2/3 - (n^2-1) = n^2/6+1$. One can check that irreducible representations are an open subset, and thus that the space of irreducible representations has the same dimension. So certainly there are uncountably many, because they're parameterized by a positive-dimensional manifold! However, it's clear from this analysis that this space should be one of the primary objects of study in the representation theory here, as it is for representations of surface groups and in some other cases of interest. REPLY [9 votes]: $\DeclareMathOperator\SL{SL}$The issue is that $\SL(2,\mathbb{Z})$ is very close to a free group, so it is not hard to map it to other groups, and in particular to produce lots of varied representations of it. You might be interested to know that things are quite different for $\SL(n,\mathbb{Z})$ when $n$ is at least $3$. I wrote a note The representation theory of $\operatorname{SL}_n(\mathbb Z)$ describing results of Margulis and Lubotzky giving a complete and fairly simple description of its representations. REPLY [2 votes]: Far from being a specialist on the topic, section 4 of the article $\text{SL}_2(\mathbb{Z})$, by K. Conrad discusses non-congruence subgroups.<|endoftext|> TITLE: Lifting isomorphisms between linear categories QUESTION [8 upvotes]: Let $C$ be a $\mathbb{Z}$-linear category, such that $C(x,y)$ is a free abelian group with finite rank, for every $x,y\in\mathrm {Ob}(C)$. Given a commutative ring with identity $R$, let $RC$ denote the category with the same objects of $C$, and morphisms $RC(x,y):=R\otimes_{\mathbb{Z}} C(x,y)$. Does any isomorphism in $\mathbb{F}_pC(x,y)$ lift to an isomorphism in $\mathbb{Z}^{\wedge}_pC(x,y)$? REPLY [4 votes]: Here's a simplification of Maxime Ramzi's answer. Given an isomorphism $\bar f \in \mathbf F_p\mathscr C(x,y)$, the claim is that any lift $f \in \mathbf Z_p\mathscr C(x,y)$ of $\bar f$ is an isomorphism. Indeed, if $g \in \mathbf Z_p\mathscr C(y,x)$ is a lift of $\bar g = \big(\bar f\big)^{-1}$, we know that $gf = 1 - p\phi$ for some $\phi \in \mathbf Z_p\mathscr C(x,x)$. The series $\psi = \sum_{i=0}^\infty p^i\phi^i$ converges in $\mathbf Z_p\mathscr C(x,x)$ and gives a two-sided inverse of $gf$, so $f$ has a left inverse $\psi g$ (and $g$ has a right inverse $f\psi$). Swapping the roles of $f$ and $g$ shows that $f$ has a right inverse as well, so $f$ is invertible. $\square$ Remark. A note on the hypotheses: flatness is never used; the only role it plays in Ramzi's answer is uniqueness of $\phi$ such that $gf = 1-p\phi$, but this is inconsequential. On the other hand, it is crucial that the $\mathscr C(x,y)$ are finitely generated, as we need $\mathbf Z_p\mathscr C(x,y)$ to be $p$-adically complete [Tag 00MA(3)].<|endoftext|> TITLE: Lifting quasi-nilpotent elements in C$^*$-algebras QUESTION [9 upvotes]: Let $A$ be a C$^*$-algebra with closed two-sided ideal $I$. Set $B=A/I$ and let $\pi:A\to B$ be the quotient map. Suppose that $b\in B$ is quasi-nilpotent. Does there exist quasi-nilpotent $a\in A$ such that $\pi(a)=b$? REPLY [3 votes]: I am rather certain this is unknown. The issue is that procedures we know to lift the relation $a^{n+1}=0$ do not mesh well with the $a^n=0$ case. This means writing the given quasinilpotent as a limit of nilpotents does not seem to help.<|endoftext|> TITLE: On functions with strict Lipschitz constant QUESTION [8 upvotes]: We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure. Let $n \geq 2$, and let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$. That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$. Question: Is it possible that $|Df| = L$ on a measure dense set? Note: Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map. REPLY [12 votes]: I guess it suffices to give an example for $n = 1$. If $f: \mathbb{R} \to \mathbb{R}$ is an example then $g(x_1, \ldots, x_n) = f(x_1)$ will be an example for any $n \geq 1$. All we need is a measurable set $A \subseteq \mathbb{R}$ such that both $A$ and its complement have positive measure in every interval. See here, for example. Then define $$ f(x) = \int_0^x 1_A = \begin{cases} m(A \cap [0,x])&x \geq 0\cr -m(A\cap [x,0])&x < 0 \end{cases}. $$ It should be clear that 1 is a strict Lipschitz constant, but $f'(x) = 1$ at every Lebesgue point of $A$, so the derivative is $1$ on a measure dense set.<|endoftext|> TITLE: Relative and absolute Ext groups QUESTION [6 upvotes]: Given a homomorphism of rings $S \rightarrow R$, for a pair of $R$-modules $M, N$ the machinery of relative homological algebra defines relative $Ext$-groups $Ext_{R, S}(M, N)$. These can be defined, for example, by mapping the relatively projective resolution $\ldots \rightarrow R \otimes_{S} R \otimes_{S} M \rightarrow R \otimes_{S} M \rightarrow M$ into $N$ and taking cohomology. I have two questions: Is there a general framework for relating these relative $Ext$-groups with the absolute ones, for example through a long exact sequence or a spectral sequence? Can these relative $Ext$-groups be identified with homotopy classes of maps in some appropriate relative derived $\infty$-category? Can the latter be expressed in terms of the absolute derived $\infty$-categories? REPLY [3 votes]: Let $K(S)$ be the category of chain complexes of $S$-modules; this category has Hom-complexes $hom_S(-,-)$ making it a dg-category, and thus produces a quasicategory via the construction in Higher Algebra section 1.3.1. Take $f: S \to R$ a ring homomorphism, $M \in K(S)$, and $N \in K(R)$. Then there is a natural adjunction isomorphism between hom-complexes $$ hom_S(M, f^* N) \cong hom_R(R \otimes_S M, N) $$ which therefore induces an adjunction $K(S) \rightleftarrows K(R)$, and a monad $f_* f^*$ on $K(S)$ whose underlying functor is $R \otimes_S (-)$. This gives us a lift of the functor $f^*$ to the category $Alg(f_* f^*)$ of algebras for this monad (HA, 4.7.3.3). For any object $M$ and $N$ in $K(R)$, we thus get a simplicial resolution $$ M \leftarrow R \otimes_S M \leftleftarrows \dots $$ The resulting complex $$ Hom_R(R \otimes_S M,N) \to Hom_R(R \otimes_ R \otimes_S M, N) \to \dots $$ can therefore be interpreted as calculating $Hom_{Alg(f^* f_*)}(f^* M, f^* N)$, because that's how one calculates homotopy classes of maps in the category of algebras over a monad. In the case where $M$ and $N$ are discrete complexes concentrated in fixed degrees, this recovers the relative Ext-groups you were interested in. This adjunction between $K(S)$ and $K(R)$ is very unlikely to be monadic except in special circumstances. But there is a localizing subcategory $V \subset K(R)$ consisting of those complexes whose images under the forgetful functor are contractible, and in total a factorization of $f^*$ into maps of stable $\infty$-categories as follows: $$ K(R) \to K(R) / V \to Alg(f^* f_*) \to K(S) $$ Here $K(R) / V$ is the Bousfield localization / Verdier quotient / nullification of $V$. This is roughly a coimage / image factorization and one could ask whether the middle map is an equivalence; this I don't know. Whether there are useful spectral sequences I also do not know (quotient maps $S \to S / I$ indicate that there is a cap on how much nice behavior we should expect). We can certainly construct filtrations of $R$ as an object of $K(S)$ and try to run with that, but it is typically much easier to reduce calculations in the derived category $D(S)$ to calculations in $K(S)$ and not the other way around.<|endoftext|> TITLE: Is there any fast implementation of four color theorem in Python? QUESTION [12 upvotes]: I'm now using scipy.spatial.Voronoi to generate a Voronoi graph, as shown here: voronoi graph. I'd like to apply the four color theorem on it, so that no adjcent regions share the same color. I converted it into a NetworkX graph, and used the method posted in this answer Do you know a faster algorithm to color planar graphs? to color the graph. from sage.graphs.graph_coloring import vertex_coloring coloring = vertex_coloring(G, 4, solver = "Gurobi", verbose = 10) My operation system is Win10 with SageMath 9.3 installed. However, it only worked when the coloring number is equal or greater than 5, and the result is good: 5 color result. Changing the number to 4 caused the following error: Traceback (most recent call last): File "five_color_sage.py", line 33, in coloring = vertex_coloring(G, 4, solver = "Gurobi", verbose = 0) File "sage/graphs/graph_coloring.pyx", line 570, in sage.graphs.graph_coloring.vertex_coloring (build/cythonized/sage/graphs/graph_coloring.cpp:9000) File "sage/graphs/graph_coloring.pyx", line 587, in sage.graphs.graph_coloring.vertex_coloring (build/cythonized/sage/graphs/graph_coloring.cpp:9277) File "sage/numerical/mip.pyx", line 441, in sage.numerical.mip.MixedIntegerLinearProgram.__init__ (build/cythonized/sage/numerical/mip.c:3989) File "sage/numerical/backends/generic_backend.pyx", line 1640, in sage.numerical.backends.generic_backend.get_solver (build/cythonized/sage/numerical/backends/generic_backend.c:20569) File "sage/numerical/backends/generic_backend.pyx", line 1783, in sage.numerical.backends.generic_backend.get_solver (build/cythonized/sage/numerical/backends/generic_backend.c:20254) ModuleNotFoundError: No module named 'sage_numerical_backends_gurobi' I've tried to install sage_numerical_backends_gurobi, but result in the following error: Building wheel for sage-numerical-backends-gurobi (setup.py) ... error ERROR: Command errored out with exit status 1: command: /opt/sagemath-9.3/local/bin/python3 -u -c 'import sys, setuptools, tokenize; sys.argv[0] = '"'"'/tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/setup.py'"'"'; __file__='"'"'/tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/setup.py'"'"';f=getattr(tokenize, '"'"'open'"'"', open)(__file__);code=f.read().replace('"'"'\r\n'"'"', '"'"'\n'"'"');f.close();exec(compile(code, __file__, '"'"'exec'"'"'))' bdist_wheel -d /tmp/pip-wheel-vggmxrhi cwd: /tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/ Complete output (23 lines): /bin/sh: line 0: .: gurobi.sh: file not found GUROBI_HOME is not set, or it does not point to a directory with a Gurobi installation. Trying to link against -lgurobi Checking whether HAVE_SAGE_CPYTHON_STRING... Checking whether HAVE_ADD_COL_UNTYPED_ARGS... Using compile_time_env: {'HAVE_SAGE_CPYTHON_STRING': True, 'HAVE_ADD_COL_UNTYPED_ARGS': True} running bdist_wheel running build running build_py creating build creating build/lib.cygwin-3.2.0-x86_64-3.7 creating build/lib.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi copying sage_numerical_backends_gurobi/__init__.py -> build/lib.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi copying sage_numerical_backends_gurobi/gurobi_backend.pxd -> build/lib.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi running build_ext building 'sage_numerical_backends_gurobi.gurobi_backend' extension creating build/temp.cygwin-3.2.0-x86_64-3.7 creating build/temp.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi gcc -Wno-unused-result -Wsign-compare -DNDEBUG -g -fwrapv -O3 -Wall -ggdb -O2 -pipe -Wall -Werror=format-security -Wp,-D_FORTIFY_SOURCE=2 -fstack-protector-strong --param=ssp-buffer-size=4 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/build=/usr/src/debug/python37-3.7.10-2 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/src/Python-3.7.10=/usr/src/debug/python37-3.7.10-2 -ggdb -O2 -pipe -Wall -Werror=format-security -Wp,-D_FORTIFY_SOURCE=2 -fstack-protector-strong --param=ssp-buffer-size=4 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/build=/usr/src/debug/python37-3.7.10-2 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/src/Python-3.7.10=/usr/src/debug/python37-3.7.10-2 -I/opt/sagemath-9.3/local/lib/python3.7/site-packages/sage/cpython -I/opt/sagemath-9.3/local/lib/python3.7/site-packages/cysignals -I/opt/sagemath-9.3/local/lib/python3.7/site-packages -I/usr/include/python3.7m -I/opt/sagemath-9.3/local/lib/python3.7/site-packages/numpy/core/include -I/opt/sagemath-9.3/local/include -I/usr/include/python3.7m -c sage_numerical_backends_gurobi/gurobi_backend.c -o build/temp.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi/gurobi_backend.o sage_numerical_backends_gurobi/gurobi_backend.c:639:10: fatal error: gurobi_c.h: No such file or directory 639 | #include "gurobi_c.h" | ^~~~~~~~~~~~ compilation terminated. error: command 'gcc' failed with exit status 1 ---------------------------------------- ERROR: Failed building wheel for sage-numerical-backends-gurobi Running setup.py clean for sage-numerical-backends-gurobi Failed to build sage-numerical-backends-gurobi Installing collected packages: sage-numerical-backends-gurobi Running setup.py install for sage-numerical-backends-gurobi ... error ERROR: Command errored out with exit status 1: command: /opt/sagemath-9.3/local/bin/python3 -u -c 'import sys, setuptools, tokenize; sys.argv[0] = '"'"'/tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/setup.py'"'"'; __file__='"'"'/tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/setup.py'"'"';f=getattr(tokenize, '"'"'open'"'"', open)(__file__);code=f.read().replace('"'"'\r\n'"'"', '"'"'\n'"'"');f.close();exec(compile(code, __file__, '"'"'exec'"'"'))' install --record /tmp/pip-record-ibpz889r/install-record.txt --single-version-externally-managed --compile --install-headers /opt/sagemath-9.3/local/include/site/python3.7/sage-numerical-backends-gurobi cwd: /tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/ Complete output (23 lines): /bin/sh: line 0: .: gurobi.sh: file not found GUROBI_HOME is not set, or it does not point to a directory with a Gurobi installation. Trying to link against -lgurobi Checking whether HAVE_SAGE_CPYTHON_STRING... Checking whether HAVE_ADD_COL_UNTYPED_ARGS... Using compile_time_env: {'HAVE_SAGE_CPYTHON_STRING': True, 'HAVE_ADD_COL_UNTYPED_ARGS': True} running install running build running build_py creating build creating build/lib.cygwin-3.2.0-x86_64-3.7 creating build/lib.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi copying sage_numerical_backends_gurobi/__init__.py -> build/lib.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi copying sage_numerical_backends_gurobi/gurobi_backend.pxd -> build/lib.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi running build_ext building 'sage_numerical_backends_gurobi.gurobi_backend' extension creating build/temp.cygwin-3.2.0-x86_64-3.7 creating build/temp.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi gcc -Wno-unused-result -Wsign-compare -DNDEBUG -g -fwrapv -O3 -Wall -ggdb -O2 -pipe -Wall -Werror=format-security -Wp,-D_FORTIFY_SOURCE=2 -fstack-protector-strong --param=ssp-buffer-size=4 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/build=/usr/src/debug/python37-3.7.10-2 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/src/Python-3.7.10=/usr/src/debug/python37-3.7.10-2 -ggdb -O2 -pipe -Wall -Werror=format-security -Wp,-D_FORTIFY_SOURCE=2 -fstack-protector-strong --param=ssp-buffer-size=4 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/build=/usr/src/debug/python37-3.7.10-2 -fdebug-prefix-map=/pub/devel/python/python37/python37-3.7.10-2.x86_64/src/Python-3.7.10=/usr/src/debug/python37-3.7.10-2 -I/opt/sagemath-9.3/local/lib/python3.7/site-packages/sage/cpython -I/opt/sagemath-9.3/local/lib/python3.7/site-packages/cysignals -I/opt/sagemath-9.3/local/lib/python3.7/site-packages -I/usr/include/python3.7m -I/opt/sagemath-9.3/local/lib/python3.7/site-packages/numpy/core/include -I/opt/sagemath-9.3/local/include -I/usr/include/python3.7m -c sage_numerical_backends_gurobi/gurobi_backend.c -o build/temp.cygwin-3.2.0-x86_64-3.7/sage_numerical_backends_gurobi/gurobi_backend.o sage_numerical_backends_gurobi/gurobi_backend.c:639:10: fatal error: gurobi_c.h: No such file or directory 639 | #include "gurobi_c.h" | ^~~~~~~~~~~~ compilation terminated. error: command 'gcc' failed with exit status 1 ---------------------------------------- ERROR: Command errored out with exit status 1: /opt/sagemath-9.3/local/bin/python3 -u -c 'import sys, setuptools, tokenize; sys.argv[0] = '"'"'/tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/setup.py'"'"'; __file__='"'"'/tmp/pip-install-b3w83q4_/sage-numerical-backends-gurobi_297d070e1a474477b24a1f3236b14fa2/setup.py'"'"';f=getattr(tokenize, '"'"'open'"'"', open)(__file__);code=f.read().replace('"'"'\r\n'"'"', '"'"'\n'"'"');f.close();exec(compile(code, __file__, '"'"'exec'"'"'))' install --record /tmp/pip-record-ibpz889r/install-record.txt --single-version-externally-managed --compile --install-headers /opt/sagemath-9.3/local/include/site/python3.7/sage-numerical-backends-gurobi Check the logs for full command output. Is there any solution to the problem? Or any other method of four color theorem implemented in Python? Grateful for your help and suggestions. REPLY [13 votes]: If you have have some specific, moderately large graphs that you want to color with four colors, you could try using a SAT solver. For each vertex $v$ and each integer $i\in \{1,2,3,4\}$, let $x_{v,i}$ denote a binary variable that is 1 if $v$ is assigned the color $i$ and 0 otherwise. Then for every vertex $v$, introduce the clauses $$x_{v,1} \vee x_{v,2} \vee x_{v,3} \vee x_{v,4}$$ to ensure that every vertex gets some color, and $$(\neg x_{v,i}) \vee (\neg x_{v,j}) \qquad \forall i \ne j$$ to ensure that no vertex gets assigned more than one color. The proper coloring constraint means that for every edge $(u,v)$ and every color $i$ you need a constraint $$(\neg x_{u,i}) \vee (\neg x_{v,i})$$ This might seem like a lot of variables and clauses, but I would expect that modern SAT solvers would have no trouble with graphs with 3000 vertices and 10000 edges. I usually use SAT solvers written in C, but I'm sure there are Python SAT solvers out there.<|endoftext|> TITLE: Is every set with finite $\mathcal{H}^{n-1}$ measure a set of locally finite perimeter? QUESTION [5 upvotes]: Given a measurable set $E \subset \mathbb{R}^d$, with $\mathcal{H}^{d-1} (\partial E) < +\infty$, is it true in general that $E$ is a set of locally finite perimeter? that is, is it true that $\int_B |D \chi_E| dx$ is finite, for every bounded ball $B \subset \mathbb{R}^d$? It is well-known in geometric measure theory that, in general, the perimeter $P(E)$ of a measurable set $E \subset \mathbb{R}^d$ does not equal to $\mathcal{H}^{d-1} (\partial E)$; unless $E$ has some nice regularity properties, for example when it has $C^2$ boundary. In any case, by De Giorgi's structure theorem, there is a set $\partial ^\ast E$, called the reduced boundary of $E$, which the equality $P(E)= \mathcal{H}^{d-1} (\partial^\ast E)$ holds. Recall that by definition, $P(E)<+\infty$ if the characteristic function $\chi_E$ belongs to the space $BV$ of functions with bounded variation. Thus, my question is about the existence of reduced boundary $\partial^ \ast E$ for a set $E$ with with $\mathcal{H}^{d-1} (\partial E) < +\infty$; rather than any claim about equivalence between the two boundaries. Thus it maybe true that $P(E)$ exists, but $P(E) \not = \mathcal{H}^{d-1} (\partial E)$. It must be said that, I guess the answer is negative, but I have no idea to prove it. REPLY [4 votes]: The reduced boundary can be defined for just about any (measurable) subset $E \subset \mathbf{R}^n$, whether it is a Caccioppoli set or not. The precise result you seem to be after should be Theorem 4.5.11 in Federer's book; in my edition this is on page 506. Let me just quickly restate it here. Define two sets $Q$ and $R \subset \mathbf{R}^n$ respectively as containing those points where the densities of $\mathcal{H}^n$ restricted to $E$ and $\mathbf{R}^n \setminus E$ respectively are zero. Paraphrasing slightly, Federer's theorem states that if $\mathcal{H}^{n-1}(K \setminus (Q \cup R)) < \infty$ for all compact $K \subset \mathbf{R}^n$, then $E$ is a Caccioppoli set. I haven't yet untangled the proof that Federer gives. Given that the statement seems a bit hard to find in other texts, I suspect it might be a bit technical. Remark 1. Note that the interior of $E$ belongs to $R$ and that of its complement is in $Q$, so that $\mathbf{R}^n \setminus (Q \cup R) \subset \partial E$. If I am not mistaken this gives the result you're looking for. Remark 2. A more precise re-telling of Federer's result would be stated in terms of what is called the $(n-1)$-dimensional integral-geometric measure with exponent $1$, which he denotes $\mathscr{I}_1^{n-1}$. However, this is comparable to the Hausdorff measure, up to a dimensional constant $\beta = \beta_1(n,n-1)$: $\mathcal{H}^{n-1} \geq \beta \mathscr{I}_1^{n-1}$. The immediately relevant sections in Federer's books are 2.10.5 and 2.10.6, on pages 172-174 in my edition.<|endoftext|> TITLE: Best approximation of L1 function by Lipschitz function QUESTION [7 upvotes]: Fix constant $L,C>0$ and $k\geq 1$ and let $f\in W^{1,k}(\mathbb{R}^d,\mathbb{R}^n)$ with $\|f\|_{W^{1,k}}\leq C$. Is there a known estimate on the distance $$ \|f - \operatorname{Lip}_L(\mathbb{R}^d,\mathbb{R}^n)\|_{L^1(\mathbb{R}^d,\mathbb{R}^n)}, $$ depending on the constants $L,C,$ and $k$, where $\operatorname{Lip}_L(\mathbb{R}^d,\mathbb{R}^n)$ is the set of Lipschitz functions from $\mathbb{R}^d$ to $\mathbb{R}^n$ with Lipschitz constant at-most $L$? REPLY [2 votes]: Let $(\rho_\epsilon)_{\epsilon>0}$ be a standard family of mollifiers, with $\rho_\epsilon$ supported in the ball $B_\epsilon(0)$. Since $\|\rho_\epsilon\|_{L^1}=1$ and $\|\rho_\epsilon\|_{L^\infty}=c_d\epsilon^{-d}$, by interpolation we get $\|\rho_\epsilon\|_{L^{k'}}=c_d^{1/k}\epsilon^{-d/k}$ (for the dual exponent $k'=\frac{k}{k-1}$). Hence, $$\|\nabla(\rho_\epsilon*f)\|_{L^\infty}=\|\rho_\epsilon*\nabla f\|_{L^\infty}\le c_d^{1/k}\epsilon^{-d/k}C$$ for your function $f$ (by Holder). In order to get an $L$-Lipschitz function you can take $$\epsilon:=c_d^{1/d}(C/L)^{k/d}=c_d'(C/L)^{k/d}$$ for another constant $c_d'$ depending only on $d$. Then you can bound the distance of $\rho_\epsilon*f$ from $f$ as follows: $$\|\rho_\epsilon*f-f\|_{L^{k}}\le\epsilon\|\nabla f\|_{L^{k}}\le c_d'(C/L)^{k/d}.$$ Note: the question of measuring the distance in $L^1$ seems ill-posed, since $W^{1,k}$ does not embed into $L^1$ (unless $k=1$).<|endoftext|> TITLE: "Burnside ring" of the natural numbers and algebraic K-theory QUESTION [12 upvotes]: The construction of the Burnside ring $A(G)$ of a group $G$ (usually, but not always, finite) is given by taking the Grothendieck group of the commutative semi-ring of isomorphism classes of finite $G$-sets, in which the addition is given by disjoint union and multiplication is given by cartesian products. However, it seems to me that the same construction appears to work for monoids $M$, where we consider now sets with (left) $M$-action. In particular, consider the case when $M = \Bbb N$ is the monoid of natural numbers. Then an $\Bbb N$-set $X$ whose underlying set is finite amounts to a set equipped with an endomorphism. Question 1 Has $A(\Bbb N)$ been computed? If so, how may it be described? Consider the category $\text{End}_S$, consisting of pairs $(K,f)$ in which $K$ has the homotopy type of a finite complex and $f: K \to K$ is a self-map. This is a Waldhausen category. Let $K(\text{End}_S)$ denote its $K$-theory. (It seems to me that there is a homomorphism $A(\Bbb N) \to K_0(\text{End}_S)$.) Question 2 What is the relationship, if any, between $A(\Bbb N)$ and (some sort of equivariant) stable homotopy? More precisely, consider the category $C_{\Bbb N}$ consisting of pairs $(X,f)$ in which $X$ is a finite set and $f$ is a self map of $X$. We consider the isomorphisms of such objects as the morphisms of $C_{\Bbb N}$. Then the classifying space (realized nerve) $|C_{\Bbb N}|$ is a topological monoid and we can form the group completion $$ \Omega B|C_{\Bbb N}|\, . $$ Question 2' does $\pi_0(\Omega B|C_{\Bbb N}|)$ coincide with $A(\Bbb N)$? What is the homotopy type of $\Omega B|C_{\Bbb N}|$? It seems that by considering finite set as a discrete space there is a map $$ \Omega B|C_{\Bbb N}|\to K(\text{End}_S) $$ Problem 3 Give a $K$-theoretic interpretation of the homotopy fiber of this map. REPLY [7 votes]: This is more an extended comment on @MaximeRamzi's answer. His comment reminded me of one of my favorite MO questions Distinguishing combinatorial maps by their linearizations and it explains the kernel of his map. (See also the paper Steinberg - Linear conjugacy inspired by this question which gives more details.) If $M$ is a monoid, let $A(M)$ be the Burnside ring of $M$ (as defined in this post) and let $R(M)$ be the representation ring of $M$, that is, the Grothendieck ring of the semiring of isomorphism classes finite dimensional $\mathbb CM$-modules under direct sum and tensor product. There is an obvious mapping $A(M)\to R(M)$ sending an $M$-set $X$ to the transformation module $\mathbb CX$. Note that cardinality induces a ring homomorphism $A(M)\to \mathbb Z$ and dimension induces a ring homomorphism $R(M)\to \mathbb Z$ and the map $A(M)\to R(M)$ is a homomorphism over $\mathbb Z$. I claim that the kernel of the map proposed by @MaximeRamzi in his answer is exactly the kernel of the map $A(\mathbb N)\to R(\mathbb N)$. Indeed, it is shown in my answer to the above-linked question using Fitting's decompositions and Brauer's lemma that two permutation matrices are conjugate as matrices iff they are conjugate as permutations, that if $f$, $g$ are maps on a finite set $X$, then the corresponding linear representations are equivalent iff the essential ranges of $f$ and $g$ are equal in $A(\mathbb Z)$ (i.e., have the same cardinality and cycle structure) and $\lvert f^n(X)\rvert=\lvert g^n(X)\rvert$ for all $n\geq 1$. It follows immediately from this that the kernel of the map $A(\mathbb N)\to R(\mathbb N)$ is generated by all $[f]-[g]$ where $f$, $g$ are maps on the sets of the same cardinality with the same cardinality and cycle structure essential ranges and the images of $f^n$ and $g^n$ have the same cardinality for all $n\geq 1$. But this is the kernel of the map at the end of @MaximeRamzi's answer. Added further commment There is a different ring you could associate to a monoid that is the $M$-set analogue of doing $G_0$ instead of $K_0$. For groups this gives the usual Burnside ring but for monoids it is different. I strongly suspect this is what is done in the paper Erdal and Ünlü - Semigroup Actions on Sets and the Burnside Ring I linked to in the comments but I don't know. The idea is that in a $G$-set $X$ for a group $G$, if $Y\subseteq X$ is $G$-invariant, then so is $X\setminus Y$ and you can write $X=Y+X\setminus Y$ in the Burnside ring. But this doesn't work for monoids. You can have noncomplemented invariant subsets. So one work around it to define quotient of $A(M)$ by the relations that if $Y$ is an $M$-invariant subset of $X$, then $[X] = [Y]+[X/Y]$. But I think this doesn't quite work well because $X/Y$ is a pointed $M$-set. So my feeling is that one should work in the category of pointed $M$-sets and that in the associated semiring, the action consisting of just a base point should be considered $0$. The product of pointed $M$-sets is the usual direct product mod the invariant subset of points with one or more coordinate a base point. The sum identifies base points. For groups, a pointed $G$-set is basically the same thing as a $G$-set by adding a base point fixed by the group. If we do this version of things for $\mathbb N$, then finite pointed $\mathbb N$-sets are partial maps on a finite set. If you do the Grothendieck ring of the these things with pointed-direct sum and pointed direct product, you will get something complicated. But if you factor by the relations $[X]=[Y]+[X/Y]$ for invariant $Y$, then you will get $A(\mathbb Z)\times \mathbb Z$ as an abelian group and the $\mathbb Z$ copy is a two-sided ideal, generated by the pointed set $A$ consisting of the base point and one other point that is mapped to the base point. The product of a cycle with $n$ vertices (plus a base point thrown in) with $A$ is the sum of $n$ copies of $A$. From the partial mapping view point this is the fact that a strongly connected partial function digraph is either a single point with no edge or a cycle.<|endoftext|> TITLE: Cohomology ring of mapping torus QUESTION [15 upvotes]: A mapping torus, $M \rtimes_\varphi S^1$, is a fiber bundle over $S^1$ with fiber $M$, where $\varphi$ is an element of mapping class group of $M$, describing the twist around $S^1$. For $M=S^1\times S^1 = T^2$, where the two $S^1$ are parametrized by $x$ and $y$, the map $\varphi$ is given by $$ \begin{pmatrix} x\\ y\\ \end{pmatrix} \to \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} ,\ \ \ \ a,b,c,d \in \mathbb{Z}, \ ad-bc =\pm 1. $$ What is the cohomology ring of the mapping torus $M \rtimes_\varphi S^1$ in terms of $ \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$? added: I mean to ask the ring $H^∗(M;\mathbb{Z})$ and $H^∗(M;\mathbb{Z}_n)$. REPLY [7 votes]: Here are some more partial observations, overlapping some of the others. The result won't be completely explicit, as there are too many case distinctions for a clean answer, but hopefully at least the outline is right. The additive structure can be obtained from the Mayer–Vietoris sequence associated to the covering of your mapping torus $Y$ by two cylinders $U$ and $V$ homeomorphic to $T \times (0,1)$, which on rearrangement will yield (the cohomological analogue of) the result quoted from Hatcher. The map $H^1(U) \oplus H^1(V) \to H^1(U \cap V)$ is given by $f\colon (a,b) \mapsto (\varphi(b)-a,b-a)$, if we identify $H^1$ of all the solid tori in sight with $\mathbb Z^2$. The contribution of this kernel to $H^1(Y)$ is given by pairs $(a,b)$ with $a = b = \varphi(b)$; in any event, this summand is free abelian. There is another $\mathbb Z$ summand arising as the cokernel of $\delta\colon H^0(U \cap V) \to H^1(Y)$. The map $H^2(U) \oplus H^2(V) \to H^2(U \cap V)$ is similarly given by $g\colon (m,n) \mapsto (\det(\varphi)n-m,n-m)$ if we identify $H^2$ of each solid torus with $\mathbb Z$. Thus $H^3(Y) = \mathrm{im}\, \delta$ is $\mathbb Z$ if $\det \varphi = 1$ and $\mathbb Z / 2\mathbb Z$ if $\det \varphi = -1$. Now $H^2(Y)$ fits into a short exact sequence $$0 \to \mathrm{coker}\, f \to H^2(Y) \to \ker g \to 0.$$ As $\ker g$ is free abelian (either $\mathbb Z$ or $0$ depending on the sign of $\det g$, this sequence will split, with all torsion coming from $\mathrm{coker}\, f$. The product $H^2(Y) \times H^1(Y) \to H^3(Y)$ will be determined on the free summands by Poincar'e duality if $\det \varphi = 1$. The product of any element with an element $\delta a$ of $\mathrm{im}\,\delta \leq H^1(Y)$ will also be determined by $b \smile \delta a = \delta (b \cdot a)$, where the action of $H^*(Y)$ on $H^*(U \cap V)$ is by restriction followed by cup product (I may be getting the left- vs. right-sidedness wrong—otherwise one picks up a sign—but the Mayer--Vietoris connecting map is, in general, $H^*(Y)$-linear, for $Y$ the covered space.). The same reasoning also covers products with an element of the image of $g$. I think these considerations together determine most of the product structure, although they do not say anything about $H^1 \times H^2$ on the free parts arising from $\ker f$ and $\ker g$, if $Y$ is nonorientable, nor much about $\ker f \times \ker f$. It might be worth working out a specific example or two to see how bad the remaining ambiguities are.<|endoftext|> TITLE: Pullback and pseudoelements QUESTION [8 upvotes]: Let $\mathcal{A}$ be an abelian category, and let $X$ an object of $\mathcal{A}$. Recall that a pseudoelement of $X$ is an equivalence class of arrows $X_1 \to X$, where $x_1 \colon X_1 \to X$ and $x_2 \colon X_2 \to X$ are equivalent if there is an object $P$ and epimorphisms $p_1 \colon P \to X_1$ and $p_2 \colon P \to X_2$ such that $x_1 \circ p_1= x_2 \circ p_2$. We write $x \in^\ast X$ to denote that $x$ is a pseudoelement of $X$. If $x \in^\ast X$ and $f \colon X \to Y$, then $f(x) \in^\ast Y$ is obtained by composition. Pseudoelements are pretty nice, for example a morphism is equal to $0$ if and only if it sends all pseudoelements to $0$, and we can characterize monomorphisms (resp. epimorphisms) in terms of pseudoelements, as one expects. Similarly we can check if a sequence is exact by testing exactness on pseudoelements. Of course there are limitations, for example we can not test if two morphisms are equal on pseudoelements (in contrast with the Freyd–Mitchell embedding theorem). A natural problem is what happens with pullbacks. Let $f \colon X \to Z$ and $g \colon Y \to Z$ be morphisms, and let $x \in^\ast X$ and $y \in^\ast Y$ such that $f(x) = g(y)$. It's easy to construct $s \in^\ast X \times_Z Y$ such that $\pi_1(s) = x$ and $\pi_2(s) = y$, by universal property of the pullback. Now, we would like $s$ to be unique, and indeed Francis Borceux in his "Handbook of Categorical Algebra: Volume 2", Proposition 1.9.5, claims so. The proof is only sketched, and essentially Borceux says "All the relevant epimorphisms can, by successive pullbacks, be replaced by epimorphisms with the same domain, from which the claim follows". It's not completely clear how to write down the details, see for example this stackexchange question, with a partial answer. I've tried to prove the theorem by myself, essentially following Borceux's idea of constructing several pullbacks, but I am unable to finish the proof. To simplify the notation, let's consider the case of the product, so $Z=0$ in the above discussion. Let $a \colon A \to X \times Y \in^\ast X \times Y$ and $b \colon B \to X \times Y \in^\ast X \times Y$ be such that $\pi_1(a)=\pi_1(b)$ and $\pi_2(a)=\pi_2(b)$. So there are epimorphisms $a_1 \colon Z_1 \to A$, $b_1 \colon Z_1 \to B$, $a_2 \colon Z_2 \to A$ and $b_2 \colon Z_2 \to B$ such that $\pi_1 \circ a \circ a_1 = \pi_1 \circ b \circ b_1$ and $\pi_2 \circ a \circ a_2 = \pi_2 \circ b \circ b_2$. To construct the required epimorphisms we consider $P_a = Z_1 \times_A Z_2$ and $P_b = Z_1 \times_B Z_2$. We finally set $P = P_a \times_{Z_1} P_b$, where $P_a \to Z_1$ and $P_b \to Z_1$ are the first projections. We write $f_a : P \to A := a_1 \circ \pi_1 \circ \pi_1$ (two different $\pi_1$) and $f_b : P \to B := b_1 \circ \pi_1 \circ \pi_2$. Then $f_a$ and $f_b$ are epimorphisms. We need to prove that $a \circ f_a= b \circ f_b$ and it is enough to prove that $\pi_1 \circ a \circ f_a= \pi_1 \circ b \circ f_b$ and $\pi_2 \circ a \circ f_a= \pi_2 \circ b \circ f_b$. The first one follows immediately by $\pi_1 \circ a \circ a_1 = \pi_1 \circ b \circ b_1$ and the commuting square of the pullback. I don't see how to prove the second equality, I tried essentially all the combinations of the two equalities we have and all the pullback squares, but it didn't work. Of course one can try to construct different epimorphisms, but all the constructions I tried (even those more symmetric then the one above) end up with exactly the same equality to prove. Does someone know a complete proof of this result? Thanks! PS: I realized I am not able to prove it trying to formalize the result in Lean, in a lemma for the Liquid Tensor Experiment, you can see the actual Lean code here. REPLY [8 votes]: The claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear, but seems wrong. To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times_Z Y$ with $\pi_1[p] = [x]$, $\pi_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.) Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x_1 \colon X_1 \to X$, $x_2 \colon X_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, note that in this case the projections $X_1 \times_X X_2 \to X_i$ are epi.) So pseudo-elements of $X$ correspond to subobjects/subgroups of $X$. Borceux notes this fact in the closing discussion of §1.9. But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same. In terms of pseudo-elements, the maps $s_1, s_2 : \Q \to \Q \times \Q$ given by $s_i(x)=(x,ix)$ are not equal as pseudo-elements of $\Q \times \Q$, but their images under each projection are equal as pseudo-elements of $\Q$.<|endoftext|> TITLE: When is the rank of $AB+BA$ equal to one? QUESTION [7 upvotes]: For two arbitrary matrices $A$ and $B$, are there any known conditions for the rank of $AB+BA$ to be equal to one? REPLY [2 votes]: Consider the function $f:X\in M_n(\mathbb{C})\mapsto AX+XA$, where $A\in M_n(\mathbb{C})$ and $spectrum(A)=(\lambda_i)$. Then $f=A\otimes I_n+I_n\otimes A^T$, where the matrices are stacked into vectors rows by rows, cf. https://en.wikipedia.org/wiki/Kronecker_product and $spectrum(f)=\{\lambda_i+\lambda_j;i,j\}$. Finally, $f$ is a bijection iff $A$ satisfies the property $\mathcal{P}$: for every $i,j$, $\lambda_i+\lambda_j\not= 0$. Here, we want to satisfy an equation in the form $AB+BA=U$, where $U$ is a given matrix with rank $1$. The algebraic set $W=\{Y;rank(Y)\leq 1\}$ is Zariski closed and has dimension $2n-1$; $V=\{Y;rank(Y)=1\}$ is a Zariski open dense subset of $W$ and therefore is a quasi-affine variety (or a constructible set or simply a variety). Let $E_n=\{(A,B)\in M_n;rank(AB+BA)=1\}$. Consider a generic matrix $A$; then $A$ satisfies $\mathcal{P}$ and, for every $U\in V$, there exists exactly one matrix $B_U$ s.t. $AB_U+B_UA=U$. Note that $B_U$ is simply the solution of a linear equation, and can be easily computed. Then the local dimension of $E_n$ in such a point is $n^2+2n-1$ and the dimension of $E_n$ is $d_n=n^2+2n-1$; yet, the maximal dimension may be reached using matrices $A$ that do not satisfy $\mathcal{P}$. For example, assume that $n=4$ ($d_4=23$) and consider $A=diag(u,-u,v,-v)$, where $u,v$ are distinct non-zero; here $f$ is diagonalizable and $dim(\ker(f))=4$; more precisely, the matrices in $\ker(f)$ are in the form $\begin{pmatrix}0&a_1&0&0\\a_2&0&0&0\\0&0&0&b_1\\0&0&b_2&0\end{pmatrix}$. $\textbf{Remark.}$ Note that we can see the dimension of the algebraic set $F_n=\{(A,B);AB+BA=0\}$ in the neighborhood of such matrices; the stabilizer $\{P;P^{-1}AP=A,P^{-1}BP=B\}=\{diag(a,a,b,b);a,b\}$ has dimension $2$; then the dimension of the orbit is $4^2-2$; we add the variations of the entries $u,v$ and $a_1,a_2,b_1,b_2$ where the products $a_1a_2,b_1b_2$ are fixed; the total is $dim(F_4)=14+2+2=18$ and, more generally, $dim(F_n)=floor(n^2+n/2)$. Now we take $A=diag(1,-1,2,-2),\tilde{B}=\begin{pmatrix}0&1&0&0\\2&0&0&0\\0&0&0&3\\0&0&4&0\end{pmatrix}$. To obtain an element of $E_4$, we choose $B=\tilde{B}+H$, where $rank(H)=1$ and $AH=HA$; for example $H=diag(0,0,0,1)$. Using the Grobner basis theory , we can show -with the help of a computer- that the local dimension of $E_4$ in $(A,B)$ is also $23$.<|endoftext|> TITLE: Topological vector spaces (reference request) QUESTION [13 upvotes]: In his book Topological Function Spaces Arhangel'skii says that "it is well known that every nontrivial locally convex linear topological space $X$ is homeomorphic to a space of the form $Y \times \mathbb{R}$, for some space $Y$". I've been looking for a proof of this result but haven't found anything, which leads me to believe that this is a standard fact that everyone knows. Could any of you please help me with a reference in which this result appears? REPLY [17 votes]: I guess that non-trivial means that the locally convex space $X$ is not endowed with trivial topology $\{\emptyset,X\}$. This implies that $X\neq \overline{\{0\}}$ (since this closure does have the trivial topology). For $y\notin \overline{\{0\}}$, the Hahn-Banach theorem (applied to $L=\{ty: t\in\mathbb K\}$ and the linear functional $ty\mapsto t$ which is continuous because there is a continuous seminorm $p$ on $X$ with $p(y)>0$) implies that there is a continuous linear functional $\varphi\in X'$ with $\varphi(y)=1$. Then $X$ is isomorphic (in the category of topological vector spaces which is much more than homeomorphic) to $Y\times \mathbb K$ for the kernel $Y$ of $\varphi$, an isomorphism is given by $x\mapsto (x-\varphi(x)y,\varphi(x))$. If, for complex locally convex spaces, you insist on a homeomorphy to $Y\times \mathbb R$ identify $\mathbb C=\mathbb R\times \mathbb R$. Since this is such a simple application of Hahn-Banach, I doubt that it is worthwhile to search for an explicit reference. REPLY [9 votes]: This fact is a consequence of Hahn-Banach separation theorem (HBST). Such a space admits a continuous linear functional. Let us be more specific: let $V$ be a Hausdorff HLCTVS (Hausdorff locally convex topological vector space) over $\mathbb{R}$. Let $v\in V\setminus \{0\}$ and $U$ be a convex neighbourhood of zero which does not contain $v$. From HBST, it exists a linear fonctional $\varphi$ such that $\varphi(U)=I\subset \mathbb{R}$ is an open interval and $\varphi(v)\notin I$. Then $\varphi$ is continuous and $H=\ker(\varphi)$ is closed. Then the map $p:\ V\to V$ such that $p(x):=x-\varphi(x)/\varphi(v).v$ is a continuous projection $V\to H$. The map $x\mapsto (p(x),\varphi(x))$ is an isomorphism $V\to H\times \mathbb{R}$. Late edit: Jochen is right, Hausdorff is too strong, non-trivial is sufficient. One has just to modify the proof above considering that, if $V$ is non-trivial $\overline{\{0\}}\neq V$, then choose your $v\notin \overline{\{0\}}$, the remainder of the proof is unchanged. Remark 1 The converse is true in the following manner. A LCTVS (locally convex topological vector space) $V$ can be decomposed as $V\simeq H\times \mathbb{R}$ iff its topology is not the coarsest one. Remark 2 As regards the "reference-request" aspect of the question, you can read N. Bourbaki, Topological Vector Spaces, Chapters 1-5 (13 november 2002, Springer). Where, as the treatment is systematic, you get: Non-Hausdorff LCTVS, connection with coarse topology (treated in General Topology of the same treatise) and two forms of Hahn-Banach theorem as well.<|endoftext|> TITLE: A lecture by Rudin QUESTION [7 upvotes]: Is it available a written version of this lecture by Rudin on the relation between Fourier analysis and the birth of set theory? https://youtu.be/hBcWRZMP6xs If not Rudin himself, maybe someone else has collected these ideas in an article? REPLY [8 votes]: I extracted the written text from Rudin's lecture, it starts out like this (see below). If there is an interest, with some effort I can provide the rest of the text, but it requires some editing and I'm unsure this is what the OP wants... Perhaps just listening to the talk is good enough? It is a real honor to be invited to give this lecture and I'm sorry professor Marden couldn't be here to listen to it. I'm a bit out of my depth here today because I intend to give a talk which is more historical than mathematical. I was told that there would be many non mathematicians here in the audience and that I should not write too many formulas. A warning: the history of mathematics is a difficult field and it's one that has been neglected and I think there are good reasons for that. Mathematicians don't pay much attention to history. We name theorems after someone and sometimes that someone has something to do with the field and sometimes he didn't. Sometimes somebody later proved what the first guy should have proved if you see what he was really doing, and we don't ever go to the first original sources. I think we read what someone said about it, so mathematicians are not really qualified to do the history of mathematics. Historians are maybe even less qualified because you can't write about it unless you understand it and once you start studying mathematics for its own sake you become interested in it and you don't want to take time to go to the history.<|endoftext|> TITLE: Extending sets to extensional sets QUESTION [14 upvotes]: Let us say a set $X$ is extensional if, for all distinct $a,b$ in $X$, $X$ contains a member of the symmetric difference of $a$ and $b$. (id est $\langle X,=,\in\rangle$ is a model of the axiom of extensionality). My question is: ``Does every finite set have a finite extensional superset?'' I thought the answer should be an easy `yes' but i haven't been able to find a proof (even allowing myself both foundation and choice!) and i'm now beginning to doubt it. Of course now that i have posted this question i shall see the answer immediately! At some point, if people are interested, i shall narrate how this question arose.... REPLY [23 votes]: No, here is a counterexample (I thought about this question some time ago and also expected it would be easy to prove, but in the end couldn't do it, and arrived at this counterexample): We will construct a set $p = \{x, y\}$ with no extensional finite superset. First define a set $n^*$ for $n<\omega$ recursively by: $0^* = \emptyset$, $1^* = \{0^*\}$, $2^* = \{1^*\}$, $3^* = \{0^*, 2^*\}$, $4^* = \{1^*, 3^*\}$, etc, so $(2n+1)^* = \{0^*, 2^*, \ldots, (2n)^*\}$, $(2n+2)^* = \{1^*, 3^*, \ldots, (2n+1)^*\}$. Then let $x = \{0^*, 2^*, \ldots, (2n)^*, \ldots \}$ $y = \{1^*, 3^*, \ldots, (2n+1)^*, \ldots \}$ and $p = \{x, y\}$. Suppose that $p \subseteq q$ and $q$ is finite and extensional. There must be some $n^*$ in $q$, in order to distinguish between $x$ and $y$. Let $N$ be largest such that $N^*$ is in $q$. If $N = 2n+1$ then $N^* \cap q = x \cap q$, but then $q$ is not extensional. Likewise with $y$ if $N = 2n$. Contradiction. (Remark: e.g. let $q = \{x, y, 0^*\}$. Then $q$ is "weakly extensional", meaning that there are no membership-automorphisms of $q$.)<|endoftext|> TITLE: Algebraic varieties with certain topological properties QUESTION [7 upvotes]: Does there exist a smooth, projective, complex algebraic variety $X$, with two cohomology classes $\alpha,\beta \in H^{*}(X,\mathbb{Z})$ neither $\alpha$ nor $\beta$ is torsion but the product $\alpha \cup \beta$ is non-trivial and torsion? REPLY [8 votes]: Let $Y$ be a smooth complex projective surface whose cohomology group $H^2(Y;\mathbb{Z})$ has a nonzero torsion element $\gamma$ and whose torsion-free quotient has rank at least $2$, e.g., this holds for an Enriques surface. Let $D$ and $E$ be nontorsion elements whose cup product is zero. By the Hodge Index Theorem, such elements exist. Let $X$ be $Y\times Y$. Let $\alpha$ be $\text{pr}_1^*\gamma +\text{pr}_2^*D$. Let $\beta$ be $\text{pr}_2^*E$.<|endoftext|> TITLE: Relating the holomorphic Euler characteristic of a family of algebraic varieties to properties of the base and fibers QUESTION [7 upvotes]: Let $f : X\rightarrow Y$ be a proper flat morphism (of schemes) with connected fibers over a smooth projective curve $Y$ over $\mathbb{C}$. Let $X_{y_0}$ denote a smooth fiber over $y_0\in Y$. If $f$ is smooth, then the topological Euler characteristic is multiplicative: $\chi(X) = \chi(Y)\cdot\chi(X_{y_0})$. In general, there is a formula for the topological Euler characteristic which includes some correction terms for the singular fibers: $$\chi(X) = \chi(Y)\chi(X_{y_0}) + \sum_{y\in Y}(\chi(X_y) - \chi(X_{y_0})$$ Does there exist an analogous formula for the holomorphic/algebraic Euler characteristic? I'm in particular interested in the case where $X\rightarrow Y$ is a family of curves over $Y$ (so $\dim X = 2$). To be precise, I'm looking for a formula for $\chi(X,\mathcal{O}_X)$ in terms of geometric invariants of $Y$ and some fibral data. By fibral data I mean geometric properties which are local on the base $Y$ (e.g. geometric invariants of fibers, monodromy around singular fibers acting on the cohomology of a smooth fiber...etc). As a nonexample, Riemann-Roch relates this Euler characteristic to the intersection number of $X_{y_0}$ with $K - X_{y_0}$ (which can be understood locally on $Y$) plus $\chi(X,\mathcal{O}_X(X_{y_0}))$ (which does not appear to be local on $Y$). Added in edit: To state a more precise question -- Is the holomorphic Euler characteristic $\chi(X,\mathcal{O}_X)$ determined by $Y$ and data which is local on $Y$? Stated another way, do there exist maps $f : X\rightarrow Y$ and $f' : X'\rightarrow Y$ (satisfying the above conditions) such that there is a covering in the etale topology $\{U_i\rightarrow Y\}_{i\in I}$ and isomorphisms $\phi_i : X_{U_i}\rightarrow X'_{U_i}$, and $\chi(X,\mathcal{O}_X)\ne \chi(X',\mathcal{O}_{X'})$. Here I'm happy to assume that $X,X'$ are smooth over $\mathbb{C}$ (but I still want to allow singular fibers of $f,f'$). If it changes the answer, I'm also interested in the variation where $\{U_i\}$ is instead an open covering in the analytic topology. I am also interested in results where we assume that $X,X'$ are moreover general type. If the euler characteristic is not determined by $Y$ and data local on $Y$, then I'm also interested in any contraints such data might impose on $\chi(X,\mathcal{O}_X)$. REPLY [5 votes]: I'm not entirely sure what would constitute an answer. But here a few simple observations. Let me focus on what you seem be interested in, namely a projective family of connected curves over smooth projective curve $f:X\to Y$. Let me also assume $X$ smooth. Let $h$ be the genus of $Y$, and $g$ the genus of the general fibre. Then from the Leray spectral sequence, Riemann-Roch and Grothendieck duality $$\chi(\mathcal{O}_X)= (1-g)(1-h) - \deg R^1f_*\mathcal{O}_X=\chi(\mathcal{O}_Y)\chi(\mathcal{O}_{X_{y_0}})+\deg f_*\omega_{X/Y}$$ A theorem of Fujita then gives an inequality $$\chi(\mathcal{O}_X)\ge \chi(\mathcal{O}_Y)\chi(\mathcal{O}_{X_{y_0}})$$ which is something. To say more, one would need to compute $\deg f_*\omega_{X/Y}$. But I think this depends on more than local information at the bad fibres, which is what I think you are asking. The reason I say this, is that this degree can be positive and variable for Kodaira surfaces, which have no bad fibres at all. [Added Comment I guess there are various things called Kodaira surfaces. The ones I have in mind are of general type, and hence algebraic. See page 220 of Compact Complex Surfaces by Barth, Hulek, Peters and Van de Ven for further details.]<|endoftext|> TITLE: Is it known whether a closed simply-connected manifold of non-negative curvature admits positive Ricci? QUESTION [6 upvotes]: It is discussed in this question whether a simply-connected closed Riemannian manifold with non-negative Ricci curvature admits positive Ricci curvature, and the answer appears to be "no, there are counter-examples and known obstructions". My question is this: are there any known examples of closed, simply-connected manifolds of non-negative sectional curvature that do not admit positive Ricci curvature? Are there known obstructions? REPLY [3 votes]: No. Every simply connected compact manifold of nonnegative sectional curvature admits positive Ricci curvature. This has been proven by Boehm--Wilking: https://link.springer.com/article/10.1007/s00039-007-0617-8 Theorem A. Let $(M^n,g)$ be a compact Riemannian manifold with finite fundamental group and nonnegative sectional curvature. Then $M^n$ admits a metric with positive Ricci curvature.<|endoftext|> TITLE: On a variant of Carlson’s theorem QUESTION [10 upvotes]: My question is on whether or not there exists some monotone strictly decreasing sequence of positive numbers $c_1>c_2>\ldots$ such that given any $f$ which is a uniformly bounded holomorphic function in the right half of the complex plane with $$ |f(k)|\leq c_k \quad \forall\, k\in \mathbb N,$$ there holds $ f(z)=0$ on the right half plane. REPLY [8 votes]: Condition $$\lim_{n\to\infty}\frac{\log|c_n|}{n}=-\infty$$ is sufficient for $f=0$. Since $f(z)=e^{-cz}$ and $c_n=e^{-cn}$ satisfy all conditions, we see that this is best possible in certain sense. This follows for example from a (much more general) theorem of N. Levinson, Gap and density theorems, AMS, 1940, page 121. Levinson's theorem allows some growth of $F$, and much more general class of sequences instead of integers. Remark. In fact Levinson generalizes a theorem of Vladimir Bernstein 1932 (Theorem 32 in Levinson's book), which also implies the result that I stated.<|endoftext|> TITLE: Primes in arithmetic progressions: weak version of Linnik's theorem with prime power modulus? QUESTION [5 upvotes]: Looking at a problem in representation theory I ran into a question on small primes in arithmetic progressions. Let me begin with a short summary of results on small primes in arithmetic progressions. By Linnik's theorem there are constants $c,L$ such that for every $d \geq 2$ and $1\leq a 0$. A folklore conjecture (sometimes attributed to Chowla, sometimes to Heath-Brown) states that $L = 1+\epsilon$ for all $\epsilon$. Fix a prime number $p$ and fix $a$, say $a=1$. For my purposes the only relevant case is when $d$ a power of the fixed prime number $p$. In this case stronger results are known. Let $L(p)$ be defined as $$ L(p) = \limsup_{j \to \infty} \frac{\log(p_{\text{min}}(p^j,1))}{j \log(p)}. $$ In other words $L(p)$ is the infimum over all real numbers $L> 0$ such that $p_{\text{min}}(p^j) \leq c_L p^{jL}$ for some $c_L > 0$ and all $j \geq 1$. Barban, Linnik and Tshudakov proved $L(p) \leq \frac{8}{3}$. Gallagher established $L(p) < 2.5$ and Huxley improved this to $L(p) \leq 2.4$. The best bound I am aware of can be found in a paper of Banks-Shparlinski: $L(p) < 2.1115$. My question is: what can be said if $\limsup$ is replaced by $\liminf$? Let's define $$ K(p) = \liminf_{j \to \infty} \frac{\log(p_{\text{min}}(p^j,1))}{j \log(p)}.$$ Clearly, $K(p) \leq L(p)$. So according to the strongest conjectures on $L(p)$ one would have $K(p) =1$. However, it seems possible that one can approach $K(p)$ with different methods. Put differently: one "only" needs to show that for infinitely many $j$ there is a small prime in the arithmetic progression $\equiv 1 \bmod p^j$ Question: Are there upper bounds for $K(p)$ which are better than the known bounds for $L(p)$? (For me the case $a = 1$ is sufficient, but I don't see how this might be useful.) REPLY [2 votes]: With $p_{\min}(d,a)$ as the OP defines it, let us take $$p_{\min}(d)=\max_{(a,d)=1}p_{\min}(d,a).$$ Li, Pratt, and Shakan proved (see their Theorem 1.1) that for all $0<\varepsilon<\frac{1}{2}$, there exists $d(\varepsilon)>0$ such that if $d>d(\varepsilon)$ and $d$ has no more than $$\exp\Big(\Big(\frac{1}{2}-\varepsilon\Big)\frac{(\log\log d)(\log\log\log\log d)}{\log\log\log d}\Big)$$ distinct prime factors, then $$p_{\min}(d)\gg \varphi(d)\frac{(\log d)(\log\log d)(\log\log\log\log d)}{\log\log\log d}.$$ Take $d=p^j$. Then $d$ has one distinct prime factor, and $\varphi(d)=p^j-p^{j-1}$. From this, it follows that $K(p)\geq 1$.<|endoftext|> TITLE: Why is this space contractible? QUESTION [10 upvotes]: Is the following space, obtained by glueing a Cantor set worth of "hairs" to a closed disk in $\Bbb R^2$ contractible? The obvious attempt of contracting the hairs to the root and then contracting the disk doesn't look continuous on the boundary of the disk. Motivation: While looking for an example providing an answer to a question I posted a couple of days ago I stumbled on this construction of a compact contractible space which is nowhere locally connected, attributed to Robert Edwards but only published as an abstract of the AMS that I cannot find online. Showing that the space constructed by Edwards is contractible boils down to showing that the space depicted above is. REPLY [8 votes]: Calling the space $X$, you can consider the homotopy $f_t:X\to X$ such that $f_t$ rotates every point an angle of $-t$ around the origin. For points of the hairs you rotate them keeping them inside their hair, and after they reach the 'singular point', they keep rotating in the boundary of the disk. Then $f_0$ is the identity and $f_{2\pi}(X)$ is contained in the disk, so $X$ is contractible.<|endoftext|> TITLE: Factorization of an irreducible polynomial in the field extension it defines QUESTION [7 upvotes]: In field theory, the following fact is used in the construction of splitting fields: Given a field $F$ and an irreducible polynomial $f \in F[x]$, the quotient $F[\alpha]/(f(\alpha))$ is a field extension of $F$ which contains a root of $f$ (namely the congruence class of $\alpha$). Let $n$ be a positive integer and let $\lambda_{1} + \dotsb + \lambda_{k} = n$ be a partition of $n$. Does there exist a field $F$ and an irreducible polynomial $f \in F[x]$ of degree $\deg f = n+1$ such that, if we define $K := F[\alpha]/(f(\alpha))$, the factorization of $f$ in $K[x]$ is of the form $f = (x-\alpha) \cdot f_{1} \dotsb f_{k}$ where each $f_{i} \in K[x]$ is irreducible and $\deg f_{i} = \lambda_{i}$? REPLY [14 votes]: Let us show that for the partition $2+1+1=4$, there is no such $f$. If $f$ were inseparable, then over $K$ it would factor as a constant times $(x-\alpha)^5$. If $f$ were separable, its Galois group $G$ would be a transitive subgroup of $S_5$ containing a transposition, so $G=S_5$, but then the stabilizer of a point would act transitively on the other four points, so $f$ would factor over $K$ into $x-\alpha$ and an irreducible polynomial of degree $4$.<|endoftext|> TITLE: Mathematical life of Friedrich Ulmer QUESTION [12 upvotes]: In the little galaxy of Category Theory, Friedrich Ulmer is known for being one of the authors of Lokal Präsentierbare Kategorien, a book that laid the foundations for the theory of locally presentable categories. Unlike his coauthor, Pierre Gabriel, we do not have much information about him, or at least I can't find much. From the Genealogy Project we know that he did his PhD in Zürich in 1964 under Dold and van der Waerden. According to Scopus, we only have 7 documents by Ulmer, even though other documents can be found with a bit of struggle on the internet. With a bit of help from google and semanticscholar we find a couple of more entrencies. zbMath lists a total of 12 publications by Ulmer. I can't find a complete bibliography of his, nor I can tell whether he spent his life in Academia or outside of it. Does anybody have better luck then me? Or happens to have more information? What I was looking for is something like this one or this one. Of course, there might be nothing more to say. REPLY [13 votes]: I am not sure for how long and where you looked, but it takes less than 30 minutes to figure much more than you mention in your post. Apparently, he continued his academic career as Fritz Ulmer : if you look at the affiliation of the last indexed MathReviews paper Localizations of endomorphism rings and fixpoints, Journal of Algebra 43 Issue 2 (1976) Pages 529–551, https://doi.org/10.1016/0021-8693(76)90125-3, you see that he was in Wuppertal, and then a quick Google search shows that "Fritz Ulmer = Friedrich Ulmer": Look at https://idw-online.de/de/news52962 and examine the part "Kontakt", and Look at https://www.researchgate.net/profile/Fritz-Ulmer . The website http://www.wahlprognosen-info.de/ seems to contain some of his interviews on matters of statistics etc., see http://www.wahlprognosen-info.de/index2.htm?/archiv/BB7.htm .<|endoftext|> TITLE: Elegant proofs of $\bar{\partial}z^{-1} = 2\pi \delta_0$ QUESTION [6 upvotes]: For a function $f(x,y)$ on $\mathbb{R}^2,$ defined possibly outside the origin, write $$\int_\epsilon ' f \,dx\,dy : = \int_{\mathbb{R}^2\setminus D_\epsilon}f \, dx\,dy,$$ (the integral on the complement to the $\epsilon$-disk) and $$\int' f \,dx \,dy : = \lim_{\epsilon\to 0} \int'_\epsilon f \,dx\,dy,$$ when defined. We view $\mathbb{R}^2$ as the complex line with coordinate $z = x+iy$. Then I claim that the assignment $$\phi:f(x,y) \mapsto \int_{\mathbb{R}^2}' f\cdot z^{-1} \,dx\,dy$$ makes sense as a functional on compactly supported, smooth functions (indeed, if $f$ is rotationally symmetric then $\phi(f) = 0$ even before taking the limit, and if $f(0) = 0$ then $f\cdot z^{-1}$ is bounded, hence integrable, at $0$; now in the space of smooth compactly supported functions, any function can be written as a rotationally symmetric function plus a function that vanishes at the origin). We can thus formally define a distribution $z^{-1}\in C^{-\infty}(\mathbb{R}^2)$ as $$z^{-1}: = \frac{\phi}{dx\,dy}.$$ Now write $\bar{\partial} = \partial_x + i \partial_y$ .Like any vector field, this acts on distributions, and so we have a distribution $\bar{\partial} z^{-1}.$ Since $z^{-1}$ is holomorphic where it is smooth, we must have $\bar{\partial} z^{-1}$ be a distribution supported at the origin. In fact, it is known to mathematical physicists that the result is a delta function at the origin: $$\bar{\partial} z^{-1} = 2\pi \delta_0,$$ and this fact is useful in conformal field theory. I would like to see a proof of this result (the physics sources I have seen do not prove this). In fact, I know how to give one using a direct calculation with polar coordinates: but this seems ad hoc and is not very satisfying to me. I suspect there might be "nicer" proofs using one or more of the following three techniques, and I am hoping that more analytically literate MO users can provide them. If one can show that $\bar{\partial} z^{-1}$ is determined by its values on holomorphic (near the origin) functions, the Cauchy residue formula would imply that $\bar{\partial} z^{-1} = 2\pi \delta_0.$ I suspect there should be a proof using the stationary phase approximation. This is probably overkill, but it would be nice if there were a proof using pseudodifferential operators. REPLY [5 votes]: Following Gelfand-Graev, Grothdieck, and Schwartz: The right-most pole of $u_s=|z|^{2s}=z^s\cdot \overline{z}^s$ (or, properly, the distribution given by integration-against that function) is at $s=-1$, when the function ceases to be $L^1_{\mathrm{loc}}$. As usual, $$ u_s(f) \;=\; u_s(f-f(0)\cdot e^{-z\overline{z}})+f(0)\cdot u_s(e^{-z\overline{z}}) $$ The value $u_s(f-f(0)\cdot e^{-z\overline{z}})$ is computable by integration against $|z|^{2s}$, since (by design) that difference is a Schwartz function vanishing at $0$. This vanishing does not imply divisibility by $z$, nor by $\overline{z}$, nor by $x$, nor by $y$, but still does imply, by Taylor-Maclaurin expansion with remainder, that $$ f(z) - f(0)\cdot e^{-z\overline{z}} \;=\; O(|z|) \hskip30pt \hbox{(as $|z|\to 0$)} $$ In particular, the residue of $u_s$ at $s=-1$ is a multiple of $\delta$ (because $f(0)=\delta(f)$), and the multiple can be determined by integrating against $e^{-z\overline{z}}$: $$ \int_{\mathbb C} e^{-z\overline{z}}\,(z\overline{z})^s\;dx\,dy \;=\; 2\pi \int_0^\infty e^{-r^2}\,r^{2s}\;r\,dr \;=\; 2\pi \int_0^\infty e^{-r^2}\,r^{2s+2}\;{dr\over r} \;=\; \pi \int_0^\infty e^{-r}\,r^{s+1}\;{dr\over r} \;=\; \pi\cdot \Gamma(s+1) $$ The residue at $s=-1$ is $\pi$. Then, with interchange of evaluation and differentiation justified by the Schwartz-Grothendieck vector-valued extension of Cauchy-Goursat theory, $$ \overline{\partial}\,{1\over z} \;=\; \overline{\partial}\Big(z^s \overline{z}^{s+1}\Big|_{s=-1}\Big) \;=\; \Big(\overline{\partial}(z^s\overline{z}^{s+1})\Big)\Big|_{s=-1} \;=\; \Big((s+1)z^s\overline{z}^s\Big)\Big|_{s=-1} $$ $$ \;=\; \mathrm{Res}_{s=-1}(z^s\overline{z}^s) \;=\; \pi\cdot \delta $$<|endoftext|> TITLE: Mapping torus of Klein bottle QUESTION [13 upvotes]: This got 5 upvotes but no answers on MSE (Mapping torus of Klein bottle), so I'm cross-posting to MO: The mapping torus of a Klein bottle $ K $ is a compact flat 3 manifold. The mapping class group of the Klein bottle $ K $ is the Klein four group $ C_2 \times C_2 $. See proposition 20 of Paris - Mapping class groups of non-orientable surfaces for beginners. There are exactly four compact flat non-orientable 3 manifolds, one of which is $ K \times S^1 $, the mapping torus of the trivial mapping class of $ K $. Now for the four compact flat non orientable 3 manifolds. These are distinguished by their first homology $ H_1(M;\mathbb{Z}) $ (see Wolf) $$ \mathbb{Z}^2 \times C_2, \mathbb{Z}^2, \mathbb{Z}\times C_2 \times C_2, \mathbb{Z} \times C_4. $$ They correspond to the trivial mapping torus $ S^1 \times K $, the mapping torus of the Dehn twist, the mapping torus of the Y homeomorphism and the mapping torus of the Dehn twist plus Y homeomorphism, respectively. See the Wikipedia page for Seifert fibration with positive orbifold Euler characteristic. I'm curious how these homology groups are computed. I understand the vote to close because I agree this isn't MO level. But it was asked on MSE two months ago with no answer so I feel like it is reasonable to cross post. I honestly won't be mad if you close this post. I just love MO and MSE and all the generous people here who have helped me learn so many interesting things! REPLY [4 votes]: Here is a sketch of an approach. What one needs to do is compute the induced action of the homeomorphism $f : K \to K$ on the generators of a presentation of the fundamental group of the Klein bottle $K$. One can describe the fundamental group of the mapping torus of $f$ as $\pi_1(K) \rtimes_{f_*} \mathbb{Z}$, and once one knows the action of $f_{*}$ on generators, a presentation for the fundamental group of the mapping torus follows. The first homology of the mapping torus is the Abelianization of $\pi_1$ by Hurwitz's theorem, and this can be computed from a presentation of the fundamental group by allowing letters in the relations to commute.<|endoftext|> TITLE: Is containment of images of linear maps Zariski closed? QUESTION [5 upvotes]: Consider the set $$\\\{ (A,B) \in \mathbb{P}^{n\times n-1} \times \mathbb{P}^{n\times n -1} : \text{im}(A) \subseteq \text{im}(B)\}.$$ That is, this is the set of pairs of square matrices $(A,B)$ so that the image of $A$ is contained in the image of $B$. Is this Zariski closed? I would be happy if this were at least true over an algebraically closed field. I tried to write down explicit equations for this, but my first attempt would involved using determinants of submatrices of $A$, and would fail if $A$ was singular. I had thought that this would follow from the projection theorem for for projective varieties, but am unsure. REPLY [4 votes]: Another way to see your set is not closed: it is evidently not the whole space of pairs, but yet it contains the open dense set $U\times U$ where $U \subset \mathbb{P}^{n^2-1}$ consists of (classes of) invertible matrices. So your set is different from its closure.<|endoftext|> TITLE: Infinite Galois equivalence QUESTION [5 upvotes]: I am having some trouble trying to understand the proof of Theorem 7.2.5 in Bhatt and Scholze's paper The pro-étale topology for schemes. Specifically, I don't quite understand why it was necessary to prove that $F: C \to \mathit{Sets}$ preserves connectedness of objects, and how viewing morphisms $f: X \to Y$ in $C$ as monomorphisms $\Gamma_f: X \to X \times Y$ can help us prove that $F: C \to \pi_1(C, F)\text-\mathit{Sets}$ is fully faithful (I could only show that $F: C \to \mathit{Sets}$ is fully faithful). Any help is very much appreciated! REPLY [6 votes]: When $X$ is connected, we see that any graph $\Gamma_f$ for $f \colon X \to Y$ is connected as well, so we get a bijection \begin{align*} \mathscr C(X,Y) &\to \left\{\Gamma \subseteq X \times Y \text{ connected}\ \bigg|\ \Gamma \underset{\pi_1}{\overset{\sim}\to} X \right\} \\ f &\mapsto \Gamma_f. \end{align*} The same holds in $G\text{-}\mathbf{Set}$ for any Noohi group $G$. Writing $G = \pi_1(\mathscr C,F)$ for simplicity, we see that $\mathscr C(X,Y) \to \operatorname{Hom}_G(FX,FY)$ is a bijection for $X$ connected because $F \colon \mathscr C \to G\text{-}\mathbf{Set}$ preserves connected components (and products, etc). Now I suppose the general result follows from axiom (2) or (3) of infinite Galois categories (even though it's not completely spelled out what (3) means). But at the very least we know that any object $X$ is a coproduct $\coprod_{i \in I} X_i$ of connected objects $X_i$. Since $F$ preserves colimits, we get $$\begin{array}{ccccc} \mathscr C(X,Y) & \cong & \mathscr C\Big(\coprod\limits_{i \in I} X_i,Y\Big) & \cong & \prod\limits_{i \in I} \mathscr C\big(X_i,Y\big)\\ & & & & \downarrow\wr\!\! \\ \operatorname{Hom}_G(FX,FY) & \cong & \operatorname{Hom}_G\bigg(\coprod\limits_{i \in I} FX_i,FY\bigg) & \cong & \prod\limits_{i \in I} \operatorname{Hom}_G\big(FX_i,FY\big) \end{array}$$ by the connected case presented above. $\square$ This type of argument might be considered 'standard' when dealing with Galois categories (see for instance [Tag 0BN0 (7)]), which would explain why no details were given.<|endoftext|> TITLE: Examples of a group $G$ and an $F$-representation $V$ where $\cup:H^1(G,F)\otimes H^1(G,V)\to H^2(G,V)$ is injective QUESTION [5 upvotes]: Let $G$ be a group and $F$ a field. I am particularly interested in the case where $G$ is a uniform lattice in a Lie group and $F=\mathbb{F}_2$, or in finite groups $G$ where $\operatorname{char} F$ divides $|G|$, but the discussion applies to general $G$ and $F$. Let $\rho:G\to \mathrm{GL}(V)$ be a representation of $G$ on a finite-dimensional $F$-vector space $V$. Letting $G$ act trivially on $F$, we may form the cohomology ring $H^*(G,F)$, which acts via the cup product on the graded module $H^*(G,V)$. In particular, the cup product induces a linear map $$ p: H^1(G,F)\otimes_F H^1(G,V)\to H^2(G,V). $$ My question is whether there are $G$ and $V$ for which: $p$ is injective, $\dim H^1(G,V) \geq 1$, $\dim H^1(G,F)\geq 2$. More generally, is there a recipe for constructing such examples? I am not aware of even a single example of this kind. The injectivity of $p$ is the real issue. For example, if we take $V=F$, then the cup product $\cup :H^1(G,F)\times H^1(G,F)\to H^2(G,F)$ is well-known to be anti-symmetric, which means that $\dim \ker p\geq \frac{1}{2}\dim H^1(G,F)(\dim H^1(G,F)-1)$ if the characteristic of $F$ is $2$, or $\dim \ker p\geq \frac{1}{2}\dim H^1(G,F)(\dim H^1(G,F)+1)$ if $F$ is not of characteristic $2$, so $\ker p$ would not be trivial when $\dim H^1(G,F)\geq 2$. (This special example is studied in a paper of Hillman.) REPLY [3 votes]: $\newcommand{\bZ}{\mathbb{Z}}$Let $G$ be a finite group of order divisible by $p:=\mathrm{char}\, F$ such that $\dim_F \mathrm{Hom}(G,F)\geq 2$ (e.g. $G=\bZ/p\times\bZ/p$). Take $V$ to be the kernel of the augmentation map $e(\sum a_g\cdot g)=\sum a_g$ from $F[G]$ to $F$. Since $F[G]$ is an injective $G$-module, the short exact sequence $0\to V\to F[G]\xrightarrow{e} F\to 0$ yields isomorphisms $H^i(G, F)\simeq H^{i+1}(G,V)$ for $i\geq 0$ (this is automatic for $i>0$ and for $i=0$ follows from the fact that $e:F[G]^G\to F$ is the zero map because the order of $G$ is divisible by $p$). These isomorphisms are compatible with the cup-product in the sense that the following diagram commutes for every $i$ (this follows from the associativity of cup-product using the observation that the isomorphism $H^i(G,F)\to H^{i+1}(G,V)$ is obtained by cupping with a class in $H^1(G,V)$) $\require{AMScd}$ \begin{CD} H^1(G,F)\otimes H^{i+1}(G,V) @>\cup >> H^{i+2}(G,V)\\ @| @|\\ H^1(G,F)\otimes H^{i}(G,F) @>>\cup > H^{i+1}(G,F) \end{CD} For $i=0$ the bottom map is obviously an isomorphism, hence the top map is an isomorphism, as desired.<|endoftext|> TITLE: Ring of continuous functions is a Jacobson ring QUESTION [5 upvotes]: Let $X$ be an infinite discrete topological space. Is $$C_b(X)=\{ f \colon X \to \mathbb{R} \text{ bounded }\}$$ a Jacobson ring ? REPLY [6 votes]: First, let us consider the question of when the ring $C(X)$ of all continuous real-valued functions on a topological space $X$ (not necessarily discrete) is Jacobson, keeping in mind that $C_b(X) = C(\beta X)$ where $\beta X$ is the Stone-Čech compactification. In fact, every prime ideal of $C(X)$ is contained in a unique maximal ideal: see Gillman & Jerison, Rings of Continuous Functions (1960) theorem 7.15 on page 107. So $C(X)$ is a Jacobson ring iff every prime ideal is maximal, meaning that $X$ is a P-space, a rather strong condition which, see the reference in this other answer, is equivalent to the condition “every $f\in C(X)$ which vanishes at $p$ vanishes in some neighborhood of $p$“. (Note for example that $C([0,1]) = C_b([0,1])$ is not a Jacobson ring: the ring of functions which vanish in some neighborhood of $0$, while not prime itself, is contained in a prime ideal, which itself is contained in a a unique maximal ideal, namely that of functions which vanish at $0$.) Now if $X$ is infinite discrete, it is indeed a P-space, so $C(X)$ is a Jacobson ring. However, $\beta X$ is not a P-space, as every compact P-space is finite (Gillman & Jerison, problem 4K.1 on page 63, and remarks following theorem 14.29 on page 212). So $C_b(X) = C(\beta X)$ is not a Jacobson ring.<|endoftext|> TITLE: Adjunction between topological spaces and condensed sets QUESTION [5 upvotes]: I am trying to prove that the functor \begin{align*} \mathrm{Top} &\longrightarrow \mathrm{Cond}(\mathrm{Set}) \\ X &\longmapsto \underline{X} \end{align*} admits a left adjoint and it is the functor $T \mapsto T(*)$ where $T(*)$ has the quotient topology of the map $\bigsqcup_{\underline{S} \rightarrow T} S \rightarrow T(*)$ where the disjoint union runs over all profinite sets $S$ with a map to $T$. To prove this I am trying to construct the following bijection. We have the map \begin{align*} \phi: \mathrm{Hom}(T, \underline{X}) &\longrightarrow \mathrm{Hom}(T(*), X) \\ f &\longmapsto f_* \ . \end{align*} On the other hand, if $g \in \mathrm{Hom}(T(*), X)$, then we can consider the map $\underline{g}: \underline{T(*)} \rightarrow \underline{X}$. I am trying to construct a map $i: T \rightarrow \underline{T(*)}$ to have an inverse for $\phi$: \begin{align*} \psi: \mathrm{Hom}(T(*), X) &\longrightarrow \mathrm{Hom}(T, \underline{X}) \\ g &\longmapsto \underline{g} \circ i \ . \end{align*} Is there a way to construct $i$? How to prove this adjunction? Edit: I have managed to construct $i$, but I still could not prove that $\phi$ and $\psi$ are mutually inverses. I have constructed $i$ as the following: For each $S$ profinite we have the equivalence $\mathrm{Hom}(\underline{S},T) = T(S)$ thanks to the Yoneda Lemma. We can define \begin{align*} i_S: \mathrm{Hom}(\underline{S},T) &\longrightarrow \mathrm{Hom}(S, T(*)) \\ \eta &\longmapsto \eta_* \ . \end{align*} It is easy to prove that for $g \in \mathrm{Hom}(T(*), X)$ we have $(\underline{g} \circ i)_* = g$, but have not managed to prove that for $f \in \mathrm{Hom}(T, \underline{X})$ we have $f = \underline{f_*} \circ i$. So it only remains to prove that for $\eta \in \mathrm{Hom}(\underline{S},T)$ we have \begin{align*} f_S (\eta) = f_* \circ \eta_* \end{align*} I suspect this is due to the Yoneda Lemma but I could not prove it. REPLY [2 votes]: As Wojowu noted in the comments, one should really look at $T_1$ topological spaces. Consider the functors \begin{align*} G\!: \mathbf{Top}_{T_1} &\leftrightarrows \mathbf{Cond}_\kappa:\!F\\ X &\mapsto \big(\underline X \colon S \mapsto \operatorname{Cont}(S,X)\big)\\ T(*) &\leftarrow\!\shortmid T, \end{align*} where $T(*)$ is topologised with the quotient topology via the surjection $$\pi \colon \coprod_{(S,f \in T(S))} S \to T(*)\tag{1}\label{1}$$ given on the component $f \in T(S)$ by $f \colon S \to T(*)$. By the Yoneda lemma, this really means that $f \colon h_S \to T$ is a morphism from the representable sheaf $h_S$ to $T$, and the map $S \to T(*)$ is the set-theoretic map $f_{\{*\}} \colon h_S(*) \to T(*)$. But there is the consistent abuse of notation to denote $h_S = \underline S$ as $S$. To see that (\ref{1}) is surjective, use $S = \{*\}$. As Dylan Wilson noted in the comments, the unit $\eta \colon 1 \to GF$ is given by the natural transformation \begin{align*} (\eta_T)_S \colon T(S) &\to \operatorname{Cont}(S,T(*)) \\ f &\mapsto \pi \circ \iota_f, \end{align*} where $\iota_f \colon S \to \displaystyle\coprod_{(S',f')} S'$ is the insertion of the coordinate corresponding to $(S,f)$. This gives the maps \begin{align*} \phi\!: \operatorname{Hom}_{\mathbf{Cond}}(T,\underline X) &\leftrightarrows \operatorname{Cont}(T(*),X) :\!\psi \\ f &\mapsto f_{\{*\}} \\ g \circ \eta_T &\leftarrow\!\shortmid g. \end{align*} It remains to check that these are inverses. If $g \colon T(*) \to X$ is continuous, then $\phi(\psi(g)) \colon T(*) \to X$ is given by $\psi(g)_{\{*\}}$, i.e. the map taking $t \in T(*)$ to $g \circ \pi \circ \iota_f \colon \{*\} \to X$. But $\pi \circ g \colon \{*\} \to T(*)$ is just (the constant map with value) the point $t$ (this is how we checked surjectivity of $\pi$ earlier!), so $\psi(g)_{\{*\}}$ takes $t$ to $g(t)$, i.e. $\phi(\psi(g)) = g$. Conversely, given a natural transformation $f \colon T \to \underline X$, we get another natural transformation $\psi(\phi(f)) \colon T \to \underline X$. Write $g \colon T(*) \to X$ for $\phi(f) = f_{\{*\}}$. If $S$ is extremally disconnected and $h \in T(S)$, then $\psi(g)_S$ takes $h$ to the composition $$S \overset{\iota_h}\to \coprod_{(S',h')} S' \overset\pi\to T(*) \overset g\to X.$$ By definition, the composition $\pi \circ \iota_h \colon S \to T(*)$ is the map $h_{\{*\}} \colon h_S(*) \to T(*)$. Thus $\psi(g)_S(h)$ is the composition $$h_S(*) \overset{h_{\{*\}}}\longrightarrow T(*) \overset{f_{\{*\}}}\longrightarrow X.$$ This is the same thing as $(fh)_{\{*\}} \colon h_S(*) \to \underline X(*)$, which is the continuous map $S \to X$ given by $f_S(h) \in \underline X(S)$. We conclude that $\psi(\phi(f))_S(h) = f_S(h)$, and since $S$ and $h$ are arbitrary that $\psi(\phi(f)) = f$. $\square$ Remark. Morally what's going on here is the following: since $T(*)$ has the quotient topology, a map $g \colon T(*) \to X$ is continuous if and only if each of the compositions $gf \colon S \to X$ are continuous with $S$ extremally disconnected and $f \colon S \to T$ an $S$-point of $T$. Thus, a continuous map $T(*) \to X$ is the same as continuous maps $S \to X$ for every $S \to T$, which is roughly what a natural transformation $T \to \underline X$ is.<|endoftext|> TITLE: Groupoid cardinality of the class of abelian p-groups QUESTION [15 upvotes]: $\DeclareMathOperator\Aut{Aut}\newcommand\card[1]{\lvert#1\rvert}$So, after going over the classification of finite abelian groups in a class I was teaching this winter, I got curious about whether it could be used to obtain a 'nice' value for the groupoid cardinality of the class of abelian groups of fixed size, that is if one could compute : $$ \sum_{\card G=m} \frac{1}{\card{\Aut(G)}} $$ where the sum is over the isomorphism class of abelian groups with $m$ elements. That is just pure curiosity, I had absolutely no real motivation for this. One relatively quickly see that it is enough to compute it for $m=p^n$, so I started doing it for $p$, $p^2$ and $p^3$. Unsurprisingly, the results look messy at first, but then a quite surprising number of simplifications occured and I arrived at the formula : $$ \sum_{\card G=p^n} \frac{1}{\card{\Aut(G)}} = \prod_{k=1}^n \frac{p^{k-1}}{p^k-1}. $$ I checked it by hand up to $n=5$. In any case the simplicity of the formula suggests that there is a better way to compute this than going over all isomorphism class of abelian groups and computing $\Aut(G)$ and then taking the sum. So basically I'm wondering if this formula is known, and if not if someone has an idea of how to prove it — ideally with an argument that doesn't involve summing over partitions of $n$. Note : there are known formulæ for $\card{\Aut(G)}$ see for example the very end of Hillar and Rhea - Automorphisms of finite Abelian groups. REPLY [10 votes]: Hall's original paper gives a nice combinatorial proof of this, which seems not to be mentioned in any of the follow up papers. Given a partition $\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_n)$, define the Durfee square of $\lambda$ to be the largest $\mu_1$ such that the diagram of $\lambda$ contains a $\mu_1 \times \mu_1$ square. Below is a $4 \times 4$ square, fitting inside the partition $8+7+5+4+4+3+1+1$. I took this diagram from Andrews' paper Partitions and Durfee dissection If one removes $\mu_1 \times \mu_1$ from $\lambda$, one is left with two partitions, a partition $\lambda_{\downarrow}$ below the Durfee square and a partition $\lambda_{\rightarrow}$ to its right. Let $\mu_2$ be the Durfee square of $\lambda_{\downarrow}$ and continue in this way to define $\mu_3$, $\mu_4$, .... The partition $(\mu_1, \mu_2, \ldots)$ is called the Durfee dissection of $\lambda$ and was introduced by Andrews in the paper above. Here (also from that paper) is an image showing that the Durfee dissection of $8+7+5+4+4+3+1+1$ is $4+2+1+1$. I'll write $D(\lambda)$ for the Durfee dissection of $\lambda$. Then Hall proves: Theorem For any partition $\mu$, we have $$\frac{1}{\left|\text{Aut} \prod_j (\mathbb{Z}/p^{\mu^T_j} \mathbb{Z})\right|} = \sum_{D(\lambda) = \mu} \frac{1}{p^{|\lambda|}}.$$ Here $\mu^T$ is the transpose partition to $\mu$. For example, take $\mu = (2)$. The partitions with $D(\lambda) = (2)$ are of the form $a+b$ with $a \geq b \geq 2$, and $$\sum_{a \geq b \geq 2} \frac{1}{p^{a+b}} = \frac{1}{p^4(1-p^{-1})(1-p^{-2})} = \frac{1}{(p^2-p)(p^2-1)} = \frac{1}{\left| \text{Aut}((\mathbb{Z}/p \mathbb{Z})^2 )\right|}.$$ To give another example, take $\lambda = 1+1$. The partitions with $D(\lambda) = 1+1$ are of the form $a+1$ with $a \geq 1$, and $$\sum_{a \geq 1} \frac{1}{p^{a+1}} = \frac{1}{p^2(1-p^{-1})} = \frac{1}{p^2-p} = \frac{1}{\left| \text{Aut}(\mathbb{Z}/p^2 \mathbb{Z}) \right|}.$$ Let's see how this implies the product formula in the original post. As you can see, the squares in the Durfee dissection stretch from the top of the partition to the bottom, so $|D(\lambda)|$ is the number of parts of $\lambda$. Thus, $\sum_{|G| = p^n} \frac{1}{\left| \text{Aut}(G) \right|}$ is the sum of $p^{-|\lambda|}$ over partitions $\lambda$ with $n$ parts, which is well known to be $$\frac{1}{p^{-n} \prod_{j=1}^n (1-p^{-j})} = \prod_{j=1}^n \frac{p^{j-1}}{p^j-1}$$ as desired. I am writing this up here because I rediscovered it about five years ago, and did a lot of google and MathSciNet searches on words like "Durfee", "square", "automorphism", "Cohen-Lenstra" before finally reading Hall's paper and finding it was there. It seems to be a fact that has passed out of the range of search engines, so I thought I would put it back in.<|endoftext|> TITLE: Polyomino that can cover an arbitrarily large square but not the entire plane QUESTION [23 upvotes]: https://userpages.monmouth.com/~colonel/nrectcover/index.html For a polyomino with no holes that cannot tile the plane, we may ask what are the maximal rectangles and infinite strips that it can cover without overlapping, allowing the tiles to extend beyond the region's perimeter. example from this site: But can there be such a polyomino that can cover an arbitrarily large square, but not the whole plane? And is this even possible for an arbitrary tile at all? The answer is probably no, but why exactly. In other words, can polyomino which cannot tile the plane produce infinite sequence of squares with increasing edge length? REPLY [11 votes]: Here's a reformulation of my other answer in terms of compactness and general topological abstract nonsense, for those who are interested in the general template for this style of argument: For a given finite polyomino which can cover squares of arbitrary size, consider the set $S$ of all possible ways to place this polyomino onto an infinite grid. A subset $Q$ of $S$ can be considered a placement of copies of the polyomino onto the plane, where gaps or overlap may occur. A placement $Q$ may be represented by its characteristic function $\chi_Q: S \to \{0,1\}$. The space $P := \mathcal P(S)$ of all possible placements has a canonical bijection to the space $\{0,1\}^S$ of all characteristic functions. The latter set is a cartesian product of two-element sets indexed by $S$, so we can equip it with the topology of a direct product of discrete two-point spaces indexed by $S$. Two-point spaces are compact, so the resulting topology on $\mathcal P(S)$ is also compact. The topology of this direct product is generated by the clopen sets $A_s := \{Q \subset S: s \in Q\}$ for $s \in S$ and their complements $A_s^c$. For any cell $x$ of the grid, we can define the subset $P_x$ of placements which are valid on $x$, meaning that exactly one polyomino of the placement covers the cell $x$. Observe that since the set $S_x := \{s \in S: x \in s\}$ is finite, we can describe $$P_x = \bigcup_{s \in S_x}\left(A_s \cap \bigcap_{t \in S_x\setminus\{s\}}A_t^c\right)$$ as a finite union of intersections of closed sets, so $P_x$ is a closed subset of $P$. Now let $B_1 \subset B_2 \subset \ldots$ be a sequence of larger and larger squares of grid cells whose union is the entire plane. For each $B_i$ we can define the set $Q_i := \bigcap_{x \in B_i} P_x$ of placements valid within the square $B_i$. As an intersection of closed subsets of $P$, the $Q_i$ are themselves closed subsets of $P$, and since $P$ is compact so are the $Q_i$. Observe that $B_i \subset B_j$ implies $Q_i \supset Q_j$, and note that each $Q_i$ nonempty because of our initial assumption that the polyomino we chose can cover every square $B_i$ somehow. Now the $Q_i$ are an infinite descending nested sequence of nonempty compact sets. This lets us apply Cantor's intersection theorem, the core of most compactness arguments out there, which simply tells us that the intersection of all $Q_i$ is nonempty. However, the intersection of all $Q_i$ is also the intersection of all $P_x$, and if we recall how we defined those $P_x$, namely that $P_x$ only contains placements valid on the cell $x$, we notice that we just proved the existence of a placement that is valid on every single cell of the plane, also known as a tiling of the plane by copies of the polyomino. $~~\square$ Of note here is that it is essential for this proof that the polyomino in question is finite, otherwise the $P_x$ are not necessarily closed. In fact, there are many classes of infinite polyominos which can tile arbitrary squares but not the entire plane. The simplest of these infinite polyominos is probably just a whole plane with a single cell missing, but there are many pretty counterexamples. Finding some is left as an exercise to the reader.<|endoftext|> TITLE: Does a non-simple perfect group always have a maximal subgroup whose derived subgroup has nontrivial core? QUESTION [6 upvotes]: Let $G \neq 1$ be a finite perfect group which is not simple. Is it true that $G$ necessarily has a maximal subgroup whose derived subgroup has nontrivial core in $G$? Remark 1: This holds for all such $G$ of order less than 100000. Remark 2: In case the answer is negative, I would mainly be interested in a counterexample with nontrivial Frattini subgroup. REPLY [6 votes]: The answer to the title question is 'No.' Following YCor's comment, an example is furnished by $S=J_1$, the smallest Janko sporadic group, and its complex irreducible character $\chi$ of degree 76 (in Atlas notation, 76a). I have checked by hand (hopefully correctly), using the Atlas, that $(\chi\downarrow M,1_M)>0$ for all maximal subgroups $M$ of $J_1$. For any prime $p$ not dividing $|J_1|=2^3.3.5.7.11.19$, let $N_p$ be an $F_pS$-module affording the mod-$p$ reduction of $\chi$. Then, as YCor points out, the semidirect product $N_pS$ has the property that for every maximal subgroup $H\le N_pS$, $[H,H]$ does not contain $N_p$. However, this example has trivial Frattini subgroup.<|endoftext|> TITLE: Impact of Ramanujan's Note on a set of simultaneous equations QUESTION [9 upvotes]: I had been pointed to Ramanujan's 1912 article Note on a set of simultaneous equations in this answer to my former question about the Solvability of a system of polynomial equations. While the contents of the paper seemed a bit enigmatic at first reading, I'm now convinced that Ramanujan not only left it to the reader to recover the ideas that had led to the solution but also the motivation to investigate on that special system of polynomial equations. Questions: is anything known about Ramanujan's motivation to work on that problem that somehow doesn't seem to fit his primary research interest; the situation seems analogous to Euler and his characteristic of polyhedra? did Ramanujan himself ever refer to the paper in his later work? is or was the paper of importance for mathematical research and what are remarkable examples? have there been serious attempts to extend the solution method and/or to reduce more general types of systems of polynomial equations to the special case that Ramanujan has solved? REPLY [11 votes]: This refers to the third and fourth question in the OP. Ramanujan's 1912 paper addresses a problem similar to that considered by Sylvester in 1851 [1]. The method to solve the set of algebraic equations $$\sum_{k=1}^{n}x_kz_k^j=a_j,\;\;0\leq j\leq 2n-1$$ in the unknown $x_k$'s and $z_k$'s is different in the two papers, but the final algorithm is similar. This set of equations shows up in the socalled moment problem, to find a probability measure given its moments. That connection is explored in The Sylvester-Ramanujan System of Equations and The Complex Power Moment Problem by Yuri Lyubich. [1] J.J. Sylvester, On a remarkable discovery in the theory of canonical forms and of hyperdeterminants, Phil. Magazine 2, 391–410 (1851).<|endoftext|> TITLE: What is the reason for $f_!$ not preserving discrete objects? QUESTION [5 upvotes]: Let $A$ be a finitely generated $\mathbb{Z}$-algebra and let $f: \operatorname{Spec} A \rightarrow \operatorname{Spec} \mathbb{Z}$ be the canonical map. On pg. 53, Thm. 8.2 of https://www.math.uni-bonn.de/people/scholze/Condensed.pdf one defines a functor $$f_!: D(A_\blacksquare) \rightarrow D(\mathbb{Z}_\blacksquare)$$ where $A_\blacksquare$ and $\mathbb{Z}_\blacksquare$ denotes certain categories of solid modules. On the same page, Scholze remarks that $f_!$ does not preserve discrete objects in general? Is there an easy way to see that $f_!$ can not possibly preserve discrete objects for general morphisms $f?$ REPLY [6 votes]: You can see this by a direct calculation, for example in the most basic case $A=\mathbb{Z}[T]$. When you apply $f_!$ to $A=\mathbb{Z}[T]$ itself, you get the object represented by the two-term complex $$\mathbb{Z}[T]\to \mathbb{Z}((T^{-1})).$$ with $\mathbb{Z}[T]$ in degree $0$. (This is the "compactly supported cohomology of the structure sheaf" on $\mathbb{A}^1$; the Laurent series in $T^{-1}$ represents "functions defined in a neighborhood of $\infty$"). The map in this complex is injective, and the cokernel is $T^{-1}\mathbb{Z}[[T]]$, which is not discrete. In fact, $f_!A$ is predual to an infinite discrete object, namely the usual algebraic Grothendieck dualizing complex of f. This is a general feature of the situation, and follows from the adjunction between $f_!$ and $f^!$. Actually, $f_!$ has a complementary property to preservation of discreteness: it preserves pseudocoherent objects.<|endoftext|> TITLE: Does there exist a complete algebraic invariant of the homotopy type of a finite CW-complex? QUESTION [7 upvotes]: Let $\mathrm{Cell}$ be the homotopy category of finite cell complexes. The main motive of my question Is it true that for any algebraic category $A$ there is no fully faithful functor $F: \mathrm{Cell} \to A$? The answer may depend on what exactly is meant by algebraic categories. The following version seems to me the most natural and interesting, but comments on any other versions are welcome. By algebraic category here I mean the category of all models of an algebraic theory, where an algebraic theory is given by the following data: a finite set of carriers (i.e. sorts of elements) a finite set of operations returning a tuple over a tuple, the scope of which is determined by a set of equalities between operations (including the empty number of equalities, of course) a finite set of identities imposed on operations Example: category of small categories Carriers: $\mathrm{Ob}$, $\mathrm{Mor}$ Operations: $\mathrm{dom}, \mathrm{cod}, \mathrm{id}, \circ$ (the latter is defined on those pairs of morphisms for which dom = cod) Identities: $$(f \circ g) \circ h = f \circ (g \circ h)$$ $$\mathrm{dom}(f \circ g) = \mathrm{dom}~f,\;\; \mathrm{cod}(f \circ g) = \mathrm{cod}~g$$ $$\mathrm{dom} (\mathrm{id}_A) = A, \;\; \mathrm{cod} (\mathrm{id}_A) = A$$ $$\mathrm{id}_{\mathrm{dom} f} \circ f = f,\;\; f \circ \mathrm{id}_{\mathrm{cod} f} = f$$ In this example, all operations returned one element, the ability to return tuples was not used (also, of course, natural operations can occur in identities: projections and direct products of morphisms). This notion differs from the finitary algebraic theory and Lover's theories. It doesn't even seem to fall into what is called generalized algebraic theory because of the possibility of operations not being everywhere defined. However, I think I've seen something like this somewhere on nlab, but I can't find it again. P.S. Of course, it is better not to use the composition symbol (in order to use it to write identities) and write all operations in a single syntax, but in this example I could not resist and used the traditional notation. REPLY [6 votes]: I suspected that it would be possible to prove that there is no faithful functor from finite complexes to any kind of category of finite algebraic objects by slavishly following Freyd's argument and replacing "set" with "finite set" everywhere. This turned out to be true, and indeed the argument is a little bit easier in the group theory than the general case, so I reproduce it below as an advertisement for anybody lucky enough not yet to have followed Freyd's classic argument. Proposition: There is no faithful functor from the homotopy category of finite complexes to the category of finite sets. Proof: Consider the Moore spaces $X_k=M(\mathbb Z_{2^k},2),$ which are finite complexes. Let $f_k:X_1\to X_k$ be induced by the unique nontrivial homomorphism sending $1$ to $2^{k-1}.$ If $F:\mathrm{Cell}^{\mathrm{op}}\to \mathrm{FinSet}$ is a contravariant pointed functor into finite sets, then some $F(f_k),F(f_j)$ must have the same image, since $F(X_1)$ has only finitely many subsets. Without loss of generality, $k\ge j.$ Now observe that there is a map $g:Y\to X_1$ such that $f_k\circ g=0$ but $f_j\circ g\ne 0.$ Indeed, we can let $f_k=\Sigma f_k'$ and $g$ be the cone of $f_k'$. Then $f_k$ is a weak cokernel of $g,$ so $f_k\circ g=0,$ but if $f_j\circ g$ were $0,$ then we'd have a map $h:X_k\to X_j$ with $h\circ f_k=f_j.$ But this would imply the existence of a homomorphism $\mathbb Z_{2^k}\to\mathbb Z_{2^j}$ sending $2^{k-1}$ to $2^{j-1},$ which does not exist. Thus $F(f_k\circ g)=0,$ which means $F(g):F(X_1)\to F(Y)$ is zero on the image of $F(f_k),$ so that also $F(f_j\circ g)=0.$ Therefore $F$ is not faithful, and indeed, it sends a nonzero map to a zero map. $\square$ We have proved that there is no contravariant faithful functor from $\mathrm{Cell}$ to $\mathrm{FinSet}.$ Since $\mathrm{FinSet}$ admits a faithful contravariant endofunctor given by the powerset, there is also no covariant faithful functor $\mathrm{Cell}\to\mathrm{FinSet},$ so one might say that the homotopy category of finite complexes is not finitely concrete. Any reasonable category of finite algebraic objects will admit a faithful and conservative functor into $\mathrm{FinSet},$ so we can conclude that the answer to your question is negative. Remark: Interestingly, I don't think you can carry this argument over to compact objects in Liberti-Loregian's generalization since they use $\pi_n$ instead of $H_n,$ and Eilenberg-Mac Lane spaces are less friendly to finiteness.<|endoftext|> TITLE: Swimming against the tide in the past century: remarkable achievements that arose in contrast to the general view of mathematicians QUESTION [45 upvotes]: I would like to ask a question inspired by the title of a book by Sir Roger Penrose ([1]). The germ of this is to ask about the role, if any, of the fashion in research of pure and applied mathematics. I'm going to focus the post (and modulate my genuine idea) about an aspect that I think can be discussed here from an historical and mathematical point of view, according to the following: Question. I would like to know what are examples of remarkable achievements (in your research subject or another that you know) that arose against the general view/work of the mathematical community since the year 1900 up to the year 1975. Refer the literature if you need it. Many thanks. An example is the mention that the author of [2] (as I interpret it) about Lennart Carleson and a conjecture due to Lusin in the second paragraph of page 671 (the article is in Spanish). Your answer can refer to (for the research of pure or applied mathematics, and mathematical physics) unexpected proofs of old unsolved problems, surprising examples or counterexamples, approaches or mathematical methods that defied the contemporary (ordinary, mainstream) approaches, incredible modulizations solving difficult problems,... all these in the context of the question that is: the proponents/teams of these solutions and ideas swam against the work of the contemporary mathematics that they knew at the time. *You can refer to the literature for the statements of the theorems, examples, methods,... if you need it. Also from my side it is welcome if you want to add some of your own historical remarks about the mathematical context concerning the answer that you provide us: that's historical remarks (if there is some philosophical issue also) emphasizing why the novelty work of the mathematician that you evoke was swimming against the tide of the contemporary ideas of those years. References: [1] Roger Penrose, Fashion, Faith, and Fantasy in the New Physics of the Universe, Princeton University Press (2016). [2] Javier Duoandikoetxea, 200 años de convergencia de las series de Fourier, La Gaceta de la Real Sociedad Matematica Española, Vol. 10, Nº 3, (2007), pages 651-677. REPLY [13 votes]: There are various examples of certain areas of mathematics being regarded as sterile and disconnected from the rest of mathematics, but nevertheless being pursued doggedly by some researchers, and eventually finding unexpected applications in other areas of mathematics. Computability theory, especially the study of computably enumerable sets, is an example. Though it has been pursued ever since the mid-twentieth century, computability theory is even today regarded by most mathematicians as an arcane subject with little connection with the rest of mathematics. However, as explained by Soare in his paper, Computability theory and differential geometry (non-paywalled version here), computability theory has found unexpected applications in the work of Nabutovsky and Weinberger in differential geometry. Another possible example is the use of model theory to attack number-theoretic problems. The Ax–Kochen theorem dates from 1965, but my impression is that for a long time, most number theorists did not think that model theory was a particularly powerful or insightful tool. That skepticism has presumably evaporated with (for example) the success of o-minimality and the spectacular proof of the André–Oort conjecture.<|endoftext|> TITLE: Is the decomposition of the homotopy type of a complex into a bouquet unique? QUESTION [15 upvotes]: Is it true that if $A_1 ​​\vee A_2 \vee .. \vee A_n = B_1 \vee B_2 \vee .. \vee B_m$, where $A_i, B_j$ are homotopy types of complexes not decomposable into a bouquet, then the multisets $A_i$ and $B_j$ coincide? That is, is it true that a commutative monoid of homotopy types decomposable into a bouquet of a finite number of indecomposable ones is freely generated by indecomposable ones. Is it true that any finite complex decomposes into a finite number of indecomposable ones (and thus all finite complexes are included in the monoid above)? Countable complexes are not necessarily included in it, for example, any countable bouquet. But, for example, all spaces $K(G, n)$ are included, it seems. REPLY [8 votes]: There is a substantial literature on the subject. You should probably start with C. W. Wilkerson, Genus and cancellation, Topology {\bf 14} (1975), 29-36. In particular, he shows Definition: Let $X$ be a 1-connected $p$-local space of finite type. Then $X$ is said to be $prime$ if for every self-map $ f:X \rightarrow X $, either $i$) $ f $ induces an isomorphism in mod-$p$ homology, or $ii$) for every $n$, there exists an $m$ such that the $m$-fold iterate of $f$ induces the zero map on $H_{i}(X; {\bf Z}_p)$ for $ 0 < i \leq n$. Then, we have Theorem 3: (Wilkerson) $i$) Any finite dimensional 1-connected $p$-local co-$H$-space is equivalent to a wedge of prime spaces. $ii$) If a 1-connected $p$-local space of finite type is equivalent to a wedge of primes, then the prime wedge summands are unique up to order. $iii$) A prime space which is a retract of a wedge of 1-connected $p$-local spaces of finite type is a retract of one of the wedge summands.<|endoftext|> TITLE: Prove an inequality related to sums of Legendre symbols QUESTION [8 upvotes]: $\newcommand\Legendre{\genfrac(){}{}}$Let $p\equiv 1\pmod 4$ be a prime number, and $x_{i}\ge 0$ be such that $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$ Show that $$\sum_{1\le i0$ is at most $k$ then $$\sum_{1\le i0$. Without loss of generality, we may assume $$\sum_{j \neq i_1} \Legendre{ i_1 - j}{p} x_j \geq \sum_{j \neq i_2} \Legendre{ i_2 - j}{p} x_j.$$ Let $y_j = x_j$ for $j\not\in\{i_1, i_2\}$, let $y_j = x_{i_1} + x_{i_2} $ for $j=i_1$, and let $y_j=0$ for $j =i_2$. Then $\sum_j y_j = \sum_j x_j=1$ and the number of nonzero $y_j$ is at most the number of nonzero $x_j$. Now, using the $i\neq j$ sum which is twice as large as the $i \sum_{1\leq i,j \leq p, i \neq j }\Legendre{i-j}{p}x_{i}x_{j} .$$ Thus, we have increased your sum while reducing the number of nonzero $x_i$'s. We may keep doing this until the number of nonzero $x_i$'s is at most $k$, proving the claimed upper bound. Sharpness follows from taking $x_i=1/k$ for $i$ in a clique of size $k$ and $x_i=0$ for all other $i$. By the recent breakthrough upper bound on the clique number of the Paley graph by Hanson and Petridis, we have $k \leq \lceil \sqrt{p/2} \rceil$. Plugging this in, we obtain your claimed bound for $p=5$ and do better for all larger $p$. (To do better, it suffices to have $k < \frac{p+3}{4}$.)<|endoftext|> TITLE: Families of sets with parity conditions QUESTION [7 upvotes]: While crunching some numbers for this question, I came across a different one. An answer would give an estimate for complexity of my procedure, as well as potentially give an insight to the former question itself. Let $[n] = \{1, \ldots, n\}$. Say that a family $S \subseteq 2^{[n]}$ belongs to $\mathcal{F}_n$ iff for any $X \in S$ the number of $Y \in S$ containing $X$ as a subset is odd (including $Y = X$ itself). E.g. (below $1$ and $2$ are singleton sets $\{1\}$ and $\{2\}$, and $12 = \{1, 2\}$ by violent notation abuse) $\mathcal{F}_0 = \{\varnothing, \{\varnothing\}\}$ (assuming $[0] = \varnothing$), $\mathcal{F}_1 = \{\varnothing, \{\varnothing\}, \{1\}\}$, $\mathcal{F}_2 = \{\varnothing, \{\varnothing\}, \{1\}, \{2\}, \{12\}, \{1, 2\}, \{\varnothing, 1, 2\}\}$. This question is concerned with the size of $\mathcal{F}_n$ as a function of $n$. Some trivial bounds are $2^{n \choose n / 2} \leq |\mathcal{F}_n| \leq 2^{2^n}$. Taking logarithms, we have ${n \choose n / 2} \leq \log_2 |\mathcal{F}_n| \leq 2^n$. For large $n$, the lower and upper bound are $\Theta(\sqrt{n})$ apart. First few values of $|\mathcal{F}_n|$, and entries in the logarithmic bound are listed below: $n$ $|\mathcal{F}_n|$ $\log_2{|\mathcal{F}_n|}$ ${n \choose n / 2}$ $2^n$ $0$ $2$ $1$ $1$ $1$ $1$ $3$ $ \approx 1.58496$ $1$ $2$ $2$ $7$ $ \approx 2.80735$ $2$ $4$ $3$ $43$ $ \approx 5.42626$ $3$ $8$ $4$ $1687$ $ \approx 10.72024$ $6$ $16$ $5$ $2204623$ $ \approx 21.07210$ $10$ $32$ $6$ $2809835768527$ $ \approx 41.35362$ $20$ $64$ Questions: is any of the bounds above for $\log_2 |\mathcal{F}_n|$ asymptotically tight? If no, what's the correct asymptotics for $\log_2 |\mathcal{F}_n|$? UPD: a reference provided in Tim's answer (namely Thm. 18 in Section 3.4) presents a subset of $\mathcal{F}_n$ of size $2^{2^{n - 1}}$, namely all $S \subseteq 2^{[n]}$ satisfiying a stronger condition: a set $X \subseteq [n]$ is in $S$ iff it's contained in an odd number of elements of $S$. This establishes $2^{n - 1} \leq \log_2{|\mathcal{F}_n|} \leq 2^n$. One then hopes that $\log_2{|\mathcal{F}_n|} \sim c 2^n$ for a certain constant $c \in [1/2, 1]$. Does such a $c$ exist? Can it be bounded, or even obtained precisely? For reference, here is my procedure for calculating $|\mathcal{F}_n|$. We recursively obtain all possible $S \in \mathcal{F}_n$ starting from $S = \varnothing$ by repeatedly adding $X$ not containing any prior elements of $S$ (this can not violate the premise for already added elements of $S$). Let $T$ be the set of candidates to be next added to $S$. Initially $T = 2^{[n]}$. If $X$ is any maximal set in $T$ (say, lexicographically largest), we either skip it, branching to $S \to S, T \to T - X$, or add $X$ to $S$, branching to $S \to S + X, T \to T \operatorname \triangle 2^X$. The number of successful branches from any intermediate point only depends on $T$, which allows for some efficient caching. REPLY [2 votes]: Based on your data so far, one might wildly guess that $\log_2(\mathcal{F}_n) \approx\frac232^n.$ Another wild guess would be $\approx \frac{2^n+\binom{n}{n/2}}2.$ Here is a somewhat vague argument that a lower bound on the order of $\frac122^n$ might be possible. With more care it might even give a very good estimate of $\log_2(\mathcal{F}_n).$ In brief, your branching procedure has an evolving pair of $S,T$ where $S$ will be the family and $T$ is the sets available to currently be added (i.e. in an even number of already chosen sets). Instead, go through all the subsets (in a top down order) and for each, if it is available, make a binary choice and branch. If not available, reject forever (for that particular branch). How often do we get a chance to make the binary choice? It seems reasonable that this happens about half the time. It seems pretty random (after a while) if the number of supersets of $X$ in $S$ will be even or odd. Hence $\log_2(\mathcal{F}_n) \approx 2^{n-1}.$ Here is a slight change in your procedure. It might run faster. At least I find it easier conceptually. Decide at the first step which $n-1$-sets (if any) will be in $S.$ At the next step look at all the available $n-2$-sets and pick a subset of them. Continue. So my naive argument is that the expected number of choices should be around $$\prod_0^{n-1}2^{\binom{n}{j}/2}.$$ Or a bit more since at the first step we have $2^\binom{n}{n-1}$ choices rather than the estimated $2^{n/2}$ for this size. Note that there is the family $\{[n]\}\in \mathcal{F}_n$ but, other than that, no $S \in \mathcal{F}_n$ has $[n]\in S.$ This is because the next largest member of $S$ would be contained in only itself and $[n].$ An interesting experiment, which I did not do, might be to pick an $n$ and carry this out (many times) using a (virtual) fair coin. Then look at the typical behavior. AT the first step we have $2^n$ choices. At the next step we consider the $n-2$-sets. Suppose that we previously choose $m$ of the $n$ subsets of size $n-1$. Then the number of available $n-2$-sets will be $c_m=\Sigma\binom{m}{t}\binom{n-m}{m-2-t}$ where the sum is over the $t$ with the same parity as $m$. So actually, $c_0=c_n=\binom{n}{n-2}$ and $c_1=c_{n-1}=\binom{n}{n-2}-(n-1).$ for $n=12$ the counts are $66, 55, 46, 39, 34, 31, 30, 31, 34, 39, 46, 55, 66.$ Almost all these numbers exceed half of $\binom{12}{10}=66.$ And this seems typical for other $n$. However the central numbers, which are less than half, come up (much) more often. So a weighted average is $$\frac{\sum_0^{12}\binom{12}{m}c_m}{2^{12}}.$$ That comes to exactly $33.$ In retrospect, it is easy to see that exactly half the possible choices for $n-1$-sets are such that a given $n-2$-set is then available. Then again, maybe it is better to take the $\log$ of the average rather than the average of the$\log$s. $$\log_2(\frac{\sum_0^{12}\binom{12}{m}2^{c_m}}{2^{12}})=55.0852$$ That seems too large. Past there the analysis is less trivial. At the next to last stage we have (almost always) already committed to a healthy number of members of $S.$ So we would expect the number of available singleton sets to be about $n/2.$ As long as it is not zero, half the choices of available singletons will allow us to add (or not add) $\emptyset$ to $S$ and the other half will not.<|endoftext|> TITLE: The removed statement from the original edition (Matsumura's "Commutative Ring Theory") QUESTION [8 upvotes]: From Theorem 13.7 ii) in the original edition (written in Japanese) of Matsumura's "Commutative Ring Theory," the following statement is got rid of: Let $A = \bigoplus_{n \geq 0}A_n$ be a noetherian graded ring. If $P \subset A$ is a homogeneous prime ideal of height $r$, then there exists an ideal $I = (b_1, \ldots, b_r)$ generated by $r$ homogeneous elements $b_i$ such that $P$ is a minimal prime divisor of $I$. The proof is written as follows: The case $r = 0$ (hence $I = (0)$) is no problem. Assume $r > 0$. By Theorem 13.6, we can choose $a_1, \ldots, a_r \in P$ such that $P$ is a minimal prime divisor of $(a_1, \ldots, a_r)$. Put $J = (a_1, \ldots, a_r)$, and let $a_{ij}$ be the homogeneous term of $a_i$ of degree $j$. Since $JA_P$ is generated by $\{a_{ij}\}_{i, j}$, we can take a minimal basis from $\{a_{ij}\}$. Hence there exist $r$ homogeneuous elements $b_1, \ldots, b_r \in P$ with $JA_P = (b_1, \ldots, b_r)A_P$. Then $P$ is a minimal prime divisor of $(b_1, \ldots, b_r)A$. Here is my question. Where are the mistakes in the proof above? What is a counterexample against the removed statement? REPLY [2 votes]: I think this statement is wrong, because, e.g., a general $\mathbb{N}_0$-graded Noetherian graded local ring $A$ (some texts say $*$-local) does not admit a full homogeneous system of parameters necessarily, unless $A_0$ is a field. In the case where $A_0$ is field, the existence of a full homogeneous system of parameters follows from the graded version of the prime avoidance (see, e.g. [1, Proposition 5.2.]), which it is apparent from the statement that this prime avoidance needs the assumption that the ideal is generated in positive degrees. A general example is many Rees algebras $R[\mathfrak{a}t]$, where $\mathfrak{a}$ is an ideal of a Noetherian local ring $R$, which do not admit a full homogeneous system of parameters. For a general full system of parameters of a Rees algebra, which necessarily is inhomogeneous, see [2, Proposition 3.1]. (By a full homogeneous system of parameters for an $\mathbb{N}_0$-graded ring with a homogeneous maximal ideal $\mathfrak{m}$ I mean a sequence of homogeneous elements of length $\text{height}(\mathfrak{m})$ whose generated ideal is $\mathfrak{m}$-primary). [1]: T. Marely, GRADED RINGS AND MODULES, https://math.unl.edu/faculty/Marley/905notes.pdf. [2]: W. Vasconcelos, Integral Closure Rees Algebras, Multiplicities, algorithms, https://link.springer.com/book/10.1007/b137713.<|endoftext|> TITLE: Distinguished triangle of dualizing complexes and/or determinants? QUESTION [6 upvotes]: Q1 : If $X \to Y \to Z$ are maps of schemes, is there a relation such as $$\omega_{X/Z} \overset{?}{=} \omega_{Y/Z}|_X \overset{L}{\otimes} \omega_{X/Y}$$ between their dualizing complexes? Or maybe some kind of distinguished triangle? The reason I think so is that I'm told $\omega_{X/Z}$ is the determinant $\det \mathbb L_{X/Z}$ in the sense of determinants of complexes (if $X \to Z$ is l.c.i.). The cotangent complex always sits in a distinguished triangle $$\mathbb L_{Y/Z}|_X \to \mathbb L_{X/Z} \to \mathbb L_{X/Y},$$ which leads me to my next question: Q2 : Does the determinant of complexes send distinguished triangles to tensor products? Deligne emphasizes this for split exact sequences of vector spaces in La Determinante de Cohomologie: $\det (V \oplus W) = \det V \otimes \det W$ even in the sense of complexes. It's hard for me to imagine that wouldn't extend to triangles in the derived category. The stacks project 0FJW promises to add more details later, so maybe this is obvious. This came up in a class I'm in. Let $F : X \to X^{(p)}$ be the frobenius. I think surjectivity of the map $F_* \omega_X \to \omega_{X^{(p)}}$ is one of the conditions to be "F-rational." I'd like to understand the cokernel or cone of this morphism in general to see the obstruction to F-rationality and I naively expected something like $\omega_{X/X^{(p)}} = Hom(F_* \mathcal O_X, \mathcal O_{X^{(p)}})$. REPLY [3 votes]: As for the general question about determinants and distinguished triangles, the answer is no, as stated: this is why $\infty$-categories are useful. However, it is true that the determinant is well defined in terms of $\infty$-categories and that is behaves in the way we expect. The point is that $K$-theory is a sheaf of the Zariski topology (at least if we restrict to quasi-compact and quasi-separated schemes - otherwise, we should Zariski sheafify it to make things work properly), and that as such a sheaf of (connective) spectra, and that it can be not only recovered from free modules of finite type, but from perfect complexes as well. More precisely, formula $$\mathit{det}(E\oplus F)\simeq\mathit{det}(E)\otimes_{\mathcal{O}_X}\mathit{det}(F)$$ expresses the fact that, if we let $\mathit{Free}(X)$ be the groupoid of free $\mathcal{O}_X$-modules of finite rank the functor $\mathit{det}:\mathit{Free}(X)\to \mathit{Pic}(X)$, taking values in a sheaf of Picard groupoid (i.e. group-like $E_\infty$-space $\simeq$ connective spectrum), is symmetric monoidal. Since $\mathit{Pic}$ is a sheaf of (Picard) groupoids (here `sheaf' is intended in the sense of $\infty$-categories), the determinant above factors through the sheafification of $\mathit{Free}(X)$, namely the groupoid $\mathit{Vect}$ of vector bundles over $X$. Passing to $K$-theory, this gives a map $$\mathit{det}:K(\mathit{Vect}(X))\to \mathit{Pic}(X)\, .$$ If $X$ is nice enough (has an ample line bundle), Gillet-Waldhausen Theorem says that, if $\mathit{Perf}(X)$ denotes the $\infty$-category of perfect complexes, the inclusion $\mathit{Vect}(X))\subset\mathit{Perf}(X)$ induces an equivalence of ring spectra: $$K(\mathit{Vect}(X))\cong K(\mathit{Perf}(X))\, .$$ In general, if we restrict to quasi-compact and quasi-separated schemes, the presheaf $X\mapsto K(\mathit{Perf}(X))$ turns out to be the sheafification of $X\mapsto K(\mathit{Vect}(X))$ (this is a courtesy of Thomason and Trobaugh). For general schemes, $K(\mathit{Vect}(-))$ and $K(\mathit{Perf}(-))$ have the same Zariski sheafication (since they agree on affine schemes). Finally, we thus get by Zariski descent a functorial map $$\mathit{det}:K(\mathit{Perf}(X))\to \mathit{Pic}(X)\, .$$ By definition of $K$-theory, for any short exact sequence of perfect complexes $E'\to E\to E''$, in the sense of $\infty$-category theory, there is an isomorphism $$\mathit{det}(E)\simeq\mathit{det}(E')\otimes_{\mathcal{O}_X}\mathit{det}(E'')\, .$$ But having a distinguished triangle is less information than having a short exact sequence: a short exact sequence is not only given by the maps $u:E'\to E$ and $v:E\to E''$ but also by an homotopy identifying $vu$ with $0$ (together with the property that $E'$ is the fiber of $v$ or, equivalently, that $E''$ the cofiber of $u$). As a concluding remark, everything above can be done with derived (or even spectral) schemes: the only change in the proof above is that we cannot apply the theorem of Gillet-Waldhausen. We replace it by a theorem of Lurie which expresses the fact that $K(\mathit{Vect}(X))\cong K(\mathit{Perf}(X))$ for affine spectral schemes (Theorem 5 here). If we have to deal with cotangent complexes, this is useful because the functorialities are better understood in the derived context (e.g. the compatibility with pullbacks is almost automatic: the property that the cotangent complex is perfect follows from the property that the structural map is locally of finite presentation and that the relative cotangent complex is of Tor-amplitude $[-n,1]$ for some positive integer $n$). We can get classical results by restricting our attention to that case where we have sufficiently many flat maps around. We may also observe that the description of the determinant above also gives the uniqueness of the determinant of perfect complexes through a nice universal property: in particular, if we insist on functoriality and on the fact that it is map of connective spectra from $K$-theory to the Picard stack $\mathit{Pic}$, it is completely determined by its values on free modules of finite rank and their isomorphisms.<|endoftext|> TITLE: An $L^1$ function but (really) no better? QUESTION [6 upvotes]: Question: For a smooth, bounded domain $\Omega\subset \mathbb R^d$, does there exist a function $u\in L^1(\Omega)$ such that $u\not\in L^\Phi(\Omega)$ for any Orlicz space $\Phi$? For the definition of Orlicz spaces see this wikipedia page. In a nutshell, a Young function is a convex function $\Phi:\mathbb R^+\to \mathbb R^+$ which is superlinear at infinity and sublinear at the origin, $$ \frac{\Phi(u)}{u}\xrightarrow[u\to 0]{ }0 \quad\mbox{and}\quad \frac{\Phi(u)}{u}\xrightarrow[u\to +\infty]{ }+\infty $$ and (with a slight abuse) $$ u\in L^\Phi \quad \mbox{iff}\quad \Phi(|u|)\in L^1. $$ The choice $\Phi(u)=u^p$ leads to the usual Lebesgue spaces $L^p(\Omega)$, so in some sense the Orlicz spaces allow to measure integrability on an arbitrary scale (instead of just pure powers). In particular, choosing $\Phi$ carefully allows somehow to measure fine regularity scales between $L^1$ and any other $L^p$ space. The archetypical and popular example is $L^1\log L^1$, corresponding to the choice $\Phi(u)=u[\log u]^+$ Just a few thoughts (before my question is abruptly downvoted and deemed inappropriate for MO): Of course for a given regularity scale, i-e for any fixed $\Phi$, it is easy to cook-up a function $u=u_\Phi$ that belongs to $L^1$ but not to $L^\Phi$ (just as one can easily tweak the negative exponents $\alpha=\alpha(d,p)<0$ so that $u(x)=|x|^\alpha$ is in $L^1$ but not in $L^p$ for fixed $p>1$ in dimension $d$). The question is therefore: can a (necessarily pathological) function $u$ be $L^1$ but really no better, i-e $u$ does not belong to any better $L^\Phi$ space? (I would perhaps call that such a function "essentially $L^1$".) Again, it is easy to construct $u\in L^1$ that does not belong to $L^p$ for any $p>1$, but typically this function might belong to $L^1\log L^1$. And if it doesn't then it might belong to $L^1\log\log L^1$, and so on... The game is: can one "exhaust all the regulatity scales" except for the minimal $L^1$ requirement? This seems highly unplausible, as the set of all possible regularity scales is highly uncountable, but who knows, some hidden monotonicity might save the day (going "down the scales"?) Following the reverse line of thoughts, another related question is: if $u\in L^1$, can one always find some $\Phi=\Phi_u$ such that $u\in L^\Phi$ is in fact better than merely integrable? My intuition is that this should be blatantly false, but I realized that this is vaguely similar in spirit to Lusin's theorem, stating that essentially a measurable function is continuous. So perhaps this second counterintuitive statement of mine holds in some sense? Some context: it turns out that for one of my research problems I am given a family $(u_\xi)_{\xi\in G}\in \mathcal P(\Omega)$ of Borel probability measures over $\Omega$. For $\xi$ in a smaller (but uncountable) set $F\subset G$ I know that $u_\xi=u_\xi(x) \mathcal L(dx)$ is in fact absolutely continuous w.r.t. the Lebesgue measure $\mathcal L|_\Omega$, and I would like to conclude (by a very intricate argument that I shall not discuss here) that this is the case also for $\xi\in G$. What I need for my argument to work is to find first some $\Phi$ such that $u_\xi\in L^\Phi$ somehow uniformly in $\xi\in F$, and then I would be able to control what happens for $\xi \in G$ using the same scale $\Phi$. Somehow, what I am trying to do is turning a mere pointwise absolute continuity into a globally quantified absolute continuity (the quantification being measured on a putative $L^\Phi$ scale). I am fully aware that at first glance this seems hopeless, but thinking again about Lusin's theorem gave me a vague hope. In any case, I would equally love to be proven wrong or to get some positive insight (a reference, better still). REPLY [12 votes]: There is a much more general result of Vallée-Poussin from which a negative answer to your question follows. Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1(X)$ is equi-integrable if for every $\varepsilon>0$ there is $\delta>0$ such that $$ \sup_{f\in\mathcal{F}} \int_E |f|\, d\mu<\varepsilon \quad \text{whenever } \mu(E)<\delta. $$ Theorem (de la Vallee Poussin). Let $(X,\mu)$ be a measure space with $\mu(X)<\infty$ and let $\mathcal{F}\subset L^1(X)$. Then $\mathcal{F}$ is equi-integrable if and only if there is a Young function $\Phi$, $\lim_{t\to\infty}\Phi(t)/t=\infty$ such that $$ \sup_{f\in\mathcal{F}}\int_X\Phi(|f|)\, d\mu\leq 1. $$ Clearly a family consisting of a single function is equi-integrable (by absolute continuity of the integral) so for each $f\in L^1$ there is $\Phi$ with $\Phi(f)\in L^1$. Also we can adjust the function $\Phi$ around zero arbitrarily without changing integrability of the function $\Phi(f)$ so you can have the condition $\Phi(t)/t\to 0$ as $t\to 0^+$. You can find the proof of the de la Vallee Poussin theorem in C. Dellacherie, P.-A. Meyer, Probabilities and potential. C. Potential theory for discrete and continuous semigroups. Translated from the French by J. Norris. North-Holland Mathematics Studies, 151. North-Holland Publishing Co., Amsterdam, 1988. M.M. Rao, Z.D, Ren, Theory of Orlicz spaces. Monographs and Textbooks in Pure and Applied Mathematics, 146. Marcel Dekker, Inc., New York, 1991.<|endoftext|> TITLE: Weitzenböck formula and comparison of norms QUESTION [6 upvotes]: Let $M$ be a closed Riemannian manifold with a spin$^\mathbb{C}$ bundle $S$. Now for a spin connection $A,$ and a spinor $\phi,$ it can be shown that $C\lvert\nabla_A\phi\rvert^2\geq \lvert D_A\phi\rvert^2$ for some $C>0$. My question is what's the best value of $C$ one can hope for? Ideally this should depend on the geometry and dimension of the manifold I would think. Then again for a flat manifold and with a trivial line bundle, the usual Weitzenböck formula \begin{align*} D_A^2=\nabla_A^*\nabla_A+\frac{s}{4}+\frac{1}{2}F_A \end{align*} would imply that $\int_M \lvert D_A\rvert^2=\int_M \lvert\nabla_A\rvert^2$. My hope is: can $C$ be $1$? Is there an easy counter example to see? I am also interested in a similar question for any unitary connection $B$ on $S$, one can define $D_B:\nabla_B\xrightarrow{\text{Clifford mult.}}D_B$. Again we have $C\lvert\nabla_B\phi\rvert^2\geq \lvert D_B\phi\rvert^2$, does the answer really depend on what connection we use? REPLY [2 votes]: I am not an expert in all the delicate points of clifford algebras and spin structures, but I think the following shows the constant can't be 1 in general: let $(M,g)$ be a Riemannian manifold of dimension $n$. Take the usual Dirac operator $d+\delta:\Omega^{*}(M)\rightarrow\Omega^{*}(M)$ and the covariant derivative $\nabla$. To cook up a concrete example, take $f:M\rightarrow\mathbb{R}$ such that at a point $p\in M$ one has $\nabla df|_{p}=\lambda g$ and take $\omega=df\in\Omega^{1}(M)$. Such a function exists, for example, by taking normal coordinates $(x_i)$ at $p$ and setting $f(x)=\sum_{i\leq j} \lambda\,x^{i}x^{j}$ and mulyiplying by a bump funtion so it is defined on the whole of $M$. Then, since $|g|^{2}=n$, we have $|\nabla \omega|_{p}|^{2}=n\lambda^{2}$ while $|\delta\omega|_{p}|^{2}+|d\omega|_{p}|^{2}=|\mathrm{tr}_{g}\nabla df|_{p}|^{2}=n^{2}\lambda^{2}$. So the constant can't be 1, not even if $(M,g)$ is flat. Here is a more abstract comparsion: for a general 1-form $\omega\in\Omega^{1}(M)$ we find that $|\nabla\omega|^{2}=\frac{1}{4}|d\omega|^{2}+\frac{1}{4}|\mathcal{L}_{\omega^{\sharp}}g|^{2}$ since $\frac{1}{2}d\omega$ is the projection of $\nabla\omega$ onto the anti-symmetric tensors while half the Lie derivative is the porjection onto the symmetric tensors, and symmetric and anti-symmetric tensors are orthogonal. Now, we can decompose $\mathcal{L}_{\omega^{\sharp}}g$ further into a tracless part, let us denote it as $\mu(\omega)$, and its orthogonal component which is $-\frac{2}{n}\delta\omega\,g$. Since these are orthogonal as well, we find $|\nabla\omega|^{2}=\frac{1}{4}|d\omega|^{2}+\frac{1}{4}|\mu(\omega)|^{2}+\frac{1}{n}|\delta\omega|^{2}$, where we used $|g|^{2}=n$. This shows that in general $|\nabla\omega|^{2}\geq\min{(\frac{1}{4},\frac{1}{n})}\,|d\omega+\delta\omega|^{2}$, and I think the above example demosntrates that this is the best you can hope for.<|endoftext|> TITLE: $\mathbf{P} = \mathbf{NP}$, what's the problem? QUESTION [22 upvotes]: Let's take the problem of the backpack: $A_1,\ldots ,A_n$ the weights that are integers, and we want to know if we can achieve a total weight of $V$. We take $$I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times\cdots\times(1+\exp(iA_nt)) \, dt.$$ The question then becomes what $I=0$ or $I\geq 1$. Can we not approach the value of $I$ with the method of Monte Carlo, or others? Why are these approaches not succeeding? Remark : it's not difficult to find a good approximation of $B\times t \bmod 2\pi$, where $B$ is a big integer and $t\in [0,2\pi]$, because we know excellent approximation of $\pi$. REPLY [19 votes]: So you managed to reduce an NP complete problem to calculating a single integral. Actually, in the original Garey / Johnson book, one of their 320 NP complete problems is exactly what we have here (verifying whether some integral has a value of 0; their integral can actually be solved in closed form, but to do that you have to do a lot of steps where each step makes the problem simpler but increases its size - so you end up with something that is just too big to handle in less than exponential time. And numerical calculation also fails). The problem is that your integral cannot be solved in closed form as far as I can tell, but it is also the integral of a function that changes incredibly much. Numerical integration has no chance to solve the integral with the required precision in any reasonable amount of time, and Monte Carlo approximations have no chance - they can give reasonable approximations to some class of integrals, but their precision is even lower than numerical integration.<|endoftext|> TITLE: Is there a classification of minimal algebraic threefolds? QUESTION [7 upvotes]: The minimal model program aims to find a minimal representative in the birational class of a given variety with reasonable singularities. Assuming this has been done, it seems natural to ask what these minimal models look like. Is it feasible to ask for a classification of minimal (complex) varieties, say for threefolds? If so, has this been done? Or are there known wild examples which show that seeking a classification is too ambitious? REPLY [15 votes]: It depends what you mean by classification. The key results for surfaces IMO are: 1) Any surface $S$ of general type has a canonical model given by $S_{can}:={\rm Proj} R(K_S)$ and a unique minimal model given by the minimal resolution of $S_{can}$. 2) The canonical volume ${\rm vol}(S)=K_{S_{can}}^2$ is an integer and 3) $5K_{S_{can}}$ is very ample. In particular for fixed $v$, canonical modules of surfaces of general type of volume ${\rm vol}(S)=v$ can be parametrized by a variety of finite type. On the other hand, for any fixed invariants ${\rm vol}(S)$, $h^1(\mathcal O _S)$ etc it can be impossibly hard to determine the moduli space of canonical models of such surfaces of general type. All of this actually generalizes to all dimensions (with a few caveats / changes). 1) holds by Birkar-Cascini-Hacon-McKernan: $R(K_X)=\oplus H^0(mK_X)$ is finitely generated and we take $X_{can}={\rm Proj}(R(K_X))$; there are also minimal models which have terminal singularities. One canonical model can have multiple minimal models (but only finitely many and they are connected by flops). 2-3) By a result of of Hacon-McKernan, Takayama, Tsuji, for any fixed dimension, the canonical volumes ${\rm vol}(X)=K_{X_{can}}^{\rm dim(X)}$ belong to a discrete set and for fixed dimension $d$ and volume $v$ there is an integer $m$ such that $mK_{X_{can}}$ is very ample. Explicit results are few and far apart (the work of J.Chen and M.Chen comes to mind).<|endoftext|> TITLE: Does the centralizer of a regular element in a semisimple Lie algebra act by polynomials? QUESTION [5 upvotes]: Let $\mathfrak g$ be a semisimple Lie algebra over $\mathbb C$, $\rho : \mathfrak g \to \operatorname{End}(V)$ a finite-dimensional irreducible representation and $x \in \mathfrak g$ regular with centralizer $Z$. Does $\rho(Z)$ consist of polynomials in $\rho(x)$? Examples where I know this is true: $x$ is semisimple: then $Z$ is a Cartan subalgebra, $\rho(x)$ acts by distinct scalars on the weight spaces $V_\lambda$ and the claim follows from Lagrange interpolation. (This argument doesn't work: the scalars need not be distinct.) $\mathfrak g = \mathfrak{sl}_n$ and $\rho = $ standard rep., because $x$ regular implies its minimal polynomial has degree $n$ so the (trace $0$) polynomials in $x$ give all of $Z$. REPLY [5 votes]: It is false for the same reason the first example is wrong. Take $\rho$ for example the adjoint representation. As soon as $\mathfrak g$ has rank at least $2$ you can find a regular semisimple $x$ such that some two roots (whose difference is not a root) of $Z$ coincide on $x$, while this is not true for generic $y \in Z$. The standard representation for $\mathfrak{sl}_n$ is special in that the weights of a Cartan subalgebra differ by roots (rather than linear combinations of such).<|endoftext|> TITLE: Group theory with grep? QUESTION [28 upvotes]: While reading Bill Thurston's obituary in the Notices of the AMS I came across the following fascinating anecdote (pg. 32): Bill’s enthusiasm during the early stages of mathematical discovery was infectious. Once, while sitting in his living room, Bill said to me, “I can do this group with grep,” which was sort of strange to hear at first. But being his student I knew just enough computerese to have an inkling of what he was saying: he was able to compute in that group with the UNIX utility for processing regular expressions using finite deterministic automata. From there, it was exciting to observe the quick unfolding of the theory of automatic groups. Looking through David Epstein's Word Processing in Groups I can see that there are indeed connections between automatic groups and regular expressions that should allow faster algorithms for certain computations e.g. solving the word problem. But is there a concrete example of a group-theoretic computation with grep? REPLY [18 votes]: I wrote that quote, and I'll take the hint of @SamNead and try to write an answer, although the best I can do is to write a somewhat speculative extension of the story behind the quote, laced with some mathematical musings. From my memories of that period (sometime in the mid-late 1980's), I only vaguely recall looking at any actual output of Bill Thurston's computer experiments using grep to compute in groups. My memory is not at all precise enough regarding any actual examples he might have produced, but I suspect that they must have been pretty simple examples, for the following reasons. In the theory of regular languages and FDAs (finite deterministic automata), there are reasonably simple algorithms for going back and forth between regular expressions and finite deterministic automata, producing an expression matching exactly the words accepted by a given automaton, and producing an automaton accepting exactly the words matched by a given expression. The algorithm from regular expressions to FDAs works pretty efficiently and is the basis of the grep utility. The algorithm from FDAs to regular expressions seems to be horribly inefficient, as reported by Derek Holt in his comment. So one might expect that writing an actual regular expression that matches normal forms for a certain group could be quite tedious. Nonetheless if one is clever enough one can sometimes just manually come up with a regular expression to match a given language. See the exercise Derek Holt suggests in his comment; carrying out that exercise will likely give you the most direct answer to your question. I suspect that this is what Thurston was doing during this discovery period: noticing that certain finitely generated groups have a regular language of normal forms. Thurston was also very familiar with concepts of cone types (developed by Jim Cannon around that time; Thurston and Cannon talked together a lot during that period). So at some point during this process he certainly also noticed the connections between cone types and what we now call the 2-tape "multiplier automata" which are at the heart of the theory of automatic groups. It is tempting to speculate whether Thurston's grep experiments extended to writing regular expressions for 2-tape multiplier automata, but honestly I have no idea.<|endoftext|> TITLE: Why is the category of motives generated by varieties? QUESTION [8 upvotes]: I'm reading Ayoub's paper Motifs des varietes analytiques rigides, but I'm not quite familiar with motives. In this paper, he defines the category of motives to be $\mathbf{RigDM}^{\rm eff}_{\rm Nis}(k, \mathbb{Q}) := \mathbf{Ho}_{\rm Nis, \mathbb{B}^1}(\mathbf{Compl}(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q})))$. As I get it, we consider the site of all smooth rigid spaces over $k$, then consider all complexes of presheaves over this site. Just like the construction of derived category, now we invert all quasi-isomorphisms as Nisnevich sheaves, and invert all $\mathbb{B}^1-$homotopy, namely morphisms like $\mathbb{Q}(\mathbb{B}^1\times X)[i]\rightarrow \mathbb{Q}(X)[i]$. What I'm interested in is the Theorem 2.5.35, it says that this category can be compactly generated by those $\mathbb{Q}(X^{\rm an})[i]$ with $X$ being proper smooth $k$-schemes. As I follow the proof, the first step is the fact that the category $\mathbf{Ho}_{\rm Nis}(\mathbf{Compl}(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q})))$(So hence $\mathbf{RigDM}^{\rm eff}_{\rm Nis}(k, \mathbb{Q})$) is compactly generated by those $\mathbb{Q}(X^{\rm an})[i]$ with $X$ being quasi-compact $k$-schemes. Then he uses methods of algebraic geometry to reduce $X$ to be those proper smooth ones. However, I don't quite follow this first step. In the category of modules, since we allow taking arbitrary direct sums, from $\mathbb{Q}$ we get all free modules. And taking direct summands is also allowed, so we have projective modules. Then by projective resolution, we get all modules. And in the derived category, all complex is quasi-isomorphic to its cohomology, so we get the whole derived category. I guess for sheaves, it also follows this process to generate the category of motives. As I understand, the sheaf $\mathbb{Q}(X)$ that $X$ represent is just something as a constant sheaf on $X$. However, I don't quite understand how this goes, and I don't even know if there are enough projective sheaves for this topology... REPLY [2 votes]: No, $\mathbb{Q}(X)$ is the presheaf "additively represented" by $X$; it is not constant. Now I will express my understanding of this matter; I did not check that it fits with Ayoub's definitions and notation, sorry. I believe that the category $\mathbf{Ho}_{\rm Nis}(\mathbf{Compl}(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q})))$ is just a localization of the category $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$, and the localization functor respects coproducts. Thus it suffices to study $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$. Now, this category is generated (as its own localizing subcategory) by the images of $\mathbb{Q}(X)$ essentially by definition. Being more precise, this derived category is defined as the localization of $K(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$ by the subcategory of objects orthogonal to all $\mathbb{Q}(X)$ (look at $O$ with only zero morphisms from all shifts of $\mathbb{Q}(X)$ into it). Thus there are no non-zero objects in $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$ that are orthogonal to $\mathbb{Q}(X)$. Since objects of the type $\mathbb{Q}(X)$ are compact, it follows that they generated $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$ as its own localizing subcategory indeed (this is a well-known fact; see Proposition 8.4.1 of Neeman's "Triangulated categories"). Lastly, this "orthogonality argument" gives an easier (though a "less explicit") proof of the compact generation of $D(R)$ as well.<|endoftext|> TITLE: Why is the set of Hermitian matrices with repeated eigenvalue of measure zero? QUESTION [14 upvotes]: The Hermitian matrices form a real vector space where we have a Lebesgue measure. In the set of Hermitian matrices with Lebesgue measure, how does it follow that the set of Hermitian matrices with repeated eigenvalue is of measure zero? This result feels extremely natural but I do not see an immediate argument for it. REPLY [6 votes]: Here is another argument. The idea is that in any neighbourhood of a matrix with repeated roots, there is one with distinct roots. i.e. we are arguing that matrices with repeated roots are residual, that the matrices with distinct roots are open and dense in the space of matrices. The idea is to put the matrix into its Jordan form. Then you can take a small perturbation that makes all the diagonal entries distinct (only perturbing the diagonal entries). In this form, you can compute the characteristic polynomial and you see the roots are all distinct (as they are the diagonal entries). I suppose this is essentially Mizar's argument but in a coordinate system where there is a little less work to do.<|endoftext|> TITLE: Can we use formal groups to recover Lie-theoretic representation theory in characteristic p? QUESTION [9 upvotes]: In differential geometry, Lie's theorems allow us to integrate any Lie algebra representation to a Lie group representation. The algebraic version of this is more complicated (and I'm not terribly familiar with it), but over a field of characteristic zero we can still get some results from Tannaka duality. This all fails in characteristic $p$, unfortunately. One of the algebraic motivations I've seen for the theory of formal groups is that they rectify the failure of Lie algebras to encode higher-dimensional differential information in positive characteristic, with this representation theoretic-issue given as an example of that failure. But do formal groups actually fix this particular problem? Do we have an adjunction between algebraic groups and formal groups, and can we use it to integrate (some) representations from formal groups to algebraic groups? REPLY [12 votes]: The short answer is that the characteristic $p$ picture genuinely has more depth. In particular it's not correct to think of formal groups as "better" than Lie groups. In fact, they can be put on equal footing with each other. I think it's helpful to put all the objects you're interested in in the same category. In this case it is the category of group objects in formal schemes, which I will call formal group objects. There is some notational confusion here. By rights, the category of formal group schemes should denote this category (which includes all algebraic groups as well). However, often when people say "formal group scheme" or "formal group" they mean infinitesimal groups, which are formal group schemes with a single point (and other times they mean something else more restrictive, such as formally smooth infinitesimal group schemes, which I don't want to consider separately). Here I'll hopefully avoid the notational confusion and use group objects for group objects in formal schemes and infinitesimal groups for group objects with one point. Now in characteristic zero, every algebraic group $G$ has a unique infinitesimal subgroup $\hat{G}\subset G$ (as a group object) subject to the condition of having the same tangent space, i.e., the natural map $T_e\hat{G}\to T_e G$ being an isomorphism. In characteristic $p$, this is no longer the case. If $G$ is an algebraic group in characteristic $p$ then there is still a maximal sub-group object $\hat{G}\subset G$ corresponding to the "full" formal group of $G$, but there is another smaller object $$\hat{G}^{(1)}\subset \hat{G},$$ the first Frobenius neighborhood, which also has the same tangent space (there are also objects $\hat{G}^{(2)}, \dots,$ in addition to possible other intermediate subgroups). The group $\hat{G}^{(1)}$ has the nice property that its representations are equivalent to representations of the Lie algebra $\mathfrak{g}$ as a symmetric monoidal category, and this uniquely characterizes $\hat{G}^{(1)}.$ In fact if $\mathfrak{g}$ is an arbitrary Lie algebra in characteristic $p$, it might not integrate to a smooth algebraic group $G$, but it will always integrate to a group object $G^{(1)}$ that looks like a first Frobenius neighborhood. Now even if you look at the "full" formal neighborhood $\hat{G}\subset G,$ you will not have anything resembling integration of representations. Indeed, if $G\to \text{End}(V)$ is a representation of a smooth algebraic group scheme, you do get induced representations of $\hat{G}$ and $\hat{G}^{(1)}$ (equivalently, $\mathfrak{g}$). But you can neither integrate representations of $\hat{G}^{(1)}$ to representations of $\hat{G}$ nor representations of $\hat{G}$ to representations of $G$ in any reasonable sense, even in one-dimensional cases.<|endoftext|> TITLE: Understanding an involution of the category of perverse sheaves on $\mathbb{C}$ QUESTION [6 upvotes]: It is well-known (for example: chapter 2 in [GGM] A. Galligo, M. Granger, P. Maisonobe. D-modules et faisceaux pervers dont le support singulier est un croisement normal. Ann. Inst. Fourier Grenoble, 35 (1985), 1–48) that a perverse sheaf on $X=\mathbb{C}$ with stratification $X=\mathbb{C}^\times\sqcup \{0\}$ is equivalent to the datum of $(E,F,u,v)$, where $E$, $F$ are finite dimensional vector spaces, $u:E\to F$, $v:F\to E$ are linear maps such that $1+uv$ is invertible. There is an obvious involution on the category of such data, namely $$(E,F,u,v)\mapsto (F,E,v,u).$$ My question is: what is the corresponding operation on perverse sheaves? If I understand the correspondence correctly, the case $(E,F,u,v)=(\mathbb{C},0,0,0)$ corresponds to the constant sheaf $\underline{\mathbb {C}}_X$, and the case $(E,F,u,v)=(0,\mathbb{C},0,0)$ corresponds to the shifted sheaf $\underline{\mathbb{C}}_0[-1]$ supported at the origin. (I am using the convention of [GGM], so in particular the constant sheaf $\underline{\mathbb {C}}_X$ without shift is perverse.) Is there any interesting interpretation of this involution? REPLY [7 votes]: It is the Fourier–Sato transform. You can find a detailed discussion in e.g. section 4D of this article: Bezrukavnikov and Kapranov - Microlocal sheaves and quiver varieties.<|endoftext|> TITLE: $\mathbb{E}_M$ as colimit of little cubes operads QUESTION [6 upvotes]: In Lurie's "Higher Algebra", Remark 5.4.5.2 towards the end, there is the following statement: "It follows that $\mathbb{E}_M$ can be identified with the colimit of a diagram of $\infty$-operads parametrized by $M$, each of which is equivalent to $\mathbb{E}_k$". I do not see how this follows from the preceding discussion there, but let me give a bit of context: Lurie introduces the $\mathbb{E}_M$-operad of a topological manifold $M^n$ by forming the pullback $\operatorname{BTop}(n)^\otimes \times_{\operatorname{BTop}(n)^\coprod} B_M^\coprod$, where $B_M$ is defined as the category $\operatorname{BTop}(n)_{/M}$ with $\operatorname{BTop(n)}$ the full sub-$\infty$-category on $\mathbb{R}^n$ of the category of topological manifolds and embeddings. Note that the operadic structures are the coCartesian symmetric monoidal structure $\coprod$, and disjoint unions, respectively (i.e. $\operatorname{Mul}_{\operatorname{BTop}(n)^\otimes}(\mathbb{R}^n,\mathbb{R}^n ; \mathbb{R}^n) = \operatorname{Emb}(\mathbb{R}^n \sqcup \mathbb{R}^n, \mathbb{R}^n)$ for example). To put it bluntly, $\mathbb{E}_M$ describes embeddings of (disjoint unions of) disks into $M$. In the previously mentioned remark, Lurie shows that $B_M \simeq \operatorname{Sing}(M)$ so that, since $\operatorname{BTop}(n)^\otimes$ has only one object $\mathbb{R}^n$, the underlying $\infty$-category of $\mathbb{E}_M$ is equivalent to $\operatorname{Sing}(M)$. It is also not very hard to see that if we take the full sub-operad of $\mathbb{E}_M$ on one object, i.e. one point in $M$, it is equivalent to $\mathbb{E}_n$. However, I do not see how the statement about colimits above follows from these observations, as Lurie claims - I however find the statement very interesting, as it is in my eyes much more explicit then the assembly of $\mathbb{E}_M$ by copies of $\mathbb{E}_k$ that Lurie proves later. REPLY [2 votes]: I agree this is sort of scattered around as remarks (e.g. HA.2.3.3.4, say). Let's try to spell it out more cleanly. I claim that, if $X$ is a groupoid, then the category of $X$-families of operads is the same as the category of functors $\mathsf{Fun}(X, \mathsf{Op})$. Indeed, $\mathsf{Op}$ is a (non-full) subcategory of $\mathsf{Cat}\downarrow \mathsf{Fin}_*$, so begin by observing that $\mathsf{Fun}(X,\mathsf{Cat}\downarrow \mathsf{Fin}_*)$ is equivalent, by the Grothendieck construction, to the category of diagrams $X \leftarrow \mathcal{E} \to \mathsf{Fin}_*$ where $\mathcal{E} \to X$ is a cocartesian fibration. But if $X$ is a groupoid, every functor with target $X$ is a cocartesian fibration (here I am using the 'homotopy invariant notions', otherwise you have to say that we restrict to maps which are categorical fibrations, etc.) So in fact, $\mathsf{Fun}(X, \mathsf{Cat}\downarrow \mathsf{Fin}_*)\simeq \mathsf{Cat}\downarrow(X \times \mathsf{Fin}_*)$. But now the definition of an $X$-family of operads corresponds exactly to the requirement that the functor from $X$ to $\mathsf{Cat}\downarrow \mathsf{Fin}_*$ factors through the subcategory of operads. Great. Let $\mathcal{E}$ be a generalized operad and $\mathcal{O}$ be an operad. By definition, a map $\mathrm{Assem}(\mathcal{E}) \to \mathcal{O}$ is the same as a map of generalized operads $\mathcal{E} \to \mathcal{O}$. This, in turn, is the same as a map of generalized operads over $\mathcal{E}_{\langle 0\rangle}\times \mathsf{Fin}_*$ from $\mathcal{E}$ to $\mathcal{E}_{\langle 0\rangle}\times \mathcal{O}$. By HA.2.3.2.13, this is the same as a map of families of operads over $\mathcal{E}_{\langle 0\rangle}$ from $\mathcal{E} \to \mathcal{E}_{\langle 0\rangle} \times \mathcal{O}$. Finally, if $\mathcal{E}_{\langle 0\rangle}$ happens to be a groupoid, then by (1) this is the same as a map from the diagram of operads classifying $\mathcal{E}$ to the constant diagram at $\mathcal{O}$. This completes the proof that $\mathrm{Assem}(\mathcal{E})$ is the colimit of the corresponding diagram (when $\mathcal{E}_{\langle 0\rangle}$ happens to be a groupoid.)<|endoftext|> TITLE: Does $(S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$ admit a symplectic form? QUESTION [12 upvotes]: This is a crosspost from this MSE question from a year ago. Consider the smooth four-manifold $M = (S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$. Does $M$ admit a symplectic form? If $\omega$ is a symplectic form, then the real cohomology class $[\omega]$ satisfies $[\omega]^2 = [\omega^2] \neq 0$. Note that $H^*(M; \mathbb{R})$ has such classes. Recall that a symplectic manifold admits an almost complex structure. In general, a closed four-manifold $N$ admits an almost complex structure if and only if there is $c \in H^2(N; \mathbb{Z})$ such that $c \equiv w_2(N) \bmod 2$ and $c^2 = 2\chi(N) + 3\sigma(N)$. As $M$ is spin (i.e. $w_2(M) = 0$), and $\chi(M) = \sigma(M) = 0$, the class $c = 0$ satisfies the two conditions so $M$ admits an almost complex structure. In addition to the above, the vanishing of $w_2(M)$, $\chi(M)$, and $\sigma(M)$ implies that $M$ is parallelisable by the Dold-Whitney Theorem. Moreover, it follows from our knowledge of complex surfaces that $M$ does not admit a complex structure, i.e. $M$ admits almost complex structures, but none of them are integrable. One special aspect of symplectic manifolds in dimension four is that they have a non-trivial Seiberg-Witten invariant by a theorem of Taubes. When $b^+ \geq 2$, a non-trivial Seiberg-Witten invariant implies that no metric of positive scalar curvature exists, but in this case $b^+(M) = 1$ and $M$ does admit a metric of positive scalar curvature. I'm not sure if a deeper understanding of Seiberg-Witten theory could be used to provide a negative answer to the above question. In particular, I don't know if $M$ has a non-trivial Seiberg-Witten invariant. REPLY [14 votes]: No, $M$ is not symplectic. Consider a double cover $\tilde{M}$ of $M$ along one of the $S^1$ components. Then it is not hard to prove that $\tilde{M}$ is diffeomorphic with $(S^1\times S^3)\#2(S^1\times S^3)\# 2 (S^2\times S^2)$. Now if $M$ were symplectic then you could pull back the symplectic structure on $\tilde{M}$. But notice that $\tilde M$ has a 3-sphere separating it into two copies with $b_2^+ = 1$, so $SW(\tilde M) =0$, so this in contradiction with the fact that $\tilde M$ is symplectic.<|endoftext|> TITLE: Quasi-coherent cohomology in non-commutative algebraic geometry QUESTION [5 upvotes]: In non-commutative algebraic geometry, the motto so to speak is to replace the study of a scheme $X$ with the study of the category $D_{qcoh}(X)$ of quasi-coherent sheaves and study the properties thereof. My question is what is the correct analogue for the cohomology of quasi-coherent sheaves in that setting? For a quasi-coherent sheaf $\mathcal{F}$ on $X$, we can write $$H^n(X,\mathcal{F})\cong \text{Ext}^n(\mathcal{O}_X,\mathcal{F})$$ which would suggest that quasi-coherent sheaf cohomology of a category $\mathcal{C}$ of an object $\mathcal{F}\in \text{Ob}(\mathcal{C})$, we have $$H^\bullet(\mathcal{C},\mathcal{F}):= \text{Hom}_{\mathcal{C}}(G,\mathcal{F})$$ for the "correct" analogue of $\mathcal{O}_X$. The problem is, I don't know what the correct analogue is.... Any pointers? REPLY [7 votes]: The structure sheaf is not intrinsic to the derived category (for instance, the standard derived equivalence of the abelian variety and its dual takes the structure sheaf of the former to a skyscraper sheaf on the latter). Therefore, there is no preferred choice for $G$, and thus, if you want to talk about sheaf cohomology, you should specify such $G$ as an additional piece of data.<|endoftext|> TITLE: Finite set of numbers whose powers sum up to irrational number QUESTION [6 upvotes]: It is well-known that $e/\sqrt{2}$ is irrational. Indeed, if it was rational, i.e. $p/q$ then $e^2/2 =p^2/q^2.$ Thus, $q^2e^2=2p^2,$ which would imply that $e$ is a root of $q^2x^2=2p^2.$ Now my question is: Does there exist a finite number of complex numbers $a_1,...,a_N$ all different from zero such that for every $n \in \mathbb N$ $$\sum_{i=1}^N a_i^n \in \mathbb Q \frac{e}{\sqrt{2}}?$$ REPLY [8 votes]: Such numbers do not exist. Indeed, let $p_n$ (resp. $e_n$) be the $n$-th power sum (resp. elementary symmetric polynomial) of the $a_i$'s. Since $e_n=0$ for $n>N$, Newton's identities show that $$\sum_{m_1 + 2m_2 + \cdots + nm_n = n \atop m_1 \ge 0, \ldots, m_n \ge 0} \prod_{i=1}^n \frac{(-p_i)^{m_i}}{m_i ! i^{m_i}}=0,\qquad n>N.$$ Here $p_i^{m_i}$ is a rational multiple of $(e/\sqrt{2})^{m_i}$. As $e/\sqrt{2}$ is transcendental, the contribution of the $n$-tuples $(m_1,m_2,\dotsc,m_n)$ with fixed $m_1+m_2+\dotsb+m_n$ is zero. There is only one $n$-tuple with $m_1+m_2+\dotsb+m_n=n$, namely $(n,0,\dotsc,0)$, which shows that $p_1=0$. Using this information, we can restrict the sum to $m_1=0$. By choosing $n>N$ to be even, a similar reasoning yields that $p_2=0$. By choosing $n>N$ to be divisible by $3$, we can infer that $p_3=0$, and so on. So all the $p_i$'s are zero, which then yields (again by Newton's identities) that all the $a_i$'s are zero.<|endoftext|> TITLE: Great polyhedra: What does "great" signify? QUESTION [10 upvotes]: Great Cubicuboctahedron Great Icosacronic Hexecontahedron Great Rhombic Triacontahedron Great Snub Icosidodecahedron Great Stellated Dodecahedron Great Triakis Octahedron ... There are many polyhedra with a rich variety of beautifully baroque names, often prefixed with "Great." Q. What does "great" signify?            The Great Rhombic Triacontahedron. REPLY [19 votes]: There are two things “great” can refer to. The first, as Sam explained, is a specific kind of stellation. The second is to distinguish conjugates. Since they’d otherwise have the same name, one of them (usually the spikier one) is prefixed with “great” and the other with “small”. Note that not all pairs are distinguished this way. This usually happens when the polytopes in question are snubs, or one of them is convex, but there are other exceptions too, such as the disprismatotesseract and the prismatocubintercepted tesseract. REPLY [15 votes]: Quoting Wikipedia: https://en.wikipedia.org/wiki/Stellation#Naming_stellations John Conway devised a terminology for stellated polygons, polyhedra and polychora (Coxeter 1974). In this system the process of extending edges to create a new figure is called stellation, that of extending faces is called greatening and that of extending cells is called aggrandizement (this last does not apply to polyhedra). This allows a systematic use of words such as 'stellated', 'great', and 'grand' in devising names for the resulting figures. For example Conway proposed some minor variations to the names of the Kepler–Poinsot polyhedra.<|endoftext|> TITLE: An omission in K. Conrad's notes on the conductor ideal QUESTION [11 upvotes]: I am referring to the very useful K. Conrad's notes on the conductor ideal of an order in a Dedekind domain: https://kconrad.math.uconn.edu/blurbs/gradnumthy/conductor.pdf $\DeclareMathOperator\Cl{Cl}$On page 13 it is claimed that every ideal class of the order contains a representative coprime with the conductor, and this would be crucially needed in a paper I am writing. Unfortunately, the proof is omitted (Theorem 5.2). By Lemma 5.1, the statement is valid in the Dedekind domain, and the document suggests that this could be extended to the order by some work. For sure, one of the classes in the preimage via the extension map from $\Cl(\mathcal{O})$ to $\Cl(\mathcal O_{K})$ of an element of $\Cl(\mathcal{O}_{K})$ has the property, but I could not prove why this should also occur with the remaining ones. Has anyone succeeded in completing this proof? REPLY [15 votes]: I can give a non-overkill argument. Let $I$ be an invertible ideal of $\mathcal O$, and $J$ its inverse. Then for each $a \in J$ we obtain an ideal $aI$ in the ideal class of $I$. The ideal $aI$ is prime to $\mathfrak p$ if and only if $ a b \notin \mathfrak p$ for at least one $b\in I$, and $aI$ is prime to the conductor if and only if this is satisfied for each prime $\mathfrak p$ dividing the conductor. For each $\mathfrak p$, this condition clearly only depends on the congruence class of $a$ in $J / \mathfrak p J$. Now by the Chinese remainder theorem for $J$, the map $J \to \prod_{\mathfrak p \mid \mathfrak c} J/ \mathfrak p J$ is a surjection (its proof is the same as the usual Chinese remainder theorem). So it suffices to check for each $\mathfrak p$ that there is $a \in I$ and $b \in J$ with $ab \notin \mathfrak p$. But this is obvious since $IJ = \mathcal O \not\subseteq \mathfrak p$.<|endoftext|> TITLE: Vanishing of a sum of roots of unity QUESTION [15 upvotes]: In my answer to this question, there appears the following sub-question about a sum of roots of unity. Denoting $z=\exp\frac{i\pi}N$ (so that $z^N=-1$), can the quantity $$\sum_{k=0}^{N-1}z^{2k^2+k}$$ vanish ? The answer is clearly No for $N\le7$. I suspect that it never vanishes, which should answer definitively the above matricial question. Any idea of a general argument ? REPLY [14 votes]: For general $N$, we can reason by induction on the $2$-adic valuation of $N$. If $N$ is odd, GH from MO's answer shows that $S_N :=\sum_{k=0}^{N-1} \zeta_{2N}^{2k^2+k} \neq 0$, where $\zeta_{2N} = z = e^{\pi i/N}$ is a primitive $2N$-th root of unity. The same argument shows that for any odd $N$ and any $b \neq 1$, the following variant of $S_N$ is nonzero: \begin{equation*} S_{N,b} :=\sum_{k=0}^{N-1} \zeta_{2N}^{2^b k^2+k} \neq 0. \end{equation*} For convenience of the reader, the proof goes as follows. In the ring $\mathbb{Z}[z]/(2)$, we have \begin{equation*} S_{N,b}^{2^{b+3}} = \sum_{k=0}^{N-1} z^{2^{b+3} (2^b k^2+k)} = \sum_{k=0}^{N-1} \zeta_N^{2^{2b+2} k^2+2^{b+2} k} = \sum_{k=0}^{N-1} \zeta_N^{(2^{b+1} k + 1)^2 - 1} = \zeta_N^{-1} \sum_{\ell=0}^{N-1} \zeta_N^{\ell^2}, \end{equation*} and one conclude as in GH from MO's answer. Now write $N=2^a M$ with $M$ odd, and assume $a \geq 1$. The Galois group of $\mathbb{Q}(\zeta_{2N})/\mathbb{Q}(\zeta_N)$ is of order 2, generated by the automorphism $\sigma : \zeta_{2N} \mapsto \zeta_{2N}^{1+N} = - \zeta_{2N}$. Moreover, we have \begin{equation*} \sigma(S_{N,b}) = \sum_{k=0}^{N-1} (-\zeta_{2N})^{2^b k^2+k} = \sum_{k=0}^{N-1} (-1)^k \zeta_{2N}^{2^b k^2+k}, \end{equation*} so that \begin{equation*} \frac12 (S_{N,b}+\sigma(S_{N,b})) = \sum_{\substack{k=0 \\ k \textrm{ even}}}^{N-1} \zeta_{2N}^{2^b k^2+k} = \sum_{k=0}^{N/2-1} \zeta_{2N}^{2^{b+2} k^2+2k} = \sum_{k=0}^{N/2-1} \zeta_N^{2^{b+1} k^2+k} = S_{N/2, b+1}. \end{equation*} By induction, we have $S_{N/2,b+1} \neq 0$, which implies $S_{N,b} \neq 0$.<|endoftext|> TITLE: About the structure of unit groups appearing in number theory QUESTION [5 upvotes]: I think the following statement is not true in the general situations, but consider it: $R$ is a ring, $\mathfrak{p}$ is a prime ideal, then the unit group of $\dfrac{R}{\mathfrak{p}^nR}$ is isomorphic to $(\dfrac{R}{\mathfrak{p}R})^{*}\times\dfrac{R}{\mathfrak{p}^{n-1}R}$. This statement holds if $R=\mathbb{Z}$, and $\mathfrak{p}=p\mathbb{Z}$ for some odd prime number. What can we say if we consider $R=\mathcal{O}_K$, where $\mathcal{O}_K$ is the ring of integers of a number field? Can we say anything similar, if we consider the localization of $\mathcal{O}_K$ at some prime ideal? I mean are there some sufficient conditions under which the above statement holds? Is there any relation with the roots of unity? I mean can we do something similar to the following: Let $K$ be a number field, and assume that the order of torsion elements in the multiplicative group $K^*$ divides $n$. Then for any prime ideal $\mathfrak{p}$ with $\gcd(\mathfrak{p}, n)=1$ we have: $$(\dfrac{\mathcal{O}_K}{\mathfrak{p}^n})^* \cong (\dfrac{\mathcal{O}_K}{\mathfrak{p}})^{*}\times\dfrac{\mathcal{O}_K}{\mathfrak{p}^{n-1}}.$$ REPLY [7 votes]: In the case of the ring of integers of a number field, this problem is studied and solved in complete detail in Section 4.2 of my book "Advanced topics in Computational number theory", Springer GTM 193. In particular, your statement holds if $e TITLE: Decidability of completeness in propositional logic QUESTION [6 upvotes]: Propositional logic can be presented as in Mendelson’s book, with the sole inference rule of modus ponens, and with the following three axioms: $$B \Rightarrow (C \Rightarrow B)$$ $$(B \Rightarrow (C \Rightarrow D)) \Rightarrow ((B \Rightarrow C) \Rightarrow (B \Rightarrow D))$$ $$((\neg C) \Rightarrow (\neg B)) \Rightarrow (((\neg C) \Rightarrow B) \Rightarrow C)$$ This is a sound and complete theory, as are several other theories for propositional logic. I have questions about similar theories for propositional logic: Given a complete set of logical connectives and a finite set of sound axioms and inference rules, can we algorithmically determine if the resulting theory is complete? Given a finite set of axioms and inference rules $X$ (not necessarily sound) and a formula $\alpha$ in the underlying language of propositional connectives, can we algorithmically determine if $\alpha$ can be derived from $X$? REPLY [6 votes]: It is undecidable, because it is even undecidable to recognize whether a finite set of axioms together with the rule of modus ponens axiomatizes exactly classical propositional logic by the Post-Linial theorem. This was shown in 1948 by Linial and Post, see their announcement (p. 50), but the first published proof is by Yntema. There are many similar results for other propositional logics.<|endoftext|> TITLE: Surgery along knots and connected sum QUESTION [8 upvotes]: Denote $S^3_{p/q}(K)$ by performing $p/q$-surgery along a knot $K$ in $S^3$. Let $K$ and $J$ be two arbitrary oriented non-trivial knots in $S^3$. Is there a nice relation between surgery on the connected sum of knots and the connected sum of their surgeries? More explicitly, let say, do we have the following homeomorphism: $$S^3_{p/q}(K \# J) \approx S^3_{p/q}(K) \ \# \ S^3_{p/q}(J).$$ It might be false. Does this homeomorphism hold for integral surgeries? REPLY [10 votes]: If $p \neq \pm 1$ then this can't happen, because the homology groups of the two sides would be different. I suspect it can't happen at all unless one of the knots is trivial; this would follow from the well-known cabling conjecture. If both knots are non-trivial, then the right side is a reducible manifold, so you would need that at least one summand is the $3$-sphere, which doesn't happen by the resolution of the property P conjecture. The cabling conjecture asserts that a knot with a reducible surgery is a $(p,q)$ cable, the surgery would be a $pq/1$ surgery, and one of the summands would be a lens space $L(pq,1)$. But a non-trivial connected sum is not a cable of another knot. In general, surgery doesn't commute with connected sums, so one wouldn't expect an equality of the sort you're asking about. The sphere along which the connected sum is done doesn't yield a sphere in the surgered manifold; it yields a torus.<|endoftext|> TITLE: Possible values of the determinant for matrices with elements $\{1, 0, -1\}$ QUESTION [15 upvotes]: For matrices with elements $\{-1, 1\}$ it is known from here that the possible absolute values of determinants of $n \times n$, $n \leq 6$ matrices with entries $\{-1, 1\}$ are as follows: n=1: 1 n=2: 0,2 n=3: 0,4 n=4: 0,8,16 n=5: 0,16,32,48 n=6: 0,32,64,96,128,160 Are there any known such results for matrices with elements $\{-1, 0, 1\}$? I have found the following resource here, but unfortunately, it does not go beyond $n=5$. I am interested in such results and methods for $n>5$. One motivating question I am trying to answer is the following: is there a $\{1, 0, -1\}$ matrix of order $6$ with the absolute value of the determinant equal to 27? REPLY [24 votes]: Here's a large subset of possible determinants, which may be enough for your application. With $n\times n$ matrices, one can achieve any integral determinant in the interval $[-2^{n-1},2^{n-1}]$. Specifically, as @nathaniel-johnston notes, the answer to your example question is yes. There's actually straightforward way to construct the matrix (without a computer search). Consider matrices of the form $$M_b= \begin{bmatrix} b_0& 1& 1& 1& 1& 1 \\ b_1& -1& 1& 1& 1& 1 \\ b_2& 0& -1& 1& 1& 1 \\ b_3& 0& 0& -1& 1& 1 \\ b_4& 0& 0& 0& -1& 1 \\ b_5& 0& 0& 0& 0& -1 \end{bmatrix}. $$ Then $-\det(M_b)=b_0+b_1+2b_2+4b_3+8b_4+16b_5$. By picking $b_j\in\{0,1\}$ and using binary, one can get $\det(M_b)$ to be any integer in $[-32, 32]$. So concretely for your example, $$\det\left(\begin{bmatrix} 1& 1& 1& 1& 1& 1 \\ 0& -1& 1& 1& 1& 1 \\ 1& 0& -1& 1& 1& 1 \\ 0& 0& 0& -1& 1& 1 \\ 1& 0& 0& 0& -1& 1 \\ 1& 0& 0& 0& 0& -1 \end{bmatrix}\right)=-27.$$ This technique generalizes to any size matrix. Place $-1$s on the diagonal, $1$s above the diagonal, $0$s below the diagonal, then replace the first column with the binary expansion of your target determinant. The proof of this fact notes that the first row of the inverse of $M_{10\cdots 0}$ are the powers of two, then the cofactor expansion of the determinant along the first column means you get to sum powers of two to obtain possible determinants. An almost identical proof is given in more detail in this paper (shameless plug of my own work, which is was also linked to by @sandeep in the comments of your post.)<|endoftext|> TITLE: Connected vertex-transitive graph with the fixed-point property QUESTION [5 upvotes]: Many connected vertex-transitive graphs $G=(V,E)$ have the property that some of their automorphisms other than the identity have fixed points. To point out two simple examples: If $G = K_3$ then the automorphism swapping the points of an edge and leaving the remaining point intact has $1$ fixed point. For $G = C_4$, the "mirror map" along one of the diagonals has those diagonal points as fixed points. But for both graphs, there are automorphisms that do not have any fixed points. (For both examples, consider a rotation map.) Is there a connected vertex-transitive graph $G=(V,E)$ with $|V| \geq 2$ such that every automorphism has a fixed point? REPLY [9 votes]: Let me convert my comments into an answer. There is no such [EDIT:] finite graph $G$. Indeed, something stronger can be said. Suppose that a group $\Gamma$ acts transitively by permutations on a finite set $X$ , with $\#X \geq 2$. Then there is some $\gamma \in \Gamma$ for which $\gamma\colon X \to X$ has no fixed points. The proof is a simple application of Burnside's Lemma, see e.g. https://math.stackexchange.com/questions/106158/every-transitive-permutation-group-has-a-fixed-point-free-element. For infinite graphs, see this other question: Infinite vertex-transitive graph where every automorphism has a fixed vertex<|endoftext|> TITLE: Does every smooth map of rank at most d factor through a d-manifold? QUESTION [11 upvotes]: Suppose $d≥0$, $m≥0$, $n≥0$, and $\def\R{{\bf R}} f\colon \R^m→\R^n$ is a smooth map whose rank at any point of $\R^m$ is at most $d$. Here and below, smooth means infinitely differentiable. Can we find an open cover $\{U_i\}_{i∈I}$ of $\R^m$ such that for any $i∈I$ the restriction of $f$ to $U_i$ is a smooth map $U_i→\R^n$ that factors through $\R^d$ as the composition of smooth maps $U_i→\R^d→\R^n$? If the rank of $f$ is constant, this follows immediately from the constant rank theorem, so the interesting case is when the rank is not constant. The question is nontrivial only when $00$ for $t\not=0$ and $g^{(k)}(0) = 0$ for $k\ge 0$. Now, define a smooth mapping $f:\mathbb{R}^2\to\mathbb{R}^2$ by the rules $$ f(x,y) = \begin{cases}\bigl(0,yg(y^2{-}x^2)\bigr) & y^2>x^2 \\ \bigl(0,0\bigr) & y^2 = x^2\\ \bigl(xg(x^2{-}y^2),0\bigr) & x^2>y^2\end{cases} $$ Note that the differential of $f$ vanishes identically on the locus $x^2=y^2$, but it has rank $1$ everywhere else. The image of $f$ lies in the union of the $x$ and $y$ axes. In fact, note that the mapping $(x,0)\mapsto f(x,0) = (xg(x^2),0)$ is injective, with nonvanishing derivative except at $x=0$, with a similar statement for $(0,y)\mapsto f(0,y) = (0,yg(y^2))$. I claim that the point $(0,0)$ does not have an open neighborhood $U$ on which there exists a smooth map $\pi:U\to\mathbb{R}$ and a smooth map $\iota:\pi(U)\to\mathbb{R}^2$ such that $f(x,y) = \iota\bigl(\pi(x,y)\bigr)$ for all $(x,y)\in U$. To see this, assume that $\pi$ and $\iota$ exist with the specified properties and note that we can assume without loss of generality that $\pi(0,0) = 0$ and that $U$ is the open disk $x^2+y^2 < r^2$ for some $r>0$ and $\pi(U) = (-s,s)$ for some $s>0$. Because of the above noted behavior of $f$ restricted to the $x$ or $y$ axis, it follows that $\iota$ must be a injection that is an immersion except at $0\in\pi(U)$. From this it is easy to see that $\iota(\pi(U))$ for $r$ sufficiently small must be contained in both the $x$-axis and the $y$-axis, which is clearly impossible.<|endoftext|> TITLE: What is the Todd class *really*? QUESTION [18 upvotes]: My question is about how to think about the Todd class. Usually this is presented via Grothendieck Riemann Roch (GRR): if $X$ is a smooth projective scheme over a field $\mathbf{C}$, the chern character does not commute with integration/proper pushforward: $\require{AMScd}$ \begin{CD} K_0(X) @>\int >> K_0(\text{pt})=\mathbf{Z}\\ @V \text{ch} V V @VV \text{ch} V\\ \text{H}^*(X) @>\int >> \text{H}^*(\text{pt})=k \end{CD} and the Todd class $\text{Td}(\mathcal{T}_X)\in \text{H}^*(X)$ is the correction term to make the above commute. You can then do a computation to explicitly write down what $\text{Td}$ is in terms of chern roots. I will take this story as the definition of the Todd class. However, on its own the content of the above is just: there exists a correction factor, and here is an explicit formula. It doesn't do a good job of explaining what this correction factor really is, or why it should show up in GRR. Ideally there would be a couple more examples where the Todd class shows up, or ways of understanding the Todd class, so that you could understand the correction factor as "that thing that shows up in $\ldots$". Notice that the quality of this answer improves the further away these examples are from GRR. For instance, if the folk result about the Todd class being an Euler class had been written down and made precise, it would be a good answer. My question: in what other contexts (fundamentally different from GRR) does the Todd class appear? REPLY [9 votes]: To my knowledge, currently the best "motivation" for the Todd class comes from the so called "orientation theory" and the formal group laws associated to "oriented" cohomology theories. As far as I know, these ideas go back to Quillen's work on complex cobordism, although current formulation is mainly due to Panin-Smirnov's works (here, here and here). I suggest to read this exposition, which briefly summarizes the main ideas. Let me summarize the explanation without going into every detail. We say that a cohomology theory $H\colon \mathrm{Var}_k\to \mathrm{Rings}$ is oriented if it has a theory of Chern classes with all usual properties one might expect (invariance under pullback, Withney sum and projective bundle formula, nilpotence...). (I am cheating here, people define orientations differently, but there is a theorem stating that both things, orientations and reasonable Chern classes, are in fact equivalent under mild conditions). It is worth noticing two facts about this definition: the cohomology is not assumed to be graded (so that the $K$-theory is such a cohomology) For $L,L'\to X$ two line bundles $c_1(L\otimes L')$ need not to be equal to $c_1(L) + c_1(L')$. Such cohomologies are called additive. For example, $K$-theory satisfies that $c_1(L)=1-[L^*]$ so that $c_1(L\otimes L')=c_1(L) + c_1(L')-c_1(L)\cdot c_1(L')$, which is said to be multiplicative (the name comes from the multiplicative formal group law, where the set of coordinates is taken with respect to 1. That is to say, (1-x)(1-y)=(1-x-y+xy)). The target of the Chern character is always and additive cohomology. The fun thing is that a given cohomology theory $H$ may have different orientations, in other words, different theories of Chern classes. However two orientations can be related concretely: if $H$ is oriented, with some first Chern classes $c_1$, and we have a new orientation, with a new first Chern classes $c_1^{\mathrm{new}}$, then it is easy to check that $$ c_1^{\mathrm{new}}(L)=F(c_1(L))\cdot c_1(L) $$ where $F(t)\in H(pt)[[t]]$ is an invertible series and with coefficients in the cohomology ring of a point. This $F$ is the (inverse) Todd series associated to the new orientation and can be applied, through the usual multiplicative extension, to any virtual vector bundle. Let $\varphi \colon H\to H'$ be a morphism of cohomology theories (i.e., natural transformation between contravariant functor of rings) it is also easy to check that the orientation on $H$ provides an orientation on $H'$ or, in other words, new Chern classes on $H'$ and therefore, an associated (inverse) Todd series. Main example: Consider the Chern character $\mathrm{ch}\colon K\to H_{\mathbb{Q}}$, which is a morphism of cohomologies where $H_{\mathbb{Q}}$ is additive. Let $c_1^K$ denote the Chern classes with values in $K$-theory, to avoid confussion. Let us check the associated (inverse) Todd series to $\mathrm{ch}$: $$ \mathrm{ch}(c^K_1(L))=\mathrm{ch}(1-[L^*]))=1-e^{c_1(L^*)}=1-e^{-c_1(L)}=\frac{1-e^{-c_1(L)}}{c_1(L)}\cdot c_1(L) $$ Hence, its associated series is $\frac{1-e^{-t}}{t}$, which is precisely the inverse of the classical Todd series. Now, to concretely answer your question, here you have: Other examples: The universal map from cobordsim to any oriented cohomology theory $\Omega\to H$ also has its associated (inverse) Todd series, but it is trivial (the series is $F(t)=1$). However, the associated Riemann-Roch type theorem is not at all trivial, is the so-called Conner-Floyd theorem. On any oriented cohomology theory $H$ you can consider a new theory of Chern classes: $c_1^\mathrm{new}(L):=-c_1(L^*)$, its associated (inverse) Todd series is $F(t)=1$ if $H$ is additive, but it is the formal inverse otherwise. -----0----- In the aformentioned exposition you have these kind of arguments with details and "without cheating" (orientations are defined properly, and not just as a "theory of Chern classes"). The Riemann-Roch theorem in this context is as corollary of the universal property of $K$-theory (which is the universal multiplicative oriented cohomology theory).<|endoftext|> TITLE: Is a $k[G]$-module which is free on every cyclic subgroup free? QUESTION [7 upvotes]: Let $G$ be a finite group and let $M$ be a representation of $G$ over a field $k$. Suppose that, for every cyclic subgroup $C$ of $G$, we have $M|_{C} \cong k[C]^{\oplus [G:C]}$. Can we conclude that $M \cong k[G]$? In characteristic zero, this is immediately true by character theory, but I don't know what happens in finite characteristic. Motivation: I was thinking about the normal basis theorem in Galois theory, and I remembered that many sources do the cyclic case separately. Suppose that $M/k$ is a Galois extension with Galois group $G$. Then, for each cyclic subgroup $C$, we know that $M/\text{Fix}(C)$ is a Galois extension with Galois group $C$ so, if we knew the normal basis theorem for cyclic extensions, we would know that $M \cong \text{Fix}(C)[C] \cong k[C]^{\oplus [G:C]}$ as a $k[C]$-module. I was wondering whether this might already be enough to finish the proof with no more field theory input. REPLY [10 votes]: Suppose that $E = C_p \times \dots \times C_p$ be an elementary abelian $p$-group, say of rank $r$, and let $k$ be a field of characteristic $p$. Then the group algebra $kE$ is isomorphic to $k[x_1, \dots, x_r]/(x_1^p, \dots, x_r^p)$. If $v$ is in the vector space spanned by the $x_i$, then $(1+v)^p=1$, so $1+v$ generates a subgroup of the group of units of $kE$. This is called a cyclic shifted subgroup of $kE$. A result of Dade originally, then independently later proved by Carlson and Avrunin-Scott, says the following: A $kE$-module $M$ is free if and only if the restriction of $M$ to each cyclic shifted subgroup is free. (Dade: Lemma 11.8 in Endo-Permutation Modules over p-Groups II, https://www.jstor.org/stable/1971169. Carlson: Theorem 4.4 in The varieties and the cohomology ring of a module, https://www.sciencedirect.com/science/article/pii/0021869383901217.) I believe that there are examples of modules which restrict to free modules over every honest cyclic subgroup but not over every cyclic shifted subgroup, but I don't have one at my fingertips. There should be an example with $G = C_2 \times C_2$, I think.<|endoftext|> TITLE: Ultracategories with one object QUESTION [13 upvotes]: Historically, the theory of ultracategories was invented by Makkai to prove a strong conceptual completeness theorem for first-order logic, roughly: if $T$ and $S$ are two first-order theories such that the ultracategory of models of $T$ is equivalent to the ultracategory of models of $S$, then the theories $T$ and $S$ are "essentially the same" (their pretopos completions are equivalent). There is some more recent material about ultracategories by Lurie, see here and here, which uses a simplified definition of the notion of an ultracategory (which is, for the purposes of proving a strong conceptual completeness theorem, equivalent to Makkai's original definition). The definition (see Definition 7) is still technical, though. So it is natural to ask if one can formulate this notion in more familiar terms when considering special cases of the notion of a category. Indeed, an ultraset turns out to be the same as a compact Hausdorff space (Theorem 9). I wonder if the notion of an ultracategory simplifies to something more familiar if we restrict to categories which are not a set, but a monoid or a preordered set: What is an ultracategory with exactly one object? (Call these things ultramonoids.) What is an ultracategory with at most one morphism between any pair of objects? Coming back to the original motivation of ultracategories, an ultracategory usually consists of structures in the sense of model theory. Let $\mathbf A$ be a structure. Can we equip, say, the endomorphism monoid $\mathrm{End}(\mathbf A)$ of $\mathbf A$ or the monoid $\mathrm{End}'(\mathbf A)$ of elementary embeddings $\mathbf A\hookrightarrow\mathbf A$ with the structure of an ultramonoid? More generally, if $\mathcal M$ is any ultracategory and $M\in\mathcal M$ an object, is $\mathrm{End}(M)$ an ultramonoid? If $F\colon\mathcal M\to\mathcal N$ is an ultrafunctor, is for each $M\in\mathcal M$ the induced map $\mathrm{End}(M)\to \mathrm{End}(F(M))$ an ultrafunctor between ultramonoids? REPLY [11 votes]: $\newcommand{\cat}{\mathrm} \newcommand{\St}{\cat{Stone}^\cat{fr}} \newcommand{\Cat}{\cat{Cat}} \newcommand{\Cart}{\cat{Cart}} \newcommand{\Fun}{\cat{Fun}} \newcommand{\Mon}{\cat{Mon}} \newcommand{\Set}{\cat{Set}} \newcommand{\Po}{\cat{Poset}}$ EDIT : I misread the definition of ultracategory fibration, apparently only certain locally cartesian edges are closed under composition. After a thought about it, what I said for monoids remains true, but not for posets. I just need to modify the arguments. I would argue that a monoid is not the same as a category with one object, namely a monoid is the same as a pointed category with one object. This is not really relevant if you look at the category of categories as a $1$-category (because then $\hom_\Cat(BM, BM')$ is indeed isomorphic to $\hom_\Mon(M,M')$), but it does if you more naturally view it as a $(2,1)$-category ($\Fun(BM,BM')$ has nontrivial morphisms !) In particular, I'll interpret your question as : What is an ultracategory $\mathcal M$ with a morphism from the terminal ultracategory "$*$" such that for all $I$, $*_{\beta I}\to \mathcal M_{\beta I}$ is essentially surjective ? Of course, because $*_{\beta I}\simeq \prod_I*$ in a way compatible with $\mathcal M_{\beta I}\simeq \prod_I\mathcal M_*$, this is equivalent to $*\to \mathcal M_*$ being essentially surjective. The first question is : What is a pointed ultracategory ? In my original post, I made a mistake because I had misread the definition of ultrcategory. In particular, a morphism $\St\to \mathcal M$ is more than just a point in $\mathcal M_*$ : it's a point with a certain property. Namely, (see definition 1 in Lurie's notes) we ask that the comparison morphisms for base-change functors associated to all compositions $\beta I\to \beta J\to \beta K$ be isomorphisms when applied to this point (while in the definition of an ultracategory, one only requires this if $\beta I \to \beta J$ comes from a map $I\to J$). But, in particular, if this point is the only point of $\mathcal M_*$ (and thus of $\mathcal M_{\beta I}$ for all $I$), this amounts to asking that cartesian morphisms actually be closed under composition ! In other words, An ultramonoid is an ultracategory which is a cartesian fibration $\mathcal M\to \St$ with a section $\St\to \mathcal M$ sending cartesian edges to cartesian edges, such that on fibers over $*$, the map $*\to \mathcal M_*$ is essentially surjective. Now that this is clarified, the rest of my answer essentially goes through without changes. But note that there was initially a mistake, and also note that this will not work for posets ! The second thing to observe is that there is a functor $\Mon\to\Cat_*$ which implements the equivalence of $(2,1)$-categories I've mentioned above. This functor is fully faithful (in fact it has a right adjoint - given by $(C,x)\mapsto \cat{End}_C(x)$, this will be relevant later - and the unit map $M\to \cat{End}_{BM}(\bullet)$ is an isomorphism), and therefore the functor $\Fun((\St)^\cat{op}, \Mon)\to \Fun((\St)^\cat{op},\Cat)$ is also fully faithful. By what was clarified above, ultramonoids can therefore be viewed as a full subcategory of $\Fun((\St)^\cat{op},\Cat_*)$, via the Grothendieck construction (while this is not the case for general ultracategories ! for them it would be certain pseudo-functors). Precisely, it is the full subcategory spanned by functors $(\St)^\cat{op}\to\Cat_*$ that send the diagrams $(\{i\}\to \beta I)_{i\in I}$ to product diagrams. Piecing things together, using the fact that $\Mon\to\Cat_*$ and $\Cat_*\to\Cat$ both reflect and preserve products, we find : The category of ultramonoids is equivalent to the full subcategory of $\Fun((\St)^\cat{op},\Mon)$ spanned by those functors that send the diagrams $(\{i\}\to \beta I)_{i\in I}$ to product diagrams. But we can be more concrete. Indeed, $\Fun((\St)^\cat{op},\Mon)\simeq \Mon(\Fun((\St)^\cat{op},\Set))$, and $\Mon \to \Set$ preserves and reflects products, so that : The category of ultramonoids is equivalent to the category of monoids in the category of ultrasets. (note that for ultrasets the subtlety disappears: the fibrations must be cartesian, and not only locally cartesian) Ultrasets are compact Hausdorff topological spaces, so finally: Corollary : The category of ultramonoids is equivalent to the category of monoids in compact Hausdorff topological spaces. Now, if $\mathcal M$ is an ultracategory with a point $x : *\to \mathcal M_*$, there is, in general, no way to make an ultramonoid out of it because of this cartesian vs locally cartesian business. However, if it is a "nice" point, i.e. if it is a point for which the comparison morphisms for general compositions $\beta I\to \beta J\to \beta K$ are isomorphisms, then we can view it as a morphism $\St\to \mathcal M$ of ultracategory fibrations. In particular we can take the full subcategory $\mathcal M_x$ of $\mathcal M$ spanned by the image of $\St$, and this should still be an ultracategory fibration. But now this is a pointed ultracategory fibration with an essentially surjective point, so it is an ultramonoid, as explained above. We therefore find tha a sufficiently nice object $x$ (in more details: a point $\St\to \mathcal M$) in an ultracategory $\mathcal M$ has an ultramonoid of endomorphisms, $\cat{End}_\mathcal M(x)$, i.e. a compact Hausdorff topological space of endomorphisms. (For posets, I don't know a satisfying answer. I'm not sure you will get a more satisfying answer than to the question "what is an ultracategory ?", but I would be interested in seeing one) EDIT to answer the questions in a comment below (rather than as a long sequence of further comments): 1- I used "pointed category" to mean something more naive than in the nLab, namely a category $C$ with a point $*\to C$. Morphisms of such are functors $f : C\to D$ that preserve the point, and $2$-morphisms are natural transformations that are the identity on the point (this is an unfortunate clash of terminology: one can define a "pointed object" in an arbitrary category, and "pointed objects of $Cat$" do not coincide with the nLab's pointed categories). With this definition, the (usual) $1$-category of monoids is equivalent to the $(2,1)$-category of pointed categories for which the point $*\to C$ is essentially surjective. 2- "which question": I was adressing the question about ultramonoids (at first I thought also posets, but in the end, no). I don't know if I'm using the same definition of ultramonoid as you are, because it's unclear to me what "an ultracategory with one object" means in the same way that it's unclear to me what "a category with one object" means. If you mean in the strictest sense and using Definition 7, then no I am not using the same definition (a choice which my first paragraph tries to explain). There are 2 reasons I did not use the same definition: a- what I explained in the first paragraph of this (partial) answer and b- the strict definition is not invariant under the several presentations of "ultracategories" and not invariant under equivalence of ultracategories. I know that this means I'm not technically answering your question (in any case, I wasn't answering all of it, this was just a contribution ! I'm sorry if I made it sound like it was supposed to be everythin). But maybe in the situation you're interested in, the definition I used might be more relevant ? 3- The notation $*_{\beta I}\to \mathcal M_{\beta I}$ refers to $*^I\to \mathcal M^I$ in the language of definition 7, specifically the map that picks out this one object. Sorry again, Definition 1 was clearer to me, which is why I phrased my answer in this language (note that the two definitions are equivalent in a suitable sense). 4- I define the $(2,1)$-category of ultramonoids as a full subcategory of the category of pointed ultracategories (in the sense I described earlier). In particular, morphism categories (rather than morphism sets) are categories of ultrafunctors and, I guess but am not sure about the terminology, ultra natural transformations. 5- A "point" of a category is an object therein, equivalently a functor $*\to C$. I used this latter perspective (which may seem unnecessarily pedantic) to define a point of an ultracategory, as an ultrafunctor from the trivial ultracategory $*$ to $\mathcal M$. That's the content of the paragraph following "The first question is..." - I was trying to identify those more concretely. 6- I'm not sure what the nice objects are in the case of "models of a first order theory". Essentially these are the models $M$ such that for any two sets $J,K$, function $f: J\to \beta K$ and ultrafilter $\mathcal U$ on $J$, the canonical morphism $(\prod_K M)/\mathcal V \to \prod_J ((\prod_K M)/\mathcal f(j))/\mathcal U$ is an isomorphism, where $\mathcal V = \lim_\mathcal U f(j)$ in $\beta K$, equivalently $\mathcal U\wr \mathcal V_\bullet$ where $\mathcal V_\bullet = f$. It looks like this map is always injective by definition of $\mathcal U\wr \mathcal V_\bullet$, but most likely almost never surjective. For instance take $K =*$, then this is $\mathcal M\to (\prod_J \mathcal M)/\mathcal U$, which is almost never surjective. 7- I'm sorry, I misunderstood that these were your main questions. So, for the question of which structures, this amounts to answering 6-. The answer seems to be "almost never" - my guess would be "if and only if $M=*$". It makes sense to a certain extent : what would be the compact Hausdorff topology on the set of elementary embeddings/morphisms ? For the second question however, the way everything is set up makes it very easy to answer : a good point in $\mathcal M$ is an ultrafunctor $*\to \mathcal M$. This is clearly stable under composition of ultrafunctors : if $F:\mathcal{M\to N}$ is an ultrafunctor, so is $*\to \mathcal M\to\mathcal N$, and in particular, $F$ preserves "nice" objects. A corollary of how everything is set up is also (for free) that you get a morphism of ultramonoids (in the sense I described, so a morphism of compact Hausdorff topological monoids) $End(M)\to End(F(M))$. 8- By strongly connected I meant "only one isomorphism class". I'm not sure there is a standard terminology and what it is, if there is.<|endoftext|> TITLE: Infinite vertex-transitive graph where every automorphism has a fixed vertex QUESTION [11 upvotes]: This is a follow-up to the question Connected vertex-transitive graph with the fixed-point property. In particular, it is based on a comment by user bof. Let $G = (V,E)$ be a graph with $V$ infinite. Suppose $G$ is vertex-transitive, i.e., for every pair of vertices $u,v \in V$ there is an automorphism $\gamma$ of $G$ for which $\gamma(u) = v$. Is it possible that every automorphism of $G$ has at least one fixed vertex? The previous question dealt with $V$ finite, for which the answer is no (as long as $\#V \geq 2$) by a simple group theory argument. But here it does not seem possible to use a group theory argument-- at least not so easily-- because there are transitive actions on infinite sets in which every permutation has a fixed point. REPLY [6 votes]: Let $\Gamma$ be the graph whose vertices are the 2-dimensional subspaces of $\mathbb{R}^n$ and two will be adjacent if they intersect in a 1-dimensional subspace. Then $PGL(n, \mathbb{R})$ is in the automorphism group of $\Gamma $. I think it is the full automorphism group but that will need checking. Now any real matrix is similar to its Jordan normal form, which for real matrices is block diagonal with blocks either lower triangular or having size 2x2. Thus it fixes a vertex of $\Gamma$.<|endoftext|> TITLE: The smallest volume possible for a lattice with integer distances? QUESTION [8 upvotes]: Let $\Lambda \subset\mathbb{R}^n$ be a lattice satisfying $\|x-y\|_2^2 \in \mathbb{Z}$ for all $x,y\in\Lambda$. How small can $\text{vol}(\Lambda)=\det(\Lambda)$ be? For example, in dimension $2$, the hexagonal lattice has smallest volume of any integral lattice, equal to $\frac{\sqrt{3}}{2}$. In higher dimensions, rescaling an even unimodular lattice by a factor of $\sqrt{2}$ yields a lattice with integral distances with $\text{vol}(\Lambda) = 2^{-\frac{n}{2}}$. What is the smallest volume possible as $n\rightarrow\infty$? REPLY [3 votes]: Note that $\sqrt2\Lambda$ is even, and $\mathrm{vol}(\sqrt2\Lambda)^2=\det(A)$, where $A$ is the Gram matrix of $\sqrt2\Lambda$. Thus $\mathrm{vol}(\Lambda)=2^{-n/2}\sqrt{\det(A)}$, so the smallest possible volume in dimension $n$ is determined by the smallest determinant of an even $n$-dimensional lattice. The smallest determinant question is answered in the lemma of section 5 in Conway & Sloane's Lattices with Few Distances. The answer is $8$-periodic in $n$, given by $n\ (\mathrm{mod}\ 8)$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $\Lambda_0$ $\{0\}$ $A_1$ $A_2$ $D_3$ $D_4$ $D_5$ $E_6$ $E_7$ $E_8$ $\det A$ $1$ $2$ $3$ $4$ $4$ $4$ $3$ $2$ $1$ where the minimum for a particular $n$ is achieved (usually not uniquely!) by the orthogonal sum of $\Lambda_0$ with some number of copies of $E_8$. That one cannot do better follows from the classification of even forms of small determinant, SPLAG 15.8, table 15.4: if $\det A<4$ then the mod-8 signature is $\pm (\det A-1)$. Returning to the original question, this shows \begin{align} \liminf_{n\to\infty}\ 2^{n/2}\mathrm{vol}(\Lambda) =1,\\ \limsup_{n\to\infty}\ 2^{n/2}\mathrm{vol}(\Lambda) =2. \end{align}<|endoftext|> TITLE: Equality of two $q$-series. Proof? QUESTION [11 upvotes]: Recall the notation $(z;q)_n=(1-z)(1-zq)(1-zq^2)\cdots(1-zq^{n-1})$. My earlier MO question did not find enough interest or yield an answer. Perhaps the modulo $2$ part might have thrown people off. So, I now can write another question which, if proved, would certainly capture the modular problem immediately. So, I would like to propose: QUESTION. Are these two $q$-series equal to each other? $$\sum_{n\geq1}\frac{q^{\binom{n+1}2}}{(q;q)_n}\sum_{m=1}^n\frac{q^m}{1-q^m} =\sum_{n\geq1}\frac{q^{\frac{n(3n+1)}2}(1+q^{2n+1})(-q;q)_n}{(q;q)_n} \sum_{j=1}^n\frac{1+q^{2j}}{1-q^{2j}}.\tag1$$ Postscript. This is a response to Vladimir Dotsenko's request. For the LHS of my earlier post use (1) above, for the RHS use the identity $$\sum_{k\geq0}x^{k+2}q^{k+1}\prod_{j=1}^k(1-xq^j)=\sum_{n\geq0}(-1)^n\left(x^{3n+2}q^{\frac{(n+1)(3n+2)}2}+x^{3n+3}q^{\frac{(n+1)(3n+4)}2}\right)$$ and replace $x\rightarrow q^{-1}, q\rightarrow q^4$. REPLY [17 votes]: I take it from looking at the previous problem that you are familiar with the Dyson rank on partitions with distinct parts. Let's denote by $Q(r,n)$ the number of partitions of $n$ into distinct parts that have rank $r$. The following expression for the generating function of $Q$ is well known: $$F(x,q)\mathrel{\mathop:}= \sum_{r,n} Q(r,n)x^rq^n=\sum_{n\geq 0}\frac{q^{\binom{n+1}{2}}}{(xq;q)_n}.$$ This comes from grouping the partitions according to the number of parts (or equivalently the largest staircase partition that fits in the diagram). Another expression from the generating function comes from grouping the partitions according to the largest pentagonal partition that fits in the diagram (the ones that appear as the fixed points of Franklin's involution mentioned by Vladimir Dotsenko in the comments). More specifically this means that for any partition with distinct parts we find the unique $n$ such that the partition either has: $n$ parts that are $\geq n+1$ and all other parts are $\le n$ $n$ parts that are $\geq n+2$, one part $=n+1$ and all other parts are $\le n$ The contribution of the first group to the generating function of $Q$ is given by $\frac{(-x^{-1}q;q)_n}{(xq;q)_n}x^nq^{\frac{n(3n+1)}{2}}$ and the contribution of the second group is given by $\frac{(-x^{-1}q;q)_n}{(xq;q)_n}x^n q^{\frac{(n+1)(3n+2)}{2}}$. So putting everything together we get $$F(x,q)=\sum_{n\geq 0} x^n q^{\frac{n(3n+1)}{2}}(1+q^{2n+1})\frac{(-x^{-1}q;q)_n}{(xq;q)_n}.$$ Using these two expressions, you can obtain your identity by evaluating $$\frac{d}{dx}F(x,q)\Bigg\rvert_{x=1}$$ in two ways.<|endoftext|> TITLE: Does every simply connected, orientable, non-compact, 3-manifold embed in $\mathbb{R}^3$? QUESTION [6 upvotes]: Let $M$ be a simply connected, (orientable), non-compact, 3-manifold without boundary. Must $M$ be homeomorphic with a topological subspace of $\mathbb{R}^3$? REPLY [4 votes]: When the manifold is the universal cover of a compact $3$-manifold $M$ (to begin with, lets say without boundary) then you construct the embedding by hands, using geometrization. In your question let's replace $\Bbb R^3$ with $S^3$ and we will see the embedding into $\Bbb R^3$ comes for free in the situation you are interested in. An instructive case comes from the case where your manifold is a connect-sum of lens spaces. This has been the subject of a few recent threads: Universal covers of non-prime 3-manifolds The basic idea is that you choose a collection of reducing spheres for the connect sum decomposition, call them $\Sigma$. Then $M \setminus \Sigma$ is a disjoint union of punctured lens spaces. Each of these have universal covers diffeomorphic to punctured spheres, so they embed in $S^3$. You then glue the embedded punctured spheres together (in $S^3$) so that the appropriate boundary spheres are glued together. If you view the embedded punctured spheres from the perspective of their complements in $S^3$ we are essentially doing the canonical construction of a Cantor set. There is the exceptional case of $\Bbb RP^3 \# \Bbb RP^3$ where we are constucting the standard embedding $S^0 \to S^3$. But this is the basic idea. The remaining geometric 3-manifolds have universal covers that are also subsets of $S^3$, so you similarly glue these together along the (lifted) torus decomposition or sphere decompositions. The tori (being incompressible) will lift to copies of $\Bbb R^2$, so those gluings are a little easier to visualize.<|endoftext|> TITLE: Number of resampling until obtaining a uniform list QUESTION [5 upvotes]: Let $A_0$ be a list of $ n$ distinct elements. By sampling with replacement the elements of $A_0$, we obtain a new list $A_1$ of $n$ elements that are not necessarily distinct. Repeat the same process on $A_1$ we obtain $A_2$ and so on. After $ T_n$ steps we first obtain a list of same elements. What do we know about the law of $ T_n$, notably when $n \rightarrow \infty$? Here is the empirical distribution of $T_{10}$ from $50000$ samples. REPLY [3 votes]: One can show the bounds $n \leq \mathbb{E}[T_n] \leq n(n-1)$, and that the law of $T_n$ asymptotically decays exponentially at rate $\lim\limits_{t \to \infty} -\frac{\log\mathbb{P}(T_n > t)}{t} = -\log\left(1 - \frac{1}{n}\right) ( \approx \frac{1}{n}$ for large $n$) via a martingale argument. Let $p^{k}_{t}$ be the frequency of the $k^{\text{th}}$ element after $t$ iterations, and define $q_t := (p^{1}_t, p^{2}_t, \ldots, p^{n}_t)$. Let $\mathcal{F_t}$ be the sigma algebra generated by $\{q_s\}_{s \leq t}$. By the definition of the resampling process, we have: $$ q_{t+1} \sim \frac{1}{n} \cdot \text{Multi}(n, q_t) $$ and initial condition $$ q_0 = \left(\frac{1}{n}, \frac{1}{n}, \ldots, \frac{1}{n}\right) $$ We may now calculate: $$ \mathbb{E}[(p^{k}_{t+1})^2 | \mathcal{F}_{t}] = \text{Var}(p^{k}_{t+1} | \mathcal{F}_{t}) + \mathbb{E}[p^{k}_{t+1} | \mathcal{F}_{t}]^2 \\ = \frac{1}{n^2} \left( n p^k_t (1 - p^k_t) + (n p^k_t)^2 \right) = \left(1 - \frac{1}{n}\right)(p^k_t)^2 + \frac{1}{n}p^k_t $$ where we used the first two moments of the multinomial distribution that are given e.g. here. Summing over all $k$, this becomes: $$ \mathbb{E}[||q_{t+1}||^2 | \mathcal{F}_{t}] = \left(1 - \frac{1}{n}\right)||q_t||^2 + \frac{1}{n} $$ Defining $X_t := \left(1 - \frac{1}{n}\right)^{-(t+1)} \cdot \left(1 - ||q_t||^2 \right)$, we can rewrite this as: $$ \mathbb{E}[X_{t+1} | \mathcal{F}_{t}] = X_t $$ Hence, $X_t$ is a martingale with respect to the filtration $\{\mathcal{F}_{t}\}_{t \in \mathbb{N}}$. Now, note that $||q_t||^2 = \sum_k (p^{k}_t)^2 \leq \sum_k p^{k}_t = 1$ with equality only when $q_t$ is supported on only 1 element. This corresponds to the desired termination condition. Thus, $T_n$ can be characterized as the stopping time $T_n = \min \{t \in \mathbb{N} | X_t = 0\}$. Define the additional stopping time $\tau := \min(T_n, t)$. As $\tau$ is bounded, we may apply the optional stopping theorem to $X$ and $\tau$ to obtain: $$ 1 = \mathbb{E}[X_0] = \mathbb{E}[X_{\tau}] = 0 \cdot \mathbb{P}(T_n \leq t) + \left(1 - \frac{1}{n}\right)^{-(t+1)} \cdot \mathbb{E}\left[ \left(1 - ||q_t||^2 \right) \bigg| T_n > t \right] \cdot \mathbb{P}(T_n > t) $$ Note that we have the upper bound: $$\mathbb{E}\left[ \left(1 - ||q_t||^2 \right) \bigg| T_n > t \right] \leq 1 - \frac{1}{n}$$ which follows from Jensen's inequality, as $\frac{1}{n} \sum (p^{k}_t)^2 \geq \left( \frac{\sum_k p^{k}_t}{n}\right)^2 = \frac{1}{n^2}$. Rearranging gives the desired bound. We also have the lower bound: $$\mathbb{E}\left[ \left(1 - ||q_t||^2 \right) \bigg| T_n > t \right] \geq \frac{1}{n}$$ This follows from the following considerations. Note that $T_n > t$ implies that $||q_t|| \neq 1$, and hence $p^{k}_t < 1$ for all $k$. As the $p^{k}_t$ are normalized frequencies of $n$ items, this further implies that $p^{k}_t \leq \frac{n-1}{n}$, and hence $(p^{k}_t)^2 \leq \frac{n-1}{n} p^{k}_t$. Summing over $k$ and rearranging gives us $1 - ||q_t||^2 \geq \frac{1}{n}$, as desired. Substituting these two bounds gives: $$\left(1 - \frac{1}{n}\right)^{t}\leq \mathbb{P}(T_n > t) \leq n \left(1 - \frac{1}{n}\right)^{t+1} $$ Taking logs, dividing by t, and taking a limit gives the desired rate of decay $\lim\limits_{t \to \infty} -\frac{\log\mathbb{P}(T_n > t)}{t} = -\log\left(1 - \frac{1}{n}\right)$. Summing over all $t \geq 0$ and using the identity $\mathbb{E}[Y] = \sum_{t \geq 0} \mathbb{P}(Y > t)$ gives: $$ n \leq \mathbb{E}[T_n] \leq n(n-1)$$ as desired.<|endoftext|> TITLE: Is this semi-direct product residually finite? QUESTION [5 upvotes]: I've copied over this question from what I asked on Mathematics Stack Exchange, in the hope that some experts here can help me find a way to check the residual finiteness of this group. Consider the group $G=K\rtimes \mathbb{Z}$ defined as follows: The subgroup $K$ is generated by elements $x_i,y_k$ with $i,k \in {\mathbb Z}$ and $k > 0$, and it has defining relations \begin{eqnarray*} x_i^2 &=& y_j^2= 1\ \mbox{for all}\ i,j,\\ [x_j,x_i] &=& y_{j-i}\ \mbox{for}\ j>i,\\ [y_k,x_i] &=& 1\ \mbox{for all}\ i,k, \end{eqnarray*} The action of $({\mathbb Z},+)$ on $K$ is defined by the automorphism $1 \in {\mathbb Z}$ maps $x_i$ to $x_{i+1}$ for all $i \in {\mathbb Z}$. Question: Is group $G$ residually finite? The progress: My idea is to check if $K$ is residually finite first (because if $K$ is not residually finite, then $G$ can't be). So far, if a word $w$ from $K$ satisfies the following condition, then there is a homomorphism from $K$ to a finite group that doesn't send $w$ to the identity. if there exists $x_i$ in $w$, and the total power of $x_i$ is odd. (we can map $K$ to some direct product of $\mathbb{Z}_2$) if $w= y_j$ and $j$ is odd. (We can map $K$ to the Heisenberg group over $\mathbb{Z}_2$) I am not sure how to show such homomorphism exists for any general word. (e.g. a string of $y_i$'s). REPLY [6 votes]: Yes, it's residually finite. This group maps onto the residually finite wreath product $C_2\wr\mathbf{Z}$ and the kernel is central, free over the $y_k$, $k\ge 0$, as 2-elementary abelian group. So one needs to show that for every non-empty subset $J$ of the set $J$ of positive integers, the element $y_J=\prod_{j\in J}y_j$ survives in some finite quotient. Let $t$ be the generator of $\mathbf{Z}$ and kill $t^n$. This kills $y_n$ and identifies $y_i$ to $y_{i+n}$ for all $n$ (and identifies $x_i$ to $x_{i+n}$. At the level of the quotient, the central kernel of the homomorphism onto $C_2\wr C_n$ is freely generated by the $y_i$, $0 TITLE: Harmonic polynomials on the sphere QUESTION [12 upvotes]: Let $\mathbb{S}=\{x\in\mathbb{R}^n|x_1^2+\ldots +x_n^2=1\}$ be the unit sphere in $\mathbb{R}^n$, $\mathbb{C}[x]=\mathbb{C}[x_1,\ldots ,x_n]$ the complex-valued polynomial functions on $\mathbb{R}^n$, and $\Delta=\partial_1^2+\ldots +\partial_n^2$ the Laplacian. Let $H_n=\ker(\Delta)$ be the space of harmonic polynomials. Consider the restriction to the sphere $$\rho:\mathbb{C}[x_1,\ldots ,x_n]\to\mathbb{C}[x_1,\ldots ,x_n]/(x_1^2+\ldots +x_n^2-1)$$ and the image $\bar{H}_n=\rho(H_n)\subseteq\mathbb{C}[x_1,\ldots ,x_n]/(x_1^2+\ldots +x_n^2-1)$. Now, there is the curious fact that $\bar{H}_n$ is in fact the whole of $\mathbb{C}[x_1,\ldots ,x_n]/(x_1^2+\ldots +x_n^2-1)$. In other words, for every $f\in\mathbb{C}[x_1,\ldots ,x_n]$, the polynomial restriction $f|_{\mathbb{S}}$ on the sphere has a harmonic representative $h\in H_n$ such that $f|_{\mathbb{S}}=h|_{\mathbb{S}}$. This follows from a $\sum(\mathrm{radial})\cdot (\mathrm{harmonic})$ decomposition Theorem. Let $f$ be a polynomial. Then $f(x)=\sum_i f_i(x_1^2+\ldots +x_n^2)\cdot h_i(x)$ where $h_i\in H_n$ and $f_i\in\mathbb{C}[t]$. Indeed: $f|_\mathbb S =(\sum_i f_i(x_1^2+\ldots+x_n^2) \cdot h_i)|_\mathbb S=\sum_i f_i(1)\cdot h_i|_\mathbb S = (\sum_i f_i(1)h_i)|_\mathbb S$ and now $h:=\sum_i f_i(1)h_i$ is harmonic. In particular, harmonic restrictions on the sphere form an algebra (equal to the algebra of all polynomial restrictions). From this, I think, one can prove the spherical harmonics are a Hilbert basis on the sphere by using Stone-Weierstrass (and density of continuous functions in $L^2$) without proving that they form a basis of eigenfunctions of the Laplace-Beltrami operator $\Delta_\mathbb{S}$ on $\mathbb S$ (which by the way they do, and it's kinda interesting in itself). Q. Is this stuff part of some theory, or just a coincidence for the usual Laplacian? For example, what if $D$ is a differential operator with symbol $\sigma_D(x)$ and $X_D\subseteq\mathbb{R}^n$ the variety $\{x|\sigma_D(x)=1\}$. Is there sometimes a polynomial decomposition $$f=\sum_i f_i(\sigma_D(x))\cdot h_i(x)$$ with $Dh_i=0$? Or an "intrinsic" operator $P$ on $X_D$, related to $D$ in a similar way as $\Delta_\mathbb{S}$ is related to $\Delta$, such that the homogeneous $h\in\ker D$ are eigenfunctions of $P$? REPLY [2 votes]: I view this as a concatenation of two facts: Fact 1: Let $k$ be a field, let $I$ be an ideal of $k[x_1, \ldots, x_n]$ and let $J$ be associated graded ideal of $I$, meaning that a degree $d$ homogenous polynomial $g(x)$ is in $J$ if and only if there is an element of $I$ of the form $g(x) + (\text{terms of degree $\leq d$})$. Let $\{ p_a \}$ be a set of homogenous polynomials which maps to a basis of $k[x]/J$. Then $\{ p_a \}$ also maps to a basis of $k[x]/I$. The proof of Fact 1 is just an upper triangularity argument. Apply Fact 1 with $I$ the ideal $\langle \sum x_i^2 -1 \rangle$ and $J$ the ideal $\langle \sum x_i^2 \rangle$. So this reduces us to the question of why harmonic polynomials are a basis for $\mathbb{R}[x]/\langle \sum x_i^2 \rangle$. Turn $\mathbb{R}[x_1, x_2, \ldots, x_n]$ into a module over itself by $$g(x_1, \ldots, x_n) \ast f(x_1, \ldots, x_n) := g(\tfrac{\partial}{\partial x_1}, \ldots, \tfrac{\partial}{\partial x_n}){\big(} f(x_1, \ldots, x_n) {\big)}$$ To be clear, this is a differential operator, encoded by $g$, acting on the polynomial $f$. This action induces an inner product on the degree $d$ homogenous polynomials by $\langle g,f \rangle := g \ast f$, since $g \ast f$ is a degree $0$ polynomial and can be thought of an element of $\mathbb{R}$. Let $J \subseteq \mathbb{R}[x_1, \ldots, x_n]$ be any graded ideal of $\mathbb{R}[x_1, \ldots, x_n]$. We write $J = \bigoplus J_d$ for its composition into graded pieces. Let $J^{\perp} = \bigoplus J_d^{\perp}$, where the orthogonal complement is taken with the above inner product on $\mathbb{R}[x_1, \ldots, x_n]_d$. By generalities of inner products, the map $J^{\perp} \hookrightarrow R[x_1, \ldots, x_n] \twoheadrightarrow R[x_1,\ldots, x_n]/J$ is an isomorphism. Fact 2 We have $$J^{\perp} = \{ f : g \ast f=0 \ \forall g \in J \}.$$ Moreover, if $g_1$, $g_2$, ..., $g_N$ is a list of generators for $J$, it is sufficient to impose that $g_i \ast f=0$ for each generator $g_i$. Thus, the set of polynomials with $\sum (\tfrac{\partial}{\partial x_i})^2 f =0$ is a basis for $\mathbb{R}[x]/(\sum x_i^2)$. I learned this by reading Mark Haiman's papers on "diagonal harmonics", and trying to figure out why he was calling them harmonic functions. I'll try to find some better references.<|endoftext|> TITLE: When do multiple polynomials have a common root? QUESTION [5 upvotes]: I was wondering if it is well understood under what circumstances say three univariate polynomials $f(x),g(x),h(x)$ have a common root. In this situation, I can see that the resultant of each pair must vanish but that only ensures that each pair has a common root. Is there a way to generate a finite set of polynomials in the coefficients of $f,g,h$ which tells you when all 3 share at least one common root? Would be interested in an answer for the more general (more than 3 polynomials) case too. EDIT: After thinking about it a tad more here is a possibly interesting observation. If you have $n$ polynomials of degree at most $n$ then you can write it as a linear system in $x,x^2,...,x^n$. Using determinants, minors, etc you will be able to get $n-1$ necessary and sufficient relations between the coefficients which tells you when the polynomials share a common root. I would suspect that means in general if you have $k$ polynomials it might be possible to give $k-1$ polynomials in the coefficients which will be necessary and sufficient conditions for having a common root. REPLY [5 votes]: Assume that $f_n$ is monic. Then for indeterminates $u_1,\ldots,u_{n-1}$, all coefficients of the polynomial $Res_x(f_n,u_1f_1+\ldots+u_{n-1}f_{n-1})\in k[u_1,\ldots,u_{n-1}]$ vanish if and only if $f_1,\ldots,f_n$ have a common root [expanding out this resultant then gives the list of polynomials]. This is how elimination theory works - if $f_1(x_1,\ldots,x_m)=0,\ldots, f_n(x_1,\ldots,x_m)=0$ set-theoretically define a variety $V$, $f_n$ is monic in $x_m$ (which one ensure by applying a Noether normalization coordinate change), and $\pi:\mathbb{A}^m\to \mathbb{A}^{m-1}$ is the projection away from the last coordinate, then $\pi(V)$ is a variety in $\mathbb{A}^{m-1}$ defined by the coefficients of $Res_x(f_n,u_1f_1+\ldots+u_{n-1}f_{n-1})$. Repeatedly eliminating variables like this eventually gives you a finite morphism surjecting $V$ to $\mathbb{A}^{\dim(V)}$.<|endoftext|> TITLE: Decoupling inequality/counterexample QUESTION [11 upvotes]: I am embarrassed to be stuck on this seemingly simple question. Suppose that $X,Y$ are mean-zero real-valued random variables and $\tilde X,\tilde Y$ are their "independent copies": $\tilde X,\tilde Y$ are mutually independent and independent of $(X,Y)$, and $\tilde X$ (resp., $\tilde Y$) is distributed identically to $X$ (resp., $Y$). Here is the inequality I am trying to prove/disprove: for some universal constant $c>0$, $$ \mathbb{E}|\tilde X-\tilde Y| \le c\left( \mathbb{E}|X-Y| + \sqrt{|\mathbb{E}XY|} \right). $$ Update. Note that the related inequality, $$ \mathbb{E}|\tilde X-\tilde Y|^2 \le \mathbb{E}|X-Y|^2 + 2|\mathbb{E}XY| , $$ is trivially true. REPLY [11 votes]: Let $\varepsilon \to 0$ and $M= \sqrt{(1-\varepsilon)/\varepsilon}$. Define $X$ and $Y$ as follows: w.p. $\frac{1-\varepsilon}2$: $X=Y=1$ w.p. $\frac{1-\varepsilon}2$: $X=Y=-1$ w.p. $\frac{\varepsilon}2$: $X=M$ and $Y=-M$ w.p. $\frac{\varepsilon}2$: $X=-M$ and $Y=M$ Then $\mathbf{E}[XY] = (1-\varepsilon) - \varepsilon M^2 = 0$. Also, $\mathbf{E}|X-Y| = \varepsilon \cdot (2M) = 2 \sqrt{\varepsilon(1-\varepsilon)}$. Finally, $\mathbf{E}[X] = \mathbf{E}[Y] = 0$. However, $\mathbf{E}|\tilde X- \tilde Y| \geq \mathbf{Pr}(\tilde X = - \tilde Y) \cdot 2\geq \frac{(1-\varepsilon)^2}{2} \cdot 2 = (1-\varepsilon)^2$. On a separate note, observe that $\mathbf{E}|\tilde X - \tilde Y| = \Theta(\mathbf{E}|\tilde X| + \mathbf{E}|\tilde Y|)$.<|endoftext|> TITLE: (Asymmetric) matrix power series in closed form: $\sum_{i=0}^{\infty} A^i \left(A^i\right)^{\top}={?}$ QUESTION [10 upvotes]: Let $A\in \mathbb{S}^{N\times N}$ be a symmetric, real and stable matrix, i.e., $\rho(A)<1$, where $\rho(A)$ stands for the spectral radius of $A$. Then, $$\sum\limits_{i=0}^{\infty} A^{2i}=\left( I-A^2 \right)^{-1}.\tag{$\star$}\label{star}$$ Now, let $A$ be asymmetric. It is not clear whether the power series $$F(A):=\sum_{i=0}^{\infty} A^i (A^i)^{\top}$$ admits a compact closed-form expression as in equation \eqref{star} (when it converges). If the matrix $A$ is normal — i.e., it commutes with its transpose — and stable, then we have that $F(A)=\left(I-AA^{\top}\right)^{-1}$. In general, that is not the case. Framework. Let $A$ be a random matrix generated as follows. Let $\widetilde{A}\in \mathcal{G}(N,p)$ be the adjacency matrix associated with the realization of a random graph over $N$ nodes with probability $p$ of arrow drawing, i.e., the entries $\widetilde{A}_{ij}$ of $\widetilde{A}$ are i.i.d. $\mathsf{Bernoulli}(p)$ for all $i\neq j$ — this is often referred to as a binomial random graph. Set $\widetilde{A}_{ii}=0$ for all $i$. Let $A$ be obtained from $\widetilde{A}$ by: i) [off-diagonal] normalizing the entries of $\widetilde{A}$ as $A_{ij}:=\alpha_1 \widetilde{A}_{ij}/d_{\max}$, for all $i\neq j$, where $d_{\max}$ is the maximum in-flow degree of the random graph and $0<\alpha_1<1$ is some constant; ii) [diagonal] setting the diagonal terms as $A_{ii}:=\alpha-\sum_{k\neq i} \widetilde{A}_{ik}$, where $0<\alpha_1\leq\alpha<1$. In other words, the rows of $A$ sum to $\alpha$ and its support is given by the realization of the binomial random graph on $N$ nodes. Question. Is it true that $$\mathbb{P}\left(\lVert\sum_{i=0}^{\infty}A^i \left(A^i\right)^{\top} - \left(I-AA^{\top}\right)^{-1}\rVert_\text{max}>\varepsilon\right)\overset{N\rightarrow \infty}\longrightarrow 0?$$ Is there any known result in this direction? Any reference would be appreciated. The specifics of the random nature of $A$ are not important other than that of $A$ being stable and randomly generated (so that it is more natural to consider the limit $N\rightarrow\infty$). Some numerical experiments are pointing in the direction of a concentration of $\sum_{i=0}^{\infty} A^i \left(A^i\right)^{\top}$ about $\left(I-AA^{\top}\right)^{-1}$ as the dimension $N$ grows large (I may attach them later, if necessary) under the referred framework. Context. $F(A)$ is the covariance matrix associated with the time-series reflecting the state-evolution of certain linear stochastic dynamical systems. Having a closed form expression for the covariance in terms of $A$ is helpful for model identification purposes. REPLY [7 votes]: Let me answer the first question: Q: Does $F=\sum_{i=0}^{\infty} A^i (A^i)^{\top}$ for non-symmetric $A$ have a closed-form solution analogous to the solution $\sum_{i=0}^{\infty} A^{2i}=\left( I-A^2 \right)^{-1}$ for a symmetric $A$? A: Yes, in terms of the vectorization operation: $$\operatorname{vec}(F)=(I-{A} \otimes A)^{-1}\cdot\operatorname{vec}(I).\tag{$\ast$}\label{ast}$$ This does require the inversion of an $N^2\times N^2$ matrix, rather than the inversion of an $N\times N$ matrix as in the case of a symmetric $A$. The vectorization formula follows because $F$ solves the Lyapunov equation $$AFA^\top=F-I.$$ I do not have an answer to the second question, whether \eqref{ast} converges for large $N$ to $(I-AA^\top)^{-1}$.<|endoftext|> TITLE: Dihedral extension unramified at primes dividing order of group? QUESTION [5 upvotes]: Consider dihedral Galois extensions $L/\mathbb{Q}$ of degree $n$ (and we know they exist thanks to Shafarevich), can we show there always exists an extension $L/\mathbb{Q}$ unramified at $p$, for all $p \mid 2n$? REPLY [3 votes]: Since the question mentions (the somewhat heavy machinery) "Shafarevich" as a reason for the existence of dihedral extensions, it might be worth adding that the same question has a positive answer (now, indeed due to Shafarevich's method) for all solvable groups (with the condition $p|2n$ replaced by "primes dividing the group order", or in fact even "primes in any prescribed finite set").<|endoftext|> TITLE: Weil height vs Moriwaki height QUESTION [7 upvotes]: Let $X$ be a projective veriety over a number field. After fixing an embedding into $\mathbb P^n$ (i.e. a very ample line bundle $L$), one can define the Weil height $\hat h_{L}$ by restriction of the naive height on $\mathbb P^n$ to $X$. Around 2000 Moriwaki defined a different notion of height in the following way: fix a couple $(X,L)$ where $X$ is as above and $L$ is a line bundle on $X$. Moreover choose a model $(\mathcal X,\overline{\mathcal L})$ where $\mathcal X$ is an arithmetic variety over $O_K$ and $\overline{\mathcal L}$ is a hermitian line bundle on $\mathcal X$ extending $L$. Then by means of Arakelov intersection theory (actually Gillet-Soule' theory) one can define the height function $h_{(\mathcal X,\overline{\mathcal L})}$. What is the advantage of working with the Moriwaki height $h$ instead of $\hat h$? Why Weil height was not enough? As far as I understand, if one fixes $L=O(1)$ and operates with the Fubini-Study metric restricted to $X$ it is possible to recover $\widehat h$ as a special case of the Moriwaki height... (this is shown in chapter 9 of Moriwaki's book on Arakelov Geometry) Why do we need the whole machinery of arithmetic intersection theory in order to obtain "just" a generalisation of the Weil height? REPLY [8 votes]: One can define a height with respect to a special metric so that some particular useful equation involving the height holds exactly, instead of approximately as for the Weil height. For example, one can define the canonical height using a metrized line bundle. This puts it on the same footing as the Weil height and not defined only as a limit of Weil heights composed with various maps. More substantially, the Faltings height was defined this way, using a metrized line bundle on the moduli stack of abelian varieties. The invariance of the Faltings height under isogenies associated to p-divisible groups was crucial in Faltings' proof of the Mordell and Shafarevich conjectures. Edit: Probably the answer to the second question is one doesn't need the full machinery of arithmetic intersection theory, i.e. one can define the height associated to a metrized line bundle without invoking all the techniques needed to handle arithmetic intersection theory in full generality. One just needs to take a section and add up its valuation at each place.<|endoftext|> TITLE: An inequality in C*-algebras QUESTION [5 upvotes]: Let $A$ be a non-unital C*-algebra, and let $\pi: A\to B(A)$ defined by $\pi(a)(x)=ax$ be the left representation of A. Is the following inequality correct? $$\lVert I+ \pi(a) \rVert\ge 1$$ for all $a \in A$. ($I$ is the identity operator and $\lVert\cdot\rVert$ is the operator norm.) If the inequality is correct, is it a known inequality? REPLY [3 votes]: $\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$This is true. Let $a \in A$. Then $|a|$ also belongs to $A$, and since $A$ is nonunital a little bit of functional calculus shows that there exists $x \in A$ with $\norm x = 1$ and $\norm{\abs a x}$ arbitrarily small. (I guess you have to consider the cases where $0$ is or is not isolated in the spectrum of $\abs a$ separately, unless there's an easier argument I'm missing.) Supposing $A$ is sitting inside $B(H)$, we can write $a = u\abs a$ where $u \in B(H)$ is a partial isometry, and then calculate $$\norm{(I + a)x} = \norm{(I + u\abs a)x} \geq \norm x - \norm{u\abs a x} = 1 - \norm{\abs a x}.$$ Since $\norm{\abs a x}$ can be made arbitrarily small, this shows that the norm of $I + a$ in $B(A)$ is at least $1$. I'm sure this is something many people know, but I don't have any reference.<|endoftext|> TITLE: Can every smooth space curve be realized as an origami curved crease? QUESTION [6 upvotes]: Many years ago, Ron Resch told me that he proved that every smooth simple space curve $C$ could be realized as a curved crease $\gamma$ in the interior of a piece of paper. He never published this (as far as I know), and perhaps never wrote it down formally. I would like to know if some form of this claim has been, or can now be, proved. By a curved crease in a piece of paper, I mean what David Huffman achieved with his remarkable curved-crease origami "sculptures": Huffman proved1 that at any point $x$ on $\gamma$, the osculating plane of $\gamma$ at $x$ bisects the two planes tangent to the surface at $x$ from either side. Aside from that, I do not know if further structural properties of $\gamma$ are known and would help prove Resch's claim. 1David A. Huffman. Curvature and creases: A primer on paper. IEEE Transactions on Computers, C-25(10): 1010–1019, Oct.1976. REPLY [6 votes]: Note: I'm revising my answer to make the argument/construction more transparent. In the previous version, I stated an existence result about flat surfaces, but didn't indicate a proof (because, at the time, I didn't remember a simple proof). Now, having thought about it (or maybe just remembered the original construction), I can just provide the flat surface explicitly. Here is a straightforward proof that works as long as the curvature of $\gamma$ is nonzero. I don't know where it might be proved in the literature, but it's really just a standard `curves and surfaces' argument. Suppose that one has an embedded, unit speed space curve $\gamma:I\to \mathbb{R}^3$ as well as a unit vector field $\nu:I\to S^2$ `that is normal along $\gamma$', i.e. $\gamma'(s)\cdot \nu(s) =0$ and also has the property that $\gamma'\cdot \nu'$ is non-vanishing. Then the triple $(e_1,e_2,e_3) = (\gamma', \nu\times\gamma', \nu)$ is an orthonormal frame field along $\gamma$, so there are functions $a$, $b$, and $c$ on $I$ such that $$ \gamma'' = a\,\nu + c\,\nu\times\gamma',\qquad \nu' = - a\,\gamma' - b\,\nu\times\gamma',\qquad (\nu\times\gamma')' = b\,\nu - c\,\gamma'. $$ Of course, $a=-\gamma'\cdot\nu'$ is nonvanishing. It is then not difficult to show that the mapping $X:I\times\mathbb{R}\to\mathbb{R}^3$ defined by $$ X(s,t) = \gamma(s) + t\,\bigl(b(s)\,\gamma'(s) - a(s)\,\nu(s)\times\gamma'(s)\bigr) $$ is a smooth embedding on an open neighborhood $U$ of $t=0$ and that the image is a ruled surface with vanishing Gauss curvature, i.e., it is locally isometric to the plane. Moreover, calculation shows that, in this surface, oriented so that $\mathrm{d}s\wedge\mathrm{d}t$ is positive, the function $c$ is the geodesic curvature of $\gamma(I)$ in this surface $X(U)$. Finally, although $(e_1,e_2,e_3)$ is not (usually) the Frenet frame of $\gamma$, one can compute the Frenet frame, and one finds that the curvature $\kappa$ and torsion $\tau$ of $\gamma$ as a space curve are given by the following formulae: $$ \kappa = \sqrt{a^2 + c^2},\qquad \tau = \frac{c\,a'-a\,c' + b\,(c^2-a^2)}{a^2+c^2} $$ Now, suppose that one is only given a unit speed space curve $\gamma:I\to\mathbb{R}^3$ that has nonvanishing curvature $\kappa$ and torsion $\tau$ (which might or might not vanish). Choosing a function $\theta:I\to\mathbb{R}$ such that $\sin\theta$ and $\cos 2\theta$ are nonvanishing, we can write down a solution of the above equations for $a$, $b$, and $c$ as $$ a = \kappa\sin\theta,\quad b = (\tau-\theta')\sec2\theta,\quad c = \kappa\cos\theta $$ If $T=\gamma'$, $N$ and $B$ are the Frenet frame of $\gamma$, this solution corresponds to choosing $\nu_\theta = \sin\theta\, N + \cos\theta\, B$, as is easy to check. Now, consider the two ruled surfaces $X$ that one gets by using $\nu_\theta$ and $\nu_{-\theta}$. They both have the same geodesic curvature along $\gamma$, and so joining the $t\ge0$ part of the $\theta$-surface and the $t\le 0$ part of the $(-\theta)$ surface, we get a flat surface that is creased along $\gamma$ at a normal angle of $2\theta$ (which is the angle between their normals along $\gamma$), but whose induced metric is flat in a neighborhood of $\gamma$. Thus, we have a creased flat surface (i.e., a 'piece of paper') whose crease is the given space curve $\gamma$.<|endoftext|> TITLE: Comprehension axiom who helps in the opposite direction QUESTION [8 upvotes]: Usually, having more comprehension axiom means the more you can prove. We wonder if the converse can be the case. Is there a natural problem $\mathsf{P}$ so that $\mathsf{P}+\neg(\Gamma-\mathsf{ComprehensionAxiom})$ implies $\mathsf{Q}$, but $\mathsf{P}$ does not implies $\mathsf{Q}$. Also I'd like to exclude some trivial case such as $\mathsf{Q}$ is $\neg(\Gamma-\mathsf{ComprehensionAxiom})$. We can accept $\mathsf{Q}$ to be some Comprehension Axiom, but not the negation of them. The implication is not necessary over $\mathsf{RCA}$ and $\Gamma$ can be any set of formulas. REPLY [4 votes]: There are many such examples in higher-order Reverse Mathematics. I discuss a couple and indicate a relevant paper of mine related to David's. As usual, let RCA$_0^\omega$ be Kohlenbach's base theory which is conservative over RCA$_0$. First of all, here are some examples: RCA$_0^\omega$ (and much stronger systems) cannot prove the Lindeloef lemma for uncountable coverings as follows: For any $\Psi:\mathbb{R}\rightarrow \mathbb{R}^+$, there is a sequence $x_0, x_1,x_2, \dots$ of reals such that $\cup_{n\in \mathbb{N}}B(x_n, \Psi(x_n))$ covers $\mathbb{R}$. However, RCA$_0^\omega + \neg$ACA$_0$ can prove the previous statement. RCA$_0^\omega +$ WKL (and much stronger systems) cannot prove the Heine-Borel theorem for uncountable coverings as follows: For any $\Psi:[0,1]\rightarrow \mathbb{R}^+$, there is a finite sequence $x_0, \dots, x_k$ of reals in $[0,1]$ such that $\cup_{n\leq k}B(x_n, \Psi(x_n))$ covers $[0,1]$. However, RCA$_0^\omega + $ WKL $+ \neg$ACA$_0$ can prove the previous statement. RCA$_0^\omega +$ WKL (and much stronger systems) cannot prove the (essence of the) Vitali covering theorem for uncountable coverings as follows: For any $\Psi:[0,1]\rightarrow \mathbb{R}^+$ and $\epsilon>0$, there is a finite sequence $x_{0}, \dots, x_k$ of reals in $[0,1]$ such that $\cup_{n\leq k}B(x_n, \Psi(x_n))$ has measure at least $1-\epsilon$. However, RCA$_0^\omega + $ WKL $+ \neg$ACA$_0$ can prove the previous statement. (It is an interesting question whether WWKL also suffices.) There are many many more such examples. Secondly, since Noah mentions David Belanger's paper, I should mention my "follow-up" paper which appeared in the NDJFL (and arxiv): Sam Sanders, Splittings and disjunctions in Reverse Mathematics, Notre Dame J. Formal Logic 61(1): 51-74 (January 2020). DOI: 10.1215/00294527-2019-0032. One will find many examples based on the results that paper (directly or indirectly). Thirdly, for those who wish to know more: let $(\exists^2)$ be Kleene's arithmetical quantifier, i.e. $$ (\exists E^2)(\forall f^1)\big((\exists n^0)(f(n)=0)\leftrightarrow E(f)=0 \big). $$ It is well-known that RCA$_0^\omega + (\exists^2)$ is conservative over ACA$_0$. One generally refers to $E^2$ as in $(\exists^2)$ as a `comprehension functional', for obvious reasons. As shown by Kohlenbach (by formalising so-called Grilliot's trick in the base theory), $(\exists^2)$ is equivalent over the base theory to there exists an $\mathbb{R}\rightarrow \mathbb{R}$-function that is not everywhere (epsilon-delta) continuous. Clearly, $(\exists^2)\rightarrow$ACA$_0$ and contraposition tells us that $\neg$ACA$_0$ implies Brouwer's theorem in that all functions on the reals are continuous. In this light, the above items go through.<|endoftext|> TITLE: If we replace the spectrally ringed space in the definition of a spectral scheme with an arbitrary infinity-topos, what objects do we get? QUESTION [7 upvotes]: I'll phrase this in terms of spectral AG, but I'm curious about the same question in the classical context. We define a nonconnective spectral Deligne-Mumford stack to be a spectrally-ringed topos which admits a cover by the étale spectra of $E_{\infty}$-rings. The definition of a nonconnective spectral scheme, on the other hand, is a spectrally-ringed space which admits a cover by the Zariski spectra of $E_{\infty}$-rings. Suppose we instead look at spectrally-ringed topoi which are locally isomorphic to the Zariski topos of an $E_{\infty}$-ring. Clearly, these generalize nonconnective spectral schemes; and, on the other hand, these objects admit a natural functor to nonconnective spectral DM-stacks, which should be a fully faithful embedding extending the inclusion of schemes as schematic stacks. My question is this: are these objects actually more general than nonconnective spectral schemes, or is the underlying topos always generated by a topological space? If the former, how much more general are they? (For instance, a nonconnective spectral DM-stack is schematic iff it admits a cover by $(-1)$-truncated affine objects. Is there a similar characterization of these "generalized schematic stacks"?) REPLY [13 votes]: Yes, they are more general. This is in fact already the case with ordinary rings. Let's call a classically-ringed $\infty$-topos which is locally the Zariski $\infty$-topos of an affine scheme an $\infty$-scheme. A classical scheme is then the same as a $0$-localic $\infty$-scheme. To construct an $\infty$-scheme which is not classical, let $F$ be any object in the Zariski $\infty$-topos $\mathrm{Shv}(X)$ of a classical scheme $X$. Then the slice $\infty$-topos $\mathrm{Shv}(X)_{/F}$ with the restricted sheaf of rings is an $\infty$-scheme (it is covered by open subschemes of $X$), which is classical iff $F$ is $0$-truncated. This same construction works to define spectral $\infty$-schemes that are not $0$-localic. A theorem of Lurie (Theorem 2.3.13 in DAG V) implies that every $\infty$-scheme is such a slice over a $1$-localic $\infty$-scheme. I do not know if there exist $1$-localic examples that are not slices over classical schemes. If one works with the étale topology, however, this theorem implies that every DM $\infty$-stack is a slice over a classical DM stack. $\infty$-schemes can also be identified with the full subcategory of $\mathrm{Shv}_{\mathrm{Zar}}(\mathrm{CRing}^\mathrm{op})$ consisting of those Zariski sheaves (of spaces) $F$ that admit an effective epimorphism from a coproduct of representable sheaves $R_i$ such that each $R_i\to F$ is "representable by slice $\infty$-topoi" (see Proposition 2.4.17(6) in loc. cit.). Unlike with classical schemes, however, such sheaves do not usually satisfy descent with respect to the étale topology, because higher Zariski and étale cohomology generally disagree. ETA: This last remark implies that the natural functor from $\infty$-schemes to DM $\infty$-stacks is not fully faithful. If it were, then the functor of points of any $\infty$-scheme would be the same as the functor of points of the associated DM $\infty$-stack, which is an étale sheaf.<|endoftext|> TITLE: Shelah's "Can you take Solovay's inaccessible away?" QUESTION [6 upvotes]: I was wandering if there was a book, thesis or some notes where Shelah's argument for $\mathtt{ZF}+\mathtt{DC}+$"All sets of reals are Lebesgue measurable" is equiconsistent with $\mathtt{ZFC} + \exists \kappa$ inaccessible $\mathtt{ZF}+\mathtt{DC}+$"All sets of reals have the Baire property" is equiconsistent with $\mathtt{ZFC}$ contained in Can you take Solovay's inaccessible away? is explained in a newer and/or "more digestible" way. Is there? Thanks! REPLY [9 votes]: Chapter 9.5 of the book by Bartoszynski-Judah presents Raisonnier's proof (answering your question 1): Assume that $\aleph_1$ is not inaccessible in $L$, hence a successor in $L$. So there is a real $x$ which knows that the $L$-predecessor is countable, hence $\aleph_1^{L[x]}=\aleph_1$, so $X:=\mathbb R \cap L[x]$ is uncountable. Now assume that all $\Sigma^1_2(x)$-sets are Lebesgue measurable. Then (this needs some work) the "Raisonier filter" $F_X$ built from $X$, which is a $\bf \Sigma^1_3$-set, is a rapid filter and therefore (again some work, known earlier) not measurable.<|endoftext|> TITLE: Why are we interested in spectral gaps for Laplacian operators QUESTION [10 upvotes]: Let $M$ be a Riemannian manifold and let $\Delta$ be its Laplacian operator. There is a large literature on a spectral gap for such a $\Delta$, that is, finding an interval $(0,c)$ which does not contain any eigenvalues of $\Delta$. Why are people interested in producing such spectral gaps? What interesting information do they give about the manifold? Are there interesting applications in the `real work', i.e. in physics? REPLY [5 votes]: A spectral gap gives information on geometry of the manifold via Cheeger's inequality, https://en.wikipedia.org/wiki/Cheeger_constant See also Buser's inequality discussed there. More directly, a spectral gap for the Laplacian yields exponential decay for the heat kernel and determines the asymptotic rate of decay. The heat kernel represents transition probabilities for Brownian motion on the manifold, so a spectral gap means the motion mixes rapidly (in the finite volume case) and dissipates quickly (in the infinite volume case). https://en.wikipedia.org/wiki/Heat_kernel<|endoftext|> TITLE: Counting monomials and the Catalan numbers QUESTION [5 upvotes]: Given a multi-variable polynomial $F$, denote the number of monomials by $N(F)$. Take for instance, \begin{align*}N(x(x+y)+(x+y)^2-(x-y)^2)=N(x^2+5xy)&=2 \qquad \text{and} \\ N((x+z)(x+y)^2)=N(x^3 + 2x^2y + x^2z + xy^2 + 2xyz + y^2z)&=6.\end{align*} Denote the symmetric group on $n$ letters $\{1,2,\dots,n\}$ by $\mathfrak{S}_n$. Define the action of $\mathfrak{S}_n$ on a function $F(x_1,\dots,x_n)$ in a natural way: given $w\in\mathfrak{S}_n$, then $w\cdot F(x_1,\dots,x_n)=F(x_{w(1)},\dots,x_{w(n)})$. Introduce the (multi-variable) rational functions $$G(\mathbf{x},\mathbf{z})=\prod_{k=1}^n\frac{x_1z_1+x_2z_2+\cdots+x_kz_k}{x_k-x_{k+1}}.$$ Assuming that the symmetric groups act only on the $x$-variables, let's compute the polynomial $$G_{\mathfrak{S}_{n+1}}=\sum_{w\in\frak{S}_{n+1}}w\cdot G.$$ I would like to ask: QUESTION. Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Is there a combinatorial proof that $N(G_{\mathfrak{S}_{n+1}})=C_n$? REPLY [9 votes]: Any monomial $P:=\prod x_i^{c_i}$ of degree $\sum c_i=n$ maps to a non-zero constant after symmetrization $$ P\to \Phi(P):=G_{\mathfrak{S}_{n+1}}\frac{P}{(x_1-x_2)(x_2-x_3)\ldots (x_n-x_{n+1})}. $$ Indeed, $\Phi(P)$ is a constant by a degree consideration, to prove that this constant is non-zero you may use, for example, Theorem 2 here. Thus any monomial $\prod (x_iz_i)^{c_i}$ maps to a non-zero constant times $\prod z_i^{c_i}$. Thus, you simply count the number of monomials which may arise, when you multiply the linear forms $\prod_{k=1}^n(x_1z_1+x_2z_2+\cdots+x_kz_k)$. The only condition is that $c_1+\ldots+c_i\geqslant i$ for all $i$. Such sequences are indeed enumerated by Catalan numbers, a bijection with lattice paths below diagonal is straightforward.<|endoftext|> TITLE: Is there a conjectured dependence on $n$ in van der Waerden's conjecture? QUESTION [6 upvotes]: Bhargava 2021 proves van der Waerden's conjecture about Galois groups of random integer polynomials: over all $x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 0$ with $a_k \in \{-H, \ldots, H\}$, the number of polynomials $E_n(H)$ with Galois group not equal to the full symmetric group $S_n$ is $O(H^{n-1})$. First (and this is a very basic question), am I correct that the $O$ here is in terms of $H$ only, so that in detail we have $E_n(H) < c_n H^{n-1}$ for some constants $c_n$ that depend on $n$? Second, is there a conjectured growth rate for $c_n$? From Bhargava 2021 it seems like in the worst case $c_n$ might have a factor related to the number of nonisomorphic groups of order $n$ (possibly with other factors), but presumably most of the count concentrates on fewer of the groups. REPLY [10 votes]: There are two different questions you could ask here. One is the dependence on $n$ in the conjectured asymptotic best constant, i.e. the $c_n$ such that we have $E_n (H) < c_n H^{n-1} + o(H^{n-1})$ with the little $o$ depending on $n$. This one has a precise answer, because we do not need to worry about all the groups that are proven or expected to occur with frequency $o(H^{n-1})$. So we just need to deal with the reducible polynomials, where it was computed by Chela that we can take $$ c_n = 2^n (\zeta(n-1) - 1/2) + 2 k_n $$ where $k_n$ is the volume of the region in $\mathbb R^{n-1}$ consisting of tuples $x_1,\dots, x_{n-1}$ with $$-1 \leq x_1,\dots, x_{n-1}, \sum_{i=1}^{n-1} x_i \leq 1 .$$ It is straightforward to check that this has growth rate proportional to $2^n$ in $n$. The other is the dependence on $n$ of the conjectured best constant valid for all $H$. This is a somewhat annoying question because it will depend heavily on small values of $H$. For example, we must take $c_n \geq 3^{n-1}$ to deal with the case where $H$ is $1$ (and the constant coefficient vanishes). It's possible that we can take $c_n \leq 3^n$ - I don't see a reason against it.<|endoftext|> TITLE: Is equation $xy(x+y)=7z^2+1$ solvable in integers? QUESTION [31 upvotes]: Do there exist integers $x,y,z$ such that $$ xy(x+y)=7z^2 + 1 ? $$ The motivation is simple. Together with Aubrey de Grey, we developed a computer program that incorporates all standard methods we know (Hasse principle, quadratic reciprocity, Vieta jumping, search for large solutions, etc.) to try to decide the solvability of Diophantine equations, and this equation is one of the nicest (if not the nicest) cubic equation that our program cannot solve. REPLY [20 votes]: Here is a short proof, inspired by Michael Stoll's nice arguments. Assume that $x,y,z\in\mathbb{Z}$ solve the equation. Then $xy(x+y)$ is positive and not divisible by $7$. Without loss of generality, $x$ is positive and not divisible by $7$. Using the identity $$\left(x^2 + \frac{2}{x+y}\right)^2 + 7\left(\frac{2z}{x+y}\right)^2 = x (4 + x^3),$$ we see that $x(4+x^3)$ is a norm in $\mathbb{Q}(\sqrt{-7})$. Hence both $x$ and $4+x^3$ are norms in $\mathbb{Q}(\sqrt{-7})$, because the only prime factor they can share is $2$, which splits in $\mathbb{Q}(\sqrt{-7})$. In particular, $4+x^3$ is a quadratic residue mod $7$, contradiction.<|endoftext|> TITLE: What is the smallest size of a shape in which all fixed $n$-polyominos can fit? QUESTION [24 upvotes]: Let $n$ be an integer and consider all fixed $n$-polyominos, i.e., without rotation or reflection. I am interested in finding a shape in which all polyominos can embed. (It is OK if multiple polyominos overlap.) For instance, for $n=3$, the fixed 3-polyominos are: ### #.. ##. ##. #.. .#. ... #.. #.. .#. ##. ##. ... #.. ... ... ... ... and these polyominos all embed in the following shape with 5 cells, which is the best possible: .#. ### .#. More generally, a suitable shape for arbitrary $n$ is the $n \times n$ square (with $n^2$ cells) and a naive lower bound would be $2n-1$ cells (necessary to embed the horizontal and vertical line $n$-polyomino). I define an integer sequence $S_n$ to be the minimal number of cells of a shape in which all $n$-polyominos embed, and I am interested in understanding this sequence. In particular, specific questions are: Can we always find an optimal shape that fits into an $n \times n$ square? (this seems intuitively reasonable but I do not know how to prove it) Can we prove that, asymptotically, $S_n = \Theta(n^2)$? (the challenge is to show an $\Omega(n^2)$ lower bound -- maybe this is already possible by simply looking at a subset of the polyominos, but I couldn't see how to do it) More generally, has this sequence already been studied? To understand what happens here, I was able to compute by bruteforce computer search the first values of $S_n$, making the assumption that optimal shapes always fit in an $n$ by $n$ square (first point above) -- these values may turn out not to be optimal if this assumption is wrong: We have $S_1 = 1$, $S_2 = 3$ (easily), and $S_3 = 5$ (see above) We have $S_4=9$ with a surprising shape: ..#. .##. #### .##. We have $S_5 = 13$, with the unsurprising shape: ..#.. .###. ##### .###. ..#.. We have $S_6 = 18$ with a surprising shape: ..##.. ..##.. ###### #####. ..##.. ..#... We have $S_7 = 24$, the shape is similar to $n=5$ but with a hole: ...#... ..###.. .#.###. ####### .#####. ..###.. ...#... I do not know $S_8$ There are matching sequences for these terms in OEIS, but their definitions do not seem relevant... Edit: maybe https://oeis.org/A203567 https://www.sciencedirect.com/science/article/pii/S0012365X01003570 would be worth investigating. Acknowledgement: This question is by Thomas Colcombet and Antonio Casares. Edit: fixed the values of $S_5$ and $S_7$, many thanks to @RobPratt for noticing and reporting the errors! REPLY [4 votes]: It is actually $\ge cn^2$ with some $c>0$. The value of $c$ I'll obtain is pretty dismal but I tried to trade the precision for the argument simplicity everywhere I could, so it can be certainly improved quite a bit. I have no doubt that it is written somewhere (perhaps, in the continuous form: the $2$-dimensional measure of a set containing a shift of every rectifiable curve of length $1$ is at least some positive constant) but I'll leave it to more educated people to provide the reference. We shall work on the 2D $n\times n$ lattice torus $T$ whose size $n$ is a power of $2$. Clearly, wrapping around makes the set only smaller. Define $K$ to be the integer such that $2^{K^3+K}\le n< 2^{(K+1)^3+K+1}$ (I assume that $n$ is large enough, so $K$ is not too small either). Put $\varepsilon_k=2^{-k}, M_k=2^{k^3}$ ($k\ge 4$). Note that $\frac 12+3\sum_{k\ge 4}\varepsilon_k=\frac 78<1$. Put $\mu_k=\frac 12+3\sum_{m=4}^{k}\varepsilon_m$ (so $\mu_3=\frac 12$). Now take any set $E\subset T$ of density $d(E,T)=\frac{|E|}{|T|}=1/2$. Our aim will be to construct a connected set $P$ of cardinality $|P|\le Cn$ such that no its lattice shift of $E$ on $T$ contains $P$. Start with dividing $T$ into $M_4^2$ equal squares $Q_4$. Notice that the portion of squares $Q_4$ with density $d(E,Q_4)=\frac{|E\cap Q_4|}{|Q_4|}>\mu_3+\varepsilon_4$ is at most $\mu_3/(\mu_3+\varepsilon_4)\le 1-\varepsilon_4$. Now choose $N_4=\frac{2\log_2 (M_4/\varepsilon_4)}{\varepsilon_4}=2\cdot(4^3+4)\cdot 2^4$ squares $Q_4$ independently at random. The probability that none of them has density $d(E,Q_4)\le \mu_3+\varepsilon_4$ is at most $(1-\varepsilon_4)^{N_4}< \left(\frac{\varepsilon_4}{M_4}\right)^2$. This means that if we consider not only the standard partition but also all its shifts $E'$ by multiples of $\varepsilon_4 n/M_4$, then there exists a configuration $P_4$ of $N_4$ squares $Q_4$ such that for each such shift, the density of $E'$ in at least one square $Q_4$ in $P_4$ is $d(E',Q_4)\le \mu_3+\varepsilon_4$. However, every lattice shift can be approximated by such shifts with precision $\varepsilon_4 n/M_4=\varepsilon_4\ell(Q_4)$, so we conclude that for any shift $E'$ of $E$, the configuration $P_4$ contains a square $Q_4$ with density $d(E',Q_4)\le\mu_3+3\varepsilon_4=\mu_4$. Our $P$ will be essentially contained in $\bigcup_{Q_4\in P_4}Q_4$. Notice that we can construct some set in each square $Q_4$ and the cost of joining them afterwords will be at most $ 2n N_4 $. Notice also that the sidelength $\ell(Q_4)$ of each $Q_4$ is $n/M_4$. Now partition the torus into $M_5^2$ equal squares $Q_5$ and consider shifts by multiples of $\varepsilon_5 n/M_5$. Fix one square $Q_4\in P_4$ and choose $N_5=\frac{2\log_2 (M_5/\varepsilon_5)}{\varepsilon_5}=2\cdot(5^3+5)\cdot 2^5$ independent random squares in it creating some configuration $P'_5$. Repeat the same configuration in all other squares $Q_4$ to get a configuration $P_5$ of $N_4N_5$ squares $Q_5$ with sidelength $\ell(Q_5)=n/M_5$. Since for all such shifts at least one square $Q_4$ in $P_4$ satisfies $d(E',Q_4)\le\mu_4$, the same probabilistic argument results in the conclusion that one can choose $P_5'$ so that for every shift $E'$ by multiples of $\varepsilon_5n/M_5$ there will be a square $Q_5$ in $P_5$ with $d(E',Q_5)\le\mu_4+\varepsilon_5$, which, by approximation, yields again that for every shift $E'$ we shall have some $Q_5\in P_5$ with $d(E',Q_5)\le\mu_5$. The extra joining cost is now $2nN_4N_5/M_4$. Continue the same way until we reach $P_K$ consisting of $N_4\dots N_K$ squares $Q_K$ of sidelength $n/M_K\le 2^{3K^2+4K+2}$. Now just fill these squares completely. This will create $$ 2^{O(K^2)}N_4\dots N_K\le 2^{O(K^2)}N_K^K=2^{O(K^2)}[2(K^3+K)2^K]^K=2^{O(K^2)} TITLE: Is the minimal polynomial of an algebraic formal Laurent series always separable? QUESTION [5 upvotes]: Let $f(x)\in K((x))$ be an algebraic formal Laurent series and let $P(x,y)\in K(x)[y]$ be its minimal polynomial. Is $P(x,y)$ always separable? An example of non separable polynomial comes from Puiseux series: the polynomial $y^2-x$ has a double root $y=\sqrt{x}$. So I wonder what happen if we restrict to Laurent series. Thank you in advance. REPLY [3 votes]: Yes. Let $p=char(K)$ and $\alpha\in \overline{K(x)}\cap K((x))$ assumed to be inseparable over $K(x)$. Let $L= K^{1/p^\infty}$ which is perfect. If $\alpha$ is inseparable over $L(x)$ then $\alpha$'s monic $L(x)$-minimal polynomial is $$f(y)=g(y^p)=h(y)^p$$ with $h(y)\in L(x)^{1/p}[y]=L(x^{1/p})[y]$ so $$[L(x^{1/p},\alpha):L(x)] = p \deg(h) =[L(x,\alpha):L(x)]$$ ie. $x^{1/p}\in L(x,\alpha)$. This gives that $x^{1/p}\in L((x))[\alpha]= L((x))$ which is a contradiction. Whence $\alpha$ is separable over $L(x)$ $\implies f(y)$ is separable. There is a finite extension $E/K$ such that $f(y)\in E[y]$. Take a basis $E=\bigoplus_{j=1}^q b_j K$ with $b_1=1$. We get that $$E(x)=\bigoplus_{j=1}^q b_j K(x),\qquad E((x))=\bigoplus_{j=1}^q b_j K((x))$$ $$f(y)=\sum_{j=1}^q b_j f_j(y), \qquad f_j(y)\in K(x)[y]$$ $f(\alpha)=\sum_{j=1}^q b_j f_j(\alpha) = 0$ in $E((x))$ gives that $$f_1(\alpha)=0$$ $f_1(y)\in K(x)[y]$ being monic and of degree $= \deg f$ it must be that $f(y)=f_1(y)$. $f_1$ is in $K(x)[y]$, separable, irreducible, which proves that $\alpha$ is in fact separable over $K(x)$.<|endoftext|> TITLE: Freeman Dyson's approach to string theory QUESTION [6 upvotes]: Context: In celebrating the centenary of Ramanujan's birth, Freeman Dyson presented the following career advice for talented young physicists [1]: My dream is that I will live to see the day when our young physicists, struggling to bring the predictions of superstring theory into correspondence with the facts of nature, will be led to enlarge their analytic machinery to include not only theta-functions but mock theta-functions … But before this can happen, the purely mathematical exploration of the mock- modular forms and their mock-symmetries must be carried a great deal further. —Freeman Dyson Question: Was Freeman Dyson guided by physical intuitions that could have convinced top-notch quantum field theorists of his generation, such as Richard Feynman? Though I am aware that Freeman Dyson and Richard Feynman collaborated on Feynman's approach to quantum field theory, the precursor to string theory, I doubt that Feynman would have advanced the hypothesis that Ramanujan's work had any important consequences for theoretical physics. Complementary insights: In parallel, I wonder whether it may not be equally sensible to reconcile quantum theory with the facts of probabilistic number theory where probabilistic events are of a deterministic and frequentist nature. Upon closer inspection, this would be a complementary effort but I don't know of a systematic research program aimed at this particular objective although a large number of physicists appear to have a strong interest in the pair correlation conjecture which emerged from a tea-time discussion between Freeman Dyson and Hugh Montgomery. These are related observations, which may be relevant for a couple reasons: (1) The theory of modular forms potentially enters mathematical physics via the analysis of the Pair-Correlation conjecture. (2) By John Bell's own admission, his 1964 theorem known as Bell's theorem was motivated by the super-deterministic theory proposed by De Broglie and Bohm. Furthermore, I suspect that Erdős is often quoted saying: God may not play dice with the universe, but something strange is going on with the prime numbers. because all mathematical systems may be constructed from Peano Arithmetic, and the prime numbers are the atomic units of the integers, so the distribution of the prime numbers may be viewed as fundamental scientific data. Based on a recent discussion with Max Tegmark [7], who believes that a physicist can only understand the mathematical relations between things, this perspective is worth consideration if we assume that the mathematical structure of the Universe emerged from an information-theoretic singularity(i.e. Big Bang Cosmology). Note: Contrary to those who are voting to close this question, I believe that if there are fundamental physical insights which motivated Freeman Dyson's hypothesis then this question is of interest to the MathOverflow community. References: Jeffrey A. Harvey. Ramanujan’s influence on string theory, black holes and moonshine. 2019. Hardy, G. H.; Ramanujan, S. “The normal number of prime factors of a number n”, Quarterly Journal of Mathematics. 1917. Erdős, Paul; Kac, Mark. “The Gaussian law of errors in the theory of additive number theoretic functions”. American Journal of Mathematics. 1940. Montgomery, Hugh L. "The pair correlation of zeros of the zeta function", Analytic number theory, Proc. Sympos. Pure Math. 1973. Bell, J.S.“On the Einstein-Podolsky-Rosen paradox,” Physics. 1964. Tegmark, Max. "The Mathematical Universe". Foundations of Physics. Arxiv. 2008. Email discussion with Max Tegmark on tabletop experiments for the Mathematical Universe Hypothesis via Probabilistic Number Theory. Dec 18 2021. REPLY [10 votes]: I don't think it would have convinced Feynman because he didn't like the rabbit hole that string theory seemed to be going down. That instead of trying to explain some phenomenon, that they were involved in mathematical gamesmanship that was too divorced from physical phenomena. (In contrast to how BCS and Feynman raced to explain superconductivity; itself, very similar to how Watson and Crick feared and raced Linus Pauling for a structure of DNA.) So coming up with an even more obscure special function, than the already obscure special function was not the way to win Feynman over. If anything, the last 20 years or so with string theory falling out of fashion has somewhat validated Feynman's skepticism. But I don't think he was dogmatic about it...sure if the noodling around produced something, great. But he was skeptical. And seems to have been borne out, so far. But so far it hasn't. Has just been a mathematical circle jerk and a fundraising, job-getting side track in theoretical physics.<|endoftext|> TITLE: Functions $\mathbb{R}^2\to\mathbb{R}^2$ that preserve lines QUESTION [6 upvotes]: The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1] The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (Proof: let $L_1$ and $L_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L_1]$ and $f[L_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L_1]$ and $f[L_2]$. This line intersects $f[L_1]$ in some point, say $f(p_1)$, and it intersects $f[L_2]$ in some point, say $f(p_2)$. The image under $f$ of the line through $p_1$ and $p_2$ must be the line through $f(p_1)$ and $f(p_2)$, which contains $p$, and therefore $p$ is in the image of $f$.) My question is: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation? Added later: Thanks to @Wojowu in the comments for explaining how to construct a function $\mathbb{R}^2\to\mathbb{R}$ that maps every line to the whole of $\mathbb{R}$, which answers the second question I asked. So: what if we also insist that there be three non-collinear points in the image of $f$? It follows that there must be two distinct lines in the image of $f$, and then the argument above (more or less) shows that $f$ is surjective. But must it be injective? [1] For a proof of the general case see e.g. Andrew Putman, The fundamental theorem of projective geometry. REPLY [3 votes]: Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a function that maps lines to lines, and suppose that there are three non-collinear points in the image of $f$. Lemma 1: If $f[\ell_1]$ and $f[\ell_2]$ are distinct parallel lines, then $\ell_1$ and $\ell_2$ are distinct parallel lines. Proof: Suppose otherwise. Then $\ell_1$ and $\ell_2$ intersect at a point $x$, and $f(x)$ must lie on both $f[\ell_1]$ and $f[\ell_2]$; this is clearly impossible since the lines are parallel. Lemma 2: Let $a, b, c, d \in \mathbb{R}^2$ be four points such that $f(a), f(b), f(c), f(d)$ form the vertices of a non-degenerate parallelogram (in that order). Then $a, b, c, d$ also form the vertices of a non-degenerate parallelogram in the same order, so in particular $a + c = b + d$. Proof: Apply Lemma 1 twice, once to each pair of opposite edges. Lemma 3: If four points in the image of $f$ form the vertices of a non-degenerate parallelogram, then they each have one preimage under $f$. Proof: Suppose otherwise. Then there are two points $a, a' \in \mathbb{R}^2$ with $f(a) = f(a')$; let $f(b), f(c), f(d)$ be the other three vertices of a non-degenerate parallelogram (in that order). Then we have $a + d = b + c$ and $a' + d = b + c$ by applying Lemma 2, whence it follows that $a = a'$. Lemma 4: $f$ is surjective. Proof: Let $x,y,z$ be three points such that $f(x), f(y), f(z)$ are non-collinear points. It follows that $x,y,z$ are also non-collinear. The line through $x,y$ maps to the line through $f(x),f(y)$. Given any point $P$ in the plane of $f(x), f(y), f(z)$, we can take the line through $f(z)$ and $P$ and let it intersect the line through $f(x)$ and $f(y)$ at a point $Q$, which is in the image of $f$ and is therefore $f(w)$ for some point $w$. The line through $w$ and $z$ maps to the line through $f(z)$ and $f(w)$, which contains $P$, so $P$ is indeed in the image of $f$. Lemma 5: $f$ is injective. Proof: Since $f$ is surjective by Lemma 4, every point $f(a)$ in the image of $f$ is the vertex of a non-degenerate parallelogram with its other three vertices in the image of $f$. By Lemma 3, there cannot be a point $a' \ne a$ with $f(a) = f(a')$. The result follows. Corollary: $f$ is a bijection, so is an affine transformation by the claim at the beginning of the OP. Can we weaken the condition even further? In particular, if we weaken 'maps lines to lines' to 'preserves collinearity' (the difference being that a line can map to a proper subset of a line), then does the result still hold? Lemma 2 holds (and so does Lemma 3), so the difficult part is proving surjectivity.<|endoftext|> TITLE: Reflection quotients of Coxeter groups QUESTION [5 upvotes]: I am interested in a concept somehow dual to reflection subgroups. A reflection quotient of a Coxeter system $(W, S)$ shall be a surjective homomorphism $W \to W'$ to a Coxeter group $W'$ such that the images of $S$ are reflections. I did not find any literature on the topic, so it's possible that this notion has a name different from the one I just gave it. My question basically is the following: Given a Coxeter system $(W, S)$, is there a way to find all its reflection quotients? For $S_n$ (the Coxeter group with graph $A_{n-1}$), the answer is quite straightforward because $S_n$ has almost no normal subgroups: The only reflection quotients are $S_n$ and $S_2$. I do not know an answer for any other Coxeter groups, not even the finite ones. I am particularly interested in the Coxeter groups with graph $D_n$ (I even asked a specific question on it on MSE, but it did not receive any attention so far). The general problem however seems interesting enough for this forum, I hope. One would hope that the problem has a graph-theoretic answer, similar to the one requested in the dual question to this one. I do not know how realistic this hope is. REPLY [3 votes]: It turns out that for finite Coxeter groups $W$, this was solved by Maxwell in 1996 in the article The normal subgroups of finite and affine Coxeter groups. His Theorem 0.1 asserts that Theorem. If $W$ is an irreducible finite Coxeter group and $H$ is a normal subgroup of $W$, then either $H = \{ \pm 1 \}$, or there exists a graph homomorphism $\psi \colon W \to W'$ such that $H = \ker(\psi)$. Here, a graph homomorphism maps standard generators of $W$ either to $1$ or to a standard generator of $W'$ (so a graph homomorphism is a reflection quotient map if and only if none of the generators map to $1$). Together with the classification of finite Coxeter groups, this makes it rather straightforward to list the reflection quotients (in fact, this is done in the same article in Table 3). The answer in the case of affine Coxeter groups seems more complicated.<|endoftext|> TITLE: How to pass on research posthumously QUESTION [31 upvotes]: I've been independently researching math, for a while in the homeless community. While I am generally safe, situational realities (weather, equipment stress, health and especially how cities can discard possessions) are weighing on me more heavily as I prepare to publish work. I have as much on paper as digitally, and I'm working on getting notebooks photographed. What is a responsible/automatic way, should I (e.g.) die suddenly, to pass on research work? I'm a bit unusual in the sense that people don't understand why I'm homeless (so...more isolated, I have no peers it would be responsible to use), and it seems dubious a university would take me -- an unknown -- seriously. I know, e.g., that gmail can transfer ownership to another entity after an inactivity timer expires, but that could easily go wrong. A lawyer doesn't make sense; I have no estate and the city would get my stuff first. I'm not concerned about credit as much as passing on the work. Collaboration could work; I just don't know how to escrow / dead-man switch this so someone actually sees it. REPLY [7 votes]: Create a public repository on Github and upload your files. Github was designed with software in mind (it is primarily a version control system) but it can be used for pretty much anything. It works best with plain text files (e.g. txt, md, tex); but will happily handle images and pdf. Here's a maths example from a (randomly selected) user called arnaugamez. There may be a limit on storage, but if there is I haven't found it yet. As for longevity, Github seems to take this pretty seriously as evidenced by the Arctic Code Vault which beats any back up plan I have...<|endoftext|> TITLE: Čech cohomology is isomorphic to singular cohomology QUESTION [6 upvotes]: The singular cohomology with integer coefficients of a projective variety is isomorphic to the Čech cohomology of the constant sheaf of integers on this variety. If the above statement is correct then consider the following example: Look at $\mathbb{P}^1$ and the standard open cover $U_0,U_1$. Then the Čech complex for any sheaf $\mathcal{F}$ on $\mathbb{P}^1$ would be $$ 0 \rightarrow \mathcal{F}(U_0)\oplus\mathcal{F}(U_1) \rightarrow \mathcal{F}(U_0\cap U_1) \rightarrow 0. $$ If we consider the constant sheaf of integers this becomes $$ 0 \rightarrow \mathbb{Z}\oplus\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0 $$ with non-trivial map $(a,b)\mapsto a-b$. But the singular cohomology of $\mathbb{P}^1$ is $$H^i\mathbb{P}^1=\begin{cases} \mathbb{Z},\quad i=0,2 \\ 0,\quad \text{else.}\end{cases}$$ Where am I going wrong? REPLY [23 votes]: The Cech complex only computes cohomology if the subspaces have vanishing cohomology themselves: you need $$ H^i(U_0; \mathcal{F}) \cong H^i(U_1; \mathcal{F}) \cong H^i(U_0 \cap U_1; \mathcal{F}) \cong 0 $$ for all $i > 0$. In the example you've written, this is not true for $U_0 \cap U_1$. In general there is actually a Mayer-Vietoris long exact sequence relating the cohomology of $X$, $U_0$, $U_1$, and $U_0 \cap U_1$, or more generally a Mayer-Vietoris spectral sequence if the cover has more open sets. (As a remark, you may have seen this constraint pushed under the rug because it is automatically true in special cases. If $X$ is a scheme and the subspaces $U_0$, $U_1$, and $U_0 \cap U_1$ are affine---the last automatic if $X$ is separated---then Serre's vanishing theorem says that these higher cohomologies vanish whenever $\mathcal{F}$ is a quasicoherent sheaf of $\mathcal{O}_X$-modules.)<|endoftext|> TITLE: Is there a configuration of 5 points on the plane where any two can be covered by an axis aligned rectangle? QUESTION [33 upvotes]: I'm trying to figure out the question in the title for a project that I'm working on. My goal is to find a configuration of five integer points on the plane, where we can overlap any pair of them without it covering the other three points. For $n=2$ it's trivial, for $n=3$ I believe any acute triangle works. $n=4$ is slightly trickier, but still easily solvable: Unfortunately $n=5$ is exactly where I'm stuck on. From some doodling around it seems impossible, but maybe I'm missing something. It feels like any configuration of four will be equivalent to the one above, and I don't think there's any way to add another point to it, but I can't tell for sure. If it is possible, what would be the "smallest" configuration that satisfies this restriction? In other words, the one with the smallest total bounding box. In case it is not possible, a proof would be nice for closure, but if there's some "dumb" way to brute-force this it'd also be ok. REPLY [50 votes]: One cannot arrange 5 points in that manner, even if the coordinates are allowed to be any real numbers. Indeed, assume that the 5 points are $(x_n,y_n)$ with $x_1\leq\dotsb\leq x_5$. By the Erdős-Szekeres theorem, there are $1\leq i TITLE: What is $\mathbb Q^{\mathrm{hypoab}}$? QUESTION [6 upvotes]: $\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gal{Gal}\newcommand{\ab}{\mathrm{ab}}$Let $G(\mathbb Q) = \Gal(\overline{\mathbb Q} / \mathbb Q)$ be the absolute Galois group. It's well-known that the abelianization $G(\mathbb Q)^{\ab}$ of $G(\mathbb Q)$ is isomorphic to $\Aut(\mathbb Q / \mathbb Z) = \widehat {\mathbb Z}^\times$, and that the fixed field $\mathbb Q^{\ab}$ of the commutator subgroup of $G(\mathbb Q)$ may be constructed by adjoining all roots of unity to $\mathbb Q$. Abelian Galois groups are generalized by solvable Galois groups, or more generally hypoabelian Galois groups (recall that a group is hypoabelian if its derived series stabilizes at the trivial group, possibly after transfinitely many steps). Question 1: What is the hypoabelianization of $G(\mathbb Q)$? Question 2: What is the fixed field of the maximal perfect normal subgroup of $G(\mathbb Q)$? (Recall that in general, the maximal perfect normal subgroup is the subgroup at which the derived series stabilizes, possibly after transfinitely many steps; the hypoabelianization of a group is its quotient by its maximal perfect normal subgroup.) REPLY [7 votes]: My comment, Wojowu's answer, YCor's comment, and Z. M's comment already contain everything we need. Let me provide a little more detail here. I will shift the indices by $1$ for reasons that will become apparent: Definition. Set $K_0 = \mathbf Q$ and let $K_1 = \mathbf Q(\boldsymbol \mu)$ be the field obtained by adjoining the roots of unity $\boldsymbol \mu \subseteq \bar{\mathbf Q}$. Inductively define $K_{i+1}=K_i\big(\sqrt[\infty\ \ ]{K_i^\times}\big)$, and set $K_\infty = \underset{\substack{\longrightarrow \\ i}}{\operatorname{colim}} K_i$. We claim that this is the extension we're after. We first introduce some notation: Definition. Given a profinite group $G$, its (profinte) derived series is the transfinite chain of closed subgroups $$G = G^{(0)} \trianglerighteq G^{(1)} \trianglerighteq \cdots \trianglerighteq G^{(\alpha)} \trianglerighteq \cdots$$ defined by $G^{(\alpha+1)} = \overline{[G^{(\alpha)},G^{(\alpha)}]}$ and $G^{(\beta)} = \bigcap_{\alpha < \beta} G^{(\alpha)}$ for any limit ordinal $\beta$ (which is already closed as each $G^{(\alpha)}$ is closed). One could alter the notation to distinguish it from the abstract derived series, but I will never use the latter (the same goes for the Kronecker–Weber theorem: it computes the topological abelianisation, not the abstract one!). Note that for any continuous surjective homomorphism $G \to H$ of profinite groups, the image of $G^{(\alpha)}$ is $H^{(\alpha)}$. Lemma. Let $G$ be a profinite group. Then $G^{(\omega + 1)} = G^{(\omega)}$, and this group is trivial if and only if $G$ is pro-soluble¹. Proof. For any finite group $G$, the descending chain $G^{(i)}$ stabilises after finitely many steps, so $G^{(\omega + 1)} = G^{(\omega)}$. The same statement for profinite groups follows since any closed normal subgroup $H \trianglelefteq G$ is the intersection of the open normal subgroups $U \trianglelefteq G$ containing it. Similarly, $G^{(\omega)} = 1$ if and only if the same holds in every finite quotient $G/U$, i.e. if and only if all $G/U$ are soluble. $\square$ Let's denote $G^{(\omega)}$ by $G^{(\infty)}$. For $n \in \mathbf N \cup \{\infty\}$, we will say that $G$ is $n$-soluble if $G^{(n)} = 1$, and we write $G^{n\text{-}\!\operatorname{sol}} = G/G^{(n)}$ for its maximal $n$-soluble quotient (in which we omit $n$ if $n = \infty$). For instance, $G$ is $1$-soluble if and only if it is abelian, and $\infty$-soluble if and only if it is pro-soluble (equivalently, hypoabelian as profinite group). Theorem. Let $\Gamma = \Gamma_{\mathbf Q}$ be the absolute Galois group of $\mathbf Q$. For $n \in \mathbf N \cup \{\infty\}$, the fixed field of $\Gamma^{(n)}$ is $K_n$ (i.e. $K_n$ is the maximal pro-soluble extension of derived length $\leq n$); For $n \in \mathbf N \setminus \{0\}$, the Galois group $\operatorname{Gal}(K_{n+1}/K_n) = \Gamma^{(n)}/\Gamma^{(n+1)}$ is isomorphic to $$\operatorname{Hom}_{\operatorname{cont}}\!\big(K_n^\times,\hat{\mathbf Z}(1)\big),$$ where $K^\times$ has the discrete topology and $\hat{\mathbf Z}(1) = \lim_m \boldsymbol \mu_m$ is the Tate module of $\bar{\mathbf Q}^\times$. Proof. Statement (1) is trivial for $n=0$, and is the Kronecker–Weber theorem for $n=1$. Statements (1) and (2) for finite $n \geq 2$ follow inductively by Kummer theory (see the corollary below). Finally, statement (1) for $n = \infty$ follows from the statement at finite levels, since $K_\infty = \bigcup_n K_n$ and $G^{(\infty)} = \bigcap_n G^{(n)}$. $\square$ Note also that the Galois group $\operatorname{Gal}(K_1/K_0)$ is isomorphic to $\operatorname{Aut}(\boldsymbol \mu) = \hat{\mathbf Z}^\times$. However, explicitly computing $\operatorname{Gal}(K_{n+1}/K_n)$ in a meaninful way is pretty hard, let alone saying anything about how the various pieces fit together. Edit: After writing this answer, I became aware of the following two striking results: Theorem (Iwasawa). The Galois group $\operatorname{Gal}(K_\infty/K_1) = \Gamma^{(1)}/\Gamma^{(\infty)}$ is a free pro-soluble group $\widehat{F_\omega}^{\operatorname{sol}}$ on countably infinitely many generators. So we know that $\Gamma^{\operatorname{sol}}$ sits in a short exact sequence $$1 \to \widehat{F_\omega}^{\operatorname{sol}} \to \Gamma^{\operatorname{sol}} \to \hat{\mathbf Z}^\times \to 1.$$ I find it hard to imagine that this sequence splits as a semi-direct product (but I am more optimistic about the derived length $\leq 2$ situation). Theorem (Shafarevich). Any finite soluble group $G$ occurs as a quotient of $\operatorname{Gal}(K_\infty/\mathbf Q) = \Gamma^{\operatorname{sol}}$. A modern reference is Neukirch–Schmidt–Wingberg's Cohomology of number fields, Corollary 9.5.4 (Iwasawa) and Theorem 9.6.1 (Shafarevich). (This is a truly great book, but even at $>800$ pages it can be a bit terse at times.) We used the following general result: Lemma (Kummer theory). Let $m \in \mathbf Z_{>0}$, and $K$ be a field of characteristic not dividing $m$ that contains $\boldsymbol \mu_m$. The maximal abelian extension of exponent $m$ of $K$ is $L=K\big(\sqrt[m\ \ ]{K^\times}\big)$; The map \begin{align*} \operatorname{Gal}(L/K) = \Gamma_K^{\operatorname{ab}}/m &\to \operatorname{Hom}_{\operatorname{cont}}\!\big(K^\times,\boldsymbol \mu_m\big) = \left(K^\times/(K^\times)^m\right)^\vee \\ \sigma &\mapsto \left(a \mapsto \frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right) \end{align*} is an isomorphism of profinite groups, where $K^\times/(K^\times)^m$ has the discrete topology and $A^\vee$ denotes the Pontryagin dual of a locally compact abelian group $A$. We avoid the notation $\widehat A$ for Pontryagin duals, since it clashes with the notation for profinite completions. (Note that Z. M's comment uses $(-)^\vee$ for a linear dual, which differs from my notation by a Tate twist.) Because it's not very hard, let's include a proof. Proof. For (2), by Pontryagin duality it suffices to show that the dual map \begin{align*} K^\times/(K^\times)^m &\to \operatorname{Hom}\!\big(\Gamma_K,\boldsymbol \mu_m\big) = \left(\Gamma_K^{\operatorname{ab}}/m\right)^\vee \\ a &\mapsto \left(\sigma \mapsto \frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right) \end{align*} is an isomorphism. Note that it is well-defined since any two $m$-th roots of $a$ differ (multiplicatively) by an element of $\boldsymbol \mu_m \subseteq K$, on which $\sigma$ acts as the identity. Since $\boldsymbol \mu_m \subseteq K$, the $\Gamma_K$-module $\boldsymbol \mu_m$ has trivial action, so $\operatorname{Hom}_{\operatorname{cont}}(\Gamma_K,\boldsymbol \mu_m) = H^1(K,\boldsymbol \mu_m)$. The Kummer sequence $$1 \to \boldsymbol \mu_m \to \mathbf G_m \stackrel{(-)^m}\to \mathbf G_m \to 1$$ and Hilbert's theorem 90 compute $K^\times/(K^\times)^m \stackrel\sim\to H^1(K,\boldsymbol \mu_m)$ via the map above. Now (1) follows since $\sigma \in \Gamma_K$ is in the kernel of $\Gamma_K \to \big(K^\times/(K^\times)^m\big)^\vee$ if and only if $\sigma$ fixes all $m$-th roots of elements in $K$. $\square$ Corollay. Let $K$ be a field of characteristic $0$ containing $\boldsymbol \mu$. The maximal abelian extension of $K$ is $L=K\big(\sqrt[\infty\ \ ]{K^\times}\big)$. The map \begin{align*} \operatorname{Gal}(L/K) &\to \operatorname{Hom}_{\operatorname{cont}}\!\big(K^\times,\hat{\mathbf Z}(1)\big) \\ \sigma &\to \left(a \mapsto \left(\frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right)_{m \in \mathbf Z_{>0}}\right) \end{align*} is an isomorphism of profinite groups. Proof. Take inverse limits over all $m \in \mathbf Z_{>0}$ in the lemma above, noting that the inverse limit pulls out of $\operatorname{Hom}(K^\times,-)$. $\square$ ¹Linguistic footnote: soluble and solvable mean the same thing. I used to think that this is one of those BrE vs AmE things (for instance, my Oxford Advanced Learner's dictionary does not contain the word solvable at all). But I think some folks in the UK also use solvable, so it's not entirely clear to me.<|endoftext|> TITLE: Properties of Jech's hierarchy of stationary sets (Exercise 8.13, 8.14 of Jech) QUESTION [6 upvotes]: I must first preface that while this is indeed a question on an exercise, I believe this is advanced enough for MathOverflow. Let $\kappa$ be a regular uncountable cardinal. Recall that the notion of a stationary subset makes sense for subsets of limit ordinals of uncountable cofinality. In Jech's Set Theory (Third Millenium Edition), he defined an ordering $<$ on stationary subsets $S,T \subseteq \kappa$ by (Definition 8.18, or (2.6) of the linked paper below): $$ S < T \iff S \cap \alpha \text{ is stationary for almost all } \alpha \in T $$ Note that we implicitly assume that almost all $\alpha \in T$ are limit ordinals of uncountable cofinality. The standard examples of such stationary sets are of the form: $$ E_\lambda^\kappa := \{\alpha < \kappa : \operatorname{cf}(\alpha) = \lambda\} $$ This was first introduced in his paper Stationary subsets of inaccessible cardinals. He proved in Lemma 8.19 (Theorem 2.4 of the paper) that $<$ is a well-founded relation, so it makes sense to define a rank function on this relation, $o(S)$, which he calls the order of the set. He then proceeds to give two exercises on this matter: (Exercise 8.13) If $\lambda < \kappa$ is the $\alpha^\text{th}$ regular cardinal, then $o(E_\lambda^\kappa) = \alpha$. (Exercise 8.14) $o(\kappa) \geq \kappa$ if and only if $\kappa$ is weakly inaccessible; $o(\kappa) \geq \kappa + 1$ if and only if $\kappa$ is weakly Mahlo. I have no idea how to solve either exercise. It may be helpful to note that in his paper, he defined the notion of a canonical stationary set (of order $\nu$), and mentioned without proof that the set $E_\lambda^\kappa$ is the canonical stationary set of order $\lambda$. However, I do not see why this is true. REPLY [3 votes]: Here were my solutions from a few years ago, when I worked through Jech's 1984 paper on this topic. Exercise 8.13: If $\lambda<\kappa$ is the $\alpha$th regular cardinal, then $o(E_{\lambda}^{\kappa})=\alpha$. Proof. Now, $o(E_{\lambda}^{\kappa})\geq \alpha$ follows from the fact that $E_{\nu}^{\kappa}\alpha$ and $E_{\alpha}$ exists, then $S\cap {\rm Tr}(E_{\alpha})\neq \emptyset$. Proof of Lemma 3.1. The proof is by contradiction, letting $S$ be a minimal counterexample. Let $T\alpha$, if it is non-empty modulo $I_{\rm NS}$. In other words, for any stationary set $S$, if $o(S)>\alpha$ then $S>E_{\alpha}$. Proof of Corollary 3.2. Assume not, so there is some stationary set $S$ with $o(S)>\alpha$ but $S$ is not contained in ${\rm Tr}(E_{\alpha})$ (modulo $I_{\rm NS}$). Let $T:=S-{\rm Tr}(E_{\alpha})$, which is a stationary set by the assumption. Further, $T$ is disjoint from ${\rm Tr}(E_{\alpha})$ and $o(T)\geq o(S)$ (from order reversal from $T\subset S$), a contradiction to the previous lemma. $\square_{\rm Corollary\ 3.2}$ Lemma 3.3 from Jech's paper: The set $E_{\alpha}$ exists if and only if there exists a largest stationary set $M_{\alpha}$ of order $\alpha$. Then \begin{equation}\tag{A}\label{Eq:3.3fromJech} M_{\alpha}=E_{\alpha}\cup {\rm Tr}(E_{\alpha}) \end{equation} and \begin{equation}\tag{B}\label{Eq:3.4fromJech} E_{\alpha}=M_{\alpha}-{\rm Tr}(M_{\alpha}). \end{equation} The union in (\ref{Eq:3.3fromJech}) is disjoint, and ${\rm Tr}(M_{\alpha})\subset M_{\alpha}$. (Both statements are taken modulo $I_{\rm NS}$.) Moreover, ${\rm Tr}(E_{\alpha})={\rm Tr}(M_{\alpha})$. Proof of Lemma 3.3. $(\Rightarrow)$: Assume $E_{\alpha}$ exists. Note that $E_{\alpha}$ and ${\rm Tr}(E_{\alpha})$ are disjoint modulo $I_{\rm NS}$ since $o({\rm Tr}(E_{\alpha}))>\alpha$ and the heredity property on $E_{\alpha}$. Putting $M:=E_{\alpha}\cup {\rm Tr}(E_{\alpha})$, the order of $M$ is $\alpha$. [Prove by induction, using minimal counterexample, that $o(S\cup T)=\min\{o(S),o(T)\}$ for stationary sets. We call this the union property.] If $o(S)=\alpha$ write $S=A\cup B$ where $A=S\cap E_{\alpha}$ and $B=S-A$. If $B$ is nonstationary, then $S=A\subset M$ modulo $I_{\rm NS}$. If $B$ is stationary, by the second defining property on $E_{\alpha}$ and the union property we have $o(B)>\alpha$, and so $B\subset {\rm Tr}(E_{\alpha})\mod I_{\rm NS}$ by the previous corollary. Hence in this case we also have $S\subset M\mod I_{\rm NS}$; proving that $M$ is the largest set of order $\alpha$. $(\Leftarrow)$: Assume $M_{\alpha}$ exists. By the union property [or by trivial considerations when ${\rm Tr}(M_{\alpha})$ is nonstationary] it follows that $M_{\alpha}\cup {\rm Tr}(M_{\alpha})$ has order $\alpha$, so ${\rm Tr}(M_{\alpha})\subset M_{\alpha}$ (modulo $I_{\rm NS}$) by maximality. Let $E=M_{\alpha}-{\rm Tr}(M_{\alpha})$. The order of $E$ is $\alpha$ since $M_{\alpha}=E\cup {\rm Tr}(M_{\alpha})$, by the union property, etc. We first show that $E$ has order $\alpha$ hereditarily. Let $S\subset E$ be stationary. If $o(S)>\alpha$ then there is a stationary set $T\alpha$ so by Corollary 3.2 $M_{\alpha+1}\subset {\rm Tr}(E_{\alpha})={\rm Tr}(M_{\alpha})\mod I_{\rm NS}$. $\square_{\rm Lemma\ 3.6}$ As diagonal intersection gives inf for sets of cardinality $<\kappa^{+}$, the previous two lemmas complete an induction, so we see that we can find the maximal sets $M$ inductively for $\alpha<\kappa^{+}$ (when the height of $\kappa$ is bigger than $\alpha$) by: $M_{0}$ is the limit points below $\kappa$, $M_{\alpha+1}={\rm Tr}(M_{\alpha})$, and for limits $\alpha$ we have $M_{\alpha}=\triangle_{\beta<\alpha}M_{\beta}$. In the first paragraph we showed that the height of $\kappa$ is at least the ordinality of the sequence of regular infinite cardinals $<\kappa$. It is easy to check (via induction) using these definitions that $E_{\alpha}=E_{\lambda}^{\kappa}$ as claimed.$\square_{\rm Exercise\ 8.13}$ Exercise 8.14: The height of $\kappa$ is at least $\kappa$ if and only if $\kappa$ is weakly inaccessibly; the height of $\kappa$ is at least $\kappa+1$ if and only if $\kappa$ is weakly Mahlo. Proof. First suppose $\kappa$ is an uncountable, regular, successor cardinal. Say $\kappa$ is the $\alpha$th regular cardinal. We have $\alpha<\kappa$. If $\alpha$ is a successor ordinal, say $\alpha=\beta+1$, let $\lambda$ be the $\beta$th regular cardinal. Then $E_{\beta}=E_{\lambda}^{\kappa}$ and ${\rm Tr}(E_{\beta})=\emptyset$ by Exercise 8.12, so the height of $\kappa$ is $\alpha$ by Lemma 3.6 in the solution to the previous problem. If $\alpha$ is a limit ordinal, then since $\triangle_{\beta<\alpha}E_{\alpha}=\triangle_{\lambda<\kappa:\lambda\text{ regular}}E_{\lambda}^{\kappa}=\emptyset$, by Lemma 3.5 in the solution of the previous problem we have the height of $\kappa$ is $\alpha$. Second suppose $\kappa$ is an uncountable, regular, limit cardinal. This means the sequence of limit cardinals $<\kappa$ is a cofinal sequence, so the height of $\kappa$ is at least $\kappa$, again by the previous problem. Putting these together, we see $h(\kappa)\geq \kappa$ iff $\kappa$ is weakly inaccessible. Now, fix $\kappa$ to be weakly inaccessible. The diagonal intersection of the sequence $E_{\lambda}^{\kappa}$ consists of regular cardinals $<\kappa$, so this is stationary if and only if $\kappa$ is weakly Mahlo. Hence by Lemma 3.5 of the previous solution, $h(\kappa)\geq \kappa+1$. Conversely, if $h(\kappa)\geq \kappa+1$, the diagonal intersection of $E_{\lambda}^{\kappa}$ is stationary, hence there is a stationary sequence of regular cardinals $<\kappa$, so $\kappa$ is weakly Mahlo.$\square_{\rm Exercise\ 8.14}$<|endoftext|> TITLE: Coefficient of the top Pontryagin class in $L$-genus QUESTION [5 upvotes]: The $L$ genus can be expressed as combinations of the Pontryagin classes with the first few terms as follows: $$L_1=\frac{1}{3}p_1,$$ $$L_2=\frac{1}{45}(7p_2-p_1^2),$$ $$L_3=\frac{1}{945}(62p_3-13p_1p_2+2p_1^3)$$ Let $a_k$ denote the coefficient of $p_k$ in $L_k$. Is there any estimate of $a_k$? Is it true that $a_k$ is positive and decreasing? Checking the first 14 $a_k$ using the formula in http://www.mctague.org/carl/blog/2014/01/05/computing-L-polynomials/, it is true that $a_k$ is decreasing (and exponentially fast). I am wondering if this is true in general. REPLY [11 votes]: The coefficient of $p_k$ is given by $$2^{2n}(2^{2n-1}-1)\frac{B_n}{(2n)!} = \zeta(2n)\frac{2^{2n}-2}{\pi^{2n}},$$ see e.g. Appendix A of this older version of Weiss (warning: for Weiss, the convention is that the Bernoulli numbers $B_n$ are positive for all positive $n$). From this it is clear that it is always positive. You can presumably obtain growth estimate from growth estimates for the zeta function (or equivalently, Bernoulli numbers). In fact, all coefficients of the $L$-polynomial can be expressed in terms of multiple zeta values, by the results of Berglund-Bergström (and by their Corollary all $a_k$ are positive).<|endoftext|> TITLE: Is there a classification of homomorphisms $S_n \to S_{n+k}$ for small $k$? QUESTION [6 upvotes]: Homomorphisms $B_n \to B_{2n}$ and $B_n \to S_{2n}$ have been classified in Chen–Kordek–Margalit - Homomorphisms between braid groups and Lin - Braids and permutations respectively. I am interested in the corresponding question for symmetric groups: what are the homomorphisms $S_n \to S_{n+k}$ (up to conjugation)? I know from Explicit description of all morphisms between symmetric groups. that such a classification is difficult in full generality—I am willing to restrict my attention to suitably small $k$. The answer to the linked question discusses the classification of maximal subgroups of $S_m$ isomorphic to $S_n$, but I do not see how this is directly useful, as the image of a map $S_n \to S_{n+k}$ needn't be maximal. When $k = 1$ and $n + 1 \neq 6$, the only index $n+1$ subgroups of $S_{n+1}$ are point stabilizers. Thus, any non-cyclic map $S_n \to S_{n+1}$ is conjugate to the obvious inclusion. When $k > 1$, we may combine* the identity map $S_n \to S_n$ with a sign representation $S_n \to \mathbb{Z}_2 \to S_{k}$ to produce a map $S_n \to S_{n+k}$ not conjugate to an inclusion. Following Chen–Kordek–Margalit, we might hope that these are all such homomorphisms (for small $k$). Is anything known about these homomorphisms? *In case it wasn't clear: this combination is found by letting $S_n$ act on $[1,\ldots, n]$ and $S_k$ act on $[n+1,\ldots,n+k]$ REPLY [7 votes]: As spin observed in a comment below, if you know all (conjugacy classes of) subgroups of $S_n$ of index up to $m$, then you can determine the equivalence classes of homomorphisms $\phi:S_n \to S_m$, because the subgroups tell you the possible actions on the orbits of the image of $\phi$. The subgroups of index up to $m=n^2$ are listed explicitly in Mikko Korhonen's answer to the the MSE post Large subgroups of Symmetric Group. As an illustration, I will answer the question on the assumption that $k6$, the only subgroups of index less than $2n$ are $S_n$, $A_n$, and the point stabilizers, which are isomorphic to $S_{n-1}$. So if $S_n$ is acting faithfully on the set $\Omega := \{1,2,\dotsc,n+k\}$ with $k TITLE: Resources where I can find open problems in number theory along with their level of difficulty QUESTION [13 upvotes]: I have completed my master's in mathematics a couple of years ago and due to very strong personal and professional reasons I couldn't get admitted to grad school despite having a good academic background. I have a really good background in number theory. I don't have guidance of any professor right now and I want to try working on an open problem. Can you please let me know of resources( websites/ blogs/books) where I can find open problems in number theory to work on along with their estimated level of difficulty ( if possible)? Thanks! REPLY [3 votes]: Some open problem collections from my bookmarks: Number Theory @ Open Problem Garden List of unsolved problems in number theory @ Wikipedia Unsolved problems @ MathWorld Unsolved Problems and Rewards by Clark Kimberling Number theory problems by Peter J. Cameron MathPages Wanted List Unsolved Problem of the Week Archive<|endoftext|> TITLE: Is every virtually free group residually finite? QUESTION [7 upvotes]: Question: Is every (finitely generated) virtually free group residually finite? A well-known question asks whether every hyperbolic group is residually finite (Mladen Bestvina. Questions in geometric group theory. http://www.math.utah.edu/ ~bestvina/eprints/questions- updated.pdf.). My question is a very speciall case, and I wonder if it has been done. It is known that residual finiteness is not a quasi isometry invariant (Is residual finiteness a quasi isometry invariant for f.g. groups?). REPLY [2 votes]: To add to the list, here's a proof that is probably actually longer than the others, but is immediate "modulo standard facts". Standard fact 1: All finitely generated free groups embed into $SL_2(\mathbb{Z})$. Indeed, famously the congruence subgroup $\Gamma(2)$ of $PSL_2(\mathbb{Z})$ is free of rank 2, all finitely generated free groups embed in $F_2$, and by the universal property we can embed into $SL_2(\mathbb{Z})$. Standard fact 2: $SL_2(\mathbb{Z})$ is residually finite. Indeed, given a non-identity matrix, map to $SL_2(\mathbb{Z}/n)$ for large enough $n$.<|endoftext|> TITLE: Linear logic and linearly distributive categories QUESTION [6 upvotes]: I asked this question ten days ago on MathStackexchange (see here). Despite having placed a bounty on the question, I have not received any answers or comments until now. Following Nick Champion's advice, I therefore have decided to cross-post the question on this site. 1. Context On page two of the introduction to their paper Weakly distributive categories on linearly distributive categories Cockett and Seely write: It turns out that these weak distributivity maps, when present coherently, are precisely the necessary structure required to construct a polycategory superstructure, and hence a Gentzen style calculus, over a category with two tensors. The weak distributivity maps allow the expression of the Gentzen cut rule in terms of ordinary (categorical) composition. We call categories with two tensors linked by coherent weak distribution weakly distributive categories. They can be built up to be the proof theory of the full multiplicative fragment of classical linear logic by adding the following maps. $$\top \rightarrow A ⅋ A^{\perp}$$ $$A \otimes A^{\perp} \rightarrow \bot$$ Similarly, Blute and Scott write in Category theory for Linear Logicians: Roughly, linearly distributive categories axiomatize multiplicative linear logic in terms of tensor and par, as opposed to tensor and negation. I am trying to understand the sentences marked in bold. It seems that they refer to a relationship between linearly distributive categories and linear logic which I do not understand. 2. Questions What is meant by the sentence "[weakly distributive categories] can be built up to be the proof theory of the full multiplicative fragment of classical linear logic"? How do you make this statement (formally) precise? Is it related to the adjunction $Theories \rightleftarrows Categories$ described in the nLab-article syntactic category? How do the linear distributivity maps "allow the expression of the Gentzen cut rule in terms of ordinary (categorical) composition"? Besides an explanation, I would also happily accept a reference to a text that discusses the relationship in question in detail. REPLY [6 votes]: Yes, Cockett and Seely's comment about proof theory is a reference to the theory/category adjunction. Each kind of theory corresponds to a kind of category, for instance: Theory Category simply typed lambda-calculus cartesian closed category intuitionistic multiplicative linear logic closed symmetric monoidal category classical multiplicative linear logic with $\otimes,⅋$, and negation $\ast$-autonomous category = linearly distributive category with duals classical multiplicative linear logic with $\otimes$ and ⅋ only linearly distributive category In between the theories and the above "categories with structure", there are notions like multicategories and polycategories that represent the judgmental structure of a theory only, and in which connectives can be characterized by universal properties yielding a structure equivalent to the above categories with structure. For instance, IMLL has sequents of the form $A_1,\dots,A_m \vdash B$, corresponding to the homsets of a multicategory, in which $\otimes,\multimap$ can be given universal properties yielding a CSMC. Similarly, CMLL has sequents of the form $A_1,\dots,A_m \vdash B_1,\dots,B_n$, corresponding to the homsets of a polycategory, and in which $\otimes,⅋,(-)^\perp$ can be given universal properties. A polycategory in which only $\otimes,⅋$ exist is equivalent to a linearly distributive category, while if it also has duals $(-)^\perp$ it is equivalent to a $\ast$-autonomous category. I think this adding of duals is what Cockett and Seely are referring to "build up" -- starting from a linearly distributive category you can add an assumption of duals to obtain the structure corresponding to full CMLL (namely, a $\ast$-autonomous category). Your quote from Blute and Scott is, I believe, a reference to this connection between linearly distributive and $\ast$-autonomous categories. The usual definition of a $\ast$-autonomous category involves only $\otimes$ and duals, from which ⅋ can be constructed; so I think they are contrasting this with linearly distributive categories, which have $\otimes,⅋$, in terms of which duals can be characterized (though they may not always exist). Finally, the sentence about the Gentzen cut rule is a reference to the equivalence between polycategories with $\otimes$,⅋ and linearly distributive categories. Given a linearly distributive category $\cal C$, we define the polycategorical morphisms $(A_1,\dots,A_m) \to (B_1,\dots,B_n)$ to be the morphisms $A_1 \otimes \cdots \otimes A_m \to B_1 ⅋ \cdots ⅋ B_n$ in $\cal C$. The linear distributivities are then used to construct the polycategorical composition on these morphisms. The simplest nontrivial case is composing a morphism $A \to (B,C)$ with a morphism $(C,D) \to E$, which should yield a morphism $(A,D) \to (B,E)$. This is an instance of polycategorical composition, which is a categorical representation of the cut rule for CLL: $$ \frac{A \vdash B,C \qquad C,D \vdash E}{A,D \vdash B,E}$$ To construct such a composition in a linearly distributive category, we have morphisms $A\to B⅋C$ and $C\otimes D\to E$, and we can form the composite $$A\otimes D \to (B⅋C) \otimes D \to B ⅋(C\otimes D) \to B ⅋E$$ using the linear distributivity in the middle.<|endoftext|> TITLE: Existence of nontrivial categories in which every object is atomic QUESTION [8 upvotes]: An object $X$ of a cartesian closed category $\mathbf C$ is atomic if $({-})^X \colon \mathbf C \to \mathbf C$ has a right adjoint (hence is also internally tiny). Intuitively, atomic objects are "very small" (as the name suggests), and consequently there aren't usually many tiny objects in $\mathbf C$. However, is this necessarily the case? More precisely, do there exist any non-posetal cartesian closed categories in which every object is atomic? If there are no non-posetal examples, are there any nontrivial posetal examples? (The terminal category forms a trivial example of a posetal example.) REPLY [11 votes]: Building on Maxime's answer -- if $C$ is cartesian closed and has an initial object $0$, and if $0$ is atomic (or even just tiny), then $C$ is the terminal category. For $1 = 0^0 = 0$ (the former equation holds in any cartesian closed category with an inital object $0$; the latter holds because $(-)^0$ preserves initial objects). That is, $C$ is pointed, and the only pointed cartesian closed category is the terminal one.<|endoftext|> TITLE: Are entire functions “essentially” determined by their maximum modulus function? QUESTION [28 upvotes]: (Note: This has been asked on Math SE, but without an answer after almost two years and one offered bounty.) For an entire function $f$ let $M(r,f)=\max_{|z|=r}|f(z)|$ be its maximum modulus function. $M(r, f)$ does not change if $f$ is replaced by $e^{i\varphi}f(e^{i\theta}z)$, if $f$ is replaced by by $\overline{f(\overline z)}$, or combinations thereof. My question is: What can be said about entire functions $f$ and $g$ which satisfy $M(r, f) = M(r, g)$ for all $r > 0$? Are $f$ and $g$ necessarily related by the above listed transformations? Some thoughts: If $f(0) = 0$ with multiplicity $k$ at the origin then $M(r, f) \sim r^k$ for $r \to 0$, so that $g(0) = 0$ with the same multiplicity. We can divide both functions by $z^k$ and therefore assume that $|f(0)| = |g(0)| \ne 0$. After multiplying both functions with suitable factors $e^{i\varphi}$ we can assume that $f(0) = g(0) = 1$. If $f(z) = 1 + a_m z^m + \cdots$ at the origin with $a_m \ne 0$ then $M(r, f) \sim 1 + |a_m| r^m$ for $r \to 0$, so $g(z) = 1 + b_m z^m + \cdots$ with $|b_m| = |a_m|$. After suitable rotations $z \to e^{i\theta}z$ in the argument we can assume that $a_m = b_m > 0$. After this normalization the question is whether necessarily $g = f$ or $g(z) = \overline{f(\overline z)}$, or if no such conclusion can be drawn. REPLY [30 votes]: This is a classical problem, but only partial results are available: MR3155684 Hayman, W. K.; Tyler, T. F.; White, D. J. The Blumenthal conjecture, in Complex Analysis and Dynamical Systems V, 149–157. On the latest results see: MR4348902 Evdoridou, Vasiliki; Pardo-Simón, Leticia; Sixsmith, David J. On a result of Hayman concerning the maximum modulus set. Comput. Methods Funct. Theory 21 (2021), no. 4, 779–795, and references there.<|endoftext|> TITLE: Can the Bousfield class of projective space be computed directly? QUESTION [11 upvotes]: Recall that the Bousfield class of a spectrum $E$, written $\langle E\rangle$, is the class of spectra $X$ such that $X\wedge E$ is not contractible. For example the Bousfield class of any of the spheres $\mathbb{S}^k$ is the class of all noncontractible spectra. Now take the complex projective space $\mathbb{CP}^n$, choose a basepoint and consider the suspension spectrum $\Sigma^\infty\mathbb{CP}^n$. I think it follows from the thick subcategory theorem of Hopkins-Smith that the Bousfield class $\langle \Sigma^\infty\mathbb{CP}^n\rangle$ is equal to $\langle \mathbb{S}^k\rangle$ (it's a "type 0" finite spectrum). But that theorem seems rather high-powered for the job it's doing here. So my $\textbf{question}$ is: can the the Bousfield class $\langle \Sigma^\infty\mathbb{CP}^n\rangle$ be computed directly? For example for $n=1$, $\Sigma^\infty\mathbb{CP}^1\simeq \mathbb{S}^2$ so things are looking good. For $n=2$, since $\Sigma^\infty\mathbb{CP}^2$ is the cone $C(\eta)$ of the Hopf map $\eta:\mathbb{S}^3\rightarrow\mathbb{S}^2$ one can use the facts that $C(\eta^k)$ can be gotten as the cofiber of (suspensions of) $C(\eta^{0$. In the unstable category, cellular approximation gives $$ [\mathbb{C}P^n,\mathbb{C}P^n] = [\mathbb{C}P^n,\mathbb{C}P^\infty] = [\mathbb{C}P^n,K(\mathbb{Z},2)] = H^2(\mathbb{C}P^n) = \mathbb{Z}. $$ If we write $f_q$ for the map corresponding to $q\in\mathbb{Z}$, then the effect on the cohomology ring $$ H^*(\mathbb{C}P^n)=\mathbb{Z}[x]/x^{n+1} = \mathbb{Z}\{1,x,\dotsc,x^n\} $$ is given by $f_q^*(x^k)=q^kx^k$. Put $\lambda(n,q)=\Lambda(\Sigma^\infty f_q)$. When calculating this, it is the trace of $f_q^*$ on reduced cohomology that is relevant, giving $$ \lambda(n,q) = q + q^2 + \dotsb + q^n = q(q^n-1)/(q-1). $$ If $p\nmid n$ then $\lambda(n,1)\neq 0\pmod{p}$ Now suppose that $p\mid n$ but $p-1\nmid n$ (so $p>2$). Let $q$ be a primitive root mod $p$, so in particular $q-1$ is invertible mod $p$, so there is no problem with interpreting the formula $\lambda(n,q)=q(q^n-1)/(q-1)$ modulo $p$. We then find that $\lambda(n,q)\neq 0\pmod{p}$ Now suppose that $p(p-1)\mid n$. By considering the cases $q=0\pmod{p}$, $q=1\pmod{p}$ and $q\neq 0,1\pmod{p}$ separately, we find that $\lambda(n,q)=0\pmod{p}$ for all $q$. In conclusion: If $p(p-1)\not\mid n$ then the above method proves that $\langle\mathbb{C}P^n\rangle=\langle S\rangle$ If $p(p-1)\mid n$ then we still know from nilpotence theory that $\langle\mathbb{C}P^n\rangle=\langle S\rangle$, but the above method does not suffice to prove it.<|endoftext|> TITLE: Similarity of a matrix with its transpose QUESTION [5 upvotes]: An $n\times n$ matrix $A$ over a number field is similar to its transpose $A^T$. Is there any natural way to construct a nonsingular matrix $P$ such that $P^{-1}AP=A^T$? REPLY [5 votes]: As mentioned in another answer the fact that $A$ and $A^T$ are similar corresponds to a commutative diagram of modules over the polynomial ring $F[\lambda]$: $$ \begin{array}{c} 0 & \to & F[\lambda]\otimes V & & \stackrel{\lambda I - A}{\longrightarrow} & F[\lambda]\otimes V & \to & V_{A} & \to & 0\\ && \downarrow~R &&& Q~\downarrow & & \downarrow P \\ 0 & \to & F[\lambda]\otimes V & & \stackrel{\lambda I - A^T}{\longrightarrow} & F[\lambda]\otimes V & \to & V_{A^T} & \to & 0 \end{array} $$ where: $V$ denotes that $n$-dimensional column vector space over $F$ For a square matrix $B$ we use $V_B$ to denote $V$ as a module over $F[\lambda]$ on which $\lambda$ acts by $B$. The maps $P$, $Q$, $R$ are isomorphisms of $F[\lambda]$ modules. The matrix $P$ is what we need to determine. The idea is that once we determine $Q$ and $R$, then $P$ can be easily calculated. The Wikipedia page for the Smith Normal Form sketches an algorithm from which one can compute invertible matrices $F$ and $G$ over $F[\lambda]$ such that $F(\lambda u - A)G$ is a diagonal matrix over $F[\lambda]$. It follows that $G^T(\lambda u -A^T)F^T$ is the same diagonal matrix. We can now take $R^{-1}=G(F^T)^{-1}$ and $Q=(G^{T})^{-1}F$ to get the identity $Q(\lambda I - A)=(\lambda I - A^{T})R$ as required. As mentioned above, now that we have calculated $Q$ and $R$, we can calculate $P$. Update: Since it may not be immediately obvious how to calculate the matrix of $P$ with entries in $F$, here is an explicit calculation which uses the matrix $Q$ over $F[\lambda]$. Note that if $e_i$ denotes the $i$-th standard basis vector of $V$, then under the horizontal map from $F[\lambda]\otimes V$ to $V_A$ or $V_B$, $1\otimes e_i$ goes to $e_i$. Under $Q$, the image of $1\otimes e_i$ in $F[\lambda]\otimes V$ is $E_i=\sum_{k=1}^n Q_{k,i} \otimes e_k$, where $Q_{k,i}$ is the $(k,i)$-th entry of $Q$ and is a polynomial in $\lambda$. The image of $E_i$ in $V_B$ is $f_i=\sum_{k=1}^n Q_{k,i}(A)e_k$. We then write $f_i=\sum_{l=1} P_{l,i} e_l$ to get the matrix $P$. (It is a bit curious that it only depends on $Q$!) Note that to actually calculate $P$, we need to calculate the matrix entries of $Q_{k,i}(A)$ for all $i$ and $k$.<|endoftext|> TITLE: Does entropy of the random walk control the return probability QUESTION [7 upvotes]: Given an infinite connected graph $G$ of bounded degree with vertex set $X$, let $P_x^n$ the time $n$ distribution of the simple random walk started at the vertex $x$ (so $P^n_x(y)$ is the probability that a simple random walk started at $x$ ends at $y$ after $n$ steps). Let further $$ H_n :=\displaystyle \sup_{x \in X} h(P_x^n), \qquad \eta_n := \displaystyle \inf_{x \in X} h(P_x^n) \quad \text{ and } \quad r_n := \displaystyle \sup_{x,y \in X} P_x^n(y). $$ Here $h(\mu) = \sum_y -\mu(y) \ln \mu(y)$ is the entropy of $\mu$. Given that for any measure $\mu$ one has $\displaystyle \sup_{y \in X} \mu(y) \geq e^{-h(\mu)}$, one gets the bound $r_n \geq e^{-H_n}$. One question is: Question 1: what are upper bounds on $r_n$ in terms of $H_n$ and $h_n$? Note that in the case of [Cayley graphs of] groups, there is such a bound (the sharpest version known to me is by Peres & Zheng). My question is for "generic" [infinite connected] graphs [of bounded degree]. There is a natural counterpart to the question (which is hopefully easier to get): Question 2: what are [family of] examples where upper bounds of $r_n$ in terms of $H_n$ and $\eta_n$ are "bad"? Of course "bad" is not well-defined, but in Cayley graphs it may happen that $r_n$ behaves roughly like $C_1e^{-C_2\sqrt{H_n}}$. I would expect much worse for a graph. PS: if "generic" needed to be made explicit, then I would say something like: "no finitely generated subgroup of the subgroup of self quasi-isometries of the graph acts co-compactly". PPS: any result with "lazy random walk" in place of "simple random walk" is also OK. REPLY [2 votes]: Too long for a comment, but I wanted to mention that we encountered some related issues in my recent preprint with Noah Halberstam https://arxiv.org/abs/2203.01540 and thought that the work-arounds we found might be useful to you too. We would have liked to have argued that if $d(X_0,X_n)$ grows super-diffusively then the Greens function $G(X_0,X_n)$ decays quickly, but this does not seem to be true in general. (Actually we did not find a counterexample, although on page 23 we discuss some constructions related to the comment I wrote, where one has positive speed but slow heat kernel decay.) For our purposes, we were able to avoid this problem with a nice trick: Introducing spatially-dependent killing to the random walk. If the walk is superdiffusive and one makes the rate of killing decay at a well-chosen rate as a function of the distance from the starting point then return probabilities on the killed network decay quickly but the walk has positive probability to never be killed (see Section 3 of our paper). As such, if you want to use heat kernel estimates to prove some almost-sure property of the random walk, you can run the argument on the network with the spatially-dependent killing to get that the property holds almost surely for the killed walk, then deduce that it also holds almost surely for the original walk by absolute continuity.<|endoftext|> TITLE: Why is fast matrix multiplication impractical? QUESTION [12 upvotes]: I am wondering why fast matrix multiplications are impractical, especially for Boolean matrix multiplication. I read some content saying fast matrix multiplications are impractical because of large constant factors. These constant factors are because of algebraic techniques. I do not understand where these constant factors come from. Some references also say $O(n^{3-\omega})$ may become practical when $O(n^{3-\omega})$ combinatorial algorithm exists. What is the large constant factor and combinatorial algorithm for Boolean matrix multiplication? REPLY [3 votes]: Addressing the Boolean part. Usually, fast matrix multiplication relies heavily on the element type being a ring; in particular, that every element has an additive inverse. For example, Strassen's algorithm contains subtractions. This works fine, for example, with reals, integers, rationals, and finite fields. In particular, you could easily do fast matrix multiplication on $\mathbb{F}_2$, that is, elements are bits with addition defined modulo two (so $1+1=0$). However, in Boolean matrix multiplication the addition of elements is the Boolean disjunction: $1+1=1$ instead of zero. This innocent change means that subtraction no longer works: from $x+1=1$ you cannot know whether $x=0$ or $x=1$. Thus Strassen's algorithm, unmodified, does not work with Booleans. Yes, this is a slight "impracticality", but it is well known that this can be circumvented relatively easily, by embedding the Booleans into a suitable ring. You can use $b$-bit integers modulo $2^b$, with $b$ chosen large enough so that overflows (wraparounds) will not happen in the fast matrix multiplication. Generally $b$ will be something like a logarithm of the matrix size. In fact there have been implementations of Strassen-like algorithms for Booleans. One example is Karppa & Kaski (2019), Engineering Boolean Matrix Multiplication for Multiple-Accelerator Shared-Memory Architectures (arxiv).<|endoftext|> TITLE: Sum of reciprocals of Sophie Germain primes QUESTION [15 upvotes]: A prime $p$ is called a Sophie Germain prime if $2p+1$ is also prime: OEIS A005384. Whether there are an infinite number of such primes is unsolved. My question is: If there are an infinite number of Germain primes, is the sum of the reciprocals of these primes known to converge, or diverge? Of course if there are only a finite number of Germain primes, the sum is finite. And a lower bound on any infinite sum can be calculated. But it is conceivable that it is known that the sum either converges or is a finite sum. And maybe even an upperbound is known? (My connection to this topic is via this question: "Why are this operator's primes the Sophie Germain primes?".) REPLY [23 votes]: Googling "sum of reciprocals of Sophie Germain primes" brings up the very recent paper: Wagstaff, Samuel S. jun., Sum of reciprocals of germain primes, J. Integer Seq. 24, No. 9, Article 21.9.5, 10 p. (2021). ZBL1482.11122. Here's a link to the paper on the journal's website. In particular, Wagstaff proves that the sum of the reciprocals of the Sophie Germain primes is between 1.4898 and 1.8027. REPLY [22 votes]: Here is a general result. For a sequence of nonnegative numbers $\{a_n\}$, let $A(x) = \sum_{n \leq x} a_n$. For example, if $S \subset \mathbf Z^+$ and we set $a_n = 1$ when $n\in S$ and $a_n = 0$ when $n \not\in S$, then $A(x)$ is the number of elements of $S$ that are $\leq x$. Exercise: If $A(x) = O(x/(\log x)^r)$ for a positive integer $r$ and all $x \geq 2$, then $\sum_{n \leq x} a_n/n$ converges as $x \to \infty$ if $r \geq 2$ and $\sum_{n \leq x} a_n/n = O(\log \log x)$ for $r = 1$. Example: if $f_1(T), \ldots, f_r(T)$ are polynomials with integer coefficients that fit the hypotheses of the Bateman-Horn conjecture (twin primes are $f_1(T) = T$ and $f_2(T) = T+2$, while Sophie Germain primes are $f_1(T) = T$ and $f_2(T) = 2T+1$), then Bateman and Stemmler showed $60$ years ago that the number of $n \leq x$ such that $f_1(n), \ldots, f_r(n)$ are all prime is $O(x/(\log x)^r)$, where the $O$-constant depends on $f_1, \ldots, f_r$. Therefore if above we take $S$ to be the $n \in \mathbf Z^+$ such that $f_1(n), \ldots, f_r(n)$ are all prime and define $a_n$ to be $1$ or $0$ according to $n \in S$ or $n \not\in S$, then the exercise above says the sum of all $1/n$ for $n \in S$ converges if $r \geq 2$. So for any sequence of pairs of primes $p$ and $ap+b$ that are expected to occur infinitely often ($p$ and $p+2$, or $p$ and $2p+1$, or $\ldots$), the sum of $1/p$ for such primes converges. That the sum of the reciprocals of the twin primes converges indicates that this summation is the wrong thing to be looking at. We want a strategy to prove the infinitude of twin primes, and that suggests a better sum. The Bateman-Horn conjecture predicts the number of $n \leq x$ such that $f_1(n), \ldots, f_r(n)$ are all prime is asymptotic to $Cx/(\log x)^r$ where $C$ is a positive constant depending on $f_1, \ldots, f_r$, and if $A(x) \sim cx/(\log x)^r$ as $x \to \infty$ for some $c > 0$ then $\sum_{n \leq x} a_n(\log n)^{r-1}/n \sim c\log\log x$. Therefore we expect (but have never proved) that the sum of $(\log p)/p$ over prime $p \leq x$ such that $p$ and $p+2$ are prime should grow like $c\log\log x$ for some constant $c > 0$, and a similar asymptotic estimate (for a different constant $c$) should hold for the sum of $(\log p)/p$ over all prime $p \leq x$ such that $p$ and $2p+1$ are prime.<|endoftext|> TITLE: Writing $1-xyzw$ as a sum of squares QUESTION [5 upvotes]: Can you write $1 - xyzw$ in the form $p + q (1 - x^{2}-y^{2}-z^{2}-w^{2})$ where $p$ and $q$ are polynomials that are of the form $\sum g_{i}^{2}$ where $g_{i}$ $\in$ $\mathbb{R}[x,y,z,w]$? For instance, in the two variable case, $1 - xy = \frac{1}{2} + \frac{1}{2}(x-y)^{2} + \frac{1}{2}(1-x^{2}- y^{2})$. In this example, $q = \frac{1}{2}$, so we have a very simple expression. I'm looking for an analogous expression in four variables ($p$ and $q$ here can be anything as long as they are sums of squares). Also can we say anything in general for $2n$ variables? REPLY [4 votes]: The question would be even more interesting if the factor $1-x^2-y^2-w^2-z^2$ was replaced by $4-x^2-y^2-w^2-z^2$. The reason is that a necessary condition for the existence of such $p,q$ is that whenever $xywz>1$, the last factor is also negative. And it turns out that $$(xywz>1)\Longrightarrow(x^2+y^2+w^2+z^2<4),$$ by AM-GM inequality. Notice that if the answer is positive with this constant $4$, then it is so with your constant $1$. Now the answer: $p$ and $q$ exist for the sharp constant $4$. You may take $q=\frac1{16}(4+x^2+y^2+w^2+z^2)$ and $$p=\frac1{16}(x^2+y^2+w^2+z^2)^2-xywz.$$ To see $p$ is a sum of squares, just notice $$48p=(x^2+y^2-w^2-z^2)^2+\cdots+4(x-y)^2(w+z)^2+\cdots,$$ where the first sum consists of $3$ terms, and the second one consists in $6$ terms.<|endoftext|> TITLE: Techniques for debugging proofs QUESTION [55 upvotes]: After writing many proofs, most of which contained errors in their initial form, I have developed some simple techniques for "debugging" my proofs. Of course, a good way to detect errors in proofs is to send them to a colleague for review. But even friendly and helpful colleagues tend to focus on the more interesting proofs and ignore the more technical ones, which are prone to errors. So it is important to have techniques for self-debugging. Some simple techniques that I use are: Verify that every assumption made in the theorem is used at least once in the proof. Mathematically this is not strictly required, since if "A" implies "C", then "A and B" implies "C" too. But, if an assumption is made and not used, it may indicate an error [there is an analogous technique in programming: many compilers will raise a warning if they detect that a variable is declared but not used.] Rewrite the proof in the opposite direction. For example, if the proof is by contradiction, rewrite it as a direct proof, and vice-versa. Mathematically it should not matter, but the process of rewrite may help to discover hidden errors. Read the proof in print. I have no rational explanation for this, but I found out that, when I read proofs in print, I often detect errors that evaded my eyes when I read them on the computer screen. Question: what other techniques do you find useful for detecting errors in mathematical proofs? I am looking for general techniques, that you would recommend to your research students. REPLY [6 votes]: Explain the proof to your cat. Indeed, it is often difficult to find a willing collegue listening for hours to all details without falling asleep or getting lost (both things happen simultaneously quite often). Cats are infinitely more enduring and are eager listeners. Do not expect them to point paws at your errors but simply trying to explain things carefully makes you generally aware of errors. Cats can of course be replaced by PHD-students. Coffee is however generally more expensive than food for cats. PS: During my studies I had no cat. I replaced it with my grandmother.<|endoftext|> TITLE: Invariants for the isotropy representation of a Riemannian symmetric space QUESTION [5 upvotes]: Statement: Let $M = G/K$ be a Riemannian symmetric space of compact type, and $V = T_o M$ be its isotropy representation (of $K$ acting on the tangent space of $M$). Then the Hilbert–Poincaré series $\sum_{n=0}^{+\infty} \dim(\mathcal{S}^n V)^K\,t^n$ (encoding the dimension of the invariants on the $n$-th symmetric power) of $K$ acting on $V$ coincides with Hilbert–Poincaré series $\sum_{n=0}^{+\infty} \dim(\mathcal{S}^n \mathfrak{a})^W\,t^n$ of the restricted Weyl group $W$ acting on the underlying space $\mathfrak{a}$ of the restricted root system for $G/K$. This statement is hopefully correct, and certainly very standard (I knew it in the special case $M$ is the underlying space $(G\times G)/G$ of a Lie group, but understood it in the more general case following some useful comments 1 2 3 by Robert Bryant on Invariants for the exceptional complex simple Lie algebra $F_4$). Where can I find it, or as close as possible to it, written in the literature, preferably in a textbook or survey article? (Note: The reason I have worded it in terms of Hilbert–Poincaré series is that I hope whatever reference comes up will include this as part of a general discussion in the form “the following statement is useful to compute the Hilbert–Poincaré series of certain representations of semisimple groups by reducing them to representations of finite reflection groups; let us now discuss the question of which representations $V$ can be attained that way, and what can be done to compute the Hilbert–Poincaré series of those that cannot”. But maybe I'm asking for too much!) REPLY [5 votes]: One reference is in Helgason's 1984 book Groups and Geometric Analysis. The result you want appears there as Corollary 5.12. The notation he uses is $X=G/K$ is a symmetric space where $G$ is connected and semisimple and $K$ is a maximal compact. As usual, write ${\frak{g}} = {\frak{k}} + {\frak{p}}$ as the Cartan decomposition, let ${\frak{a}}\subset {\frak{p}}$ be a maximal abelian subspace, and let $W$ be the associated Weyl group, i.e., the quotient of the normalizer of ${\frak{a}}$ in $K$ by the centralizer of ${\frak{a}}$ in $K$. Here is what Helgason says: Corollary 5.12 Every $W$-invariant polynomial $P$ on ${\frak{a}}$ can be uniquely extended to a $K$-invariant polynomial on ${\frak{p}}$. (Note that the restriction of a $K$-invariant polynomial on ${\frak{p}}$ to ${\frak{a}}$ is clearly $W$-invariant, so we get an isomorphism, and, in particular, an equality of Hilbert-Poincaré series.) Helgason attributes this result to Chevalley. He calls it "the Chevalley restriction theorem", though he says in the Notes at the end of the Chapter that it is an "unpublished result of Chevalley". He also includes an Exercise (D1) at the end of the Chapter in which he outlines Harish-Chandra's proof of Chevalley's result.<|endoftext|> TITLE: Examples of 6-manifolds without an almost complex structure QUESTION [10 upvotes]: Question: I am searching for examples for closed (hence orientable ), smooth $6$-manifolds without an almost complex structure. Finding such an example is equivelant to finding a manifold where the image of the Bockstein homomorphism $H^{2}(X,\mathbb{Z}_2) \rightarrow H^{3}(X,\mathbb{Z})$ maps the second Stiefel-Whitney class to something non-zero (this is due to Wall). REPLY [9 votes]: Turning comments into answer: An example of a closed 6-manifold not admitting an almost complex structure is $S^1 \times (SU(3)/SO(3))$. From the obstruction theory for lifting the map $M \to BSO(6)$ classifying the tangent bundle of an oriented 6-manifold through $BU(3) \to BSO(6)$, one sees that the unique obstruction to the existence of an almost complex structure is the Bockstein of $w_2(M)$, also known as the third integral Stiefel-Whitney class $W_3(M)$. This is generally, in all dimensions, the unique obstruction for an orientable manifold to admit what is called a spin$^c$ structure. The manifold $SU(3)/SO(3)$ does not admit a spin$^c$ structure (see e.g. Friedrich's "Dirac operators in Riemannian geometry" p.50). Also, a calculation shows that for orientable manifolds $M$ and $N$, the product $M\times N$ is spin$^c$ if and only if each factor is. Hence $S^1 \times (SU(3)/SO(3))$ is not spin$^c$, and thus not almost complex. To create a simply connected example, one can surger out e.g. any circle of the form $S^1 \times pt$ in $S^1 \times (SU(3)/SO(3))$; note that $SU(3)/SO(3)$ is simply connected as can be seen from the homotopy long exact sequence for $SU(3)/SO(3) \to BSO(3) \to BSU(3)$. The process of taking a manifold, crossing with a circle, and then surgering out such a circle, is sometimes referred to as spinning the original manifold. Spinning a closed orientable manifold produces a spin$^c$ manifold iff the original manifold was spin$^c$, see Proposition 2.4 here https://arxiv.org/abs/1805.04751. To create more examples for free, you can use the fact that the connected sum $M\# N$ of orientable manifolds is spin$^c$ iff each factor is.<|endoftext|> TITLE: Given that $n > 3$ and $z$ is a Gaussian integer, when can $z^n \pm z$ be a rational integer? QUESTION [7 upvotes]: I came across the following conjecture. If you have any thoughts on how to approach it, let me know. Conjecture. For any integer $n > 3$ and any Gaussian integer $z$ that is not a unit, if $z^n - z$ or $z^n + z$ is a rational integer, then $z$ is a rational integer. REPLY [2 votes]: The question can be rephrased: Find the integers $n > 3$ for which there exist Gaussian integers $z$ such that $z^n \pm z = \overline{z}^n \pm \overline{z}$. Rewriting the equation as $(z^n -\overline{z}^n)/(z-\overline{z}) = \mp 1$, one recognizes an instance of the problem of finding the terms of a Lucas sequence with no primitive divisors. The complete answer to an even more general problem (Lehmer sequences instead of Lucas sequences) was given by Bilu, Hanrot, Voutier J. Reine Angew. Math. 539, 75-122 (2001) and Abouzaid J. Théor. Nombres Bordx. 18, No. 2, 299--313 (2006) . To settle the conjecture, it suffices to look up the tables in these papers .<|endoftext|> TITLE: Harmonic functions on complete Riemannian manifolds QUESTION [7 upvotes]: I have started reading a paper of Colding and Minicozzi, where they prove that on a complete Riemannian manifold $M$ of non-negative Ricci curvature, the space of harmonic functions of growth order at most $d$ is finite dimensional. It is known that these $M$ have at most polynomial volume growth. I am wondering how much the growth of $M$ is related to the dimensionality of harmonic functions. For example, is it known what happens when the volume growth is known to be not polynomial? To be more specific, suppose we ask the analogous question on spaces of negative sectional curvature: are the harmonic functions of polynomial growth there infinite dimensional? This is mainly a reference request. Note: Edited after R W's reply below. REPLY [5 votes]: For the $n$-dimensional hyperbolic space it is already the space of bounded harmonic functions that is infinitely dimensional, which follows from the integral Poisson formula.<|endoftext|> TITLE: Harmonic flow on the Young lattice QUESTION [7 upvotes]: Let me begin with some preliminary concepts: A positive real-valued function $\varphi: P \rightarrow \Bbb{R}_{>0}$ on a locally finite, ranked poset $(P, \trianglelefteq)$ is harmonic if $\varphi(\emptyset)=1$ and \begin{equation} \varphi(u)=\sum_{\stackrel{\scriptstyle u \, \triangleleft \, v}{|v| \, = \, |u| + 1}} \varphi(v) \end{equation} where $\emptyset$ is the unique bottom element of $P$ (which we require to exist) and where $|u|$ denotes the rank of an element $u \in P$. The poset $P$ is 1-differential if in addition $\bullet$ The number of elements covered by both $u$ and $v$ equals the number of elements in $P$ covering both $u$ and $v$ whenever $u \ne v$. $\bullet$ If $u \in P$ covers exactly $k$ elements then $u$ is covered by exactly $k+1$ elements. In (https://arxiv.org/abs/math/9712266) Goodman and Kerov introduced a semi-group flow on the space of harmonic functions ${\frak{H}}(P)$ when $P$ is 1-differential. Specifically, for $\tau \in [0,1]$ and $\varphi \in {\frak{H}}(P)$ \begin{equation} C_\tau(\varphi)(v) \, := \, \sum_{k=0}^{|v|} {\tau^k (1-\tau)^{|v|-k} \over {(|v| -k )! }} \sum_{|u| = k} \varphi(u) \dim(u,v) \end{equation} where $\dim(u,v)$ is the number of saturated chains $p_{|u|} \! \lhd \cdots \lhd p_{|v|}$ in $P$ starting at $p_{|u|} = u$ and ending at $p_{|v|}= v$. A fairly easy calculation reveals that $C_\tau(\varphi)$ is harmonic whenever $\varphi$ is and that $C_\tau (C_\sigma(\varphi)) = C_{\tau \sigma}(\varphi)$. Furthermore we recover the original function $\varphi$ when $\tau = 1$ while we obtain the function \begin{equation} v \mapsto {1 \over {|v|!}} \dim(\emptyset, v) \end{equation} when $\tau = 0$, which is known to be harmonic whenever $P$ is 1-differential. Now consider the Young lattice $(\Bbb{Y}, \subseteq)$ of all integer partitions, ordered by inclusion of their respective Young diagrams. In virtue of the Pieri rule we know that the function \begin{equation} \varphi(\lambda) \, := \, {s_\lambda({\bf x}) \over {s^n_{\Box}({\bf x})}} \quad \text{where $\lambda \vdash n$} \end{equation} is a harmonic function on $\Bbb{Y}$ where $s_\lambda({\bf x})$ is the Schur function associated to $\lambda$ and $s_\Box({\bf x}) = x_1 + x_2 + x_3 + \cdots < \infty$ is the Schur function associated to the partition $(1)$. Let's apply the Goodman-Kerov flow to this function: For $\lambda \vdash n$ we get \begin{equation} C_\tau (\varphi)(\lambda) \, = \, {1 \over {s^n_\Box({\bf x})}} \, \underbrace{\sum_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} s_\Box^{n-k}({\bf x}) \sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda)}_{\text{call this $s_\lambda({\bf x};\tau)$}} \end{equation} We may now expand $s_\lambda({\bf x}; \tau)$ as $\sum_{\rho \vdash n} a_{\lambda, \rho}(\tau) s_\rho({\bf x})$. Question: What can be said about the polynomials $a_{\lambda, \rho}(\tau)$? Have they already been identified/considered in the literature? Sub-question: If the coefficient polynomials $a_{\lambda, \rho}(\tau)$ are (in general) messy, does anything nice happen with $s_\lambda({\bf x}; \tau)$ when we perform either the principal or content specializations, i.e. \begin{equation} \begin{array}{ll} x_i \mapsto \ \ q^{i-1} \ \ \text{for all $i \geq 1$} & \\ x_i \mapsto \left\{ \begin{array}{ll} 1 &\text{for all $i \leq d$} \\ 0 &\text{for all $i > d$} \end{array} \right. &\text{for some fixed but far out $d \geq 1$} \end{array} \end{equation} thanks, ines. REPLY [4 votes]: Let us identify $s_\lambda$ with the character of the irreducible representation $S^\lambda$ of the symmetric group $S_n$ indexed by the partition $\lambda$. Then $$ s_\Box^{n-k}({\bf x}) \sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda) $$ defines the character of a certain representation of $S_n$. We can explicitly describe it as follows. Since the Young lattice is the branching graph of representations of $S_n$, $\dim(\mu,\lambda)$ is exactly the multiplicity of $S^\mu$ in the restriction of $S^\lambda$ from $S^n$ to $S^k$ (where $n = |\lambda|$ and $k = |\mu|$), so $\sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda)$ is nothing but the character of the restriction of $S^\lambda$ to $S_k$. Similarly, it is a well known fact that multiplication by $s_\Box$ corresponds to induction from $S_m$ to $S_{m+1}$. So the quantity displayed above is the character of $$ \mathrm{Ind}_{S_k}^{S_n}\left( \mathrm{Res}_{S_k}^{S_n}\left(S^\lambda \right)\right) = \mathrm{Ind}_{S_k}^{S_n}(1) \otimes S^\lambda. $$ So this is the same thing as tensoring $S^\lambda$ with the representation obtained by inducing the trivial representation from $S_k$ to $S_n$. The induced representation $\mathrm{Ind}_{S_k}^{S_n}(1)$ that we are tensoring with might be more familiar under the notation $M^{(k, 1^{n-k})}$ (it is a permutation module). Its character expresses as $h_k s_\Box^{n-k}$, where $h_k$ is the complete symmetric function of degree $k$. Now we will use the internal product of symmetric functions defined by $p_\mu * p_\nu = \delta_{\mu, \nu} z_\mu p_\mu$. So in particular, homogeneous symmetric functions of different degrees multiply to zero. It is convenient for us because it describes the tensor product of representations of symmetric groups: $s_\mu * s_\nu = \sum_{\lambda} k_{\mu, \nu}^\lambda s_\lambda$, where $k_{\mu, \nu}^\lambda$ is a Kronecker coefficient. So now we may express $$ s_\lambda({\bf x}; \tau) = \sum_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} s_\Box^{n-k}({\bf x}) \sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda) = \left( \sum_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} h_k s_\Box^{n-k}\right) * s_\lambda. $$ To tidy up this equation, we sum over all possible values of $n$ and $k$ (there is no harm in doing this because the introduced terms will become zero upon taking the internal product with $s_\lambda$), which gives us a generating function $$ \left(\left( \sum_{k} \tau^k h_k \right)\left( \sum_l \frac{(1-\tau)^l s_\Box^{l}}{l!}\right)\right) * s_\lambda = \left(H(\tau) \exp((1-\tau)s_\Box) \right) * s_\lambda, $$ where $l = n-k$, and $H(\tau) = \sum_{k \geq 0} \tau^k h_k$ is the generating function of complete symmetric functions. Now, we can express $H(\tau)$ in terms of power-sum symmetric functions as $$ H(\tau) = \exp\left( \sum_{i \geq 1} \frac{p_i \tau^i}{i} \right). $$ Using the fact that $s_\Box = p_1$, we are left with $$ s_\lambda({\bf x}; \tau) = \exp\left( p_1 + \sum_{i \geq 2} \frac{p_i \tau^i}{i} \right) * s_\lambda = \sum_{|\mu| = n} \frac{\chi_\mu^\lambda p_\mu}{z_\mu} \tau^{n-m_1(\mu)}, $$ where $m_1(\mu)$ is the number of parts of size 1 in $\mu$ and we used the equation $s_\lambda = \sum_\mu \frac{\chi_\mu^\lambda p_\mu}{z_\mu}$. But the quantity that was asked about, $a_{\lambda, \rho}(\tau)$, is the coefficient of $s_\rho$ in this expression: $$ a_{\lambda, \rho}(\tau) = \langle s_\rho, \exp\left( p_1 + \sum_{i \geq 2} \frac{p_i \tau^i}{i} \right) * s_\lambda \rangle = \sum_{|\mu| = n} \frac{\chi_\mu^\rho \chi_\mu^\lambda}{z_\mu} \tau^{n - m_1(\mu)}, $$ So this is like the inner product between two Schur functions with respect to a deformed inner product where power sums $p_i$ are weighted by a factor of $\tau$ if $i \geq 2$. As a function of $\tau$, this interpolates between the case $\tau = 1$, where we just get $\langle s_\rho, s_\lambda \rangle = \delta_{\rho, \lambda}$, and the case where $\tau = 0$ (where only $\mu = (1^n)$ has a nonzero contribution to the sum) which gives $\frac{\dim(\rho)\dim(\lambda)}{n!}$. I do not recall seeing this particular interpolation before. Regarding the evaluations you mention, they can be described using homomorphisms $\varphi: \Lambda \to \mathbb{C}$ defined by $$ \varphi(p_n) = 1^n + q^n + q^{2n} + \cdots = (1-q^n)^{-1} $$ (or $\varphi(p_n) = d$ in the latter case). The upshot here is that we can describe the result of applying $\varphi$ to $s_\lambda({\bf x}; \tau)$ as the result of applying a different homomorphism to the Schur function $s_\lambda$. Specifically, $\varphi(s_\lambda({\bf x}; \tau)) = \psi(s_\lambda)$, where $$ \psi(p_n) = \frac{\tau^n}{1-q^n} $$ for $n \geq 2$ and $\psi(p_1) = (1-q)^{-1}$. (In the latter case we take $\psi(p_n) = d\tau^n$ for $n\geq 2$ and $\psi(p_1) = d$.) The fact that these descriptions are not uniform (have a different case for $n=1$) makes me suspect there might not be nice formulas for the evaluations.<|endoftext|> TITLE: Is this a counterexample to Reineke's conjecture on total stability conditions for Dynkin type quivers? QUESTION [8 upvotes]: Let $A=KQ$ be a path algebra over a field $K$ with finite connected quiver $Q$. A slope function $\mu$ is a function of the form $\mu=\sigma/dim$ defined on the Grothendieck group $K_0(A) \setminus 0$ (without the zero module), where $\sigma$ is linear and dim is just the sum of entries or equivalently the dimension on modules. A module $M$ of $KQ$ is called $\mu$-stable if $\mu(N) < \mu(M)$ for every non-zero submodule $N$ of $M$. Note that a slope function $\mu$ on a dimension vector $[a_1,...,a_n]$ is simply a function given by $\mu([a_1,...,a_n])=x_1 a_1 + ... x_n a_n/(a_1+...+a_n)$ for some $x_1,...,x_n$. Reineke's conjecture (see conjecture 7.1 in https://link.springer.com/article/10.1007/s00222-002-0273-4) can be stated as follows: On every Dynkin type quiver $Q$ there exists a slope function $\mu$ having the property that every indecomposable module is $\mu$-stable. I made a program that verified this conjecture for all Dynkin quiver with at most 6 vertices. But surprsisingly, it seems it is wrong for Dynkin type $Q=E_7$ with the following orientation (or I have a stupid thinking error which is very likely since I started with this only 2 weeks ago): Quiver( ["1","2","3","4","5","6","7"], [["2","1","a_1"],["3","2","a_2"],["4","3","a_3"],["5","4","a_4"],["6","5","a_5"],["3","7","a_6"]] ) Fix this $Q$ for the rest now and assume a slope function $\mu$ given by the $x_i$ exists. Now we have the following five indecomposable $KQ$-modules $M_i$ with submodules $N_i$ for $i=1,...,5$ (note that indecomposable modules are uniquely determined by their dimension vectors). The inclusions are in fact irreducible maps: $[0,0,1,1,1,0,1]-> [1,1,2,2,2,1,1]$ Applying $\mu$ leads to the inequality $0<4x_1+4x_2-2x_3-2x_4-2x_5+4x_6-6x_7$. $[0,1,1,1,0,0,0] -> [0,1,2,2,1,0,1]$ Applying $\mu$ leads to the inequality $0<-4x_2-x_3-x_4+3x_5+3x_7$. $[0,0,1,0,0,0,0] ->[0,0,1,1,0,0,0]$ Applying $\mu$ leads to the inequality $0<-x_3+x_4$. $[0,1,1,0,0,0,0] -> [0,1,2,1,0,0,1]$ Applying $\mu$ leads to the inequality $0<-3x_2-x_3+2x_4+2x_7$. $[1,1,1,1,1,1,1] -> [1,2,2,1,1,1,1]$ Applying $\mu$ leads to the inequality $0 < -2x_1 +5x_2 + 5x_3 - 2x_4 -2x_5-2x_6-2x_7$. Here a proof by hand that the system of those five inequalities has no solution: (1)$0<4x_1+4x_2-2x_3-2x_4-2x_5+4x_6-6x_7$ (2)$0<-4x_2-x_3-x_4+3x_5+3x_7$ (3)$0<-x_3+x_4$ (4)$0<-3x_2-x_3+2x_4+2x_7$ (5)$0 < -2x_1 +5x_2 + 5x_3 - 2x_4 -2x_5-2x_6-2x_7$ adding 1/2 *(1) to (5) gives the inequalities: (1)$0<4x_1+4x_2-2x_3-2x_4-2x_5+4x_6-6x_7$ (2)$0<-4x_2-x_3-x_4+3x_5+3x_7$ (3)$0<-x_3+x_4$ (4)$0<-3x_2-x_3+2x_4+2x_7$ (5)$0 < 7x_2 + 4x_3 - 3x_4 -3x_5-5x_7$ adding now (2) and (4) to (5) gives: (1)$0<4x_1+4x_2-2x_3-2x_4-2x_5+4x_6-6x_7$ (2)$0<-4x_2-x_3-x_4+3x_5+3x_7$ (3)$0<-x_3+x_4$ (4)$0<-3x_2-x_3+2x_4+2x_7$ (5)$0<2x_3-2x_4$ But now clearly inequalities (3) and (5) give a contradiction. Thus since there is no solution, no slope function can exist for this $Q$ of Dynkin type $E_7$. REPLY [3 votes]: Your argument looks correct to me. Note that the corresponding question for the derived category has been answered positively, and a parameterisation of total stability conditions given by QiuYu and ZhangXiaoting. In the case of $\mathsf{E}_7$, this space is $7$-dimensional (over $\mathbb{C})$.<|endoftext|> TITLE: Group of exponential growth always contains a free sub-group? QUESTION [7 upvotes]: I am not very conversant with the growth of a group, so this may be a very silly question. It is known that $F_2$, the free group of rank $2$, has exponential growth. I was wondering whether the following is true: If a group has exponential growth does it contain a free subgroup? REPLY [12 votes]: A famous theorem of Wolf shows that the growth of a solvable group is either polynomial or exponential. So no intermediate growth among solvable groups. And a famous theorem of Gromov shows that having polynomial growth implies being virtually nilpotent. Consequently, any solvable group that is not virtually nilpotent provides an example of a group with exponential growth but no non-abelian free subgroups. Of course, using big theorems is not necessary to find explicit examples, but it gives some general perspectives, and it justifies that many examples exist. One simple example is the Baumslag-Solitar group $BS(1,2)$. It has a Cayley graph that is sufficiently simple to be drawn. The pictures are taken from Wikipedia, where there is also a nice animation.<|endoftext|> TITLE: Non-trivial automorphisms and descent QUESTION [9 upvotes]: In this expository paper by Low it says: Roughly speaking, a topos in the sense of Grothendieck is the category of sheaves on a kind of generalised space whose “points” may have non-trivial automorphisms. Question 1: What does that mean? In a footnote it says: More precisely, every Grothendieck topos is equivalent to the topos of equivariant sheaves on a localic groupoid; see [Joyal and Tierney, 1984]. Still, I wonder how this is related to points having non-trivial automorphisms. The MathSciNet review of the paper by Johnstone says: Basically, it is about descent theory: the question of whether, given a morphism of toposes $f: \mathcal F → \mathcal E$, it is possible to reconstruct objects $X$ of $\mathcal E$ from objects $f^* X$ of $\mathcal F$ equipped with “descent data”. Question 2: In simple terms, what is the motivation for that kind of problem, and what is descent data? Also, why does it say "objects $f^* X$" - isn't $f^* X$ a single object? Further in the review: The main theorem (VIII 2.1) asserts that every open surjection is an effective descent morphism. Question 3: What is the idea of an effective descent morphism? Coupling this with the result, essentially due to R. Diaconescu[Comm. Algebra 4 (1976), no. 8, 723–729; MR0414658], that every Grothendieck topos $\mathcal E$ admits an open surjection $\mathcal F → \mathcal E$ where $\mathcal F$ is localic, yields the authors’ second main theorem (VIII 3.2) which is a representation theorem asserting that an arbitrary Grothendieck topos is equivalent to the topos of “equivariant sheaves” on a groupoid in the category of locales. Question 4: Why does VIII 2.1 together with the theorem of Diaconescu imply the second main theorem (the representation theorem)? (I'm only interested in getting an idea how these statements fit together, i.e., a very rough proof sketch suffices.) In the introduction to [Joyal and Tierney, 1984] the authors write: In fact, our first descent theorem for modules is completely analogous to the usual descent theorems of commutative algebra. Question 5: What are the usual descent theorems in commutative algebra? I did not find anything comprehensible googling "descent theorems in commutative algebra". Caveat: I expect that some people will react to this question saying "just read the books and papers yourself", but I find that it is quite hard to go through tons of technical definitions and lemmas without knowing the main idea in advance. This is why I am asking these questions: in order that I am able to read the papers and books myself. :-) I also tried this MO thread but I can't make head nor tail of it. There it says the main questions of descent theory are: When an object $G$ in $C_Y$ is in the image via $f^*$ of some object in $C_X$ ? Classify all forms of object $G\in C_Y$, that is find all $E\in C_X$ for which $f^*(E)\cong G$. This setting seems to be different from the setting in the third quote above, because here we consider $G\in C_Y$ which is in the codomain of $f^*$ whereas above we are given $X\in\mathcal E$, and $\mathcal E$ is the domain of $f^*$. This increases my confusion. REPLY [6 votes]: Briefly, descent is an analogue of taking quotients. In the category of sets, we have the following familiar facts: an equivalence relations on a set is a relation $R \subseteq A \times A$ satisfying certain properties; a quotient for an eq. rel. $R$ is a map $[-] : A \to Q$ that is universal among maps sending $R$ to equality; we can construct a quotient for any equivalence relation (using equivalence classes); for $x, y \in A$, we have $[x] = [y]$ if and only if $(x,y) \in R$; every surjection $e : A \twoheadrightarrow A'$ is a quotient of $A$ by the relation “$e(x)=e(y)$”. We can generalise these to arbitrary categories, ending up with the ingredients of a regular or exact category. The possibly-unexpected bit is the importance of property (4) — that elements become related in a quotient only if they were related by the given relation; this is not automatic from being a quotient, and is called an effective quotient. Lots of the ways we use quotients rely on effectivity (and not on anything else about the specific construction of the quotient). The practical takehome is: If you can present an object as an effective quotient of a well-understood object, that can be very useful for understanding it. So it’s useful to have theorems for giving/identifying effective quotients, like fact (5) above. Now since toposes are categorical objects, we have to generalise this to a 2-categorical version. When we “take a quotient” of an object of a 2-category, we want to “identify its elements/points/objects”, i.e. add new “isomorphisms”. But as ever, objects can be isomorphic via multiple different isomorphisms — equivalently, they can have non-trivial automorphisms. So when we “identify” them, we have to keep track of these extra automorphisms. So we want not just a “relation”, but extra data explaining what identifications should happen. This is the idea of the truncated simplicial object that appears in the descent construction: it’s a generalisation of an equivalence relation. Effective descent morphisms of toposes are the analogue of effective quotients (and the category of descent data is a particular construction of such a quotient). And the motivation of the theorem is to give a tractable concrete presentation for an arbitrary topos, by expressing it as an “effective quotient” of a particularly nice and familiar kind of topos. This tackles your questions 1–3. Questions 4 and 5 are a bit separate from these, and would be much better asked as separate questions.<|endoftext|> TITLE: Is there a geometric interpretation of the second derivative of the Alexander polynomial at $1$? QUESTION [9 upvotes]: For an (oriented) knot in $S^3$ the number $\Gamma(K) := \Delta_K’’(1)$ shows up in a number of places in knot theory, for example the Casson-Walker-Lescop invariant. Here $\Delta_K(t)$ is the Alexander-Conway polynomial. One explanation is that it’s the unique nontrivial order $2$ finite-type invariant, so it’s going to show up anywhere there’s something finite-type of small degree. Is there a more geometric interpretation than this? What does $\Gamma$ really mean? This question discusses some related invariants but not $\Gamma$. REPLY [6 votes]: Given a knot in $S^3$, think of it as an embedding $$f : S^1 \to S^3.$$ The configuration space of $5$ distinct points in $S^3$ is denoted $C_5(S^3)$, this is a $15$-dimensional manifold and it consists of all $5$-tuples of distinct points in $S^3$. Similarly, we can talk about $C_5(S^1)$, but since $S^1$ has a cyclic order, this space has $5!/5 = 4!$ diffeomorphic path-components, so consider the subspace where the points are in cyclic order. There is a subspace of $C_5(S^3)$ consisting of $5$ points that sit on a round circle. I.e. these are points that sit on an affine $2$-dimensional subspace of $\Bbb R^4$, and they also happen to be in $S^3 \subset \Bbb R^4$. Since any three points in $S^3$ sit on a unique round circle, the subspace of $5$ points on a circle has co-dimension 4 in $C_5(S^3)$. So our map $$ f : S^1 \to S^3 $$ has an induced map $$ f_* : C_5(S^1) \to C_5(S^3)$$ the domain is $5$-dimensional. Consider the pre-image of the circular subspace under $f_*$. Generically, it is a $1$-dimensional submanifold of $C_5(S^1)$. We consider only the components where when you compare the circular ordering of the $5$ points in $C_5(S^1)$ with the circular ordering of the circle in $S^3$, you get a pentagram. i.e. if any pair of points are adjacent in $S^1$, they are not adjacent in the circle in $S^3$. This manifold is canonically an oriented manifold, inherited by the normal orientation induced by $f_*$. So we can consider its projection to $S^1$, i.e. forget $4$ of the $5$ coordinates in this manifold. This gives us a map from a $1$-manifold (the pentagrammic 5-tuples in the preimage of $f_*$) to $S^1$, so we can take the degree. This degree is the invariant you are discussing. So this is perhaps a "very geometric" interpretation of what you are looking for. Is this more or less what you want, or are you looking for something else? This was written up by Garrett Flowers in JKTR, 2013. https://doi.org/10.1142/S021821651350017X<|endoftext|> TITLE: What must a set of $n$ points in 2D space fulfill so that it is possible to connect them through tangent circles QUESTION [8 upvotes]: In high-school I learned how to find the circles that connect points in 2D space forming a curve made out of tangent circles like this: (The green line shows the "initial direction" of the curve) You could imagine trying to close the curve with one last circle. However, this doesn't always close the curve "nicely": And sometimes it does: My question is what conditions must these points follow such that they can be connected with a "nice" curve? What I have found messing around is that, for an odd number of points, it seems to be always possible to find an "initial direction" that closes the curve nicely. But for an even number of points, this is only possible for some set of points: Sets that do close nicely for one "initial direction", close nicely too for all "initial directions". For example, points that can be all put in one circumference always close nicely. I think I have some intuition for what is going on but I can't find a nice mathematical answer. I also can't find anything like this on the internet, so I'm sorry if this isn't very rigorous. Is there an official name for this kind of curves? Where could I find more information and theory on this? REPLY [3 votes]: The comment by mlk is right on the mark. The idea of this solution is basically what he wrote, but there a lot of messy details coming from the question of which direction the arcs are curving. Let the points be $P_1$, $P_2$, ... $P_n$ in that order, and assume that the curve is a simple closed curve. Let $\alpha_i$ be the angle $\angle(P_{i-1} P_i P_{i+1})$, measured inside the simple closed polygon. I will show that: For $n=2m$ even, the curve closes if and only if $\alpha_1+\alpha_3+\cdots+\alpha_{2m-1} = \alpha_2 + \alpha_4 + \cdots + \alpha_{2m}$. In this case, we can start at any angle. (Since the sum of the angles of a $2m$-gon is $(2m-2) \pi$, the common value of the sum of $(m-1) \pi$.) If $n$ is odd, there is exactly one angle we should start at. Let $O_i$ be the center of the arc from $P_i$ to $P_{i+1}$. I found the following sign convention useful: If $O_i$ is inside the curve, I set $\theta_i = \angle(O_i P_i P_{i+1}) = \angle(O_i P_{i+1} P_i)$; these angles are equal because they are the base angles of an isosceles triangle. If $O_i$ is outside the curve, I put $\theta_i = \pi-\angle(O_i P_i P_{i+1}) = \pi-\angle(O_i P_{i+1} P_i)$. The condition that the arcs $P_{i-1} P_i$ and $P_i P_{i+1}$ are tangent says that $O_{i-1}$, $O_i$ and $P_i$ are collinear with $O_{i-1}$ and $O_i$ on the same (respectively opposite) side of $P_i$ if $O_i$ and $O_{i+1}$ are on the same (respectively opposite) side of the curve. I claim that the following relation holds: $$\alpha_i = \theta_{i-1} + \theta_i. \qquad (\ast)$$ We need to check all four cases for whether $P_{i-1}$ and $P_i$ are inside or outside of the curve: For example, if both are inside, then this holds because $$\angle(P_{i-1} P_i P_{i+1}) = \angle(P_{i-1} O_{i-1} P_i)+ \angle(P_{i} O_{i} P_{i+1}) = \angle(P_{i-1} O_{i-1} P_i)+ \angle(P_{i} O_{i+1} P_{i+1}).$$ In the other cases, we need to similarly check that all the $\pi$'s and $-$'s work out. We can use $(\ast)$ to compute all of the $\theta_i$ in terms of $\theta_1$: We have $\theta_2 = \alpha_2-\theta_1$, $\theta_3 = \alpha_3-\alpha_2+\theta_1$, etcetera. If $n$ is even, we see that the curve closes if and only if $\alpha_1+\alpha_3+\cdots+\alpha_{2m-1} = \alpha_2 + \alpha_4 + \cdots + \alpha_{2m}$. If $n$ is odd, we see that the key equation is $\theta_1 = \alpha_1 - \alpha_2+\alpha_3-\alpha_4 +\cdots +\alpha_n - \theta_1$. This will hold exactly for one value of $\theta_1$ modulo $\pi$. $\square$ I tried to work out the case of a self intersecting curve, but it got too messy. There is a method called "directed angles" which often cleans up situations like this -- see Section 1.7 of Geometry Unbound -- but I didn't find it helpful.<|endoftext|> TITLE: Cancellation of irreducibility for Galois conjugates QUESTION [6 upvotes]: Motivation: Take an algebraic number $\lambda$. In my research, I've stumbled upon the question in which cases the expression $\sum_{\sigma \in S} \sigma(\lambda)$, where $S$ is a subset of the field embeddings of $K=\mathbb{Q}(\lambda)$, can be 'less irrational' then $\lambda$ itself. Question: Take a finite Galois extension $L/\mathbb{Q}$ with Galois group $G$, let $H \leq G$ be a subgroup and let $K$ be the field fixed by $H$. Assume that $K$ has no proper subfields except for $\mathbb{Q}$. Consider a subset $S$ of $G/H$ and consider the $\mathbb{Q}$-linear vector space homomorphism $$ f_S: K \to L, x \mapsto \sum_{ \sigma \in S} \sigma(x). $$ Is it true that this map is injective unless $S=G/H$? . . . Outdated Earlier version without the assumption that $K$ has no subfields (I leave this up only so Will's reply still makes sense) If $SH$ is a subgroup of $G$ strictly bigger than $H$, this map will not be injective because $f_S$ takes values in the subfield of $K$ fixed by $SH$. Intuitively, it seems likely that $SH$ being (the coset of) a subgroup is the only case in which this 'reduction of irrationality' can happen. Is this true, i.e. is $f_S$ injective under the condition that $SH$ is not the coset of a subgroup? EDIT: As Will's example below shows below, one needs to exclude union of cosets, not only cosets. Indeed, if $U \subset G/H$ such that $UH$ is a subgroup, there is a linear relation which will carry over to any union of cosets of $UH$ as for $SH=UH \cup \sigma UH$,$f_S=f_U+\sigma \circ f_U$. Apologies, I updated the question. REPLY [2 votes]: The answer to your new question is still no. I mentioned this problem (or rather, the group-theoretic reformulation given by Will Sawin) to my colleague Steve Humphries, and he found the following two examples. Let $G$ be the 9th group of order 36 as indexed by the Magma code "SmallGroup(36,9)". It is a Frobenius group with generators $a,b,c,d$ subject to the relations $$a^2 = b, b^2 = Id(G), c^3 = Id(G), d^3 = Id(G), c^a = c d^2, c^b = c^2, d^a = c^2 d^2, d^b = d^2.$$ Let $H=\{1,a,b,ab\}$, which is the subgroup generated by $a$ and $b$. It is a maximal subgroup of index $9$ in $G$. Let $s=Id(G) + c d^2 + c^2 d$. The three elements in the support of $s$ belong to distinct left cosets in $G/H$. Let $t=\sum_{h\in H}h$ be the sum over the four elements of $H$, and let $$ u=-4Id(G) - 7d - 7d^2 + 2c + 3c d + b d + b d^2 + b c d^2 + b c^2 + b c^2 d + b c^2 d^2 + a + a d + a d^2 + a c d^2 + a b d + a b d^2 + a b c^2 d^2. $$ One can check (or have Magma check) that $tu\neq 0$, but $stu=0$. For a bigger, but perhaps conceptually simpler, example work in the group $A_5$. Take $$s=(1, 4, 2, 5, 3) + (1, 5, 4) + (2, 4)(3, 5) + (1, 3, 5, 2, 4),$$ take $$t=Id + (1, 2)(3, 4) + (3, 4, 5) + (1, 2)(4, 5) + (3, 5, 4) + (1, 2)(3, 5),$$ and take $$ u=-4*Id - 6*(1, 2, 3, 4, 5) + 2*(1, 5, 4, 3, 2) - 6*(1, 3, 2) - 10*(1, 4, 2, 5, 3) + (1, 2, 4, 3, 5) + (1, 5, 4, 2, 3) + (1, 3)(2, 4) + (1, 4)(3, 5) + (1, 2)(3, 4) + (1, 3, 2, 5, 4) + (1, 2, 3, 5, 4) + (1, 5, 2) + (2, 3)(4, 5) + (1, 2, 4) + (1, 3)(4, 5) + (1, 3, 2, 4, 5) + (2, 5, 3) + (1, 4, 2, 3, 5) + (3, 5, 4) + (1, 2, 3) + (1, 3, 5, 4, 2) + (1, 4)(2, 5) + (2, 3, 5) + (1, 2, 4, 5, 3) + (1, 5)(3, 4) + (1, 3, 5, 2, 4) + (2, 5)(3, 4) + (1, 5)(2, 4). $$ Then $tu\neq 0$ but $stu=0$. Also notice that the support of $t$ is a maximal subgroup of $A_5$. The elements in the support of $s$ belong to distinct left cosets.<|endoftext|> TITLE: Fibrant replacement of an injective model category of enriched diagrams QUESTION [5 upvotes]: Take a topologically enriched small category $\mathcal{P}$ and the category of enriched diagrams of spaces $[\mathcal{P},\mathrm{Top}]_0$. We work with the category of $\Delta$-generated spaces equipped with the mixed model structure. Suppose that the injective model structure exists (the paper http://dx.doi.org/10.4310/HHA.2019.v21.n2.a15 gives some sufficient conditions). Is there an explicit description of a fibrant replacement somewhere ? I can only understand that the injective fibrant diagrams are some kind of cofree enriched diagrams. EDIT: by explicit, I mean which enables us to make some calculations. REPLY [5 votes]: Section 8 of my paper All (∞,1)-toposes have strict univalent universes shows that under fairly general conditions, injective fibrant replacements can be given by cobar constructions (e.g. the dual of Corollary 8.16). I think this will apply to your situation if the hom-objects of $\mathcal{P}$ are cofibrant and the inclusions of identity morphisms are cofibrations. I don't know what sort of calculations you want to do, but cobar constructions come with a filtration that sometimes gives rise to calculational tools like spectral sequences.<|endoftext|> TITLE: Exercise 8.13 - Brezis QUESTION [6 upvotes]: Let $1 \leq p < \infty$ and $u \in W^{1,p}(\mathbb{R}$). Set $$ D_{h}u(x) = \frac{1}{h}(u(x+h) - u(x)), \ \ x \in \mathbb{R}, h> 0 $$ Show that $D_{h}u \to u'$ in $L^{p}(\mathbb{R}$) as $h \to 0$. I'm trying to use the fact that $C_{c}^{1}(\mathbb{R}$) is dense in $W^{1,p}(\mathbb{R}$) REPLY [18 votes]: The proof is not short, because it is done from first principles, without using any theorems about Sobolev space except its definition. By the definition of $W^{1,p}$, there exist $v_n \in C_{c}^{1}(\mathbb{R})$ and $w \in L^p(\mathbb{R})$ such that $v_n \to u$ in $L^p(\mathbb{R})$ and $v_n' \to w$ in $L^p(\mathbb{R})$. In this case we write $u'=w$. Note that $v_n'$ is a classical derivative, so $$D_h v_n(x)=\frac{v_n(x+h)-v_n}{h}= I_h(v_n')(x) \,, \tag{1} $$ where for $f\in L^p(\mathbb{R})$, we write $$I_h(f)(x):=\int_0^h \frac{f(x+t)}{h} \,dx \,.$$ By Jensen's inequality [1], for all $n \ge 1 $ and $x \in \mathbb{R}$, we have $$|I_h(v_n')(x)-I_h(u')(x)|^p \le \frac{1}{h} \int_0^h |v_n'(x+t)-u'(x+t)|^p \,dt \,. $$ Integrating both sides $\,dx$ and using Fubini on the right-hand side, we obtain $$\|I_h(v_n') -I_h(u') \|^p \le \frac{1}{h}\int_0^h \|v_n'(\cdot+t)-u'(\cdot+t)\|_p^p \,dt= \|v_n' -u' \|_p^p \,.\tag{2} $$ Given $\epsilon>0$, find $k$ such that $$ \|v_k'-u'\|_p<\epsilon \,. \tag{3} $$ Let $M$ denote the Lebesgue measure of the support of $v_k$. Since $v_k'$ is uniformly continuous, there exists $h_0\in(0,1)$ such that $$\forall t\in [0, h_0], \quad \sup_{x \in \mathbb{R}} |v_k'(x+t)-v_k'(x)|<\epsilon/(M+1) \,,$$ so for $h\in [0, h_0]$ and all $x$, we have $|I_h (v_k')(x)-v_k'(x)|<\epsilon/(M+1)$, whence $$\|I_h (v_k') -v_k'\|_p^p \le (M+h) (\epsilon/(M+1))^p <\epsilon^p \,.$$ In conjunction with $(2)$ and $(3)$, this gives $$\|I_h(u')-u' \|_p \le \|I_h(u')-I_h(v_k') \|_p + \|I_h(v_k')-v_k' \|_p + \| v_k' -u'\|_p <3\epsilon \,. \tag{4}$$ Next, fix $h\in [0, h_0]$, and choose $m=m(h,\epsilon)$ such that $$\|u-v_m \|_p<\epsilon h \quad \text{and} \quad \|u'-v_m'\|_p<\epsilon \,. $$ The first inequality implies that $\|D_h(u) -D_h(v_m) \|_p<2\epsilon$. Therefore, by $(1),\, (2)$ and $(4)$, \begin{eqnarray} \|D_h(u)-u'\|_p &\le& \|D_h(u) -D_h(v_m) \|_p+\|I_h(v_m')-I_h(u')\|_p+\|I_h(u')-u' \|_p \\ &<& 2\epsilon+\epsilon+3\epsilon=6\epsilon \,. \end{eqnarray} This completes the proof. [1] https://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form<|endoftext|> TITLE: Can an odd number of marbles jump to infinity? QUESTION [22 upvotes]: Loosely inspired by the game Abalone, I've encountered the following simple problem I cannot solve. Suppose that we are given a finite set of marbles on an infinite chessboard. One move consists of one marble jumping over another to an empty space. For example, if (0,0) and (1,0) have marbles, but (2,0) doesn't, then we can move (0,0) to (2,0). It is easy to see that there are starting configurations of an even number of marbles that we can "march off" to infinity using such moves. But is this possible to do with an odd number of marbles? It feels like this should have a simple explanation, but I don't see any right now. More on motivation: In Abalone it is useful to start by moving to the center of the board. If one doesn't make sideways moves, then the triangular Abalone board reduces to a chess board. Or almost, because in Abalone it is also allowed to move adjacent marbles in any direction, not just in the direction of the row they are contained in. If one allows that too, then there are more configurations that can do an infinite march. Also, in Abalone one can move at most 3 marbles, so to move as fast as possible, one would always move 3 marbles, and not 2 at a time, but whatever argument works for parity, probably also works for divisibility with 3. REPLY [33 votes]: With $5$ you can using the following moves: ..... ..... ..... ..... ..... ..... ..... ..oo. ...oo ..... ..o.. ..o.. ..o.. ..oo. ...oo ..ooo ..ooo ..ooo .oo.. .oo.. .oo.. ..oo. ..oo. ..oo. ..oo. ..... ..... ooo.. oo... .oo.. ..oo. ..o.. ..o.. ..... ..... ..... So the only numbers of pieces for which no configuration can go to infinity are $1$ and $3$.<|endoftext|> TITLE: Is $\mathit{Topos}^\text{op} \to \mathit{Pr}^L$ monadic? QUESTION [6 upvotes]: $\newcommand\Logos{\mathit{Logos}}\newcommand\Topos{\mathit{Topos}}\newcommand\op{^\text{op}}\newcommand\Pr{\mathit{Pr}}$Let $\Logos = \Topos\op$ be the $\infty$-category of $\infty$-topoi and geometric morphisms, where a geometric morphism points in the direction of its inverse image functor. Then $\Logos$ is a non-full subcategory of the $\infty$-category $\Pr^L$ of presentable $\infty$-categories and left adjoint functors. Question 1: Is the inclusion $\Logos \to \Pr^L$ monadic? Question 2: If so, is the induced monad lax-idempotent? I believe this functor preserves limits and filtered colimits. It doesn't preserve coproducts. I'm not sure if it actually has a left adjoint. If the answer is "yes, up to size issues", that would be interesting too. I think this might be one of those questions which is cleaner to consider in the $\infty$-categorical context than in the 1-categorical context, but I could be wrong. I'd be interested to hear about the 1-categorical case as well (where I suppose one would consider the $(2,1)$-categories of 1-logoi and locally presentable 1-categories). REPLY [5 votes]: Regarding monadicity (rather than comonadicity), the (2-categorical variant of the) question is answered in Bunge–Carboni's The symmetric topos. In their paper, $\mathbf A$ denotes the 2-category of locally presentable categories and cocontinuous functors (i.e. left adjoint functors), and $\mathbf R$ denotes the 2-category of logoi. There is a 2-adjunction $\Sigma : \mathbf A \rightleftarrows \mathbf R : U$ (Theorem 3.1) and the induced 2-monad is lax idempotent (Theorem 4.1 and the following discussion). Presumably everything works out similarly in the $(\infty, 2)$-categorical setting.<|endoftext|> TITLE: Explicit triples of isomorphic Riemann surfaces QUESTION [11 upvotes]: Inspired by a discussion with Neil Strickland I am very interested to hear of explicit examples (one per answer, please), as follows. A compact Riemann surface can be presented in many different ways. For example: A smoothly embedded embedded surface in the three-sphere $S^3$. A smooth projective curve (say cut out of $\mathbb{CP}^2$ by a single equation). A quotient of the hyperbolic plane by some fuchsian group. For each of these "ways" we can accept some minor modifications. We prefer embeddings into $S^3$ (or the three-torus $\mathbb{T}^3$) because we want to actually "see" the surface. However giving the surface as a level-set of a nice function, or via some other nice analytical construction (for example as a minimal surface with symmetries), is also welcome. I am not algebraic enough to deform the condition given in (2) - I hope some reader will suggest the correct modifications. We also accept quotients of $\mathbb{C}$ by a lattice $\mathbb{Z} + \mathbb{Z}\omega$. We also accept tilings of the upper half-plane as long as the tiling has no "moduli" (or has enough explicit side conditions) so that the fuchsian group can be deduced, with sufficient amount of hyperbolic trig). We may also modify (3) in another way - for example giving square-tiled surfaces or more generally surfaces given by gluing explicitly described polygons in $S^2$, $\mathbb{E}^2$, or $\mathbb{H}^2$. Question: Give explicit examples of pairs (or preferably triples) of isomorphic Riemann surfaces of the above types. We first dispose of the trivial example of the sphere. Here (1) and (3') are addressed by saying "the round sphere". (2) is addressed by saying (for example) "$x + y + z = 0$". As an actual example, we have tori. For (3), we specify $\omega$ and thus the lattice $\mathbb{Z} + \mathbb{Z}\omega$. For (2), we have the Eisenstein series giving the modular invariants. For (1) we have the Hopf tori embedded in $S^3$ (and thus, after stereographic projection, embedded in three-space). Pinkall [Inventiones, 1985] defines these, and has explicit constructions (see Figure 4b (reproduced below) and Figure 6b). See also Figure 5 of Sullivan [Bridges, 2011]. I give further examples in the comments here but please see, in addition, Strickland's talk. REPLY [8 votes]: A particular example is that of the Lawson surface $\xi_{g,1}$ of genus $g$. As defined here, it is a compact minimal surface in the 3-sphere, obtained by reflecting the solution of the Plateau problem for a specific geodesic 4-gon. The surface has many symmetries, so it is easy to deduce that it is given by the algebraic equation $$y^2=z^{2g+2}-1.$$ Finally, the Riemann surface has so many symmetries that you can write down its uniformisation oper (and its monodromy) explicitly: the former is given by the desingularisation of the pull-back (by the holomorphic map to the projective line obtained by quotienting out all symmetries) of the Fuchsian system $$\nabla\,=\,d+\begin{pmatrix}\frac{1}{8}&0\\0&-\frac{1}{8}\end{pmatrix}\frac{dz}{z}+ \begin{pmatrix}-4\rho^2&1\\ \rho^2-16\rho^4&4\rho^2\end{pmatrix}\frac{dz}{z-1}$$ where $\rho=\tfrac{g}{2g+2}.$ The Fuchsian group can be computed explicitely from the monodromy of the Fuchsian system, which is (conjugated to) the representation of the 3-punctured sphere with monodromies $$ M_0\,=\,\begin{pmatrix}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$$ $$M_1\,=\,\begin{pmatrix} -\cos{\frac{\pi}{k}}-\sqrt{(1+\cos{\frac{\pi}{k}}) \cos\frac{\pi}{k}}& 1 + \cos{\frac{\pi}{k}}+\sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}\\ -1 -\cos{\frac{\pi}{k}}+\sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}& -\cos{\frac{\pi}{k}}+\sqrt{(1+\cos{\frac{\pi}{k}}) \cos\frac{\pi}{k}} \end{pmatrix}$$ $$M_\infty\,=\,\begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{-1- 2\cos{\frac{\pi}{k}}-2 \sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}}{\sqrt{2}}\\ \frac{1+ 2\cos{\frac{\pi}{k}}-2\sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}}{\sqrt{2}}& \frac{1}{\sqrt{2}} \end{pmatrix},$$ where $k=g+1.$ A 3D print of the (stereographic projection of the) genus two Lawson surface. The (conformal) parameterisation is due to Sebastian Heller and Nicholas Schmitt - the print is due to Nicholas Schmitt and Wjatscheslaw Kewlin, and the photo is due to Wjatscheslaw Kewlin. REPLY [7 votes]: There are indeed very few pairs (except spheres with 3 or 4 singularities, or tori, and what can be obtained from them by finite coverings, where correspondence 2)-3) is completely explicit. See: H. P. de Saint-Gervais, Uniformisation des surfaces de Riemann, ENS Editions, 2010 (there is an English translation), Chap. IX. Concerning pairs 1)-3), many beautiful examples occur in the theory of minimal surfaces; they can be seen in the Bloomington Virtual Minimal Surface Museum. REPLY [6 votes]: A classical, wonderful example in which is possible to explicitly see all the three descriptions is the Klein quartic.<|endoftext|> TITLE: Can $\kappa^\lambda$ be large if $2^\lambda$ is small and $\lambda<\mathrm{cof}(\kappa)$? QUESTION [14 upvotes]: We work in ZFC throughout. The following question was posed to me by a friend: Can there exist cardinals $\kappa,\lambda$ such that $\lambda<\mathrm{cof}(\kappa)$ and $2^\lambda<\kappa<\kappa^\lambda$? Originally I thought this should be easy, after all many minor variants are almost immediate: if we allow $\lambda=\mathrm{cof}(\kappa)$, then Konig's theorem implies $\kappa^\lambda>\kappa$ and we can easily arrange $2^\lambda$ to be small (either by choosing a model where $2^\lambda$ is small, or picking $\kappa$ large enough), and if we don't require $2^\lambda<\kappa$, then we can for instance let $\kappa=\aleph_{\omega_1}$ and take a model in which $2^{\aleph_0}$ is larger than that. However, thinking about this problem further made me realize this is probably quite close to some unsolved problems in set theory. The smallest viable counterexample could be given by $\kappa=\aleph_{\omega_1}$ and $\lambda=\aleph_0$. It is not hard to see that for $\kappa^\lambda>\kappa$ to hold, we need $\mu^\lambda>\kappa$ for some $\mu<\kappa$ (this is because of the inequality $\lambda<\mathrm{cof}(\kappa)$). Therefore one idea to get the result is to pick a model in which $2^{\aleph_0}<\aleph_{\omega_1}$ and $\aleph_\omega^{\aleph_0}>\aleph_{\omega_1}$. I have no idea whether this is expected to be possible, but some results and conjectures in Shelah's PCF theory suggest the answer might be negative - specifically, if we ask for a stronger inequality $2^{\aleph_0}<\aleph_{\omega}$, then it is known $\aleph_\omega^{\aleph_0}<\aleph_{\omega_4}$, and it is conjectured that $<\aleph_{\omega_1}$ should also be true then. Of course, those results are not strong enough to exclude it, and they show that our tools are not good enough to disprove that result. Is it then, against all odds, consistent with ZFC that such a pair of cardinals exists? Some comments indicate the answer should be positive under large cardinal (consistency) assumptions. I am primarily interested in whether this is relatively consistent with ZFC alone, but if no such results are available I am happy to accept an answer conditional on higher consistency strength. REPLY [14 votes]: It is consistent that such a pair exists, see my paper Singular cofinality conjecture and a question of Gorelic. To show that some large cardinals are needed, suppose for example $\lambda=\aleph_0 < \aleph_1=cf(\kappa)$ and $\kappa^\omega > \kappa > 2^\omega.$ Then for some $\mu < \kappa, \mu^\omega > \kappa,$ and without loss of generality $2^\omega < \mu$ and $cf(\mu)=\aleph_0$ (otherwise pick some $\mu'$ in the interval $(\mu, \kappa)$ as required). Thus SCH (the singular cardinals hypothesis) fails and hence we need some large cardinals. Indeed $\mu^\omega \geq \kappa^+ \geq \mu^{+\omega_1+1}$, so it sems a large cardinal $\theta$ with $o(\theta) \geq \theta^{+\omega_1+1}$ is needed. Suppose in general $\lambda < cf(\kappa)$ and $2^\lambda < \kappa < \kappa^{\lambda}$. We may assume that $\lambda$ is the least cardinal with this property. Let me first show that $\lambda$ is regular. Otherwise let $(\lambda_i: i< cf(\lambda))$ be an increasing sequence of regular cardinals cofinal in $\lambda$. By the choice of $\lambda,$ for all $i, \kappa^{\lambda_i} \leq \kappa$ (as clearly for all $i, \lambda_i< cf(\kappa)$ and $2^{\lambda_i}<\kappa$), thus $\kappa^\lambda= \prod_{i TITLE: How wild can an open topological 3-manifold be if it has a compact quotient? QUESTION [11 upvotes]: Let $M$ be an open, simply connected, 3-manifold. Suppose $M$ admits a properly discontinuous, co-compact topological action by a finitely generated group. Question 1: If $M$ is 1-ended, must it be homeomorphic with $\mathbb{R}^3$? More generally: Question 2: Is $M$ determined up to homeomorphism by its number of ends? (By a classical result of Hopf, this number is 1, 2, or uncountably infinite, I think.) More generally, I'm interested in results of the form: if a 3-manifold $M$ covers a compact manifold/orbifold, then it cannot be as wild as a generic open 3-manifold such as e.g. the Whitehead manifold. Edit: Having discussed this with an expert, I believe that my questions above boil down to the following conjecture: Conjecture: Let $G$ be the fundamental group of a closed 3-manifold $M$. Then $G$ is simply connected at infinity. (Equivalently, the universal cover of $M$ is simply connected at infinity.) This conjecture is implicit in this paper by Funar & Otera: https://www-fourier.ujf-grenoble.fr/~funar/funote.pdf REPLY [7 votes]: The possible universal covers of closed 3-manifolds are $S^3-C$, where $|C|=0, 1, 2$ or $C$ is a tame Cantor set, corresponding to the space of ends of the fundamental group as you suspect. This follows from the geometrization theorem, known to experts but might not be written down. I gave a survey talk on this once, you can find the notes here. If $M$ is a closed orientable connected 3-manifold, and $\pi_1 M$ is finite, then its universal cover is $S^3$ by the Poincaré conjecture. If $\pi_1M$ is infinite and $\pi_2 M=0$, then the universal cover $\tilde{M} \cong \mathbb{R}^3$. In this case, $M$ has a geometric decomposition. If the decomposition is trivial, then $M$ is modeled on one of the six geometries homeomorphic to $\mathbb{R}^3$, and hence the universal cover is $\mathbb{R}^3$. Otherwise, $M$ has an essential torus, hence is a Haken manifold. Waldhausen proved that Haken manifolds have universal cover $\mathbb{R}^3$ - see Theorem 8.1. If $\pi_1 M$ is infinite but $\pi_2 M\neq 0$, then either $M$ is modeled on the $S^2\times \mathbb{R}$ geometry and $M$ is homeomorphic to $\mathbb{RP}^3\#\mathbb{RP}^3$ or $S^2\times S^1$. In this case the number of ends of $\pi_1 M$ is 2. Otherwise, $M$ is a non-trivial connect sum by the sphere theorem and its universal cover is $S^3-C$ where $C$ is a tame Cantor set. The connect summands have universal cover either $S^3$, $S^2\times \mathbb{R}$ or $\mathbb{R}^3$. When you take connect sums, you remove open balls from each manifold and glue the sphere boundaries together. The universal cover is obtained by gluing the universal covers of each summand punctured along balls, either finitely many in $S^3$ or infinitely many in $S^2\times\mathbb{R}$ or $\mathbb{R}^3$. One can see that such manifolds are built out of thrice punctured spheres, and hence the universal cover can be decomposed into thrice punctures spheres. Such a manifold is homeomorphic to $S^3-C$.<|endoftext|> TITLE: $2n \times 2n$ matrices with entries in $\{1, 0, -1\}$ with exactly $n$ zeroes in each row and each column with orthogonal rows and orthogonal columns QUESTION [9 upvotes]: I am interested in answering the following question: Question For a given $n$, does there exist a $2n \times 2n$ matrix with entries in $\{1, 0, -1\}$ having orthogonal rows and columns with exactly $n$ zeroes in each row and column? Conjectures For $n=2^k$, $k\ge0$ such a matrix always exists. For $n=3$ such a matrix does not exist. For $n=5$ such a matrix exists, for example: $$ \left( \begin{array}{cccccccccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & -1 & 0 & -1 & -1 & -1 & 0 \\ 1 & 0 & 0 & -1 & 0 & 0 & -1 & 1 & 1 & 0 \\ 0 & -1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & -1 \\ 1 & -1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 & 1 & -1 & 0 & -1 & 0 & 1 \\ 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1 & 1 & -1 \\ 0 & -1 & 1 & 0 & 0 & -1 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & -1 & 1 & 1 & 0 & 0 & -1 & -1 \\ \end{array} \right) $$ For $n=7$ such a matrix does not exist. REPLY [10 votes]: There is no such matrix if $n\equiv 3\pmod 4$. Suppose otherwise. Each column represents a vector of length $\sqrt n$. Since those vectors are pairwise orthogonal, their sum is a vector whose scalar square is $2n^2$. On the other hand, the sum of all columns has odd entries, so their squares are all congruent to $1$ modulo $8$. Hence the sum of those squares is congruent to $2n$ modulo $8$. Hence we should have $2n^2\equiv 2n\pmod 8$, or $8\mid 2n(n-1)$ which does not hold.<|endoftext|> TITLE: Second Skorokhod embedding in high dimensions QUESTION [5 upvotes]: The first Skorokhod embedding theorem says that any random variable $X$ with $\mathbb E X=0$ and $\mathbb E X^2<\infty $ can be written as $X=B_{\tau }$ where $B$ is a Brownian motion and $\tau$ is a stopping time. The second Skorokhod embedding (which follows easily from the first one) says that a random walk $S_n:=\sum _{k=1}^n X_i$ where $X_i$ are normalized i.i.d. can be coupled with a Brownian motion such such that almost surely, for all large $n$ we have $|S_n-B_n| \le n^{1/4}\log ^2 n$. In dimension $d\ge 2$ the first embedding theorem is not true but the second one is true. Is there a simple proof of this fact? I am aware of the fact that there are better quantitative results like KMT. These results hold also in higher dimenstions and they say that we can couple the R.W. and the B.M. up to a logarithmic distance. The problem is that these results are very complicated. I wonder if there is a simpler proof if we only want to couple up to a distance of $n^{1/4}$. REPLY [3 votes]: For higher dimensions there is a paper of J. R. Baxter, R. V. Chacon(1974) "Potentials of stopped distributions" giving the first embedding under some conditions, see it at https://doi.org/10.1215/ijm/1256051015 . In particular, under these conditions, they show that $\mathbb E\tau<\infty$ when the second moment of $X$ is finite, which should imply the second embedding theorem. Conditions are weakened in reference [1] below. There is also a nice one-dimensional version of the paper, which explains the construction by Chacon and Walsh (1976) One-dimensional potential embedding. Other 2 references I know are N. Falkner. On Skorohod embedding in n-dimensional Brownian motion by means of natural stopping times. In Seminaire de Probabilitees, XIV, volume 784 of Lecture Notes in Math., pages 357–391. Springer, Berlin, 1980. MR580142 D. Heath. Skorokhod stopping via potential theory. In Seminaire de Probabilites, VIII, pages 150–154. Lecture Notes in Math., Vol. 381. Springer, Berlin, 1974. MR368185<|endoftext|> TITLE: How often are forcing extensions of countable computably saturated models of $\mathsf{ZFC}$ computably saturated? QUESTION [6 upvotes]: Recall that given a finite language $\mathcal{L}$, we say that an $\mathcal{L}$-structure is computably saturated (or recursively saturated) if for any computable set $\Sigma(\bar{x},y)$ of $\mathcal{L}$-formulas in the variables $\bar{x}y$ and any $\bar{a} \in M^{\bar{x}}$, if $\Sigma(\bar{a},y)$ is finitely satisfiable in $M$, then it is satisfied in $M$. An easy inductive argument shows that every countable $\mathcal{L}$-structure has a countable computably saturated elementary extension. Another easy argument shows that any expansion of a computably saturated structure by finitely many constants is still computably saturated. One important property of countable computably saturated structures is resplendence, which for our purposes can be defined like this: An $\mathcal{L}$-structure $M$ is resplendent if for any finite extensions $\mathcal{L}' \supseteq \mathcal{L}$ and any $\mathcal{L}'$-sentence $\varphi$ that is consistent with $\mathrm{Th}(M)$, there is an expansion $M'$ of $M$ that is a computably saturated model of $\mathrm{Th}(M)\cup\{\varphi\}$. All countable computably saturated structures are resplendent. Moreover, the same is true after any expansion by any finite number of constants. Forcing is typically defined in terms of well-founded models of $\mathsf{ZFC}$, but, as discussed in the answers to this question, forcing ultimately makes sense over arbitrary models of $\mathsf{ZFC}$: Given an model $M$ of $\mathsf{ZFC}$ and a forcing poset $\mathbb{P} \in M$, we can consider the language that contains $\in$, $\mathbb{P}$, and a unary predicate $G$. There is a single sentence $\chi(\mathbb{P},G)$ in this language that says that $\mathbb{P}$ is a forcing poset and $G$ is a generic filter on $\mathbb{P}$. As Emil Jeřábek points out in his answer to that question, $\mathsf{ZFC} \cup \{\chi(\mathbb{P},G)\}$ is a conservative extension of $\mathsf{ZFC}$. (The rest of the work of a forcing argument is showing that $(M,\mathbb{P},G)$ interprets an end extension of $M$ modeling $\mathsf{ZFC}$ in which $G$ is a set. We don't really need to worry about that for the sake of this question.) We have, furthermore, that the theory $\mathrm{eldiag}(M) \cup \{\chi(\mathbb{P},G)\}$ is consistent. If $M$ is a countable computably saturated model of $\mathsf{ZFC}$, then, by resplendence, we have that for any forcing poset $\mathbb{P}\in M$, there is a $G \subseteq \mathbb{P}$ which is $M$-generic such that $(M,\mathbb{P},G)$ is computably saturated. (In particular, this means that the actual forcing extension $M[G]$ will be computably saturated as well.) My question is about whether this always happens, 'usually' happens, or 'usually' doesn't happen for various choices of $G$. Question 1: Let $M$ be a countable computably saturated model of $\mathsf{ZFC}$, and let $\mathbb{P} \in M$ be a forcing poset. If $G \subseteq \mathbb{P}$ is an $M$-generic filter, does it follow that $(M,\mathbb{P},G)$ is computably saturated? I doubt that this does in fact always hold, but I don't know how to build a counterexample. What's less certain to me is whether a 'sufficiently generic' $M$-generic filter results in a computably saturated expansion. To state this carefully, we need the following observation: Fix a countable computably saturated model $M$ of $\mathsf{ZFC}$ and a forcing poset $\mathbb{P} \in M$. Let $A$ be the set of all elements $\alpha$ of $\mathbb{P}^\omega$ such that for each $i<\omega$, $\alpha(i+1) \leq \alpha(i)$. $\mathbb{P}^\omega$ can easily be identified with Baire space, and $A$ is a $G_\delta$ subset of $\mathbb{P}^\omega$ and therefore itself a Polish space. It's easy to see that the set $$B = \{\alpha \in A : \alpha\text{ generates an }M\text{-generic filter}\}$$ is comeager in $A$. (This is essentially the Rasiowa–Sikorski lemma.) For any $\alpha \in A$, let $G_\alpha = \{p \in \mathbb{P} : p \geq \alpha(i)\text{ for some }i\}$. (This is what we mean by 'generating' a filter.) Question 2: Is the set $$\{\alpha \in B : (M,\mathbb{P},G_\alpha)\text{ is computably saturated}\}$$ always comeager in $A$? I would also be interested in answers involving a weaker set theory than $\mathsf{ZFC}$. REPLY [8 votes]: The answer to Question 1 is positive (thus the answer to Question 2 is also positive). More explicitly, the positive answer to Question 1 follows from the following well-known facts: Lemma 1. $(M,\mathbb{P},G)$ is parametrically definable in $M[G]$. Lemma 2. $M[G]$ is recursively saturated. Lemma 3. Any structure that is parametrically definable in a computably saturated structure is also computably saturated. Lemma 1 is due independently to Laver and Woodin, who proved that $M$ is parametrically definable in $M[G]$; see this MO post of Hamkins for more detail. Lemma 2 was proved as part of the proof of Theorem 2.6 of this paper of mine. You can also find a proof of Lemma 2 in this blogpost of Kameryn Williams (see the third proposition). The same blogpost also includes a reference to another result of mine which shows that Lemma 2 can fail if $\mathbb{P}$ is a proper class notion of forcing in $M$. Lemma 3 follows from the definitions involved. Finally, let me point out that Lemma 2 is true for all models $M$ of ZF, but I do not know the status of Lemma 1 for a model of ZF in which AC fails.<|endoftext|> TITLE: Mathematicians learning from applications to other fields QUESTION [16 upvotes]: Once upon a time a speaker at the weekly Applied Mathematics Colloquium at MIT (one of two weekly colloquia in the math department (but the other one is not called "pure")) said researchers in a certain area of mathematics thought that their work could be of value to some field other than mathematics—maybe it was some kind of engineering, so I'll just call it "engineering"—but then it was found that interactions between engineers and mathematicians made substantial contributions to mathematical research but not to engineering. I don't remember what it was about, beyond that. So my question is: What are the most edifying examples in recent centuries, of applications to fields other than mathematics greatly benefitting mathematical research when mathematicians had expected to be only the benefactors of those other fields? REPLY [12 votes]: Not sure if this is what you had in mind, but experiments on the random packing of tetrahedral dice were able to achieve denser packings than had been constructed by mathematicians at the time. Both engineering and mathematics have played a role in this subject, but arguably mathematics has been more enriched than engineering.<|endoftext|> TITLE: Much weaker condition for Kakeya sets over finite fields QUESTION [5 upvotes]: What is the minimum size of a subset $S \subseteq \mathbb{F}_p^n$ such that for all directions $a \in \mathbb{F}_p^n$, there is a line in direction $a$ that intersects $S$ in at least $C$ points? If $C = p$ this is exactly the finite field Kakeya problem, and we know $|S| \ge C_n p^n$ by the proof of Dvir in On the size of Kakeya sets in finite fields. However, I am curious what bounds we can show if e.g. $C$ is much smaller than $p$, eg. $C = \log^{O(1)}(pn)$. I believe by random sampling, one can show an upper bound of something like $\lvert S\rvert \le p^{(1-1/C)n}$, which is trivial for $C = \log^{O(1)}(pn)$. A lower bound of $\lvert S\rvert \ge p^{(n-1)/2}$ is also clear, because the points in $S$ determine at most $\lvert S\rvert^2$ lines. Also, in Dvir's paper, he also handles the case where $C = p^{1-\epsilon}$, but I am asking for even much smaller values of $C$. So is the truth closer to $p^{n/2}$ or $p^n$? REPLY [8 votes]: As claimed in equation (2) of this recent paper Dhar, Manik; Dvir, Zeev; Lund, Ben, Simple proofs for Furstenberg sets over finite fields, ZBL07471814 [arXiv]. one can get a lower bound of $2^{-n} C^n$ by adapting the arguments from Bukh, Boris; Chao, Ting-Wei, Sharp density bounds on the finite field Kakeya problem, ZBL07471810. [arXiv] For $C=p$ the bound is tight up to a factor of $2$. I don't know what to conjecture for other values of $C$. EDIT: for $C=2$ I would imagine by the usual union bound argument that a random set of cardinality $p^{n/2} n^{10} \log^{10} p$ (say) should work to give a upper bound, though I haven't checked this carefully. So maybe the truth is something like $p^{n/2} C^{n/2}$ up to lower order terms?<|endoftext|> TITLE: Projective tensor product of injective operators QUESTION [7 upvotes]: I've seen claims that it is known that for a pair of bounded injective linear operators $T\colon X\to Y, S\colon W\to V$, their tensor product $T\otimes S\colon X \otimes_\pi W\to Y \otimes_\pi V$ need not be injective. Here $\otimes_\pi$ stands for the projective tensor product of Banach spaces. Can this happen when $T = {\rm id}_X$, the identity operator on some Banach space $X$? If so, can it happen for $T = {\rm id}_{L_1}$, the identity operator on $L_1$? Question 2 has negative answer when $S$ is an isomorphism onto its range. Not surprisingly, the answer would be always positive for the injective tensor product. REPLY [5 votes]: $\require{AMScd}\newcommand{\id}{\operatorname{id}}$I use a common characterisation of the approximation property as found in e.g. Ryan's book Zbl 1090.46001. A Banach space $X$ has the approximation property if and only if for each Banach space $Y$ (it is enough to take $Y=X^*$) the natural map $$ X \widehat\otimes Y \rightarrow X \check\otimes Y $$ is injective. Here I write $\widehat\otimes$ and $\check\otimes$ for the completed projective, respectively, injective tensor products. We can now answer (2) in the negative. Let $X$ have the approximation property, and let $S:W\rightarrow V$ be injective. Consider the commutative diagram $$ \begin{CD} X\widehat\otimes W @>>> X \check\otimes W \\ @V{\id\otimes S}VV @VV{\id\otimes S}V \\ X\widehat\otimes V @>>> X \check\otimes V \end{CD} $$ The map $\id\otimes S: X \check\otimes W \rightarrow X \check\otimes V$ is injective, and the horizontal arrows are injective as $X$ has AP, so $\id\otimes S: X \widehat\otimes W \rightarrow X \widehat\otimes V$ is injective. In particular $X=L_1$ has the AP, showing the negation of (2). As Jochen Wengenroth noted, Q1 can be reduced to the $T\otimes S$ case which the OP stated has a positive answer. However, here is a concrete example, following Chapter 5, Corollary 4 of Defant and Floret Zbl 0774.46018. Let $X$ be any Banach space, and let $B_{X^*}$ be the unit ball of the dual space $X^*$, consider $\ell_\infty(B_{X^*})$ and define $j:X\rightarrow \ell_\infty(B_{X^*})$ by evaluation: $j(x) = ( \phi(x) )_{\phi\in B_{X^*}}$. Then $j$ is an isometry onto its range. We know that $\ell_\infty(B_{X^*})$ has AP so $$ X^* \widehat\otimes \ell_\infty(B_{X^*}) \rightarrow X^* \check\otimes \ell_\infty(B_{X^*}) $$ is injective. Consider now the commutative diagram $$ \begin{CD} X^* \widehat\otimes X @>>> X^* \check\otimes X \\ @V{\id\otimes j}VV @VV{\id\otimes j}V \\ X^* \widehat\otimes \ell_\infty(B_{X^*}) @>>> X^* \check\otimes \ell_\infty(B_{X^*}) \\ \end{CD} $$ The bottom arrow is injective, and the right-hand down arrow is. If $X$ does not have AP then the top arrow is not injective, and so the left-hand down arrow must fail to be injective, which gives an example of (1). (There is nothing special about $\ell_\infty$ here: any Banach space $F$ with the AP and any injection $j:X\rightarrow F$ would work.)<|endoftext|> TITLE: Converse to Hopf degree theorem QUESTION [15 upvotes]: Below, I mean smooth oriented closed connected manifolds and smooth maps (but am happy to hear about the topological category, or unoriented manifolds, etc instead). Say that $X^n$ has the Hopf property if two maps $f_0,f_1 : M^n\to X^n$ are homotopic if and only if they have the same degree. Say that $X$ has the self-Hopf property if the Hopf property holds for $M=X$. The Hopf degree theorem says that $S^n$ has the Hopf property. It's easy to see that $T^n$ doesn't have the (self-)Hopf property. My question is: If $X$ has the (self-)Hopf property is it homeomorphic to $S^n$? REPLY [7 votes]: A manifold $M$ of dimension $>1$ with $H^1(M) \neq 0$ does not have the self-Hopf property. For if so, then there is a map $f: M \to S^1$ that induces a non-trivial homomorphism $f_*: H_1(M) \to \mathbb{Z}$. Compose this with a map $g: S^1 \to M$ carrying a non-torsion homology class to get a map $M \to M$ that induces a non-trivial map in homology and hence is not null-homotopic. But it factors through $S^1$ and so has degree $0$. This was a counterexample to an attempted proof that a hyperbolic manifold with trivial isometry group has the self-Hopf property. (Hyperbolic manifolds with trivial isometry group exist, at least in dimension $3$.) The argument is that a map of non-zero degree must in fact have degree one (using Gromov's norm) and hence are homotopic to an isometry. (This uses Gromov's proof of Mostow rigidity; see Haagerup, Uffe; Munkholm, Hans J. Simplices of maximal volume in hyperbolic n-space. Acta Math. 147 (1981), no. 1-2, 1–11.) So by the hypothesis on the isometry group must be homotopic to the identity. But the previous paragraph says that this breaks down for degree 0.<|endoftext|> TITLE: Is there any genus one knot other than $5_2$ with Alexander polynomial $2t^2-3t+2$? QUESTION [6 upvotes]: It is known that genus one fibred knots are two trefoils and the figure-eight knot. Is there any characterization of the knot $5_2$? Specifically, is there any other genus one knot that shares the same Alexander polynomial $2t^2-3t+2$ with $5_2$? REPLY [3 votes]: Maybe a different approach but having the same flavor: Take a (untwisted) Whitehead double of any nontrivial knot $K$. Denote such a double knot $WD(K)$. Construct a satellite knot using $5_2$ knot as the pattern and $WD(K)$ as the companion. Then the Alexander polynomial of the satellite knot is the product of the Alexander polynomials of $5_2$ knot and $WD(K)$. The latter has the trivial Alexander polynomial.<|endoftext|> TITLE: Is there a Dold-Kan theorem for circle actions? QUESTION [8 upvotes]: There are several interesting equivalences of "Dold-Kan type" in the setting of stable $\infty$-categories. Namely, let $\mathcal C$ be a stable $\infty$-category. Then the following 3 stable $\infty$-categories are known to be equivalent: The $\infty$-category $Fun(\mathbb N, \mathcal C)$ of filtered objects in $\mathcal C$ (where $\mathbb N$ is the poset of natural numbers). The $\infty$-category $Fun(\Delta, \mathcal C)$ of cosimplicial objects in $\mathcal C$. The $\infty$-category $Ch(\mathcal C)_{\leq 0}$ of nonpositively-homologically-graded chain complexes in $\mathcal C$. $(1) \Leftrightarrow (2)$ is due to Lurie (see HA 1.2.3) and $(1) \Leftrightarrow (3)$ is due to Ariotta. $(2) \Leftrightarrow (3)$ is meant to be reminiscent of the classical Dold-Kan theorem. There's another place where chain complex structures can come from though, namely from actions by the circle group $S^1$. For instance, in the HKR theorem, the differential on the de Rham complex arises directly from the $S^1$-action on Hochschild cohomology. (I'm not familiar enough with the literature to have a reference for this, but I gather that the idea is to look at the map $X \oplus \Sigma X = \Sigma^\infty_+ S^1 \wedge X \to X$ coming from a circle action; the second component is a map $\Sigma X \to X$ which is exactly the data needed for a differential; that it squares to zero comes from the associativity of the circle action, since the top cell of $S^1 \times S^1$ splits off.) In other words, we are led to consider The $\infty$-cateogry $Fun(\mathbb C\mathbb P^\infty, \mathcal C)$ of $\mathcal C$-objects with $S^1$-action. Now, I think that (4) lives in the unbounded world -- the right things to compare to are ' $Fun(\mathbb Z, \mathcal C)$ (filtrations extending in both directions) ' (omitted -- but Kan's combinatorial spectra might be relevant) ' $Ch(\mathcal C)$ (unbounded chain complexes) Question: Are the $\infty$-categories (1') and (3') equivalent to (4), for an arbitrary stable $\infty$-category $\mathcal C$? REPLY [9 votes]: No, they are not equivalent, even for $C = Sp$. Indeed, the category of spectra with $S^1$-action is also the category of $\mathbb S[S^1]$-modules, and is compactly generated by a single object. On the other hand, compact objects of $Fun(\mathbb Z, Sp)$ are retracts of finite colimits of representables, and any finite set $S$ of representables cannot generate the whole thing - e.g. because if $n$ is below all the elements in $S$, then $F(n) = 0$ for any $F$ generated under colimits by $S$ . So it is not compactly generated by a single object (you have to change the proof a bit, but the same holds for $Fun(\mathbb N, Sp)$)<|endoftext|> TITLE: Rationality of field embeddings QUESTION [5 upvotes]: After my earlier question question turned out to have a negative answer (Thank you to all respondents!), here is a more modest one. Both a positive answer and a counterexample would help my work. If some context is of interest, these questions turned up while studying self-similar measures with algebraic contraction ratios. The sets $S$ I'm interested in are of the form $S_\lambda=\{\sigma: |\sigma(\lambda)|<1\}$ for some $\lambda$ generating $K$, if it makes any difference. Question: Let $K$ be a finite extension of $\mathbb{Q}$. Let $S$ be a subset of the field embeddings into $\mathbb{R}, \mathbb{C}$ (picking both or none in a pair of complex embeddings). If $$\sum_{\sigma \in S} \sigma(x) \in \mathbb{Q}$$ for all $x \in K$, does $S$ then necessarily consist of all (or no) field embeddings? REPLY [12 votes]: The answer is yes. Suppose $S$ is nonempty. Write $K=\mathbb Q(\alpha)$ (using the primitive element theorem). Applying your assumption to $\alpha^n$ for all $n\in\mathbb N$ we get that all sums $\sum_{\sigma\in S}(\sigma(\alpha))^n$ are rational. Using Girard-Newton formulas, this implies that all the elementary symmetric polynomials in $\{\sigma(\alpha)\mid\sigma\in S\}$ are rational. This implies that the polynomial $\prod_{\sigma\in S}(x-\sigma(\alpha))$ has rational coefficients. However, each $\sigma(\alpha)$ has degree $[K:\mathbb Q]$ over $\mathbb Q$, so the degree $|S|$ of this polynomial must be at least $[K:\mathbb Q]$, which means precisely that $S$ consists of all embeddings $\sigma$.<|endoftext|> TITLE: Why don't we study hyperbolic equations as elliptic and parabolic equations? QUESTION [7 upvotes]: In the research of elliptic and parabolic equations, the Schauder estimate is one of the most important issues for them. In this topic, we always bound the norm of higher regularity in the small ball by a bigger one. That is, for the elliptic equation $ \operatorname{div}(A(x)\nabla u)=0 $, we have estimates like $ \left\|u\right\|_{C^{0,\alpha}(B_1)}\leq C\left\|u\right\|_{L^2(B_2)} $, where $ B_r=B(0,r) $ is the ball with center $ 0 $ and radius $ r $. I want to ask why we do not study such estimates for hyperbolic equations. REPLY [25 votes]: Why we do not study such estimates for hyperbolic equations? Because they are false. Now: you may ask "why are they false?" This is a fairly deep question, and answers often involve discussion of propagation of singularities and characteristics. Quite a few chapters in Hörmander's Analysis of Linear Partial Differential Operators are devoted to this and similar questions. REPLY [10 votes]: For hyperbolic PDE's, we have two aspects that distinguish them from elliptic and parabolic PDE's and thus forbid one from being able to obtain estimates in the form you want: Usually the loss of derivatives is worse than in the elliptic case — in fact, even worse than in the more general subelliptic case. Typically, you lose a full (weak = Sobolev) derivative in a priori regularity estimates for linear hyperbolic PDE's: the so-called energy estimates. This loss is essentially sharp and can be tracked back to the fundamental theorem of Calculus. As pointed by Willie Wong in his answer, this can be refined if you take into account the characteristic set of the principal symbol of the operator, by means of techniques collectively called microlocal analysis. More precisely, worse than elliptic-type regularity must propagate along the bicharacteristic curves of the principal symbol, which are nonexistent in the elliptic case. Even if you account for the above loss of derivatives, the "optimal" relative shape of the regions $B_1$, $B_2$ is different due to a hallmark property of hyperbolic PDE's — namely, the finite speed of propagation of (supports of) solutions. Typically, $B_2$ will be a "cone-shaped" shadow region cast by $B_1$, whose precise form is once again dictated by the principal symbol of the operator.<|endoftext|> TITLE: Describing the Gamma-transform explicitly in terms of power series QUESTION [5 upvotes]: The Gamma transform of a measure is defined as follows. If $\alpha$ is a $\mathbf{Z}_p$-valued measure on $\mathbf{Z}_p$, then the Gamma transform of $\alpha$ is: $$\Gamma_{\alpha}(s) = \int_{\mathbf{Z}_p^{\times}} \langle x \rangle^s \, d\alpha(x). $$ So the Gamma transform takes as input a measure $\alpha$, and returns an analytic function of the variable $s$, which we call $\Gamma_{\alpha}(s)$. But we can also think of the Gamma transform in a different way: as taking as input a power series and returning as output a power series. Namely: as input, the Gamma transform takes in the power series $F_{\alpha}(T)$ corresponding to the measure $\alpha$ as output, the Gamma transform returns the power series $G$ such that $G((1+p)^s - 1) = \Gamma_{\alpha}(s)$. My question is: can one explicitly describe the Gamma transform as a map from power series to power series? That is, given a power series $f(T) = \sum a_nT^n$, is there an explicit formula for the power series expansion of $\Gamma_F(T)$ in terms of the power series expansion of $F(T)$? Here is my motivation for asking this. Washington has a very nice article, "On Sinnott's Proof of the Vanishing of the lwasawa Invariant $\mu_p$", where he gives a different proof of the Ferrero–Washington Theorem, inspired by a proof of Sinnott. On page 3, Washington does a few calculations with power series and says that "this is essentially the Gamma-transform". But I don't know what the Gamma transform looks like as a map from power series to power series, so I don't understand what that comment means. It seems to lie at the heart of the proof, so I want to ask this question to make sense of that step in the paper. REPLY [6 votes]: This is a hard problem (and one which is easily overlooked by the unwary)! Just to be clear, I'll summarize (how I think about) the problem: as a relation between additive and multiplicative Fourier transforms. We start with a measure $\mu$, i.e. a linear map $\mu$: (continuous functions on $\mathbf{Z}_p$) $\to \mathbf{Z}_p$. For any $t$ with $|t| < 1$, there is a unique character $\kappa_t$ of $\mathbf{Z}_p$ as an additive group which sends 1 to $1 + t$; and we have $\mu(\kappa_T) = F_\mu(T)$ for some power series $F_\mu(T) \in \mathbf{Z}_p[[T]]$: the additive Fourier transform of $\mu$. On the other hand, for each $u$ with $|u| < 1$, there is also a character $\chi_u$ of $\mathbf{Z}_p^\times$ as a multiplicative group, which is trivial on roots of unity and sends $1 + p$ to $1 + u$. This is given by $\mu(\chi_u) = G_\mu(u)$ for some power series $G_\mu \in \mathbf{Z}_p[[U]]$: the multiplicative Fourier transform of $\mu$. (I am deliberately using a different name for the variable here!) Your question is, then, how $F_\mu$ and $G_\mu$ are related. The answer is: "not in any straightforward way". For a given $t$, the functions $\chi_u$ and $\kappa_u$ are totally different as functions on $\mathbf{Z}_p$, so the values of $F_\mu$ and $G_\mu$ at a specific $u$ do not determine each other. On the other hand, if $\mu$ is supported in $1 + p\mathbf{Z}_p$ (so we lose nothing by restricting), then the two series $F_\mu$ and $G_\mu$ do determine each other: we get a bijection (the "Mellin transform") $$\mathfrak{M}: (1 + T) \cdot \mathbf{Z}_p[[(1+T)^p - 1]] \longleftrightarrow \mathbf{Z}_p[[U]]$$ which is a bijection of additive groups mapping $F_\mu(T)$ to $G_\mu(U)$ (but not respecting multiplication on either side). As an example, if $\mu$ is the Dirac measure sending a function to its value at $a$, for some $a \in 1 + p\mathbf{Z}_p$, then $F_\mu(T) = (1 + T)^a$, and $G_\mu(U) = (1 + U)^{log_p(a) / log_p(1 + p)}$. (This formula actually determines $\mathfrak{M}$ uniquely, since the Dirac measures are dense.) One can pull various information about the power series through this isomorphism -- for instance, the Iwasawa $\lambda$ and $\mu$-invariants of the two series are the same, and the Newton polygons coincide beyond a certain point (Sarah Zerbes and I proved this in an old paper of ours). But there is no simple formula that would allow you to read off the coefficients of $F_\mu$ from those of $G_\mu$ or vice versa.<|endoftext|> TITLE: A question regarding Cramér's proof on prime gaps under the Riemann Hypothesis QUESTION [5 upvotes]: Let $p_n$ be the $n$th prime. Assuming the Riemann hypothesis, Harald Cramér proves that $p_n-p_{n-1}\le C(\sqrt p_n \log p_n)$ for sufficiently large $n$. Is there a value known for the constant $C$ that works? In Cramer's original paper (On the order of magnitude of the difference of consecutive prime numbers, Acta Arithmetica 2 26-46) Cramér makes $C=5\lambda$, but I was unable to give an upper bound value to $\lambda$ from a first reading of the paper. Perhaps one of you knows? REPLY [2 votes]: I wanted to give some additional remarks to Will Sawin's answer and the associated comments. Carneiro, Milinovich, and Soundararajan certainly have the best result in literature (as far as I'm aware). They showed that $C=21/25$ for sufficiently large $n$, but only $C=22/25$ for $p_n>3$. It's still an open and interesting question (that I've been thinking about) to determine the lowest value of $C$ without assuming anything beyond the Riemann Hypothesis. Certainly, if one assumes Montgomery's pair correlation (or something stronger), then it's possible to take $C$ arbitrarily small as $n\to\infty$. See this paper by Goldston and Heath-Brown for instance.<|endoftext|> TITLE: Reinhardt's ultimate classes QUESTION [5 upvotes]: In the preface to Sets and Classes by Muller, several research programs are outlined that were in development concurrently with publication (or finished slightly beforehand) that he would have liked to discuss, but was unable to for various reasons. One such program is called ultimate classes, apparently an attempt to explore the outer limits of extensions of $ZF$ via axiomatic projection schemata by adding a new primitive concept to set theory (in addition to $\in$) corresponding to embeddings $\mathbb{V}\to\mathbb{V}$ of the universe into itself. Apparently, if consistent these schemata provide for the existence of extendible cardinals of high degree in addition to many other large cardinals. Where can I read about Reinhardt's work on this approach? Any relevant pointers are greatly appreciated. REPLY [5 votes]: You can find Reinhardt's philosophy of set theory in "Set existence principles of Shoenfield, Ackermann, and Powell", Fundamenta Mathematica, vol 84, pp 5-34 and "Remarks on reflection principles, large cardinals, and elementary embeddings", Proceedings of symposia in pure mathematics, vol 13, part 2, American Mathematical Society, Providence 1974, pp 189-205. Downloadable pdf in the Math Stack Exchange question, "How does Reinhardt's extension of the set-theoretical universe beyond $V_\Omega$ work?" Compare and contrast Reinhardt's philosophy of set theory with the philosophy of set theory implicit in Kunen's paper introducing the "Kunen Inconsistency". Happy Hunting!<|endoftext|> TITLE: Grothendieck's relative point of view and Yoneda lemma QUESTION [13 upvotes]: I asked this question on M.SE, but didn't get any answers. Occasionally I hear people saying that one of Grothendieck's big insights was that often when interested in an object $X$ it's better to study morphisms into that object, $-\to X$. Apparently that's called the relative point of view. First question. How is that principle applied in practice? What are some concrete examples in mathematics where the relative point of view is useful? Wikipedia mentions the Riemann–Roch theorem and a similar MSE question mentions a theorem about coherent sheaves. Unfortunately, I don't know any algebraic geometry yet. Are there more down-to-earth applications of the relative point of view that an undergraduate can understand, say, in linear algebra, group theory, ring theory, Galois theory, or maybe even in basic category theory? What are (some of) the most important theorems that feature the relative point of view? I recently heard about the Yoneda lemma in category theory (I know the statement and can prove it). I know that it can be used to prove that two objects are isomorphic whenever they have the same universal property. In Awodey's category theory book, there's a concrete application of that: in categories with enough structure, $(A\times B)+(A\times C)\cong A\times (B+C)$. That proof is elegant, I agree. But it doesn't live up with the praise many people give to the Yoneda lemma, does it? Maybe a more concrete application in non-category theory would help me to get convinced of the contrary. For instance, I read on Wikipedia (and elsewhere) that Grothendieck used the Yoneda lemma in his famous book EGA (which a lot of people seem to talk about). (In fact, it seems this was another insight of him: that Yoneda is useful.) Second question. So what were Grothendieck's main applications of the Yoneda lemma in algebraic geometry? (In contrast to the first question, here it suffices for me to just know roughly what kind of statement he proved with the Yoneda lemma---rather than understanding it in detail, because I already know one application of the Yoneda lemma.) Third question. Is the second question related to the first one, i.e., is there a connection between the relative point of view and the Yoneda lemma? (At least the Wikipedia page linked above mentions the Yoneda lemma.) REPLY [11 votes]: There is a crucial aspect of the relative point of view that I think has not been completely covered here so far. The relative point of view does not just mean that we want to look at morphisms. It says we want to look at morphisms as analogues of objects. How does this analogy work? A space $X$ (for example a topological space, manifold, or scheme) has the same information as a map from $X$ to a point. So a map between two spaces is a generalization of a space, which we can think of as a family of spaces, the fibers, parameterized by the points of the target. Families of spaces were certainly something people considered before Grothendieck, but a big part of the philosophy is to think of every morphism as a family of this type, even ones where the fibers are very different from each other. One thing this means is that for every interesting property that a space can satisfy, there should be an analogous property that a morphism can satisfy. The relative point of view tells you to look for this, and to generalize theorems about the old property to theorems about the new. There's many examples of this in scheme theory, but a more elementary field where the same insight applies is point-set topology. For example, compact topological spaces are interesting. What could the generalization to morphisms be? It must be a property of maps $X \to Y$ of topological spaces which are satisfied when $Y$ is a point if and only if $X$ is compact. One can look at maps where the inverse image of every compact set is compact (which are at least sometimes called proper maps), or universally closed maps, i.e. maps where for every topological space $Z$, the image of every closed subset of $X \times Z$ inside $Y \times Z$ is also closed.<|endoftext|> TITLE: On fixed point sets of actions of compact Lie groups QUESTION [7 upvotes]: Let a compact Lie group $G$ act smoothly on a compact smooth manifold $M$. For any compact subgroup $H\subset G$ denote by $E^H$ the image in $M/G$ of the fixed point set of $H$ in $M$. Is it true that the family of all such subsets $\{E^H\}$ is finite when $H$ runs over all compact subsgroups of $G$? REPLY [11 votes]: The quotient $M/G$ carries a stratification by orbit type (see e.g. this MO question for references). More precisely, for any closed subgroup $F\subseteq G$ the stratum $(M/G)_{(F)}$ is the set of all orbits which are isomorphic to $G/F$. The set $E^H$ is the union of all strata such that $H$ is conjugate to a subgroup of $F$. Since $M$ is compact, the stratification is finite. So also only finitely many subsets of the form $E^H$ are possible.<|endoftext|> TITLE: A constructive proof of the theorem of the cube QUESTION [5 upvotes]: Do you know a constructive proof of the theorem of the cube ? More precisely, let $X$, $Y$, $Z$ be projective varieties (e.g., over an algebraically closed field $k$) with points $x$, $y$, $z$ respectively and let $D$ be a divisor on the direct product $V := X \times Y \times Z$. Assume that the restrictions of $D$ to $$ V_0 := \{x\} \times Y \times Z, \qquad V_1 := X \times \{y\} \times Z, \qquad V_2 := X \times Y \times \{z\} $$ are the principal divisors of some functions $f_i \in k(V_i)$. In other words, $D \cdot V_i = \mathrm{div}(f_i)$. The theorem of the cube claims that $D$ is principal. How to construct explicitly a function $f \in k(V)$ such that $D = \mathrm{div}(f)$, given the functions $f_i$? I am mainly interested in the case $X = Y = Z$ is an elliptic curve and $x = y = z$ is the zero point on it. REPLY [2 votes]: This is not really an answer, but a rephrasing together with some comments on why this is difficult. In summary, the question reduces to the case where the three restrictions of $D$ are trivial as divisors (as opposed to merely principal), i.e. the $f_i$ are all $1$. It's actually convenient for me to change the notation a little: Notation. We assume throughout that $(X,x)$, $(Y,y)$, and $(Z,z)$ are pointed projective varieties over a field $k$ (for me, variety means geometrically integral). It will occasionally be convenient to denote these $(V_1,v_1)$, $(V_2,v_2)$, and $(V_3,v_3)$ respectively. Denote by $V$ the product $X \times Y \times Z$, and for $I \subseteq \{1,2,3\}$ write $$V_I = \prod_{i \in I} V_i.$$ Let $\pi_I \colon V \to V_I$ be the projection, for which we use shorthands like $\pi_{1,2} = \pi_{\{1,2\}} \colon V \to X \times Y$. It has a section $s_I \colon V_I \to V$ given by the points $v_j$ for $j \not\in I$, whose image we also denote $V_I \subseteq V$ by abuse of notation (the OP calls this $V_{I^{\text{c}}} \subseteq V$). For a subscheme $W \subseteq P$, denote by $W_I \subseteq V_I$ the intersection $W \cdot V_I$ (there are problems with intersection products on the level of divisors; see the difficulties below). Firstly, note that the theorem of the cube is equivalent to the following: Lemma. Let $D$ be a Cartier divisor on $V$. Then the divisor \begin{align*} \bar D :=& \ D - \pi_{1,2}^*D_{1,2} - \pi_{1,3}^*D_{1,3} - \pi_{2,3}^*D_{2,3} + \pi_1^*D_1 + \pi_2^*D_2 + \pi_3^*D_3 \\ =& \sum_{I \subseteq \{1,2,3\}} (-1)^{\lvert I \rvert+1} \pi_I^*D_I \end{align*} on $V$ is principal. Note that we omitted $\pi_{\varnothing}^*D_{\varnothing}$ since it is trivial (it's pulled back from $\operatorname{Spec} k$). Proof. It suffices to show this when $D$ is effective. For any $I \subseteq \{1,2\}$, the restrictions of $\pi_I^*D_I$ and $\pi_{I \cup \{3\}}D_{I \cup \{3\}}$ to $V_{1,2}$ agree, so we see that $\bar D \cdot V_{1,2}$ is trivial (as divisor!), and likewise for $\bar D \cdot V_{1,3}$ and $\bar D \cdot V_{2,3}$. The theorem of the cube says that $\bar D$ is principal. Conversely, if this lemma holds, applying it to the case where $D_I \sim 0$ on $V_I$ for any $I \subsetneq \{1,2,3\}$ gives $D \sim \bar D \sim 0$, proving the theorem of the cube. $\square$ The situation in the lemma shows what can happen: if we're presented with $D = \bar E$ for some divisor $E$, then the restrictions to $V_I$ for $\lvert I \rvert = 2$ are trivial as divisors. Thus, the $f_I \in k(V_I)$ witnessing the rational triviality of $D_I$ can be taken $1$, yet on $V$ there is something nontrivial happening. In fact, since $\operatorname{Div}(P)$ is a free abelian group, the lemma shows that there exists some homomorphism $$\phi \colon \operatorname{Div}(X \times Y \times Z) \to k(X\times Y \times Z)^\times$$ such that $(\phi(D)) = \bar D$ for all $D \in \operatorname{Div}(X \times Y \times Z)$. The task at hand is to "compute" this map: if we can find an explicit $f$ such that $(f) = \bar D$, then the assumption $D_I = (f_I)$ for all $I \subsetneq \{1,2,3\}$ gives $$D = \bar D + \sum_{U \subsetneq \{1,2,3\}} (-1)^I \pi_I^*D_I = (f) + \sum_{U \subsetneq \{1,2,3\}} (-1)^{\lvert I \rvert} (\pi_I^*f_I),$$ explicitly exhibiting $D$ as a principal divisor. Now we come to the difficulties. Difficulty 1. The first issue is that $\operatorname{Div}$ has fewer functoriality properties than $\operatorname{Pic}$. Notably, the intersection product $D \mapsto D \cdot V_I$ is not defined on $\operatorname{Div}$ in general: what to do if $D$ contains $V_I$? This problem goes away when $X$, $Y$, and $Z$ are curves: for $\lvert I \rvert = 2$, we get $D \subseteq V_I$ if and only if $D = V_I$. But since the normal bundle of a fibre is trivial, we can simply set $V_I \cdot V_I = 0 \in \operatorname{Div}(V_I)$. This also works for principal divisors: the specialisation $\operatorname{Prin}(V) \to \operatorname{Prin}(V_I)$ is then defined using the local ring $\mathcal O_{V,V_I}$ at the generic point of $V_I$. This is a DVR, so $k(V)^\times \cong \mathcal O_{V,V_I}^\times \times \mathbf Z$. The specialisation $k(V)^\times \to k(V_I)^\times$ is defined on $\mathcal O_{V,V_I}^\times$ by reduction modulo the maximal ideal, and identically zero on the factor $\mathbf Z$. Note also that neither $\operatorname{Div}$ nor $\operatorname{Pic}$ have a pushforward along proper maps, although coherent sheaves in general do (these play a role in some proofs of the theorem of the cube ― one then shows that the pushforward is again locally free). Difficulty 2. Even when $X=Y=Z=E$, it's not clear at all what the function $\phi$ above is. I don't think this is something that you can write down, for one because the function field of a product of elliptic curves is a kind of unwieldy thing to work with. The theorem of the cube predates the six functor formalism, so the original proof is not the one you find in Mumford. I personally like the proof using Picard schemes: triviality along $X \times Y \times z$ gives a map $X \times Y \to \mathbf{Pic}^0_Z$ that is trivial on $X \times y$ and $x \times Y$, hence trivial by the rigidity lemma. But this does not lift to $X \times Y \to \mathbf{Div}_Z$ because the scheme of relative effective Cartier divisors has an additional flatness hypothesis. You might be able to work around this, but it gets messy. Maybe the most suited language is the one in Weil's original proof, which you can read for instance in Lang's Abelian varieties, §III.2. But even that uses maps to Jacobians and is ultimately inexplicit. Notably, Weil uses the divisor language (as opposed to line bundles), so the fact that the old proofs are inexplicit suggests that it might be really hard. Here's a first question: Question. Can you do this for "simple" examples? For instance, if $E$ is an elliptic curve and $D = \{(a,b,c) \in E^3\ |\ a+b+c=0\}$, then do you know how to find $\phi(D)$? This will tell you why \begin{align*} \bar D =\ &\{a+b+c=0\} - \{a+b=0\} - \{a+c=0\} - \{b+c=0\} \\ &+ \{a=0\} + \{b=0\} + \{c=0\} \end{align*} is rationally trivial on $E \times E \times E$.<|endoftext|> TITLE: Finite index subgroup of $\mathbb{Z}^2$ that is invariant under a non-singular matrix QUESTION [5 upvotes]: Let $M $ be a matrix in $ \operatorname{GL}(2, \mathbb{Z})$ that has at least one eigenvalue of absolute value strictly bigger than $1$. What are the finite index subgroups $H$ of $\mathbb{Z}^2$ such that for all $v \in H$, we have $M(v) \in H$? Question: One example of such subgroup has the form $H=\langle (a,0),(0,a)\rangle$, for a non-zero integer $a$, this example works for any matrix in $ \operatorname{GL}(2, \mathbb{Z})$. I was wondering if there are any other such subgroups. Thought so far: I think the finite index subgroups of $\mathbb{Z}^2$ have the form $H = \langle (a,b),(c,d)\rangle$, where $a,b,c,d \in \mathbb{Z}$ such that $ad-bc\neq 0$. Hence, in order for $H$ to be a subgroup, we need $M(a,b)= \alpha_1(a,b) + \alpha_2(c,d)$ and $M(c,d)= \beta_1(a,b) + \beta_2(c,d)$ for some integers $\alpha_i$ and $\beta_i$, but I can't see any obvious solutions for these equations. I was also thinking maybe we can use the eigenvectors, but unfortunately the eigenvectors might not have integers entries. Any ideas for constructing such a subgroup would be really appreciated. REPLY [2 votes]: To illustrate how such a question is of arithmetic nature (and what a complete answer should look like), here is a partial answer in a specific case. Namely: I specify to $M=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$ and I address the question: what are invariant finite index subgroups of prime index. The characteristic polynomial of this matrix is $X^2-3X+1$, which has discriminant 5. By the quadratic reciprocity formula, for an odd prime $p\neq 5$, 5 is a square mod $p$ iff $p=\pm 1$ mod $5$ (i.e. the decimal expansion of of $p$ terminates with $1$ or $9$). And this polynomial is irreducible mod $2$, and has a double root mod $5$. Hence, for a prime $p$: for $p=\pm 2$ mod $5$, there is no $M$-invariant subgroup of index $p$ in $\mathbf{Z}^2$; for $p=\pm 1$ mod $5$, there are exactly two $M$-invariant subgroups of index $p$ in $\mathbf{Z}^2$; there is a single $M$-invariant subgroup of index $5$ in $\mathbf{Z}^2$. Remarks: For a general matrix $M$, the characteristic polynomial is $X^2+nX\pm 1$ for some $n$ and there should be a similar discussion. For a general index $q$, one should boil down to when $q$ is a power of a prime $p$. Then in turn one should boil down to when the quotient is cyclic (the quotient being isomorphic to $C_{p^a}\times C_{p^b}$ for some $a\le b$, the subgroup is contained in $p^a\mathbf{Z}^2$ and we can then "replace" $\mathbf{Z}^2$ with $p^a\mathbf{Z}^2$ to assume $a=1$).<|endoftext|> TITLE: On permanent of a square of a doubly stochastic matrix QUESTION [7 upvotes]: Let $A = (a_{i,j})$ be a double stochastic matrix with positive entries. That is, all entries are positive real numbers, and each row and column sums to one. A permanent of a matrix $A = (a_{i,j})$ is defined as $$ \text{perm}(A) = \sum_{\pi \in S_n}\prod_{i=1}^{n}a_{i,\pi(i)}. $$ Is it true that $\text{perm}(A^2) \leq \text{perm}(A)$ with the equality achieved only by $P_n = (1/n)$? More generally, is it true that $\text{per}(A^n) \leq \text{per}(A^m)$ for $n > m$ with the equality achieved only by $P_n = (1/n)$? Van der Waerden's conjecture states that the minimum permanent among doubly stochastic matrices is achieved precisely by the matrix $P_n = (1/n)$. Since $P_n$ is the only doubly stochastic matrix with positive entries and $P_n^2 = P_n$, the conjecture above implies Van der Waerden's conjecture for matrices with positive entries, so I don't assume it's easy to prove it if it's true. I wrote a simple code (well, I mostly got it from here) in Wolfram Mathematica to test the conjecture on random doubly stochastic matrices but couldn't find any counterexamples. n := 5 While[Min[ dsm := FixedPoint[ Standardize[Transpose[Standardize[#, 0 &, Total]], 0 &, Total] &, RandomReal[1, {n, n}], SameTest -> (Norm[#1 - Transpose[#2], "Frobenius"] < 1.*^-12 &)]] < 0.1] Do[ mat = dsm; If[Permanent[mat] < Permanent[mat.mat], Print[mat]]; , {i, 1, 10000}] The intuition behind the conjecture is from the two-sided Markov chains. Since $A$ is doubly stochastic, one can run two Markov chains on both sides. The permanent can be interpreted as some statistics on collisions between particles that start at certain positions. Since the Markov chain, after many steps, "wash out" original biases, the number of collisions should get smaller and achieve the minimum on the limit, which is $P_n = (1/n)$ for a doubly stochastic matrix. Update: Relevant partial result: Sinkhorn, Richard. "Doubly stochastic matrices whose squares decrease the permanent." Linear and Multilinear Algebra 4, no. 2 (1976): 153-158. Update: Joseph Van Name showed in the answer that the conjecture is wrong: the following doubly stochastic matrix is a counterexample. \begin{pmatrix} 15/32 & 1/64 & 1/2 & 1/64 \\ 1/64 & 1/64 & 15/32 & 1/2 \\ 1/2 & 15/32 & 1/64 & 1/64 \\ 1/64 & 1/2 & 1/64 & 15/32 \end{pmatrix} REPLY [5 votes]: A related question is Conjecture 17 in Minc's catalogue of open problems on permanents. He attributes it to Foregger. The conjecture is that for any positive integer $n$, there exists an integer $k=k(n)$ such that per$(A^k)\le$ per$(A)$ for all $n\times n$ doubly stochastic matrices $A$. Chang [On two permanental conjectures, Linear Multilinear Algebra 26 (1990), 207-213] made some very modest progress on this question, but as far as I know it is still wide open.