TITLE: Noncoprime polynomial values QUESTION [10 upvotes]: Let $p_1, \ldots, p_n$ be a finite sequence of nonconstant polynomials with integer coefficients. Does there exist a finite sequence of integers $x_1, \ldots, x_n$ such that the integers $p_1(x_1), \ldots, p_n(x_n)$ have a nontrivial common divisor? REPLY [8 votes]: Here is a bit more pedestrian argument. We may assume that $p_i$ are irreducible. Take $\alpha_i$ a root of $p_i$; then $\mathbb Q[\alpha_1,\dots,\alpha_n]=\mathbb Q[\xi]$ by the primitive element theorem. Taking $q$ to be the monic minimal polynomial of $\xi$, we have $\alpha_i=r_i(\xi)$, so $q(x)\mid p_i(r_i(x))$. Now we may take any prime divisor $p$ of the values of $q$ which does not appear in the denominators of the coefficients of $q$ and $r_i$; $p$ divides appropriate values of $p_i$ as well.<|endoftext|> TITLE: Algebraicity and non-algebraicity of leaves of the characteristic foliation QUESTION [5 upvotes]: Let $X$ be a compact complex manifold equipped with a holomorphic symplectic form $\omega$. Let $D$ be a smooth divisor on $X$. At each point of $D$, the restriction of $\omega$ to $D$ has one-dimensional kernel. This gives a non-singular foliation $F$ on $D$. Is it possible that some leaves of $F$ are algebraic while some are not? REPLY [6 votes]: Suppose $X$ is projective manifold endowed with holomorphic symplectic form. Let $D$ be smooth divisor on $X$. Characteristic foliation with algebraic and non-algebraic leaves. Suppose that $X$ is a product of two abelian varieties of dimension $n$ ($n\ge 2$) $A_1 \times A_2$ in such a way that the fibers of the natural projection $\pi: X \to A_2$ are Lagrangian. Let $D$ be the pull-back under $\pi$ of an ample divisor $E$ on $A_2$. The characteristic foliation on $D$ will be everywhere tangent to $\pi$ and over each fiber is a linear foliation. The slope of the foliation on the fiber over a a point $p \in E$ is determined by the tangent of $E$ at $p$. But since $E$ is an ample divisor, it has non-degenerate Gauss map and every possible linear foliation on $A_1$ will appear among the restriction of the characteristic foliation $D$ to fibers of $\pi$. If $A_1$ itself is a product of elliptic curves (or isogeneous to the product of an elliptic curve and an abelian variety of dimension $n-1$) then $A_1$ carries linear foliations with all leaves algebraic. This shows the existence of divisors with the requested property. Characteristic foliation with all leaves algebraic. If every leaf of the characteristic foliation is algebraic then it was proved by Amerik and Campana (refining previous result by Hwang and Viehweg) that either every leaf is rational or $X$ is, up to étale coverings, the product of a symplectic surface $S$ with a symplectic manifold $Y$ and $D$ is the product of a curve $C\subset S$ with $Y$. In the latter case the leaves of the characteristic foliation are the fibers of the projection to $Y$. In particular, the Kodaira dimension of $D$ is at most $1$. Characteristic foliation on ample divisors. One interesting problem on the subject (already raised by Hwang and Viehweg) is whether or not the characteristic foliation on an ample divisor can have one algebraic leaf. To the best of my knowledge this problem is wide open.<|endoftext|> TITLE: Are two equivariant maps between aspherical topological spaces homotopic? QUESTION [5 upvotes]: Let $f: X \rightarrow Y$ be continuous, X,Y pathwise connected and aspherical (i.e. trivial higher homotopy groups). Then $\pi_1(X)$ acts on the universal cover of $X$ via deck transformations, and on the universal cover of $Y$ via $$([\sigma],y)\mapsto [\sigma \circ f]y$$ If g is another continuous map whose lift to the universal covers is equivariant under these actions, is g homotopic to f? REPLY [10 votes]: In the language of equivariant homotopy theory, your question is as follows. You have a group homomorphism $\phi\colon G\to H$, which you use to make $EH$ into a $G$-space, and then you ask whether $[EG,EH]^G$ is a singleton. In fact, it is true more generally that $[EG,Y]^G$ is a singleton whenever $Y$ is a $G$-space that is nonequivariantly contractible. The reason is as follows. We can write $EG$ as the colimit of a sequence $X_k$, where $X_k$ is obtained from $X_{k-1}$ by attaching $G\times B^{n+1}$ along $G\times S^n$ for some $n$. Thus, to extend an equivariant map $X_{k-1}\to Y$ equivariantly over $X_k$ is the same as to extend a nonequivariant map $S^n\to Y$ nonequivariantly over $B^{n-1}$, and this can be done in an essentially unique way because $Y$ is assumed to be nonequivariantly contractible. (This kind of argument is called "obstruction theory".)<|endoftext|> TITLE: What is the time complexity of the matrix exponential? QUESTION [12 upvotes]: While trying to compute the Matrix Exponential of an $n \times n$ array I decided to take advantage of a Python function called scipy.linalg.expm(). According to the documentation, this function adopts the Padé's Approximant to perform a calculation that, otherwise, is said to require a lot of resources and is still a matter of discussion in the mathematics community (many of you might be aware of this). The documentation itself references a work by Al-Mohy and Higham (doi: 10.1137/04061101X), in which I was not able to find any indication regarding the Time Complexity (either T or Theta would work) of this procedure. Has anyone come across the same problem? What is your suggestion for determining said Time Complexity? REPLY [13 votes]: The exponential of a matrix is not, in general, computable explicitly. Strictly speaking, the Complexity of Matrix Exponential does not exist. Instead, you have an approximate calculation and you are asking about its complexity. Now, a few comments : there is not just one approximate exponential, but actually a sequence of approximations. Therefore, the complexity must involve the tolerance parameter. One cannot overlook the stability of the approximation. At least two ingredients are at stake. Either the diagonalisation of the matrix can be ill-conditionned (the matrix is far from normal). Or (even worse) its spectrum is moderately large. Say that $30$ and $-30$ are eigenvalues, then the magnitude of $\exp A$ is $e^{30}$, a large number. But $e^{-30}$ is also an (extremely small) eigenvalue, and the action of the approximate $\exp A$ on the corresponding eigenvector will be totally eroneous. More generally, a moderate condition number of $A$ causes huge errors in the modes associated with the smallest eigenvalues. These phenomena are well known by numerical analysts when dealing with ODEs. And they must choose a version of the Euler scheme adapted to the matrix itself.<|endoftext|> TITLE: Thorough reference on regular homotopy QUESTION [7 upvotes]: I would like to learn this topic of algebraic topology but I cannot find a relevant reference to answer my basic questions on the subject (for example, is there a Hurewicz theorem for regular homotopy groups? under which conditions can one homotop a non-zero class in a non-trivial higher homotopy group to an immersion?). REPLY [14 votes]: The Smale-Hirsch theorem says that the existence of an immersion in a given homotopy class is equivalent to the existence of certain bundle map, in turn equivalent to the existence of a section of a certain bundle. See Hirsch, Morris W. Immersions of manifolds; the Mathscinet review (MR0119214)by Kervaire is a good summary. Thus these questions are `reduced' to algebraic topology, in particular to obstruction theory. This is of course a great thing, but I doubt there's an answer to your question in any great generality, because obstructions are generally hard to compute. The best hope for a general answer would be in the so-called stable range where the dimension of the source is less than 2/3 the dimension of the target; see Haefliger, André; Hirsch, Morris W. Immersions in the stable range. Ann. of Math. (2) 75 1962 231–241.<|endoftext|> TITLE: Is the exponent $2$ sharp in the Balog-Szemerédi-Gowers Theorem? QUESTION [6 upvotes]: The Balog-Szemerédi-Gowers theorem can be stated in the following form: let $A,B$ be subsets of $\mathbb{Z}/n\mathbb{Z}$ (say) with equal cardinality, such that $$ \|1_A*1_B\|_2 \ge K^{-1} \|1_A\|_1 \|1_B\|_2. $$ Then there exist $A'\subset A$, $B'\subset B$ with $|A'|\ge K^{-C}|A|$, $|B'|\ge K^{-C}|B|$ such that $|A'+B'|\le K^C |B'|$. Here $C$ is an absolute constant. My questions is: what happens if the $2$-norm is replaced by a $q$-norm? If $q\in (1,2)$, then it follows from Hölder that $$ \|1_A*1_B\|_q \ge K^{-1} \|1_A\|_1 \|1_B\|_q $$ implies $$ \|1_A*1_B\|_2 \ge K^{-\frac{q}{2(q-1)}} \|1_A\|_1 \|1_B\|_2, $$ so the conclusion still holds, with a constant $C$ that depends on $q$ (blowing up as $q\downarrow 1$). What about $q>2$? More precisely: Is there any $q>2$ such that the conclusion of the B-S-G Theorem continues to hold if one assumes that $\|1_A*1_B\|_q \ge K^{-1} \|1_A\|_1 \|1_B\|_q $ (possibly with a constant $C$ that depends on $q$)? REPLY [2 votes]: Sorry to answer my own question, but I realized the answer is easy, and similar to the case $q\in (1,2)$: yes, the B-S-G theorem holds for any $q\in (2,+\infty)$: if $q>2$, $$ \|1_A*1_B\|_q^q \le \|1_A*1_B\|_2^2 \|1_A*1_B\|_\infty^{q-2} \le \|1_A*1_B\|_2^2 |A|^{q-2}, $$ so if $\|1_A*1_B\|_q \ge K^{-1} \|1_A\|_1 \|1_B\|_q$, it follows that $$ |A|^q |B| K^{-q} \le \|1_A*1_B\|_q^q \le \|1_A*1_B\|_2^2 |A|^{q-2}, $$ and then $$ \|1_A*1_B\|_2 \ge K^{-q/2} \|1_A\|_1 \|1_B\|_2, $$ and we can apply the Balog-Szemerédi-Gowers Theorem. This is simpler than I expected, but since the question got a few upvotes it might be useful to someone.<|endoftext|> TITLE: Does the rank (=height) of a well partial order bound its type (=length, =stature)? QUESTION [5 upvotes]: Terminology and context (This should all be standard, but is recalled because terminology sometimes varies, and also to put the question into perspective.) A partially ordered set is called well-founded iff it has no infinite decreasing sequence. It is called well-partially-ordered (=wpo) iff it is well-founded and also has no infinite antichain; equivalently, every linearization (=total order extending the given order) is a well-order; equivalently, the lattice its downsets (=downwards-closed subset, =initial segments), partially ordered by inclusion, is well-founded. The (well-founded) rank of a well-founded partially ordered set $P$ is the ordinal defined inductively by $\mathop{\mathrm{rk}}P = \sup\{\mathop{\mathrm{rk}}(x)+1 : x\in P\}$ where $\mathop{\mathrm{rk}}x = \sup\{\mathop{\mathrm{rk}}(y)+1 : y TITLE: On a parallelizable manifold, is there always a frame satisfying $[X_i,X_j]=0$? QUESTION [5 upvotes]: [This question was asked on MSE, but got no answers, I thought it could be more appropriate here] Let $M$ be a parallelizable manifold. Is there always a global frame $(X_i)$ such that $[X_i,X_j]=0$ for all $i,j$ ? If the answer is no, what kind of obstruction there is to find such a frame ? and what kind of general (topological ?) condition on $M$ makes it possible to find such a frame ? REPLY [10 votes]: Suppose that $M$ is compact, such a frame exists implies that the commutative group $R^n$ acts transitively on it, this implies that $M$ is a torus. But a compact Lie group is parallelizable, so that is not always possible. REPLY [4 votes]: On a compact manifold, you have a global frame such that $[X_i, X_j]=0$ if and only if your manifold is the torus. Starting from dimension 3, there are parallelisable manifolds different from the torus, possibly the simplest example is $S^3$. Indeed, the flows of the vector fields generate an action of $R^n$ on the manifold, the manifold is therefore $R^n$ quotient by the stabilisor, and the stabilisor is a discrete subgroup of $R^n$ which preserves the vector fields and is therefore a lattice. REPLY [3 votes]: Every Lie group is parallelizable, but a compact manifold with such a global frame is an abelian Lie group, so a torus. Hence lots of counterexamples, but I don't know the classification. For the noncompact, you can take out a Cantor set, or something worse, from Euclidean space or from a torus, so there is no classification possible. REPLY [3 votes]: If $M$ is connected and non-compact of dimension $m$, then parallelizability implies that $M$ can be immersed in $\mathbb R^m$, and this implies existence of such a framing.<|endoftext|> TITLE: When is a mapping the proximity operator of some convex function? QUESTION [9 upvotes]: Is there a characterization of mappings $p : \mathbb R^n \rightarrow \mathbb R^n$ which are proximity operators (in the sense of Moreau) of l.s.c (extended) real-valued functions ? That is, given $p : \mathbb R^n \rightarrow \mathbb R^n$, under what sufficient conditions does there exist an extended-valued l.s.c convex function $g:\mathbb R^n \rightarrow (-\infty, +\infty]$ such that $$p(x) \equiv \mathrm{prox}_g(x) := \underset{z \in \mathbb R^n}{\text{argmin }}\frac{1}{2}\|z-x\|_2^2 + g(z) \;?$$ N.B: Of course it's necessary that $p$ be firmly-nonexpansive, and have other classical properties of prox operators. Motivation: In regularization techniques for signal / image processing, one usually proposes to minimize an energy of the form $f(x) + g(x)$, where $x=x^*$ is the image to be recovered from noisy / corrupted measures, $f(x)$ is a data fidelity and measures the "fit" of the model, while $g(x)$ is a regularization term that imposes some structural constraints. For example, one can take $f(x) = \frac{1}{2}\|y-Ax\|_2^2$, under a additve Gaussian-noise assumption, where $y$ is the observed image and $A$ is a sensing linear operator, so that $y \approx Ax + \text{ noise}$, etc., etc. A brilliant idea that has been proposed in Social Sparsity! is to impose the penalty $g$ only implicitly, by instead constructing its proximal operator $p(x)$, i.e by stating the intended shrinkage action of $g$ on the model coefficients $x_j$. For a concrete example, think of a (fictional) world in which we didn't know about the $\ell_1$ norm, but instead decided to invent the Lasso by stating that the prox of the (unknown) $\ell_1$ penalty should shrink the coefficients according to the soft-thresholder $$(p(x))_j = st_{\lambda}(x_j) = sign(x_j)(|x_j| - \lambda)_+,$$ where $\lambda > 0$ is a regularization parameter and $sign(x_j): = -sign(-x_j) = 1$ if $ x_j > 0$ and $0$ else. Note that the above prox would correspond to a penalty $g(x) = \lambda \|x\|_1$, and acts component-wise only because we're assuming (in this example) a separable penalty. The question is then: How to show that $st_{\lambda}$ actually corresponds to the proximal operator of some penalty function. REPLY [4 votes]: You may be interested in the characterization of proximity operators (of convex or nonconvex functions) provided in the paper https://hal.inria.fr/hal-01835101 In fact, the Moreau paper (Corollary 10.c) shows that $f: \mathcal{H} \to \mathcal{H}$ is the proximal map associated to some convex lsc $\varphi: \mathcal{H} \to \mathbb{R}$ if, and only if, the following two conditions hold: $f$ is the subdifferential of some convex lsc function $\psi: \mathcal{H} \to \mathbb{R} \cup \{+\infty\}$; $f$ is non-expansive Here $\mathcal{H}$ is any Hilbert space, finite or infinite dimensional. To answer your question, one also needs to consider the case where $\varphi$ may be nonconvex. In https://hal.inria.fr/hal-01835101 it is shown (Theorem 1) that: $f: \mathcal{Y} \subset \mathcal{H} \to \mathcal{H}$ is the proximal map associated to some (possibly nonconvex) $\varphi: \mathcal{H} \to \mathbb{R}$ if, and only if, $f$ is the subdifferential of some convex lsc function $\psi : \mathcal{H} \to \mathbb{R} \cup \{+\infty\}$. In practice, it may not always look straightforward to check that $f$ is a subdifferential (this is related to notions such as cyclic monotonicity). Luckily this is easier when $f$ is $C^1$ (Theorem 2): when $f$ is $C^1$, it is the proximity operator of some $\varphi$ if, and only if, its differential $Df(y)$ is symmetric positive semi-definite. The case of social sparsity shrinkage operators is precisely considered in Section 1.4: using Theorem 2, it is shown (Corollary 5) that a family of social shrinkage operators are not proximity operators, except in classical cases where they match group-sparsity shrinkage with disjoint groups. As far as I know, Moreau fully characterizes proximity operators of convex $\varphi$ are fully characterized by Moreau -a function f<|endoftext|> TITLE: Minimum negative eigenvalue of zero-one matrices QUESTION [6 upvotes]: The following question must have been answered decades ago. For $n$ fixed, what is the most negative eigenvalue among all trace zero zero-one matrices (that is, all entries are either zero or one, and all diagonal entries are zero) of size $n$? By most negative eigenvalue, I mean the least (that is, biggest absolute value) among the negative eigenvalues (if any). There is an obvious candidate. Let $A_n$ be the $n\times n$ matrix whose entries are $1$ in the $(i,j)$ position whenever $i+j$ is odd, and zero otherwise. This is rank two, trace zero, and it is easy to see that its two nonzero eigenvalues are $\{ \pm n/2\}$ if $n $ is even, and $\{ \pm\sqrt{n^2-1}/2\}$ if $n$ is odd. Not only do I expect that the most negative value is $-n/2$ if $n$ is even and $-\sqrt{n^2-1}/2$ if $n$ is odd, but I also expect that if $B$ is a trace zero zero-one $n \times n$ matrix whose most negative eigenvalue achieves this bound, then $B$ is permutation conjugate to $A_n$. This is trivial for $n=2$, easy for $n=3$, and requires a Cayley-like$^*$ argument for $n=4$, which I actually didn't complete (except in the symmetric case). When $n$ is even, $A_n$ is the adjacency matrix of the complete bipartite graph with $n$ vertices, so I expect this type of question (most negative eigenvalue of a symmetric zero-one matrix) has been answered for symmetric matrices (which of course correspond to undirected graphs). Anyway, I am looking for a reference for this problem (and if my expectations are false, I'd like to know that too). $*$ Cayley-like means substituting for each variable, and in this case, finding the smallest root—but not directly, but by finding the smallest $\alpha >0$ to arrange that the determinant of $I + \alpha B$ is zero. For $n=4$, this means minimizing a fairly simple polynomial in twelve variables, many of which turn out to be zero. This refers to Cayley's delightfully naive proof of what we call the Cayley-Hamilton theorem. Edit To summarize—after hectic soul-searching—the conjectured result (complete with the identification of the optimal choices) is true (even without the trace zero hypothesis), with an elegant argument given in Christian's answer, based on Brendan's now-deleted answer. A rather surprising consequence (at least to me) is given near the bottom of my answer. REPLY [2 votes]: I wish to repost Brendan's answer temporarily, which I believe clarified matters completely and in an easy and elegant way, until Brendan undeletes his own version or someone explains to me what I'm missing here. Let $A$ be an arbitrary $01$ matrix, with smallest (real) eigenvalue $\lambda_-(A)$ and associated normalized eigenvector $v\in\mathbb R^n$. Let's also assume that $v_j\le 0$ for $j=1,\ldots , k$ and $v_j>0$ for $j>k$. Now to minimize $\langle v, Bv\rangle$ for this $v$ over all $01$ matrices $B$, we must make $(Bv)_j$ as large as possible for $1\le j\le k$ and as small as possible for $j>k$. Clearly, this happens for $$ B= \begin{pmatrix} 0_{kk} & 1\\1& 0_{n-k,n-k}\end{pmatrix} \quad\quad\quad\quad (1) $$ Luckily for us, $B$ happens to be symmetric, so $$ \lambda_-(A) = \langle v, Av\rangle \ge \langle v, Bv\rangle \ge \lambda_-(B) . $$ In other words, it suffices to restrict our search to matrices of the form (1). We observe that $(1,0)^t$, $(0,1)^t$ are both eigenvectors of $B^2$ with eigenvalue $k(n-k)$. Since $B$ has rank $2$ and trace zero, this implies that the eigenvalues of $B$ are $\pm \sqrt{k(n-k)}$. Finally, we maximize this over $k$ to obtain the claim.<|endoftext|> TITLE: Does every dimension $n\geq4$ admit a manifold with an exotic smooth structure? QUESTION [12 upvotes]: It is known that $\mathbb{R}^4$ has exotic smooth structures, and there are many such examples in higher dimensions, such as the famous 7-sphere. My (probably very naive) question is, for every $n\geq4$, does there exist an $n$-manifold with exotic smooth structures? In other words, for every $n\geq4$, does there exist topological $n$-manifolds which admit more than one diffeomorphism class of smooth structures? REPLY [17 votes]: Yes. For every $n\ge 5$ there are exotic tori. In fact, the PL-structures on $T^n$ are in one-to-one correspondence with $H^3(T^n;\mathbb{Z}/2)$, and every one of these is smoothable (Reference: "Surgery on Compact Manifolds" by C. T. C. Wall, Chapter 15A). Since any smooth manifold admits a unique PL-structure up to PL-isomorphism, it follows that there are many manifolds homeomorphic but not diffeomorphic to the standard torus.<|endoftext|> TITLE: Does the Nash inequality hold on manifolds with Lipschitz boundary? QUESTION [5 upvotes]: Let $N$ be a smooth manifold without boundary of dimension $n$. $M$ is a manifold with Lipschitz boundary if $M \subset N$, $M$ and $N$ are of the same dimension, and in the charts of $N$, the boundary of $M$ is Lipschitz. The Nash inequality (sometimes also called $L^2$-type Nash inequality) reads $$ ||f||_2^{2+\frac{4}{n}} \le (C_1||f||_2^2+C_2||\nabla f ||_2^2)||f||_1^{\frac{4}{n}}$$ It is closely related to the Gagliardo-Nirenberg-Sobolev inequality $$||f||_p^2 \le C_1||f||_2^2+C_2||\nabla f||_2^2$$ where $\frac{1}{p}+\frac{1}{n}=\frac{1}{2}$. The Sobolev inequality only makes sense if the dimension is at least $3$ and it holds for manifolds with Lipschitz boundary, see for example Aubin's book 'Nonlinear analysis on manifolds'. If the dimension is at least $3$, the Nash and Sobolev inequality are equivalent to each other, see for example Carlsen, Kusuoka, Stroock 'Upper bounds for symmetric Markov transition functions'. This means the only remaining case is dimension $2$. The Nash inequality holds for smooth manifolds without boundary, see for example Hebey 'Sobolev spaces on Riemmanian manifolds'. Some papers like Humbert 'Best constants on the $L^2$-type Nash inequality' say that the Nash inequality also holds for manifolds with boundary, but I assume this means smooth boundary and the proof just says something like 'analogous to the Sobolev inequality'. His theorem also requires an injectivity radius bounded from below, which using the classic definition, does not make sense on manifolds with boundary (I had a question on injectivity radius on manifolds with boundary here recently, the upshot was that this is a nontrivial issue with multiple distinct reasonable definitions). Edit: I forgot to mention that the manifolds are compact. Most of the theorems quoted also hold for noncompact manifolds with some suitable global curvature bounds but the compact case is all I'm interested in. REPLY [2 votes]: I think the answer to my question is yes, proved using a version of Stein's extension theorem. Theorem: Let $M$ be a manifold with Lipschitz boundary sitting inside $N$. Then there exists an extension operator E from functions on $M$ to functions on $N$ such that a) $Ef|_{M}=f$, that is $E$ is an extension b) $E: W^p_k(M) \rightarrow W^p_k(N)$ continuously for all $1 \le p \le \infty$ and all non-negative integral $k$ where $W^p_k$ is the Sobolev space of weak derivatives up to order $k$ lying in $L^p$ This is proved in Stein's book 'singular integrals and differentiability' for domains sitting in $\mathbb{R}^n$, the proof already uses charts, carrying it over to manifolds is fairly straightforward. Using this theorem and the fact that the Nash inequality holds on the $N$, one can prove the Nash inequality on $M$ with a simple chain of inequalities. \begin{align*} &||f||_2^{2+\frac{4}{n}} \\ \le &||Ef||_2^{2+\frac{4}{n}} \\ \le & (C_1||Ef||_2^2+C_2||\nabla Ef ||_2^2)||Ef||_1^{\frac{4}{n}} \\ \le & \max(C_1,C_2) ||Ef||_{W_1^2}^2 ||Ef||_1^{\frac{4}{n}} \\ \le & \max(C_1,C_2)C^2 ||f||_{W_1^2}^2 ||f||_1^{\frac{4}{n}} \\ \le &(\max(C_1,C_2)C^2 ||f||_2^2+\max(C_1,C_2)C^2 ||\nabla f ||_2^2)||f||_1^{\frac{4}{n}} \end{align*}<|endoftext|> TITLE: Integral cohomology of $G/N(T)$ QUESTION [14 upvotes]: Let $G$ be a compact connected simple Lie group, $T$ a maximal torus, $N(T)$ the normalizer of $T$, and $W=N(T)/T$ the Weyl group. It is well-known that $H^*(G/T,\mathbb{Q})$ is the regular representation of $W$ induced by the $W$-action on $G/T$, and hence $H^*(G/N(T), \mathbb{Q})\cong H^*(G/T, \mathbb{Q})^W\cong\mathbb{Q}$. I would like to know whether the integral cohomology $H^*(G/N(T), \mathbb{Z})$ has been known before. I am particularly interested in the case $G=SU(n)$. REPLY [3 votes]: Here are some comments about how this calculation can be organised. First, let $T(n)$ be the standard maximal torus in $U(n)$, and let $N(n)$ be its normaliser, so $W(n)=N(n)/T(n)$ is the symmetric group. Put $F(n)=U(n)/T(n)$ and $E(n)=F(n)/W(n)=U(n)/N(n)$, so we are interested in the (co)homology of $E(n)$. We know that $H^*(E(n);\mathbb{Q})=\mathbb{Q}$. I think it will be best to study $H_*(E(n);\mathbb{Z}/p)$ (where $p$ is prime) and recover information about $H^*(E(n);\mathbb{Z})$ at the end, using the universal coefficient theorems. From now on I'll write $H_*(X)$ for $H_*(X;\mathbb{Z}/p)$. It will also be best to study all the groups $H_*(E(n))$ together as a bigraded object $H_*(E(*))$. Note that there are sum and tensor product operations $\oplus\colon U(n)\times U(m)\to U(n+m)$ and $\otimes\colon U(n)\times U(m)\to U(nm)$. These are group homomorphisms, so they induce maps $BU(n)\times BU(m)\to BU(n+m)$ and $BU(n)\times BU(m)\to BU(nm)$. They preserve monomial matrices, so they give maps $BN(n)\times BN(m)\to BN(n+m)$ and $BN(n)\times BN(m)\to BN(nm)$. We can regard $E(n)$ as the fibre of the map $BN(n)\to BU(n)$, so we also get maps $E(n)\times E(m)\to E(n+m)$ and $E(n)\times E(m)\to E(nm)$. This will give two different products on the object $H_*(E(*))$, which will make it into a Hopf ring. Also, using block matrices we get a standard inclusion $\Sigma_n\wr U(m)^n\to U(nm)$. This gives an action of an $E_\infty$ operad on $BU(*)$ and $BN(*)$ and $E(*)$, and so gives Dyer-Lashof operations on the corresponding homology groups. Now $BN(*)$ is essentially the free $E_\infty$ space generated by $BN(1)=\mathbb{C}P^\infty$, which implies that $H_*(BN(*))$ is the free commutative algebra with Dyer-Lashof operations generated by $H_*(\mathbb{C}P^\infty)$. On the other hand, $H_*(BU(*))$ is just the free commutative algebra generated by $H_*(\mathbb{C}P^\infty)$. This implies that the map $H_*(BN(*))\to H_*(BU(*))$ is surjective (which can also be seen using the Becker-Gottlieb transfer map). Note also that the Dyer-Lashof operations in $H_*(BU(*))$ were worked out by Priddy. It seems likely that $H_*(E(*))$ is just the kernel (in the Hopf algebra sense) of the map $H_*(BN(*))\to H_*(BU(*))$. Some care is needed here because the relevant Hopf algebras do not have antipodes; they correspond to commutative monoids (rather than groups) in the category of coalgebras. Anyway, the message is that we expect that $H_*(E(*))$ will have structure closely related to that of the Dyer-Lashof algebra, and of similar size and complexity. So far we have not used the tensor product maps $U(n)\times U(m)\to U(nm)$, or equivalently, the second product in the Hopf ring structure. It is a theorem of Paul Turner that when $p=2$, the object $H_*(BW(*))$ has a simple presentation as a Hopf ring, which is in some ways more tractable than the presentation using Dyer-Lashof operations. I expect that there is a similar story for $H_*(BN(*))$, although I do not know whether that has ever been written down. It is also straightforward to write down the Hopf ring structure on $H_*(BU(*))$, and this may well give a nice description of $H_*(E(*))$. I am not sure whether a similar approach would work for $p>2$.<|endoftext|> TITLE: Group cohomology of fundamental group of a curve QUESTION [8 upvotes]: The question should be an elementary result in the theory of etale cohomology, but I failed to understand it because I am a complete beginner of the theory. So, I should apologise in advance for this rudimenatry question. Let $C$ be a connected hyperbolic curve defined over an algebraically closed field of characteristic 0. Let us write $\pi_1$ for the maximal pro-$p$ quotient of the etale fundamental group of $C$ with some choice of base-point. There is a statement which claims that, for any integers $i,r$, there exists a natural morphism $H^i(\pi_1,\mathbb{Z}_p(r))\to H^i(C,\mathbb{Z}_p(r))$ which in fact an isomoprhism. My questions are: How is the natural map defined? Why is it an isomorphism? Surely any comment would helpful, but if anyone can explain it with details, that would be greatly helpful. REPLY [2 votes]: The best thing you could do in my opinion is to have a look at Appendix A Algebraic $K(\pi, 1)$ Spaces in Stix, Jakob Projective anabelian curves in positive characteristic and descent theory for log-étale covers. Dissertation, Rheinische Friedrich-Wilhelms-Universität Bonn, Bonn, 2002. http://www.ams.org/mathscinet-getitem?mr=2012864 http://www.math.uni-frankfurt.de/~stix/research/preprints/STIXdissB.pdf where this issue (for the whole fundamental group and for the pro-$\ell$ quotient) is treated with great care.<|endoftext|> TITLE: Unexpected regularity of the distance from a $C^2$ submanifold QUESTION [12 upvotes]: Let $\Gamma$ be a $C^2$ compact submanifold of $\mathbb{R}^n$. Consider the distance function $\delta$ from $\Gamma$. It is well known that, for sufficiently small $\varepsilon>0$, $\delta$ is $C^2$ on $\{ 0<\delta < \varepsilon\}$, and that it satisfies the eikonal equation $$ \| \nabla \delta \| = 1, \qquad \text{with} \qquad \delta|_{\Gamma} = 0. $$ Now recall Bochner's formula, valid for a smooth function $u \in C^\infty(M)$ on a general Riemannian manifold. It reads: $$ \frac{1}{2} \Delta\left( \| \nabla u\|^2 \right) = \nabla u \cdot \nabla \Delta u + \| \mathrm{Hess} (u) \|^2_{\mathrm{HS}} + \mathrm{Ric}(\nabla u, \nabla u). \qquad (\star)$$ Here $\| \cdot \|_{\mathrm{HS}}$ is the Hilbert-Schmidt norm and $\Delta$ is the Laplace-Beltrami operator. When we specify $(\star)$ to $M = \mathbb{R}^n$ and $u$ is a smooth solution of the eikonal equation, we have the following: $$ \nabla u \cdot \nabla \Delta u = - \| \mathrm{Hess} (u) \|^2_{\mathrm{HS}}. \qquad (\star\star)$$ Observe that the r.h.s. requires only $C^2$ regularity of $u$, while the l.h.s., a priori, requires a further derivative ($\nabla u \cdot \nabla \Delta u$ is indeed the directional derivative of $\Delta u$ in the direction $\nabla u$). Q: Is it true, even if $\delta$ is a priori only a $C^2$ solution of the eikonal equation, that $\Delta \delta$ admits a directional derivative in the direction of $\nabla \delta$? Or, in other words, is it true that $(\star\star)$ holds for $\delta$? The same question indeed can be posed in the Riemannian setting. P.S: A direct computation with the distance from the $C^2$ (but not $C^3$) surface in $\mathbb{R}^2$ given by $y= x^{5/2}$ seems to support this claim. This could be a standard fact about the regularity of solutions of the eikonal equation, but I have not been able to find any reference on this precise point. REPLY [3 votes]: Since this question seems to have attracted some interest, I will post my own answer (which is a proof of the statement in the comments by Anton). If $u : M \to \mathbb{R}$ is a $C^2$ solution of the Eikonal equation $$ \| \nabla u\| = 1, $$ then $\mathrm{Hess}(u)$, which a priori is only continuous, is smooth along integral lines of $\nabla u$ (and analytic, in the Euclidean case). The proof boils down to show that the Riccati equation for $\mathrm{Hess}(u)$ holds true in the distributional sense (any "simpler" proof is welcome). Assume $M = \mathbb{R}^n$. The eikonal equation is (sum over repeated indices is assumed) $$(\partial_i u) (\partial_i u) = 1.$$ Its first derivative gives $$(\partial_{ij} u) (\partial_i u) = 0.$$ We can indeed differentiate again, but not use Leibnitz since $\partial_{ij} u$ might not be differentiable. Nevertheless, as distributions we obtain that $$\partial_k (\partial_i \partial_j u) (\partial_i u)= - (\partial_{k\ell} u)( \partial_{\ell j} u) .$$ (To be precise one must consider $\partial_{i}\partial_j u \in C_c^0(\mathbb{R}^n)^*$ so that its distributional derivative is in $C_c^1(\mathbb{R}^n)^*$, and so the multiplication for $C^1(\mathbb{R}^n)$ functions, such as $\partial_i u$, is well defined). We can interchange the derivatives w.r.t. $k$ and $i$ in the distributional sense (indeed we can do it for test functions in $C^\infty_c(\mathbb{R}^n)$, and then by density also for test functions in $C^1_c(\mathbb{R}^n)$): $$\partial_i (\partial_k \partial_j u) (\partial_i u) = - (\partial_{k\ell} u)( \partial_{\ell j} u). $$ Since the r.h.s. is continuous, the identity holds also in the strong sense. In particular, the Hessian $H_{kj} := \partial_{k}\partial_j u$ admits derivative in the direction of $v = \nabla u$ and it satisfies the Riccati equation $$ D_v H = -H^2 .$$ One can compute explicit solutions of this along the integral lines of $v=\nabla u$, and they are analytic. Riemannian case. The argument generalizes in a straightforward way to the Riemannian setting, with the non-trivial occurrence of the curvature due to the interchange of covariant derivatives. Denoting with $S$ the symmetric $(1,1)$ tensor associated with $\mathrm{Hess}(u)$ (the shape operator of the level sets of $u$): $$ \nabla_{v} S + S^2 + R = 0,$$ where $R$ is the curvature operator $g(R X,Y) = \mathrm{Rm}(x,\nabla u,\nabla u,Y)$. In particular, $S$ is smooth along integral lines of $\nabla u$. Taking the trace, we get: $$(\nabla u) ( \Delta u) + \|\mathrm{Hess}(u)\|_{\mathrm{HS}}^2 + \mathrm{Ric}(\nabla u,\nabla u) = 0,$$ which is the particular case of Bochner's formula for $C^2$ solutions of the Eikonal equation. Comments. These equations are indeed classical, the only non-trivial part (at least to me), was the unexpected extra regularity of $\mathrm{Hess}(u)$ along the integral curve $\gamma(t)$ of $\nabla u$ (which is indeed a geodesic). Remark. This means that the mean curvature $\Delta u$ of a $C^2$ surface $\{u = 0\}$ is actually smooth in the normal direction (and analytic if the ambient space is Euclidean).<|endoftext|> TITLE: Do we have a braided tensor category for vertex algebra modules by using conformal blocks on an arbitary compact Riemann Surface? QUESTION [6 upvotes]: In Huang & Lepowsky's series of papers A theory of tensor products for module categories for a vertex operator algebra, they defined for a rational vertex algebra $V$ the $P(z)$ tensor product of two modules $W_1$ and $W_2$ to be $$ W_1\boxtimes_{P(z)} W_2=\coprod_k{(\mathcal{M}[P(z)]^{M_k}_{W_1~W_2})}^*\otimes M_k $$ where $\mathcal{M}[P(z)]^{M_k}_{W_1~W_2}$ is the (finite dimensional) vector space of conformal blocks for modules $M'_k$, $W_1$ and $W_2$ at points $\infty$, $z$ and $0$ on the compact Riemann surface $\mathbb{P}^1$. Here $M'_k$ is the module contragedient to $M_k$. They showed that under certain conditions this tensor product gives a vertex tensor category. Now if we define the tensor product by using conformal blocks on an arbitary compact Riemann surface, can we get a vertex tensor category (and hence a braided tensor category) under certain reasonable conditions? If this is true, can we show furthermore the rigidity and the modularity for this category? REPLY [7 votes]: In general, you won't get a vertex tensor category, because you don't get well-defined unit behavior when you use conformal blocks on higher genus surfaces. Huang-Lepowsky assume the vertex operator algebra is rational and $C_2$-cofinite, and this is conjecturally strong enough to obtain strong factorization properties for conformal blocks in higher genus. In particular, we expect the higher genus conformal blocks to be derived from genus zero blocks by a decomposition into pants. You can define a product structure by fixing a Riemann surface, and moving marked points in a small neighborhood to get a braiding, but all of the spaces of conformal blocks will be too large to describe a tensor category with unit if the original representation category has non-identity simple modules. On the other hand, if your vertex operator algebra is holomorphic and $C_2$-cofinite, then the tensor category derived from genus zero conformal blocks is equivalent to the category of finite dimensional vector spaces, and you should get essentially the same tensor category if you look at higher genus blocks.<|endoftext|> TITLE: Vector with many non-zero coordinates QUESTION [6 upvotes]: Given finite field $\mathbb{F}_q$, positive integers $n$ and $k TITLE: Finite realization of irrational transfer functions QUESTION [9 upvotes]: In the field of digital signal processing, linear time-invariant systems play a distinguished role. These are the systems for which there exists an impulse response, a function $h:\mathbb{Z}\to\mathbb{C}$ such that for any input $x:\mathbb{Z}\to\mathbb{C}$ the output is just the convolution $x\ast h$ given by $(x\ast h)(n) = \sum_{m=-\infty}^\infty x(n-m)h(m)$. We restrict to cases where such sums converge, e.g. square-summable sequences. Typically we focus on causal systems, those for which $h(n)=0$ when $n<0$, so future inputs do not affect the present output. The tools for analyzing these systems are equivalent to tools used in non-engineering math, but often known by different names, e.g. $z$-transforms instead of generating functions. There are two subclasses of causal linear time-invariant systems which are commonly implemented in practice. 1) Finite impulse response systems are those for which $h$ has finite support. Each output value is a finite linear combination of input values, which can be computed directly. 2) Infinite impulse response systems with rational transfer functions are those for which $H(z) := \sum_{-\infty}^\infty z^{-n}h(n)$ is a rational function of the complex number $z$ (and which do not have finite impulse responses). This rationality is often implicit in the phrase "infinite impulse response". The input-output relations of these systems can be expressed in terms of difference equations of finite order. These give a way of computing the present output in terms of past outputs as well as present and past inputs such that the amount of computation, the requisite memory, and the delay are all bounded. Question: Are there other causal linear time-invariant systems which can be implemented? I fully expect that the answer is no -- just because engineers never talk about other kinds of systems -- but I've never seen such a statement proven or even formalized. I would be happy with any "reasonable" model of computation such that the computational load of computing the output at index $n$ does not grow with $n$. In particular the number of arithmetic operations, the amount of memory needed, and the delay from the time a new input is available to the time the corresponding output is available should all be bounded independent of how much of the input sequence has been processed so far. Schmidlin's 2013 paper Realization of Irrational Transfer Functions, from which I took the title of this question, shows an implementation for which I believe the amount of computation is quadratic in the length of the input sequence (number of nonzeros), and so violates the delay constraint imposed here. REPLY [3 votes]: There is at least one (rather trivial) way to formalize "implementability" that has the property that the only implementable functions are rational. Suppose that the implementation is defined as a continuous mapping $f:\mathbb R^{N+1}\to \mathbb R^N$ such that the last coordinate of the image of the current state concatenated with the next input produces the next output. This model makes a reasonable (IMHO) compromise between allowing the infinite computation precision and disallowing some clever encoding of the entire past into a single number. The proof of the rationality of the implementable functions is then very simple. Since there is no continuous injection from $\mathbb R^{N+1}$ to $\mathbb R^N$, there are two beginnings of length $N+1$ that result in the same state of the computer. If we continue with $0$ input after that, we will have equal outputs from there on, resulting in a finite recurrence relation for $h$.<|endoftext|> TITLE: Periodicity of KdV equation in relation to zero-curvature equation QUESTION [6 upvotes]: In most of the resources that I have read, integrable systems described by a PDE posses a zero-curvature equation $$ \partial_t U - \partial_x V + [U,V] = 0 $$ which gives rise to the monodromy matrix $$ T = \mathcal{P} \exp \int\limits_0^{L} \mathrm{d} x \; U $$ Then the following quantities $$ I_n(\lambda) = \mathrm{Tr} (T^n(t,\lambda)) $$ are independent in time and so the system has an infinite amount of conserved charges, as required. However, to proof this, it is always (at least in all the notes and books I've read) assumed that all the fields are periodic in $x$ with period $L$ such that $$ V(0,t,\lambda) = V(L,t,\lambda) $$ However, I really don't understand what is the justification for this assumption. For the KdV equation, for instance, it seems clear to me that $$ u(x+L,t) \neq u(x,t) \forall x $$ which can be seen by simply looking at a plot of the (for instance 3-soliton solution) of the KdV equation I might be misunderstanding something really simple because I've read many resources and they all make this assumption of periodicity without justifying it, but I hope someone can explain it to me. REPLY [2 votes]: The fact that integrable evolutionary partial differential systems have infinitely many integrals of motion (conserved quantities) is indeed of (differenial-)algebraic nature and is not related to the choice of boundary conditions. In the KdV case this can be established in a very simple fashion using the (generalized) Miura transformation, see e.g. here. For more general results on constructing conservation laws from zero-curvature-type representations, see e.g. this survey in the case of two independent variables or this paper and references therein for the case of dispersionless (i.e. first-order quasilinear homogeneous) integrable systems. One general idea here is that if we have the (generalized) Lax equations $$L(\psi)=0, \quad \psi_t=A(\psi),$$ where $L$ and $A$ are linear differential operators involving the spectral parameter $\lambda$ then the conservation laws, and hence the integrals of motion, for the associated integrable system can be obtained from the formal expansion of $\psi$ with respect to $\lambda$.<|endoftext|> TITLE: j-invariants for isogenous elliptic curves QUESTION [5 upvotes]: Let $E$ be a smooth complex elliptic curve, and $\sigma$ translation of $E$ given by a point $p$ on $E$ of finite order, with respect to some fixed origin. What are the $j$-invariants related with $E$ and $E/\langle \sigma\rangle$?. Or to be more precise of my question, if one knows the Weierstrass equation or Hesse equation for $E$ and the point $p$, can one write down that for $E/\langle \sigma\rangle$? REPLY [10 votes]: The j-invariants satisfy a relation $F_n(j(E),j(E/\sigma))=0$ where $n$ is the order of $p$ and $F_n$ is the modular polynomial (see e.g. Silverman's "Advanced Topics ..." book pg 146). To get the Weierstrass equation of $E/\sigma$ there are the Vélu formulas (thanks Noam).<|endoftext|> TITLE: Explicit calculations of small homotopy limits of CDGAs QUESTION [9 upvotes]: I would like to carry out explicit calculations of homotopy limits of certain simple diagrams of CDGAS. My set-up is the following : I have a finite graded poset $R$ with minimal element $0$ and a diagram of rational CDGAs $$s:R^\mathrm{op}\to\mathrm{CDGA}$$ These diagrams shouldn't be Reedy fibrant. The two simplest examples of such shapes $R^\mathrm{op}$ (represented by their Hasse diagram) are * -----> 0 <----- * and * / \ * ---- * * ---- * \ \ / / * ---- 0 ---- * / / \ \ * ---- * * ---- * \ / * (all arrows pointing inwards) The first diagram shape calculates the homotopy pullback, and there are explicit formulas for expressing it : according to the nLab, since all objects in CDGA are fibrant, the homotopy pullback of $$ \begin{array}{ccc} &&A^*\\ &&\downarrow\\ B^*&\rightarrow & C^* \end{array} $$ can be calculated as the standard pullback of the associated diagram $$ \begin{array}{ccc} &&(C^*)^I\\ &&\downarrow\\ A^*\times B^*&\rightarrow & C^*\times C^* \end{array} $$ where $(C^*)^I$ is a path object for $C^*$, which apparently can be taken to be $\mathbb{Q}\oplus\big(\overline{C^*}\otimes S(t,dt)\big)$. My question is three-fold : Is it possible to explicitely calculate the homotopy limit of a diagram of CDGAs of the second shape? (By which I mean obtain generators, relations and the value of the differential on those generators.) Can the calculation be split up into subcalculations? I.e. can I calculate homotopy limits of subpieces of the diagram and break the calculation into small steps? I have a conjectured description of this homotopy limit, how can I verify that it is correct? All CDGAs that appear in these diagrams are finite dimensional and their differentials are zero (they are CGAs, actually cohomology algebras of certain manifolds); furthermore they all have explicit presentations in terms of generators and relations, and the presentations are compatible with the maps of the diagram. Also, the outmost layer of those diagrams is populated by copies of $\mathbb{Q}$. I wrote "explicit calculation" in the bounty note; by that I mean, for instance, a simple associated diagram (finite if possible) whose ordinary limit can be identified with the homotopy limit I'm after. REPLY [2 votes]: (3) is the easiest to answer. Once you have a conjectured description, just compute the honest limit of it and check that it's weakly equivalent to the limit of your original diagram. You can also check that your description is weakly equivalent in the Reedy model structure (i.e. entrywise weakly equivalent) to the original diagram. To check that your conjectured description is Reedy fibrant, use the dual of Proposition 10.6 from Dwyer-Spalinski, i.e. check that the maps in the diagram are fibrations and the pullback corner maps as well. (1) is asking for quite a lot. For the diagram in your picture, I would guess that one Reedy fibrant replacement would involve replacing all the maps pointing at 0 by fibrations (this can be done via factorization into trivial cofibration followed by fibration) and then replacing the remaining objects by products of themselves with pullbacks of the new diagrams (i.e. the ones with fibrations). So, for example, if the top part of your diagram was $A\to 0 \gets B$ and $A\gets D\to B$, then we'd replace $A$ and $B$ by $A'$ and $B'$ so that $A'\to 0\gets B'$ are both fibrations, and then replace $D$ by $D\times P$ where $P$ is the pullback of the diagram $A'\to 0 \gets B'$. The map $D\to A'$ is the composition $D\to A\to A'$, and same for $B$. The maps from $P$ are the pullback maps. This is complicated, but it's not impossible. In fact, you can work your way up the poset one vertex at a time, simply replacing the next vertex by the pullback of itself with everything new you've created so far. The dual story is presented beautifully in this paper by Luis Pereira. See Lemma 4.8. His $Q_{T_e}$ is the colimit of everything up to the new vertex $e$. This inductive process also answers your (2). In the bounty note you mention being interested in simplicial methods. A great reference is Model Categories and Simplicial Methods by Goerss and Schemmerhorn, section 4. It turns out to be easy to make a simplicial resolution (Reedy fibrant replacement) using the Dold-Kan equivalence, i.e. normalized chains functor and it's inverse $\Gamma$. A detailed study of this equivalence and its relationship to simplicial mapping spaces is given in Schwede-Shipley Equivalences of monoidal model categories. A lot of this story goes back to lecture notes of Curtis from the 70s, and they remain very readable and relevant. Good luck!<|endoftext|> TITLE: Characters of simply connected semsimple algebraic groups over local fields QUESTION [9 upvotes]: Let $G$ be a semisimple algebraic group over $\mathbb{Q}_p$. Then by definition $G$ admits no non-trivial algebraic characters, i.e. homomorphisms $G \to \mathbb{G}_m$. However, it is quite possible that $G(\mathbb{Q}_p)$ admits topological characters. E.g. take $G=\mathrm{PGL}_n$ and consider the composition $$\mathrm{PGL}_n(\mathbb{Q}_p) \to \mathbb{Q}_p^*/\mathbb{Q}_p^{*n} \to S^1, \quad g \mapsto \chi(\det(g)),$$ where $\chi: \mathbb{Q}_p^*/\mathbb{Q}_p^{*n} \to S^1$ is some character. In this special case $G$ is adjoint, however. I can also do similar constructions for other adjoint groups. So I'm wondering whether this can also happen for simply connected $G$. Let $G$ be a simply connected semisimple algebraic group over $\mathbb{Q}_p$. Is any continuous homomorphism $$G(\mathbb{Q}_p) \to S^1$$ trivial? REPLY [6 votes]: As I have written in a comment, the answer is YES (any abstract homomorphism into an abelian group is trivial) when $G$ is an isotropic, simply connected, simple algebraic group over a nonarchmedean local field $k$. For a proof see the book by Platonov and Rapinchuk, Section 7.2, Theorems 7.1 and 7.6. Note that any simply connected anisotropic simple group is isomorphic to $\mathrm{SL}(1,D)$, where $D$ is a central simple algebra over a finite separable extension $K$ of $k$. However, the answer is NO when $k=\mathbb{Q}_2$, $G=\mathrm{SL}(1,D)$, and $D$ is the quaternion division algebra over $k$. EDIT of 18.11.2018: As Arkandias explains in his comments below, for the group $G=\mathrm{SL}(1,D)$ as above, it follows from the Corollary to Theorem 21 of Carl Riehm's paper The norm 1 group of a p-adic division algebra that the abelianization $G^{\rm ab}:=G/[G,G]$ is a group of order 3. Since the commutator subgroup $[G,G]$ is open and hence, closed in $G$, we see that $G$ admits a non-trivial continuous homomorphism to $S^1$.<|endoftext|> TITLE: Auslander-Reiten theory for Gorenstein algebras QUESTION [6 upvotes]: In the paper "Cohen-Macauley and Gorenstein artin algebras", Auslander and Reiten have a short section about Auslander-Reiten theory in Gorenstein algebras (I always assume we have an artin algebra here). For Gorenstein algebras $A$ with Gorenstein dimension $g$, the category of Gorenstein projective modules has almost split sequences and the Auslander-Reiten translate of an object $C$ is given by the nonprojective part of the module $\Omega^gD\Omega^g Tr(C)$. Here are my questions: Are there examples (of larger classes of Gorenstein algebras) in the literatur, where Auslander-Reiten quivers of the category of Gorenstein projectives has been calculated? If there are only finitely many Gorenstein projective modules, one can build the direct sum $M$ of all such modules and calculate the endomorphism ring $B=End_A(M)$. Is there a relation between the Auslander-Reiten quiver of the Gorenstein projectives and the quiver of $B$ as is the case for "normal" Auslander algebras? Are there useful tools to calculate the middle term in such an almost split sequence? REPLY [4 votes]: Concerning Q3: In QPA one can compute (although I have not tested it extensively) almost split sequences in $^\perp T = \{ X\mid\operatorname{Ext}^i_\Lambda(X,T) = 0 \textrm{ for } i >0\}$ for a cotilting module $T$. When $\Lambda$ is an admissible quotient of a path algebra and $\Lambda$ is Gorenstein, then letting $T = \Lambda$ gives you a way of computing in the category of Cohen-Macaulay modules. I hope that this is helpful. Best regards, the QPA-team. Command: AlmostSplitSequenceInPerpT( T, M ); where $T$ is the cotilting module and $M$ is an indecomposable module in $^\perp T\setminus \operatorname{add T}$.<|endoftext|> TITLE: Does the injection $\text{Aut}(F_n) \hookrightarrow \text{Aut}(F_{n+1})$ split? QUESTION [16 upvotes]: Let $F_n$ be the free group on $n$ letters. The question is as in the title: letting $i:\text{Aut}(F_n) \hookrightarrow \text{Aut}(F_{n+1})$ be the natural injection, does there exist a homomorphism $\phi: \text{Aut}(F_{n+1}) \rightarrow \text{Aut}(F_n)$ such that $\phi \circ i = \text{id}$? My guess is "no", but I have no idea how to prove it. As "evidence", here are two similar situations in other analogous contexts: Let $S_n$ be the symmetric group on $n$ letters. Then the injection $S_n \hookrightarrow S_{n+1}$ does not split (at least for $n$ sufficiently large). This follows easily from the simplicity of the alternating group. The injection $\text{SL}(n,\mathbb{Z}) \hookrightarrow \text{SL}(n+1,\mathbb{Z})$ does not split. One way to see this is to use the fact that for $n \geq 3$, all normal subgroups of $\text{SL}(n,\mathbb{Z})$ are either finite or finite-index (this is a consequence of the Margulis normal subgroup theorem, but it can be proved in more elementary ways as well; I do not know who to attribute it to). The above two proofs work because we understand normal subgroups of $S_n$ and $\text{SL}(n,\mathbb{Z})$. Such an understanding seems entirely out of reach for $\text{Aut}(F_n)$. REPLY [21 votes]: Bridson and Vogtmann proved a much stronger result. From the abstract: 'If $m$ is less than $n$ then [the image of] a homomorphism $\mathrm{Aut}(F_n)\to\mathrm{Aut}(F_m)$ can have cardinality at most 2.'<|endoftext|> TITLE: In $(\mathbb{R}^4,\omega_{std})$ is positive symplectic area enough to guarantee a pseudoholomorphic disc representative? QUESTION [5 upvotes]: I will present my question in the context that I encountered it, although I believe it probably applies in general context. Consider $\mathbb{R}^4 \cong \mathbb{C}^2$ with the standard symplectic form $\omega = \sum dp_k\wedge dq_k$. Consider a Lagrangian $L\cong S^1 \times S^1$ given by the product of two circles, $A$ and $B$, each lying in an $\mathbb{R}^2$ factor of $\mathbb{R}^4$. Consider a class $\alpha\in\pi_2(\mathbb{R}^4,L)$, such that $\omega(\alpha)>0$. 1) Does there exists an almost complex structure $J$ compatible with $\omega_{std}$ such that $\alpha$ is represented by a $J$-holomorphic disc $u$ with boundary mapped to $L$? Note that for the integrable complex structure $J_{std}$ that comes from $\mathbb C$ this is true only for $\alpha$ of the form $kA+mB$, where $k,m\ge0$ (here I abused notation and referred to the discs bounded by the circles $A$ and $B$ using the same letters) 2) If the answer to 1 is positive, can $J$ be chosen to be regular? (in the sense that all the moduli spaces of curves are being cut out transversally). Thank you! :) REPLY [2 votes]: 1) Yes. The boundary of a holomorphic disc on $S^1 \times S^1$ is subcritical isotropic, and we can use Gromov's h-principle to move its boundary (by a Hamiltonian isotopy) to the sought curve on $a S^1 \times b S^1$, given that the latter curve bounds the same symplectic area. (I.e. we must first rescale the initial disc appropriately.) The new disc has boundary on $a S^1 \times b S^1$ and is pseudoholomorphic for the push-forward of the standard complex structure under the Hamiltonian diffeomorphism (this complex structure is also compatible with $\omega_0$!). 1b) Since there is an h-principle also for open Lagrangian submanifolds, under the additional condition that Maslov index of the two curves on the respective tori coincide, one can arrange so that even an open neighbourhood of the initial Lagrangian $S^1 \times S^1$ is mapped into target Lagrangian $aS^1 \times bS^1$ by the Hamiltonian diffeomorphism. 2) In some explicit cases it is obviously not possible to achieve regularity. For instance, the symplectic embedding $z \mapsto (a \overline{z},bz)$ from the unit disc to a disc of Maslov index zero with boundary on $aS^1 \times bS^1$, $b>a>0$, can be made pseudoholomorphic, but is never regular. The refined construction in (1b) does however produce regular discs, given that the initial disc was regular.<|endoftext|> TITLE: Nice things that can be proved easily with characteristic classes QUESTION [16 upvotes]: A bit of context for this question: as a project for my master's degree my supervisor asked me to understand the construction of Milnor's exotic spheres. After learning the heavy material (I knew very little algebraic topology so learning about characteristic classes counted as "heavy" for me) the proof of existence of these spheres is surprisingly short and easy. I have finished early and I would like to go on and explore other ideas, ideally results which are not unreasonably difficult and preferably are "surprising" i.e. the conclusion is something other than the classification of vector bundles over a thing. Any pointers? :) REPLY [9 votes]: I like this example. The Spheres $S^{2n}$ cannot be complex manifolds unless $n=0,1,3$. One proves that $TS^{2n}$ does not have the structure of a complex vector bundle in these cases. If $TS^{2n}$ were a complex vector bundle, then $c_{n}(TS^{2n}) = e (TS^{2n})$, so $c_n (TS^{2n})$ is twice a generator of $H^{2n} (S^{2n}; \mathbb{Z})$. Case 1: $n=2m$ even. Then $p_m (TS^{2n}) = \pm c_{2m} (TS^{2n} \otimes \mathbb{C}) = c_{2m} (TS^{2n} \oplus \overline{TS^{2n}})$. By the product formula for Chern classes, this is $= \pm c_{2m}(TS^{2n}) + c_{2m} \overline{TS^{2n}}) =\pm 2 c_{2m }(TS^{2n}) \neq 0$. This is a contradiction since $TS^{2n} \oplus \mathbb{R}$ is trivial. Case 2: $n \geq 4$. This is more difficult and relies on the Bott periodicity theorem, one of whose corollaries states that a the top Chern class $c_n (V)$ of a complex vector bundle $V\to S^{2n}$ is divisible by $(n-1)!$<|endoftext|> TITLE: Zero-dimensional $F$-space which is not strongly zero-dimensional QUESTION [7 upvotes]: Does anyone know of an example of a (Tychonoff) $F$-space which is zero-dimensional but not strongly zero-dimensional? By an $F$-space we mean every cozeroset is $C^*$-embedded. By zero-dimensional we mean has a base of clopen sets. By strongly zero-dimensional we mean every cozeroset is a countable union of clopen sets. REPLY [9 votes]: Alan Dow and I constructed an example. It can be found by following this link. The construction is based on the locally compact modification of Dowker's example in this answer. The final version of the example is now available on Arxiv.org and in Topology and its Applications<|endoftext|> TITLE: Why do we denote (co)ends with integral notation (beyond Fubini's Theorem)? QUESTION [19 upvotes]: I know that (co)ends (i.e. universal wedges) follow Fubini-like relation, i.e. $$ \int_{\langle c,d\rangle} F(c,d,c,d) \cong \int_c\int_d F(c,c,d,d) \cong \int_d\int_c F(c,c,d,d) $$ where we regard $F$ as both a functor $(C\times D)^{op}\times (C\times D)\to X$ and a functor $C^{op}\times C \times D^{op} \times D \to X$. Is there any other reason we use this notation? E.g. is their formulation as a limit in some way analogous to Riemann sums, do they have some kind of linearity relation, etc? REPLY [5 votes]: Since coends commute with colimits, in particular coproducts, we have some kind of "linearity" $$\int^i \bigl(X(i,i) \oplus Y(i,i)\bigr) \cong \int^i X(i,i) \oplus \int^i Y(i,i).$$ Every functor $F : C \to D$ into a cocomplete category $D$ may be decomposed as $$F \cong \int^{x \in C} F(x) \otimes \hom(x,-),$$ this is basically a reformulation of the Yoneda Lemma. Here, $\otimes$ denotes the copower $D \times \mathsf{Set} \to D$. This looks very similar to the equation (in fact, this is how I remember this) $$f(y) = \int_{x \in \Omega} f(x) \cdot d \delta_y(x),$$ where $\delta_y$ denotes the dirac measure concentrated at $y \in \Omega$ and $f : \Omega \to \mathbb{R}$ is a measurable function, as well as to the equation $$\mu(A) = \int_{x \in \Omega} d \mu(x) \cdot \chi_A(x)$$ for measures $\mu$ on $\Omega$. However, the analogy between coends and integrals, in particular between coproducts and sums doesn't go too far: The sum of real numbers has no analogue of a universal property. There are no analogues of coproduct inclusions. In category theory there are no nontrivial objects which are invertible with respect to coproducts. And sums in analysis are not really dual to products. Finally notice that Fubini's theorem in category theory is a formal consequence of the definitions, whereas Fubini's theorem in measure theory is not.<|endoftext|> TITLE: Horizontal distribution of a totally geodesic foliation QUESTION [7 upvotes]: Let $\mathbb{M}$ be a $n+m$ dimensional manifold. Consider on $\mathbb{M}$ a rank $n$ sub-bundle $\mathcal{H}$ of the tangent bundle. We assume that $\mathcal{H}$ is endowed with a fiber wise inner product $g_\mathcal{H}$. Under which conditions on $(\mathbb{M}, \mathcal{H}, g_\mathcal{H})$ can we find a totally geodesic Riemannian foliation $(\mathcal{F},g)$ on $\mathbb{M}$ such that: 1) $g$ is bundle-like, 2) $g$ restricts to $g_\mathcal{H}$ on $\mathcal{H}$ 3) The leaves of the foliation are orthogonal to $\mathcal{H}$ ? REPLY [2 votes]: Here is a necessary condition. I have not checked if it is sufficient. If $\mathcal F$ is totally geodesic, then translation along horizontal paths give local isometries, so you have the condition $$\mathcal L_H g^{\mathcal F}=0\;.$$ Here, $H$ is a horizontal vector field that looks like a horizontal lift of a vector field on the base, so it is parallel along leaves with respect to the Levi-Civita connection, and $g^{\mathcal F}$ is the restriction of the total metric to $\mathcal F$. We have of course used that the flow of $H$ maps local leaves to local leaves because $g$ is bundle-like.<|endoftext|> TITLE: Pop's proof that $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})=\mathrm{Aut}\underline\pi^{alg}_{\overline{\mathbb{Q}}}$ QUESTION [18 upvotes]: I've heard of this result in a paper on which Yves André proves the p-adic analogue (that is, $\mathrm{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)=\mathrm{Aut}\underline\pi^{temp}_{{\mathbb{C}}_p}$), conditional on Pop's theorem. In there, it is referenced as unpublished at the time (2002): F. Pop, A combinatorial description of $G_\mathbb{Q}$, talk at Oberwolfach, July 1998 (cf. Tagungsbericht 28), manuscript in preparation. A report of the 1998 Oberwolfach conference survives. The relevant section reads: Let $\mathcal{Var}_\mathbb{Q}$ be the category of all $\mathbb{Q}$-varieties and morphisms of such varieties. Taking the fundamental group functor we get $$\overline{\pi}_1: \mathcal{Var}_\mathbb{Q} \to \mathcal{G}$$ $$X \mapsto \pi_1(\bar{X})$$ into the category $\mathcal{G}$ of all profinite groups and outer homomorphisms. Let $\mathcal{G}_\mathbb{Q}$ be the image of $\mathcal{Var}_\mathbb{Q}$ under $\overline{\pi}_1$. For every $X$ there exists a canonical representation $$\rho_X:G_\mathbb{Q} \to \mathrm{Out}(\pi_1(\bar{X}))=\mathrm{Aut}_\mathcal{G}(\pi_1(\bar{X}))$$ which behaves functorially. Then we get a homomorphism $$\imath_\mathbb{Q} : G_\mathbb{Q} \to \mathrm{Aut}(G_\mathbb{Q})$$ $$\sigma \mapsto (\rho_X(\sigma))_X$$ We gave a positive answer to the question of Oda-Matsumoto, asking whether $\imath_\mathbb{Q}$ is an isomorphism. It seems like Pop didn't publish anything else about this result. Is the proof (or a sketch of it) available anywhere? REPLY [7 votes]: Florian Pop has a manuscript on his web page, On I/OM, dated 29 May 2014, where he proves a pro-$\ell$-abelian by central version of the Ihara/Oda-Matsumoto conjecture over any field, and explains how it implies the initial conjecture over the rational numbers. I did not read the paper, though, and I don't know its status with regard to publication.<|endoftext|> TITLE: Group cohomology question, trivial Galois action on discrete Galois module means we can say what about kernel of map QUESTION [7 upvotes]: Say we have a number field $K$. Let $G_K = \text{Gal}(\overline{K}/K)$. Let $M$ be a discrete $G_K$-module. We know that $H^1(K, M) := H^1(G_K, M)$, i.e. profinite group cohomology. For each place $v$ of $K$, let $K_v$ be the completion of $K$ at $v$; restriction to a decomposition group $G_v$ at $v$ defines a homomorphism $H^1(K, M) \to H^1(K_v, M)$. Define$$F^1(K, M) := \text{Ker}\left(H^1(K, M) \to \prod_v H^1(K_v, M)\right).$$If the $G_K$-action on $M$ is trivial, what can we say about $F^1(K, M)$? REPLY [7 votes]: If the $G_K$-action on $M$ is trivial, then $$H^1(K,M)=\mathrm{Hom}(G_K,M),$$ and by Chebotarev's density theorem $$ F^1(K,M)=0.$$ For details see Lemma 1.1(i) of Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80. REPLY [2 votes]: For a refined statement, see also https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/index-de.html (9.1.9) (i).<|endoftext|> TITLE: What is the smallest positive integer for which the congruent number problem is unsolved? QUESTION [17 upvotes]: The congruent number problem is the problem of figuring out whether a given positive integer $N$ is the area of a right-angled triangle with all side lengths rational. According to Dickson's "History of the theory of numbers" (volume 2 chapter 16) the problem was explicitly raised in an anonymous Arab manuscript which dates back to 972AD or before. Some relatively straightforward algebra shows that $N$ is a congruent number iff the elliptic curve $E_N:=y^2=x^3-N^2x$ has positive (algebraic) rank. Tunnell in 1983 gave a relatively efficient algorithm for computing the special value of the $L$-function of $E_N$ at 1 (I don't know how efficient it is in practice but it's certainly going to run in time polynomial in $N$ which I know is atrocious if $N$ is very large, however I think our problems are going to start before then). If the value of this $L$-function at 1 is non-zero then by the Coates-Wiles theorem the curve has algebraic rank zero and hence $N$ is not congruent. If BSD is true then Tunnell's algorithm gives a necessary and sufficient condition for $N$ to be congruent. However BSD is not yet known, and my guess is that as $N$ increases, the fact that we don't know BSD for $E_N$ will trouble us before Tunnell's algorithm starts slowing up. In theory there could exist some $N$ for which $E_N$ had analytic rank 2 or more, and algebraic rank zero. Tunnell would compute the special value at 1 to be zero and then we would be faced with the problem of computing the algebraic rank, which can very occasionally be hard in practice, especially if Sha is non-trivial etc etc. Even if the algebraic rank is positive, finding a non-torsion point can be hard in practice; the question is like Pell where relatively innocuous values of $N$ like 157 can give rise to elliptic curves for which the height of every non-torsion point is very large. In fact general theory can deal with 157 because the analytic rank is 1, but certainly the analytic rank can be higher than 1 in general (for example if $N=41$ then the analytic rank is 2). Because of this issue, I think that (until a new major advance is made with this problem) there must surely be some explicit smallest positive integer $N$ for which we don't know whether or not $N$ is a congruent number. My guess is that this value of $N$ will actually be a congruent number in practice, and the elliptic curve $E_N$ will be computationally checked to have analytic rank at least 2, but no-one will be able to find a rational point (not because there isn't one, but because it's out of computational range). My guess is that there will be papers containing results of the form "we have solved the congruent number problem for all $N\leq 10^{10}$" or some such statement -- but I have no idea what the state of the art is, and perhaps my google-fu is weak today. This does not change the fact that I am giving a talk on the congruent number problem tomorrow :-) and I thought that it might be nice to mention the state of the art, if I can find it out by then! Does anyone have any leads? REPLY [9 votes]: Kazuo Matsuno writes (personal communication): "I verified (10 years ago) by using mwrank and magma that E_N:y^2=x^3-N^2x has a non-torsion point if N<=10^6 is congruent to 1,2,3 modulo 8 and the analytic rank of E_N is positive. For all N<=10^6 congruent to 5,6,7 modulo 8, Elkies proved rank(E_N)>0. http://arxiv.org/abs/math/0208056 " If no-one can better this, then the answer to my question is hence $10^6+2$ (given that I (KB) just checked $10^6+1$ earlier today).<|endoftext|> TITLE: The homotopy category is not complete nor cocomplete QUESTION [29 upvotes]: I understand that the homotopy category of (pointed) topological spaces and continuous maps is not complete. Nor is it cocomplete. In particular it neither has all pullbacks nor all pushouts. What are some simple examples of spans $Z\leftarrow X\rightarrow Y$ and cospans $Z\rightarrow X\leftarrow Y$ that cannot be completed to pushout and pullback squares, respectively? REPLY [6 votes]: Here's an example of missing pullback in the homotopy category of spaces, related to Neil's and Qiaochu's answers. I'll work in pointed spaces, though we'll see that the argument also works in unpointed spaces. Let's say we work in the usual homotopy category (i.e., the localization with respect to weak homotopy equivalences). Assume that $$\require{AMScd} \begin{CD} P @>>> K(\mathbb{Z},2)\\ @VVV @VV {n} V\\ \ast @>>> K(\mathbb{Z},2)\\ \end{CD}$$ is a pullback in the homotopy category. This says that maps $[X,P]$ consist of the subset of $[X,K(\mathbb{Z},2)]$ of maps becoming null-homotopic after composing with $n \colon K(\mathbb{Z},2) \to K(\mathbb{Z},2)$, in other words, the $n$-torsion in $[X,K(\mathbb{Z},2)]$. Taking $X = S^0$ yields that $P$ is path-connected. Taking spheres $X = S^i$ yields that $P$ is weakly contractible. In particular, $P$ has the singular cohomology of a point. Now let $\alpha \colon K(\mathbb{Z}/n,1) \to K(\mathbb{Z},2)$ be a generator of $H^2( K(\mathbb{Z}/n,1); \mathbb{Z}) = \mathbb{Z}/n$. The equality $n \alpha = 0$ implies that $\alpha$ factors through $P$. This is a contradiction, since $\alpha$ induces a non-zero map on cohomology $H^2(-;\mathbb{Z})$. Remark 1. The argument also works in the naive homotopy category (i.e., spaces and homotopy classes of maps). The space $P$ might not have the homotopy type of a CW complex and we don't know that singular cohomology $H^2(P;\mathbb{Z})$ is represented by maps $P \to K(\mathbb{Z},2)$; we only used $H^2(P;\mathbb{Z}) = 0$. For the fact that a weak homotopy equivalence induces an isomorphism on singular (co)homology, see Hatcher Proposition 4.21. Remark 2: Let's adapt the example to unpointed spaces. Recall that for path-connected $Y$, the forgetful map $[X,Y]_* \to [X,Y]$ from pointed homotopy classes of pointed maps to unpointed homotopy classes of unpointed maps is surjective and induces a bijection $[X,Y] = [X,Y]_* / \pi_1(Y)$ to the set of $\pi_1(Y)$-orbits in $[X,Y]_*$; see Hatcher Proposition 4A.2. In the example above, taking $X = \ast$ yields that $P$ is path-connected. Taking $X = S^1$ yields that $\pi_1(P)$ has only one conjugacy class, hence is the trivial group. The rest of the argument goes through.<|endoftext|> TITLE: Simplest example of failure of finite Galois descent in algebraic $K$-theory? QUESTION [14 upvotes]: Let $E \to F$ be a $G$-Galois extension of fields. What is the simplest example where the natural map $K(E) \to K(F)^{hG}$ is not an equivalence on connective covers (i.e., where finite Galois descent fails)? REPLY [8 votes]: First of all, by the long exact sequence in homotopy, it suffices to give a $G$-Galois extension $E \subset E'$ such that the homotopy fiber of $K(E) \to K(E')^{hG}$ contains a class in degree $-1$. This means that the map $\pi_0 K(E) \to \pi_0 K(E')^{hG}$ is not surjective. The spectrum $\mathrm{fib}( K(E) \to K(E')^{hG} )$ is annihilated by the group order $|G|$. This follows from a classical argument with the transfer: anything in the image of the transfer $K(E')_{hG} \to K(E)$ annihilates the homotopy fiber. (In the language of $G$-spectra: if $R$ is a $G$-ring spectrum, then any class in $\pi_0 R^G$ which is in the image of $\pi_0 R_{hG} \to \pi_0 R^G$ annihilates the fiber of $R^G \to R^{hG}$). Note that the order of $G$ is the image of $1$ under the transfer map for $K$-theory. It follows that $\mathrm{fib}( K(E) \to K(E')^{hG} )$ splits as a product of its $l$-adic completions for various primes $l$. It thus suffices to show that $\pi_{-1}$ of some $l$-adic completion is nonzero, or equivalently that $l$-adically completed $K$-theory is not a sheaf of connective spectra. If I'm not mistaken, the following (based on an idea of Dustin Clausen) should give an example. Let $E$ be a field of characteristic $\neq l$ with the following two properties: 1) $E$ contains all the $l$-th power roots of unity 2) $H^2( \mathrm{Gal}(E); \mathbb{Z}_l(1)) \neq 0$. Then I claim the functor $E' \mapsto K(E')$ is not a sheaf of connective spectra on the etale site of $E'$ (equivalently doesn't satisfy Galois descent). It suffices to work after $l$-adic completion, and for the rest of this post I'll write $\hat{K}$ for the $l$-adic completion of $K$. In fact, I claim that there exists a finite Galois extension $E'$ of $E$ such that $\hat{K}(E')^{hG}$ is bigger than $\mathbb{Z}_l$. To see this, choose a finite Galois extension $E'$ of $E$ with group $G$ such that $H^2(G; \mathbb{Z}_l) \neq 0$. I claim that $\hat{K}(E')^{hG}$ has a $\pi_0$ larger than $\mathbb{Z}$. To see this, we observe that there is a map $\mathbb{Z}_l \to \pi_2 ( \hat{K}(E)) \to \pi_2 (\hat{K}(E'))$ (recall that we're $l$-completing everywhere) and the composite to $\mathbb{Z}_l \to \pi_2( \hat{K}(E')) \to \pi_2( \hat{K}(E^{bar}))$ is an isomorphism (the latter by Suslin). By our assumptions, we thus get a cycle in $H^2( G, \pi_2 \hat{K}(E'))$ which maps to a nonzero cycle in $H^2(E, \mathbb{Z}_l(1))$. To see that it survives the HFPSS, it suffices to work at the level of spaces. But there is a $G$-equivariant map (where the hat denotes completion) $\widehat{BGL_1(E')} \to \Omega^\infty \hat{K}(E')$. On $\pi_2$, we get exactly a $\mathbb{Z}_l$ hitting the class desired. Since $BGL_1(E')$ has no higher homotopy, we find that the class in $H^2( G, \pi_2 \hat{K}(E'))$ must survive as it comes from $(\widehat{BGL_1(E')})^{hG}$. Since this permanent cycle is detected nontrivially in the descent ss for the full Galois group, it must be nonzero. Thus, $\pi_0 \hat{K}(E')^{hG}$ is bigger than $\mathbb{Z}_l$.<|endoftext|> TITLE: Algebraic points of uniformly bounded degree on an algebraic variety QUESTION [8 upvotes]: Let $k$ be a perfect field, and let $\bar k$ be a fixed algebraic closure of $k$. Let $\overline{X}$ be a nonempty smooth algebraic variety over $\bar k$. Does there exist a natural number $d=d(\overline{X})$ with the following property: For any $k$-form $X$ of $\overline{X}$, the variety $X$ has a $K$-point over some finite field extension $K$ of $k$ of degree $[K:k]\le d$ ? The answer YES would imply Theorem 2 of my answer to this question. This question and the references in comments to it may be relevant. REPLY [13 votes]: No: this fails already when $X=E$ is an elliptic curve and $k=\mathbb{Q}$. This would imply that every element of $H^1(\mathbb{Q},E)$ has order at most $d$, and I'm pretty sure that this cohomology group has elements of arbitrarily large order. Otherwise the Tate-Shafarevich conjecture would be rather trivial, as it is well-known that $Sha(E)[d]$ is finite for every $d$.<|endoftext|> TITLE: What's the supersymmetric analogue of the Monster group? QUESTION [31 upvotes]: Bosonic string theory lives in 26 dimensions, and it gives a conformal field theory where the field is a map from a Riemann surface to $\mathbb{R}^{24}$. The Leech lattice $L$ is an even unimodular lattice in $\mathbb{R}^{24}$. We can form a conformal field theory where the field is a map from a Riemann surface to the torus $T = \mathbb{R}^{24}/L$, and this theory almost has the Monster group as its symmetry group. In fact we need to go one step further and replace $T$ by the orbifold where we mod out by the involution of $T$ coming from the transformation $x \mapsto -x$ of $\mathbb{R}^{24}$. In this case Frenkel, Lepowsky and Meurman showed we get a conformal field theory, or more technically a vertex operator algebra, whose symmmetry group includes the Monster group. There could be a supersymmetric analogue of this, and it's probably been studied. What group does that give? More precisely: Superstring theory lives in 10 dimensions, and it should give a conformal field theory where the field is a map from a Riemann surface to $\mathbb{R}^{8}$, or actually a super-vector space $V$ with $\mathbb{R}^8$ as its even part. The $\mathrm{E}_8$ lattice is an even unimodular lattice in $\mathbb{R}^8$. I suspect we should be able to form a form a conformal field theory where the field is a map from a Riemann surface to the 'supertorus' $T_\mathrm{super} = V/\mathbb{E}_8$. Is the symmetry group of the corresponding vertex operator algebra known? We may have to replace $T_\mathrm{super}$ by a super-orbifold, e.g. mod out by an involution, to get a really interesting group. REPLY [3 votes]: This may be of interest: Monstrous BPS-Algebras and the Superstring Origin of Moonshine by Natalie M. Paquette, Daniel Persson, Roberto Volpato http://arxiv.org/abs/1601.05412<|endoftext|> TITLE: Abstract result on partitions of unity? QUESTION [9 upvotes]: A motivation: The classical Stone-Weierstrass theorem says that polynomials are dense among continuous functions (say, on the unit interval), while the abstract Stone-Weierstrass theorem (and also the related Kakutani-Krein theorem) give sufficient conditions for a set of continuous functions to be dense. The classical partitions of unity are continuous ones or smooth ones. I ask if there are abstract theorems on the existence of partitions of unity generalizing the classical ones. Specifically, I'm looking for "known" or "natural" statements of the following form: If a topological space $X$ satisfies [yadda yadda] and a set $F$ of real-valued functions on $X$ satisfies [blah blah] then any open cover of $X$ has a subordinate partition of unity composed of functions in $F$. As an example of such a statement, replace [yadda yadda] with [locally compact Hausdorff] and [blah blah] with [$F$ is an algebra of continuous functions that satisfies Urysohn's Lemma]. (Proof: the same as in Rudin's RCA book, page 40.) Is there a cleaner formulation? Related questions: Is there such an "abstract" version of Urysohn's Lemma? Of Tietze Extension Theorem? Please provide references, if available. :) REPLY [14 votes]: I will leave to Yemon Choi discussing the answer from Gelfand-Raikov-Shilov's book (Commutative Normed Rings, I suppose?), and restrict myself to more recent discussions on the matter... There is an extensive discussion on abstract partitions of unity in Chapter III of the book of Andreas Kriegl and Peter Michor, The Convenient Setting of Global Analysis (AMS, 1997). There they define precisely what you want - the concepts of $\mathcal{S}$-regularity, $\mathcal{S}$-normality and $\mathcal{S}$-paracompactness. To wit, let $X$ be a Hausdorff topological space and $\mathcal{S}\subset C(X,\mathbb{R})$ be a subalgebra. Then $X$ is: $\mathcal{S}$-regular if for every neighborhood $U$ of a point $p\in X$ there is a $f\in\mathcal{S}$ such that $f(p)=1$ and $f\equiv 0$ in $X\smallsetminus U$, i.e. $X$ has enough bump functions in $\mathcal{S}$ (this should actually be called $\mathcal{S}$-completely regular or $\mathcal{S}$-Tychonoff, but it is called that way in Kriegl-Michor, so whatever); $\mathcal{S}$-normal if given two disjoint closed subsets $A_0,A_1\subset X$ there is a $f\in\mathcal{S}$ such that $f|_{A_i}=i$, $i=0,1$, i.e. it satisfies the Urysohn lemma with elements of $\mathcal{S}$ (this, I believe, should be the same definition as the one in Gelfand-Raikov-Shilov's book pointed by Yemon Choi's comment above). Equivalently (Theorem 16.2 (1) $\Leftrightarrow$ (3), pp. 165-166 of Kriegl-Michor), every locally finite open covering of $X$ has a partition of unity subordinate to it comprising only of elements of $\mathcal{S}$; $\mathcal{S}$-paracompact if every open covering of $X$ has a partition of unity subordinated to it comprising only of elements of $\mathcal{S}$. Equivalently (Theorem 16.2, pp. 165-166 of Kriegl-Michor), $X$ is paracompact and $\mathcal{S}$-normal. There are a few more characterizations of $\mathcal{S}$-normality in Theorem 16.2, pp. 165-167 of Kriegl-Michor besides the one cited above. Moreover, under the assumption that $\mathcal{S}$ is local (i.e. if $f\in C(X,\mathbb{R})$ is such that there is an open covering $\mathscr{U}$ of $X$ such that for every $U\in\mathscr{U}$ there is $f_U\in\mathcal{S}$ such that $f|_U=f_U$, then $f\in\mathcal{S}$) and that $h\circ f\in\mathcal{S}$ for all $h\in C^\infty(\mathbb{R},\mathbb{R})$, $f\in\mathcal{S}$, we have the following results (Edit: the previous assumptions were missing in a previous version of this answer): R. Bonic and J. Frampton showed (Smooth Functions on Banach Manifolds, J. Math. Mech. 15 (1966) 877-898, see also Theorem 14.8, pp. 155 in Kriegl-Michor) that $X$ is $\mathcal{S}$-regular iff the topology of $X$ is initial with respect to $\mathcal{S}$, i.e. the inverse images of open subsets of $\mathbb{R}$ under elements of $\mathcal{S}$ form a subbasis for the topology of $X$; J.C. Wells showed (Differentiable Functions on Banach Spaces with Lipschitz Derivatives, J. Diff. Geom. 8 (1973) 135-152, see also Theorem 16.10, pp. 171-172 of Kriegl-Michor) that if $X$ is Lindelöf and $\mathcal{S}$-regular, then it is $\mathcal{S}$-paracompact (actually, for this result we do not need $\mathcal{S}$ to be local or closed under composition with any $h\in C^\infty(\mathbb{R},\mathbb{R})$, only that $h\circ f\in\mathcal{S}$ for all $f\in\mathcal{S}$ and all $h\in C^\infty(\mathbb{R},\mathbb{R})$ such that $h(\mathbb{R})\subset[0,1]$, $h(t)=0$ for $t\leq 0$ and $h(t)=1$ if $t\geq 1$); It was not known at the time Kriegl-Michor's book was published if paracompactness and $\mathcal{S}$-regularity entail $\mathcal{S}$-paracompactness, even with the above hypotheses. As for paracompactness itself, if $X$ is also locally compact, there is a nice algebraic characterization of it by R. Bkouche, whose proof was later simplified by R.L. Finney and J. Rotman (Paracompactness of Locally Compact Spaces, Michigan Math. J. 17 (1970) 359-361): A locally compact Hausdorff topological space $X$ is paracompact iff the ideal $C_c(X,\mathbb{R})\subset C(X,\mathbb{R})$ of continuous functions with compact support is a projective $C(X,\mathbb{R})$-module. The beautiful and simple argument of Finney and Rotman carries over to smoothly paracompact manifolds, and presumably should be adaptable to general $\mathcal{S}$.<|endoftext|> TITLE: What is a field [Körper] really? QUESTION [14 upvotes]: The notion of a field (a commutative ring $R$ with $0\neq 1$ and $R^\times=R-\{0\}$) seems to fit uncomfortably into modern algebra. To see what I mean, consider the following statements: The concept of a field seems necessary for the foundation of geometry. For example, the "Steinitz Exchange Lemma" (proving that dimension of a vector space is well-defined) seems to require a field. Dimension of a vector space, in turn, seems to underlie all other notions of dimension in mathematics. The category of fields is not much of a category. It doesn't have products or coproducts. Every morphism is a monomorphism. It's not connected (there is one connected component for each characteristic). Free fields don't exist. In general, fields do not form an "equational theory" so they don't behave like an "algebraic category". So what is a field, really? Here are a couple suggestions to get you started: A single field $K$ is something like the category of sets. This is suggested by the analogy between the bifunctor $\mathrm{Hom}_{\mathcal{C}}(-,-):\mathcal{C}^{\mathrm{op}}\times\mathcal{C}\to\mathsf{Set}$ on a category $\mathcal{C}$ and a bilinear form $\langle -,-\rangle:V\times V\to K$ on a $K$-vector space $V$. It also agrees with the use of the "ground field" in algebraic geometry, which is something like the "universe of discourse". The construction of the splitting field doesn't have an UMP because fields have too many automorphisms. So it seems that "Galois theory" is what makes fields special. Perhaps a formalization of Galois theory can help explain what fields really are. REPLY [4 votes]: Related to the previous answer but different - in abelian categories, (not necessarily commutative) fields are endomorphism rings of simple objects (just as local rings are endomorphism rings of indecomposable objects). From this point of view it is even somehow not natural to consider their homomorphisms: I cannot think of any kind of natural data between two objects which would induce a homomorphism between their endomorphism rings. The natural kind of morphisms to consider in this context are bimodules - hom between objects is a bimodule over their endomorphism rings.<|endoftext|> TITLE: Globalizing local field extensions with controlled ramification QUESTION [9 upvotes]: Let $K_1/k_1, \ldots, K_r/k_r$ be "separable cyclic" extensions of degree $n$ where each $k_i$ is a local field of characteristic 0 (archimedean or not). By separable cyclic of degree $n$, I mean the localization of some global cyclic extension of degree $n$, so a given $K_i/k_i$ may be split or not. (E.g., if $n=2$, either $K_i/k_i$ is a quadratic field extension or $K_i = k_i \oplus k_i$.) Does there exist an extension of number fields $E/F$ globalizing all $K_i/k_i$ and unramified everywhere else? That is, is there a global $E/F$ with a set $S = \{v_1, \ldots v_r \}$ of places of $k$ such that $E_{v_i} = K_i$, $F_{v_i} = k_i$ and $E_v/F_v$ is unramified for $v \not \in S$. Results are known about globalizing local field extensions when one fixed a global field $F$, e.g. Grunwald-Wang, but I don't want to do that. Indeed, a positive answer is not always possible if one, say, requires $F=\mathbb Q$, even if each $F_{v_i}$ specified is a distinct local completion of $\mathbb Q$. (For instance, take $n=2$, specify that $E$ should be $\mathbb C$ at infinity, ramified at 3 and unramified everywhere else---there is a unique choice globally $E=\mathbb Q(\sqrt{-3})$, but locally there are 2 ramified extensions at 3 one can choose.) Rather I want to prescribe stronger local conditions than what one gets from Grunwald-Wang. A positive answer (even with some conditions) should allow one to use global methods to prove results about local representation theory via trace formulas. Note this is related to finding number fields $F$ such that a ray class field has prescribed local behaviour. I imagine such questions are hard, so partial results are welcome. (I'm not even sure whether the answer is yes when $n=2$ and $r=1$, though I think it is doable.) Edit: Based on the answer and comments, I want to clarify that when $r=1$, the question is asking for a global extension $E/F$ which is unramified at all but 1 place, where it is $K_1/k_1$. The current comments and answer seem to give an extension $E/F$ that at all ramified places $v \in R$ looks like $K_1/k_1$, but at least a priori we do not have $|R| = 1$. REPLY [4 votes]: See my paper "Extensions de corps globaux à ramification et groupe de Galois donnés", C.R. Acad. Sci. Paris 311 (1990), 273–276. I hope this link works.<|endoftext|> TITLE: Is polynomial convexity a topological invariant? QUESTION [6 upvotes]: Is the property of being polynomially convex a topological invariant? In other words, let $M$ and $N$ be two homeomorphic, compact subsets of $n$-dimensional complex Euclidean space, and assume in addition that $M$ is polynomially convex. Is $N$ necessarily polynomial convex? REPLY [10 votes]: The answer is no. In fact, Kallin has shown in [Kal64] that the union of three disjoint closed balls is polynomially convex, but the union of three disjoint closed polydisks needs not to be polynomially convex. Actually, it turns out that polynomial convexity is not even preserved by (real) linear transformations. For instance, Khudaĭberganov and Kytmanov have given in [KK84] an example of three closed disjoint ellipsoids the union of which is not polynomially convex. References [Kal64] E. Kallin: Polynomial convexity: The three spheres problem, Proc. Conf. Complex Analysis (Minneapolis, 1964), Springer, Berlin, pp. 301–304, doi:10.1007/978-3-642-48016-4_26. [KK84] G. Khudaĭberganov and A. M. Kytmanov: An example of a nonpolynomially convex compact set consisting of three non-intersecting ellipsoids, Sibirsk. Mat. Zh. 25 (5) (1984), 196-198 (in Russian).<|endoftext|> TITLE: Weak* continuity of positive parts, again QUESTION [6 upvotes]: Bill Johnson pointed out to me yesterday that the map $$f \mapsto f^+ = \max(f,0)$$ is not weak* continuous on $l^\infty$. Nonetheless, I think I can prove that if $V$ is a linear subspace of $l^\infty$ which is stable under this operation, then its weak* closure is too. In other words, the weak* closure of any vector lattice in $l^\infty$ is a vector lattice. The proof is by transfinite induction! Does anyone know an easier proof of this simple fact? REPLY [2 votes]: The most high-tech way (and probably the most enlightening way) of proving this is via the Mackey or bounded weak$^*$ topologies on $\ell^\infty$, which happen to coincide in this case. The Mackey topology is the topology of uniform convergence on weakly compact absolutely convex subsets of $\ell^1$, which by the Mackey-Arens theorem is the finest vector topology on $\ell^\infty$ whose continuous linear functionals (and thus closed convex sets) are the same as the weak$^*$ topology. The bounded weak$^*$ topology is the topology of uniform convergence on norm compact absolutely convex sets of $\ell^1$, which by the Mackey-Arens theorem has the same continuous linear functionals as the weak$^*$ topology. Since weak compactness and norm compactness agree on $\ell^1$, the Mackey topology and bounded weak$^*$ topology on $\ell^\infty$ are identical. The characterization of the bounded weak$^*$ topology as the finest linear topology that agrees with the weak$^*$ topology on bounded sets is the key point here. Given that, we can show that the bounded weak$^*$ topology is actually the topology of bounded pointwise convergence, i.e. the topology defined by sequential convergence $(f_n) \to f$ whenever $\|f_n - f\|$ is uniformly bounded and $(f_n) \to f$ pointwise. Since this defines a finer vector topology that agrees with the weak$^*$ topology on bounded sets, it is equal to the bounded weak$^*$ topology. The map $f \mapsto f^+$ is continuous with respect to the topology of bounded pointwise convergence, so the weak$^*$ closure of any convex set invariant under this function is also invariant under this function. You can also give a direct proof that the Mackey topology coincides with the topology of bounded pointwise convergence by using the Schur property and the characterization of norm compactness in terms of convex hulls of null sequences. I suspect (but haven't verified) that you should be able to unravel every use of the Hahn-Banach Theorem so that you are only using it for the $(\mathrm{c}_0, \ell^1)$ dual pair, and thus Dependent Choice suffices. In the category of Banach spaces (with either contractive or bounded morphisms), the dual of a product of spaces may be much larger than the sum of the duals of the spaces, as evidenced by $\ell^\infty$ itself. If you instead consider spaces defined by mixed topologies, this flaw is removed. A Saks space is Banach space equipped with a norm and a weaker locally convex topology in which the norm's unit ball is complete. The mixed topology of the norm and the locally convex topology is defined to be the finest vector topology that agrees with the locally convex topology on norm-bounded sets. The morphisms are norm-bounded linear maps that are continuous for the locally convex topologies on bounded sets. The dual of a Saks space is not naturally a Saks space, but rather a co-Saks space, with a dual definition. Given a family of Banach spaces considered as Saks spaces (with the norm defining both the bounded sets and the topology), the dual of the product of Saks spaces is equal to the sum of the co-Saks duals, and this dual pairing gives the Mackey topology on the product. The proof essentially reduces to the case of $(\ell^\infty, \ell^1)$, because any weakly compact subset of a sum has countable support. The unit ball of a Saks space is compact for the mixed topology precisely when the Saks space is the bounded weak$^*$ topology for some predual, or when the Saks space is a projective limit of finite-dimensional spaces.<|endoftext|> TITLE: Elements of arbitrary large order in the first Galois cohomology of an elliptic curve QUESTION [15 upvotes]: Let $E$ be an elliptic curve over $k=\mathbb{Q}$. Consider $H^1(k,E)$. In this answer Daniel Loughran writes: "I'm pretty sure that this cohomology group has elements of arbitrarily large order". I would be happy to have an explanation of this fact and/or a reference. (I usually work with linear algebraic groups. For an abelian linear algebraic group $A$ there exists a natural number $d=d(A_{\bar k})$ such that the order of any element of $H^1(k,A)$ is $\le d$.) EDIT: Cassels, § 10 and § 27, writes that this result was first proved by Shafarevich in 1957. REPLY [3 votes]: Theorem 6 of this paper gives a generalization of Shafarevich's Theorem: Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$. As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof quickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.<|endoftext|> TITLE: The Hypercomplex Structure of $SU(3)$ QUESTION [7 upvotes]: (A) In this really stylish answer it is shown that one can define a family of complex structures $J_{\lambda}$ on the Lie group SU(3), dependent on the parameter $\lambda \in {\mathbb C}\backslash {\mathbb R}$. (B) Wikipedia can tell us that $SU(3)$ has a hypercomplex structure. I am searching a description of (B) using the language of (A). REPLY [10 votes]: I'm assuming that you have a copy of Dominic Joyce's 1992 JDG article, "Compact hypercomplex and quaternionic manifolds" handy. Write $\frak{g} = \frak{su}(3)$ as a direct sum $$ \frak{su}(3) = \frak{b}\ \oplus\ \frak{d}\ \oplus\ \frak{f}\ , $$ where $$ {\frak{b}} = \left\{\begin{pmatrix}ia&0&0\\0&ia&0\\0&0&-2ia\end{pmatrix}\ \Biggl|\ a\in\mathbb{R}\ \right\}\simeq \mathbb{R} $$ $$ {\frak{d}} = \left\{\begin{pmatrix}ip&q+ir&0\\-q+ir&-ip&0\\0&0&0\end{pmatrix}\ \Biggl|\ p,q,r\in\mathbb{R}\ \right\}\simeq{\frak{su}}(2) $$ $$ {\frak{f}} = \left\{\begin{pmatrix}0&0&-\overline z\\0&0&-\overline w\\z&w&0\end{pmatrix}\ \Biggl|\ z,w\in\mathbb{C}\ \right\}\simeq \mathbb{C}^2\simeq\mathbb{H}. $$ One can easily check that this decomposition satisfies all of the conditions of Lemma 4.1 on page 751 (where, since $n=1$ in this case, I'm omitting the subscripts and summation signs). Now, in Theorem 4.2, we have $k=m=0$, and we can define the complex structures $I_1$, $I_2$, $I_3$ on ${\frak{su}}(3)$ satisfying $I_1I_2=I_3$ as on pages 753–754. (This depends on choosing a basis of $\frak{b}$, so there is a $1$-parameter family of such choices.) By Lemma 4.3, these extend by left-invariance to integrable almost complex structures on $G=\mathrm{SU}(3)$, and so they define a left-invariant hypercomplex structure on $\mathrm{SU}(3)$. Addendum: The OP's request was for a 'description of (B) using the language of (A)', and the above answer doesn't quite do that. Moreover, while thinking about the best way to honor the original request, I took another look at Joyce's paper and realized that there is a sign mistake in one of his formulae that is liable to confuse whoever tries to carry out that description (maybe the OP?) and lead to some frustration. First, the sign error: When Joyce goes to construct the three complex structures $I_1$, $I_2$, and $I_3$ on ${\frak{f}} \subset {\frak{su}}(3)$ at the top of page 754, he gives the formula $I_a(v) = \bigl[v,\phi_j(i_a)\bigr]$, but this would lead to $I_1I_2=-I_3$ on $\frak{f}$ instead of the desired $I_1I_2=I_3$, so his definition should instead be $I_a(v) = \bigl[\phi_j(i_a),v\bigr]$, and then everything works out correctly. The upshot is that, if we then introduce a basis $\theta_i$ for the left-invariant $1$-forms on $\mathrm{SU}(3)$ in such a way that the canonical left-invariant form is $$ \gamma = g^{-1}\,\mathrm{d}g = \begin{pmatrix} i(\theta_1+\theta_2) & i\theta_3-\theta_4 & \theta_5 + i\theta_6\\ i\theta_3+\theta_4 & i(\theta_1-\theta_2) & i\theta_7+\theta_8\\ -\theta_5+i\theta_6 & i\theta_7 - \theta_8 & -2i\theta_1 \end{pmatrix}, $$ then the $\mathbb{H}^2$-valued left-invariant $1$-form $$ \theta = \bigl(\theta_1+i\theta_2 + j\theta_3 +k\theta_4\,,\ \theta_5+i\theta_6 + j\theta_7 +k\theta_8 \bigr) $$ satisfies $\theta(I_1v) = i\theta(v)$ and $\theta(I_2v) = j\theta(v)$ for all $v\in {\frak{su}}(3)$. As a result, the $\mathbb{C}^4$-valued $1$-forms $$ \alpha = \bigl(\ \theta_1{+}i\theta_2\ \ \theta_3{+}i\theta_4\ \ \theta_5{+}i\theta_6\ \ \theta_7{+}i\theta_8\ \bigr) =(\alpha_1\ \alpha_2\ \alpha_3\ \alpha_4) $$ and $$ \beta = \bigl(\ \theta_1{+}i\theta_3\ \ \theta_2{-}i\theta_4\ \ \theta_5{+}i\theta_7\ \ \theta_6{-}i\theta_8\ \bigr) =(\beta_1\ \beta_2\ \beta_3\ \beta_4) $$ satisfy $\alpha(I_1v) = i\alpha(v)$ and $\beta(I_2v) = i\beta(v)$ for all tangent vectors $v\in T\mathrm{SU}(3)$. Consequently the $\alpha_i$ are a basis for the $(1,0)$-forms on $\mathrm{SU}(3)$ with respect to the almost-complex structure $I_1$ while the $\beta_i$ are a basis for the $(1,0)$-forms on $\mathrm{SU}(3)$ with respect to the almost-complex structure $I_2$. Finally, using the identity $\mathrm{d}\gamma = -\gamma\wedge\gamma$ to get the formulae for $\mathrm{d}\theta_i$, we find that $$ \mathrm{d}\alpha_i\equiv 0\ \mod\ \alpha_1,\alpha_2,\alpha_3,\alpha_4 $$ while $$ \mathrm{d}\beta_i\equiv 0\ \mod\ \beta_1,\beta_2,\beta_3,\beta_4\,. $$ Thus, $I_1$ and $I_2$ (and hence $I_3 = I_1I_2=-I_2I_1$) are integrable complex structures on $\mathrm{SU}(3)$, and thus the triple $(I_1,I_2,I_3)$ defines a hypercomplex structure on $\mathrm{SU}(3)$.<|endoftext|> TITLE: Interior periodic points of area preserving homeomorphisms of a pair of pants QUESTION [8 upvotes]: A celebrated result of Franks shows that any area preserving homeomorphism of the closed annulus $A$ with at least one periodic point (possibly along the boundary) has infinitely many interior periodic points. Let $\phi$ be an area preserving homeomorphism of the pair of pants P (viewed as a sphere minus three open discs). Question: What are some natural conditions which guarantee that $\phi$ has infinitely many periodic points? For example is there some behavior near the boundary which guarantees this? An obvious example would be that it agrees with an irrational rotation near one of the boundary components so that we could extend it to A by a rotation. Perhaps there are more subtle statements. REPLY [6 votes]: It always has infinitely many periodic points, since you can collapse one of the boundary components to get a homeomorphism of the closed annulus with a fixed point, and then you can use Franks' theorem (the area-preserving condition can be replaced by the existence of a non-atomic invariant measure of full support; Franks proves the results using these assumptions, and in any case this is equivalent to being area-preserving after a topological conjugation due to the Oxtoby-Ulam theorem).<|endoftext|> TITLE: Biorthogonal functionals QUESTION [5 upvotes]: If $X$ is a separable Banach space and $(x_n)$ is a basic sequence, then we can define biorthogonal functionals $(x^{*}_n)$ in $X^{*}$ such that $x^{*}_n(x_k)=\delta_{nk}$. What about conversely? If $(x^{*}_n)$ is a basic sequence in $X^{*}$, can we always find vectors $(x_n)$ in $X$ such that $x^{*}_n(x_k)=\delta_{nk}$? This is true when $X$ is reflexive, but I cannot see a proof or a counterexample when $X$ is not reflexive. REPLY [3 votes]: Yes, there is such a sequence in any non-reflexive Banach space. It is known$^\#$ that if $Y$ is not reflexive, then $Y$ contains a basic sequence that is not boundedly complete. So if $X$ is not reflexive, then there is a semi-normalized basic sequence $(x_n^*)_{n=1}^\infty$ in $X^*$ s.t. $\sup_n \|\sum_{k=1}^n x_k^*\| < \infty$. Assume that $(x_n) \subset X$ is biorthogonal to $(x_n^*)$ (else you are done). Let $x_0^*$ be a weak$^*$ cluster point of $(\sum_{k=1}^n x_k^*)_n$. Necessarily $\langle x_0^* , x_n\rangle =1 $ for all $n\ge 1$, but $x_0^*$ is not in the normed closed span of $(x_n^*)$ because $(x_n^*)_{n=1}^\infty$ is basic, and hence $(x_n^*)_{n=0}^\infty$ is basic and not biorthogonal to any sequence in $X$. # It is inconvenient right now for me to look for a reference for this result. It certainly is in Ivan Singer's book on bases; probably in volume one. It is almost proved in Albiac-Kalton but not stated there.<|endoftext|> TITLE: Does viewing an Eisenstein series as a sum over cusps explain the antagonism between Eisenstein serieses and cusp forms? QUESTION [5 upvotes]: I'm trying to understand the relationship between various aspects of the concept of "Eisenstein series" (as discussed for example in Diamond & Shurman's "A First Course in Modular Forms"), in particular: A lowest-level Eisenstein series can be seen as a sum over cusps; that is, there's a bijection given by matching each term in the summation to the cusp at which it blows up; and moreover, this way of indexing the terms in the summation has the conceptual advantage of automatically giving a good normalization of the series without artificial introduction of a normalizing factor. Eisenstein serieses span the Petersson-orthogonal complement of the ideal of cusp forms (the "antagonism" that I referred to). Thus I want to know: Is there a direct connection from the "sum over cusps" aspect to the "orthogonality to cusp forms" aspect here, that I can re-use in other situations where the role played by the "cusps" is taken over by some sort of "generalized cusps"? (This seems to me more of an entry-level question than a research-level one but I didn't have luck with it on lower-level forums. I'm open to (hearing) suggestions about possible better places for discussion of such a question, if there are any.) In repsonse to comments: To try to clear up any miscommunication about what I mean by "cusp" here: a cusp is a point where two arcs meet at a zero-degree angle, so joining into a single "direction-reversing arc". The arcs in this case are the mirror arcs of the (2,3,infinity) Coxeter group, pictured via the Poincare upper half-plane or the Poincare disk; thus the cusps are essentially the extended rational numbers on the boundary. A modular form can be thought of as a function on the upper half-plane with a certain automorphicity property, and as a special case of this a lowest-level Eisenstein series is a sum over the cusps, with the term corresponding to a given cusp blowing up at that cusp; thus a term with cx+d in the denominator blows up at the extended rational number x = -d/c. I'm using this (pretty reasonable and etymologically prior) meaning of "cusp" because it facilitates the description of the sort of conjectural pattern that I'm asking about: There's a Lie group G with a discrete subgroup L, and a G-space X with a G-orbit A in X and an L-orbit C in the closure of A. L-automorphic forms (wrt some automorphy factor) on A are considered, and the ones vanishing at the C-points are called the "C-forms". And forms that are "complementary" to the C-forms are created by summing over the C-points with the term corresponding to a C-point blowing up at that C-point. Maybe this isn't quite the right pattern, or maybe the idea of looking for such a pattern at all is wrongheaded, but in any case my question is, roughly, whether there's some interesting pattern like this, generalizing the example of lowest-level Eisenstein serieses, which are modular forms "complementary" to the cusp forms and created by summing over the cusps. There's still the issue of exactly what "complementary" should mean here, as from Paul Garrett's comment it seems that my first stab at it was too naive; I may have garbled something that I read somewhere. Anyway, I have lots more follow-up questions, especially relating to Paul Garrett's comment, but I'll try to post those in a separate comment or comments, as it's already taking me long enough to write this. [I should perhaps mention that although I'm not very clear on this forum's posting guidelines, I'm clear enough on them to know that I don't like them; in particular I'm not at all sympathetic to the official "no discussion, best answer floats to the top" philosophy of the forum. I'd be interested to know of possible good places for actual discussion of this sort of topic, where "good place for discussion" doesn't just mean "discussion is officially permitted".] REPLY [4 votes]: I'd say the approximate answer to the question in the title is "not exactly, but partly". But I'd also claim that clarity here is made considerably more difficult by the context in which the question is posed, and in which, apparently, an answer is desired to be. That is, although for elliptic modular forms we can draw attractive pictures of fundamental domains, and although there are compactifications (due to Satake, Bailey-Borel, Borel-Serre, et al) in various generality, and although the analogy with the elliptic modular case is attractive, for understanding decompositions of various spaces of automorphic forms it seems (by this year...) that this viewpoint is not generally the most efficient. Namely, a more group-theoretic description of automorphic forms seems (for many purposes) to be more direct, and to show the how/why of "orthogonality" [sic] of cuspforms and Eisenstein series (noting that this needs qualification). Even at an elementary level, the holomorphic Poincare series, also written as "sums over cusps" (literally $\Gamma_\infty\backslash\Gamma$) construct elliptic modular forms that are easily shown to be cuspforms. So it's not just the indexing set $\{\hbox{cusps}\}\approx \Gamma_\infty\backslash\Gamma$ (given by $\gamma\cdot\infty\leftarrow \gamma$). I would claim that the happy entry-level notion of (holomorphic) cuspform, while indeed conveniently straightforward for elliptic modular forms, gives several misleading impressions about how things work more generally, both outside the holomorphic class, and on larger groups, whether situations like $Sp_n(\mathbb Z)\subset Sp_n(\mathbb R)$ where there are holomorphic modular forms, or $SL_n(\mathbb Z)\backslash SL_n(\mathbb R)$ where (for $n>2$) there are "waveforms", nothing holomorphic, (although "cohomological" things exist...). Specifically, somewhat in contrast to the holomorphic case, more generally a/the "constant term" is the zeroth Fourier component, and is a function, not just a number. The condition of holomorphy often (though not always) constrains that function completely. Without such a constraint, that function can be almost anything (invariant under the unipotent radical along which the Fourier expansion is taken...). In particular, the condition for a "cuspform" is somewhat more complicated, since it's no longer the vanishing of a number, but identical-vanishing of a function. (The Gelfand condition.) This vanishing is perhaps most sanely characterized distributionally as "vanishing when integrated against any (unipotent-radical-invariant) test function". When the latter are "automorphized" (summed over the proper analogue of $\Gamma_\infty\backslash\Gamma$) they produce pseudo-Eisenstein series, which are in $L^2$, and in fact are test functions on the quotients $\Gamma\backslash G$. That is, the condition of being a cuspform is literally orthogonality to pseudo-Eisenstein series (with test-function data). (Then, generally, pseudo-Eisenstein series themselves are spectrally decomposed in terms of genuine Eisenstein series.) Unsurprisingly, since test functions are never holomorphic, such a discussion falls outside of the world of holomorphic things. In any case, I'd say that the "cuspform" condition is best generally understood in terms of the Gelfand condition, rather than "vanishing at a cusp", which becomes progressively incorrect in general.<|endoftext|> TITLE: $id:A\to A^{op}$ is completely positive iff $A$ is abelian QUESTION [6 upvotes]: Let $A$ be a $C^*$-algebra and $A^{op}$ it's opposite $C^*$-algebra. Let $id:A\to A^{op}$ be the identity map. $id$ is positive. The claim is: $id$ is completely positive iff $A$ is abelian. I need this statement for further studies. The direction $\Leftarrow$ is easier to understand (I hope that everything is correct): If $A$ is abelian, then $A^{op}$ is abelian too. One can identify $A^{op}$ with $C_0(X)$ for a locally compact Hausdorff space $X$ and consider $id$ as a map $A\to C_0(X)$. This map is positive, hence completely positive (it's a standard fact about completely positive maps). It follows that $id:A\to A^{op}$ is completely positive. But I'm stuck to prove $\Rightarrow$. Can anybody give me some hints or does anybody know good references? It will be greatly appreciated. REPLY [5 votes]: In the unital case, it is a result of Choi that a unital 2-positive map $f:A\to B$ is a $*$-homomorphism if and only if $f(a^2) = f(a)^2$ for all self-adjoint $a\in A$. In the present case, this implies that if $\mathrm{id}:A\to A^{\mathrm{op}}$ is 2-positive, then it is a $*$-homomorphism, i.e. $\mathrm{id}(ab)=\mathrm{id}(a)\mathrm{id}(b)$, which means that $ab=ba$, so that $A$ is abelian. By the way, the $\Leftarrow$ direction does not need Gelfand duality. $A$ being abelian means precisely that $A=A^{\mathrm{op}}$, and $\mathrm{id}:A\to A$ is trivially completely positive. Conclusion: If $A$ is unital, then $A$ is abelian if and only if $\mathrm{id}:A\to A^{\mathrm{op}}$ is 2-positive. I don't know whether the above argument can be generalized to cover the non-unital case as well. One way to go about this would be to show that the extension of $\mathrm{id}:A\to A^{\mathrm{op}}$ to the unitization is still 2-positive, or otherwise to go through Choi's arguments and see to what extent they actually require unitality.<|endoftext|> TITLE: Vector bundles with exactly one nonzero SW-class QUESTION [14 upvotes]: I am interested in seeing examples of a space $X$ (preferably a closed smooth manifold, but any finite-dimensional CW-complex would also be of interest) with a vector bundle $\xi\colon E \to X$ on it, so that there is exactly one index $i$ with $w_i(\xi) \neq 0$, and $i$ is bigger than $8$. Here are some remarks: (1) First, this could only happen if $i$ is a power of 2: As a module over the Steenrod algebra, $H^{\ast}(BO;\mathbb F_2) = \mathbb F_2[w_1,w_2,w_3,\dots]$ is generated by $w_{2^k}$, so the first nonzero SW-class is always of degree $2^k$ (this is also an exercixe in Milnor-Stasheff). (2) For $i=1,2,4,8$ one could take $S^1 = \mathbb RP^1, S^2 = \mathbb CP^1, S^4 = \mathbb HP^1$ and $S^8 = \mathbb OP^1$ with the canoncial bundles on these spaces. (3) Beyond dimension 8, a sphere (or even a connected sum of products of spheres) does not give rise to any such bundle. This can be seen from analyzing $$w\colon \tilde{KO}(S^{d_1} \times \dots \times S^{d_r}) \to H^{\ast}(S^{d_1} \times \dots \times S^{d_r};\mathbb F_2),$$ the main input for understanding this map is Adams' Hopf invariant one theorem. REPLY [12 votes]: For $i = 2^k$ you can get an example with $X = \Bbb{RP}^{m}$ for any $m \geq 2^k$. If $L$ is the canonical line bundle on $\Bbb{RP}^{m}$ then let $E = \bigoplus_{i=1}^{2^k} L$ be the a sum of several copies of it. Then it has total Stiefel-Whitney class $$ w(E) = \prod_{i=1}^{2^k} w(L) = (1+x)^{2^k} = 1 + x^{2^k}, $$ where $x$ is the generator of $H^1(\Bbb{RP}^{2^k};\Bbb F_2)$.<|endoftext|> TITLE: Infiniteness of the Galois cohomology over a number field with coefficients in a finite Galois module QUESTION [13 upvotes]: Let $k$ be a number field and $M$ be a nonzero finite discrete $\mathrm{Gal}(\bar k/k)$-module. Is it true that $H^1(k,M)$ is infinite? This would complete the answer of Daniel Loughran. There is a comment of nfdc to that answer, but I think that this question deserves an answer rather than just a comment! REPLY [2 votes]: Let $k$ be a number field. Let $M$ be a nonzero finite $\mathrm{Gal}({\bar k}/k)$-module, i.e., a nonzero discrete finite abelian group with a continuous action of $\mathrm{Gal}({\bar k}/k)$. Theorem. The group $H^1(k,M)$ is infinite. In this proof I use Chebotarev's theorem (of course!) and a finiteness/surjectivity result from Sansuc's paper. Let $M'=\mathrm{Hom}(M,\mathbb{G}_{m,k})$ denote the Cartier dual of $M$. The action of $\mathrm{Gal}({\bar k}/k)$ on $M'$ defines a continuous homomorphism $\varphi\colon \mathrm{Gal}({\bar k}/k)\to\mathrm{Aut}\, M'$. We denote by $\Gamma$ the image of $\varphi$ and by $K\subset {\bar k}$ the subfield corresponding to $\ker\varphi$, then $\Gamma=\mathrm{Gal}(K/k)$. For a finite set $S$ of places of $k$ we consider the localization homomorphism $$ \lambda_S\colon H^1(k,M)\to\prod_{v\in S} H^1(k_v,M), $$ where $k_v$ denotes the completion of $k$ at $v$. We set $$Ч_S=\mathrm{coker}\,\lambda_S$$ ( Ч is to be read "Cheh" as in "Chebotarev"). Let $S_0$ denote the set of places $v$ of $k$ for which a decomposition group $\Gamma_v\subset \Gamma$ (defined up to conjugation) is not cyclic. All $v\in S_0$ are ramified in $K$. Lemma. For any finite set $S$ of places of $k$, the canonical surjective map $$ Ч_S\to Ч_{S\cap S_0} $$ is bijective. See Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80, the formula after Lemma 1.5. It follows from the lemma that if $S\cap S_0=\emptyset$, then $Ч_S=0$ and hence, the homomorphism $\lambda_S$ is surjective in this case (see also Theorem 9.2.3(vii) from the book by Neukirch, Schmidt, and Wingberg "Cohomology of number fields"). Proof of the theorem. By the Chebotarev density theorem, for any natural $n$ there exists a finite set $S$ of cardinality $n$ consisting of finite places $v$ of $k$ with $\Gamma_v=1$. Clearly $S\cap S_0=\emptyset$, hence the localization homomorphism $\lambda_S$ is surjective. For $v\in S$, the local Galois group $\mathrm{Gal}({\bar k}_v/k_v)$ acts on $M$ trivially. Since $\mathrm{Gal}({\bar k}_v/k_v)$ has $\widehat{\mathbb{Z}}$ as a quotient group, we have $$\# H^1(k_v,M)=\#\mathrm{Hom}(\mathrm{Gal}({\bar k}_v/k_v),M)\ge \#\mathrm{Hom}(\widehat{\mathbb{Z}},M)=\# M.$$ Since $\lambda_S$ is surjective, $$\mathrm{Card}\, H^1(k,M)\ge \# \prod_{v\in S} H^1(k_v,M)\ge(\#M)^n.$$ Since this is true for any natural $n$, and $\# M>1$, we conclude that $H^1(k,M)$ is infinite.<|endoftext|> TITLE: Gerbes on the multiplicative group QUESTION [8 upvotes]: Let $k$ be an arbitrary field with absolute Galois group $\Gamma$. The group $\text{Hom}(\Gamma,\mathbb{Q}/\mathbb{Z})$ injects into $H^2(\mathbb{A}^1 \setminus \{ 0 \},\mathbb{G}_m)$, as one can see e.g. by computing $\text{Ext}^2(\mathbb{G}_m,\mathbb{G}_m) = H^2(\Gamma,\mathbb{Z})$. If such a gerbe coming from $\Gamma \to \mathbb{Q}/\mathbb{Z}$ extends to $\mathbb{A}^1$, is it necessarily trivial? It's not clear to me because there are some weird gerbes on $\mathbb{A}^1$ if $k$ is not perfect. REPLY [3 votes]: I believe the answer is "yes, such a gerbe is necessarily trivial". We first note that for any field $K$ and open subscheme $U$ of $\mathbb{A}^{1}_{K}$, the inclusion $\operatorname{Br}(U) \subseteq \mathrm{H}_{et}^{2}(U,\mathbb{G}_{m}) $ is an isomorphism. Also the coboundary $\operatorname{Hom}(\Gamma,\mathbb{Q}/\mathbb{Z}) \simeq \mathrm{H}^{1}(\Gamma,\mathbb{Q}/\mathbb{Z}) \to \mathrm{H}^{2}(\Gamma,\mathbb{Z})$ is an isomorphism since $\mathrm{H}^{i}(\Gamma,\mathbb{Q}) = 0$ for $i \ge 1$. The Leray spectral sequence for $\mathbb{A}^{1}_{k} \to \operatorname{Spec} k$ is \begin{align} \mathrm{E}_{2}^{p,q} = \mathrm{H}^{p}(\Gamma,\mathrm{H}_{et}^{q}(\mathbb{A}^{1}_{k^{\mathrm{sep}}},\mathbb{G}_{m})) \implies \mathrm{H}_{et}^{p+q}(\mathbb{A}^{1}_{k},\mathbb{G}_{m}) \end{align} with differentials $\mathrm{E}_{2}^{p,q} \to \mathrm{E}_{2}^{p+2,q-1}$. This and the analogous spectral sequence for $\mathbb{A}^{1}_{k} \setminus \{0\} \to \operatorname{Spec} k$ gives a commutative diagram $\require{AMScd}$ \begin{CD} 0 @>>> \operatorname{Br}(k) @>\xi_{1}>> \operatorname{Br}(\mathbb{A}^{1}_{k}) @>\xi_{2}>> \mathrm{H}^{0}(\Gamma,\operatorname{Br}(\mathbb{A}^{1}_{k^{\mathrm{sep}}})) \\ @. @V\rho_{1}VV @V\rho_{2}VV @V\rho_{3}VV \\ 0 @>>> \operatorname{Br}(k) \oplus \mathrm{H}^{2}(\Gamma,\mathbb{Z}) @>>\xi_{1}'> \operatorname{Br}(\mathbb{A}^{1}_{k} \setminus \{0\}) @>>\xi_{2}'> \mathrm{H}^{0}(\Gamma,\operatorname{Br}(\mathbb{A}^{1}_{k^{\mathrm{sep}}} \setminus \{0\})) \\ \end{CD} where the $\rho_{i}$ are induced by restriction along the open immersion $\mathbb{A}^{1}_{k} \setminus \{0\} \subset \mathbb{A}^{1}_{k}$. Here both rows are exact because $\operatorname{Pic}(\mathbb{A}^{1}_{k^{\mathrm{sep}}}) = 0$ and $\operatorname{Pic}(\mathbb{A}^{1}_{k^{\mathrm{sep}}} \setminus \{0\}) = 0$ (so that $\mathrm{E}_{2}^{p,1} = 0$ for $p \ge 0$ for both spectral sequences). With respect to the above commutative diagram, the question may be rephrased as follows: Suppose $\alpha_{1}' \in \mathrm{H}^{2}(\Gamma,\mathbb{Z})$ is such that $\xi_{1}'(\alpha_{1}') = \rho_{2}(\alpha_{2})$ for some $\alpha_{2} \in \operatorname{Br}(\mathbb{A}^{1}_{k})$. Is $\xi_{1}'(\alpha_{1}')$ necessarily $0$? For such $\alpha_{1}',\alpha_{2}$ we have $\rho_{3}(\xi_{2}(\alpha_{2})) = \xi_{2}'(\rho_{2}(\alpha_{2})) = \xi_{2}'(\xi_{1}'(\alpha_{1}')) = 0$. Since $\mathbb{A}_{k}^{1}$ and $\mathbb{A}_{k^{\mathrm{sep}}}^{1}$ are regular Noetherian (and also since $\mathrm{H}^{0}(\Gamma,-)$ is left exact), the restriction maps $\rho_{2}$ (and hence also $\rho_{1})$ and $\rho_{3}$ are injective. Thus $\xi_{2}(\alpha_{2}) = 0$, thus there exists some $\alpha_{1} \in \operatorname{Br}(k)$ such that $\xi_{1}(\alpha_{1}) = \alpha_{2}$; then $\xi_{1}'(\rho_{1}(\alpha_{1})) = \rho_{2}(\xi_{1}(\alpha_{1})) = \rho_{2}(\alpha_{2}) = \xi_{1}'(\alpha_{1}')$, but injectivity of $\xi_{1}'$ implies $\alpha_{1}' = \rho_{1}(\alpha_{1})$; thus both $\alpha_{1} = 0$ and $\alpha_{1}' = 0$ since $\alpha_{1}',\rho_{1}(\alpha_{1})$ are in different summands of $\operatorname{Br}(k) \oplus \mathrm{H}^{2}(\Gamma,\mathbb{Z})$; thus $\xi_{1}'(\alpha_{1}') = 0$ as well.<|endoftext|> TITLE: Cohomology of tangent sheaf of a singular hypersurface QUESTION [7 upvotes]: Let $X\subset\mathbb{P}^n$ be a hypersurface singular at finitely many points $p_i\in X$. We may assume that $X$ has ordinary singularities at the $p_i$'s. Does there exists a formula, perhaps in terms of $deg(X)$ of the multiplicities of $X$ at the $p_i$'s, for the dimension of $H^1(X,T_X)$, where $T_X = \mathcal{H}om_{\mathcal{O}_X}(\Omega_X,\mathcal{O}_X)$ ? If not, does anyone know a good way to proceed in order to compute $h^1(X,T_X)$ ? REPLY [10 votes]: Put $d:=\deg(X)$. From the exact sequence $$0\rightarrow \mathcal{O}_X(-d)\rightarrow \Omega ^1_{\mathbb{P}^n|X}\rightarrow \Omega ^1_X\rightarrow 0$$you get an exact sequence $\ 0\rightarrow T_X\rightarrow T_{\mathbb{P}^n|X}\rightarrow \mathcal{O}_X(d)\rightarrow \mathcal{E}xt^1(\Omega ^1_X,\mathcal{O}_X)\rightarrow 0$. From this you find easily $H^i(X,T_X)=0$ for $i\geq 2$. The sheaf $\mathcal{T}^1_X:=\mathcal{E}xt^1(\Omega ^1_X,\mathcal{O}_X)$ is a skyscraper sheaf concentrated at the $p_i$; its rank at $p_i$ is the Tyurina number $\tau _{p_i}$, which is easily computed in terms of a local equation for $X$. Taking Euler-Poincaré characteristics you get $$h^1(T_X)-h^0(T_X)= -\chi (T_{\mathbb{P}^n|X})+\chi (\mathcal{O}_X(d))-\sum \tau _{p_i}\ ,$$where each term on the right hand side is easily computable. If $X$ has no infinitesimal automorphisms, $h^0(T_X)=0$; I believe this is the case if $d\geq 3$ and $X$ is not a cone. Edit : As pointed out by @Benjamin Tighe, my assertion that $H^{i}(X,T_X)=0$ for $i\geq 2$ is incorrect. Using $H^1(T_{\mathbb{P}^n|X})=0$, one should add to the right hand side of the formula $\dim \operatorname{Coker}\bigl( H^0(\mathscr{O}_{\mathbb{P}^n}(d))\rightarrow H^0(\mathcal{T}^1_X)\bigr)$. Even when the singular points are ordinary double points, this is delicate to compute: it is zero iff the singular points impose independent conditions on hypersurfaces of degree $d$.<|endoftext|> TITLE: Noetherian Rings in Constructive Mathematics QUESTION [8 upvotes]: These definitions are largely from pages 92-93 of Ingo's notes. All rings are commutative with 1. I'm interested in understanding the extent to which discussion of Noetherian rings can be carried over into constructive settings. Consider a standard definition of a Noetherian ring: A ring $R$ is Noetherian if for every ascending chain of ideals stabilizes; given $I_1 \subseteq I_2 \subseteq \dotsi$, there exists $n$ so that $I_n = I_m$ for all $m \ge n$. This definition is constructively untenable; the claim that $\mathbb Z$ (or even $\mathbb Z / n \mathbb Z$) is Noetherian implies LPO. This definition will also run into problems if dependent choice fails. We can solve the issue with dependent choice by introducing ascending processes. Let $M = (M,\preceq)$ be a poset. An ascending process $(S_n)$ with values in $M$ is a sequence $S_0,S_1,S_2,\dotsc$ of inhabited subsets of $M$, such that for each $x \in S_i$, there exists $y \in S_{i+1}$ so that $x \preceq y$. $(S_n)$ is said to halt if $S_n \cap S_{n+1}$ is inhabited for some $n$. We say that $M$ satisfies the ascending process condition of every ascending process with values in $M$ halts. We say that $M$ satisfies the weak ascending chain condition if for every every ascending chain $(x_i)$, we have $x_{i+1} \preceq x_i$ for some $i$. We say that $R$ is Process-Noetherian if the poset of finitely generated ideals in $R$ satisfies the ascending process condition. Let $R$ be a ring. Let $M \subseteq \mathcal P(R)$ be the poset of ideals, and let $M_F \subseteq M$ be the poset of finitely generated ideals. Assuming dependent choice and excluded middle, the following candidate definitions of Noetherianity are easily shown to be equivalent: $M = M_F$ $M$ satisfies the weak ascending chain condition $M_F$ satisfies the weak ascending chain condition $M$ satisfies the ascending process condition $M_F$ satisfies the ascending process condition ($R$ is process-Noetherian) For every sequence $(x_i)_{i \ge 1}$ of elements of $R$, there exists $n \ge 0$ so that $x_{n+1} \in (x_1,\dotsc,x_n)$. If excluded middle fails, then properties $1,2,4$ all fail when $R = \mathbb Z$, and so are completely hopeless. We have $5 \Rightarrow 3 \Rightarrow 6$ unconditionally, and it doesn't seem reasonable to expect that $3 \Rightarrow 5$ without dependent choice, and so $3$ and $6$ seem unlikely work in the absence of dependent choice. Question 1: Is the implication $3 \Rightarrow 6$ reversible? What if we assume dependent choice? Question 2: Can we constructively prove the analogue of Hilbert's basis theorem using any of the three not-completely-hopeless definitions? What changes if we assume dependent choice? REPLY [6 votes]: I cannot answer off the top of my head but but you should definitely consult the literature: A Course in Constructive Algebra, by Mines, Richman and Ruitenburg has a chapter on Notherian rings. Commutative Algebra: Constructive Methods by Lomardi and Quitté is also worth looking at.<|endoftext|> TITLE: A clarification on pointed Gromov-Hausdorff convergence QUESTION [9 upvotes]: According to Burago-Burago-Ivanov, one says that the sequence of pointed metric spaces $(X_n,d_n,p_n)$ GH-converges to $(X,d,p)$ if for every $R>0,\varepsilon>0$, there exists a $N$ such that for $n\geq N$ one can find $f:B_{X_n}(p_n,R)\to X$ such that : $f(p_n)=p$, for every $x,y\in B_{X_n}(p_n,R)$, $|d_{X_n}(x,y)-d_X(f(x),f(y))|\leq\varepsilon$, the $\varepsilon$-neighborhood of the image of $f$ contains $B_X(p,R-\varepsilon)$. Before stating this definition, the authors give an informal one : "Roughly speaking, a sequence $\{X_n\}$ of metric spaces converges to a space $X$ if for every $r > 0$ the balls of radius $r$ in $X_n$ centered at some fixed points converge (as compact metric spaces) to a ball of radius $r$ in $X$." Then they go on to say : The actual definition (Definition 8.1.1 below) is more complicated, but in most cases it is equivalent to this description. My question is : Under which assumptions is the informal definition equivalent to the rigorous one ? I know that the metric on the limit space being intrinsic should play a role, as exemplified by the sequence $(1-\tfrac{1}{n})\mathbb{Z}$, which GH-converges to $\mathbb{Z}$ according to the formal definition, but fails to do so for the informal one. Actually Ex 8.1.3 in Burago-Burago-Ivanov show that the formal definition implies that for every $R>0$, $B_{X_n}(p_n,R)$ GH-converges to $B_X(p,R)$ if $X$ is a length space. I tried to prove a converse to this statement but was stuck at one point, building maps $f$ which satisfy conditions (2) and (3) of the definition is not complicated when $(X,d,p)$ is a length space. However I am not sure if one can the points $f(p_n)$ from staying too far away from $p$. REPLY [4 votes]: In other words you want to remove the condition $f(p_n)=p$. In this case, instead of sequence $X_n$ one can take a fixed space $Y$ which is not isometric to $X$, but such that for some points $p\in X$ and $q\in Y$ the balls $B(p,R)_X$ and $B(q,R)_Y$ are isometric for any radius $R<\infty$. One example is given here, let me describe an other way to think about it. Imagine that you cut the vertical side of right half plane in the $\ell_\infty$-plane using the following pattern. Equip the obtained paper with induced intrinsic metric; this way you get space $X$. Now you do the same for the positive quadrant to get space $Y$. You can take the origin as the marked points in both spaces.<|endoftext|> TITLE: Algebro-geometric version of {vector fields} $\longleftrightarrow$ {flows} correspondence? QUESTION [9 upvotes]: Main Question: What Is the correpondence between flows and vector fields in algebraic geometry? Here is a more precise statement could be an answer If it was true (I have no idea it is): "Proposition": Let $X$ be a nice enough (intentionally ambiguous) scheme over a field $k$ with tangent sheaf $\mathcal{T}_X$. For every point $x \in X$ There is a 1-1 correpsodence between germs of vector fields through $x$ and "germs" of formal flows. More precisely for every $v \in \mathcal{T}_{X,x}$ there corresponds a (continuous) morphism of complete local rings $\phi_v \in Hom_{cont}(\widehat{\mathcal{O}_{X,x}},\widehat{\mathcal{O}_{X,x} \otimes_k k[[t]]})$ and vice versa. This "Proposition" was mentioned to me as a brief comment by a very respected mathematician (who's name I won't mention since I'm not completely convinced that the above is a fiathfull interpretation of his original statement). Unfortunately he didn't have time to elaborate on this and couldn't point me to a relevant reference. Reference request: What are some good sources to turn to considering questions about formal geometry? Specifically the formulation of the classical theory of differential equations (of which this question is a particular case) in formal geometric terms. Even the "proposition" above isn't really precise since I'm not sure what meaning I should give to the tensor product (some kind of completed tensor product perhaps). Which leads me to the following minor question. Consider the category whose objects are topological rings $A$ whose topology can be generated by an $I$-adic filtration for some ideal $I \subset A$ and let the morphisms be continuous homomorphisms of rings. (minor) Question: For arbitrary $A$ and $B$ in this category what is pushout of $A \leftarrow \mathbb{Z} \to B$? should this be the correct interpretation of the tensor product above? REPLY [10 votes]: You need a characteristic zero assumption, and you need some additional axioms on the homorphism to make it a bijection. The map from derivations $D$ to homomorphisms sends a function $y \in \mathcal O_{X,x}$ to $$e^{t D} y = y + (Dy) t + (D^2 y) t^2/2 + (D^3 y) t^3/6 + \dots$$ The inverse map is just going to send a homomorphism $f$ to the derivation that sends $y$ to the coefficient of $t$ in $f(y)$. However we need some axioms to ensure this is actually a derivation and that its exponential is $f$. You can find the right axioms by thinking of a flow as a (formal) group action by the (formal) group of time translations. The axioms are: Composing $f$ with the projection from $\mathcal O_{X,x}[[t]]$ (which is equivalent to the completed tensor product you wrote down) to $\mathcal O_{X,x}$ should give the identity. Composing $f$ with the map $\mathcal O_{X,x}[[t]] \to \mathcal O_{X,x}[[t_1,t_2]]$ that sends $t$ to $t_1+t_2$ should equal applying $f$ twice, once adding the variable $t_1$ and once adding the variable $t_2$. These axioms immediately imply that the first coefficient of $f(y)$ is $y$, the second is a derivation of $y$, and the other ones are inductively given by powers of the derivation in this way.<|endoftext|> TITLE: Holomorphic vector bundles over $\mathbb{C}^{n}\setminus 0$ QUESTION [8 upvotes]: Is it true that every holomorphic vector bundle over $\mathbb{C}^{n}\setminus 0$ is trivial? If not true, how can one construct a counterexample? And just a small note here (wrong): For $n\leq 2$, we can push the bundle $\mathcal{F}$ over $\mathbb{C}^{n}\setminus 0$ to be defined on $\mathbb{C}^{n}$, denoted by $\bar{\mathcal{F}}$. By taking a double dual $\bar{\mathcal{F}}^{**}$ from which we can get reflexive sheaf which is isomorphic to $\mathcal{F}$ away from $0$. For a coherent analytic reflexive sheaf, we know that the set where it fails to be locally free has codimension $\geq 3$. So for $n\leq 2$, $\bar{\mathcal{F}}^{**}$ is indeed a bundle. But for $n\geq3$, I have no clue. This note turns to be wrong because I mistook $\bar{\mathcal{F}}$ to be a coherent sheaf, while the behavior of $\mathcal{F}$ near 0 is unknown. A further comment here: For the line bundle case, based on the reference provided in the comments by pgraf, for $q\geq 1$, $H^{q}(\mathbb{C}^{n}\setminus 0, \mathcal{O})\neq 0$ iff $q=n-1$ which implies there exists nontrivial line bundle on $\mathbb{C}^{2}\setminus 0$ which cannot be extended to be on $\mathbb{C}^{2}$. In other cases, all holomorphic line bundles over $\mathbb{C}^{n}$ for $n\geq 3$ and $n=1$ would be trivial. Note when $n=1$, $\mathbb{C}^{*}$ is a Stein manifold, or we can directly solve $\bar{\partial}$ equation by integration. REPLY [11 votes]: For $n\geq 3$, there exists such non-trivial bundles. For example, take the bundle on $\mathbb{C}^n-\{0\}$ given by the quotient of $\mathcal{O}^n$ by the subbundle generated by the vector $(x_1,\ldots,x_n)$, where the $x_i$s are the coordinate functions. Then, for $n\geq 3$, this bundle is not trivial. The non-triviality will essentially follow from an argument like Hartog's theorem.<|endoftext|> TITLE: Distributional equation X+Y=2X QUESTION [7 upvotes]: Let $X$ be a positive real-valued random variable. Let $Y$ be an independent copy of $X$ and assume that the equality $X+Y=2X$ holds in distribution. Does this imply that $X$ is constant? REPLY [13 votes]: Yes, it does. As the random variable is positive, consider the Laplace transform $f(\lambda)=E \exp(-\lambda X)$ (instead of the Fourier one). Then, one has $f^2(\lambda)= f(2\lambda)$. Taking the logarithm, one gets $2\log f(\lambda)=\log f(2\lambda)$, and for the function $g(\lambda)=\frac{\log f(\lambda)}{\lambda}$ this implies $g(2\lambda)=g(\lambda)$. Finally, pass to the limit as $\lambda\to\infty$. Then $g(\lambda)$ tends to the essential infimum of the law of $X$. On the other hand, $g(e^s)$ is $\log 2$-periodic; as it has a limit as $s\to\infty$, this is a constant function. Hence, $g$ is a constant function, and thus $X$ is constant almost surely.<|endoftext|> TITLE: On progress towards inverse Galois problem over rationals QUESTION [5 upvotes]: I have heard that Progress towards the Inverse Galois problem over $\mathbb{Q}$ is very well documented for sporadic groups ($M_{23}$ is the only case open) and for $PSL_n(q)$. From where I can read more about these results in details. Like papers, articles, books, etc. REPLY [11 votes]: All the sporadic groups except for $M_{23}$ and $M_{24}$ are realized over $\mathbb{Q}$ using the rigidity criterion. This technique is explained in all three main textbooks on inverse Galois theory: Jean-Pierre Serre - Topics in Galois theory Helmut Volklein - Groups as Galois Groups Gunter Malle & B. H. Matzat - Inverse Galois theory For example Volklein explains some of the details applied to the case of $M_{12}$ and the Monster group $M$, while Malle & Matzat present simplified proofs of much more sporadic groups. In the case of the the Monster, you might want to look at Thompson's classic paper: John G. Thompson - "Some finite groups which appear as $\mathrm{Gal} (L/K)$, where $K ⊆ \mathbb{Q}(μ_n)$" (1984) There's also this very informative answer here at MO by user Wanderer. The case of $M_{24}$, which escapes rigidity, is realized in: B. H. Matzat - "Rationality Criteria for Galois extensions" (1987) The progress on $\mathrm{PSL}_n(q)$ is less uniform. Volklein in his book covers his own regularity result in the case of $n$ even and $n \geq q$, using weak rigidity and braiding action. He also looks at the $\mathrm{PSL}_2(q)$ case, for which Serre (chapter 5) is also a good reference. There's of course plenty of other results and special cases known. Malle-Matzat is by far the more comprehensive and includes most of the references that you might need.<|endoftext|> TITLE: Is this weak form of $V=L$ (in)consistent with large cardinals? QUESTION [8 upvotes]: I have been considering a (definability-free) weak form of the constructibility axiom, which is intended to capture the coarse structure of the constructible hierarchy. This means that this weak form is intended to capture the ordinal pattern abstracted from the constructible hierarchy if the fine behavior of the ordinal values of the constructible rank is covered up. It turns out that basic consequences of the constructibility axiom remain valid in the resulting theory: Let $ZF_{\rho}^-$ be the following theory: Its language has, in addition to $\in$, an unary function symbol $\rho$, and the axioms include all axioms of $ZF^-$ ($ZF$ minus the axiom of foundation) along with replacement and separation axioms for formulas containing $\rho$. Now, add to $ZF_{\rho}^-$ the following axiom: The function symbol $\rho$ is such that $\forall x, \rho(x)$ is an ordinal. $\forall \alpha, (\rho(\alpha)=\alpha)$. $\forall x, y, (x\in y\rightarrow \rho(x)<\rho(y))$. (i.e., $\rho$ preserves membership.) $\forall \alpha\: \exists f;(f:\alpha \cup \omega\rightarrow\left\{x:\rho(x)<\alpha\right\}$ is surjective). (i.e., $|\left\{x:\rho(x)<\alpha\right\}|\leq|\alpha\cup \omega|.)$ For every set $x$, $(i)$ if $x\in V_{\omega}$, then $\rho(x)=rk(x)$, and $(ii)$ if $x\notin V_{\omega}$, then given a transitive set $T$ containing $x$ and $r:T\rightarrow T$ satisfying 1-4 above, $\rho(x) TITLE: Clarification on the weak BSD conjecture QUESTION [11 upvotes]: It is usually told that Birch and Swinnerton-Dyer developped their famous conjecture after studying the growth of the function $$ f_E(x) = \prod_{p \le x}\frac{|E(\mathbb{F}_p)|}{p} $$ as $x$ tends to $+\infty$ for elliptic curves $E$ defined over $\mathbb{Q}$, where the product is defined for the primes $p$ where $E$ has good reduction at $p$. Namely, this function should grow at the order of $$ \log(x)^r $$ when $x$ tends to $+\infty$, where $r$ is the (algebraic) rank of $E$. Question 1. Why is it natural to look at these kind of products? Nowadays, people usually state the BSD conjecture as the equality $$ r = \text{ord}_{s=1}L(E,s)\text{.} $$ Question 2. Are these two statements equivalent? REPLY [3 votes]: If $E$ is an elliptic curve over $\mathbf{Q}$, define $L_p(E,s) = \frac{1}{1-ap^{-s} + p^{1-2s}}$, where $a$ is $p-(|E_p(\mathbf{F}_p)| - 1)$. The L-function of E is defined as the product over the $L_p(E,s)$ over all primes not dividing $2\Delta$ (called “good primes”): $$ L(E,s) = \prod_{p} L_p(E,s)$$ Since we wish to look at $L(E,1)$, we notice that $\displaystyle L_p(E,1) = \frac{p}{p+1-a}$. From the definition of $a$, we see that $p+1-a=|E_p(\mathbf{F}_p)|$, and so $\displaystyle L_p(E,1) = \frac{p}{|E_p(\mathbf{F}_p)|}$. Taking the product over all good primes $p$ shows that $$L(E,1) = \prod_{p}\frac{p}{|E_p(\mathbf{F}_p)|}.$$<|endoftext|> TITLE: What are Motivic homotopy types? QUESTION [8 upvotes]: There are suggestions that says that Grothendieck developed (in some sense) a theory of Motivic homotopy types or at least named it. I would like to know the reference in which Grothendieck did it, and some references on subsequently development. REPLY [13 votes]: Not sure about later developments, but the idea is mentioned in a famous passage of Grothendieck's Récoltes et Semailles. I quote from Roy Lisker's translation: Thus, the motive presents itself as the deepest "form invariant" which one has been able to associate up to the present moment with an algebraic variety, setting aside its "motivic fundamental group". For me both invariants represent the "shadows" projected by a "motivic homotopy type" which remains to be discovered (and about which I say a few things in the footnote: "The tower of scaffoldings- or tools and vision" (R&S IV, #178, see scafolding 5 (Motives), and in particular page 1214)). It is the latter object which appears to me to be the most perfect incarnation of the elusive intuition of "arithmetic form" ( or "motivic"), of an arbitrary algebraic variety.<|endoftext|> TITLE: Product of limit $\sigma$-algebras QUESTION [7 upvotes]: Let $X$ and $Y$ be Polish (i.e. Borel subsets of separable completely metrizable) spaces. For a Polish space $Z$, let $\mathscr{S}(Z)$ denote the limit $\sigma$-algebra on $Z$, i.e. the smallest $\sigma$-algebra on $Z$ that contains the Borel subsets of $Z$ and is closed under the Suslin operation. Question: Does $\mathscr{S}(X) \otimes \mathscr{S}(Y) = \mathscr{S}(X \times Y)$? REPLY [4 votes]: Every member of your limit $\sigma$-algebra is both Lebesgue measurable and has the Baire property (for a proof, see section 29.B in Kechris book). A result of Mansfield and Rao implies that the universal analytic set in plane is not in the sigma algebra generated by rectangles with measurable (resp. Baire property) sides - For a short proof see Theorem 1 in this paper of Arnold Miller.<|endoftext|> TITLE: Does a Kähler manifold always admit a complete Kähler metric? QUESTION [19 upvotes]: Every smooth manifold admits a complete Riemannian metric. In fact, every Riemannian metric is conformal to a complete Riemannian metric, see this note. What about in the Kähler case? Does a Kähler manifold always admit a complete Kähler metric? Of course, every metric on a compact manifold is complete, so the question is only of interest in the non-compact case. One might hope that the proof in the aforementioned note will still be of use, but as is shown in this question, a metric conformal to a Kähler metric cannot be Kähler (with respect to the same complex structure) except in complex dimension one. The only condition that I am aware of that ensures the existence of complete Kähler metrics is that the manifold is weakly pseuodoconvex (i.e. it admits a plurisubharmonic exhaustion function), see Demailly's Complex Algebraic and Analytic Geometry, Chapter VIII, Theorem 5.2. REPLY [4 votes]: I give a related answer for the following non-compact case which we can get complete Kähler metric Take $\overline M$ be a compact Kähler manifold and $Y\subset \overline M$ be the simple normal crossing divisor and take $M=\bar M\setminus Y$ now we can define complete Kähler metric $\omega_P$ on non-compact manifold $M$ as follows Since $Y$ is simple normal crossing divisor , so it can be defined by the equation $z_1^{\alpha}\cdots z_{n_\alpha}^\alpha=0$ Take a cover for $\overline M=U_1\cup\cdots\cup U_p\cup \cdots \cup U_q$ such that $\overline{U_{p+1}}\cup\cdots \cup \overline{U_{q}}=\phi$ Let $\{η_i\}_{1≤i≤q}$ be the partition of unity subordinate to the cover $\{U_i\}_{1≤i≤q}$. Let $\omega$ be a Kähler metric on $M$ and let $C$ be a positive constant. Then for $C$ enough large, the following Kähler form is complete Kähler metric $$\omega_P=C\omega+\sum_{i=1}^p\sqrt{-1}\partial\bar\partial\left(\eta_i\log\log\frac{1}{z_1^{i}\cdots z_{n_i}^i}\right)$$ See the paper https://projecteuclid.org/download/pdf_1/euclid.jdg/1214448444 Moreover Let $X$ be a singular subvariety of the compact Kähler manifold $M$ and let $\omega$ be a Kahler $(1,1)$ form on $M$ then the Saper-form $$\omega_{Saper}=\omega-\frac{\sqrt{-1}}{2\pi}\partial\bar\partial \log(\log F)^2$$ is a complete Kähler metric on $M-X_{sing}$<|endoftext|> TITLE: Should I quit the PhD? QUESTION [13 upvotes]: I am not sure whether this is the right place to post this question. I am at the end of my seventh year. I won't have funding neither from my department nor from my advisor next year and I do not have a thesis. I can enroll as a student if I can support myself, which will be possible but not easy for me. My advisor made it clear from day one that he won't be much help to me. He said: PhD is taken not given. I started studying seriously two years ago. I could not much talk to him about what I learnt. Our conversation ended in a minute or two, if I asked immature questions. He didn't even suggest a problem. He said good students find their own problem. I went astray at times. I spent more than necessary stuck on problems. I felt like Buridan's ass: I could not decide between learning new material or sticking to the same because of my limited time. I talked to his last student and a friend of mine, over the phone for six months. He is playing the role of my advisor now. I feel good about this. If I do not earn a PhD my life will be much different. But at the same time, I am not sure how much more time I will need to get a result and get "done". Two years ago I thought I will have a result in six months. This was over-optimistic. Same thing happened a year ago. I am not sure whether continuing is overoptimistic at this point. Beside the fact that I feel lonely, and stressed, and worn... REPLY [25 votes]: I do not think you are the only one in this position. Personally I knew one 9th-year PhD, one 8th-year PhD, and one 16th-year PhD in our department. Not long ago, I talked to an 8th-year PhD who was fortunate to graduate. But he said he was so exhausted by the work that he did not want to do math anymore. He could have dropped out and done something he enjoyed in the mean time. One website you might find some help is https://academia.stackexchange.com/. Fake proofs or fake original ideas are surprisingly common if you get yourself isolated by working in a cubicle for a long time. I would encourage you to speak out your ideas to more people (conferences could be a good idea). Unfortunately, mathematicians are of measure zero density in the society. A lot of us do not socialize and cannot form an intellectual conversation outside of our working field. I am one of them as well. But life is not all about mathematics. Put away your thesis for a while, try to do something different, then look back to see if you are interested. If not, I would not suggest you to stay in PhD. Otherwise I think you might have found the answer already.<|endoftext|> TITLE: Kazhdan's property (T) vs. residual finiteness QUESTION [10 upvotes]: I have asked this question already on mathstackexchange but got no answer (see https://math.stackexchange.com/questions/1795795/kazhdans-property-t-vs-residual-finiteness) and it was suggested that I may ask it here too. There is a theorem that states that a discrete group $G$ with Kazhdan's Property $(T)$ and Property $F$ (so called factorisation property) is residually finite (see Kirchberg, Discrete groups with Kazhdan' s property T and factorization property are residually finite), i.e. Kazhdan's Property $(T)$ + Property F $\Rightarrow$ Residual finiteness. For the definitions of Kazhdan's Property $(T)$ and residual finiteness see e.g. the corresponding wiki-articles. I am wondering if some kind of "converse" is true. More precisely, I am looking for some property, let us call it Property X, such that: Residual finiteness + Property X $\Rightarrow$ Kazhdan's Property $(T)$. Maybe there is something similar in the literature? Definition for Property $F$ of the cited paper. REPLY [9 votes]: Perhaps it's worth mentioning that Property (T) seems to repel certain strengthenings of residual finiteness. An open question of Long and Reid asks: Is there an infinite finitely generated group which is LERF and has Property (T)? Recall that LERF stands for Locally Extended Residually Finite, and means that every finitely generated subgroup is closed in the profinite topology. (Residual finiteness means the trivial subgroup is closed) Long and Reid's question may have a positive answer, but the point of it is that such groups seem in any case to be extremely rare and/or difficult to construct.<|endoftext|> TITLE: Digital physics and "Gandy-like" machines QUESTION [14 upvotes]: Various physicists, famously John Wheeler, have asserted that physical information is the central object of study in physics, in the sense that an object or concept is "physically meaningful" if it makes a difference to the information that can be extracted from (actual or hypothetical) physical experiments. For example, gauge transformations are "unphysical" in this sense. Some have gone further and proposed that the Universe is "computational in essence": that, at the most fundamental level, it consists of the processing of information according to fixed rules; see e.g. the Stanford Encyclopedia of Philosophy's article "Computation in Physical Systems" and Wikipedia's article on digital physics. I am trying to understand the possible interpretation and coherence of this proposal. Robin Gandy, in "Church's Thesis and principles for mechanisms", formulated a broad model for parallel computation subject to "locality" constraints. The model subsumes Turing machines and cellular automata, and is given in terms of hereditarily finite sets, although John Byrnes and Wilfried Sieg have recast it in terms of locally finite labelled graphs, in "A graphical presentation of Gandy's parallel machines". Since the information processing in a Gandy machine is local, it seems plausible to me that this formulation is already consistent with relativity; if not, I can't imagine it being difficult to modify the definitions to make them so. The "pancomputationalist" proposal might then be formalized as claiming that the Universe is (equivalent to) a "Gandy-like" machine. My question is: Is it possible for this proposal to be consistent with current physical observations? More precisely, are there any known obstacles to the possibility that some Gandy-like machine, with certain initial conditions, could exhibit "large-scale" behaviour consistent with current physical models? I am aware of several possible objections: The Church-Turing thesis may be false, in that there could be physical systems ("hypercomputers") capable of performing computational supertasks. An example is given by Oron Shagrir and Itamar Pitowsky in "Physical Hypercomputation and the Church-Turing Thesis". Whether such a device is physically possible is obviously unknown, but if so it would refute the proposal. Bell's theorem essentially* rules out local hidden variables, so the proposed "locality" of computation may not correspond simply to locality in spacetime (*another possibility is to reject counterfactual definiteness; this position is known as superdeterminism). Even the state of a simple quantum mechanical system resides in a Hilbert space and so contains an infinite amount of classical information. That said, the amount of information which may actually be extracted from a system (and is hence "physically meaningful") is finite. In particular, there is a bound due to Jacob Bekenstein which arises when considering thermodynamics together with general relativity. It states that the entropy (or, to my understanding, "classical information") enclosed in a region of space is bounded from above by the surface area of the region. I have tagged this question as soft since it may not have a definite answer; nevertheless I would find relevant literature and discussion very useful. Some may contend that this question belongs on physics.stackexchange rather than here; however, my perspective is that the question - at least, the more precise version of it stated above - is ultimately a mathematical one. REPLY [6 votes]: This line of thought seems to be the essence of the research program of Gerard 't Hooft, as exposed in a series of papers culminating in the monograph The Cellular Automaton Interpretation of Quantum Mechanics. A cellular automaton is a system with localised, classical, discrete degrees of freedom, typically arranged in a lattice, which obey evolution equations. The evolution law for the data in every cell only depends on the data in the adjacent cells, and not on what happens at larger distances. This is a desirable form of locality, which indeed ensures that information cannot spread faster than the speed of light. The Cellular Automaton Theory assumes that, once a universal Schrödinger equation has been identified that encapsulates all conceivable phenomena in the universe (a Grand Unified Theory, or a Theory for Everything), it will feature an ontological basis that maps the system into a classical automaton. The Cellular Automaton Interpretation of quantum mechanics suggests to us what it is that we actually do when we solve a Schrödinger equation. We thought that we are following an infinite set of different worlds, each with some given amplitude, and the final events that we deduce from our calculations depend on what happens in all these worlds. This is an illusion. There is no infinity of different worlds, there is just one, but we are using the “wrong” basis to describe it, because the basis we are using is not an ontological one. So, at least according to 't Hooft, the answer to the question of the OP "Is it possible for [a local cellular automaton] to be consistent with current physical observations?" is Yes.<|endoftext|> TITLE: Homotopy groups of Moore spaces QUESTION [11 upvotes]: Is there anything known about the homotopy groups of the Moore spaces $M(\mathbb Z_m,n)$ if $m\neq 2$ and $n \geq 2$? REPLY [5 votes]: Let's consider $M(\mathbb{Z}/p^r, n)$ for large $n$, so that $\pi_{i+n}(M(\mathbb{Z}/p^r, n) \cong \pi_{i+n+1}(M(\mathbb{Z}/p^r, n+1)$ i.e. we are in the stable range. I think, the homotopy groups of $M(\mathbb{Z}/p^r, n)$ stabilizes for $n \geq 4$. Then the problem of computing the homotopy group of $M(\mathbb{Z}/p^r, n)$ is essentially computing the stable homotopy groups of a spectrum, called the $r$-th Moore spectrum and is denoted by $M_p(r)$. By a Theorem of Hopkins and Smith (see, nilpotence and stable homotopy theory II) there exists a periodic family of self-maps $$ (v)^k: \Sigma^{2kt(p-1)}M_p(r) \to M_p(r) $$ where $t$ is a function of $r$ and $k$ is any positive integer. This map has the property that the composite $$\overline{v}(k):S^{2kt(p-1)} \to \Sigma^{2kt(p-1)}M_p(r) \to M_p(r) $$ is non-trivial element in stable homotopy groups of $M_p(r)$ for all $k$. Moreover, $\overline{v}(k)$ is $p^r$-torsion for all $k$. It is also known that $t$ gets arbitrarily large as $r$ gets large. Therefore, we can conclude that $$ \pi_{n+2kt(p-1)}(M(\mathbb{Z}/p^r, n)) \neq 0 $$ for all $n \geq 4$ and $k \geq 1$ as there is an infinite family of generators $$ v(k) \in \pi_{n+2kt(p-1)}(M(\mathbb{Z}/p^r, n)) $$ which are $p^r$-torsion. I do not know if $t$ is known for all values of $r$, but is known for small values of $r$. For example at prime $2$, $t=4$ when $r=1$ and at odd primes $t=1$ when $r=1$.<|endoftext|> TITLE: Normalization of complete intersection QUESTION [6 upvotes]: Let $A$ be an integral complete local ring over a field which is complete intersection. Let $B$ be a normalization of $A$. Q. Is $B$ Gorenstein? I guess that even the normalization of Gorenstein local ring should be Gorenstein. REPLY [10 votes]: Consider $A=k[[x^3,x^2y,y^3]]\subset k[[x^3, x^2y, xy^2, y^3]]=B$. $B$ is the integral closure of $A$, $A$ is a hypersurface, but $B$ is not Gorenstein.<|endoftext|> TITLE: What are types of coalgebras that are more naturally described by cooperads? QUESTION [12 upvotes]: Some background. Let $\mathsf{C}$ be a symmetric monoidal category. An object $X \in \mathsf{C}$ has two operads "naturally" (the two constructions aren't functorial) associated to it: the operad of endomorphisms and the operad of coendomorphisms $$\mathtt{End}_X(r) = \operatorname{Mor}(X^{\otimes r}, X), \; \mathtt{coEnd}_X(r) = \operatorname{Mor}(X, X^{\otimes r}).$$ An algebra over an operad $\mathtt{P}$ is then a morphism $\mathtt{P} \to \mathtt{End}_X$, and a coalgebra over $\mathtt{P}$ is a morphism $\mathtt{P} \to \mathtt{coEnd}_X$. If one likes, everything can be done in the enriched setting, e.g. dg-modules over a field, and a coalgebra structure on $X$ is equivalent to coactions: $$\mathtt{P}(r) \otimes X \to X^{\otimes r}$$ that satisfy associativity, equivariance, unitality... relations. Often, one hears that "coalgebras are encoded by cooperads". The definition of cooperads is formally dual to the definition of operads, and a coalgebra over a cooperad $\mathtt{C}$ is given by morphisms $$X \to \mathtt{C}(r) \otimes^{\Sigma_r} X^{\otimes r}$$ satisfying relations formally dual to the relations defining an algebra over an operad. My problem. But it seems to me that, in algebra at least, most notions of coalgebras encountered in nature are more naturally seen as coalgebras over operads, rather than coalgebras over cooperads. For example, a coassociative coalgebra is more naturally seen as an operator $\Delta : C \to C \otimes C$ (so, something like $\mathtt{Ass}(2) \otimes C \to C \otimes C$), rather than a map $C \to \mathtt{Ass}^*(2) \otimes^{\Sigma_2} (C \otimes C)$. I think it's even more evident with e.g. Poisson coalgebras. IMO it's natural to think there's a cobracket and a coproduct both acting on $C$. If $\mu$ and $\lambda$ are the two generators of $\mathtt{Poiss}(2)$, this is something like $\mathtt{Poiss}(2) \otimes C \to C \otimes C$ mapping $\mu \otimes x$ to the coproduct $\Delta(x)$ of $x$ and $\lambda \otimes x$ to the cobracket $\delta(x)$). But it's not natural at all (IMO) to think of a coaction that maps $$x \in C \mapsto \mu^* \otimes \Delta(x) + \lambda^* \otimes \delta(x) \in \mathtt{Poiss}^*(2) \otimes^{\Sigma_2} (C \otimes C);$$ everything is jumbled together. Of course, when one deals with finite-type dg-modules over a field (of characteristic zero to deal with invariants vs. coinvariants), everything can be dualized and there's no difference (AFAIK) between coalgebras over $\mathtt{P}$ and coalgebras over $\mathtt{P}^*$. Question. Are there examples of coalgebra types that are more naturally given by cooperads (without requiring to look at the dual notion first)? And what about topological operads, say, where things cannot be dualized as easily? (Asked in February on math.SE) REPLY [6 votes]: This is an example of something that is natural to do from the dual point of view, but not of something that is easier to do with cooperads than operads. It is a little too long for a comment. The dual point of view is likely to be useful for defining the notion of a "divided power coalgebra". Suppose C is a cooperad. Then you may define a divided power coalgebra over C to be an object X equipped with morphisms $X \to C(n)\otimes_{\Sigma_n} X^{\otimes n}$, subject to compatibility conditions. You can equally well define divided power coalgebras over an operad $O$ in this way, via structure maps $X \to \hom(O(n), X^{\otimes n})_{\Sigma_n}$. One reason to be interested in divided power coalgebras is that the Bar construction on an algebra over an operad $O$ is not just a coalgebra over the bar construction of $O$, but a divided power coalgebra. This is discussed in the paper of Francis and Gaitsgory. It is conjectured that under some additional hypothesis there is an equivalence of homotopy categories of algebras over $O$ and divided power coalgebras over the Koszul dual of $O$. Some version of this conjecture is proved in this paper of Ching and Harper.<|endoftext|> TITLE: Motivating the existence of Canonical Bases for Representations QUESTION [7 upvotes]: In Representation Theory, the theme of the existence of a canonical basis has been explored quite a lot. I will limit myself in this question to the kind of canonical bases that arise from the Geometric Satake Correspondence (due to Mirkovic-Vilonen, followed up in works by Anderson, Kamnitzer and many others). Here, one studies certain sheaves on the Affine Grassmannian for the group G and from this geometrical setup, one gets a canonical basis for a finite dimensional irrep of the Langlands dual group $G^\vee$. Now, to someone who has studied the theory of finite dimensional representations of compact Lie Groups (Ex : can do tensor products, find branching laws etc, maybe not by the most efficient means, but by some means), is there any striking property of this theory that one can take as a hint of existence of the kind of canonical basis recalled above ? Now, let me note here that if one takes the existence of the canonical basis, then one can prove non-trivial facts about finite dimensional representation theory in a nice way. For example Anderson (in this paper) proves a relationship between weight multiplicities and tensor product multiplicities using the MV basis. But, I don't think one can turn this argument in the other direction. That is, the existence of a relation between weight multiplicites and tensor product multiplicities is, by itself, not enough to indicate the existence of a canonical bases. I am looking for something in this direction. REPLY [6 votes]: I don't really see how to get there from just compact groups, so in that sense this is not an answer. My take on the question is something like: how might one have guessed the existence of canonical basis theories*, knowing only the Lie theory of, say, 1975? (The type A case was well under way by then, motivating standard monomial theory.) The standard way to make the canonical basis uses quantum groups, and uses $q$ in a fundamental way. The closest we can get to a quantum group in the classical theory is the semiclassical limit, the Poisson structure on the Lie group. (Still too recent.) One question that becomes meaningful now is to ask what subgroups are Poisson subgroups. There are very few: I'll focus on the Borel $B$ (in particular, this is about complex groups not compact), whose importance was well-appreciated by then. Now we can go beyond representation theory of $G$, i.e. complete linear systems over $G/B$, to talk about Demazure modules, complete linear systems over Schubert varieties $\overline{BwB}/B$, and look at the surjections $H^0(G/B; \mathcal L) \twoheadrightarrow H^0(\overline{BwB}/B;\mathcal L)$. In particular, one can ask for a basis for the irrep such that each of these kernels is spanned by a subset of the basis. Moreover, you can filter your basis of the kernel according to the order of vanishing along Schubert varieties, which suggests a means of indexing the basis. This line of thinking leads to the Lakshmibai-Seshadri conjecture (about characters, but motivated by the putative existence of such a basis). *There's a problem with this question in that there are several bases, at the very least the canonical basis and the (different) semicanonical basis, the latter of which may or may not be the MV basis you asked about. Most of the striking properties you ask about hold for all of these. As such it's hard to find overlooked smoking guns that would pick out e.g. the MV basis over the others.<|endoftext|> TITLE: On property of monic polynomial with integer coefficients QUESTION [6 upvotes]: For a monic polynomial with integer coefficients $f$ where $\partial f = 2$, we have $$ \textrm{inf}(f(x)) > 0 \implies \textrm{inf}(f(x)) \geq \frac{3}{4} . $$ Could we generalize this (for common $\partial f$)? REPLY [9 votes]: No. If $p,q$ satisfy Pell's equation $p^2-2q^2=1$, then minimum of $f(x)=(x^2-2)^2+(qx-p)^2$ is at most $f(p/q)=1/q^4$.<|endoftext|> TITLE: Density of prime divisors of $a^n + b$ QUESTION [7 upvotes]: Suppose $a > 1, b \neq 0$ be two rational numbers. Is it known in general that the set of prime divisors of (the numerator of) $a^n + b$ has a positive relative density? REPLY [7 votes]: This is only known conditional on the generalized Riemann hypothesis. A prime $p$ dividing the numerator of $a^{n} + b$ is more or less equivalent to the statement that the subgroup of $\mathbb{F}_{p}^{\times}$ generated by $a$ contains $-b$. This problem is addressed in the 2000 Journal of Number Theory paper by Moree and Stevenhagen (titled A two-variable Artin conjecture) where it is proved (in a similar manner to Hooley's conditional proof of the Artin conjecture) that the set of primes has a positive density. The paper contains explicit formulas for that density.<|endoftext|> TITLE: Does the Turaev-Viro theory for the generalized $E_6$ subfactor for $\mathbb{Z}/7$ distinguish $L(7,1)$ and $L(7,2)$? QUESTION [13 upvotes]: In the paper Sato-Wakui "COMPUTATIONS OF TURAEV-VIRO-OCNEANU INVARIANTS OF 3-MANIFOLDS FROM SUBFACTORS" they compute certain Turaev-Viro-Ocneanu invariants of certain lens spaces. One of the results is that the generalized $E_6$ subfactor for $\mathbb{Z}/p$ distinguishes the lens spaces $L(p,1)$ and $L(p,2)$ for $p=3,5$, and they conjecture (or at least raise the question) of whether this holds for $p=7$. I am wondering if any progress has been made on this question since this paper appeared? REPLY [10 votes]: I think this paper of Wakui says that the answer is "No". The Turaev-Viro invariants associated to the generalized $E6$ subfactors for $\mathbb{Z}/7$ don't seem to distinguish $L(7,1)$ and $L(7,2)$.<|endoftext|> TITLE: Text for studying group representations in the context of (abstract) harmonic analysis QUESTION [11 upvotes]: I would like to study elements of representation theory as I often encounter it when reading texts on harmonic analysis. I was therefore curious if someone could recommend a book for this. When looking into the books that have "representation theory" as their title, it seems there is a variety of books written by people from different fields. For example, it seems there is almost no overlap in the content of Naimark's "Theory of Group representations", and the books by Serre and Fulton & Harris on representation theory. I was therefore wondering which "type of representation theory" is used in harmonic analysis, and how it relates to other "types of representation theory" such as the one studied in abstract algebra. Any advice or explanation is highly appreciated. REPLY [5 votes]: I'd like to make a couple of recommendations at a slightly lower or comparable level than Varadarajan's excellent treatise that I wished I had read before that book. First, Folland's 'A course in abstract harmonic analysis' gives a clear road map from functional analysis to the representation theory of topological groups. It treats compact groups carefully with many worked examples, it emphasizes the peculiarities of non-unimodular groups (and even non type I groups, which I found very illuminating), it has a complete treatment o systems of imprimitivity (finally!) and in general touches upon all the subtle analytical issues sometimes glossed over in representation theory textbooks. This is why I recommend it before Varadarajan. Secondly, the book by Howe and Tan Non-abelian harmonic analysis has a completely different feel: it is driven by the representation theory of a single group, but because of that it has the luxury of going pretty deep into the analytical aspects of its representation theory. It is much more readable (but definitely not as complete) than Lang's $\textrm{SL}(2,\mathbb{R})$ and chapter 4 is a spectacular demonstration of the power of representation theory in analysis! (although the one I found most useful was chapter 5 on decay of matrix coefficients). I recommend these two books, the first for the solid analytical foundations it gives you for further study, and the second for a remarkable example with all sorts of applications that gives an honest, if rank $1$, glimpse into the general theory of representations of semisimple groups.<|endoftext|> TITLE: Why is this new result such a big deal? QUESTION [38 upvotes]: This popular article reports a recent result in reverse mathematics, showing that a certain theorem in Ramsey theory is provable from RCA$_0$, the base theory in SOSOA. Then there are a bunch of surprised remarks, since (according to the article) the theorem itself is infinitary. Does that just mean that it uses unbounded quantifiers, or maybe that it refers to infinite sets? Isn't that the whole point of second-order arithmetic? And aren't there tons of theorems of the same sort? I don't know much about the subject, but I thought one of the basic discoveries in RM was that lots of the familiar results of calculus and analysis can be reached from RCA$_0$. I get that $RT^2_2$ has a Friedman-like flavour that sounds like it might need strong axioms to prove, and I get that proving its strength (that was lower than expected) was difficult, but I'm having trouble understanding why there is such excitement over this, or what far-reaching consequences it's supposed to have. I'd appreciate any enlightenment. REPLY [37 votes]: The statement in question, frequently denoted $\mathsf{RT}^2_2$ in the context of reverse mathematics, is the instance of the infinite Ramsey theorem for unordered pairs and two colors. Specifically, the theorem is that given any graph on the set of natural numbers contains an infinite clique or an infinite anticlique. This is an infinitary theorem since the graph is infinite and so is the clique or anticlique. One of the reasons why this result is of deep interest is that it has many finitary consequences. For example, the finite Ramsey theorem follows from the infinite case by a compactness argument. The theorem has many other seemingly unrelated consequences, for example Ramsey's theorem is a key step in the proof of termination of some algorithms. The recent results of Patey and Yokoyama show (among other things) that $\mathsf{RT}^2_2$ is $\Pi^0_2$ conservative over $\mathsf{PRA}$, i.e. that every $\Pi^0_2$ statement provable using the infinitary theorem $\mathsf{RT}^2_2$ is already provable using only primitive recursive arithmetic. This is surprising because primitive recursive arithmetic is a completely finitary theory: primitive recursive functions have a finite description and they always terminate in finite time. Thus $\mathsf{PRA}$ is considered among the "safest" theories of arithmetic. It is much weaker than Peano Arithmetic, in fact Peano Arithmetic is far from conservative over $\mathsf{PRA}$. $\Pi^0_2$ statements include statements of the form "this algorithm terminates on every input" so we now know that all algorithms that have been shown to terminate assuming $\mathsf{RT}^2_2$ have an alternate proof of termination that only uses $\mathsf{PRA}$ and therefore that these algorithms are primitive recursive themselves! REPLY [17 votes]: They show that $\DeclareMathOperator{\WKL}{WKL}\DeclareMathOperator{\RT}{RT}\DeclareMathOperator{\RCA}{RCA} \RT^2_2$ is $\Pi^0_3$-conservative over $\RCA_0$. Thus, there is no way that $\RT^2_2$ can be essential in a proof of simple first-order statements like for instance the twin prime conjecture, that have a $\Pi^0_3$ form: $$ (\forall n)(\exists p>n)(p\text{ is prime and }p+2\text{ is prime.}) $$ (This is $\Pi^0_2$ which is, in particular, $\Pi^0_3$.) First-order means you quantify over numbers only, whereas $\RT^2_2$ itself involves quantifying over sets of numbers, making it second order. For comparison, Leo Harrington (1978) showed that $\WKL_0$ (weak König's lemma) is $\Pi^1_1$-conservative, hence in particular also $\Pi^0_3$-conservative, over $\RCA_0$.<|endoftext|> TITLE: $G$-orbits in Springer resolution (or, stabilizer actions on Springer fibers) QUESTION [5 upvotes]: This may be an elementary question, but I'm having trouble coming up with an answer: Let $\tilde{N} = T^*(G/B)$ be the Springer resolution of the nilpotent cone. Does it have finitely many $G$-orbits? If not, what's the "first" example where it doesn't? The answer for $SL_2$ is obviously yes (finitely many orbits). The answer for $SL_3$ also seems to be yes. There are three nilpotent orbits. The Springer fiber above the zero nilpotent orbit is $G/B$ and has one orbit associated to it. The fiber above the regular orbit is a point, so it has one orbit associated to it. Above a subregular element, say $x = \left(\begin{matrix} 0&1&0\\0&0&0\\0&0&0\end{matrix}\right)$, the fiber is given by flags (the notation here is: take the span of the first column, then the span of the first two columns etc.) $$\left(\begin{matrix} a&1&0\\0&0&1\\b&0&0\end{matrix}\right), \left(\begin{matrix} 1&0&0\\0&a&1\\0&b&0\end{matrix}\right)$$ i.e. two $\mathbb{P}^1$ intersecting nodally. However, the torus $$\left(\begin{matrix}t&0&0\\0&t&0\\0&0&t^{-2}\end{matrix}\right)$$ stabilizes $x$, and acts on this fiber by five orbits. This seems to indicate to me that there are only finitely many orbits for $SL_3$. Same seems to be true for $SL_4$, though I'm less certain about my calculation there. On the other hand, I can't think of a reason why this would be the case, and can't seem to find anything about it in the literature (which leads me to believe it's possibly untrue, and I'm missing something obvious). REPLY [7 votes]: The question is equivalent to studying the $B$-orbits in $\mathfrak u$. Then Kashin proved that for $G$ simple the number of $B$-orbits in $\mathfrak u$ is finite iff $G$ is of type $A_n$, $n\le4$ or $B_2$. See "Orbits of an adjoint and co-adjoint action of Borel subgroups of a semisimple algebraic group. Problems in group theory and homological algebra (Russian), 141–158, Matematika, Yaroslav. Gos. Univ., Yaroslavlʹ, 1990". So a counterexample to your question is $SL(6,\mathbb C)$. The corresponding question for generalizided flag varieties was considered by Gerhard Röhrle and developed to a small industry in the '90s. For the start, see "Röhrle, Gerhard: Parabolic subgroups of positive modality. Geom. Dedicata 60 (1996), no. 2, 163–186".<|endoftext|> TITLE: Critical case of Sobolev Embedding QUESTION [5 upvotes]: I got stuck in the following lemma: Lemma: Let $B$ be the unit ball in the 4 dimensional Euclidean space. Suppose that $u\in W^{2,2}(B)$, then $e^{u}\in L^{q}$ for any $q>1$. As we know this is exactly the critical case of Sobolev's imbedding where we fail to get $L^{\infty}$ bounds. Any suggestion and help would be appreciated. My idea of this that fails: Note $\int_{B} e^{pu}\leq \sum_{k=1}^{\infty}\frac{p^{k}}{k!}\int_{B} |u|^{k}$. By Sobolev imbedding we know that $W^{2,2}(B)\hookrightarrow L^{k}$ for any $k\geq 1$. So we know that $$\parallel u\parallel_{L^{k}(B)}\leq C_{k} \parallel u \parallel_{W^{2,2}} $$ We can then get : $\int_{B} e^{pu}\leq \sum_{k=1}^{\infty}\frac{p^{k}}{k!} (C_{k})^{k} \parallel u \parallel_{W^{2,2}}^{k}$. So that it suffices to control $C_{k}$. But based on my calculation, $C_{k} \sim k$, from which it seems this argument would fail. REPLY [3 votes]: As pointed in my comment above, what you seem to be looking for is Trudinger's inequality (see e.g. formula (7.40), pp. 162 of the book Elliptic Partial Differential Equations of Second Order by D. Gilbarg and N.S. Trudinger (Springer-Verlag, 1998)), with the proviso that in fact you need $u\in W^{2,2}_0(B)=$ closure of $\mathscr{C}^\infty_c(B)$ in $W^{2,2}(B)$. Since $B$ has Lipschitz boundary, Trudinger's inequality states that for all $u\in W^{2,2}_0(B)$ there are constants $C_1,C_2>0$ such that $$\tag{1}\label{e1}\int_B\exp\left(\left(\frac{|u(x)|}{C_1\|D^2 u\|_2}\right)^2\right)\mathrm{d}x\leq C_2|B|\ ,$$ where $|B|$ stands for the Lebesgue measure of $B$. Notice that if $u$ is constant on $B$ then $\|D^2 u\|_2=0$ and in this case the left hand side of the above inequality equals $+\infty$, hence the need for taking $u\in W^{2,2}_0(B)$. It is not difficult to infer from inequality \eqref{e1} that $e^u\in L^q(B)$ for all $q>0$. To that end, set $a=C_1\|D^2 u\|_2$ ($>0$!) and notice that whenever $|u(x)|\geq a^2 q$ we have that $e^{q|u(x)|}\leq e^{(\frac{|u(x)|}{a})^2}$ and whenever $|u(x)|\leq a^2 q$ we have that $e^{q|u(x)|}\leq e^{a^2q^2}$. Therefore $$\int_B |e^{u(x)}|^q\mathrm{d}x\leq\int_B e^{q|u(x)|}\mathrm{d}x\leq e^{a^2q^2}|B|+\int_B e^{(\frac{|u(x)|}{a})^2}\mathrm{d}x\leq (e^{a^2q^2}+C_2)|B|<\infty\ ,$$ as desired. Trudinger's inequality can indeed be seen as a sharpening of Sobolev's embedding theorem. Interestingly, Trudinger's original argument (On Imbeddings into Orlicz Spaces and Some Applications, J. Math. Mech. 17 (1967) 473-483) is not far off what you were trying to do - it consisted in controlling each individual term in the power series for the exponential but that involved estimating the action of certain Riesz potentials.<|endoftext|> TITLE: What's wrong with Advanced Studies in Contemporary Mathematics (Kyungshang)? QUESTION [10 upvotes]: By some reason the Journal mentioned in the title is no longer covered by the AMS Math. Reviews. On the MathSciNet web page it says: Last Issue: 24, no. 1  2014 Indexed cover-to-cover Status: No longer indexed The journal, however, still exists and it is still covered by the Zentralblatt. I have written to the journal and the AMS several times already and asked for the reason why the AMS doesn't review this journal any more and I have got no answer from either side! Does someone know why this journal is no longer indexed? REPLY [17 votes]: There has been an explosion in new allegedly "scholarly" journals because they are profitable and many university libraries have lost control over their inventory due to the "big deal" they were trapped into a decade ago where they were lured by lower prices into relinguishing autonomy over subscription decisions, currently made in a centralized fashion. The result is a rapid decline in quality and a rapid increase in quantity. MathSciNet had to deal with this by terminating their coverage of numerous journals and not merely this particular item from Kyungshang. Just because the title the folks in Kyungshang chose is very similar to an AMS publication (Contemporary Mathematics), which is a common practice among the new crop of so-called scholarly journals, is not a reason to suspect foul play on the part of the AMS.<|endoftext|> TITLE: Density of non-algebraic leaves in the characteristic foliation QUESTION [5 upvotes]: Let $X$ be a compact complex manifold equipped with a holomorphic symplectic form $\omega$. Let $D$ be a smooth divisor on $X$. At each point of $D$, the restriction of $\omega$ to $D$ has one-dimensional kernel. This gives a non-singular foliation $F$ on $D$. Is it true that if there exists at least one non-algebraic leaf then the union of all non-algebraic leaves can not be contained in a closed subvariey of $D$? REPLY [3 votes]: For a foliation on a projective manifold $Y$, the set of points belonging to invariant subschemes with a given Hilbert polynomial is a Zariski closed subset of $Y$. Therefore the set of algebraic leaves is a countable union of Zariski closed subsets. In particular, the set of points belonging to nonalgebraic leaves, if not empty, is dense both in Zariski and in Euclidean topology. If the ambient is compact Kahler then the analogue statement probably holds true. Without the Kahler assumption I am not sure what can be said.<|endoftext|> TITLE: Measuring the failure of pushforward to commute with Steenrod squares QUESTION [15 upvotes]: Let $f \colon X \rightarrow Y$ be a map of topological spaces. Let's say that they are (closed) manifolds (not necessarily orientable), though to be honest I'm really interested in the more general setup of $\mathbb{Z}/2$-Poincare duality spaces. (From now on all cohomology coefficients are $\mathbb{Z}/2$.) Poincare duality induces, from the usual pullback, a pushforward map $$ f_* \colon H^*(X) \rightarrow H^*(Y) .$$ While pullback commutes with Steenrod square, the pushforward unfortunately does not. Question: can one "quantify" the failure of this commutativity? (i.e. some nice expression for the difference $f_* Sq - Sq f_*$?) I have an answer to my question in terms of Wu classes, but I want something that doesn't involve them. Here's an example of the kind of thing that I'm looking for. Suppose now that $X \hookrightarrow Y$ is a closed embedding of smooth manifolds. A reliable friend told me the following nice formula measuring the failure of this commutativity. If $N_{X/Y}$ is the normal bundle of $X$ in $Y$, then by the Gysin map, tubular neighborhood, excision etc. we get a composition $$H^*(X) \xrightarrow{\text{Gysin}} H^{*+r}(N_{X/Y}, N_{X/Y}-X) \cong H^*(Y, Y-X) \xrightarrow{g^*} H^*(Y)$$ and the claim is that this composition is $f_*$. In particular, this can be used to write down a formula for the "error term" in commuting Sq with pushforward: if $u$ is the Thom class of $N_{Y/X}$ then $$ f_*(Sq \alpha) = g^* (Sq \alpha \cup u) = Sq (g^* \alpha \cup Sq^{-1} u) $$ Now by definition $Sq^{-1} u = w^{-1} u$ where $w$ is the (total) Stiefel-Whitney class of $N_{Y/X}$, so $$ f_* (Sq \alpha) = Sq ( g^* \alpha \cup w^{-1} u ) = Sq (f_*(\alpha \cup w^{-1})) $$ Unfortunately this doesn't make literally make sense if $X,Y$ are not smooth (since there's no $N_{X/Y}$). However, some things make sense; for instance, we can say that $g^* u = f_*(1)$. In my situation $f_*(1)$ is very simple; for instance I know that $Sq (f_*(1)) = f_*(1)$. (This is basically why I suspect that ``$w"=1$ in my case, so that $f_*$ and Sq should commute.) Question: is there a "salvage" of the above formula to the case where $X,Y$ are not smooth? I would prefer an answer that is formulated at the level of Poincare duality spaces, since that's more like my situation (and the question makes sense at that level). REPLY [10 votes]: If $f: X \longrightarrow Y$ is a map whose homotopy fibers are all Poincaré duality spaces then one gets a wrong-way map in the category of spectra: $\Sigma_+^{\infty}Y \longrightarrow X^{\nu}$ where $\nu$ is the stable spherical fibration given by the Spivak normal bundle of this map. On cohomology we get $g: H\mathbb{F}_2^*(X^{\nu}) \longrightarrow H\mathbb{F}_2^*(Y)$ which commutes with all Steenrod operations since it comes from a map of spectra. Composing with the Thom isomorphism $H\mathbb{F}_2^*(X) \cong H\mathbb{F}_2^*(X^{\nu})$ (where there's a degree shift) gives your umkehr map coming from Poincaré duality. The Thom isomorphism comes from cupping with the Thom class $U_{\nu}$, so you could define some notion of Stiefel-Whitney class by $Sq^{-1}U_{\nu}$ and get a formula similar to the one you've got for an embedding of smooth manifolds. Here's a sketch of a construction of that pushforward map. If $Y$ is a point we'd like a stable spherical fibration $\nu$ on $X$ together with a map $S^0 \longrightarrow X^{\nu}$ of spectra sending the generator in homology to the fundamental class under the Thom isomorphism (which we can arrange if $X$ is simply connected so that $\nu$ is orientable). Such a thing exists and is unique up to fiberwise stable homotopy and it's called the Spivak normal fibration. You can also get something when $X$ isn't simply connected, and you can arrange for these to vary in a family over a base $Y$ and get the map I've indicated. In the case of a map of manifolds this goes back at least to Atiyah: factor $f$ as $X \hookrightarrow \mathbb{R}^N \times Y \longrightarrow Y$ and let $\nu_N$ be the normal bundle to the first inclusion. Then Pontryagin-Thom collapse gives a map $S^N \wedge Y_+ \longrightarrow X^{\nu_N}$, and then let $N \rightarrow \infty$ to get the map of spectra above.<|endoftext|> TITLE: Regularization by mean curvature flow QUESTION [6 upvotes]: I have a $C^{1,\alpha}$ surface defined as the graph of some function $\varphi : B \to \Bbb{R}_+$ ($B$ is a ball). This surface has positive and bounded mean curvature in the weak sense (since the curvature is not necessarily well defined): $$ K\geq \int_B \frac{\nabla \varphi \cdot \nabla \psi}{\sqrt{1+|\nabla \phi|^2}} \geq 0 \text{ for every } \psi \in C^1(B),\psi \geq 0. $$ I want to uniformly approximate this surface with more regular surfaces, at least $C^2$, with non-negative mean curvature. I guess this should be possible and it is natural to consider a mean-curvature flow $(\phi_t)$ with initial condition $\phi_0=\varphi$. A result of Ecker and Huisken tells us that $\phi_t$ is smooth for small $t>0$ if the initial condition is uniformly Lipschitz. In this case the hypothesis is verified since $\phi_0$ is $C^{1,\alpha}$. The fact that the initial condition has non-negative mean curvature should be preserved under the flow, so $\phi_t$ will have non-negative mean curvature. On the other hand the bound on the curvature of the initial condition should give a finite upper bound for the curvature of $\phi_t$ (for small $t$). We can then deduce that for $t \in [0,T]$ $\phi_t$ is a uniform approximation of $\phi_0$. All the above should be correct if the initial condition is $C^2$ (so that the curvature is well defined). Does the above reasoning still apply in the $C^{1,\alpha}$ case when we only have some information on the "weak" curvature? Do you know any references for this case? REPLY [4 votes]: The statement you want (and a bit more I think) can be found in Lemma 3.8 by Metzger--Schulze in their article "No mass drop for mean curvature flow of mean convex hypersurfaces".<|endoftext|> TITLE: Categorification of covering morphisms QUESTION [5 upvotes]: Given a category $\mathsf{A}$, let $\mathsf{Fam}(\mathsf{A})$ be its free coproduct cocompletion (which is always extensive). This means every object has a unique up to iso presentation as a coproduct of connected objects. This category is really part of the data of a fibration $\Pi_0:\mathsf{Fam}(\mathsf{A})\longrightarrow \mathsf{Set}$ assigning to each object its set of connected components. There's also an adjunction $\Pi_0\dashv H$ where $H$ is the "discrete" functor taking a set $A$ to $A\cdot \mathbf{1}=\coprod_A\mathbf{1}$. In the book Galois Theories by Borceux and Janelidze, a neat process of abstraction leads to the following definition. Definition 6.5.9. An arrow $\alpha:A\longrightarrow B$ in $\mathsf{C}=\mathsf{Fam}(\mathsf A)$ is said to be a covering morphism if there exists an effective descent morphism $p:E\longrightarrow B$ such that the square below is a pullback. $$\require{AMScd} \begin{CD} E\times_BA @>{\eta_{E\times_BA}}>> H\Pi_0(E\times_BA)\\ @V{p^\ast\alpha}VV @VV{H\Pi_0(p^\ast\alpha)}V\\ E @>>{\eta_E}> H\Pi_0(E) \end{CD}$$ Concretely, the unit $\eta$ takes a point to its connected component. Since I'm having a hard time visualizing this definition as it's presented, I thought of abstracting the definition of a fiber bundle, and the trying to abstract the definition of a covering space as a fiber bundle with discrete fibers. Definition 1. A trivial fiber bundle with fiber $F$ is an arrow which is isomorphic to $\pi_1:B\times F\longrightarrow B$ in $\mathsf{C}/B$. Definition 2. Let $\mathcal M$ be a class of arrows in a site $\mathsf{C}$. An arrow $\pi:E\longrightarrow B$ is said to be locally in $\mathcal M$ if there's a covering $\left\{u_i:U_i\rightarrow B \right\}$ such that $u_i^\ast\pi$ are all in $\mathcal M$. Definition 3. A fiber bundle with fiber $F$ a locally trivial fiber bundle with fiber $F$, i.e it is locally in the class of projections $X\times F\longrightarrow X$. If $\mathsf{C}$ is complete, trivial fiber bundles are stable under base change (with invariant fiber). If $\mathsf{C}$ is a complete superextensive site, fiber bundles are stable under base change, also with invariant fiber. So complete superextensive sites look like especially good settings for working with fiber bundles. Definition 4. A covering morphism is a fiber bundle such that $F$ is in the essential image of $H$. Why is this definition poor, and why does it not capture what we want a covering morphism to capture? How does it compare to definition 6.5.9 (e.g if we take the extensive topology on $\mathsf{C}$)? What do we want a covering morphism to capture? REPLY [2 votes]: First, definition 3 must be mended to allow varying fibers, so a fiber bundle is locally some product projection. Then, definition 4 also allows for varying discrete fibers. Whenever both definitions (this and 6.5.9) are applicable, they coincide. We'll work our way into increasing generality starting with spaces. $\mathsf{Top}$ with its usual topology is a superextensive site. This allows to replace every covering family with a single arrow - its associated cover (see below). Being an extensive category, $\mathsf{Top}$ has universal coproducts, which ensures the the squares on the left below are pullbacks iff the right one is. $$\require{AMScd} \begin{CD} U_i\times F_i @>>> A\\ @VVV @VV{\alpha}V\\ U_i @>>> B \end{CD}\forall i\iff\require{AMScd} \begin{CD} \coprod_iU_i\times F_i @>>> A\\ @V{p^\ast\alpha}VV @VV{\alpha}V\\ \coprod_iU_i @>>{p}> B \end{CD}$$ This allows us to encapsulate the local triviality definition for fiber bundles in terms of a single base change. Next, for spaces, note the associated cover $p$ is an étale surjection. Étale surjections are effective descent morphsims of spaces, which suggests generalizing to such morphisms in general contexts. The problem is that we can't even write a pullback square with varying fibers if we replace $\coprod_iU_i$ by $E$. However, if we assume $\mathsf{C}$ is actually a free coproduct cocompletion, we may write $E\cong \coprod_iE_i$ where $E_i$ are the connected components of $E$. This yields a notion of a "generalized fiber bundle" - an arrow $\alpha$ for which there exists an effective descent morphism making the square below a pullback. $$\begin{CD} \coprod_iE_i\times F_i @>>> A\\ @VVV @VV{\alpha}V\\ \coprod_iE_i @>>{p}> B \end{CD}$$ Following the setting of spaces, say an arrow is a trivial covering morphism if it's a trivial fiber bundle with discrete fiber (in the essential image of the copower functor $H:A\mapsto \coprod_A\mathbf 1=A\cdot \bf 1$. Since $\mathsf{C}$ is extensive and has products, it is distributive, hence $E_i\times F_i\cong |F_i|\cdot E_i$. This shows the connected component decomposition of the total space of a trivial covering morphism simply has duplicates of the connected compnents of the base space, with multiplicity equal to the cardinality of the fiber (which is always given by $|\Pi_0(F_i)|$). Proposition. Suppose $\mathsf C$ is a free coproduct cocompletion. Then $\alpha$ is a trivial covering morphism if and only if the square below is a pullback. $$\require{AMScd} \begin{CD} A @>{\eta_A}>> H\Pi_0(A)\\ @V{\alpha}VV @VV{H\Pi_0(\alpha)}V\\ B @>>{\eta_B}> H\Pi_0(B) \end{CD}$$ Proof. Let $B\cong\coprod_iB_i$ be the connected components decompositio of $B$. The pullback $B\times_{H\Pi_0(B)}H\Pi_0(A)$ is $\coprod_i | (\Pi_0\alpha)^\ast \left\{ i \right\}|\cdot B_i$ with projection on $B$ given componentwise by identities. The description for $\alpha$ as a trivial fiber bundle with discrete fibers is equivalent, since the components $\alpha:\coprod_i | (\Pi_0\alpha)^\ast \left\{ i \right\}|\cdot B_i\rightarrow \coprod_iB_i$ correspond by connectedness of $B_i$ to the identity. $\square$<|endoftext|> TITLE: Witness to a failure of Fubini/Tonelli QUESTION [12 upvotes]: Is it provable in ZFC that there is a subset of the plane all of whose vertical cross sections have Lebesgue measure zero and all of whose horizontal cross sections are complements of sets of Lebesgue measure zero? There are such sets in models in which every set of reals of cardinality less than the continuum is Lebesgue measurable. REPLY [6 votes]: In 1980 H. Friedman proved (in A consistent Fubini-Tonelli Theorem) that after adding $\mathfrak{c}^+$-many random reals the equality $\int f dx dy=\int f dy dx$ holds for any non-negative function for which the iterated integrals exist. In particular, a subset of the plane with the required properties does not exist in such model.<|endoftext|> TITLE: 1-st cohomology of multiplicative group in a vector space QUESTION [14 upvotes]: Let $\mathbb k$ be a field of characteristic $p$ and let $\mathbb k_n$ be a 1-dimensional representation of $\mathbb k^\times$, where the action is given by $t\circ v= t^n v$. Is it known what are the Ext-groups $\mathrm{Ext}^1(\mathbb k_n,\mathbb k_m)$ in the category of $\mathbb F_p[\mathbb k^\times]$ modules? I'm mostly interested in the case when $\mathbb k$ is algebraically closed and $n=0$ if this makes things simplier REPLY [3 votes]: So returning to the question I think I can prove that $\mathrm{Ext}^1_{\mathbb F_p[\mathbb k^\times]}(\mathbb F_p, \mathbb k_n)=0$ for $n\neq 0$ and $\mathbb k=\mathbb F_q$ or $\mathbb k=\overline{\mathbb F_p}$. First general remark: $\mathrm{Ext}^1_{\mathbb F_p[\mathbb k^\times]}(\mathbb F_p, \mathbb k_n)$ is naturally a subgroup of $\mathrm{Ext}^1_{\mathbb Z[\mathbb k^\times]}(\mathbb F_p, \mathbb k_n)$, which is a subgroup of $H^1(\mathbb k^\times, \mathbb k_n):=\mathrm{Ext}^1_{\mathbb Z[\mathbb k^\times]}(\mathbb Z, \mathbb k_n)$ (considering the short exact sequence $0\rightarrow \mathbb Z \rightarrow \mathbb Z \rightarrow \mathbb F_p \rightarrow 0$ and using the fact that $\mathbb k^{\mathbb k^\times}_n=0)$. So it is enough to prove that $H^1(\mathbb k^\times, \mathbb k_n)=0$. If $\mathbb k=\mathbb F_q$, we have $\mathbb F_q^\times\simeq \mathbb Z/(q-1)\mathbb Z$ and $H^1(\mathbb k^\times, \mathbb k_n)=H^1(\mathbb Z/(q-1)\mathbb Z, \mathbb k_n)=0$ as it should be killed by $p$ and $q-1$ simultaniously. If $\mathbb k =\overline{\mathbb F_p}$, then any element is a root of unity of degree not divisible by $p$ and it is easy to see that $\mathbb k^\times=\mathrm{colim}_{p\nmid m} \mathbb Z/m \mathbb Z = \mathbb Z_{(p)}/\mathbb Z$. For each $n$, we have $ \mathbb k_n^{\mathbb Z/m \mathbb Z}=0$, from this it follows that $H^1(\mathbb k^\times,\mathbb k_n)=H^1(\mathrm{colim}_{p\nmid m} \mathbb Z/m \mathbb Z ,\mathbb k_n)=\lim_{p\nmid m}H^1(\mathbb Z/m \mathbb Z ,\mathbb k_n)=0$ by the same argument, as $(m,p)=1$ for all $m$ in the limit. Now in general situation if $\mathbb k$ is algebraically closed we have $(\mathbb k^\times)_{\mathrm{tors}}=\overline{\mathbb F_p}^\times=\mathbb Z_{(p)}/\mathbb Z$, moreover the quotient $\mathbb k^\times/(\mathbb k^\times)_{\mathrm{tors}}$ is divisible and torsion free abelian group, so is isomorphic to some vector space $V_{\mathbb Q}$ over $\mathbb Q$ (as an abelian group). Moreover $\mathrm{Ext}_{\mathbb Z}^1(V_{\mathbb Q},\mathbb Z_{(p)}/\mathbb Z)=0$, so $\mathbb k^\times=\mathbb Z_{(p)}/\mathbb Z\times V_{\mathbb Q}$. So from Kunneth it would be enough to know that $H^1(V_{\mathbb Q}, \mathbb k_n)=0$. Seems like this should be true for any $p$-torsion module in general. So does anybody know by any chance if it is true that $H^1(\mathbb Q, M)$ for $\mathbb F_p[\mathbb Q]$-module $M$?<|endoftext|> TITLE: Regular minimal model of $X_0(p^2)$ QUESTION [5 upvotes]: Consider the compactified modular curve $X_0(p^2)$ and the corresponding algebraic curve over $\mathbb{Q}$. My questions are the following: Where do the cusps of $X_0(p^2)_{\mathbb{Q}}$ live? That is for which extension $K$ of $\mathbb{Q}$, $X_0(p^2)(K)$ contains all the cusps? Let $K$ be as above and $\mathcal{O}_K$ the ring of integers of $K$. Consider the regular minimal model of $X_0(p^2)$ over $\mathcal{O}_K$. What are the singular fibers and what do they look like? What is a good reference for this material? REPLY [3 votes]: The cusp divisor of $X_0(N)$ is given (at least in principle) by the “torsion” of the degenerate elliptic curves and (in principle) described as such in Katz-Mazur's book, Arithmetic moduli of elliptic curves, Annals of math. studies, vol. 108 (1985), Princeton Univ. Press. An answer to question 2 (and much more) can be found in the paper “Minimal resolution and stable reduction of modular curves”, by Bas Edixhoven published in Annales de l'Institut Fourier, tome 40, n° 1 (1990), p. 31-67. It can be downloaded from Numdam. By the way, more sophisticated methods allowed Jared Weinstein to compute a stable reduction of general modular curves. See “Semistable models for modular curves of arbitrary level”, Inventiones Math., to appear (and arXived). These 3 excellent references (read in that order) probably furnish the best way to get informed on the arithmetic properties of modular curves.<|endoftext|> TITLE: Irreducible polynomial $p_{n}(x)=\sum_{k=0}^{n}\frac{x^k}{k!}$ for all positive integers $n$ QUESTION [6 upvotes]: Let $n$ be a positive integer greater than $1$, and define the polynomial $$p_{n}(x)=\sum_{k=0}^{n}\dfrac{x^k}{k!}$$ Is $p_{n}(x)$ irreducible in $\mathbf{Q}[x]$? I can show it when $n$ is a prime number, since $$n!p_{n}(x)=x^n+nx^{n-1}+n(n-1)x^{n-2}+\cdots+n!x+n!$$ using Eisenstein's criterion. REPLY [20 votes]: That follows from a theorem of Schur saying that any polynomial $\sum_{k=0}^nc_k\frac{x^k}{k!}$ with $c_i\in\mathbf{Z}$, $c_0,c_n\in\{1,-1\}$, $n\ge 1$, is irreducible over $\mathbf{Q}$. I. Schur, Einige Sätze über Primzahlen mit Anwendungen auf Irreduzibilitätsfragen I, Sitzungsberichte Preuss. Akad. Wiss. Phys.-Math. Klasse 14 (1929), 125–136. Also in Gesammelte Abhandlungen, Band III, 140–151 See also here.<|endoftext|> TITLE: Learning roadmap for algebraic number theory QUESTION [10 upvotes]: I have read some elementary number theory from David Burton's text and I know groups and rings from Herstein's book Topics in Algebra and some field theory from different sources online. I am currently learning commutative algebra from Atiyah and Macdonald's engaging book on commutative algebra. I would really like to learn some algebraic number theory and I was wondering if someone could provide me a sequence of books/ steps so that I can understand enough number theory to be a strong graduate student about to begin research. I apologize sincerely if my question is off-topic here or partly vague. REPLY [4 votes]: Two good books for an introduction to global algebraic number theory (i.e., number fields) are: Algebraic Theory of Numbers: Translated from the French by Allan J. Silberger (Dover Books on Mathematics) May 19, 2008 by Pierre Samuel (less than $8 in paperback) A Classical Introduction to Modern Number Theory (Graduate Texts in Mathematics) (v. 84) by Kenneth Ireland & Michael Rosen, Springer<|endoftext|> TITLE: Does the ring $R = \mathbb{Z}[X^{\pm1}]$ of Laurent polynomials over $\mathbb{Z}$ satisfy $SL_2(R) = E_2(R)$? QUESTION [19 upvotes]: Let $R = \mathbb{Z}[X^{\pm1}]$ be the ring of Laurent polynomials on one indeterminate over $\mathbb{Z}$. Let $E_2(R)$ be the subgroup of $GL_2(R)$ generated by the matrices that differ from the identity by a single off-diagonal element. The question whether $SL_2(R) = E_2(R)$ is a problem of S. Bachmuth and H. Y. Mochizuki [2]. It is also an open problem in [3, MA1] and a conjecture in [4]. I am very curious about the status of this problem/conjecture or any progress on it. A complement prompted by the first answer: P. M. Cohn proved that $SL_2(R) \neq E_2(R)$ for $R = \mathbb{Z}[X]$ by considering the matrix $\begin{pmatrix} 1 + 2X & 4 \\ -X^2 & 1 - 2X \end{pmatrix}$ after his [Lemma 8.4, 1]. This matrix is easily seen to sit in $E_2(\mathbb{Z}[X^{\pm 1}])$ by cancelling one coefficient and then using Whitehead's lemma. P. M. Cohn also proved using [Proposition 7.2, 1] that the matrix $\begin{pmatrix} 1 + XY & X^2 \\ -Y^2 & 1 - XY \end{pmatrix}$ doesn't lie into $E_2(k[X, Y])$ for any field $k$. This matrix is usually referred to as Cohn's matrix and admits the following generalization discussed by T. Y. Lam and T. Dorsey [Example VI.3.5, 5]: $C_{m, n}(X, Y) = \begin{pmatrix} 1 + XY & X^n Y^m \\ (-1)^r X^m Y^n & 1 - XY + X^2 Y^2 - \cdots + (-1)^r X^r Y^r \end{pmatrix}$ where $m, n \ge 0$ and $r = m + n - 1 \ge 0$. The first answer's author legitimately asks whether $C_{0, 3}(X - 1, p)$ lies into $E_2(\mathbb{Z}[X^{\pm 1}])$ for $p$ a prime number (it is known that $C_{m, n}(x, y) \in E_3(R)$ for any elements $x, y$ in a commutative ring $R$). [1] "On the structure of the $GL_2$ of a ring", P. M. Cohn, 1966. [2] "$E_2 \neq SL_2$ for most Laurent polynomial rings", S. Bachmuth, H. Y. Mochizuki, 1982. [3] "Open problems in combinatorial group theory", G. Baumslag et al., 2000. [4] "On finite and elementary generation of $SL_2(R)$", P. Abramenko, 2007, http://arxiv.org/abs/0808.1095. [5] "Serre's problem on projective modules", T. Y Lam, 2006. REPLY [8 votes]: This is only a long comment. I would think of $\mathbb{F}_q[x,t^{\pm 1}]$ as the function field analogue of $\mathbb{Z}[t,t^{-1}]$. For the latter, it is known that $SL_2\neq E_2$ from the results in the following paper: S. Krstic and J. McCool. Free quotients of $SL_2(R[x])$. Proc. Amer. Math. Soc. 125 (1997), 1585-1588. Applying the main result of Krstic-McCool to $R=\mathbb{F}_q[t,t^{-1}]$ implies that the matrices $$ \left(\begin{array}{cc} 1+(t-1)x^k & x^{3k}\\ (t-1)^3 & 1-(t-1)x^k+(t-1)^2x^{2k} \end{array}\right) $$ map to a basis of a free quotient of $SL_2(\mathbb{F}_q[x,t,t^{-1}])/E_2$. In particular, they witness that $SL_2$ is far from elementarily generated. It is also possible to see that the above matrices are not in the elementary subgroup using Park's realization algorithm. The naive translation to the ring in question suggests the following candidates for matrices in $SL_2(\mathbb{Z}[t,t^{-1}])$ which are not products of elementary matrices: $$ \left(\begin{array}{cc} 1+(t-1)p & p^{3}\\ (t-1)^3 & 1-(t-1)p+(t-1)^2p^{2} \end{array}\right) $$ As I said, this is not an answer because I have no idea how to show that these matrices are in the elementary subgroup (or to factor them as product of elementary matrices).<|endoftext|> TITLE: Main term in the number of sign changes of $\psi(x) - x$ QUESTION [8 upvotes]: Define $N_\Delta(T)$ to be the number of sign changes of $\psi(x) - x$ in the interval $[1, T]$. Landau's Theorem says $N_\Delta(T)$ is $\Omega(\log T)$ [1]. But perhaps that estimate is too crude. Is the main term of $N_\Delta(T)$ known? Or are only strict upper and lower bounds known? What type of machinery is used to determine something of this nature? [1] Emil Grosswald "Oscillation Theorems of Arithmetical Functions" Transactions of the American Mathematical Society 126 (1967) pp. 7. REPLY [10 votes]: An asymptotic for the number of sign changes is not known, and indeed only a lower bound of $c \log T$ is known. There have been small improvements in the constant $c$ that is allowed here (see Kaczorowski). For recent work related to this, see Montgomery and Vorhauer, who show that the remainder term changes sign in any interval $[x,2.02x]$ once $x$ is large enough. They also speculate that the right order for the number of sign changes up to $T$ is $\sqrt{T}$ (every once in a while $\psi(x)-x$ will be very small, and near that point there'll be many sign changes). One more recent reference is a paper of Kaczorowski and Wiertelak. Also one shouldn't really expect an asymptotic formula for the number of sign changes -- if $\psi(x)-x$ is large and positive (on scale $\sqrt{x}$) then it is likely to stay positive at $1.01x$ as well. Montgomery and Vorhauer speculate that $$ 0< \liminf_{T\to \infty} \frac{N_{\Delta}(T)}{\sqrt{T}} < \limsup_{T\to\infty} \frac{N_{\Delta}(T)}{\sqrt{T}} < \infty. $$<|endoftext|> TITLE: Multiplication in universal enveloping algebra in terms of PBW isomorphism QUESTION [6 upvotes]: Let $\mathfrak g$ be a Lie algebra. Consider the multiplication map $m:\mathfrak g\otimes U(\mathfrak g)\to U(\mathfrak g)$ and $i:S(\mathfrak g)\to U(\mathfrak g)$ -- Poincare-Birkhoff-Witt isomorphism. Is there an explicit formula for $$i^{-1}\circ m\circ id\otimes i:\mathfrak{g}\otimes S(\mathfrak g)\to S(\mathfrak g)$$ in terms of, say, multiplication in $S(\mathfrak g)$ and its $\mathfrak g$-module structure? REPLY [2 votes]: $\newcommand{\g}{\mathfrak g}$ There are several (equivalent) answers to this question. Let $BCH$ be the Baker-Campbell-Hausdorff formula and write $$BCH(x,y)=\sum_{p,q\geq 0} BCH_{p,q}(x,y)$$ where $BCH_{p,q}(x,y)$ is the homogenous part of $x$ degree $p$ and $y$ degree $q$ in the Free Lie algebra on $x,y$. Then for any $x,y \in \g$, the product $x^py^q$ is given by: $$\sum_{k\geq 0}\frac{p!q!}{k!} \sum_{(p_1,\dots,p_k)\in Part_k(p),(q_1,\dots,q_k)\in Part_k(q)} i(BCH_{p_1,q_1}(x,y)\dots BCH_{p_k,q_k}(x,y))$$ Note that for $p=1$ (which is the case you are interested in) there are well known explicit formulas for $BCH_{1,q}$ in terms of Bernouilli numbers, see e.g. equation 2.22 in http://arxiv.org/abs/math/9905080 The "tree part" of Kontsevich star product coincide with the PBW multiplication. An answer to your question using this formalism is in section 3 of http://arxiv.org/abs/math/9905080. Again you get an explicit formula involving Bernouilli numbers. This product is also known as the 'Gutt star product" in the litterature. Indeed, Gutt showed that there is a fairly explicit star product on $T^*G$ whic in turn can be used to compute the PBW star product on $S(\g) $(http://link.springer.com/article/10.1007/BF00400441) All those formulas are thought of quantization of the Poisson manifold $\g^*$ where $\g$ is a finite-dimensional Lie algebra. However, at the end of the day the formulas make sense in the infinite-dimensional case as well (which would not be the case, e.g., if you did not throw away the wheels in 2).<|endoftext|> TITLE: How much mathematics has been formally verified? QUESTION [37 upvotes]: That's a vague question so allow me to tighten it up a bit. I recently noticed that there is a formal machine verified proof of the Central Limit Theorem (CLT) implemented with Isabelle. This requires a substantial amount of machinery that is taught in undergraduate courses on calculus, measure theory and probability theory. As Williams' textbook Probability with Martingales culminates with CLT it seems like it might be fair to conservatively estimate that maybe half of an undergraduate level probability theory course has been formalised. So would it be fair to say that half of the material (in general) that is taught to mathematics undergraduates has been formally verified by machine? If not, what similar proposition is true? REPLY [46 votes]: A couple of years ago I made a database of all the formalization files that I could find at that point, in several different systems. You can view it here: http://bim.shef.ac.uk/formal/list_formalizations.php I would guess that the proportion of a typical undergraduate curriculum that has been formalized in at least one framework is substantially greater than 50%. However, the general usability of these systems remains poor. Even in a single proof assistant, there are problems with incompatible versions, incompatible libraries and incompatible approaches to the same mathematical material in a single library, and various kinds of bitrot. Documentation is often poor. Typically it is aimed at people who are primarily interested in the technology of proof assistants, and if not, then it is aimed at people who already know the relevant mathematics very well. I once gave a talk about this sort of thing. The slides are here: http://neil-strickland.staff.shef.ac.uk/talks/pa_talk.pdf and there is video here: https://www.youtube.com/watch?v=4O84o1hk1Qs<|endoftext|> TITLE: Errata on Rezk's paper QUESTION [17 upvotes]: I am reading this paper of Rezk's http://arxiv.org/abs/0901.3602 A cartesian presentation of weak n-categories, and as it is pointed out in the introduction, it contained a wrong statement (2.19 in the old version) and now there is a new proof for the main result, which is 6.6. Now, section 6.9 of the new proof refers to Prop. 6.4, which uses 2.19 towards the end. Am I missing something? How is this apparent circularity resolved? Thanks in advance! REPLY [30 votes]: It looks like I completely missed this. Here's what I guess happens: although the original 2.19 was wrong, there is a weaker version that is true (I'll just state it for simplicial sets): If $X$ is a simplicial set, and $\mathcal{P}$ a finite collection of subobjects of $X$ which is closed under intersections, and the $\bigcup_{K\in\mathcal{P}} K=X$, then $\mathrm{hocolim}_{\mathcal{P}} K \approx X$. I think the original non-proof of 2.19 that I gave was really trying to prove this weaker version (I'll try to get that straight when I have some time). Now the weak version of 2.19 applies to the poset $\mathcal{P}_K$. Almost: $\mathcal{P}_K$ isn't necessarily closed under intersection (finite limits), but only because it fails to contain an initial object, corresponding to $\varnothing \to K$. If you append this object to $\mathcal{P}_K$, the argument seems to go through (note that $V_\varnothing(A_1,\dots,A_m)$ is itself empty). Working this out ... Let $\{S_i\}_{i\in I}$ be an indexed collection of (distinct) subsets of a set $S$, implicitly making $I$ into a poset by set-inclusion, whence a functor $F\colon I\to Set_{/S}$. Suppose (*) every element of $S$ is contained in a minimal $S_i$, i.e., for each $x\in S$, there exists $i(x)\in I$ such that $x\in S_j$ iff $S_{i(x)}\subseteq S_j$. (This holds in particular if the indexed collection is closed under intersection, but is a weaker condition in general.) Condition (*) implies that the functor $F$ is a coproduct of free functors: $F\approx \coprod_{x\in X} \mathrm{Hom}_I(i(x),-)$. In particular, $\mathrm{colim} F \approx S$. Now suppose $\{S_i\}_{i\in I}$ is an indexed collection of subobjects of a simplicial set $S$, satisfying (*) in each degree. Using the above observation (applied inductively by degree to non-degenerate simplicies of $S$), we see that the functor $F\colon I\to sSet$ is cofibrant in the projective model structure. As a consequence, $\mathrm{hocolim} F \xrightarrow{\sim} \mathrm{colim} F\approx S$. We can extend the claim to replace $sSet$ with $Fun(C,sSet)$, since hocolims in functor categories are computed termwise. This gives a version of 2.19 which is true, and which should apply to the $\mathcal{P}_K$ of 6.4. I think.<|endoftext|> TITLE: Differentiability of geodesics in Alexandrov subspaces of Riemannian manifolds QUESTION [8 upvotes]: Let $M$ be a smooth Riemannian manifold. Let $X\subset M$ be a closed path connected subset which has curvature bounded below in the sense of Alexandrov with respect to the induced intrinsic metric. Let $\gamma\colon [a,b]\to X$ be a shortest path in $X$ (parameterized by its length). Let $\iota\colon X\to M$ denote the natural imbedding. Question. Is it true that $\iota\circ\gamma\colon [a,b]\to M$ has one sided derivatives everywhere? Remark. The question has positive answer at almost every point of $[a,b]$ since $\iota\circ \gamma$ is a Lipschitz map, and such maps have first derivative almost everywhere. ADDED: In the special case when $M=\mathbb{R}^n$ is the Euclidean space, and $X$ is a convex hypersurface, the answer is positive and due to I.M. Liberman (1941). REPLY [4 votes]: No. If $M = R^2$ and $X$ is the sawtooth below, then the map has no derivative at the central point. Note that the induced intrinsic metric on $X$ is just proportional to distance on the $x$-axis, and in particular is flat. Similarly if $M = R^3$ and $X$ is the sawtooth below crossed with $R$, then the induced intrinsic metric is still flat, and there is still no derivative at any of the central points.<|endoftext|> TITLE: Product of Bruhat Cells QUESTION [6 upvotes]: Fix a $(B,N)$ pair (Tits system) of a semisimple Lie group $G$. Let $u$ and $v$ be two Weyl group elements such that $l(uv)=l(u)+l(v)$. It is known that $BuvB=(BuB)(BvB)$ (see for example Humphreys's Linear Algebraic Groups Section 29.3 Lemma A). Let $x$ be an element of the Bruhat cell $BuvB$ and I would like to factor $x$ as a product $yz$ with $y\in BuB$ and $z\in BvB$. Is it true that $y$ is unique up to a right multiple of an element in $B$ and $z$ is unique up to a left multiple of an element in $B$? If so, how do I see it? Thanks a lot. REPLY [3 votes]: Here is a (probably equivalent) way to see this. I'll give an argument for $(BsB)(BwB)$, where $s$ is a simple reflection. You should be able to expand this argument to $(BvB)(BwB)$ as requested in the question. The expansion will use the fact that $\ell(vw) = \ell(v) + \ell(w)$ implies $\ell(s_t \cdots s_1 w) = t + \ell(w)$, where $s_n \cdots s_1 = v$ is a reduced decomposition for $v$ and $n \geq t$. Throughout, let $T\subset B$ be a fixed maximal torus and $U \subset B$ the unipotent subgroup of $B$. Recall that $B = TU$. For any positive root $\alpha$, let $U_\alpha$ be the corresponding root subgroup of $U$. Recall that for any ordering of positive roots, we can write $U$ as a produce of $U_\alpha$ in the chosen order. Fix an odering of positive roots so that $U = U_{w{-1}} U_{w_0w^{-1}}$, where $w_0$ is the longest element of the Weyl group. Then, the elements of $BwB$ are uniquely written as $U_{w^{-1}}wB$, where \begin{equation} U_{w^{-1}} = \prod_{\alpha \in R(w^{-1})} U_\alpha. \end{equation} Here, $R(w^{-1}) =\{\alpha \in R^+\;|\; w^{-1} \alpha <0\;\}$. This decomposition follows from the Bruhat decomposition because if $w^{-1} \alpha >0$, then $w^{-1} u_{\alpha} w = u_{w^{-1} \alpha}$, and so $ u_{\alpha} w = wu_{w^{-1} \alpha}$ and $u_{w^{-1} \alpha} \in B$. So, this element is absorbed to the right. Then, every element $ b s b_1b_2 w \tilde{b} \in (BsB)(BwB)$ can be written in the form $BsuwB$, where $u \in U_{w^{-1}}$ as above. Explicitly, $b_1b_2 = t(uu') \in B$, where $t \in T$ and $(uu')\in U_{w{-1}} U_{w_0w^{-1}}$ according to the decomposition given above. Then, $ t(uu')w \tilde{b} = u b'$ for some $u \in U_{w^{-1}}$ and $b' \in B$. For each $(b_1b_2)$, we get a unique $u b'$. If we can show that $sR(w^{-1}) \subset R^+$, then for $u = u_{\alpha_1} \cdots u_{\alpha_k}$ (i.e, $\alpha_i \in R(w^{-1})$) we have \begin{equation} su = s (u_{\alpha_1} \cdots u_{\alpha_k})=(u_{s\alpha_1} \cdots u_{s\alpha_k})s, \end{equation} and $(u_{s\alpha_1} \cdots u_{s\alpha_k}) \in B$. Then, every $b s b_1b_2 w \tilde{b}= b(u_{s\alpha_1} \cdots u_{s\alpha_k})sw b'$. This tells us that every $b_3 sw b_4 = b(u_{s\alpha_1} \cdots u_{s\alpha_k})sw b'$ (sorry for the excessive indexing), where $(b_1b_2)$ can be arbitrary because we need only adjust $b$ and $b'$ to make this equation work. However, once $(b_1b_2)$ is fixed, $b$ and $b'$ are determined by this value. To show $s R(w^{-1}) \subset R^+$, we use the assumption that \begin{equation} 1+ \ell(w^{-1}) = 1 + \ell(w) = \ell(sw) = \ell(w^{-1}s) . \end{equation} So, $|R(w^{-1}s)| = |R(w^{-1})| + 1$ since the cardinality of $R(w)$ is just the length of $w$. Using a basic formula for $R(w^{-1}s)$ (see, e.g., Springer's Linear Algebraic Groups, section 8.3.1) this implies $s R(w^{-1}) \subset R^+$ as desired.<|endoftext|> TITLE: Graphs with prescribed numbers of k-cliques QUESTION [8 upvotes]: Let $(a_1,a_2,\dots, a_n)$ be a sequence of non-negative integers. Q. When does there exists a simple graph $G$ such that its number of $k$-cliques is $a_k$ (that is $G$ has $a_1$ vertices, $a_2$ edges, $a_3$ trinagles, etc)? Probably, it is hopeless to get a complete description of such sequences, so I am interested in any necessary or sufficient conditions on $\{a_i\}$. Upd. As Steve Huntsman mentions in the comments there are trivial necessary conditions: $a_k\le \binom{a_0}{k}$. There are also more complicated inequalities like $$ \binom{a_1}{3}-a_3\le \bigl(\binom{a_1}{2}-a_2\bigr)(a_1-2). $$ REPLY [2 votes]: The case $n=3$ was addressed in Razborov's On the Minimal Density of Triangles in Graphs, where he used his Flag Algebra technique to provide an asymptotically tight lower bound on the number of triangles contained in a graph with a given number of vertices and edges. Let $\rho=\frac{a_2}{\binom{a_1}{2}}$ be the density of edges in the graph. If $\rho=1-\frac{1}{t}$ for some integer $t$, then the minimum number of triangles comes from a balanced, complete $t$-partite graph. If $\rho \in (1-\frac{1}{t}, 1-\frac{1}{t+1})$, then the optimal graph ends up being a complete $(t+1)$-partite graph, with $t$ equal sized parts, and one part smaller than the rest. Optimizing over the size of the parts gives the (complicated) bound in Razborov's paper.<|endoftext|> TITLE: Roughly equal number of swimmers in teams QUESTION [6 upvotes]: $b^2$ swimmers are to be put into one of the teams $1,2,\dots,b$. A team $i$ has a value function $f_i$, so that if they get swimmer $k$, they get value $f_i(k)$. The value $f_i(k)$ is randomized uniformly from $[0,1]$, independently of this value for other $i,k$. (So, there are $b^3$ different values in total.) The value that a team has for a set of swimmers is simply the sum of the individual values. Suppose that we divide the swimmers to maximize the product of the team values. For large enough $b$, is it true that each team will get a number of swimmers in the range $[b/2, 2b]$ with high probability? In other words, is it highly likely that the swimmers will be divided roughly equally between the teams? REPLY [2 votes]: This is not exactly a solution, but what seems a good way to approach the problem. 1) Consider for each player $k$ the team $I(k)$ to which he brings a maximal value $$ v(k)=\max_i f_i(k) = f_{I(k)}(k). $$ BTW: note, that most probably for most of the players $v(k)$ is quite close to 1, as it is a maximum of $b\gg 1$ independent $R[0,1]$'s; to be more precise, the expectation of $v(k)$ is $1-\frac{1}{b+1}$. 2) Take a "greedy" way of forming the teams: put each player $k$ into the team $I(k)$ where he plays the best. Then, each player equiprobably goes to any of the teams, hence the number of players in any team is equal to the sum of $b^2$ expectation-$1/b$ Bernoulli variables, and thus is roughly normal with expectation and dispersion $\sim b$. In particular, the teams formed in this way are roughly equal, differing from $b$ by something like $\sqrt{b}$. If I'm not mistaken, this means that the product for this case differs (in average, most probably) from the theoretical maximum of $b^b$ by a factor of constant. (This could be surprising, as in the theoretical upper bound 1 is used instead of all the values, but in fact the greedy method almost gets 1 everywhere, and $(1-1/b)^b$ is $1/e$ -- a constant) 3) The idea now is that if you divide the players in a strongly non-equal way, you will get a value that will be lower than the ``greedy'' one. 4) Namely: imagine, that one of the teams is much smaller than the average, that it has less than $\epsilon b$ players (where, say, $\epsilon=\frac{1}{10}$). Then, you get an upper bound for the product by $\epsilon b$ times $(b+(1-\epsilon)b\cdot \frac{1}{b-1})^{b-1}$ (it is a bit rough: we're again setting all the players to values 1). But this is a theoretical maximum of $b^b$, multiplied by a constant, which is approximately equal to $\epsilon \cdot \exp(1-\epsilon)$, and the smaller is $\epsilon$, the smaller it becomes. In particular, for sufficiently small $\epsilon$ it becomes smaller, than the constant that we are getting in the greedy algorithm. The same applies if one of the teams is too large, larger than $Ab$. The upper bound is again $b^b$ times $A\cdot \exp(1-A)$, and the factor $A\cdot \exp(1-A)$ tends to zero as $A$ tends to infinity. 5) Finally, it looks quite plausible that one can make the above arguments work for $\epsilon$ and $A$ arbitrarily close to $1$, but for that one should improve the above arguments in two points: *) First, take $S=\sum_k v(k)$ to be maximal possible sum of values, and use $(S/b^2)^b$ as a reference point instead of $b^b$. *) Second, re-equilibrate the teams: for every player, consider the team which is second-best for him, and try moving $\sim \sqrt{b}$ players from large teams to smaller ones, which are second-best for them. It looks plausible that with this improvement of the greedy algorithm you get a product that is equivalent to our new maximum-reference point $(S/b^2)^b$, while any $\epsilon<1$ or $A>1$ reduce maximum possible value by a constant factor.<|endoftext|> TITLE: Can you prove Givental's conjecture on wavefronts and the icosahedron? QUESTION [26 upvotes]: In his remarkable book The Theory of Singularities and its Applications, Vladimir Arnol'd discussed a conjecture of A. B. Givental, which asserts that the symmetry group of the icosahedron is secretly lurking in the problem of finding the shortest path from one point to another in a region of the plane. Via Huyghen's principle, this problem is connected to the motion of waves in such a region. Arnol'd nicely expresses the awe mathematicians feel when they discover connections like this: Thus the propagation of waves, on a 2-manifold with boundary, is controlled by an icosahedron hidden at an inflection point at the boundary. This icosahedron is hidden, and it is difficult to find it even if its existence is known. Unfortunately, even now a full proof of Givental's conjecture seems hard to find! Can you find one, or give one? Let me sketch the idea here. For more details, see: Involutes of a cubical parabola, Visual Insight, 1 May 2016. Discriminant of the icosahedral group, Visual Insight, 15 May 2016. This image, created by Greg Egan, shows the discriminant of the symmetry group of the icosahedron, a 120-element group known as $\mathrm{H}_3$. This group acts as linear transformations of $\mathbb{R}^3$, and thus also $\mathbb{C}^3$. By a theorem of Chevalley, the space of orbits of this group action is again isomorphic to $\mathbb{C}^3$. Each point in the surface shown here corresponds to a nongeneric orbit: an orbit with fewer than the maximal number of points. More precisely, the space of nongeneric orbits forms a complex surface in $\mathbb{C}^3$, called the discriminant of $\mathrm{H}_3$, whose intersection with $\mathbb{R}^3$ is shown above. The following image, created by Marshall Hampton, shows the involutes of the curve $y = x^3$: Loosely speaking, an involute of a plane curve $C$ is a new plane curve $D$ obtained by attaching one end of a taut string to a point $p$ on $C$ and tracing the path of the string's free end as you wind the string onto $C$. There are different involutes for different choices of $p$ and different lengths of string. I am ignoring some important nuances here, some of which are discussed on my Visual Insight post. But here's the point: the involutes, shown in blue, look like slices of the discriminant of the icosahedral group! Indeed, Givental conjectured this is true, and Arnol'd says it's been proved. In The Theory of Singularities and its Applications, he wrote: The discriminant of the group $\mathrm{H}_3$ is shown in Fig. 18. Its singularities were studied by O. V. Lyashko (1982) with the help of a computer. This surface has two smooth cusped edges, one of order 3/2 and the other of order 5/2. Both are cubically tangent at the origin. Lyashko has also proved that this surface is diffeomorphic to the set of polynomials $x^5 + ax^4 + bx^2 + c$ having a multiple root. The comparison of this discriminant with the patterns of the propagation of the perturbations on a manifold with boundary (studied as early as in the textbook of L'Hopital in the form of the theory of evolutes of plane curves), has led A. B. Givental to the conjecture (later proven by O. P. Shcherbak) that this discriminant it locally diffeomorphic to the graph of the multivalued time function in the plane problem on the shortest path, on a manifold with boundary, which is a generic plane curve. (Figure 18 is a hand-drawn version of the picture at the top of this post.) This seems like an exciting claim worthy of a nice conceptual proof. Unfortunately, I haven't been able to find a complete proof the literature, not even in these promising-looking papers: O. P. Shcherbak, Singularities of a family of evolvents in the neighbourhood of a point of inflection of a curve, and the group $\mathrm{H}_3$ generated by reflections, Funktsional. Anal. i Prilozhen. 17:4 (1983), 70–72. English translation in Functional Analysis and its Applications 17:4 (1983), 301–303; free version available here. O. P. Shcherbak, Wavefronts and reflection groups, Uspekhi Mat. Nauk 43:3 (1988), 125–160. English translation in Russian Mathematical Surveys 43:3 (1988), 1497–194. The Russian versions are open-access; the English versions are not, but the first paper is currently available for free online. Perhaps an expert could construct a full proof based on the ideas in these papers! For more clues and references, see my Visual Insight posts. REPLY [8 votes]: In http://www.sciencedirect.com/science/article/pii/S0167278998900057 (Remarks on quasicrystallic symmetries) Arnold so describes the idea of Shcherbak's proof: Now the proof of this theorem depends on the relation of the icosahedron symmetry group $H_3$ to the so-called crystallographic (or Weyl) group $D_6$ (associated with the simple Lie algebra $D_6\approx O(12)$ and with the simple hypersurface singularity $x^2y+y^5+z^2$). The Weyl group $D_6$ is a reflection group, acting in an Euclidean 6-space and conserving a crystallographic lattice (the set of integer linear combinations of 6 independent vectors). In his proof of the above stated theorem, Scherbak constructed a decomposition of this 6-space into two orthogonal 3-spaces, invariant under the two irreducible real representation of the isohedral symmetry group $H_3$ in $R^3$. These two 3-spaces are irrational with respect to the $D_6$ integer lattice. The restrictions of the $D_6$-periodic functions to these irrational subspaces are the desired quasiperiodic functions in 3 variables having icosahedral symmetries. Arnold cites Shcherbak's paper http://link.springer.com/article/10.1007%2FBF01076721 (can be found also here http://www.inp.nsk.su/~silagadz/Shcherbak.pdf). For me it is rather cryptic how what Arnold writes is related to the content of the Shcherbak's paper but hope it will be clear for experts in the field. Maybe the following well-illustrated review (in French) https://eudml.org/doc/110042 (Caustique mystique, by Daniel Bennequin) will be also somewhat useful.<|endoftext|> TITLE: Groupoid cardinality and Egyptian fraction representations of 1 QUESTION [7 upvotes]: It is well-known that any rational number can be represented using a sum of distinct Egyptian fractions (that is, rational fractions of the form $1/n$ with $n\in\mathbb{N}$). This may be proven by establishing a greedy algorithm that constructs a sequence of such decompositions. For instance, \begin{align} 1 &= \frac12+\frac13+\frac16 \\ &= \frac12+\frac13+\frac17+\frac1{42}\\ &= \frac12+\frac13+\frac17+\frac1{43}+\frac1{1806}\\ &= \cdots\\&=\sum_{k=1}^\infty \frac{1}{x_k} \tag{1} \end{align} where $x_1=2$ and $x_k = 1+\prod_{j=1}^{k-1}x_j$ for $k>1$ (i.e. Sylvester's sequence A000058). As noted by Bergner & Walker [1], such Egyptian fractions are relevant in the context of groupoid cardinality. To briefly review: Given a groupoid $G$, the groupoid cardinality is defined [2] by $|G|=\sum_{[\bullet]\in G}\frac{1}{\#\text{Aut}(\bullet)}$ where $[\bullet]$ is a component of $G$ and $\#\text{Aut}(\bullet)$ is the order of the automorphism group of $\bullet$. If $G$ is a group, then this reduces to $|G| = \frac{1}{\# G}$, and if we form the groupoid $G\coprod H$ as a direct union of groups $G,H$ we have $\textstyle |G\coprod H| =\frac{1}{\#G}+\frac{1}{\#H}$. Here the link to Egyptian fractions arises: Since every rational number can be represented in terms of Egyptian fractions, so too can every rational fraction be obtained as the cardinality of some groupoid. As a key example, since $|\mathbb{Z}/n|=1/n$ the representations in $(1)$ imply $$1 = |\mathbb{Z}/2\coprod \mathbb{Z}/2|=|\mathbb{Z}/2\coprod \mathbb{Z}/3\coprod \mathbb{Z}/6|=\cdots.$$ But this construction, while valid, is quite artificial. In particular, it is far from obvious to me why each successive groupoid has unit cardinality. This suggests the following question: Is there a natural sequence of groupoids $\{G_k\}$, each of unit cardinality, which generate $(1)$? References: [1] John Baez, James Dolan, "From Finite Sets to Feynman Diagrams" (arXiv) [2] Julia Bergner, Christopher Walker, "Groupoid Cardinality and Egyptian Fractions" (JSTOR) REPLY [8 votes]: First a notational issue: you shouldn't write $G$ for the one-object groupoid corresponding to $G$. A much better name for this groupoid is $BG$, or $\text{pt} / G$. A natural way to write down a groupoid whose groupoid cardinality is $1$ is to write down a groupoid / homotopy quotient $X/G$ where $|X| = |G|$. In turn, a natural class of such quotients are the adjoint quotients $G/G$ where $G$ acts on itself by conjugation. This is equivalent to the groupoid $$\coprod_{[g]} BC_G([g])$$ where $C_G(-)$ denotes the centralizer and the coproduct runs over the conjugacy classes of $G$. We get a potential Egyptian fraction representation $$1 = \sum_{[g]} \frac{1}{|C_G([g])|}$$ if the sizes of the centralizers are distinct. The largest $n$ such that $\frac{1}{n}$ appears in this sum will always come from the conjugacy class of the identity, where $C_G(e) = G$, so $n = |G|$, and everything else will divide $|G|$. This identity is just the usual class equation divided by $|G|$. To get the first sum this way we want a group of order $6$ with $3$ conjugacy classes having centralizers of orders $2, 3, 6$, or equivalently $3$ conjugacy classes of sizes $3, 2, 1$. There is a unique such group, namely $D_3$. To get the second sum this way we want a group of order $42$ with $4$ conjugacy classes having centralizers of orders $2, 3, 7, 42$, or equivalently $4$ conjugacy classes of sizes $21, 14, 6, 1$. Such a group necessarily has elements of orders $2, 3, 7$, so these must be the non-identity conjugacy classes (in the same order as before, since everything centralizes itself). There are $6$ elements of order $7$, so the Sylow $7$-subgroup $C_7$ is normal, and so by Schur-Zassenhaus our group $G$ must be a semidirect product $C_7 \rtimes H$ where $|H| = 6$. We must have $H = D_3$ or else $G$ has an element of order $6$, so $G$ must be the semidirect product $C_7 \rtimes D_3$. There is a unique nontrivial such semidirect product because the only nontrivial action of $S_3$ on $C_7$ is the one where $\sigma \in S_3$ acts by $\text{sgn}(\sigma)$; unfortunately, this group has an element of order $21$. So in fact no group with the desired condition on conjugacy classes exists. Nevertheless, the second sum is still related to some geometry, specifically to the geometry of the (2,3,7) triangle group, or rather to the orbifold quotient of the upper half plane by this group. I don't know about the third sum, though.<|endoftext|> TITLE: The geometric median of a triangle QUESTION [18 upvotes]: Let $\Omega\subset \mathbb R^n$ be a compact domain of dimension $n$. Define the geometric median on $\Omega$ as the point $m_{\Omega}\in \mathbb R^n$ such that the integral $\int_{\Omega}|x-m_{\Omega}|dx$ attains its minimum. Question Suppose that the domain $\Omega$ is a triangle $\Delta$ in $\mathbb R^2$. Is there a closed formula for the geometric median of $\Delta$? Disclaimer. The name geometric median is taken from the Wikipedia article https://en.wikipedia.org/wiki/Geometric_median . There is huge amount of articles, in particular in statistics, probability, location theory, ect, that use this notion. It is clear as well that this notion has a lot of different names (some of which are given in the Wikipedia article). This notion is mainly applied to the case when $\Omega$ is a finite set. However, after an extensive search on Google, MathSciNet, Google Scholar, etc. I was not able to find any reasonable source treating the above question. REPLY [5 votes]: Here is some evidence that there is no formula. I took a right isoceles triangle as the simplest non-trivial example. For the triangle on $(0,0), (1,-1), (1,1)$, the integral of distances to $(h,0)$ is $$ \frac{h^3 - h^2s + 2s}{6} +\frac{\sqrt{2}\ h^3}{12}\log \dfrac{2-h+\sqrt{2}s}{(\sqrt{2}-1)h} +\frac{(1-h)^3}{3} \log\dfrac{1+s}{1-h} $$ where $s=\sqrt{2-2h+h^2}$. Numerically I find this minimized at $h \simeq 0.648863$. But any explicit formula for this median would have to minimize this function -- and more complicated functions with three more parameters for the general case.<|endoftext|> TITLE: Is there a divisor in $\mathbb P^2$ such that all analytic maps into its complement algebraize? QUESTION [10 upvotes]: Is there a closed subscheme $D$ in $\mathbb P^2_{\mathbb C}$ pure of codimension one such that, for all algebraic varieties $X$ over $\mathbb C$, any analytic map $$ \phi: X(\mathbb C) \to \mathbb P^2(\mathbb C) - D(\mathbb C)$$ algebraizes? (That is, is there a morphism $f:X\to \mathbb P^2_{\mathbb C}- D$ whose analytification is $\phi$?) REPLY [11 votes]: On page 73 of Kobayashi's book Hyperbolic Complex spaces he shows that if D is a certain configuration of 6 lines in the plane then its complement is complete hyperbolic and hyperbolically embedded in the projective plane.By theorem 6.3.24 page 290 of the same book your map from X to the complement extends meromorphically to an algebraic compactification of X .The algebraicity should follow from Serre's GAGA .<|endoftext|> TITLE: Real-valued measurable cardinals QUESTION [10 upvotes]: A cardinal $\kappa$ is real-valued measurable if there is a probability measure on the $\sigma$-algebra of all subsets of $\kappa$ which is zero on singletons and additive on disjoint families of fewer than $\kappa$ subsets. What if I weaken this to: $\kappa$ is uncountable and there is a finitely additive probability measure which is zero on singletons and such that the union of any family of fewer than $\kappa$ null sets is null. Has this notion been studied? What is the consistency strength of the existence of a cardinal with this property? Edit: A paper has been written on the project which provoked this question. It has been posted on the arXiv. Ashutosh's nice result is discussed in Section 7. (I am adding an "operator algebras" tag because the paper is about large cardinal aspects of von Neumann algebras.) REPLY [12 votes]: Suppose $m:\mathcal{P}(\kappa) \to [0. 1]$ is a finitely additive measure whose null ideal $I$ is $\kappa$-additive. Then $I$ is an $\omega_1$-saturated $\kappa$-additive ideal over $\kappa$. Solovay showed this implies that $\kappa$ is measurable in $L[I]$. If we start with a measurable cardinal $\kappa$, and force with a finite support iteration of random forcing of length $\kappa$, then in the resulting model there is a finitely additive measure on $\mathcal{P}(\kappa)$ whose null ideal is $\kappa$-additive and there is no real valued measurable cardinal below the continuum. You can find a proof of this fact in A. Kumar, K. Kunen, Induced ideal in Cohen and random extensions, Topology and its Applications, Vol. 174, Sept. 2014, 81-87.<|endoftext|> TITLE: Schur positivity on 2 letter alphabets implies Schur-positivity on n letters? QUESTION [7 upvotes]: Suppose we have a symmetric polynomial $P$ in $n$ variables. We can partition this alphabet into sets with one or two letters, e.g. ${ {x_1}, {x_2, x_3}}. We can thus see $P$ as an element in $Q[x_1][x_2,x_3]$, and expand it in the Schur basis in each sub-set of variables. E.g, $P=5 s_{2}(x_1)s_{32}(x_2,x_3)$ or similar. Suppose now that this expansion always have non-negative coefficients, for every choice of partition of letters into 1-or-2-element subsets. Can we conclude that $P$ itself is Schur-positive in $n$ variables? Note that the Littlewood-Richardson rule tells us that the converse is true. If not, is there some $k = k(n)$ such that Schur-positivity on subsets of size $\leq k$ implies Schur-positivity in $n$ variables? The intuition behind why 2-element sets might be enough is as follows: If we want to show that a sum over some combinatorial objects is Schur-positive, it suffices to create a map to SSYTs, or equivalently, to reading-words which are Knuth-equivalent to the SSYTs. To completely describe a reading-word, it suffices to know how many times $i$ appear before $j$, for every pair $i$, $j$. So, assume that I have a set of objects in bijection with SSYTs of shape $\lambda$. By expanding the sum over these it in different choices of 2-variable subsets, we can almost figure out this data above. But I cannot make this formal, at least without using type A crystals or similar... REPLY [3 votes]: My intuition is that if $P$ is in three variables then your requirements only force $P$ to be unimodal in each pair of variables, while being Schur positive is much more restrictive. For example, take $$P(x_1,x_2,x_3)=s_3(x_1,x_2,x_3)+s_{21}(x_1,x_2,x_3)-s_{111}(x_1,x_2,x_3).$$ Equivalently, $$P(x_1,x_2,x_3)=m_3(x_1,x_2,x_3)+2m_{21}(x_1,x_2,x_3)+2m_{111}(x_1,x_2,x_3).$$ Now, partition the variables as $\{\{x_1\},\{x_2,x_3\}\}$, this partition is essentially unique. And then we have $$P(x_1,x_2,x_3)=s_\emptyset(x_1)(s_3(x_2,x_3)+s_{21}(x_2,x_3))+2s_1(x_1)s_2(x_2,x_3)+2s_2(x_1)2s_1(x_2,x_3)+s_3(x_1)s_\emptyset(x_2,x_3).$$<|endoftext|> TITLE: Essential surfaces in knot complements QUESTION [6 upvotes]: Given any knot $K \subset \mathbb{S}^3$, one can find a closed oriented embedded surface $S$ such that $K \subset S \subset \mathbb{S}^3$. Moreover, pick such an $S$ that has minimal genus. One can prove that, if $S \setminus K$ is connected, then it's essential in the knot complement $M=\mathbb{S}^3 \setminus K$. On the other hand, when not, then it provides two copies of a Seifert surface of minimal genus. As it has no reason to be connected, i wonder if one can determine precisely when it is connected. Similarly, take a checkerboard surface of a diagram $D$ of $K$, then it's a theorem from Ozawa that it's essential in $M$ iff $D$ is $\textit{semi-adequate}$. But it has no reason to be oriented, and again there is two cases : if it is, then it provides a Seifert surface, and gluing two copies of it along $K$ will give a surface $S$ as in the first paragraph (is it of minimal genus?), with $K \subset S$ separating. If not, then the boundary of its orientation covering will be an essential, oriented surface with two boundary components, and again gluing those components together will give an $S$ as in the first paragraph, with $K \subset S$ non-separating. Are those two constructions equivalent? What happens for non semi-adequates knots? Is there a characterization of knots for which checkerboard surfaces are oriented or not? Thanks for any answer or comment. REPLY [3 votes]: It is not true in general that if $S\setminus K$ is connected, then it is essential, even if $S$ is assumed to have minimal genus, since every knot embeds in a torus $S$ such that $S\setminus K$ is a boundary parallel annulus. In fact, the Neuwirth conjecture, which is still open, asks whether every non-trivial knot in $S^3$ can be embedded in a closed surface $S$ such that $S\setminus K$ is both connected and essential in $M$. A stronger version of the conjecture asked whether every non-trivial non-torus knot bounds an essential non-orientable spanning surface, but counterexamples to this statement were found by Dunfield. Although a weaker version, sometimes also called the Neuwirth conjecture, was proven by Culler and Shalen. Another way to see that the two constructions are not equivalent, is to notice that for homological reasons any spanning surface must have even integral boundary slope, whereas it is possible for $S\setminus K$ to have odd boundary slope. A checkerboard surface is orientable if and only if its Tait graph is bipartite, and every knot admits a diagram with this property.<|endoftext|> TITLE: When are generic models not too wild? QUESTION [11 upvotes]: This is a question related to ideas raised in http://arxiv.org/abs/1410.1224 and http://arxiv.org/pdf/1405.7456.pdf. Basically, the idea is the following: Suppose I have a first-order theory $T$. Under what conditions are there "few" models of $T$ across all possible forcing extensions of the universe? Maybe "few" is the wrong word - what I really mean is, when does the collection of "models of $T$ you can get by forcing" have some nontrivial structure? There are a bunch of aspects of this question; the one I'm interested in right now is the following: Say that a theory $T$ is generically embeddable if - whenever $\mathbb{P}$ is a forcing notion and $\nu$ is a $\mathbb{P}$-name such that $\Vdash_\mathbb{P}``\nu\models T"$ - there is some forcing notion $\mathbb{Q}$ and $\mathbb{Q}$-name $\mu$ such that $$\Vdash_{\mathbb{P}\times\mathbb{Q}} ``\mbox{$(\mu[G_1]\models T)$ and there is a homomorphism $h\colon \nu[G_0]\rightarrow\mu[G_1]$}."$$ (Note that this homomorphism of course exists in $V[G_0\times G_1]$.) To get a sense of what this means, let's look at a counterexample. Take $T$ to be the true theory of second-order arithmetic - that is, $T=Th(\omega\sqcup 2^\omega; +, \times, \in)$. Then models of $T$ code reals, and this is bad. Specifically, let $\mathbb{P}$ be Cohen forcing, and let $\nu$ be the $\mathbb{P}$-name picking out the "true" model $(\omega\sqcup 2^\omega; +, \times, \in)^{V[G]}$. Then if $G_0\times G_1$ is $\mathbb{P}\times\mathbb{Q}$ generic - for any $\mathbb{Q}$! - since the real $G_0$ won't be in $V[G_1]$, no model $N$ of $T$ in $V[G_1]$ will embed $\nu[G_0]$, since such a homomorphism would have to biject $\nu[G_0]$ with the well-founded segment of $N$, and from this we could recover $G_0$ in $V[G_1]$. More generally, theories which let you code sets are bad. My question is: When is a theory generically embeddable? Specifically, are there model-theoretic niceness properties which guarantee this? For instance, I can't come up with a stable example of a non-generically embeddable theory $T$, but I also can't prove there isn't one. I can prove that if $T$ is totally categorical, then $T$ is generically embeddable, but this isn't very interesting (in fact, it's hard not to prove this). EDIT: Note that there are many natural strengthenings of this: e.g. We may demand that $\mathbb{Q}=\mathbb{P}$, or even $\mathbb{Q}=\mathbb{P}$ and $\mu=\nu$! We may ask for the homomorphism $h$ to be an elementary embedding, or satisfy some other strong property (e.g. the papers linked above looked at isomorphisms, not homomorphisms). We may restrict attention to certain classes of forcing notions, or certain names (e.g. the papers linked above looked only at $\nu$ such that $\Vdash_\mathbb{P}$``$\nu[G]$ is countable"). For now, though, I'm interested in the question as it stands. REPLY [5 votes]: For an example of a super-stable theory failing the property, you can take infinitely many unary predicates (so $Th(2^\omega, U_n: n \in \omega)$ where $U_n(\eta)$ holds iff $\eta(n) = 1$). Then if $\nu$ and $\mu$ satisfy that if $\nu[G_0]$ embeds into $\mu[G_1]$ in $\mathbb{V}[G_0 \times G_1]$ then every quantifier-free type realized in $\nu[G_0]$ is also realized in $\mu[G_1]$, hence is in $\mathbb{V}[G_0] \cap \mathbb{V}[G_1] = \mathbb{V}$. So let $\mathbb{P}$ adjoin a Cohen real $x$ and let $\nu$ be a name for the model of $T$ corresponding to $\{\eta: \mbox{supp}(\eta) \mbox{ is finite}\} \cup \{x\}$. I can give an exact characterization if we replace homomorphism with elementary embedding (which isn't much of a loss, since we can just Morleyize) and require $T$ to be complete, in a countable language: Say that $T$ is generically elementarily embeddable (g.e.e.) if for all $\mathbb{P}$-names $\nu$ for a model of $T$ there is a $\mathbb{Q}$-name $\mu$ for a model of $T$ such that $\mathbb{P} \times \mathbb{Q} \Vdash \nu[G_0]$ elementarily embeds into $\mu[G_1]$. Claim. If $T$ is a complete theory in a countable language, then $T$ is g.e.e. iff $T$ is small. Proof. It follows from the previous example essentially that g.e.e. implies $T$ is small. (If $T$ is not small we can arrange $\nu \models T$ realizes a type not in $\mathbb{V}$, which is enough.) For the reverse direction, let $\mathbb{P}, \nu$ be given. Let $\mathbb{Q}$ force that $\nu$ (i.e., the name in $\mathbb{V}$) becomes countable and let $\mu$ be a name for a countable saturated model of $T$. Then in $\mathbb{V}[G_0 \times G_1]$, $\nu[G_0]$ is countable, $\mu[G_1]$ is countable and saturated (since being $\aleph_0$-saturated is absolute for small theories) and so $\nu[G_0]$ elementarily embeds into $\mu[G_1]$. P.S. Chris Laskowski, Richard Rast and myself independently investigated several similar things to the papers you quoted, see A New Notion of Cardinality for Countable First Order Theories<|endoftext|> TITLE: Fusion Classification of $U_q(sl_N)$ Categories QUESTION [5 upvotes]: Frohlich and Kerler classify categories with $SU(2)_k$ fusion rules and Kazdhan-Wenzl expand this to $SU(N)_k$ categories. In both cases, unless I am missing something, the classification are determined by the dimension of the generating object which seems to imply that there's a hidden pivotal structure lurking about. Is this true? If so, how does the classification refine down to the fusion level? REPLY [2 votes]: The classification says that a fusion category $\mathcal{C}$ with the same fusion rules of $SU(N)_k$ is a twist of $SU(N)_k$ (the twists are determined by an $N$-th root of unity). The point is that $SU(N)_k$ is pivotal (even more is spherical), and a twist of a pivotal fusion category is again pivotal (same for spherical).<|endoftext|> TITLE: When does a quadratic algebra have zero divisors? QUESTION [6 upvotes]: Consider two quadratic algebras based on the vector space $\mathbb{R}^3$ with basis $x,y,z$. The antisymmetric tensor algebra $\Lambda \mathbb{R}^3$ obviously has zero divisors, e.g. $(x)(xy)=0$, but the symmetric tensor algebra $S \mathbb{R}^3$ has not, as it is isomorphic to the polynomial functions on $x,y,z$. For $\Lambda \mathbb{R}^3$ the existence of zero divisors is obvious at degree 1, as $x^2=0$. We could ask this more generally for a possibly noncommutative quadratic algebra with relations $R\subset V\otimes V$ - when is the quadratic algebra $Q(V,R)$ an integral domain? To give a more specific question, if $Q(V,R)$ an not integral domain, then are there always examples of zero divisors in degree 1, i.e. $x,y\in V$ so that $xy\in R$? Apologies if this is well known - could you give me a reference for a proof or counterexample if it is! The question arose from trying to generalise the behaviour of the symbols of differential operators, where classically $SV$ is an integral domain, to operators with other relations. REPLY [2 votes]: Partial answer: when $V$ is 4-dimensional there is a commutative counter-example. In other words, there is a subset of relations $R \subseteq V \otimes V$ containing the commutators and such that the associated quadratic algebra has zero divisors but not in degree 1. Let $V$ be spanned by $W,X,Y,Z$, and let $R \subseteq V\otimes V$ be the subspace spanned by the commutators and by the two polynomials $Y^2-XZ$ and $XW-YZ$. The ideal generated by $R$ corresponds geometrically to the intersection of the two quadrics, in this case the union of a line and a twisted cubic. So the quadratic algebra has zero divisors, namely $$X(YW-Z^2) = (Y^2 - XZ)Z + (XW-YZ)Y = 0.$$ Now suppose that $UV = 0$ for $U, V$ both of degree 1. This would imply that there is a hyperplane containing the twisted cubic, which is not the case.<|endoftext|> TITLE: Connectedness of semialgebraic sets via CAD QUESTION [5 upvotes]: I do not know whether there is a standard or some traditional ways to decide whether a semialgebraic set is connected or not. One way I know is the cylindrical algebraic decomposition (CAD) algorithm. I have read some papers on the CAD algorithm, and they mentioned connectedness can be done by using CAD. However, I have not seen any explanation or example. Can CAD do the connectedness of semialgebraic sets and how? Or any other theoretical way to decide the connectedness of semialgebraic sets? REPLY [2 votes]: We can use the CAD algorithm to decompose a semialgebraic set $S$ into a finite set of connected semialgebraic components $S_i$. We can test whether two such components are adjacent, meaning $\bar{S_i}\cap\bar{S_j}\cap S\neq\emptyset$, using the standard decision procedure for real-closed fields. Then we can form a graph $G$ using the components $S_i$ as vertices, with an edge between $S_i$ and $S_j$ iff they are adjacent in the above sense. Finally we can test whether $G$ is connected, and this will be equivalent to whether $S$ is connected.<|endoftext|> TITLE: Has Dedekind's proof of existence of infinite sets been analyzed by historians? QUESTION [12 upvotes]: This pdf by David Joyce notes that in paragraph 66 of his famous essay, Dedekind claims to prove the existence of an infinite set. The proof exploits the assumption that there exists a set $S$ of all things, and that a mathematical thing is an object of our thought. Then if $s$ is such a thing, then the thought, denoted $s'$, that $s$ is a mathematical object is a thing distinct from $s$. Denoting the passage from $s$ to $s'$ by $\phi$, Dedekind gets a self-map $\phi$ of $S$ which is some kind of blend of the successor function and the brace-forming operation. From this Dedekind concludes that $S$ is infinite, QED. This is quite remarkable. It would be interesting to analyze the hypotheses/axioms that would be used in a formalisation of such a proof. Ferreiros alludes to this proof on page 111 of his book Labyrinth of thought but does not analyze it. Question. Has this proof been analyzed by historians, mathematicians, or philosophers? REPLY [3 votes]: You might want to take a look at Greg Oman's preprint "Unifying Some Notions Of Infinity In $ZC$ and $ZF$", It was to appear in Reports on Mathematical Logic, but you can find the preprint in the Publications section of his Homepage. In that preprint, you might want to take a look at his "Problem 1": Let $ZC$$-$$I$ denote Zermelo set theory with Choice minus Infinity (extensionality, pairing, union, power set, and choice remain). Study the sentences ($\exists$$x$)$\varphi$($x$) for which $ZC$$-$$I$$+$($\exists$$x$)$\varphi$($x$)$\vDash$"There exists a Dedekind (Tarski) infinite set." As regards your question (and Dedekind's 'proof' of the existence of an infinite set), you might want to make '$\varphi$' resemble what you describe in your synopsis of Dedekind's 'proof' and see if it (plus $ZC$$-$$I$) satisfies "There exists a Dedekind (Tarski) infinite set." If your '$\varphi$' does satisfy "There exists a Dedekind (Tarski) infinte set.", then one might rightfully claim Dedekind's 'proof' was successful (the 'realm of his thoughts' notwithstanding). Oman (in his preprint) also states Zermelo's Axiom of Infinity: There exists a set $Y$ with the following properties: (1) $\emptyset$$\in$$Y$, and (2) For all $y$, {$y$}$\in$$Y$. In what sense does $Y$ resemble Dedekind's 'realm of thoughts'?<|endoftext|> TITLE: Metrizability of the space of probability measures endowed with the topology of setwise convergence QUESTION [5 upvotes]: Let $X$ be a separable completely metrizable space, let $\mathscr{B}(X)$ denote the Borel $\sigma$-algebra on $X$, and let $\mathscr{P}(X)$ denote the space of all probability measures on $(X, \mathscr{B}(X))$. Let $\tau$ denote the topology of setwise convergence on $\mathscr{P}(X)$, i.e. the smallest topology on $\mathscr{P}(X)$ such that for every $B \in \mathscr{B}(X)$ the mapping $\nu \mapsto \nu(B)$ on $\mathscr{P}(X)$ is continuous. Question: Is $\tau$ metrizable? REPLY [5 votes]: It's not metrizable. The simple argument given in this answer shows that the finitely supported measures are dense in $\tau$. However, assuming $X$ is uncountable, they are not sequentially dense. Let $\mu_n$ be any sequence of finitely supported probability measures, and let $\mu$ be any atomless Borel probability measure. (Such a $\mu$ exists for any uncountable $X$; for instance, $X$ contains a copy of the Cantor set, which we can equip with Cantor measure. Or for an explicit example, just consider $X = [0,1]$ and $\mu$ Lebesgue measure.) Let $A_n$ be the finite set which is the support of $\mu_n$ and let $A = \bigcup_n A_n$, which is a countable set. We have $\mu_n(A) = 1$ for every $n$, but $\mu(A) = 0$ since $\mu$ is atomless. So $\mu_n$ does not converge setwise to $\mu$. Thus $\mu$ is in the closure of the finitely supported measures, but isn't the limit of any sequence of them. This shows $\tau$ is not first countable.<|endoftext|> TITLE: Why is this character expression an integer? QUESTION [11 upvotes]: Let $\gamma$ be an $n$-dimensional complex representation of a finite group $G$ with character $\chi$ and let $e=c_0, c_1, ..., c_{\ell}$ be a set of conjugacy class representatives for $G$. In the case where $\gamma$ is faithful, I recently obtained the formula $$ \frac{1}{|G|} \prod_{i=1}^{\ell} (n-\chi(c_i))$$ as the size of a certain abelian group associated with $\gamma$ (see Theorem 3 in this paper if interested). In particular, this implies that this quantity is always an integer (if $\gamma$ is not faithful, then the product is 0). Is there a nice way to see that this is an integer just using basic character theory? REPLY [26 votes]: Yes, this is an observation of H. Blichfeldt ( I think J-P. Serre also attributes it to Minkowski), which has been rediscovered many times over the years. Letting $1$ denote the trivial character, note that $\theta = \prod_{i = 1}^{\ell} ( \chi - \chi(c_{i})1) $ is an algebraic integer combination of characters of $G$, and also vanishes everywhere on $G$ except the identity, where its value is $\prod_{i = 1}^{\ell} ( n - \chi(c_{i})) .$ Hence $\theta$ is an algebraic integer multiple of the regular character of $G$, and $\langle \theta,1 \rangle = \frac{1}{|G|}\prod_{i = 1}^{\ell} ( n - \chi(c_{i})) $ is an algebraic integer ( which is easily seen to be rational by Galois theory, hence is a rational integer). (Edit: In fact, it is only necessary to take the distinct values of $\chi(c_{i})$ without repetition in the product which is what Blichfeldt did).<|endoftext|> TITLE: Is there a nice explanation for this curious fact about cyclic subgroups? QUESTION [32 upvotes]: Here's something that I noticed that quite surprised me. Let $G$ be a finite abelian group. Consider the following expression. $$ \nu(G) = \sum_{\substack{H \leq G \\ H \text{ is cyclic}}} |H| $$ It is easy to see that for cyclic groups, we have that $\nu(G) = \sigma_1(|G|)$. What is significantly more surprising is the following. Theorem For every finite abelian group, we have that $$ \nu(G) = \sigma_1(|G|) $$ That is, this only depends on the order of the group. Now, one can see pretty quickly that this is a multiplicative function, and so proving this reduces to studying Abelian $p$-groups. But even there it isn't obvious: consider the groups $\mathbb{Z}/p \times \mathbb{Z}/p$ and $\mathbb{Z}/p^2$. The first of these has lots of small cyclic subgroups, which the second one just has one large one, and amazingly enough these work out to contribute the same amount. So is there a nice explanation for this? This definitely surprised me, and the only way I can prove this is with not-so-pretty computations that don't enlighten me much. REPLY [3 votes]: Another proof of Strickland's result. Let $G$ be a group of order $p^n$ and $\nu(G)$ be the sum of orders of its cyclic subgroups. To prove that $\nu(G)=\sigma_1(G)$, we proceed by induction on $|G|$. If $G$ is cyclic, then, obviously, $\nu(G)=\sigma_1(G)$. Therefore assume that $G$ is noncyclic. If $H TITLE: Are there some intrinsic invariants of surfaces other than Gaussian curvature? QUESTION [16 upvotes]: The principal curvatures of a surface is denoted by $\kappa_{1}, \kappa_{2}$. Let $P(x,y)$ be a polynomial with real coefficients. Assume that $P(\kappa_{1}, \kappa_{2})$ is an intrinsically invariant quantity of all surfaces in $\mathbb{R}^{3}$(It is invariant under isometries of surfaces). Is it true to say that $P(x,y)$ is in the form $P(x,y)=F(xy)$ for some one variable polynomial $F$? In fact this question, which is motivated by "Gauss theorema egregium", asks: Are there some "Theorema Egregiums" other than "Gauss theorema Egregium"? REPLY [15 votes]: As others have pointed out, it's not hard to show that any function $F(\kappa_1,\kappa_2)$ that is intrinsic to the surface metric must be a function of $K = \kappa_1\kappa_2$, so that settles what one might call the 'lowest-order' case. However, there are certainly higher-order versions. For example, the expression $|\nabla K|^2$ is an intrinsic invariant, and it can be expressed as a polynomial in the second fundamental form and its first covariant derivative (a good exercise in a curves and surfaces course). One might want to think of this as a 'higher-order' version of Gauss' theorem, but it's not very exciting because, in some sense, it's a derivative of Gauss' theorem. A natural question that arises (and the one that I thought the OP wanted to ask, based on the title of the question) is whether there is any higher-order theorem of this kind that is not just a derivative (of some order) of Gauss' theorem. The answer to this question is 'no', in the following more precise sense: Suppose given a surface described locally as a graph $z = f(x,y)$ where $f(0,0) = f_x(0,0) = f_y(0,0) = 0$, so that $f$ has a Taylor series expansion of the form $$ f = \tfrac12 c_{20} x^2 + c_{11} xy + \tfrac12 c_{01} y^2 + \tfrac16 c_{30} x^3 + \cdots = \sum_{i+j\ge2} \tfrac1{i!j!} c_{ij}\, x^iy^j. $$ Then Gauss' theorem says that $K(0,0) = c_{20}c_{02}-{c_{11}}^2$. In fact, as is not difficult to show, if one takes the Taylor series of $K$ to be of the form $$ K = \sum_{i+j\ge0} \tfrac1{i!j!} b_{ij}\, x^iy^j, $$ that there exist formulae of the form $b_{ij} = B_{ij}(c)$ where $B_{ij}$ is a universal polynomial in the $c_{kl}$ for which $k+l\le i+j+2$. In fact, for each order $d$, one can collect these to define polynomial mappings $$ B_d: \oplus_{k=2}^{d+2} S^k(\mathbb{R})\longrightarrow \oplus_{k=0}^{d}S^k(\mathbb{R}) $$ that represent the formula giving the Taylor series of $K$ to order $d$ in terms of the Taylor series of $f$ to order $d{+}2$. The version of the question that I have in mind is whether every formula expressing an intrinsic invariant of the induced metric of finite order in terms of the second fundamental form and its covariant derivatives must factor through some $B_d$ at the series level. (Gauss' Theorem and the above arguments show that the answer is 'yes' for intrinsic invariants of order $0$.) The answer is that, indeed, every finite order intrinsic invariant function on the domain of $B_d$ must factor through $B_d$. This is a consequence of the usual proofs of the isometric embedding theorem for real-analytic surfaces.<|endoftext|> TITLE: Dissecting Ramanujan´s Cuboid: 1729 = 19 x 13 x 7 QUESTION [9 upvotes]: Consider the cuboid of dimensions 19 x 13 x 7 whose volume is 1729, the Hardy-Ramanujan number. What is the least number of smaller cuboids into which it can be dissected so that the resulting pieces can be reassembled, first as a unit cube and a cube of side 12, and next as a cube of side 10 and another of side 9? REPLY [3 votes]: This is a partial answer, as I have not fully explored assembly into a cube of side 12. It is a decomposition into 9 cuboids to get the Ramanujan cuboid and the cubes of sides 9 and 10. Maybe someone can break the large piece into 10 or fewer cuboids to finish the answer and give an upper bound. As in the paper in Gerry Myerson's answer, I start with two large pieces of the cuboid, of sizes 9x7x13 and 10x7x13, and break a large piece off each of these (9x9x7 and 10x10x7), leaving 9x7x4 and 10x7x3. The latter piece I leave alone, and the former I divide in half leaving a 9x7x2 piece from which I need to make a 3x3x10 cuboid and a 9x2x2 cuboid. The latter can be cut as a single piece from the 9x7x2 cuboid, giving the third piece for the 9-cube. The remaining 9x5x2 block can be split into two 3x5x2 and two 1x5x3 cuboids to complete the 6 piece dissection of the 10-cube. All the small pieces from the 10-cube dissection fit in the shell formed by removing a corner 9-cube from a 12-cube. If we need to cut the 10x10x7 cuboid to get the remaining pieces, we will need at least five pieces: four to extract the 1-cube, and as one of those pieces is too thick to share room with a 9x9x7 cuboid in a 12-cube, it will need to be cut. Seeing a published dissection having 12 non cuboid pieces, I will be surprised if any cuboidal dissection exists with 14 pieces. I expect a dissection of the above (all the listed pieces above, except breaking the 10x10x7 into smaller cuboids) may be achieved with 18 pieces, but I haven't gotten there yet. Gerhard "Is Feeling Somewhat Cubist Today" Paseman, 2X16.-6.X4<|endoftext|> TITLE: Global section of universal bundle on Grassmanian QUESTION [7 upvotes]: Let $G=G(k, V)$ be the Grassmanian of $k$-dimensional subspaces of the $n$th dimensional vector space $V$, regarded as a smooth algebraic variety over $\mathbb{C}$. Denote with $S$ the tautological (universal) bundle over $G$. On Kapranov's "Coherent sheaves on Grasmann manifold" the following result are stated: $H^0(G, S^*) \simeq V^*$ and $H^0(G, V/S) \simeq V$ The author claims that "These facts are well known". However after a lot of research, I could not find this statement in any reference where I looked for it. It is reasonable that the proof has to be done "by hands", like in the case of the tautological sheaf $\mathcal O (-1)$ over $\mathbb{P}^n$. Do you have any suggestion? REPLY [5 votes]: These are simple instances of the Bott-Borel-Weil theorem. For a complex semsimple group $G$ and a parabolic subgroup $P$ and a complex irreducible representation $W$ of $P$ consider the homogeneous vector bundle $G\times_P W\to G/P$. In this situation the BBW theorem computes the cohomology of the shaef of local holomorphic sections of this bundle as a representation of $G$. The case you need here is that the highest weight of $W$ already is $G$-dominant and integral, in which case the cohomology is concentrated in degree zero and is the $G$-irreducible representation of the same highest weight. (Observe that $S^*$ is the $P$-irreducible quotient of $V^*$, while $V/S$ is the $P$-irreducible quotient of $V$.) The classical Borel-Weil theorem handles the case where $P=B$, the Borel subgroup of $G$, and states that the finite dimensional irredcible representations of $G$ corresponding to a dominant integral weight can be realized as the space of holomorphic sections of the homogeneous line bundle on the full flag manifold $G/B$ induced by the one-dimensional representation of $B$ defined by that weight. A nice exposition of the BBW-theorem can be found in the book on the Penrose transform by Baston and Eastwood.<|endoftext|> TITLE: $p | f(x)$ if and only if $p^k | x$. QUESTION [9 upvotes]: Given a prime number $p$ and a positive integer $k$. Consider integer-valued polynomials $f$ satisfying the property that $p | f(x) \Leftrightarrow p^k | x$. Question. What is the smallest degree of such $f$? I can construct such an $f_k$ with degree $2p^{k-1} - 1$. The idea is to use induction with $f_1 = x$ and $$f_{k+1} = f_k + \sum_{j = 0}^{p^k-1} (-1)^{j} \binom{x}{p^k+j}.$$ Using Lucas' theorem, it can be shown that the sum $\sum_{j = 0}^{p^k-1} (-1)^{j} \binom{x}{p^k+j}$ is 0 $mod~p$ when $p^k \nmid x$, and it is $r~mod~p$ when $p^k | x$, where $r$ is defined to be the residue of $\frac{x}{p^k}$ $mod~p$. Thus when $p^k \nmid x$, $f_{k+1}(x) \equiv f_{k}(x)$ is not divisible by $p$ by induction hypothesis. And when $p^k \mid x$, $f_{k+1}(x) \equiv f_k(x) + r \equiv r~(mod~p)$. It is divisible by $p$ precisely when $p | r$, or $p^{k+1} | x$. I can also show this construction is optimal for $k = 1, 2$. The case $k = 1$ is trivial. For $k = 2$, write any such $f$ as $\frac{g}{p^l}$ where $g$ is a primitive polynomial with integer coefficients and $l \in\mathbb{Z}$. Ignoring other possible factors in the numerator/denominator is without loss. The condition implies $l$ cannot be negative, and $l$ cannot be zero because then we deduce from $p|g(0)$ that $p|g(p) = f(p)$. Moreover, if $l \geq 2$ then the degree of $f$ must be $2p$, due to Polya's result that $d!f(x) \in Z[x]$. Thus we can assume $l = 1$. By condition we have $p| g(x)$ for every integer $x$ but $p^2| g(x)$ only when $p^2 | x$. Hensel's lemma then yields $p | g'(x)$ whenever $p \nmid x$. Therefore $g$ has double roots at every $x \in F_p^*$ and at least one root at $x = 0$, suggesting its degree must at least be $2p-1$. I'm not sure how far this method can be extended to larger $k$, although I do suspect $2p^{k-1}-1$ is the right answer. In general I can only show $d \geq \frac{p^k-1}{p-1}$ by noting that $1-f^{p-1}$ vanishes at $x = 1, 2, \dots, p^{k}-1$ but not at $x = 0$. This implies for $p = 2$ the answer is indeed $2^k-1$. REPLY [5 votes]: Your construction is optimal, that is, the degree of $f$ is always at least $2p^{k-1}-1$. Note that if $m,x$ have $p$-base expansions $m=\sum_{i}m_i p^i$, $x=\sum_{i}x_i p^i$, we have $\binom{x}{m}\equiv \prod \binom{x_i}{m_i} \pmod p$ by Lucas' theorem, and any integer-valued polynomial is an integer linear combinations of binomials $\binom{x}{m}$ (whose result is it, by the way?). So your question is essentially about polynomials in $k$ variables $x_0,\dots,x_{k-1}$ over $\mathbb{F}_p$, which are reduced (have degree at most $p-1$ in each variable). Assume that degree of $f$ is strictly less than $2p^{k-1}-1$. It means that the corresponding polynomial $F\in \mathbb{F}_p[x_0,\dots,x_{k-1}]$ is at most linear in $x_{k-1}$ and its coefficient of $x_{k-1}x_0^{p-1}\dots x_{k-2}^{p-1}$ equals 0. Denote $F=g(x_0,\dots,x_{k-2})+x_{k-1}h(x_0,\dots,x_{k-2})$ and write $y=(x_0,\dots,x_{k-2})$. We have $g(0)=0$, $g(y)\ne 0$ for $y\ne 0$, $h(0)\ne 0$, $h(y)=0$ for $y\ne 0$ (else we may set $x_{k-1}=-g(y)/h(y)$ and find another root of $F$.) But then we may reconstruct the coefficient of $x_0^{p-1}\dots x_{k-2}^{p-1}$ in the polynomial $h(x_0,\dots,x_{k-2})$, and it is non-zero: the sum contains unique non-negative summand (I guess, it is a partial case of Alon-Füredi theorem). A contradiction.<|endoftext|> TITLE: Does 53 diverge to infinity in this Collatz-like sequence? QUESTION [17 upvotes]: This function has been explored a bit at MSE (in June 2016): \begin{eqnarray} f(n) &=& (n-1)^2 \; \textrm{if} \; (n \bmod 4) = 1\\ f(n) &=& \lfloor n/4 \rfloor \; \textrm{otherwise} \end{eqnarray} with $f^k(n) = f(f( \cdots (n) \cdots ) )$ the result of applying $f(\;)$ $k$ times to $n$. The analysis by Gottfried Helms reveals interesting patterns and anomalies, including that it requires $90$ iterations to achieve convergence (to zero) for $n=725$. It would be useful to answer this specific question, on the smallest $n$ that might possibly diverge: Is $f^k(53)=0$ for any finite $k$? The first $25$ iterations are: $$ 53,2704,676,169,28224,7056,1764,441,193600,48400,12100,3025,\\ 9144576,2286144, 571536,142884,35721,1275918400,318979600,\\ 79744900,19936225,397453027378176,99363256844544,\\ 24840814211136,6210203552784,1552550888196,\ldots $$ REPLY [4 votes]: This is no answer, just some more illustrative material triggered by the numberlist of @Stefan Kohl. I consider the numbers $m$ from Stefan's list in base-4 representation. By that representation the iterations exhibit some much simplified pattern: if the last digit is "1" subtract and square, otherwise simply cut that digit. This reduces to two operations on some number $m$: $f(m)$: cut all trailing digits in base-4-representation of $m$ after the last "1" ; return(m) $g(m)$: do $m = (m-1)^2$; do $m = f(m)$ ; return(m) In the following table I document the initial $f(m)$-transform and then the first couple of transforms by $g(m)$. (Note that a lot(!) of iterations of the original description are thus hidden). The data found in the table reduce to only two systematic patterns which are able to decrease to zero. --- m=65533 '33333331'_4 ------------------ 33333331 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=65541 '100000011'_4 ------------------ 100000011 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=131065 '133333321'_4 ------------------ 133333321 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=131069 '133333331'_4 ------------------ 133333331 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=131077 '200000011'_4 ------------------ 200000011 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=131081 '200000021'_4 ------------------ 200000021 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=262129 '333333301'_4 ------------------ 333333301 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=262132 '333333310'_4 ------------------ 33333331 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=262134 '333333312'_4 ------------------ 33333331 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=262135 '333333313'_4 ------------------ 33333331 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=262137 '333333321'_4 ------------------ 333333321 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=262141 '333333331'_4 ------------------ 333333331 3333333200000001 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=262149 '1000000011'_4 ------------------ 1000000011 10000000200000001 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 --- m=262153 '1000000021'_4 ------------------ 1000000021 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=262161 '1000000101'_4 ------------------ 1000000101 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=262164 '1000000110'_4 ------------------ 100000011 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=262166 '1000000112'_4 ------------------ 100000011 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=262167 '1000000113'_4 ------------------ 100000011 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524257 '1333333201'_4 ------------------ 1333333201 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524260 '1333333210'_4 ------------------ 133333321 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524262 '1333333212'_4 ------------------ 133333321 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524263 '1333333213'_4 ------------------ 133333321 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524273 '1333333301'_4 ------------------ 1333333301 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524276 '1333333310'_4 ------------------ 133333331 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524278 '1333333312'_4 ------------------ 133333331 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524279 '1333333313'_4 ------------------ 133333331 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524281 '1333333321'_4 ------------------ 1333333321 3333333200000001 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 1 --- m=524285 '1333333331'_4 ------------------ 1333333331 33333333000000001 3333333200000001 333333300000001 33333320000001 3333330000001 333332000001 33333000001 3333200001 333300001 33320001 3330001 332001 33001 3201 301 21 --- m=524293 '2000000011'_4 ------------------ 2000000011 100000001000000001 10000000200000001 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 --- m=524297 '2000000021'_4 ------------------ 2000000021 10000000200000001 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 --- m=524305 '2000000101'_4 ------------------ 2000000101 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524308 '2000000110'_4 ------------------ 200000011 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524310 '2000000112'_4 ------------------ 200000011 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524311 '2000000113'_4 ------------------ 200000011 1000000100000001 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524321 '2000000201'_4 ------------------ 2000000201 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524324 '2000000210'_4 ------------------ 200000021 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524326 '2000000212'_4 ------------------ 200000021 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 --- m=524327 '2000000213'_4 ------------------ 200000021 100000020000001 10000010000001 1000002000001 100001000001 10000200001 1000100001 100020001 10010001 1002001 101001 10201 1101 121 21 1 Another try with numbers near the odd powers of $\sqrt2$: $x_k=\sqrt 2 ^k $ where $k=2j+1$. Let $a_k=\lfloor x_k \rfloor $ and $b_k=\lceil x_k \rceil $ the two neighboured integers. I expected, that in cases where $x_k$ is "extremely" near to one of the bounds we have possibly special behaviour, like for instance if $m=a_{15} = 181$ and indeed this value seems to have a divergent trajectory while for $k$ in the near of that the trajectories approach zero. Unfortunately it seems that this is not so significant as expected; for $k>80$ I didn't find any convergent trajectory at all. Here is some part of the table: Part 1: take that number of $a_k$ and $b_k$ which is odd as argument $m$: these converge: k=7 m=11------------------ 23 121 21 1 k=9 m=23------------------ 11 1 k=11 m=45------------------ 231 1321 3201 301 21 1 k=13 m=91------------------ 11 1 this doesn't seem to converge k=15 m=181------------------ ( likely divergent !) 2311 133221 33220021 3310030322001 3220330213031221021 31033320111230303300302001313223101 223320031033333201331222232232023120303121021311312301030210223221 these converge k=17,19,21,23,25,27,29,35,39,47,55,63,67,71 k=75 m=194368031999 ------------------ 2311001 13333321 333332000001 33333000001 3333200001 333300001 33320001 --- known convergent pattern these don't seem to converge: k=31,33,37,41,43,45,49,51,53,57,59,61,65,69,73,77... 255 For the selections of the even number of $a_k$ and $b_k$ we have a similar picture. Part 2: take that number of $a_k$ and $b_k$ which is even as argument $m$: these converge k=7 m=12 --------------- 30 1321 3201 301 21 1 k=9 m=22 --------------- 11 1 k=13 m=90 --------------- 11 1 k=15 m=182 --------------- 231 1321 3201 301 21 1 these converge too k=17,21,25,29,31,33,37,43,51,59,75,79 k=83 m=3109888511976 --------------- 2311001 13333321 333332000001 33333000001 3333200001 333300001 33320001 // this is known to be a convergent pattern These don't seem to converge k=11,19, 23, 27, 35,39,41,45 .. 49,53 ..57 ,61..73 ,77, 81 ,85 ...255<|endoftext|> TITLE: Do we have super Plucker relations for a super Grassmannian? QUESTION [5 upvotes]: Super Grassmannians are introduced by Manin, see for example. We have Plucker relation for Grassmannian. Are there some references about super Plucker relations for super Grassmannian? Thank you very much. REPLY [2 votes]: There's a paper titled On super Plücker embedding and possible application to cluster algebras just came out this summer.<|endoftext|> TITLE: $p$-adic completeness of the ring of Witt vectors QUESTION [8 upvotes]: Let $R$ be a ring that is $p$-adically complete for a prime $p$ and let $W(R)$ denote the ring of $p$-typical Witt vectors. Is it true that $W(R)$ is $p$-adically complete? (A ring $A$ is $p$-adically complete if the map $A \rightarrow \varprojlim(A/p^nA)$ is bijective.) A reference that contains a proof or a counterexample would suffice. I am familiar with the case when $R$ is a perfect $\mathbb{F}_p$-algebra and am interested in the general case. REPLY [2 votes]: The answer is yes. A reference is $\S1$, proposition 3 of "The display of a formal $p$-divisible group" by Thomas Zink, published in Asterisque no. 278. Here is the address of the paper on Zink's website... https://www.math.uni-bielefeld.de/~zink/display.pdf<|endoftext|> TITLE: Hasse principle for rational times square QUESTION [22 upvotes]: Does a Hasse principle hold for the property of being a rational times a square ? Let $a \in \mathbb{K}$ be an element of a number field. Assume that at every place $\mathbb{K}_v$ of $\mathbb{K}$, $a$ can be written as $a=q k^2$, with $q \in \mathbb{Q}$ and $k \in \mathbb{K}_v$. Is it true that $a$ can always be written as $q k^2$ with $q \in \mathbb{Q}$ and $k\in \mathbb{K}$ ? EDIT : there's a restriction at the real places of $\mathbb{K}$. One should assume that the sign of $a$ is the same at every real place. REPLY [4 votes]: No, in general the Hasse principle for the property of being a rational number times a square does not hold. I consider the question in the form of GH from MO. I give a counter-example with a non-normal extension $K/\mathbb{Q}$ of degree 6 (the accepted answer of David Speyer gives a counter-example with a normal extension of degree 16). It suffices to take the concrete Galois extension $L/\mathbb{Q}$ of degree 12 with Galois group $G=A_4$ and with cyclic decomposition groups from the answer of Jeremy Rouse, and to take $K=L^H$, where $H$ is a subgroup of $G$ of order 2. Consider the homomorphism of $\mathbb{Q}$-tori $$ \varphi\colon\ \mathbb{G}_{m,\mathbb{Q}}\times_{\mathbb{Q}}R_{K/\mathbb{Q}}\mathbb{G}_{m,K} \ \to\ R_{K/\mathbb{Q}}\mathbb{G}_{m,K}\,,\quad (q,k)\mapsto qk^2. $$ We wish to show that $Ш^1(\mathbb{Q},\ker \varphi)\ne 0$. Set $T=R_{K/\mathbb{Q}}\mathbb{G}_{m,K}/\mathbb{G}_{m,\mathbb{Q}}$. We write $T[2]$ for the subgroup of elements of order dividing 2 in $T$. The homomorphism $$ \mathbb{G}_{m,\mathbb{Q}}\times_{\mathbb{Q}}R_{K/\mathbb{Q}}\mathbb{G}_{m,K} \ \to\ T,\quad (q,k)\mapsto \mathbb{G}_{m,\mathbb{Q}}\cdot k $$ indices a homomorphism $\ker\varphi\to T[2]$ fitting into a short exact sequence $$ 1\to \mathbb{G}_{m,\mathbb{Q}}\to\ker\varphi\to T[2]\to 1, $$ from which we obtain a canonical isomorphism $$ Ш^1(\mathbb{Q},\ker\varphi)\overset{\sim}{\to} Ш^1(\mathbb{Q}, T[2]).$$ Write $M=T[2]$. It suffices to show that $Ш^1(\mathbb{Q}, M)\ne 0$. We have a reduction $Ш^1(\mathbb{Q},M)= Ш^1(L/\mathbb{Q},M)$, see Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80, Lemma 1.1(ii). Since all the decomposition groups of $L/\mathbb{Q}$ are cyclic, we have $Ш^1(L/\mathbb{Q},M)=Ш^1_\omega(G,M)$ (see my question for the definition of $Ш^1_\omega(G,M)$ ). By the answer of Kasper Andersen $Ш^1_\omega(G,M)\ne 0$. Thus $Ш^1(\mathbb{Q},\ker\varphi)\neq 0$, and our extension $K/Q$ is a counter-example to the Hasse principle. It would be interesting to construct explicitly an element $a\in K$ for which the Hasse principle fails (i.e., $a=qk^2$ locally, but not globally.)<|endoftext|> TITLE: Is it known whether every symmetric pair of finite groups of Lie type is a Gelfand pair? QUESTION [8 upvotes]: A pair of groups $(G,H)$ is called a symmetric pair if $H$ is the group of fixed points of an involutive automorphism of $G$, for example $(GL(2n,\mathbb{F}_q),Sp(2n,\mathbb{F_q}))$ is a symmetric pair with respect to the involution $\theta(A)=J(A^{-1})^tJ^{-1}$ where: $$ J=\left(\begin{array}{cc} 0&id_n\\-id_n&0 \end{array}\right) $$ We also say that a pair $(G,K)$ where $K$ is a subgroup of $G$ is a Gelfand pair if for any irreducible representation $\pi$ of $G$, we have that $\pi^K$ (the vectors fixed by $K$) is at most one dimensional. Alternatively, $Hom_K(\pi, 1)\leq 1$. It is known that many symmetric pairs of finite groups of lie type are Gelfand pairs, as in many we have that an associated anti-involution $\sigma(x)=\theta(x^{-1})$ preserves $H-H$ double cosets. I was wondering if it was known that any such symmetric pair of finite groups of Lie type is a Gelfand pair, or if there is a known counterexample to that assertion. REPLY [9 votes]: The symmetric space ${\rm GL}(2,{\mathbb F}_q)/T$, where $T$ is the diagonal torus, is not a Gelfand pair : the Steinberg representation contains the trivial character of $T$ with multiplicity $2$. Here is a proof that the multiplicity of this character is $\geqslant 2$. The Steinberg representation may be viewed as the space of functions $f$ on the projective line ${\mathbb P}^1 ({\mathbb F}_q)$ satisfying $$ \sum_{x\in {\mathbb P}^1 ({\mathbb F}_q)}f(x)=0\ . $$ The torus $T$ fixes exactly two points $x$, $y$ of the projective line and acts transitively on the $z\not= x,y$. Let $f_0$ the functions defined by $f_0 (x)=1=-f_0 (y)$ and $f_0 (z)=0$, if $z\not= x,y$. Similarly let $f_1$ the function defined by $f_1 (x)=f_1 (y)=1$ and $f(z)=-2/(q-1)$, if $z\not= x,y$. Then ${\rm Vect} (f_0 ,f_1 )$ is a two dimensional subspace of the Steinberg representation fixed by $T$.<|endoftext|> TITLE: Harmonic analysis, compute that this integral tends to $0$ QUESTION [5 upvotes]: We have the following setting. $U$ is a bounded Lipschitz domain in the complex plane. Consider the following classical Dirichlet problem for the Laplace operator: $$\begin{align} \Delta{}u&=0 \text{ } \text{in}\text{ } U, \quad u =f \,\partial U.\end{align} $$ $F(x)=-\frac{1}{2\pi}\log|x|$ denotes the fundamental solution for the Laplacian in the complex plane. For $x\in{}U$, $B(x,\epsilon)\subseteq U$ denotes the usual ball centered at $x$ with raidus $\epsilon>0$. $U_\epsilon{}=U\backslash B(x,\epsilon)$. So what I want to get is the following $$\begin{align} \int\limits_{\partial{}B(x,\epsilon)}F(x-y)\frac{\partial{u}}{\partial{\nu}}(y)d\sigma(y)=\int\limits_{\partial{}B(x,\epsilon)}-\frac{1}{2\pi}\log|x|\frac{\partial{u}}{\partial{\nu}}(y)d\sigma(y) \end{align}$$ tends to zero as $\epsilon\rightarrow0$. Perhaps we can derive this by using the explicit formula for $F$, but I don't know how to. My thoughts so far are as follows. Note that $y \mapsto F(x - y)$ is constant on $\partial B(x, \epsilon)$, and is equal to $-1/2\pi\,\log(\epsilon)$. Hence, we can factor it, and we need a final inequality of the form$$\left|\int_{\partial B(x, \epsilon)} {{\partial \mu}\over{\partial \nu}}\right| \le c\epsilon,$$but I can not tell much about that. REPLY [2 votes]: If $u$ is smooth on $B(x,\epsilon)$, then $$\int_{\partial B(x,\epsilon)} \frac{\partial u}{\partial \nu} = \int_{B(x,\epsilon)} \Delta u = 0$$ by the divergence theorem.<|endoftext|> TITLE: Davis, Figiel, Johnson and Pełczyński factorization through spaces with a bases QUESTION [8 upvotes]: Davis, Figiel, Johnson and Pełczyński's Factorization Theorem states that each weakly compact operator $T:X \to Y$ between Banach spaces $X$ and $Y$ factors through a reflexive Banach space $Z$. In addition, they prove that $Z$ can be constructed to have a Schauder basis if $Y$ has a shrinking basis or an unconditional basis. It was later shown that if $Y=C[0,1]$ then $Z$ can be constructed with a basis. I am interested generally in positive and negative results regarding when $Z$ can be constructed to have a basis. For example: Q1: If $Y=L_1[0,1]$, can $Z$ be constructed to have a basis for any weakly compact $T:X \to Y$? Q2: Is there an example of a space $Y$ with basis and weakly compact $T:X \to Y$ so that every reflexive $Z$ which $T$ factors through does not have a basis? I'm also interested to know if anyone is aware of other results similar to the those stated above. REPLY [5 votes]: I think Q1 has a positive answer. To see this, review how the results are proved in DFJP. You have a weakly compact subset $W$ of $Y$ (the closure of the image of the unit ball of $X$ under a weakly compact operator) and you apply the factorization technique (which we now know is just a real interpolation method applied to the pair $(W, B_Y)$) to $W$ to obtain a larger weakly compact symmetric subset $C$ of $Y$. Consider the normed space that has $C$ as its unit ball. It is proved that $C$ is weakly compact in in this normed space and hence this normed space is reflexive. When $Y$ has a (always Schauder) basis, instead of applying the interpolation technique to the original weakly compact set $W$, we enlarged $W$ to a bigger weakly compact set $W_1$ that is invariant for the partial sum projections associated with the basis for $Y$. Then the interpolation technique, applied to $W_1$, produces a still larger weakly compact set $C$ that is invariant for the partial sum projections associated with the basis for $Y$. From this it is easy to see that the basis for $Y$ is also a basis for the space that has $C$ as its unit ball. For $W_1$ we used the closed convex hull of the union over $n$ of $P_n W$, where $(P_n)$ are the partial sum projections for the basis for $Y$. When the basis is shrinking, we proved that $W_1$ is weakly compact. Suppose that $Y=L_1$ and you use the Haar basis for $Y$. Even though the Haar basis is far from being shrinking, the same construction produces a weakly compact set $W_1$. To see this, recall that a weakly closed set $D$ in $L_1$ is weakly compact iff for all $\epsilon > 0$ there is $M=M_\epsilon$ s.t. $D\subset M B_{L_2} + \epsilon B_{L_1}$ (``uniform integrability") and use the fact that both $B_{L_2}$ and $B_{L_1}$ are invariant under the Haar basis projections.<|endoftext|> TITLE: $C(X)$-compact operators and families of compact operators QUESTION [5 upvotes]: In this question Operators on Hilbert $C^*$-module and families of Fredholm operators I asked about the relation between being a family of compact operators $F:X \to K(H)$ on Hilbert space $H=\ell^2$ and being $C(X)$-compact operator on Hilbert module $H_{C(X)}$ (here $X$ is a compact space). If I take a family $F$ like above then I can view it as an operator on $H_{C(X)}$ by the formula $F(\xi)(x)=F_x(\xi(x))$ where $\xi \in H_{C(X)}$ and therefore $\xi(X) \in H=\ell^2$ so $F_x(\xi(x))$ does makes sense. I wonder whether the opposite is true, namely if we have a $C(X)$-compact operator on $H_{C(X)}$ may we somehow interpret it as a continuous family of compact operators? I suspect that this no longer will be true for all operators, so in the case if the answer is affirmative for compact operators, and negative for general operators, I would also like to understand where the obstruction lies. REPLY [3 votes]: The Hilbert module $H_{C(X)}$ is the space of all continuous functions $X \to H$, and hence for each $s \in H_{C(X)}$, we can talk about $s(x) \in H$, for each $x \in X$. Assume that $F:H_{C(X)} \to H_{C(X)}$ is $C(X)$-linear. We get individual operators $F_x: H \to H$, namely $$F_x (v):= (F\tilde{v})(x), $$ where $v \in H$ yields the constant function $\tilde{v}: X \to H$. It is then not hard to show that for each $s \in H_{C(X)}$ and $x \in X$, the relation $$(Fs)(x)= F_x (s(x))$$ holds. Claim: if $F$ is $C(X)$-compact, then $F_x$ is compact for each $x \in X$ and $x \mapsto F_x$ is (norm-)continuous. Proof: let $\epsilon>0$. Then, by the definition of a $C(X)$-compact operator, there exists finitely many $s_i, t_i \in H_{C(X)}$, such that the operator $$G=\sum_i t_i \langle s_i, \_ \rangle$$ on $H_{C(X)}$ satisfies $\| G-F\| \leq \epsilon$. It is clear that $G_x$ is compact for each $ x\in X$ and $x\mapsto G_x$ is norm-continuous. Note that $\| F_x - G_x\| \leq \epsilon$. Since $\epsilon$ was arbitrary, the proof is complete. WARNING: if $F$ is a $C(X)$-Fredholm operator on $H_{C(X)}$, then you also get Fredholm operators $F_x$ on $H$, but the map $x\mapsto F_x$ is not continuous, at least not if the target has the norm topology.<|endoftext|> TITLE: References - Voevodsky motives are the derived category of Nori motives? QUESTION [6 upvotes]: First I would like to know if this has been worked out, and if the answer is affirmative I would like to know some references. REPLY [5 votes]: Example 3.20 of "Tannaka duality for enhanced triangulated categories", arXiv: 1309:0637v4 says that this would be true for rational motives if a t-structure compatible with Betti realisation exists. By Theorem 3.4 of Hanamura's "Mixed motives and algebraic cycles III" (MRL 1999), such a t-structure would follow from Grothendieck's standard conjectures, Murre's conjectures and a generalisation of the Beilinson-Soule vanishing conjecture.<|endoftext|> TITLE: Polynomial approximations of curves QUESTION [8 upvotes]: This is the 3D version of this question. The responses to that question contained a lot of complaints about fuzzy definition of the problem, so I made this new question very narrow and explicit. For a given fixed degree, $n$, I'm interested in constructing parametric polynomial approximations of degree $n$ of certain curves, for example those that arise as intersections of two surfaces. This is important in engineering and manufacturing computing, where parametric polynomials are usually the only curves available (in the form of Bézier curves). A typical example is as follows: Let $f(x,y,z) = x^2 + y^2 - 16$, and let $S(f)$ be the cylindrical surface where $f(x,y,z)=0$. Similarly, let $g(x,y,z) = y^2 + z^2 - 25$, and let $S(g)$ be the cylindrical surface where $g(x,y,z)=0$. Let's focus on the curve $C$ that is the portion of the intersection of these two surfaces lying in the positive octant where $x,y,z \ge 0$. A few simple calculations show that $C$ has end-points $\mathbf{p}_0 = (4,0,5)$ and $\mathbf{p}_1 = (0,4,3)$. I want to construct a parametric polynomial curve $t \mapsto \mathbf{x}(t) = \big(x(t), y(t), z(t)\big)$ of degree $n$ that approximates $C$ and such that $\mathbf{x}(0) = \mathbf{p}_0$ and $\mathbf{x}(1) = \mathbf{p}_1$. The error in the approximation will be measured by $$ E(\mathbf{x}) = \sup\big\{ f(\mathbf{x}(t))^2 + g(\mathbf{x}(t))^2 : 0 \le t \le 1\big\} $$ First question: for a given $n$ how can I find the polynomial curve $\mathbf{x}$ of degree $n$ that minimizes $E(\mathbf{x})$? If we were approximating a real-valued function, there would be a great deal of approximation machinery that would help us: the Weierstrass approximation theorem, the equi-oscillation criterion, the Remez algorithm, etc. If I parameterize $C$, then each of its three components will be a real-valued function, and I can apply these known techniques. But the result will depend on how I parameterize $C$, and will not be optimal. Second question: Has any of the above-mentioned approximation machinery been generalized beyond the real-valued case? REPLY [3 votes]: Do you care more about the curve or the surfaces? Replacing the first function in your example by $100(x^2 + y^2 - 16)$ or $x^2-z^2+9$ gives the same curve but a different error function. Also, consider the intersection of the cylinder $(x-11)^2+(y-11)^2=221$ and the plane $x+y-z=1.$ It is an ellipse and the portion $\mathit{C}$ in the positive octant goes from $(1,0,0)$ to $(0,1,0)$ visiting $(k,k,2k-1)$ along the way where $k=11+\sqrt{442}/2 \approx 21.51.$ I'm guessing that a polynomial curve minimizing what you want would go along the small missing section which goes through $(j,j,2j-1)$ where $j= 11-\sqrt{442}/2 \approx 0.49$ so $2j-1<0$. Of course we could specify that the entire curve should stay in the positive octant but then $(t,1-t,0)$ might be pretty hard to beat with anything we would think of as approximating $\mathit{C}.$. That said, your particular example seems well behaved. In the case of $n=2$ (chosen for illustration) you seek $9$ constants so that for $\mathbf{x}(t)=(at^2+bt+c,dt^2+et+f,gt^2+ht+i)$ the maximum of $ f(\mathbf{x}(t))^2 + g(\mathbf{x}(t))^2$ on $[0,1]$ is minimized. The extra condition on the position for $t=0$ and $t=1$ reduces to $3$ degrees of freedom. In the case of polynomial approximation of a real curve $f(t)$ by a degree $n$ polynomial $p_n(t)$ on $[0,1]$ (with or without side conditions for $t=0,1$) It is usually not possible to explicitly find the best polynomial approximation although it is known that there is one which is unique and it can be recognized by the extreme values of $f-p_n$ in $[0,1]$ being equal (say to $E$) in absolute value, alternating in sign, and sufficient in number. (In particular, $f-p_n=0$ (at least) $n+1$ times in $[0,1].$) There are iterative schemes , as I recall, which can do well. Pick a set of points $X=\{t_1,t_2,\cdots,t_n\}$ in the open interval (somehow) and solve the system $p_n(0)=f(0),p_n(1)=f(1)$ and $p_n(t_i)-f(t_i)=(-1)^iE$ for the unknown coefficients of $p_n$ and the unknown error value $E.$ Find the set $Y$ of maximum errors of $|p_n-f|.$ If by a miracle you got the right thing you can recognize it. Otherwise, replace $X$ with $Y$ and try again. I'm not sure how to adapt that. I was able to get a pretty good approximation for your example with $n=2$ $$(- 3.776\,{t}^{2}+ 7.776\,t,- 2.831\,{t}^{2}- 1.169\,t+4,- 0.813\,{t}^ {2}+ 2.813\,t+3). $$ The worst errors are about $1.1$ near $t=0.2$ and $t=0.8.$ This arose as follows: Specify that the error is $0$ (as required) at $t=0,1$ and also at $t=1/2$ This gives one degree of freedom (Say for $d$ which corresponds to picking which point on the curve corresponds to $t=1/2$.) The error increases to a maximum in the first half then decreases to $0$ at $t=1/2$ then increases then decreases again. Keep adjusting until the two maxima are roughly equal. LATER Thinking more about your particular example, you have a curve $\mathit{C}$ so that the projections onto the $xy$ and $zy$ planes are (quarter) circles. For a curve with projections onto two orthogonal planes which are easily described curves your method with the intersection of the two corresponding (generalized) right cylindrical surfaces seems like a natural choice based on the curve itself. Even better might be projection (if possible) onto three pairwise orthogonal planes. In your case the projection onto the $xz$ plane is is a portion of the hyperbola $z^2-x^2=9$. So minimize the maximum value over the interval of $ f(\mathbf{x}(t))^2 + g(\mathbf{x}(t))^2+h(\mathbf{x}(t))^2.$ Of course another three orthogonal planes would give a different error function though maybe not too different a result. Still, for an application, Bézier curves for each component might be fast and satisfactory.<|endoftext|> TITLE: Spin^c structures on manifolds with almost complex structure QUESTION [13 upvotes]: Let $M$ be a smooth even-dimensional manifold. Is it true that for each almost-complex structure $J$ on $M$ there exists a canonical spin$^c$ structure $S_J$ associated to $J$ ? (I've read this somewhere but I didn't see the actual construction). Is there a way to caracterize that spin$^c$ structure ? (What I mean by that is : is there a theorem of the form : $S_J$ is the unique spin$^c$ structure on $M$ satisfying [a property of compability with $J$]). Suppose $M$ admits an almost-complex structure. Does every spin$^c$ structure comes from an almost-complex structure on $M$ (i.e. is it true that every spin$^c$ structure is equal to a $S_J$ for an almost-complex structure $J$ ?). (If not : what happens if we restrict to parallelizable open 4-manifolds ?). Same question as 1. 2. but for spin (and not just spin$^c$) structures. EDIT (to clarifiy this question) : Does every spin structure on $M$ comes from has an associated spin^c structure that is associated to an almost-complex structure $J$ on $M$ ? REPLY [8 votes]: A $\text{Spin}^c$ structure is equivalent to (a homotopy class of) an almost complex structure on the 2-skeleton of a manifold which extends to the 3-skeleton (except for a surface or when the fiber dimension is odd, where we first need to stabilize the tangent bundle). So in the case of 4-manifolds without 4-handles (in particular 4-manifolds with non-empty boundary) there is a canonical one to one correspondence between $\text{Spin}^c$ structures and almost complex structures. The original reference for this result is Gompf's Spin^c structures and homotopy equivalences (https://doi.org/10.2140/gt.1997.1.41), but there is a more expository version available in Surgery on Contact 3-Manifolds and Stein Surfaces by Ozbagci and Stipsicz.<|endoftext|> TITLE: Imaginary quadratic fields: Euclidean if and only if norm Euclidean QUESTION [10 upvotes]: Let $K$ be an imaginary quadratic field and $O_K$ be its ring of integers. We say $O_K$ is norm Euclidean if the norm is a Euclidean function. It is known from the classification of imaginary quadratic fields with class number 1 that $O_K$ is Euclidean if and only if it is norm Euclidean. Is there a more straightforward proof of this fact? REPLY [9 votes]: Assume that $O_K$ is Euclidean. The Motzkin Sets $_j$ are defined by $E_0 = \{0\}$, $E_1 = E_0 \cup O_K^\times$, $E_2$ is the set of elements of $O_K$ such that each residue class is represented by an element in $E_1$ etc. $O_K$ is Euclidean if every element of $O_K$ is in some $E_i$ (see Sect. 2.3 here). Since $E_1$ has $3$ elements, $O_K$ must have an element of norm $\le 3$, which implies that the discriminant of $K$ is $\ge -11$. All these rings are norm-Euclidean, and the converse is trivial. This is classical and has nothing to do with the solution of the class number $1$ problem.<|endoftext|> TITLE: Quasi-split tori and algebraic groups QUESTION [10 upvotes]: Let $k$ be a perfect field. Recall that an algebraic torus $T$ over $k$ is called quasi-split if there exists some finite étale $k$-algebra $A$ such that $$T \cong \mathrm{R}_{A/k} \mathbb{G}_m.$$ A reductive group $G$ over $k$ is called quasi-split if it contains a Borel subgroup $B$ over $k$. A priori, I see no reason why these two definitions should be related; Indeed any algebraic torus is quasi-split in the second sense (take $B=T$), but generally not in the first sense. However, I have seen it claimed, or at least implicitly used, in proofs that for semisimple $G$, the latter implies the former. Namely: Let $G$ be a semisimple algebraic group over $k$ with a maximal torus $T$. If $G$ is quasi-split, then is $T$ quasi-split? Does the converse hold? REPLY [12 votes]: A torus $T$ is quasi-split if its character group is a permutation representation for the Galois group. So a counterexample to your question is: let $G$ be the quasi-split group $SO(n+1,n-1)$, $n\ge2$, $k=\mathbb R$. The weight lattice is $\mathbb Z^n$ with Galois action $$ (x_1,\ldots,x_{n-1},x_n)\mapsto(x_1,\ldots,x_{n-1},-x_n) $$ This is not a permutation module. On the other hand, the Galois action on $T\subseteq B\subseteq G$ of a quasi-split group is induced by an action on the simple roots. These form a basis if $G$ is of adjoint type. For simply connected groups one can play the same game with fundamental representations. So: If $G$ is quasi-split, and either simply connected or of adjoint type then the maximal $k$-torus sitting in a $k$-Borel is quasi-split. I have the suspicion that the two notions of "quasi-splitness" developed independently.<|endoftext|> TITLE: Example of steady Ricci soliton whith indefinite or nonpositive Ricci curvature QUESTION [6 upvotes]: I am looking for example of steady Ricci soliton with indefinite or nonpositive Ricci curvature. Any help will be appreciated. Thanks! REPLY [5 votes]: I have found doing some calculation that the metric: $$g(t)=\frac{dx^2+dy^2}{e^{-4t}-x^2-y^2}$$ satisfies $\frac{dg(t)}{dt}=-2Ric(t)$ Where $$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$ this should be a solution to the Ricci flow on $R^2$. Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$. Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive. a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$ b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$. For $t=0$,   I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$. Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$. Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$. This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.<|endoftext|> TITLE: Blow-up in family QUESTION [8 upvotes]: Let $\pi \colon X \to T$ be a flat projective morphism, and let $Y$ be a closed sub-scheme of $X$ which is flat over $T$. We can assume that everything is defined over the complex numbers, and $T$ is one dimensional, say smooth and affine. I want to consider the blow-up $\eta \colon X' \to X$ with centre $Y$. I have the following three questions: Is $X'$ flat over T? Is $(X')_t$ the blow-up of $X_t$ at $Y_t$? Where $t$ is a closed point of $T$, and the sub-index $t$ means that we are taking the fibre over $t$. In case of positive answer to the previous question, does the exceptional divisor $E$ of $X'$ restrict to the exceptional divisor of $(X')_t $? I think the answer is, thanks to the flatness of $Y$, always positive; however, I am not that sure about the proofs I have, and I could not find any reference. REPLY [15 votes]: No, that is not true. First the positive result. If you add the hypothesis that $X$ is integral, then $X'$ is integral (Proposition II.7.16, p. 166 of Hartshorne). Thus $X'$ is flat over $T$ (Proposition III.9.7, p. 257 of Hartshorne). Even in this case, typically $(X')_t$ does not equal the blowing up of $X_t$ at $Y_t$. The simplest example I know uses a general pencil of hyperplane sections of a singular quadric cone. So let $\widehat{X}$ be the singular quadric cone in $\mathbb{P}^3_k = \text{Proj}\ k[p,q,r,s]$ with defining equation $pr-q^2$. Consider the Lefschetz pencil of hyperplane sections with base locus $\text{Zero}(pr-q^2,q,s)$, and let $X\to \widehat{X}$ be the blowing up of the base locus. Equivalently, for $T=\mathbb{P}^1_k = \text{Proj}\ k[Q,S]$, let $X$ be the closed subscheme of $\mathbb{P}^3_k \times \mathbb{P}^1_k$ that is the zero scheme $\text{Zero}(pr-q^2, qS-sQ)$. The issue, of course, is that $X$ is not locally factorial at the singular point $x$ given by $([p,q,r,s],[Q,S]) = ([0,0,0,1],[0,1])$. The induced morphism $\pi:X\to T$ is flat since $X$ is Cohen-Macaulay (even locally a hypersurface), $T$ is regular, and $\pi$ has constant fiber dimension $1$. Denote by $t$ the image of $x$ under $\pi$. So now let $Y\subset X$ be the closed subscheme that is the following zero scheme $\text{Zero}(p-q,r-q,qS-sQ)$. This closed subscheme of $X$ is the image of a section of the projection morphism. In fact, $Y\subset X$ is everywhere locally a Cartier divisor except at the special point $x$. The blowing up of $Y$ in $X$ (technically the ideal sheaf of $Y$) is the same as the blowing up of $x$ in $X$. The exceptional divisor $E$ is a smooth curve isomorphic to $\mathbb{P}^1$, and it is contained in the fiber over $t$. Thus, $(X')_t$ has an extra irreducible component, $E$, beyond the blowing up of $X_t$ at $Y_t$. I believe that the argument about flatness above may work even without the hypothesis that $X$ is integral, but I need to check this . . . Edit. The flatness result above is true so long as $X$ is $T$-flat, even if $Y$ is not $T$-flat. Since $X$ is flat over $T$, the sheaf $\mathcal{O}_X$ is flat over $\pi^{-1}\mathcal{O}_T$. Since $T$ is a Dedekind scheme, this equivalent to saying that $\mathcal{O}_X$ has zero $\pi^{-1}\mathcal{O}_T$-torsion. Thus, for every quasi-coherent ideal sheaf $\mathcal{J}\subset \mathcal{O}_X$, also $\mathcal{J}$ has zero $\pi^{-1}\mathcal{O}_T$-torsion. In particular, for the defining ideal $\mathcal{I}$ of $Y$, for every integer $d\geq 0$, the ideal $\mathcal{J}=\mathcal{I}^d$ has zero $\pi^{-1}\mathcal{O}_T$-torsion. Thus the sheaf of graded $\mathcal{O}_X$-algebras, $$\mathcal{S} = \bigoplus_{d\geq 0} \mathcal{I}^d,$$ has zero $\pi^{-1}\mathcal{O}_T$-torsion. Thus, the relative Proj of $\mathcal{S}$ is $T$-flat. Second Edit. Thanks to Allen Knutson for correcting inconsistent notation.<|endoftext|> TITLE: Convergence of an implicitly defined sequence of random variables QUESTION [12 upvotes]: Let $\{X_n\}_{n\ge 1}$ be a sequence of independent identically distributed Poisson random variables with mean $\lambda^*$. Consider a sequence of random variables $\{\hat{\lambda}_{n}\}_{n\ge 1}$ defined recursively as solutions to the following equation: $$ X_{n+1} = \hat{\lambda}_{n+1} + \left(\sum_{i=1}^n\hat{\lambda}_i\right)\ln\frac{\hat{\lambda}_{n+1}}{\hat{\lambda}_{n}},\ n=1,2,\ldots$$ with $\hat{\lambda}_{1}=X_1$. That is, given $\{\hat{\lambda}_{i}\}_{i=1}^n$ and $X_{n+1}$, we solve the above equation to find $\hat{\lambda}_{n+1}.$ Question: does $\hat{\lambda}_{n}$ converge to $\lambda^*$ in some sense (e.g., in probability)? I am looking for a general argument if possible using tools like martingale convergence theory, stochastic approximation, etc., but the basic convergence can be seen in simulations. Here is an example for $\lambda^*=10$: Moreover, if $\delta=\hat{\lambda}_{n+1}-\hat{\lambda}_{n}$ is small, then $\ln\frac{\hat{\lambda}_{n+1}}{\hat{\lambda}_{n}}\approx \frac{\delta}{\hat{\lambda}_{n}}$. If also $\{\hat{\lambda}_{i}\}_{i=1}^n$ are all "close" to a particular value, then the right-hand-side is approximately $\hat{\lambda}_{n+1} + n\delta$ and $$\hat{\lambda}_{n+1}\approx\frac{X_{n+1}+n\hat{\lambda}_{n}}{n+1},$$ which is the standard recursive formula for the sample average. REPLY [2 votes]: This is just a bound from below. The derivation is a bit long for a comment. Clearly, a positive solution to the equation exists (although you need to specify in more details what happens when $X_1=0$ as $\lambda_1=0$ in this case). Now, we have \begin{align*} \sum_{j=2}^nX_{j} &= \sum_{j=2}^n\lambda_j+\sum_{j=2}^n\left(\sum_{i=1}^{j-1} \lambda_i\right)\ln (\lambda_j/\lambda_{j-1})\\ &=\sum_{j=2}^n\lambda_j+\sum_{i=1}^{n-1}\lambda_i\left(\sum_{j=i+1}^n \ln(\lambda_j/\lambda_{j-1})\right) \\ &=\sum_{j=2}^n\lambda_j+\sum_{i=1}^{n-1}\lambda_i \ln(\lambda_n/\lambda_i) \end{align*} Now since $\lambda_i>0$ we can make use of the inequality $\ln(1+x)\le x$ to obtain \begin{align*} \sum_{j=2}^nX_{j}&\le \sum_{j=2}^n\lambda_j+\sum_{i=1}^{n-1}\lambda_i (\lambda_n/\lambda_i-1)\\ &=\lambda_n-\lambda_1+(n-1)\lambda_n \end{align*} Dividing both sides by $n$ and letting $n \to \infty$ we obtain by the Law of Large Numbers, $$ \lambda \le \liminf_{n\to\infty}\lambda_n, \quad \mbox{a.s.} $$<|endoftext|> TITLE: minimal surfaces in $S^n$ QUESTION [6 upvotes]: Thanks to Choi-Schoen theorem, we know that the space of embedded minimal surfaces into $S^3$ of fixed genus is compact. My question are simples: Can we remove the embeddness assumption? Can we replace $S^3$ by $S^n$ for $n\geq3$, still for surfaces (not hypersurfaces)? Else what are the most reasonable results of "weak" compactness we can expect. For instance, is that reasonable to expect that if we have $\displaystyle \Sigma_\infty =\lim_{n\rightarrow +\infty} \Sigma_n$ in a reasonable sense, do we have $\displaystyle Area(\Sigma_\infty)= \lim_{n\rightarrow +\infty} Area(\Sigma_n)$? REPLY [9 votes]: Without embeddedness, the Choi--Schoen theorem is false. For example, there is a huge family of rotationally symmetric immersed tori in $\mathbb{S}^3$ (the only embedded one is the Clifford torus, by Brendle's solution to the Lawson conjecture). See Brendle's survey, Theorem 1.4, where he shows that there is an infinite family of immersed minimal tori of the form $$ \left(\sqrt{1-r(t)^2} \cos s, \sqrt{1-r(t)^2} \sin s, r(t) \cos t, r(t) \sin t\right) \in \mathbb{S}^3\subset\mathbb{R}^4 $$ for $r(t)\in (0,1)$ a smooth function. I'll sketch the construction and argue that compactness fails. Thanks to the symmetry, the minimal surface equation for $r(t)$ can be integrated once, to become $$ \frac{r'(t)^2}{r(t)^4(1-r(t)^2)^2}+\frac{1}{r(t)^2(1-r(t)^2)}=\frac{4}{c^2} $$ for some constant $c\in (0,1]$. The solution $r(t) = \frac{1}{\sqrt{2}}$ (i.e. $c=1$) is the embedded Clifford torus. You can argue that the maximum and minimum values of $r(t)$ (depending on $c$) are given by $$ \overline{x}(c)=\sqrt{\frac{1-\sqrt{1-c^2}}{2}}, \underline{x}(c)=\sqrt{\frac{1+\sqrt{1-c^2}}{2}}. $$ Furthermore, the period of the solution is given by $$ T(c) = \int_{\underline{x}(c)}^{\overline{x}(c)}\frac{c}{x\sqrt{1-x^2}\sqrt{4x^2(1-x^2)-c^2}}dx. $$ For this to be an immersed torus, one needs that $\frac{2\pi}{T(c)}$ is rational, so that it closes up. You can check that $T(c)\to\sqrt{2}\pi$ as $c\nearrow 1$ and $T(c)\to\infty$ as $c\searrow 0$. Now, suppose that $c$ is chosen so that $\frac{2\pi}{T(c)}$ is rational. Choose $t$ so that $r'(t) =0$ and $r(t) = \underline{x}(c)$. Then, as computed in Brendle's survey, we have that the second fundamental form at this point satisfies $$ h(\partial_s,\partial_s) = r(t)\sqrt{1-r(t)^2} $$ On the other hand, the metric satisfies $$ g(\partial_s,\partial_s) = 1-r(t)^2, $$ so the norm of the second fundamental form at this point is at least $$ \frac{r(t)}{\sqrt{1-r(t)^2}}=\sqrt{\frac{1+\sqrt{1-c^2}}{1-\sqrt{1-c^2}}}\to\infty $$ as $c\searrow 0$. As for minimal surfaces in $\mathbb{S}^n$ for $n>3$, I think that the space of even minimal spheres is quite big. See, e.g. Calabi's paper or e.g. this paper. I'm not completely sure about embeddedness of these examples, or whether or not their second fundamental forms blow up (certainly their areas do). If you have that $\Sigma_n$ converges to $\Sigma_\infty$ as varifolds (up to a subsequence, this always holds assuming a priori area estimates on $\Sigma_n$), then $area(\Sigma_n) \to area(\Sigma_\infty)$.<|endoftext|> TITLE: For which $n$ is there a permutation such that the sum of any two adjacent elements is a prime? QUESTION [25 upvotes]: For which $n$ is it possible to find a permutation of the numbers from $1$ to $n$ such that the sum of any two adjacent elements of the permutation is a prime? For example: For $n=4$ the permutation $(1,2,3,4)$ has sums $1+2=3$, $2+3=5$, $3+4=7$,and $4+1=5$. For $n=16$ the permutation $(1,2,3,4,7,6,5,14,15,16,13,10,9,8,11,12)$ has sums of adjacent elements $(3,5,7,11,13,11,19,29,31,29,23,19,17,19,23,13)$. REPLY [4 votes]: The question is not new. In fact, in 1982 Antonio Filz [J. Recreational Math. 14(1982), p.64] conjectured that for any $n = 2,4,6,\ldots$ there is a circular permutation $i_1,\ldots,i_n$ of $1,\ldots,n$ such that all the $n$ adjacent sums $$i_1 + i_2,\ i_2 + i_3,\ \ldots,\ i_{n−1} + i_n,\ i_n + i_1$$ are prime. In 2013 I made a similar conjecture involving twin primes. I conjectured that for any positive integer $n$ there is a circular permutation $i_0,i_1,\ldots,i_n$ of $0,1,\ldots,n$ such that all the $n + 1$ adjacent sums $$i_0 + i_1,\ i_1 + i_2, \ \ldots,\ i_{n−1} + i_n,\ i_n + i_0$$ belong to the set $$S=\{k ∈\mathbb N :\ 6k −1\ \text{and}\ 6k + 1\ \text{are twin primes}\}.$$ See http://oeis.org/A228917.<|endoftext|> TITLE: Are rational surface singularities $\mathbb{Q}$-Gorenstein? QUESTION [5 upvotes]: I know that, in general, rational singularities are not necessarily $\mathbb{Q}$-Gorenstein. So I ask: is there any positive result in this direction known for surfaces? REPLY [6 votes]: The answer is yes. In fact, there is the following Proposition 1. Every $2$-dimensional rational singularity $(X, \, x)$ is analytically $\mathbb{Q}$-factorial, i.e. there exists an analytic neighborhood $V$ of $x$ such that every Weil divisor on $V$ is a $\mathbb{Q}$-Cartier divisor. In particular, every $2$-dimensional rational singularity is $\mathbb{Q}$-Gorenstein. Concerning the particular case of rational Gorenstein singularities, we have Proposition 2. For a $2$-dimensional normal singularity $(X, \, x)$, the following are equivalent: $(X, \, x)$ is a rational double point (i.e., a Du Val singularity); $(X, \, x)$ is a rational hypersurface singularity; $(X, \, x)$ is a rational Gorenstein singularity; $(X, \, x)$ is a canonical singularity. See S. Ishii, Introduction to Singularities, Theorem 7.3.2 and Theorem 7.5.1.<|endoftext|> TITLE: Serre fibration vs Hurewicz fibration QUESTION [16 upvotes]: What are the simplest/ typical examples of a Serre fibration which is not a Hurewicz fibration? Is it something pathological? Sorry if the question is too elementary for MO. REPLY [18 votes]: A short paper with references to several early counterexamples proves that (in the good category of compactly generated weak Hausdorff spaces) a Serre fibration in which the total space and base space are both CW complexes is necessarily a Hurewicz fibration. M. Steinberger and J. West. Covering homotopy properties of maps between CW complexes or ANRs. Proc. Amer. Math. Soc. 92 (1984), 573–577, doi:10.1090/S0002-9939-1984-0760948-6. (The proof is corrected in R. Cauty. Sur les overts des CW-complexes et les fibr'es de Serre. Colloquy Math. 63 (1992), 1–7, link). No relationship between the covering map and the CW structures is required. This argues either that counterexamples are pathological or that it is a special property for the total space of a Serre fibration with CW base space to be a CW complex, although it has the homotopy type of a CW complex if the fiber does.<|endoftext|> TITLE: When is bar-cobar duality an equivalence? QUESTION [14 upvotes]: Let $A$ be an augmented differential graded algebra over a field $k$. I will write $BA$ for its bar construction (whose homology is $Tor^A(k, k)$). This is a co-augmented differential graded coalgebra over $k$; write $\Omega BA$ for its cobar construction. There is a natural dga map $\Omega BA \to A$ which uses the fact that $\Omega C$ is the tensor algebra on a shift of $C$, and the projection map $BA \to A$. I think I'm only going to highlight my ignorance here, but my question is: what assumptions on $A$ imply that the map $\Omega BA \to A$ is an equivalence? If I am reading any of the following references correctly: Husemoller-Moore-Stasheff's "Differential homological algebra and homogeneous spaces" (section II.4) Felix-Halperin-Thomas "Rational homotopy theory" (exercise 2, section 19) Keller "A-infinity algebras, modules and functor categories" (Theorem 4.3, attributed to Lefevre), then I think that the answer is supposed to be: "no assumptions are required at all." I seem to have two possible counterexamples to this claim. But their being counterexamples seems to rely on the positive answer to another question: when is $\Omega$ a homotopy-invariant functor? That is: if $f: C \to D$ is an equivalence of dg coalgebras, is $\Omega f$ an equivalence of dga's? Let me explain the (counter?)examples. Let $G$ be a finite group, $k[G]$ its group ring, and $k^G$ the dual Hopf algebra (the ring of functions on $G$ with values in $k$). Then both $k[G]$ and $k^G$ are augmented algebras, where the augmentation is the counit. $B(k[G])$ is isomorphic to the simplicial chain complex of the bar construction $BG$ (with coefficients in $k$). If $k$ has characteristic zero, then this is is equivalent to $k$, via a transfer argument. If $\Omega$ is a homotopy invariant, then $\Omega B(k[G]) \simeq \Omega k = k$, which is certainly not $k[G]$ unless $G$ is trivial. $k^G$ is isomorphic as a $k$-algebra to a product of fields $$k^G \cong \prod_{g \in G} k,$$ and the augmentation can be identified with the map which projects onto the factor corresponding to the identity $e \in G$. The kernel of the augmentation is a summand, and hence a projective $k^G$-module. From this, you can write down a very short resolution of $k$ over $k^G$ and compute $Tor^{k^G}(k, k) = k$, concentrated in $Tor_0$. So $B(k^G) \simeq k$. Again, if $\Omega$ is homotopy-invariant, then $\Omega B(k^G) \simeq \Omega k = k$, which is not equivalent to $k^G$. So either I have misunderstood the references above, or $\Omega$ is not a homotopy invariant functor. I'm guessing it's the latter; can anyone point me to a reference which gives criteria for this to hold? REPLY [7 votes]: See Hasegawa's thesis, available here. Lemma 1.3.2.3 is what you want, on page 35 of the pdf. To see that $\Omega B A \to A$ is a quasi-isomorphism, one filters $BA$ by its primitive elements (i.e. by the length of bar elements) which then induces on $\Omega BA$ a filtration. One gives $A$ the trivial filtration (i.e. $F^0A=0$ and all higher $F^i$ are $A$) and sees the counit is a filtered morphism. The associated graded map is the identity of $A$ on degree one, and then one shows that $\text{Gr}_i(\Omega BA)$ is contractible for $i>1$. Another way to remember why the counit is a quasi-isomorphism is that it corresponds to the trivial twisting cochain $\beta:BA\longrightarrow A$ whose total space $BA\otimes_\beta A$ is acyclic. Then one fits the counit into the following sequence $$1\otimes \varepsilon_A : BA\otimes_\gamma \Omega B A\to BA\otimes_\beta A$$ Both twisted complexes above are acyclic so $1\otimes \varepsilon_A$ is a quasi-isomorphism. Now $BA$ is simply connected so a spectral sequence argument works to show that, because $1$ is of course a quasi-isomorphism, so is $\varepsilon_A$. This argument fails for $\Omega$ because it decreases the connectivity of your space. Lemma 1.3.2.2 in the thesis above gives a sufficient criterion for $\Omega$ to preserve quasi-isomorphisms. The bar construction, on the other hand, preserves quasi-isomorphisms by a direct argument using the filtration by length and the Kunneth formulas. Again, no connectivity issues arise here, which renders the desired spectral sequence convergent.<|endoftext|> TITLE: IC sheaf of certain explicit variety QUESTION [15 upvotes]: Let $n,m$ be two positive integers. Let $Z$ denote the closed subvariety in $\mathbb A^n \times \mathbb A^m$ given by the equation $x_1...x_n=y_1...y_m$. QUESTION: What is the stalk (with the action of Frobenius) of the IC sheaf of $Z$ at the point (0,....0)? When n=m=2 you get 3-dimensional quadratic cone and the stalk is cohomology of $\mathbb P^1$ (up to the corresponding shift and Tate twist). But I don't know what happens for higher $n,m$. Note that the stalk in this case is the same as the global intersection cohomology. REPLY [2 votes]: It's a toric variety, for which intersection cohomology is computed algorithmically, see e.g. De Catldo's lectures.<|endoftext|> TITLE: Sequence of nested sets in $[0, 1]$ with bound on gaps QUESTION [9 upvotes]: What is the best possible $\epsilon$ and sequence $(a_n)_{n = 1}^\infty \subset [0, 1]$ we can find such that $$ d_{N}:=\sup_{x\in [0,1]}\inf_{n=1}^N |x-a_n|\leq \frac{1+\epsilon}{N} $$ for all $N\in \mathbb{N}$? Note that dyadic decomposition provides $\epsilon=1$. REPLY [8 votes]: Let me begin by reformulating the question a little bit, making it precise, and also avoiding what happens for the initial finite number of $N$. Given an infinite sequence $a_n$ in $[0,1]$, for every $N$ let $u_N$ denote the smallest gap between the numbers $a_1$, $\ldots$, $a_N$ (when arranged in ascending order between $0$ and $1$) and let $v_N$ denote the maximum gap. The problem is essentially interested in understanding $v_N/2$ (the point $x$ is halfway between the ends of the maximum gap). The problem as formulated above was studied by A. Ostrowski, de Bruijn and Erdos and also solved (somewhat later) by Toulmin. de Bruijn and Erdos (and Toulmin) showed that for any sequence one has $$ \limsup_{N\to \infty} Nv_N \ge\frac{1}{\log 2} = 1.4426\ldots, $$ and this value is attained for the sequence $$ a_n = \{ \log_2 (2n-1)\}, $$ where $\{ x\}$ stands for fractional part, and $\log_2$ denotes the logarithm to base $2$. They also show that $\liminf_{N\to \infty} Nu_N$ is always at most $1/\log 4$ and this too is attained in the example above. You might also find interesting this related recent paper of Chung and Graham. My first thought on seeing the problem was sequences of the form $\{ n\theta\}$ for an irrational $\theta$. I think that in this class of examples, the golden ratio $\phi = (1+\sqrt{5})/2$ might be best and gives a normalized largest gap of $\phi^3/\sqrt{5} = 1.8944\ldots$ (if I calculated correctly). This is not as good as Toulmin's example. Nevertheless, you might also find interesting this survey of the three gaps theorem (which appeared in L'Enseignement).<|endoftext|> TITLE: Vanishing of Kahler differentials vs. surjective Frobenius? QUESTION [8 upvotes]: Let $A$ be an $\mathbf{F}_p$-algebra such that $\Omega_{A/\mathbf{F}_p}=0$. Is the Frobenius map on $A$ surjective? Some context: i. The converse is clearly true. ii. The answer is yes if $A$ is a field, or of finite type (the latter with a somewhat silly interpretation). REPLY [3 votes]: Here is a counterexample for $p$ odd. We will construct a sequence of $\mathbb{F}_p$-algebras $$\mathbb{F}_p[t]=R_0 \subset R_1 \subset R_2 \subset \dotsb$$ such that each $R_i$ has the following properties: For any $f \in R_i$, there exists $j \ge i$ so that $df$ becomes zero in $\Omega^1_{R_j/\mathbb{F}_p}$. $t$ is not a $p$th power in $R_i$. The underlying set of $R_i$ is countable. $R_i$ is $\mathbb{Q}_{\ge 0}$-graded, the degree $0$ piece is $\mathbb{F}_p$, and $\deg t=1$. Let $R = \varinjlim_i R_i$. Then property 1 implies that $\Omega^1_{R/\mathbb{F}_p}=0$, while property 2 guarantees that $t$ is not a $p$th power in $R$. Properties 3 and 4 will be used to prove properties 1 and 2, respectively. It suffices to show that if we have a ring $R_i$ satisfying properties 2-4 and a homogeneous element $f\in R_i$ of positive degree, then we can find a ring $R_{i+1} \supset R_i$ so that $R_{i+1}$ also satisfies properties 2-4, and $df=0$ in $\Omega^1_{R_{i+1}/\mathbb{F}_p}$. The construction will be based on iterated applications of the following observation (which I learned from Johan de Jong): in the ring $$\mathbb{F}_p[x,y,z]/(z^p-xy^p)\,,$$ $d(x^2y^p)=0$ but $x^2 y^p$ is not a $p$th power. Proof: $$d(x^2 y^p)=y^p d(x^2)=2xy^p\,dx=2x\,d(xy^p)=2x\,d(z^p)=0$$ On the other hand, the subring of $p$th powers is generated by $x^p$, $y^p$, $xy^p$, and it is clear that $x^2y^p$ is not in this subring. So, finally, here is the construction. Choose an integer $n$ so that $(\lfloor np/(p-1) \rfloor+1) \deg f > 1$. Let $R_{i+1}$ be $R_i[x_1,\dotsc,x_n,y_1,\dotsc,y_n,z]$ modulo the relations \begin{align*} f & = x_1^2 y_1^p \\ x_1 y_1^p & = x_2^2 y_2^p \\ & \dots \\ x_{n-1} y_{n-1}^p & = x_n^2 y_n^p \\ x_n y_n^p & = z^p \,. \end{align*} Give $x_1,\dotsc,x_n,y_1,\dotsc,y_n,z$ any positive grading consistent with the relations. We can show by induction that $d(x_m^2 y_m^p)=0$ for each $m$, so in particular $df=0$. We can also show that for each $m$, $R_i[x_1,\dotsc,x_m,y_1,\dotsc,y_m] \cap R_{i+1}^p$ is generated by $R_i[x_1,\dotsc,x_m,y_1,\dotsc,y_m]^p$ and some elements of the form $(x_m y_m^p)^a$ for $a \ge \lfloor (n-m)p/(p-1) \rfloor+1$ (the $n-m+1$st positive integer not divisible by $p$), and similarly $R_i \cap R_{i+1}^p$ is generated by $R_i^p$ and some elements of the form $f^a$ for $a \ge \lfloor np/(p-1) \rfloor+1$. In particular, by our choice of $n$, all such elements have degree $>1$. So $t$ cannot become a $p$th power in $R_{i+1}$.<|endoftext|> TITLE: Needing proof of convergence for a sequence QUESTION [14 upvotes]: Let $\left\{u_i\right\}_{i=1}^\infty$ be a sequence of real vectors (i.e. $u_i\in R^n, i=1,2,... $) and $m$ an integer large enough such that $\sum_{i=1}^m u_i u_i^T$ is a positive definite matrix. Define: $$K_m:=\left(\sum_{i=1}^m u_i u_i^T\right)^{-1}u_m$$ Show that $\left\| K_m\right\| $ tends to zero as $m\longrightarrow \infty$. If $u_i$s are scalar, i.e. $n=1$, the proof seems trivial (not that trivial, but easy to get). I have not been able to find a counter-example for the cases where $u_i$s are vectors (and believe me, I have tried hard). So I think this is very likely to be true in general. REPLY [3 votes]: Let me first discuss the special case where $|u_n|$ is bounded. Let's denote our matrices by $A_n=\sum_{j=1}^n u_ju_j^*$. What we're trying to show can only fail if there are infinitely many $u_n=c_nv_n + w_n$ that have a coefficient $|c_n|\ge\delta >0$ when expanded in terms of the eigenvectors of $A_n$, such that the corresponding eigenvalue $\lambda_{n}$ stays bounded. By compactness, these eigenvectors $v_n$ converge to a vector $v$ on a subsequence. Then, however, $\langle v_n, A_n v_n\rangle$ will become large because each of the many summands $u_ju_j^*$ of $A_n$ with a $u_j$ that is as above (with a component $\ge\delta$ in almost the direction of $v_n$) makes a contribution $\gtrsim \delta^4$ to this quadratic form. So $\lambda_n$ can not stay bounded. In general, if $|u|$ is large, then consider (writing $Q=uu^*$) $$ |A^{-1/2}u|^2 = |(Q+B)^{-1/2}u|^2 =\langle u, (Q+B)^{-1} u\rangle ; $$ this is bounded because $-1/z$ is operator monotone, so $B$ in the quadratic form can be essentially dropped (more precisely, replace it by $\epsilon B$). In other words, $x_n=A_n^{-1/2}u_n$ is always bounded, and we still find ourselves in the same scenario as above: to prevent $A_n^{-1/2}x_n$ from going to zero, $x_n$ and thus also $u_n$ must have a component bounded away from zero belonging to a bounded eigenvalue of $A_n$.<|endoftext|> TITLE: bounding derivative of a sequence QUESTION [6 upvotes]: I've been banging my head against the wall on this one ... define a sequence of polynomials $q_n$ by $q_0 = 0$ and $$q_{n+1} = q_n + .5(t^2 - q_n^2).$$ If $q_n \leq t$ on $[0,1]$ then $$.5(t^2 - q_n^2) = .5(t - q_n)(t+q_n) \leq t - q_n$$ so that $q_{n+1} \leq t$, and also $$t - q_{n+1} = (t-q_n)(1 - .5(t + q_n)) \leq (t-q_n)(1 - .5t),$$ so that inductively $t - q_n \leq (1 - .5t)^n$ on $[0,1]$. Thus $q_n$ increases pointwise to $t$ on $[0,1]$. Are the derivatives $q_n'$ uniformly bounded on $[0,1]$? If $r_n = q_n'$ then the recurrence relation is $$r_{n+1} = t + (1 - q_n)r_n$$ but I don't see how this helps. REPLY [5 votes]: Your last recurrence relation indeed helps. That is: 1) Away from zero, once you consider $t>\epsilon$ and $n\ge 1$, you have $q_n$ bounded away from zero by a positive constant $c=\min_{[\epsilon,1]} q_1(t)$, and hence $r_n$ cannot jump over the fixed point of the map $r\mapsto t+(1-c) r$, which is $r=\frac{t}{c}$. 2) Now, you have to handle the neighborhood of zero. If you had the limit value $t$ or at least $\alpha t$ (with $\alpha$ a constant) instead of $q_n$ in your recurrence relation, the fixed point would be $\frac{t}{\alpha t}=\frac{1}{\alpha}$, and it would be done. 3) Now you handle different $t$'s separately. For each individual $t$, for the first $1/t$ iterations there is nothing to worry about: anyway you have $r_n\alpha t$. Indeed, on every step $j$ you add $\frac{1}{2}(t^2-q_j^2)$. Either at some $j\frac{1}{2}t$, and then you are done. Or it does not happen, and then $q_n\ge n\cdot \frac{1}{2}(t^2-\frac{1}{4}t^2) = n\cdot \frac{3}{8}t^2 \sim \frac{3}{8}t$. So you have $\alpha=\frac{3}{8}$, and the proof is complete.<|endoftext|> TITLE: Imprimitive solutions to $x^2+y^3=z^7$ QUESTION [7 upvotes]: Poonen, Schaefer, & Stoll give the primitive solutions to $x^2+y^3=z^7$: $$ (±1, −1, 0), (±1, 0, 1), ±(0, 1, 1), (±3, −2, 1), (±71, −17, 2),\\ (±2213459, 1414, 65), (±15312283, 9262, 113), (±21063928, −76271, 17). $$ I'm looking for all the solutions with $1\le z\le\ell$ for some fixed $\ell$. Clearly the primitive solutions yield infinitely many imprimitive solutions via multiplication by $\operatorname{lcm}(2,3,7)=42$nd powers, but this doesn't find them all. For example, $250^2+25^3 = 5^7$ and $832^2+112^3 = 8^7$. How can I find all the solutions? My first instinct was to deal with each special case on its own but I'm afraid I'll miss cases that way. [1] Bjorn Poonen, Edward F. Schaefer, and Michael Stoll, Twists of X(7) and primitive solutions to x^2+y^3=z^7, Duke Math. J. 137:1 (2007), pp. 103-158. REPLY [14 votes]: You can find the solutions for any given $z$ by looking for the integral points on the elliptic curve $$x^2 = (-y)^3 + z^7$$ (which would usually be written $y^2 = x^3 + z^7$). The curve is isomorphic to the curve obtained by replacing $z^7$ with $z$, so the computation is feasible for reasonable values of $z$. Magma (for example, but also SAGE) has an implementation of a procedure that finds all these points (there are finitely many in each case) in quite reasonable time. For example, there are exactly 990 solutions (up to a sign change in $x$) for $1 \le z \le 1000$. Here is the Magma code (replace ell by $\ell$): list := []; for z := 1 to ell do pts := IntegralPoints(EllipticCurve([0, z^7])); list cat:= [ : pt in pts]; end for; list; If you do not have access to Magma, you can try it out with the online Magma calculator. With ell=100, it takes about 13 seconds there. The computation for $\ell = 1000$ on my computer took about as long as writing this answer.<|endoftext|> TITLE: Can noncomputable sets be distinguishable in $RCA_0$? QUESTION [12 upvotes]: Say that a set $X\subseteq\omega$ is distinguishable if there is some Turing machine $\Phi_e$ which, when given two sets exactly one of which is $X$, can determine which set is $X$. Formally, $X$ is distinguishable if there is some Turing machine $\Phi_e$ such that for all $Y\not=X$, $$\Phi_e^{X\oplus Y}(0)=0,\quad \Phi_e^{Y\oplus X}(0)=1.$$ (Think of "$0$" and "$1$" as meaning "Left" and "Right.") Clearly every computable set is distinguishable; it is not hard to show (see my answer to https://math.stackexchange.com/questions/1189370/is-there-a-turing-machine-that-can-distinguish-the-halting-problem-among-others) that the converse also holds. My question is about the reverse mathematics of the converse. Specifically, my proof used Weak Konig's Lemma, and I don't immediately see a way to do without it. So my question is: Is it consistent with $RCA_0$ that there is a non-computable, distinguishable set? Motivation: Over the last year or so I've developed an interest in "alternate computability theories" - e.g. see Visser's delightfully-named paper "Oracle bites theory" at http://www.phil.uu.nl/preprints/lgps/authors/visser/oracle-bites-theory/pdf. I'm especially interested in "almost computable" sets in such theories, and distinguishable sets might provide such an example. REPLY [4 votes]: Extending D. Turetsky's answer: If $\mathcal{M}$ is a Turing ideal containing no PA degree, then every $\Delta_2$ degree in $\mathcal{M}$ contains a set that is computably distinguishable within $\mathcal{M}$. Lemma. If $\mathcal{M}$ contains no PA degree, then $\mathcal{M}$'s version of the Cantor space $2^\omega$ is computably homeomorphic to $\mathcal{M}$'s version of the Baire space $\omega^\omega$. Proof. Fix a computable binary tree with no path, and name its leaf nodes $\tau_1,\tau_2,\ldots$. Every $f \in 2^\omega$ is a concatenation of infinitely many $\tau_{n_1}\tau_{n_2}\cdots$ and through this maps to the string $n_1n_2\cdots \in \omega^\omega$. Proof of main claim. Fix a $\Delta_2$ set $D$ in $\mathcal{M}$. Construct a computable tree $T \subseteq \omega^{<\omega}$ whose unique path $P$ is $D$'s computation function, i.e., the map taking $n$ to the least $s$ for which $D_s\cap[0,n] = D \cap [0,n]$. Transform $T$ using the Lemma into a computable $T^* \subseteq 2^{<\omega}$. Then $T^*$ has a unique path $P^* \equiv_T P \equiv_T D$, and this $P^*$ is distinguishable by the procedure which takes $X,Y$ and checks to see if their initial segments are all in $T^*$. $\square$ I'd be curious to know whether your proof of distinguishable-equals-computable really needs $WKL$ in the form of a $PA(X)$ degree, or whether it can be finessed down to a $PA$. It would narrow the gap if you're looking for a reversal.<|endoftext|> TITLE: Bar/Cobar Adjunction Between Modules and Comodules QUESTION [9 upvotes]: There is a pretty well known, and widely written about, adjunction between augmented algebras and coaugmented coalgebras given by taking the bar construction on algebras and the cobar construction on coalgebras. Sometimes, under certain circumstances (for instance, if your algebras are group-like loop spaces), this adjunction is an equivalence (e.g. the equivalence between group-like loop spaces and connected spaces) My question is: when does this lift to categories of modules and comodules? In other words, given an augmented monoid $A$ (augmented, let's say, over $k$) and an $A$-module $X$, we can produce a $BA$-comodule $BX$ by taking a bar construction $Bar(X,A,k)$, which is effectively a kind of ``quotient" by the $A$-action. There should, I feel, be a right adjoint going from $BA$-comodules to $A$-modules, given by $Cobar(C,BA,k)$ ($C$ being a comodule). A good example is the case of $G$-spaces for a discrete group $G$. If we have a group action of $G$ on a space $X$, then we have an "action groupoid" which is basically a simplicial diagram whose bottom part looks like $G\times X\rightrightarrows X$ and whose colimit is $X//G$. In particular, we now also have a map $X//G\to BG=\ast//G$ endowing $X//G$ with a $BG$-comodule structure (using the diagonal map of $X//G$). On the other hand, given a map $Y\to BG$, we can perform a cobar construction whose effect is to take the fiber of $Y$ over the base point, which is, by elementary topological considerations, once again a $G$-space. I'm skipping a lot of the technicalities of describing this kind of structure, but I'm really interested in knowing how general this structure is. In particular, I'd really like to be able to go between modules and comodules, and I'd really like to do this in some homotopical-enough setting (e.g. quasicategories or simplicial model categories). Are there well known references for this? Note: There are TONS of references for this kind of thing for going between algebras and coalgebras. I'm very specifically interested in doing this for modules and comodules. REPLY [8 votes]: The topological picture is the following. Suppose $(X, x)$ is a based connected space. There's an equivalence of $\infty$-categories between spaces $Y \to X$ over $X$ ("$X$-comodules") and spaces with an action of $\Omega X$ ("$\Omega X$-modules"). The special case of $G$-spaces occurs when $X = BG$ as you say. You might call this, together with the equivalence between based connected spaces and grouplike $E_1$ spaces, "topological Koszul duality." One way of thinking about where this equivalence comes from is to notice that starting with a space $Y \to X$, taking the homotopy fiber at $x$ is the right adjoint of an adjunction between $\text{Space}_{/x} \cong \text{Space}$ and $\text{Space}_{/X}$, where the left adjoint is just given by composition with the inclusion of $x$ into $X$. This adjunction induces a monad on $\text{Space}$ whose functor part is $\Omega X \times (-)$, and you can hope that 1) an algebra over this monad is precisely a space with an action of $\Omega X$, and 2) this adjunction is monadic via Barr-Beck-Lurie. Dually, starting with a space $F$ equipped with an action of $\Omega X$, taking the homotopy quotient is the left adjoint of an adjunction between $\Omega X\text{-Space}$ and $\text{Space}$, where the right adjoint equips a space with the trivial action. This adjunction induces a comonad on $\text{Space}$ whose functor part is $X \times (-)$, and you can hope that 1) a coalgebra over this comonad is precisely a space over $X$, and 2) this adjunction is comonadic via Barr-Beck-Lurie. Now you can hope for the same sorts of things to be true in a more algebraic setting. For example, let $f : R \to S$ be a morphism of algebras (e.g. ring spectra). To match up with the topological picture (which you can hope is reasonable if $R, S$ are $E_{\infty}$), $\text{Spec } R$ is $X$ and $\text{Spec } S$ is $x$. Taking the (derived) tensor product $(-) \otimes_R S$ is the left adjoint of an adjunction between $\text{Mod}(R)$ and $\text{Mod}(S)$, where the right adjoint is restriction. This adjunction induces a comonad on $\text{Mod}(S)$ whose functor part is $(-) \otimes_S (S \otimes_R S)$ (again matching up with the topological picture, $\text{Spec } S \otimes_R S$ is $\Omega X$), and you can hope that this adjunction is comonadic via Barr-Beck-Lurie. $S \otimes_R S$ is a priori only a coalgebra in $(S, S)$-bimodules, but if moreover $S$ is $E_{\infty}$, $R$ is an $S$-algebra, and $f$ is a morphism of $S$-algebras (so $f$ is an augmentation), then I believe the left and right actions should match and make $S \otimes_R S$ a coalgebra in $S$-modules. Beyond this I don't have details or references but my impression is that Barr-Beck-Lurie is your friend.<|endoftext|> TITLE: Is this a semi algebraic set? QUESTION [6 upvotes]: Put $$A=\{(a_{0},a_{1},\ldots,a_{n}) \in \mathbb{C}^{n+1}\mid p(z)=a_{0}+a_{1}z+\ldots a_{n}z^{n} \;\;\text{is a one-to one function on the unit disc} \{z\in \mathbb{C} \mid |z|\leq 1\}$$ Is $\{(|a_{0}|,|a_{1}|,\ldots,|a_{n}|)\mid (a_{i})\in A\}$ a semi algebraic set in $\mathbb{R}^{n+1}$? Of course this is true for $n=2$ REPLY [16 votes]: It's fairly immediate from the Tarski-Seidenberg theorem that it's semi-algebraic. More exactly, if you can define your set in the language of real closed fields using first-order logic, then the Tarski-Seidenberg theorem guarantees that the set is semi-algebraic. As a first step, the set $A$ is semi-algebraic, as it is defined by a formula $\forall_{z, w} [z \bar{z} \leq 1 \wedge w \bar{w} \leq 1 \wedge a_0 + a_1 z + \ldots + a_n z^n = a_0 + a_1 w +\ldots + a_n w^n] \Rightarrow z = w$. Abbreviating this as $\forall_{z, w} [|z| \leq 1 \wedge |w| \leq 1 \wedge p_a(z) = p_a(w)] \Rightarrow z = w$, this is equivalent to $$\neg \exists_{z, w} (|z| \leq 1 \wedge |w| \leq 1 \wedge p_a(z) = p_a(w)) \wedge \neg (z = w)$$ where the subset $\{(z, w, a) \in \mathbb{C}^{n+3}: |z| \leq 1 \wedge |w| \leq 1 \wedge p_a(z) = p_a(w) \wedge z \neq w\}$ (before quantification) is clearly semi-algebraic. Existentially quantifying over $z, w$ corresponds to taking the image of that subset under the projection $(z, w, a) \mapsto a$; this image is semi-algebraic by Tarski-Seidenberg, and then the complement of that is also semi-algebraic. Finally, your set is $$\{(r_0, \ldots, r_n): \exists_{a \in \mathbb{C}^{n+1}} a \in A \wedge r_0^2 = a_0 \bar{a_0} \wedge \ldots \wedge r_n^2 = a_n \bar{a_n} \wedge r_0 \geq 0 \wedge \ldots \wedge r_n \geq 0\}$$ where, just as before (taking an image of a semi-algebraic set, defined in abbreviated form by $a \in A \wedge r^2 = a\bar{a} \wedge r \geq 0$, under the projection $(a, r) \mapsto r$), we easily conclude this is semi-algebraic.<|endoftext|> TITLE: Number of linear orderings of a set to have balanced frequencies of triple orders QUESTION [6 upvotes]: Let $S$ be a set of $n$ elements and let $Q = (s_1, s_2, \ldots, s_n)$ be a linear ordering of $S$. We write $s_i <_Q s_j$ when $s_i$ appears before $s_j$ in $Q$. I want to construct a set (or possibly a multi-set) of orderings $\mathcal{Q} = \{Q_1, \ldots, Q_k\}$ such that for every $a,b,c \in S$, each of the $6$ possible ordering of $a,b$ and $c$ occurs with the same frequency in $\mathcal{Q}$. Said differently, I want, for every $a,b,c \in S$, $$\Pr_{Q \in \mathcal{Q}}[a <_Q b <_Q c] = 1/6$$ Now, the set of all $n!$ orderings satisfies this property. Can we make $\mathcal{Q}$ smaller? What is the minimum number of orderings to get this property? In particular, can $|\mathcal{Q}|$ be polynomial in $n$? For $n =3$ and $n = 4$, six orderings suffice (see this stackexchange post). It is tempting to conjecture that $6$ is always enough...but that'd be a bit surprising. REPLY [5 votes]: With a colleague we are currently working on a similar problem: we want all 6 orderings to appear, though we do not care whether they happen with the same frequency. The lower bound transfers to your problem, however: Let us assume that you have a "good" set of permutations $P_1,...,P_k$, and think of the positions of the value $1$ in all these permutations. For every value different from 1, store in a binary vector whether it appears before (noted 0) or after (noted 1) in the ordering. Because all 3-orders must appear, all n-1 vectors your built must be different. Thus, $k\geq \log_2(n-1)$. This technique is used, for instance, in this paper: http://arxiv.org/abs/1502.06888 To be sure that all 3-orders appear, we currently need $\leq 2.85\cdot \log_2(n)+O(1)$ permutations, and we are trying to close the gap. The construction wouldn't work for equal frequencies, however.<|endoftext|> TITLE: Is the following identity true? QUESTION [44 upvotes]: Calculation suggests the following identity: $$ \lim_{n\to \infty}\sum_{k=1}^{n}\frac{(-1)^k}{k}\sum_{j=1}^k\frac{1}{2j-1}=\frac{1-\sqrt{5}}{2}. $$ I have verified this identity for $n$ up to $5000$ via Maple and find that the left-hand side approaches $\frac{1-\sqrt{5}}{2}$. However, this double summation has slow rate of convergence and I am unsure it is true. So I want to ask if it is true. If so, how to prove it? REPLY [12 votes]: Start with $$\sum_{n\geq1}\frac{z^{2n-1}}{2n-1}=\frac12\log\left(\frac{1+z}{1-z}\right).$$ Since $\frac1{1-z^2}\sum_{n\geq1}\frac{z^{2n-1}}{2n-1}=\sum_{n\geq1}z^{2n-1}\sum_{k=1}^n\frac1{2k-1}$, we have $$\int_0^z\left(\sum_{n\geq1}z^{2n-1}\sum_{k=1}^n\frac1{2k-1}\right)dz=\int_0^z \frac1{2(1-z^2)}\log\left(\frac{1+z}{1-z}\right)\,dz.$$ Therefore, $$\sum_{n\geq1}\frac{z^{2n}}{2n}\sum_{k=1}^n\frac1{2k-1}= \frac18\log^2\left(\frac{1+z}{1-z}\right).$$ Choose $z=i=\sqrt{-1}$ so that $$\sum_{n\geq1}\frac{(-1)^n}{2n}\sum_{k=1}^n\frac1{2k-1}= \frac18\log^2\left(\frac{1+i}{1-i}\right)=-\frac{\pi^2}{32}.$$ It follows that $$\sum_{n\geq1}\frac{(-1)^n}{n}\sum_{k=1}^n\frac1{2k-1}=-\frac{\pi^2}{16}.$$<|endoftext|> TITLE: How many chromatic polynomials of planar maps are there? QUESTION [11 upvotes]: Let P(n) be the set of polynomials that can occur as the chromatic polynomial of a planar map with n countries. What is known or conjectured about the growth of |P(n)|? PS: Thanks Gerry and Noam, can I narrow the question this way: is there an exponential lower bound on |P(n)|? REPLY [12 votes]: The following post has now been written up as Section 6 of this paper. In fact, one may show that the number of values of the chromatic polynomials at 4 grows exponentially. The number of chromatic polynomials grows exponentially. This answer is joint with Slava Krushkal. The Chromatic algebra $\mathcal{C}_n$ is an algebra over $\mathbb{C}[Q]$ (for a parameter $Q$) formed from linear combinations of (isotopy classes of) planar graphs in a rectangle $R$ with $n$ univalent points on top and bottom: Such diagrams may be composed by placing one on top of the other, fusing the $n$ univalent vertices together to get edges. Then quotient this algebra by some relations: the deletion-contraction relation and a relation to eliminate loops and kill diagrams with bridges Such diagrams represent planar graphs which agree outside of the shown pictures. These relations are satisfied by the polynomial in $Q$ which counts the number of nowhere zero flows (over an abelian group with $Q$ elemnts) in a graph. There is also a trace, where one closes up the graph top to bottom and evaluates the flow polynomial: For generic $Q\in \mathbb{C}$, this is a semisimple algebra. More generally, one can describe a planar algebra containing all of the $\mathcal{C}_n$s. To understand why the number of chromatic polynomials grows exponentially, the idea is to take some graphs in $\mathcal{C}_n$ which generate a semigroup in the algebra. Since the algebra is linear, the Tits alternative for semigroups implies that either this semigroup should contain a free subsemigroup or is virtually nilpotent. The latter is unlikely, if one takes generators of $\mathcal{C}_n$, since it is semisimple. Hence, the growth of the elements in the algebra when one multiplies these generators is exponential. Then one may extract the matrix entries by multiplying with a basis of $\mathcal{C}_n$ and taking traces (to get inner products with the basis, and hence essentially the entries of the matrix). The number of these traces grows exponentially, hence the number of flow polynomials of the graphs (hence the chromatic polynomials of the dual planar graphs) as a function of the number of vertices. For concreteness, we have an explicit example to demonstrate this approach. We want the smallest dimension subalgebra possible, so it's convenient to set $Q=\phi+1 = \frac{3+\sqrt{5}}{2}$ since $\mathcal{C}_n$ then has a non-trivial trace radical generated by the Tutte relation: For trivalent (cubic) graphs, it is convenient to use this form of the relation: Adding this relation (which doesn't affect the value of the flow polynomial for planar graphs at $Q=\phi+1$) decreases the dimension of the algebra, and hence makes it easier to compute with. We'll do a variation on the above strategy. Consider the elements $A, B$ of $\mathcal{C}_3$ in this picture: We can multiply (by stacking) $A$ and $B$ with the graphs $e_1, e_2$, which are not elements of $\mathcal{C}_3^{\phi+1}$, but of a related vector space $\mathcal{C}_{3,1}$ which is a module over $\mathcal{C}_3$ (really in the planar algebra), for $Q=\phi+1$ (obtained the same way by taking graphs modulo the deletion-contraction relations). Then we may compute $Ae_1= -\phi e_1, Ae_2=-\phi^2 e_2-\phi e_1,$ $Be_1=-\phi^2 e_2-\phi e_1, Be_2=-\phi e_2$, so the action of $A,B$ preserves the subspace spanned by $e_1,e_2$. These are represented by the matrices: $A=\left(\begin{array}{cc}-\phi & -\phi \\0 & -\phi^2\end{array}\right),$ $B=\left(\begin{array}{cc}-\phi^2 & 0 \\-\phi & -\phi\end{array}\right),$ Now, multiply $A, B$ together $n$ times, getting $2^n$ elements of $\mathcal{C}_3$. Then add on $e_1, e_2$ on bottom, and their reflections on top, to get an element of $\mathcal{C}_1$. Then take the trace, so connect by an arc running from top to bottom, and evaluate the flow polynomial. The graphs $A,B$ then have a subsemigroup in $GL(2,\mathbb{R})$ which is free (since they have distinct eigenvectors), so the various products give distinct elements of $\mathcal{C}_3$, and hence we get at least $2^{n/4}$ distinct coefficients when we pair with the $e_i$s and take the trace. Here's $ABAAB$ capped off one way:<|endoftext|> TITLE: What is the Banach-Mazur distance between $\ell_\infty$ and $L_\infty$? QUESTION [28 upvotes]: Given Banach spaces $X$ and $Y$, the Banach-Mazur distance between $X$ and $Y$ is defined as $$ d(X,Y) = \inf\{ \|\varphi\|\|\varphi^{-1}\| : \varphi\colon X\to Y \text{ isomorphism} \}. $$ We consider the Banach spaces $\ell_\infty=\ell_\infty(\mathbb{N})$ and $L_\infty=L_\infty([0,1],\lambda)$, where $\lambda$ is the Lebesgue measure on $[0,1]$. What is $d(\ell_\infty,L_\infty)$? REPLY [2 votes]: The distance is strictly greater than $2$ --- as is true for any compact, nonhomeomorphic $K_1$ and $K_2$ when $K_1$ is totally disconnected (Cohen-Chu, Stud. Math. 1995, p.6). The question of an upper bound is more puzzling. @HannesThiel comment 6/13/2016 suggests distance $\le 16$.<|endoftext|> TITLE: Existence of mixed Nash equilibria for compact strategy spaces QUESTION [6 upvotes]: In Peleg (1969) it is shown that a game with finite strategy spaces and continuous utilities has a mixed Nash equilibrium for any cardinality of players. Is the same true if the strategy spaces are compact (maybe under some extra conditions, like only countably many players)? REPLY [3 votes]: Lones Smith offers the following proof: if you have countably many players, arbitrarily enumerate them and call the original game with this enumeration $G$. Let $G_n$ be a game where the actions of players $k>n$ are fixed arbitrarily. By Glicksburg, $G_n$ has an equilibrium -- call it $x(n)$. Let $x$ be a limit point of the sequence $(x(n))_{n=1}^\infty$. Fix any player $i$. Now $x_i(n)$ is a best response to $x_{-i}(n)$ in $G_n$ for each $n$; to pass to limits, we appeal to the Theorem of the Maximum and continuity of the utility functions [1]. That shows the action $x_i$ is a best response to $x_{-i}$ in $G$. So $x$ is an equilibrium of $G$. [1] This assumes a topology on the action space at least as coarse as the product topology, so that whenever $x(n)$ converges to $x$ in the product topology, it converges in the topology in question. The utilities should be continuous under this topology for actions. This is a big assumption and easily violated!<|endoftext|> TITLE: Can you get the following asymptotic expression for an hypergeometric function? QUESTION [5 upvotes]: I conjecture that the following statement holds for large values of $N$ $$ {}_3F_1\left(-N+1,1,1;2;-\frac{1}{N}\right)\to\frac{1}{2}\bigg({}_2F_1(1,1;2;1-\frac{1}{N})+\log 2+\gamma\bigg) $$ where $\gamma$ is the Euler-Mascheroni constant. PLugging in big values for $N$ in the above formula it looks like the conjecture holds, but I am unable to prove it. Anybody can do better? REPLY [3 votes]: As the OP remarked in a comment, the ultimate purpose of the question is to derive the asymptotic large-$N$ expansion of the integral $$I_N=\int_0^1 dq\, \frac{q^2}{\log q}\left[\left(1-\frac{3}{N}\log q\right)^N-1\right]$$ According to equation (69) of this reference, $I_N$ should converge to $$I_N=\tfrac{1}{2}\left(\log N + \log 2 +\gamma_E\right)+{\cal O}(N^{-1/2})\qquad\qquad(*)$$ Noting the expansion $$(1+a/N)^N=\exp\left[a-\tfrac{1}{2}a^2/N+{\cal O}(N^{-2})\right]$$ I first transform from $q$ to $x=\log q$ and then take the large-$N$ limit as follows: $$I_N=\int_{-\infty}^0 dx\, \frac{e^{3x}}{x}\left[\left(1-\frac{3x}{N}\right)^N-1\right]=\int_{-\infty}^0 dx\, \frac{e^{3x}}{x}\left[\exp\left(-3x-\tfrac{9}{2}x^2/N+{\cal O}(N^{-2})\right)-1\right]$$ $$\quad\rightarrow \int_{-\infty}^0 dx\, \frac{1}{x}\left[\exp(-\tfrac{9}{2}x^2/N)-e^{3x}\right]$$ $$\quad=-\tfrac{1}{2}\left(\log N+\log 2+\gamma_E\right).\qquad\qquad(**)$$ This differs by a minus sign from the result $(*)$ in the reference. I checked it numerically, and it's pretty clear that (**) and not (*) is the correct asymptotics.<|endoftext|> TITLE: Relations between two definitions of non-archimedean analytic spaces QUESTION [6 upvotes]: I begin to learn some non-archimedean geometry recently, and find that there are two different definitions of analytic spaces in the literature. Let us fix a non-archimedean complete valuation field $k$ (the valuation is not necessarily non-trivial) The first definition is as in TEMKIN: An analytic space over $k$ consists of a locally Hausdorff topological space $X$, a net $\tau$ on $X$(regarded as a category), and a functor from $\tau$ to the category of $k$-affinoid spaces sending morphisms to affinoid domain embeddings, such that the underlying topological spaces of $U\subset X$ and $\tau(U)$ are naturally identified. The second definition I found is in Berkovich's original book: A $k$-analytic space is a locally ringed space $X$ with an equivalence class of collections of pairs $(U_i,\phi_i)$, where $U_i\subset X$ is open, $\phi_i$ is an open immersion of $U_i$ into a $k$-affinoid space subject to the usual compatibility conditions as in the definition of a mainfold. I'm not pretty sure what is the connection between these two definitions, I guess that the first definition axiomatizes analytic domains in the second definition, but I haven't been able to make this precise. REPLY [5 votes]: Let me give more a few more details than in nfdc23's comment. The most general definition of a analytic space is the one that you find in Berkovich's IHES paper. One requirement is that for every point $x$ and every open subset $U$ containing $x$, there exist finitely many affinoid domains $V_1,\dotsc,V_n$ of $U$ that all contain $x$ and whose union is a neighborhood of $x$. (The way I picture it is that of a brick wall, where some points lie on the edges of several bricks.) You could require something stronger, namely that every point has a basis of affinoid neighborhoods (i.e. you can always take $n=1$ above). Such a space is called a good space and this is the class of spaces that is defined in the red book. To see a typical example of a non-good space, consider the bi-disk $\mathcal{M}(k\{T,U\})$ and, inside it, the space $X$ that is the union of the two affinoid domains defined by $\{|T|=1\}$ and $\{|U|=1\}$. One can prove that the Gauss point has no affinoid neighborhood in $X$. On the other hand, it is important to have these kinds of spaces in the theory. For instance, this one appear as the generic fiber of the (non-affine) formal scheme $Spf(k^\circ\{T,U\})\setminus \{(0,0)\}$.<|endoftext|> TITLE: Del Pezzo surfaces of degree $2$ QUESTION [6 upvotes]: I'm trying to understand the relationship between the different models of del Pezzo surfaces of degree $2$. Let $k$ be a field of characteristic not equal to $2$. Usually, del Pezzo surfaces of degree $2$ are considered as hypersurfaces of degree $4$ in $\mathbb{P}(1,1,1,2)$: $$S: \quad w^2 = f(x,y,z)$$ with $\deg f = 4$. There are however other natural models, such as hypersurfaces of bidegree $(2,2)$ in $\mathbb{P}^1 \times \mathbb{P}^2$: $$q(s,t,x,y,z) = 0,$$ where $q$ is bihomogeneous of degree (2,2) in $(s,t)$ and $(x,y,z)$, respectively. Call these models of type $1$ and $2$, respectively. I'm trying to understand how one goes between these different models. Given a surface of type $2$, how does one construct a model of type $1$ (i.e. a model in weighted projective space)? I know how one should go about doing this. The anticanonial class $-K_S$ is the fibre of the projection onto $\mathbb{P}^2$. So a basis for the anticanonical sections is $x,y,z$. Now the space $H^0(S, -2K_S)$ is $7$-dimensional, so there should be a section $w$ in here which is linearly indenpendent from the obvious sections $x^2,xy,xz,y^2,yz,z^2$. With this in hand, we obtain the required map $$S \to \mathbb{P}(1,1,1,2), \quad (s,t) \times (x,y,z) \mapsto (x,y,z,w).$$ But what is $w$? I don't see how to determine $w$ explicitly. REPLY [3 votes]: One can transform birationally $S \subset \mathbb{P}^1 \times \mathbb{P}^2$ to $S' \subset\mathbb{P}(1,1,1,2)$ as follows. Choose a point $P \in \mathbb{P}^1$ such that the conic $$ C := (P \times \mathbb{P}^2) \cap S $$ is smooth. Blow up $P \times C \subset \mathbb{P}^1 \times \mathbb{P}^2$. Then blow down the proper preimage of $\mathbb{P}^1 \times C$. Then you will get $\mathbb{P}_{\mathbb{P}^2}(O \oplus O(-2))$. The strict transform of $P \times \mathbb{P}^2$ is the exceptional section and it does not intersect the strict transform of $S$. If you contract it, you get $\mathbb{P}(1,1,1,2)$ with an embedded image of $S$.<|endoftext|> TITLE: Rogers-Ramanujan continued fraction $R(e^{-2 \pi \sqrt 5})$ QUESTION [8 upvotes]: Let $$R(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^{2}}{1 + \cfrac{q^{3}}{1 + \cdots}}}}$$ It is easy to evaluate $R(e^{-2 \pi/ \sqrt 5})$ using the Dedekind eta function identity $\eta(-\frac{1}{z})=\sqrt{-iz}\eta(z)$ and one of the most fundamental properties of $R(q)$ $$\frac{1}{R(q)}-1-R(q)=\frac{f(-q^\frac{1}{5})}{q^\frac{1}{5}f(-q^5)}=\frac{\eta(\tau/5)}{\eta(5\tau)}$$ Where $q=e^{2 \pi i \tau}$ and $f(q)$ is the theta function (Ramanujan's notation). Then $$R(e^{-2 \pi/ \sqrt 5})=\sqrt[5]{\sqrt{1+\beta^{10}}-\beta^5}$$ where $$\beta=\frac{1+\sqrt{5}}{2 } $$ is the golden ratio. If $\alpha_1 , \alpha_2>0$ and $\alpha_1 \alpha_2=\pi^2$ then $$\bigg(\frac{1+\sqrt{5}}{2}+R(e^{-2 \alpha_1})\bigg)\bigg(\frac{1+\sqrt{5}}{2}+R(e^{-2 \alpha_2})\bigg)=\frac{5+\sqrt{5}}{2}$$ using this identity, I evaluate $R(e^{-2 \pi \sqrt 5})$ Since I already know the value of the $R(e^{- 2 \pi/ \sqrt5})$: $$\color{blue} {R(e^{-2 \pi \sqrt 5})= \frac{\beta+2}{\beta+\sqrt[5]{\sqrt{1+\beta^{10}}-\beta^5}}-\beta}.$$ There is another way to evaluate $R(e^{-2 \pi \sqrt 5})$. $$R(e^{-2 \pi \sqrt 5})=\sqrt{(\frac{A+1}{2})+1}-\frac{A+1}{2}$$ where $A$ satisfies the quadratic equation $$\frac{A}{\sqrt{5}V}-\frac{\sqrt{5}V}{A}=\bigg(V-V^{-1}\bigg)^2 \bigg(\frac{V-V^{-1}}{\sqrt{5}}-\frac{\sqrt{5}}{V-V^{-1}}\bigg)$$ and $$V=\frac{G_{125}}{G_5}$$ where $G_n=2^{-1/4}e^{\pi \sqrt{n}/24} \chi(e^{- \pi \sqrt{n}})$ is the Ramanujan's class invariant(algebraic), $\chi(q)$ is the Ramanujan's fuction defined by $\chi(-q)=(q;q)_\infty$, and $(a;q)_n$ is a q-Pochhammer symbol. There are my questions: $1.$ How to calculate the class invariant $G_{125}$ in order to evaluate $R(e^{-2 \pi \sqrt 5})$ as indicated above? $2.$ Does there exist another way to evaluate $R(e^{-2 \pi \sqrt 5})$ without using $R(e^{-2 \pi / \sqrt 5})$ and class invariants? REPLY [2 votes]: Let $R(q)$ be the Rogers-Ramanujan continued fraction $$ R(q):=\frac{q^{1/5}}{1+}\frac{q^1}{1+}\frac{q^2}{1+}\frac{q^3}{1+}\ldots,|q|<1 $$ Let also for $r>0$ $$ Y=Y(r):=R(e^{-2\pi\sqrt{r}})^{-5}-11-R(e^{-2\pi\sqrt{r}})^5 $$ It is easy to show someone that $$ Y\left(\frac{r}{5}\right)Y\left(\frac{1}{5r}\right)=125, : (1) $$ for all $r>0$. Hence for $r=1$ $$ Y\left(\frac{1}{5}\right)=\sqrt{125}=5\sqrt{5} $$ hence $$ R\left(e^{-2\pi/\sqrt{5}}\right)=\sqrt[5]{\frac{2}{11+5\sqrt{5}+\sqrt{250+110\sqrt{5}}}} $$ Using the modular relation $$ R\left(e^{-2\pi\sqrt{1/r}}\right)=\frac{-(1+\sqrt{5})R\left(e^{-2\pi\sqrt{r}}\right)+2}{2R\left(e^{-2\pi\sqrt{r}}\right)+1+\sqrt{5}} : (2) $$ we easily get the result. Proof of (1): The fifth degree modular equation of $R(q)$ is $$ R\left(q^{1/5}\right)^5=R(q)\frac{1-2R(q)+4R(q)^2-3R(q)^3+R(q)^4}{1+3R(q)+4R(q)^2+2R(q)^3+R(q)^4} $$ Also it holds (2), with $v_r=R\left(e^{-2\pi\sqrt{r}}\right)$, $r>0$ (see: W. Duke. 'Continued Fractions and Modular Functions'): $$ R\left(e^{-2\pi\sqrt{1/r}}\right)=v_{1/r}=\frac{-(1+\sqrt{5})v_r+2}{2v_r+1+\sqrt{5}} $$ A routine algebraic evaluation can show us that $$ \left(v_{r/25}^{-5}-11-v_{r/25}^5\right)\left(v_{1/r}^{-5}-11-v_{1/r}^{5}\right)=\ldots=125 $$ Hence the proof is complete.<|endoftext|> TITLE: Concentration bounds on weighted sum of i.i.d. Bernoulli random variables QUESTION [6 upvotes]: Let $X_1,\dots, X_n\sim\operatorname{Bern}(\frac{1}{2})$ be independent, identically distributed random variables, and $\alpha=(\alpha_1,\dots,\alpha_n)\in[0,1]^n$ a vector of non-negative weights satisfying $\sum_{k=1}^n \alpha_k = 1$. Define the random variable $X\stackrel{\rm def}{=} \sum_{k=1}^n \alpha_k X_k$. Given parameter $\delta > 0$, what is the best concentration bounds I get on $X$ in terms of the $p$-'norms' (allowing $p\in(0,1)$) of $\alpha$, i.e. $$ \mathbb{P}\{ X \leq \delta\} \leq \varphi(\delta, (\lVert\alpha\rVert_p)_{p>0}) $$ ? I was thinking of using Berry—Esseen-like approximations to reduce this to studying a Gaussian random variable (leading to a bound in terms of $\frac{1}{\lVert \alpha\rVert_2^2}$, but this falls short of what I want (as, if I am not mistaken, this requires the vector $\alpha$ to be more or less uniform). Using Chernoff/Hoeffding bounds will get me a bound expressed in terms of $\frac{1}{\lVert \alpha\rVert_\infty}$, but it feels very loose. Edit: using the conversion Bernoulli/Rademacher and the bounds for those, I get an upper bound of $e^{-\frac{(1-2\delta)^2}{\lVert \alpha\rVert_2^2}}$, similar to what the Gaussian approximation would result in. But I still don't really know if this is tight, that is if the "right" parameter for this is indeed $\lVert \alpha\rVert_2$, or something more "exotic" that just happens to coincides with $\lVert \alpha\rVert_2$ in the "uniform-weights regime." REPLY [6 votes]: An almost complete answer. First of all, indeed the $\lVert \alpha\rVert_2$-based bound mentioned in the question can be shown to be tight for many "simple" $\alpha$'s, such as balanced, or uniform/balanced on a subset of $m$ coordinates. However, it is not tight in general, and the right answer appears to be captured by a quantity, the $K$-functional between $\ell_1$ and $\ell_2$. Converting the problem to that of looking at a weighted sum $Y\stackrel{\rm def}{=}\sum_{k=1}^n \alpha_k Y_k$ of Rademacher random variables $Y_1,\dots, Y_n$ (where $Y_k = 1-2X_k$), $$ \mathbb{P}\{ X \leq \varepsilon \} = \mathbb{P}\{ Y \geq 1-2\varepsilon \}. $$ In Montgomery-Smith [1], the following is shown: Theorem. For $c\stackrel{\rm def}{=} \frac{1}{4\ln 12}\simeq \frac{1}{10}$, we have $$\begin{align} \mathbb{P}\{ Y > K_{1,2}(\alpha,t) \} &\leq e^{-\frac{t^2}{2}}, \tag{1} \\ \mathbb{P}\{ Y > cK_{1,2}(\alpha,t) \} &\geq ce^{-\frac{t^2}{c}} \tag{2} \end{align}$$ where $K_{1,2} (\alpha,t) \stackrel{\rm def}{=} \inf\{\lVert \alpha'\rVert_1+t\lVert \alpha''\rVert_2 \ \colon\ \alpha',\alpha'' \in \ell_2,\ \alpha'+\alpha'' = \alpha\}$. Moreover, the following bound holds for $K_{1,2}(\alpha,t)$, ([1], citing Theorem 4.2 of [2]): assuming wlog that $\alpha_1 \geq \alpha_2 \geq \dots\geq \alpha_n$, $$ \gamma K_{1,2}(\alpha,t) \leq \sum_{k=1}^{\lfloor t^2\rfloor} \alpha_k + \left(\sum_{k=\lfloor t^2\rfloor+1}^n \alpha_k^2\right)^{\frac{1}{2}} \leq K_{1,2}(\alpha,t) \tag{3} $$ for some absolute constant $\gamma \in (0,1)$. This gives the following corollary, that can be found in [1]: Corollary. With $c$ as above, we have that for all $\alpha\in\ell_2$ (which is trivial in our case as $\alpha$ has finite support) and $0 < t \leq c\frac{\lVert \alpha\rVert_2^2}{\lVert \alpha\rVert_\infty}$, $$\begin{align} ce^{-\frac{t^2}{c^3\lVert \alpha\rVert_2^2}} \leq \mathbb{P}\{ Y >t \} \leq e^{-\frac{t^2}{2\lVert \alpha\rVert_2^2}}, \tag{4} \end{align}$$ where the upper bound holds for all $t>0$. Since the range of interest is $t\in(0,1]$, the above shows that the $e^{-\Theta\left(\frac{t^2}{\lVert \alpha\rVert_2^2}\right)}$ dependence is tight as long as $\lVert \alpha\rVert_2^2$ and $\lVert \alpha\rVert_\infty$ are within constant factors, which happens to be the case for $\alpha$ roughly balanced or uniform on a subset of coordinates, for instance. Now, for a counter example, one can consider (a variant of) the "truncated random Harmonic series," where we set, for $1\leq k\leq n$, $$ \alpha_k \stackrel{\rm def}{=} \frac{1}{k H_n} \operatorname*{\sim}_{n\to \infty} \frac{1}{k\ln n}. $$ Then, using techniques similar as [1], one can show that $$ \mathbb{P}\{ Y > t \} \leq e^{-\Theta(n^t)} $$ for $t\in(0,1)$. However, $ \lVert \alpha\rVert_2^2 {\displaystyle\operatorname*{\sim}_{n\to\infty}} \frac{\pi^2}{6\ln^2 n} $, so (4) would only yield $$ \mathbb{P}\{ Y > t \} \leq e^{-\Theta( t^2 \ln^2 n )}. $$ Note that (sanity check) in this case, $\lVert \alpha\rVert_\infty = \frac{1}{H_n}$, so the lower bound of (4) does not apply: $\frac{\lVert \alpha\rVert_2^2}{\lVert \alpha\rVert_\infty} {\displaystyle\operatorname*{\sim}_{n\to\infty}} \frac{1}{\ln n} = o(1)$. Edit: based on results from [3] (adapting and generalizing the proof of Theorem 2.2), one can get the following refinement of (2) above: Theorem. Fix any $c>0$. Then, for any $\alpha\in\ell_1+\ell_2$, we have that for all $t\geq 1$, $$ \mathbb{P}\left\{ Y \geq \frac{1}{1+c}K_{1,2}(\alpha,t) \right\} \geq e^{-\left( \frac{2}{c}\ln\frac{\sqrt{6}(1+c)}{c}\right) (t+c)^2 }. $$ In particular, $$ \mathbb{P}\left\{ Y \geq \frac{1}{2}K_{1,2}(\alpha,t) \right\} \geq e^{-\left( \ln 24 \right) (t+1)^2 } \geq e^{-\left( 4\ln 24 \right)t^2 }. $$ [1] Montgomery-Smith S.J. (1990) The distribution of Rademacher sums. Proc. Amer. Math. Soc. 109:517522 [2] Holmstedt, Tord. (1970) Interpolation of quasi-normed spaces. Math. Scand. 26, 177–199. [3] Astashkin, S.V. (2010) Rademacher functions in symmetric spaces Journal of Mathematical Sciences, 169(6):725–886<|endoftext|> TITLE: Subgroup ranks of the symmetric group QUESTION [7 upvotes]: It's well known that every subgroup $G$ of $S_n$ has a generating set of size at most $n-1$ and that this generating set can be found algorithmically (by Jerrum's filter) I have heard many times a stronger result, which is that for all $n>3$, every subgroup $G$ of $S_n$ has a generating set of size at most $n/2$. Note that this would be tight: for example, $n/2$ disjoint transpositions cannot be minimized further. However, as far as I can tell, none of the places I have seen this theorem give a proof, nor do their references. Can anyone point me to a proper proof, or give one, for this seemingly important theorem? REPLY [9 votes]: After some googling, one finds a few references. Most point to Cameron, Peter J.; Solomon, Ron; Turull, Alexandre, Chains of subgroups in symmetric groups. J. Algebra 127 (1989), no. 2, 340–352. Which itself attributes this to Peter Neumann, private communication. They also say: "As Peter is unlikely to publish his result, we shall sketch a recipe for a proof of Neumann’s theorem here. " (And then proceed to give a sketch.) Note that the proof seems to depend on the Classification of Finite Simple Groups (to deal with the case of primitive permutation groups).<|endoftext|> TITLE: Homeomorphic but Non-Conjugate Mapping Tori QUESTION [10 upvotes]: Suppose we fix a genus $g$ closed surface $S$. Let $f, g \in Map(S)$ be conjugate, for $Map(S)$ the mapping class group of $S$. Then I know that $M_f$ (the mapping torus of $M$ with monodromy $f$) is homeomorphic to $M_g$. Now, let $M$ is a mapping torus with monodromy $f$, and $N$ is a mapping torus with monodromy $g$. Suppose we know that $M_f$ is abstractly homeomorphic to $N_g$ (not homeomorphic as surface bundles over $S^1$, just homeomorphic as 3-manifolds). If $M_f$ and $N_g$ fiber over $S^1$ with homeomorphic fibers, then need $f, g$ be conjugate? I think the answer is "no", but I'm not sure why, and I'm not sure how to construct a counterexample. Any help would be greatly appreciated! REPLY [3 votes]: To add an explicit counter-example, consider the curves $a$, $b$ and $c$ on the twice punctured torus shown below. The mapping classes $f = T_a T_a T_b T_c^{-1}$ and $g = T_a T_a T_b^{-1} T_c^{-1}$ are not conjugate (they don't even have the same dilatation). However their mapping tori $M_f$ and $M_g$ are homeomorphic. You can check this by using the latest versions of flipper, SnapPy and the Python code: >>> import flipper >>> S = flipper.load('S_1_2') >>> f = S.mapping_class('aabC') >>> g = S.mapping_class('aaBC') >>> f.is_conjugate_to(g) False >>> f.dilatation(), g.dilatation() (2.618033?, 3.254263?) >>> import snappy >>> M_f = snappy.Manifold(f.bundle()) >>> M_g = snappy.Manifold(g.bundle()) >>> M_f.is_isometric_to(M_g) True<|endoftext|> TITLE: Contraction of probability measures QUESTION [6 upvotes]: Notations. Let $(X, \mathcal{B})$ be a separable Banach space, with its Borel sigma-algebra, $\|\cdot\|$ stands for the norm in $X$, $\mathcal{P}(X)$ - the set of all probability measures on $X$. Let $P(x,B)$ be a stochastic kernel with source and target spaces both equal to $(X, \mathcal{B})$. Let also $$ L_b(X)=\left\{f\in C_b(X): \mathrm{Lip}(f):=\sup_{x_1,x_2\in X}\frac{|f(x_1)-f(x_2)|}{\|x_1-x_2\|}<\infty \right\},\\ \|f\|_L:=\|f\|_{C(X)}+\mathrm{Lip}(f). $$ By $\|\cdot\|_{TV}$, $\|\cdot\|^*_L$ we denote total variation and dual Lipschitz distances on $\mathcal{P}(X)$: $$ \|\mu_1-\mu_2\|_{TV}=\frac{1}{2}\sup_{f\in C_b(X),\\ \|f\|_{C(X)}\leq 1}\left|(f,\mu_1)-(f,\mu_2)\right|,\ \mu_1,\mu_2\in \mathcal{P}(X);\\ \|\mu_1-\mu_2\|^{*}_{L}=\sup_{f\in C_b(X),\|f\|_{L}\leq 1}|(f,\mu_1)-(f,\mu_2)|,\ \mu_1,\mu_2\in \mathcal{P}(X); $$ By $\|\cdot\|_K$ we denote Kantorovich distance on $$\mathcal{P}_1(X)=\{\mu\in\mathcal P(X):\int_X\|x-x_0\|\mu(dx)<\infty\mbox{ for some } x_0\in X\}:$$ $$ \|\mu_1-\mu_2\|_{K}=\sup_{f\in L_b(X),\\\mathrm{Lip}(f)\leq 1}|(f,\mu_1)-(f,\mu_2)|,\ \mu_1,\mu_2\in \mathcal{P_1}(X). $$ Question. If for all $x_1,x_2\in X$ we have $\|P(x_1,\cdot)-P(x_2,\cdot)\|_{TV}\leq \gamma$ for some $\gamma\in (0,1)$, then one can show that the map $$ G:\mathcal P(X)\to \mathcal P(X),\ G(\mu)(B)=\int_XP(x,B)\mu(dx) $$ is contraction $$ \|G(\mu_1)-G(\mu_2)\|_{TV}\leq \gamma \|\mu_1-\mu_2\|_{TV}. $$ Is it possible to obtain similar contraction result 1. in dual Lipschitz distance, if for all $x_1,x_2\in X$ $\|P(x_1,\cdot)-P(x_2,\cdot)\|^*_{L}\leq \gamma$ for some $\gamma\in (0,1)$? 2. in Kantorovich distance, if for all $x_1,x_2\in X$ $\|P(x_1,\cdot)-P(x_2,\cdot)\|^*_{K}\leq \gamma$ for some $\gamma\in (0,1)$? If yes, please, could you give an idea of the proof or a reference? REPLY [2 votes]: No. $P$ is not a contraction with respect to dual Lipschitz distance. As an example, take $X=\mathbb R$ and $P(x,\cdot)=\frac 12\delta_0+\frac 12\delta_{2x}$. This satisfies the contraction in $TV$ (because half of the mass is coupled in each step). But $\|G(\delta_0)-G(\delta_{\frac 14})\|_L^*= \|\delta_0-\delta_{\frac14}\|_L^*$ (as can be seen using the function $f(x)=\min(|x|,1)$). The same argument shows that there is no contraction of the Kantorovich distance. If you replace $2x$ by $3x$, then the distances are actually increased.<|endoftext|> TITLE: Is there a Fraisse limit whose automorphism group contains dense but not generic automorphisms? QUESTION [8 upvotes]: It is well known that $\mathsf{Aut}(\mathbb{Q},<)$ has generic automorphisms (i.e., a comeagre conjugacy class under the diagonal action) but does not admit ample generics. The automorphism group $\mathsf{Aut}(\omega,E)$ of the structure $(\omega,E)$ where $nEm \iff n = m \mod{2}$ does not even have a dense conjugacy class. I am looking for an example 'in the midst' of the previous two, i.e., an automorphism group of a Fraisse limit that has a dense conjugacy class but does not have a generic automorphism. I know that there are examples of Polish groups that have a dense conjugacy but do not have a generic automorphism. However, it is unknown to me or impossible to construct these examples as automorphism groups of Fraisse limits. The first example that I know of is $\mathsf{Aut}(X,\mu)$ where $(X,\mu)$ is the standard probability space, i.e., a standard Borel space $X$ with a probability measure $\mu$. We cannot express $(X,\mu)$ as a Fraisse limit since this space is not countable. Another example of a Polish group that has dense automorphisms is $U(H)$, that is, the unitary group of the infinite dimensional Hilbert space. However, the last example is not even constructed as an automorphism group. Is there an easy example of a Fraisse limit whose automorphism group has a dense conjugacy class but not a comeagre one? REPLY [7 votes]: I assume you're familiar with the paper Turbulence, amalgamation, and generic automorphisms of homogeneous structures by Kechris and Rosendal. If not, you should have a look, since it's about exactly these issues. In the language of that paper, we take a Fraïssé class $K$ and expand it to a class $K_p$ consisting of pairs $(A,\varphi)$, where $A\in K$ and $\varphi$ is a partial automorphism of $A$ (i.e. an isomorphism between two substructures of $A$). Then an embedding $(A,\varphi)\to (B,\psi)$ is an embedding $A\to B$ such that $\psi$ extends the image of $\varphi$. If $M$ is the Fraïssé limit of $K$, Kechris and Rosendal show that $\mathrm{Aut}(M)$ has a dense conjugacy class if and only if $K_p$ has the joint embedding property (JEP), and $\mathrm{Aut}(M)$ has a comeager conjugacy class if and only if $K_p$ has the weak amalgamation property (WAP) and the JEP. For a category of structures and embeddings, the WAP is the assertion that every $A$, there is an embedding $f\colon A\to B$, such that for any embeddings $g_1\colon B\to C_1$ and $g_2\colon B\to C_2$, there are embeddings $h_1\colon C_1\to D$ and $h_2\colon C_2\to D$, such that $h_1\circ g_1\circ f = h_2\circ g_2\circ f$. So to answer your question, we just need an example of a Fraïssé class $K$ such that $K_p$ has the JEP but not the WAP. (Edit: I've realized that there's a much simpler example, so I've replaced my original example). Let $K$ be the class of finite ordered graphs in the language $\{\leq,R\}$. Then $K$ is a Fraïssé class. Its Fraïssé limit is the ordered random graph. $K_p$ has the JEP: Given $(A,\varphi)$ and $(B,\psi)$, embed them in $(C,\theta)$, where $C$ is the disjoint union of $A$ and $B$, $\theta$ is the union of $\varphi$ and $\psi$, all the elements of $B$ are greater than all the elements of $A$ in the order, and no new edge relations hold. But $K_p$ does not have the WAP: Let $(A,\varphi)$ be any structure in $K_p$ such that there is some $a\in A$ such that $\varphi(a)$ is defined and $\varphi(a)> a$. Suppose $(A,\varphi)$ embeds in $(B,\psi)$. We will show that $(B,\psi)$ cannot witness the WAP for $(A,\varphi)$. Note that the sequence $\dots, \psi^{-1}(a), a, \psi(a), \psi^2(a), \dots$ is strictly increasing, so, since $B$ is finite, there is some largest $m\geq 0$ such that $b = \psi^{-m}(a)$ is defined, and there is some largest $n\geq 1$ such that $e = \psi^{n}(a)$ is defined. We will define two structures $(C_1,\theta_1)$ and $(C_2,\theta_2)$ into which $(B,\psi)$ embeds. In each, we will add a single new element $c$ and extend $\psi$ to $\theta_i$ so that $\theta_i(e) = c$. For all $d$ such that $\psi(d)$ is defined, we set $\psi(d) < c$ if and only if $d < e$ and $\psi(d)Rc$ if and only if $dRe$. This is enough to ensure that the $\theta_i$ are partial automorphisms. Now in $C_1$ we set $bRc$, and in $C_2$ we set $\lnot bRc$. This is consistent with the assignments above, since $b$ does not have a preimage under $\psi$. We may decide the rest of the order and edge relations between $c$ and elements of $B$ arbitrarily (but consistently with the assignments above). If $(C_1,\theta_1)$ and $(C_2,\theta_2)$ embed in some $(D,\rho)$ over $(A,\varphi)$, the images of $b = \theta_i^{-m}(a)$ and $c= \theta_i^{n+1}(a)$ must be equal, but $C_1$ and $C_2$ disagree about the truth of $bRc$, so we have a contradiction. Two comments: In fact, this argument shows that $K_p$ does not even have the local weak amalgamation property (i.e. there is no $A\in K_p$ such that the weak amalgamation property holds for the class of structures into which $A$ embeds), so the automorphism group of the ordered random graph does not even have any non-meager conjugacy classes. A google search for "ordered random graph" generic automorphism brought me to this paper, where essentially the same argument is used to show that the automorphism group of the ordered rational Urysohn space has no non-meager conjugacy class. At the very end of the paper it is remarked that the results also hold for the ordered random graph.<|endoftext|> TITLE: Reference request on birational invariance of Chow group of zero cycles of degree zero QUESTION [9 upvotes]: Let $CH_0(X)^0$ denote the group of zero cycles of degree zero modulo rational equivalence. I am looking for a reference for the following fact: If $X$ and $Y$ are smooth and projective varieties over a finite field and we suppose they are birational, then the groups $CH_0(X)^0$ and $CH_0(Y)^0$ are isomorphic. As a second question, let $k$ denote the base field and set $$ \pi_1^{ab}(X)^0 := \text{ker}[\pi_1^{ab}(X) \rightarrow \pi_1^{ab}(k)]. $$ Then, with the same assumptions on $X$ and $Y$, $\pi_1^{ab}(X)^0$ and $\pi_1^{ab}(Y)^0$ should be isomorphic. For the first fact, I have found a reference for the case of dimension 2 in the answer to this question (which links to a 1979 paper of Colliot-Thélène and Coray). The cases of dimension 1 and 0 are obvious. However, I have failed to find a reference for the case of general dimension. Any help is greatly appreciated. REPLY [3 votes]: Using Jason Starr's comment I was able (I think) to figure out the case of $\pi_1(X)^0$. For anyone who stumbles across this with the same question in mind I add a sketch of the proof in an answer. For the experts, if you feel like leaving a comment whether my reasoning is correct, that would be great. Let $\phi: X \rightarrow Y$ be a rational map, i.e. defined on some open $U \subset X$. Because $X$ is normal, $Z := X \backslash U$ has codimension at least two. This means the dimenson of the ring $\mathcal{O}_{X,z}$ is at least two for each point $z \in Z$. But then this local ring is pure (SGA 2, Exp. X, Thm 3.4), which in turn implies the couple $(X,Z)$ is pure (SGA 2, Exp. X, prop. 3.3), so that the categories $\operatorname{FEt}(X)$ and $\operatorname{FEt}(U)$ are equivalent. Since $X$ and $Y$ are birational, we can find an open $V \subset Y$ such that $U$ is isomorphic to $V$ and in the same way we show $V$ and $Y$ have the same \'etale coverings. This shows $\pi_1(X) = \pi_1(U) = \pi_1(V) = \pi_1(Y)$ as desired, which proves in particular what I wanted to show.<|endoftext|> TITLE: Calculate Ramanujan's class invariant by using modular equation of degree $5$ QUESTION [6 upvotes]: Let $$K(k):=\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}=\frac{\pi}{2}{ _2F_1\bigg(\frac{1}{2},\frac{1}{2},1;k^2 \bigg)}$$ where $0 TITLE: Non-vanishing of the Tate-Shafarevich kernel in group cohomology QUESTION [5 upvotes]: Let $G$ be a finite group. Let $M$ be a finite $G$-module (a finite abelian group with an action of $G$). We consider a special kind of $G$-modules; in particular, our $M$ is a finite dimensional representation of $G$ over $\mathbb{F}_p$. For our $G$-modules $M$, we ask whether it is possible that $Ш^1_\omega(G,M)\ne 0$. We ask our question with a hope to answer this hard question (in positive or negative). The relation is the following. Let $L/k$ be a Galois extension of number fields with Galois group $G$. Then we may regard $M$ as a $\mathrm{Gal}(\bar{k}/k)$-module. In this case $$ Ш^1(k,M)\subset Ш^1_\omega(G,M).$$ Moreover, if the decomposition groups of all the places of $k$ in $\mathrm{Gal}(L/k)=G$ are cyclic, then $$ Ш^1(k,M)= Ш^1_\omega(G,M).$$ Here $ Ш^1(k,M)$ is the "honest" Tate-Shafarevich group of the $\mathrm{Gal}(\bar{k}/k)$-module $M$. Following Sansuc, we define $$Ш^1_\omega(G,M)=\mathrm{ker}\left[H^1(G,M)\to\prod_C H^1(C,M)\right],$$ where $C$ runs over the set of cyclic subgroups of $G$. We write $Ш(G,M)$ for $Ш^1_\omega(G,M)$. If $G$ acts trivially on $M$, then clearly $Ш(G,M)=0$ (because then $H^1(G,M)=\mathrm{Hom}(G,M)$ ). Sansuc proves that if all the Sylow subgroups of $G$ are cyclic, then $Ш(G,M)=0$ for any $G$-module $M$. Using his method, we prove the following proposition: Proposition. Let $p$ be a prime number. If $M$ is a $G$-module such that $p^n M=0$ for some $n\ge 1$, and if a Sylow $p$-subgroup of $G$ is cyclic, then $Ш(G,M)=0$. Proof. Let $P$ be a Sylow $p$-subgroup of $G$, then the map $$ \mathrm{Cor}\circ\mathrm{Res}\colon H^1(G,M)\to H^1(P,M)\to H^1(G,M)$$ is the multiplication by $[G:P]$. Since $P$ is cyclic, this map is 0 on $Ш(G,M)$. Since $p^n M=0$, the multiplication by $p^n$ on $Ш(G,M)$ is 0 as well. Since the numbers $p^n$ and $[G:P]$ are coprime, we conclude that $Ш(G,M)=0$. Let $H$ be a subgroup of $G$ (e.g., $H=\{1\}$). We consider the $G$-set $X:=G/H$. We embed $\mathbb{F}_p$ into $\mathrm{Maps}(X,\mathbb{F}_p)$ as the subspace of constant maps, and we set $$M(G,H,p):=\mathrm{Maps}(X,\mathbb{F}_p)/\mathbb{F}_p.$$ Then $M(G,H,p)$ is a finite dimensional representation of $G$ over $\mathbb{F}_p$, hence a $G$-module. (For $p=2$, this is the Galois module $T[2]$ from this answer). Question. Do there exist $G$, $H$, and $p$, such that for $M=M(G,H,p)$ we have $Ш(G,M)\ne 0$? I would be especially interested in a counter-example with $p=2$. The proposition above shows that for a counter-example $M(G,H,p)$, the group $G$ must have a noncyclic Sylow $p$-subgroup. Proposition. If $p\nmid [G:H]$, then for $M=M(G,H,p)$ we have $Ш(G,M)=0$. Proof (due to user nfdc23). Write $X=G/H$, then $\# X$ is prime to $p$. It follows that $M=M(G,H,p)$ is isomorphic (as a $G$-module) to a direct summand of the $G$-module $\mathrm{Maps}(X,\mathbb{F}_p)$. Since $Ш(G,\mathrm{Maps}(X,\mathbb{F}_p))=0$, we conclude that $Ш(G,M)=0$. REPLY [5 votes]: I think the following is an example of $Ш(G,M(G,H,\Bbb{F}_2))\neq 0$: Take $G=A_4$ and $H$ of order $2$. Then $M$ has dimension $5$ and a (computer) calculation shows that $Ш(G,M(G,H,\Bbb{F}_2))$ has dimension $1$. The following is a sketch of how to do the computation by hand. First of all $A_4$ has two conjugacy classes of cyclic subgroups namely $H=\langle (1,2)(3,4)\rangle$ and $K=\langle (1,2,3)\rangle$. Since $\lvert K\rvert=3$ it suffices to prove that the restriction map $H^1(G,M)\rightarrow H^1(H,M)$ is not injective. Let $V_4=\langle (1,2)(3,4),(1,3)(2,4)\rangle$. The spectral sequence associated the sequence $1\rightarrow V_4\rightarrow A_4\rightarrow A_4/V_4\cong C_3\rightarrow 1$ collapses to give $H^*(A_4,M)\cong H^*(V_4,M)^{C_3}$. From this it is a straightforward (but tedious) computation to see that both $H^1(G,M)$ and $H^1(H,M)$ have dimension $1$ and that the restriction map is the zero map. Here is the Magma code I used to verify the example: G:=Alt(4); H:=sub; M:=PermutationModule(G,H,GF(2)); M:=M/Fix(M); XG:=CohomologyModule(G,M); H1G:=CohomologyGroup(XG,1); XH:=Restriction(XG,H); H1H:=CohomologyGroup(XH,1); ims:=[IdentifyOneCocycle(XH,OneCocycle(XG,H1G.i)) : i in [1..Dimension(H1G)]]; res:=homH1H|ims>; print Dimension(H1G),Dimension(H1H),Dimension(Kernel(res)); The output 1, 1, 1 confirms the computation above.<|endoftext|> TITLE: commutative diagram with $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ (for $C^*$-algebras) QUESTION [5 upvotes]: I have a question about a proof in Rosenberg and Schochet's paper "the Künneth theorem and the Universal Coefficient Theorem for Kasparov's generalized K-functor", proposition 2.6. First of all, the setting: Def.: Let $N$ be the bootstrap class of $C^*$-algebras, it's the smallest full subcategory of the seperable nuclear $C^*$-algebras which contains the separable Type I $C^*$-algebras and and is closed under strong Morita equivalence, inductive limits, extensions, and crossed products by $\mathbb{R}$ and by $\mathbb{Z}$. And if $J$ is an ideal in $A$ and $J$ and $A$ are in $N$, then so is $A/J$. And if $A$ and $A/J$ are in $N$ then so is $J$. Theorem (2.1): Let $A\in N$ and let $B$ be a $\sigma$-unital $C^*$-algebra such that $K_*(B)$ is an injective $\mathbb{Z}$-module. Then the map $$\gamma(A,B):KK_*(A,B)\to Hom(K_*(A),K_*(B))$$ is an isomorphism. One step to prove this theorem is the following Proposition (2.6): If $K_*(B)$ is injective and if $\gamma(A,B)$ is an isomorphism, then $\gamma(A\rtimes_{\rho} \mathbb{R},B)$ is an isomorphism for any continuous action of $\mathbb{R}$ on $A$. Proof: The Thom isomorphism theorems of Connes yield natural isomorphisms $$Hom(K_i(A\rtimes_{\rho} \mathbb{R}),K_j(B))\cong Hom(K_{i+1}(A),K_j(B))$$ and $$KK_i(A\rtimes_{\rho} \mathbb{R},B)\cong KK_{i+1}(A,B).$$The proposition follows immediately. $\Box$ The argument above is the following: $$\require{AMScd}\begin{CD} KK_i(A\rtimes_{\rho} \mathbb{R},B) @>\gamma(A\rtimes_{\rho} \mathbb{R},B)>> Hom(K_i(A\rtimes_{\rho} \mathbb{R}),K_j(B)) \\ @VV \eta V @VV \sigma V \\ KK_{i+1}(A,B) @>\gamma(A,B)>> Hom(K_{i+1}(A),K_j(B)) \\ \end{CD}$$ is a commutative diagram, where $\eta , \sigma$ and $\gamma(A,B)$ are isomorphisms, therefore $\gamma(A\rtimes_{\rho} \mathbb{R},B)$ is an isomorphism. My question is: why is this diagram commutative? For this, I tried to figure out how to write down the maps explicitely. The map $\gamma(A,B)$ comes from the Kasparov-product $$KK(\mathbb{C},A)\times KK(A,B)\to KK(\mathbb{C},B)\; (\epsilon_1 , \epsilon_2)\mapsto \epsilon_1 \otimes \epsilon_1 ,$$ because it is $KK_i(\mathbb{C},A)\cong K_i(A)$, $KK_i(\mathbb{C},B)\cong K_i(B)$. Hence the Kasparov-product induces a homomorphism (which should be $\gamma(A,B)$ I think) $$KK(A,B)\to Hom(K(A), K(B)),$$ $$\epsilon_2 \mapsto (\epsilon_1 \mapsto \epsilon_1 \otimes \epsilon_2).$$ The map $\sigma$ is the contravariant $Hom(-,K_j(B))$-functor apllied to the (Connes-Thom)-isomorphism $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$. But I don't know how to write down the isomorphisms $\sigma$ and $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ explicitely such that I'm stuck to prove that the diagram commutes. Can you help me to prove that the diagram commutes, or do you know how to write down the maps $\sigma$ and $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ explicitely? Best REPLY [3 votes]: I think the key idea is that Connes' Thom isomorphism is itself given by a $KK$-equivalence (see for example Blackadar's book "K-theory for Operator Algebras" - Theorem 19.3.6). This means there are classes $t \in KK_1(A, A \rtimes \mathbb{R})$ and $t^{-1} \in KK_1(A \rtimes \mathbb{R},A)$, such that the corresponding intersection products $t \otimes_{A \rtimes \mathbb{R}} t^{-1}$ and $t^{-1} \otimes_A t$ agree with the identity elements in $KK(A \rtimes \mathbb{R},A \rtimes \mathbb{R})$ and $KK(A,A)$ respectively (using Bott periodicity to identify $KK_2$ with $KK$) and the intersection product with $t$ induces the Thom isomorphism $$ K_{i+1}(A) \cong KK_{i+1}(\mathbb{C}, A) \to K_{i}(A \rtimes \mathbb{R}) \cong KK_{i}(\mathbb{C}, A \rtimes \mathbb{R}) \quad ;\quad x \mapsto x \otimes_A t $$ as well as the map $\eta$ via $$ KK_i(A \rtimes \mathbb{R},B) \to KK_{i+1}(A,B) \quad ;\quad y \mapsto t \otimes_{A \rtimes \mathbb{R}} y\ . $$ Let $x \in KK_i(A \rtimes \mathbb{R},B)$ and $y \in K_{i+1}(A)$. We have to check that $$ \sigma(\gamma(A \rtimes \mathbb{R}, B)(x))(y) = \gamma(A,B)(\eta(x))(y) $$ If we evaluate the left hand side we get $$ (y \otimes_A t)\,\otimes_{A \rtimes \mathbb{R}}\,x = y \otimes_A (t \otimes_{A \rtimes \mathbb{R}}\,x) $$ where we used the associativity of the Kasparov intersection product. But now the right hand side agrees with $\gamma(A,B)(\eta(x))(y)$ and the diagram commutes. There is an increasing amount of people with rabbit related pseudonyms/profile pictures on mathoverflow.net (I am looking at you Bugs Bunny). Is this a trend that I missed?<|endoftext|> TITLE: Morrey's inequality for Sobolev spaces of fractional order QUESTION [5 upvotes]: Let $H^s(\mathbb T)$, where $s\in\mathbb R$, be the space of $2\pi$-periodic functions (or distributions), $u(x)=\sum_{k\in\mathbb Z}\hat u_k\,\mathrm{e}^{ikx}$, such that $$ \|u\|_{H^s}^2=\sum_{k\in\mathbb Z}(1+k^2)^{s}\lvert \hat u_k\rvert^2<\infty. $$ Assume now that $s\in \big(\frac{1}{2},\frac{3}{2}\big)$. I wish to find out whether there exists a constant $c=c_s$, such that $$ \lvert u(x)-u(y)\rvert\le c \lvert u\rvert_{H^s}\lvert x-y\rvert^{s-\frac{1}{2}}, $$ where $$ \lvert u \rvert_{H^s}^2=\sum_{k\in\mathbb Z}\lvert k\rvert^{2s}\lvert \hat u_k\rvert^2. $$ If $s=1$, then this is Morrey's inequality. Any reference? Note. I have asked this question in math.stackexchange without any luck. REPLY [2 votes]: The Sobolev space $H^s(\mathbb T)$ is $W^{s,2}(\mathbb T)$ and is also the Besov space $B^{s}_{2,2}(\mathbb T)$. On the other hand the set of Hölder continuous functions with index $\alpha\in (0,1)$ is the Besov space $B^{\alpha}_{\infty,\infty}(\mathbb T)$. For $s\in(1/2,3/2),$ you have the following continuous inclusion, $$ B^{s}_{2,2}\subset B^{s-\frac12}_{\infty,\infty}, $$ which is a classical result, whose proof is immediate using the Littlewood-Paley decomposition: if $(\phi_\nu)_{\nu\in \mathbb N}$ is such a decomposition (here $\phi_\nu(\xi)=\phi(\xi 2^{-\nu})$ so that $1=\sum_\nu \phi_\nu(\xi)$, with $\phi\in C^\infty_c)$, the norm of a function $u\in B^{s}_{2,2}(\mathbb T)$ is $$ \bigl(\sum_{\nu\in \mathbb N}2^{2\nu s}\Vert{\phi_\nu(D)u}\Vert_{L^2}^2\bigr)^{1/2} $$ whereas the norm in $B^{s-\frac12}_{\infty,\infty}$ is $ \sup_{\nu\in \mathbb N}2^{\nu (s-\frac12)}\Vert{\phi_\nu(D)u}\Vert_{L^\infty} $. We have with $\psi\in C^\infty_c$ equal to 1 on the support of $\phi$, $$ 2^{\nu (s-\frac12)}\Vert{\phi_\nu(D)u}\Vert_{L^\infty} =2^{\nu (s-\frac12)}\Vert{2^\nu\hat\psi(2^\nu \cdot)\ast \phi_\nu(D)u}\Vert_{L^\infty}\\ \lesssim 2^{\nu (s-\frac12)}\Vert2^\nu\hat\psi(2^\nu \cdot)\Vert_{L^2} \Vert{\phi_\nu(D)u}\Vert_{L^2}\lesssim 2^{\nu s}\Vert{\phi_\nu(D)u}\Vert_{L^2} $$ so that the following continuous injections hold $$B^{s}_{2,2}\subset B^{s}_{2,\infty}\subset B^{s-\frac12}_{\infty,\infty}. $$<|endoftext|> TITLE: Is it inconsistent for a model of set theory to contain its own first order theory? QUESTION [9 upvotes]: I am wondering if it is inconsistent to have a model of set theory V such that V contains an $A\subset \omega$ that codes its first order theory.I.e. for all $\{\underline\epsilon\}$-sentences $\phi$, $V\models \phi \leftrightarrow \langle\phi\rangle \in A$. Tarski's theorem shows that such an A cannot be definable, but I see no reason why such an A cannot exist in general. In the situation I am considering, the prospective A would be OD. This arises in the following situation: there is an iterable model $M$, an elementary embedding $j:M\rightarrow N$, and when one compares $M$ and $N$, the iterate of $M$ is a strict initial segment of the iterate of $N$ or vice versa. REPLY [10 votes]: First, let me point out as the others have that if there are large cardinals, then indeed we expect this situation. For example, if there is a worldly cardinal, a cardinal $\kappa$ for which $V_\kappa\models\text{ZFC}$, then the theory of $V_\kappa$ will of course be an element of $V_\kappa$ and therefore this will be a model of the kind you seek. But next, although one might think at first that your situation requires large cardinals, let me point out that in fact the situation you have described is in fact exactly equiconsistent with ZFC. Theorem. There is a model of ZFC if and only if there is a model of ZFC with an object $A$ as you describe. Proof. The converse implication is immediate. For the forward implication, suppose that ZFC is consistent. Let $T$ be the theory of ZFC together with the assertions about the set $A$ that you have described, namely, the scheme of assertions that $\phi\iff\ulcorner\varphi\urcorner\in A$. My observation is that if ZFC is consistent, then every finite subtheory of $T$ is consistent. The reason is that if $M\models\text{ZFC}$, then since any finite subtheory of $T$ involves only finitely many instances of the theory scheme, it follows by the reflection theorem that that finite subtheory holds in some $(V_\alpha)^M$. So the whole theory $T$ is consistent, and any model of this is a model of your situation.QED Indeed, the same argument shows that every computably saturated model of set theory contains an element coding its own theory. Basically, the argument I give above is realizing a certain computable type. Essentially the same idea shows that ZFC is consistent just in case there is a model $M\models\text{ZFC}$ with a cardinal $\kappa$ for which $V_\kappa^M\prec M$. This is a little paradoxical at first, since you might think that $M$ would have to think that such a $V_\kappa$ is a model of ZFC itself, but that conclusion is unwarranted, since perhaps $M$ is $\omega$-nonstandard, in which case its understanding of ZFC is is not accurate. Theorem. There is a model of ZFC if and only if there is a model $M$ of ZFC with a cardinal $\kappa$ for which $V_\kappa^M\prec M$. I have used this fact in a few arguments in papers of mine, such as my paper A simple maximality principle. Finally, let me reiterate and confirm your statement (sorry for my earlier confusion) that the theory of a model can never be definable in the model. This is a consequence of Tarski's theorem on the non-definability of truth. If there were a formula $\varphi(\cdot)$ with one free variable such that a model of set theory $\langle M,\in^M\rangle\models\sigma$ just in case it satisfies $\varphi(\ulcorner\sigma\urcorner)$, then by the fixed-point lemma we can find a sentence $\sigma$ that is ZFC-provably equivalent to $\neg\varphi(\ulcorner\sigma\urcorner)$, and since $\sigma$ asserts its own falsehood, we easily get a contradiction.<|endoftext|> TITLE: Nonlinear ODE: $y'=(1+axy)/(1+bxy)$ QUESTION [5 upvotes]: Consider the first order nonlinear ODE problem: $$ y'(x)=\frac{1+ay(x)x}{1+by(x)x}, \quad x>0 $$ where $a, b>0$ are some constants. I would like to know if these kind of equations were studied somewhere else (i.e., existence of global solutions, uniqueness etc). Any references will be helpful. Remarks: Mathematica does not give me anything. I have tried to use Maple and it shows me that $y$ is a solution of an equation involving some Whittaker functions but I am not sure about this. If $a=b$ then $y=x+C$. If $a=1, b=2$ then one can take $y(x) = \frac{1}{2}\left(x-\frac{1}{x} \right)$ REPLY [2 votes]: Global existence (or not) can be clarified without any tools by just taking a good look at your equation. Clearly, the RHS $f(x,y)$ is bounded away from the curve $1+bxy=0$, where it is undefined. So global existence fails for a given solution precisely if this solution approaches $1+bxy=0$ in finite time (in particular, global existence for a solution $y(x)$ is guaranteed as soon as $y(x_0)\ge 0$ for some $x_0$). This won't happen if $a\le b$, as we see by just checking what the ODE does close to our curve. For example, if $1+bx_0y(x_0)=\epsilon>0$, then $y'(x_0)>0$, so we're moving away from the curve. If $a>b$, then, by the same argument, there are solutions that reach $1+bxy=0$ in finite time (those with sufficiently negative initial value $y(0)$).<|endoftext|> TITLE: About $C(K)$-spaces containing no copy of $l_{1}$ QUESTION [7 upvotes]: Let $K$ be a compact Hausdorff space. I wonder whether there are characterizations of $K$ such that $C(K)$ contains no copy of $l_{1}$. There are some compact Hausdorff spaces $K$ such that $C(K)$ contains no copy of $l_{1}$, for example, if $K$ is a countable compact metric space and has finite Cantor-Bendixson index. Are there more compact Hausdorff spaces $K$ such that $C(K)$ contains no copy of $l_{1}$? Thank you! REPLY [7 votes]: Yes, there is such characterisation. $C(K)$ contains no isomorphic copy of $\ell_1$ if and only if $K$ is scattered. Indeed, if $K$ is scattered then $C(K)^*$ is isometric to $\ell_1(K)$, so $C(K)$ cannot contain $\ell_1$, as then $C(K)^*$ would have contained a copy of $L_1$. Conversely, if $K$ is not scattered, then you may find a copy of $C[0,1]$ in $C(K)$.<|endoftext|> TITLE: Comparison between analytic etale cohomology and algebraic etale cohomology for affinoids QUESTION [7 upvotes]: Let $\mathcal{A}$ be an affinoid algebra over a complete non-archimedean field $K$. We have two objects we can investigate, namely the algebro-geometric spectrum $X = \operatorname{spec} \mathcal{A}$ and the non-archimedean analytic space $X^{an} = \operatorname{sp} \mathcal{A}$. The former has an etale theory from the 1960's, and the latter from V. Berkovich's '93 IHES paper. In particular for an abelian etale sheaf $\mathcal{L}$ I can analytify the sheaf to get a sheaf on the etale topology of $X^{an}$, which I'll call $\mathcal{L}^{an}$. Are there any comparison results between $H^i(X_{\acute{e}t},\mathcal{L})$ and $H^i(X^{an}_{\acute{e}t},\mathcal{L}^{an})$? The comparison theorems in that paper seem to work for algebras of finite type over $K$, or of finite type over an affinoid base. It doesn't seem directly possible to set it up so that the derived pushforward $R^q\varphi_*$ maps to $\operatorname{sp} K$. I am principally interested in smooth affinoids over a discretely valued field, and $\mathcal{L}$ a locally constant sheaf of finite abelian groups whose orders are prime to the characteristic of the residue field. I have a very convoluted argument in mind using Berkovich's most recent pre-print but I imagine there is an easier way. EDIT: I suppose I should probably indicate that I've put some thought into this: You can take a closed immersion of $X^{an}$ into some ball (e.g. higher dimensional analogues of $E(0,r) \times D(0,s)$). This closed immersion is algebraically of finite type, and so we can use the comparison theorem to compare the two push-forwards. This plus the Leray spectral sequence would reduce the problem to showing the analogous result for constructible sheaves on balls. However there doesn't really seem to be any tools that I can obviously use to attack this. REPLY [2 votes]: Not sure if this is still relevant, but the desired comparison in the case of "principal interest" to you can be proved in two different ways. On one hand, you can deduce it from some recent results of Achinger, which imply that both cohomologies in question are unchanged if you replace the etale sites by the finite etale sites. Since the finite etale sites of $X$ and $X^{an}$ are canonically equivalent, this gives what you want. This works for any affinoid $\mathcal{A}$ over a complete discretely valued $K$ and any finite locally constant sheaf of abelian groups $\mathcal{L}$ on $X_{et}$. Alternatively, and now for $K$ any complete nonarchimedean field, you can make a devissage to the case where $\mathcal{L}$ is a constant sheaf of finite abelian groups, which then is handled by Corollary 3.2.3 in Huber's book. To do this, pick a finite etale Galois cover $f:X'\to X$ with Galois group $G$ such that $f^{\ast}\mathcal{L}$ is constant, and use the Hochschild-Serre spectral sequences $H^i(G,H^j_{et}(X',f^{\ast}\mathcal{L})) \Rightarrow H^{i+j}_{et}(X,\mathcal{L})$ and $H^i(G,H^j_{et}(X'^{an},f^{an,\ast}\mathcal{L}^{an})) \Rightarrow H^{i+j}_{et}(X^{an},\mathcal{L}^{an})$. There is a canonical morphism from the first spec. seq. to the second, and it's an isomorphism on all terms of the $E_2$-page (by Huber's Corollary 3.2.3 plus some nonsense about compatibilities), so it gives an isomorphism on the abutments.<|endoftext|> TITLE: Irreducible representations containing simple actions of $\mathrm{SL}(2,\mathbb{C})$ QUESTION [5 upvotes]: Let $G$ be a complex semisimple Lie group and let $\rho: G \longrightarrow \mathrm{SL}(n,\mathbb{C})$ be a faithful irreducible representation of $G$ with $n \geq 3$. Suppose that $G$ contains a copy of $\mathrm{SL}(2,\mathbb{C})$ such that $\bullet$ its action preserves a decomposition $\mathbb{C}^n = \mathbb{C}^2 \oplus \mathbb{C}^{n-2}$ $\bullet$ its action is the natural action of $\mathrm{SL}(2,\mathbb{C})$ on $\mathbb{C}^2$ and it acts trivially on $\mathbb{C}^{n-2}$. My question is: must $\rho(G)$ be equal to the whole $\mathrm{SL}(n,\mathbb{C})$? or do you know any counterexamples? Obviously, $\mathrm{SL}(n,\mathbb{C})$ contains $\mathrm{O}(n, \mathbb{C})$ the group of automorphisms of a non degenerate quadratic form and $\mathrm{Sp}(n, \mathbb{C})$ the symplectic group if $n$ is even. So for my question not to be trivial, I make the extra assumption that $\rho(G)$ is not contained in any conjugate of these two groups. I understand that diging deep enough in the theory of representations should provide an answer to my question, but being no expert in Lie theory I was wondering if I was missing any simple argument or counter-example. Thanks for your attention. REPLY [4 votes]: The only groups $G$ of the type you are looking for are $SL(n)$, $SO(n)$ or $Sp(n)$. This is proved in a paper by Beukers and Heckman (see Proposition (6.4) in that paper). See the math review http://www.ams.org/mathscinet-getitem?mr=974906 for a reference. Some explanation: Beukers and Heckman talk about a "reflection" normalising the irreducible subgroup. In their setting, a (complex) reflection is an element $g\in GL(n)$ such that $g-1$ has kernel of dimension $n-1$. In that case, they prove that a connected irreducible algebraic subgroup of $SL(n)$ which is normalised by a complex reflection is either $SL(n)$ or $SO(n)$ or $Sp(n)$. Now the standard unipotent element $u$ in your SL(2) contained in $G$, is indeed a complex reflection and so your group $G$ is normalised by the "reflection" $u$. Incidentally, in your case the situation $SO(n)$ cannot arise since this unipotent reflection cannot lie in $SO(n)$.<|endoftext|> TITLE: Convergence of functionals on compact projections on a separable Hilbert space QUESTION [5 upvotes]: Let $H$ be a separable Hilbert space over $\mathbb{C}$, say $\ell_2$ for simplicity. Let $\mathcal{K}(H)$ denote the space of all compact operators on $H$ and $\mathcal{P}(H)$ the set of all finite rank orthogonal projections on $H$ (so $\mathcal{P}(H)\subset\mathcal{K}(H)$). Assume that $(x_n^*)_{n\in\mathbb{N}}$ is a sequence of bounded functionals on $\mathcal{K}(H)$ such that $\lim_{n\to\infty}x_n^*(P)=0$ for every $P\in\mathcal{P}(H)$. Question: Is it true that $\lim_{n\to\infty}x_n^*(T)=0$ for every $T\in\mathcal{K}(H)$? Equivalently, is $(x_n^*)_{n\in\mathbb{N}}$ norm bounded? Such a situation holds e.g. in von Neumann algebras (a result due to Darst '67) or C*-algebras of the form $C(K)$ where $K$ is the Stone space of a $\sigma$-complete Boolean algebra (Nikodym '33). References: R.B. Darst, On a theorem of Nikodym with applications to weak convergence and von Neumann algebras, Pacific J. Math. 23 (1967), no. 3, 473–477. O. Nikodym, Sur les familles bornées de fonctions parfaitement additives d’ensemble abstrait, Monatsh. Math. Phys. 40 (1933), no. 1, 418–426. REPLY [3 votes]: This is false. The dual $K(H)^*=B_1(H)$ is given by the trace class operators, and an $S\in B_1(H)$ acts on a $T\in K(H)$ by $(S,T)=\textrm{tr}\, ST$. Consider now $$ S_n = n2^{-n} \sum_{2^n\le j<2^{n+1}} \langle e_j , \cdot \rangle e_j . $$ The norm of $S_n$ as a functional is its trace norm, which equals $n$, so this sequence is unbounded. However, your condition holds: it suffices to check this for a rank one projection $T=\langle v, \cdot \rangle v$, and then $$ (S_n,T) = n2^{-n}\sum |\langle e_j, v\rangle |^2 \le n2^{-n}\to 0 , $$ as required.<|endoftext|> TITLE: Local Langlands Conjecture for p-adic SO(4), reference request QUESTION [5 upvotes]: In section 10 of Gan-Gross-Prasad's paper "Symplectic local root numbers, central critical L-values, and restriction problems in the representation theory of classical groups" http://arxiv.org/pdf/0909.2999v1.pdf, the authors stated that the full local Langlands conjectures for some small ranked classical groups are known, but no reference is given. Can anybody point out a reference for the SO(4) and SO(6) case? Do we know whether the local Langlands map preserves certain local epsilon or L-factors? Added: I appreciated the answer provided by Professor Kimball, which is very helpful. I do want to know more details about (split) SO(4). In the last section of Gross-Prasad's paper as given by Kimball, there is an exact sequence $$0\rightarrow SO_4(k)\rightarrow GL_2(k)\times GL_2(k)/\Delta k^* \rightarrow k^* \rightarrow 0.$$ To get the local Langlands conjecture for SO(4) from that of GL(2) from this exact sequence, it seems that we need to know the relationship between irreducible admissible representations of SO(4) and irreducible admissible representations of $GL_2(k)\times GL_2(k)/\Delta k^*$. Is this something standard or even trivial? On the other hand, in GGP, the authors stated that most local Langlands conjecture for classical groups should be established by the (forthcoming) book of Arthur and Moglin's papers. Arthur's books has already come out https://www.amazon.com/Endoscopic-Classification-Representations-Colloquium-Publications/dp/0821849905/ref=sr_1_1?s=books&ie=UTF8&qid=1465744787&sr=1-1&keywords=arthur+automorphic+representations . I never tried to read Arthur's book. What I would like to know is: whether they (Arthur and Moglin) proved the preservation of local epsilon and L-factors, or they just proved the existence of the local Langlands map. REPLY [2 votes]: I don't know an explicit reference for you, but I can tell you how these things are done and point you to some related reference. These groups are closely related to general linear groups via accidental isomorphism, so one can write down the Langlands parameters in terms of those for general linear groups and this should also give preservation of epsilon and L-factors. SO(4) is essentially GL(2) $\times$ GL(2). See, e.g., the last section of Gross and Prasad's original paper (Can. J, 1992). SO(6) is essentially GL(4). Both of the relations are essentially described in Gan and Takeda's paper Theta correspondences for GSp(4) (Rep Thy 2011). (Technically they work with describe slightly related groups, GSO, which contain the SO's.)<|endoftext|> TITLE: What are the orthogonal polynomials w.r.t. Maxwell distribution QUESTION [12 upvotes]: Is there a way to get a clean presentation of the orthogonal polynomials w.r.t. the Maxwell distribution https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution ? If you separate the problems to an even and odd problems, the even part of the orthogonal basis behaves nice enough: one simply takes the quotient of the odd Hermite polynomials by x (since the Maxwell distribution is given by $Cx^2\exp(-x^2/2)$, and Hermite polynomials are orthogonal w.r.t. $\exp(-x^2/2)$) ; as for the even part, I described a semi-solution in the comment below. However, I have no ide how the two parts interact. The analog question for Laguerre polynomials (i.e. when one moves from the measure $\exp(-x)$ to the measure $C_{\alpha}x^{\alpha}\exp(-x)$ yields a very satisfying result in the form of generalized Laguerre polynomials. Due to the relation between Laguerre polynomials and Hermite polynomials, I'd expect something similar here. REPLY [10 votes]: These are the socalled Maxwell polynomials $M_n^{(p)}(x)$, see page 75 and following of Spectral Methods in Chemistry and Physics. The coefficients for $p=2$, which is the case you need, are given in Table 2.14 of that reference.<|endoftext|> TITLE: When is $\mathbb C^d\setminus\mathcal Z$ simply connected? QUESTION [5 upvotes]: Let $\Delta$ be a fan in the lattice $N\cong\mathbb Z^n$ with $d$ edges $\{\rho_1,\cdots,\rho_d\}$. Consider the co-ordinate ring $\mathbb C[x_1,\cdots,x_n]$. Let $\mathcal Z=\bigcup_C\mathcal V(x_i\ |\ \rho_i\in C)$ where the union is taken over all primitive collections $C$ of edge vectors in $\Delta$. Definition - A subset $C\subseteq \Delta(1)$ (the set of edges in $\Delta$) is a primitive collection if : $C\nsubseteq \sigma(1)$ for all $\sigma \in \Delta$ For every proper subset $C'$ of $C$ there is a $\sigma\in\Delta$ with $C'\subseteq \sigma(1)$ Question - Is there a way in general to check if $\mathbb C^d\setminus\mathcal Z$ is simply connected in the analytical topology? I know that $2\le\text{codim}\mathcal Z\le d$. Suppose further that $\Delta$ is compete and simplicial then we have either $2\le \text{codim}\mathcal Z\le \lfloor \frac{n}{2}\rfloor+1$ or $d=n+1$ and $\mathcal Z=\{0\}$ It is easy to see that if $\Delta$ is simplicial and 2. is true then $\mathbb C^d\setminus\mathcal Z=\mathbb C^{n+1}\setminus\{0\}$ is simply connected. What about in general? I understand that it will not always be true, but for what codimension of $\mathcal Z$ can one expect $\mathbb C^d\setminus\mathcal Z$ to be simply connected? (Assuming there is a way to relate codimension of $\mathcal Z$ to the simply connectedness of $\mathbb C^d\setminus\mathcal Z$). Thank you. REPLY [4 votes]: There is a simple general fact about k-connectedness of the complement of an affine algebraic set, which seems not to be so well known: Theorem. If $Z \subseteq \mathbb{C}^d$ is Zariski-closed of codimension $c$, then $\pi_i(\mathbb{C}^d\smallsetminus Z) = 0$ for $0 TITLE: Definition of dense functors QUESTION [7 upvotes]: Definition. A functor $F:\mathsf C\rightarrow \mathsf D$ is dense if every $D\in \mathsf D$ is the vertex of the following colimit $$\varinjlim \left(F\downarrow D\rightarrow\mathsf C\rightarrow \mathsf D \right).$$ I would like to understand the idea behind this definition better. It seems that to imitate the spatial situation of dense subsets (or dominant functions), we could just ask every $D\in\mathsf D$ to be a colimit of some diagram in the image of $F$. Instead, we ask every object to be the colimit of an enormous diagram, because it is canonical. Being canonical is nice and all, but is there a more intuitive justification behind this definition? (Having the truncated Yoneda embedding full and faithful, in my view, is not a justification, but a consequence.) The only thing I can come up with is that in spaces, nets are hugely redundant because only their tail matters, and the real approximation using dense subsets is by "close points", whatever that means. The analogue of this is somehow looking at colimits only on objects equipped with a map from the essential image of $F$. Why is the primitive imitation of the spatial situation not interesting? REPLY [6 votes]: I don't think the topological analogue is worth paying attention to. Here's a basic reason to care: suppose $G : C \to E$ is another functor, and you'd like to compute the left Kan extension $\text{Lan}_F(G) : D \to E$. If enough colimits exist, the left Kan extension is given pointwise by $$\text{Lan}_F(G)(d) = \text{colim}_{f : F(c) \to d} G(c)$$ which can be interpreted as follows. For fixed $d$, the canonical diagram of maps $f : F(c) \to d$ is the universal way to approximate $d$ by a colimit of objects $c \in C$, and the left Kan extension figures out how to apply $G$ to $d$ by applying $G$ to this diagram, then taking the colimit in $E$. In particular, $F$ is dense iff $\text{Lan}_F(F)$ is the identity functor. The left Kan extension $\text{Lan}_F(F)$ of a functor along itself is the density comonad, dual to the codensity monad; it measures the extent to which $F$ fails to be dense. There are various notions of what it might mean for (the image of) $C$ to generate $D$; this is one of them, and there are others. See, for example, this blog post, although it doesn't discuss dense generation.<|endoftext|> TITLE: Affine GIT is an open map? QUESTION [8 upvotes]: Let $k$ be a field, $X= \text{Spec}\,A$ be an affine scheme, with $A$ a finitely generated $k$-algebra. $G=\text{Spec}\,R$ is a linearly reductive group acting rationally on A, i.e. every element of $A$ is contained in a finite-dimensional $G$-invariant linear subspace of $A$. By Nagata's theorem, $A^G$ is a finitely generated $k$-algebra. We have the affine GIT quotient $X \rightarrow X/G := \text{Spec}\,A^G$ induced by the inclusion $A^G \rightarrow A$ of $k$-algebras. Question. Is the affine GIT quotient, viewed as a map of the underlying topological spaces, necessarily an open map? It does not need to be an open immersion of schemes. If not, any counterexample? REPLY [9 votes]: Categorical quotients are in general very far away from being open. In fact, it is a theorem of Chevalley (I think) that a morphism $\pi:X\to Y$ onto a normal variety $Y$ is open if and only it is equidimensional. Unless $G$ is finite, this is a very rare condition for quotient morphisms. The idea behind this is the following: Assume that the fiber $X_y$ has bigger dimension than the generic fiber dimension $\dim X-\dim Y$. Then there is a curve $C\subseteq Y$ passing through $y$ such that the fiber dimension jumps over $C$. Let $Z$ be the closure of $\pi^{-1}(C)\setminus X_y$ in $X$. Then for dimension reasons $X_y\not\subseteq Z$. Consider the open subset $U:=X\setminus Z$ of $Z$. Then $\pi(Z)=(Y\setminus C)\cup\{y\}$ is only constrictible but not open. More formally, the openness of a morphism is expressed by the so-called going-down property of Cohen-Seidenberg. Let's try this out for the quotient of $G=\mathbf G_m$ acting on $\mathbf A^3$ by $(tx,ty,t^{-1}z)$. Then $$ \pi:\mathbf A^3\to\mathbf A^2:(x,y,z)\mapsto (u,v):=(xz,yz) $$ is the quotient morphism mapping the plane $\{z=0\}$ to $(0,0)$. Take $C=\{u=0\}$. Then $\pi^{-1}(C)=\{x=0\}\cup\{z=0\}$ and the image of $U=\mathbf A^3\setminus\{x=0\}$ is $\mathbf A^2\setminus\{u=0\}\cup\{(0,0)\}$ which is not open. PS: Quotient maps have some properties not shared by other morphism: Images of closed $G$-stable subsets are closed. A subset of $Y$ is open iff its preimage in $X$ is open. The morphism is surjective. These properties hold universally, i.e., even after base change.<|endoftext|> TITLE: Questions on J. F. Nash's answer about his errors in the proof of embedding theorem QUESTION [25 upvotes]: In the interview of John Nash taken by Christian Skau and Martin Gaussen, in EMS Newsletter, September, 2015 when asked Is it true, as rumours have it, that you started to work on the embedding problem as a result of a bet? Nash answered I began to work on it. Then I got shifted onto the $^1$ case. It turned out that one could do it in this case with very few excess dimensions of the embedding space compared with the manifold. I did it with two but then Kuiper did it with only one. But he did not do it smoothly, which seemed to be the right thing—since you are given something smooth, it should have a smooth answer. and But a few years later, I made the generalisation to smooth. I published it in a paper with four parts. There is an error, I can confess now. Some forty years after the paper was published, the logician Robert M. Solovay from the University of California sent me a communication pointing out the error. I thought: “How could it be?” I started to look at it and finally I realized the error in that if you want to do a smooth embedding and you have an infinite manifold, you divide it up into portions and you have embeddings for a certain amount of metric on each portion. So you are dividing it up into a number of things: smaller, finite manifolds. But what I had done was a failure in logic. I had proved that—how can I express it?—that points local enough to any point where it was spread out and differentiated perfectly if you take points close enough to one point; but for two different points it could happen that they were mapped onto the same point. My question is: What did Nash mean by very few excess dimensions of the embedding space compared with the manifold? It means that the they can do it in the case the dimension of the embedding space is a little bit greater than the dimension of the manifold, doesn't it? What did Nash mean by his generalization to smooth? What did Nash mean by a certain amount of metric on each portion? Does this mean that each portion has some different metrics? What did Nash mean by "I had proved that that points local enough to any point where it was spread out and differentiated perfectly if you take points close enough to one point"? How can a point be spread out and differentiated? Well, I don't want to make a cross post but this question were posted two days on MSE but there is no answer so I decide to post it here. Please explain for me. Thanks. REPLY [43 votes]: Igor already answered questions 1 and 2. What Nash wrote is an attempt to describe to a non-expert audience the solution scheme he had for non-compact manifolds. In the noncompact case he proceeds by a reduction process to the compact case. The process involves decomposing the manifold into smaller neighborhoods each diffeomorphic to disks/balls. Very roughly speaking, Nash then took a partition of unity adapted to this system of neighborhoods, and used that to cut-off the Riemannian metric on the original manifold to these disks. Now, the truncated objects are no longer Riemannian metrics (since they are not everywhere positive definite); but since the problem is not really that of geometry but that of PDEs in this setting, this is not an obstruction. When two neighborhoods overlap, the cut-off function is less than 1 for both of the pieces. So each of the neighborhood inherits "a portion of the metric". In other words, the phrase that you are asking about is just Nash trying to describe the idea of a "partition of unity" without using exactly those words. This is in regards to the second portion of the scheme. After dividing into disks and keeping track of their overlaps (using what Nash called "classes"; that's also the sense of the word in Solovay's message), it is possible to (very roughly speaking) replace the disks by spheres. You do this using that the disks can be topologically identified with the sphere with a closed disk removed, and that the "portion of the metric" on the disc is cut-off so that it approaches 0 smoothly on the boundary. This is just largely playing with cut-off functions. Now, each sphere has an isometric embedding into some Euclidean space by the theorem for compact manifolds (proven earlier in the paper). So all remains is to somehow reassemble all these spheres into one larger Euclidean space while guaranteeing that there are no self-intersections. (Remark: the method of assembly works. Full stop. So if you are happy with an isometric immersion then you are done. The problem that Solovay pointed out has to do with the embedding (non-self-intersecting) part.) The method of assembly goes something like this. a. The choice of the disks earlier means that the disks can be divided into some finitely many classes, such that two disks of the same class cannot intersect. b. For each class, construct a smooth map from $M$ such that points within the discs of that class are sent to the image of the embedding of the corresponding sphere. But points outside the disks of that class are sent to the origin. c. Take the cartesian product over the classes. This guarantees an immersion. To get non-self-intersection Nash tried to exploit the fact that his isometric embedding theorem in the compact case allows one to squeeze the manifold into arbitrarily small neighborhoods. So within each class he can arrange for the different disks to be almost disjointly embedded. He claims that this is enough. Solovay showed that there is a hole in the argument. See below the cut for more info. Incidentally, Nash's paper is available here; the portion that concerns questions 3 and 4 are all in part D, which is decidedly in the "easy" part of the paper. (The hard analytic stuff all happened in part B; here it is essentially combinatorics.) To reveal the "logic problem" with the embedding proof, we remove all unnecessary portions of the proof and focus on the argument that "ensures" non-self-intersection. This operates entirely on the level of sets and does not require any geometry or analysis. What Nash did boils down to: Given a set $M$, he showed that we can decompose $M$ as the union of a bunch of sets $U^{(i)}_j$. The index $i$ runs from $1, \ldots, n+1$ (a finite number). The index $j$ can be infinite. Each $U^{(i)}_j$ has a subset within called $V^{(i)}_j$, and the union of all these subsets $V^{(i)}_j$ is also assumed to cover $M$. For each $i$ Nash finds a space $X^{(i)}$, and a point $x^{(i)}\in X^{(i)}$, and for each $j$ there is an injective mapping $\psi^{(i)}_j: U^{(i)}_j \to X^{(i)}\setminus \{x^{(i)}\}$. Since $U^{(i)}_j$ and $U^{(i)}_k$ do not intersect unless $j = k$, for each $i$ we can extend this to a map $$ \psi^{(i)}: M \to X^{(i)}$$ by requiring $$ \psi^{(i)}(p) = \begin{cases} \psi^{(i)}_j(p), &p\in U^{(i)}_j \\ x^{(i)}, &\text{otherwise}\end{cases} $$ Let $X = X^{(1)} \times X^{(2)} \cdots \times X^{(n+1)}$. We are interested in the map $$ \psi: M \to X = \psi^{(1)}\times \psi^{(2)}\times \cdots \times \psi^{(n+1)}.$$ We want to show that $\psi$ is injective. Nash's idea: We can assume that $\psi^{(i)}(V^{(i)}_j) \cap \psi^{(i)}(U^{(i)}_k) = \emptyset$ if $k > j$. (This he achieves by the fact that isometric embedding can be "made small".) He claims this is enough to show injectivity of $\psi$, because: If $p,q\in U^{(i)}_j$ for the same $i,j$, then we are done because $\psi^{(i)}_j$ is by construction injective. If $p \in U^{(i)}_j$ and $q \not\in U^{(i)}_k$ for for any $k$, then we know $\psi^{(i)}(p) \neq x^{(i)} = \psi^{(i)}(q)$ by construction, and so we are done. So the main worry is that $p\in U^{(i)}_j$ and $q \in U^{(i)}_k$ for some different $j,k$. To deal with this, Nash argued thus (I paraphrase) Since the $V$'s cover $M$, $p$ is in some $V^{(i)}_j$ and similarly $q$. So either $q$ does not belong to any $U^{(i)}_*$ in which case we are done by point 2, or $q$ belongs to some $U^{(i)}_k$. If $k > j$ by Nash's idea their images are disjoint, so we get injectivity. If $k = j$ we are done by point 1. If $k < j$ we swap the roles of $p$ and $q$. The bold phrase was not stated as such in Nash's original paper, but it was his intent. Stated in this form, however, it becomes clear what the problem is: in the argument $p$ and $q$ is not symmetric! One cannot simply swap the roles of $p$ and $q$. It could easily be the case that $q \in U^{(i)}_k \setminus V^{(i)}_k$ and then the idea of making the images of $U^{(i)}_*$ "almost disjoint" fails to yield anything useful. Solovay's message instantiates this observation by setting up a situation where $$ p \in (U^{(1)}_1 \setminus V^{(1)}_1) \cap V^{(2)}_2 $$ and $$ q \in (U^{(2)}_1 \setminus V^{(2)}_1) \cap V^{(1)}_2 $$ and both not in any other $U^{(i)}_j$. So in terms of the turn of phrase Nash used: "points local enough to any point" == we take $p \in U^{(i)}_j$ and $q \in U^{(k)}_l$ (they are in neighborhoods of some point) "it was spread out and differentiated perfectly" == the map $\psi(p) \neq \psi(q)$ (is injective; "differentiated" in the sense of "to tell apart") "if you take points close enough to one point" == if $i = k$ and $j = l$ (in the same neighborhood) "but for two different points it could happen that they were mapped onto the same point." == injectivity may fail otherwise.<|endoftext|> TITLE: Important results with one or more than one proof QUESTION [10 upvotes]: Can you give examples of deep, important results that have only one known proof, and not just because the first proof is fairly recent, or because not many people really cared to think about it? How hard is the proof from the perspective of the non-expert in the field? In the opposite direction, can you give examples of important results, for which several genuinely different proofs were found? Are these proofs all considered equally hard, or some are much easier or, perhaps, even surprisingly easy? For example, Carleson's theorem has more than one proof and, although the latest proof is much simplified, it is probably still quite technical. Poincare conjecture has one known proof, not easily accessible to non-experts. Kadison-Singer problem has one known proof, accessible to non-experts with a little effort. Capset problem has one known proof, fully accessible to non-experts. The last three examples are relatively recent results, and some of them are likely to remain the only known approaches for a long time, while maybe not others. Instead of trying to guess the future, what are interesting examples that are less recent? REPLY [10 votes]: The first example that occurs to me is Hindman's theorem: If the set of positive integers is partitioned into finitely many pieces, then there is an infinite set $H$ such that all sums of finitely many (distinct) elements of $H$ lie in the same piece. Hindman's original proof is very complicated (Hindman himself has suggested that it could be used to torture graduate students) but it has the advantage of being elementary --- it can be formalized in a system only slightly stronger than $ACA_0$. A later, easier proof by Galvin and Glazer, now considered the standard proof, has the advantage that one can easily remember or reconstruct it, but it requires more powerful tools, including an application of Zorn's lemma to a collection of subsemigroups of a certain semigroup whose elements are ultrafilters. There's also an "intermediate" proof due to Baumgartner. REPLY [9 votes]: I believe the Atiyah-Singer Index Theorem qualifies as deep. There are several different approaches to proving it, see here. REPLY [6 votes]: I wonder whether 'deepness' is subjective or not. The Compactness Theorem of first-order logic has several proofs. Theorem. (Gödel-Maltsev) Given a language $L$ and a set $S$ of first-order sentences in that language, if every finite subset of $S$ has a model, then $S$ has model. (A language is just a set of relational, function and constant symbols, for example the language of ordered fields is $+,\times,=,\leqslant,0,1$; a first-order sentence in this language is a finite word using this language and $\exists,\forall,\land(and),\lor(or)$ and paranthesis for ease of reading only, for instance the sentence $\forall x\exists y(y^2=x)$; a model of a sentence is just a set where the sentence can be interpreted in a 'true' way. For instance $\bf R$ is not a natural model of the above sentence as $-1$ does not have a squareroot there, but $\bf C$ is a model of it.) (For one immediate application, one can deduce in a straightforewared way that there is an ordered field containing $\bf R$ as well as infinitesimal/infinite elements while having the same first order properties as $\bf R$) Some Proofs. (I believe there are many more) (1) Gödel's original proof (for Gödel, in 1929, this Theorem was stated as a 'Remark') is from his Completeness Theorem, stated in the particular case where the language $L$ is countable, in which case the axiom of choice is not needed. Hard for me to say the nature of the proof. Grammatical maybe. (2) I don't know precisely the nature of Maltsev's proof, published in German in 1936, and extending G.'s result to the case of an arbitrary signature $L$, using the axiom of choice. (3) Los' proof via the 'explicit' construction of the model as ultraproduct of finite structures, using the 'axiom of the ultrafilter', which is weaker that the axiom of choice. I would say this proof is of a topological nature. In the same (?) vein, Gromov's 'bounded version' of his Theorem stating that a finitely generated group of polynomial growth as a nilpotent subgroup of finite index has several (at least 2) proofs, apparently. Theorem. (Gromov) For any positive integers $k$, $d$, $n$, there exists a positive integer $m$ such that any $n$-generated group, in which for all $r= 1,\dots,m$ the size of the ball of radius $r$ centered at the identity is at most $kr^d$, has a subgroup of index and nilpotency class at most $m$. I read that Gromov's original proof is a Compactness argument. Van den Dries and Wilkie gave an alternative proof using Gödel's Compactness Theorem. Belegradek recently provided a third proof using yet another kind of compactness argument, using very little model theory. Edit: To match the OP's Edit, I'm just adding that the field concerning the first Theorem should be 'model theory' which is currently classified (according to Bairwise, if I am not mistaken) as the latest of the 4 branches of 'mathematical logic'. It is central/important in the sense that it more or less gave birth to the 'field', and is used more or less tacitly in pretty much every Theorem of model theory. The second Theorem belongs to the field of 'geometric group theory', and I do not know about it's being central. It seems to be important in the sense that many mathematicians seem to be interested in it. It also has a wikipedia page in 3 major mathematical languages: https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_de_Gromov_sur_les_groupes_%C3%A0_croissance_polynomiale REPLY [2 votes]: Whether you consider it a deep, very deep or shallow result: Theorem. The symplectic group ${\bf Sp}_{2n}({\mathbb R})$ is included in ${\bf SL}_{2n}({\mathbb R})$. One proof is purely algebraic and uses the Pfaffian (as a matter of fact, the same result is true when one replaces ${\mathbb R}$ by another field). The other proof is more familiar. Using the polar decomposition, one proves that ${\bf Sp}_{2n}({\mathbb R})$ is diffeomorphic to ${\mathbb R}^\ell\times{\bf U}_n$. Because the unitary group is connected, one obtains the connectedness of the symplectic group. We conclude with the fact that the determinant can take only the dicrete values $\pm1$. Edit. Here is the Pfaffian proof. Recall that the Pfaffian is a polynomial $Pf$ in the entries of the $2n\times2n$ alternate matrix, such that $$Pf(A)^2=\det A,\qquad Pf(J_n)=1,\quad J_n:=\begin{pmatrix} 0_n & I_n \\\\ -I_n & 0_n \end{pmatrix}.$$ It has the fundamental property that if $P\in{\bf M}_{2n}$, then $$Pf(M^TAM)=Pf(A)\cdot\det M.$$ Apply this identity to $A=J_n$ and to $M$ a symplectic matrix, you obtain $1=1\cdot\det M$. This proof is in my book Matrices (Springer GTM 216).<|endoftext|> TITLE: Let's keep adding once undecidable statements QUESTION [12 upvotes]: This present thread is inpired by the previous thread the true reason of the incompleteness of formal systems. I have the following intuitive idea: Gödel's second incompleteness theorem states that a "reasonably strong formal system" cannot prove its own consistency (provided it is consistent after all). Thus, for example, "ZFC is consistent" is not provable in ZFC. (I assume that ZFC is consistent, since the class of all well-founded sets is a model of ZFC.) Therefore, we can strengthen ZFC by adding the new axiom "ZFC is consistent": $$\mathrm{ZFC}_2\qquad:=\qquad \mathrm{ZFC}\quad +\quad \text{"ZFC is consistent"}$$ Now, note that we can apply Gödel's second incompleteness theorem again: "$\mathrm{ZFC}_2$ is consistent" is not provable in $\mathrm{ZFC}_2$. For that reason, we can add to $\mathrm{ZFC}_2$ the new axiom "$\mathrm{ZFC}_2$ is consistent": $$\mathrm{ZFC}_3\qquad:=\qquad \mathrm{ZFC}_2\quad +\quad \text{"$\mathrm{ZFC}_2$ is consistent"}$$ We can go on and define $\mathrm{ZFC}_4$ to be $\mathrm{ZFC}_3$ + "$\mathrm{ZFC}_3$ is consistent". Going further, we can define $\mathrm{ZFC}_5, \mathrm{ZFC}_6, \mathrm{ZFC}_7, \dots$ and so on in a similar way. Now, we can define $\mathrm{ZFC}_\omega$ to be the union of $\mathrm{ZFC}, \mathrm{ZFC}_2, \mathrm{ZFC}_3, \mathrm{ZFC}_4 \dots$, $$\mathrm{ZFC}_\omega:=\bigcup_{n}\mathrm{ZFC}_n.$$ Then, we can define $\mathrm{ZFC}_{\omega+1}$ to be $\mathrm{ZFC}_\omega+ \text{"$\mathrm{ZFC_\omega}$ is consistent"}$. And so on. We can iterate this never-ending process forever. Question time. I can image the theory $\eta:=\bigcup\{\mathrm{ZFC}_\alpha\mid\alpha\in \mathrm{Ord}\}$. Is this even a well-defined object or just fiction? Is $\eta$ decidable? Is $\eta$ complete? REPLY [5 votes]: To add to the references which have already been given, I'd like to point out that Torkel Franzén wrote an entire book (at a fairly elementary level, and with a number of IMHO rather interesting philosophical digressions) pretty much about answering this exact question: Inexhaustibility, A Non-Exhaustive Treatment, ASL Lecture Notes in Logic 16 (2004). Especially see chapters 13 ("Iterated Consistency"), 14 ("Iterated Reflection") and 15 ("Iterated iteration and inexhaustibility"): they include accounts of the results by Turing, Feferman and Franzén mentioned in Payam Seraji's answer.<|endoftext|> TITLE: On a strengthening of strong measure zero QUESTION [8 upvotes]: Recall that a set of $X$ of reals has strong measure zero (SMZ) if for every sequence $\{\epsilon_n:n<\omega\}$ of positive real numbers, there is a sequence $\{I_n:n<\omega\}$ of intervals such that $I_n$ has length at most $\epsilon_n$ and $X\subseteq \bigcup_{n<\omega}I_n$. By a theorem of Galvin, Mycielski, and Solovay, a set of reals $X$ has SMZ if and only if $X+F\neq\mathbb{R}$ for every meager $F\subseteq\mathbb{R}$. I am looking for information about the collection of sets $X\subseteq\mathbb{R}$ which satisfy the stronger property that $X+F$ is meager for every meager $F$. Clearly such sets have SMZ. My specific question is: (1) Does this family of sets admit a more "combinatorial" definition? SMZ sets can be characterized several different ways: For example, a set $X\subseteq 2^\omega$ has SMZ if and only if for every partition $\{I_n:n<\omega\}$ of $\omega$ into intervals, there is a $z\in 2^\omega$ such that $$\forall x\in X\quad\exists^\infty n\in\omega\quad x\upharpoonright I_n = z\upharpoonright I_n$$. Is anything similar known for the situation I ask about? However, I would also be satisfied with an answer to the following more general question: (2) Has this collection of sets been systematically studied before? What are some references where one might learn more? These questions are related to the field of selection principles. For example, Galvin and Miller have shown that $\gamma$-sets satisfy this stronger version of SMZ, and my graduate student Frank Ballone noted that their proof shows that sets of reals satisfying the selection principle $\binom{\Omega}{\mathcal{O}_\infty}$ (left undefined here, but due to Tsaban) do as well. REPLY [5 votes]: The sets you defined are called meager-additive, and the family of these sets is sometimes denoted $\mathcal{M}^\star$. Google suggests a good number of references, and I recommend you also look at Barotszynski and Judah's set theory book. It must contain a combinatorial characterization of the kind you seek. BTW, the result of Ballone that you cite is awesome!<|endoftext|> TITLE: sequence of graphs converge in the sense of varifold to multiplicity 2 plane QUESTION [5 upvotes]: Say in $R^3$, is there a sequence of smooth graphs $f_n$ over some plane P, such that the graphs as submanifolds in $R^3$ converge in the sense of varifold (as Radon measures on $R^3 \times Gr(2,3)$ ) to a limit V, which is multiplicity 2 of the plane P? Seems impossible but is there a proof? REPLY [2 votes]: I think that the following argument rules out getting $2[P]$: Suppose that $v_n : = graph (f_n) \rightharpoonup 2[P]$ as varifolds. Consider the open set $U = \{(x,P') \in \mathbb{R}^3\times Gr(2,3) : |x| < 1, d(P,P') <\epsilon\}$. Then, a standard result for weak convergence of measures implies that $$ \liminf_{n\to\infty} v_n(U) \geq 2[P](U) \qquad (= 2 |B_1(0)|). $$ However, it is easy to bound $v_n(U) \leq |B_1(0)| + O(\epsilon)$ as $\epsilon\to0$, since $v_n(U)$ is the $\mathcal{H}^2$ measure of the set of points in $graph(f_n) \cap B_1(0)$ whose normal vectors differ from that of $P$ by a distance $O(\epsilon)$ (the precise distance we choose on $Gr(2,3)$ is irrelevant). This is a contradiction.<|endoftext|> TITLE: finding subharmonic function on the ball with both Dirichlet and Neumann boundaries prescribed QUESTION [6 upvotes]: I have a question which looks like some sort of inverse problem. Let $B$ denote the unit ball centered at the origin in $R^N$ (take $N \ge 2$). Given any $h:\partial B \rightarrow (0,\infty)$ (smooth) we would like to find some $ f \ge 0$ in $B$ (sufficiently regular, say $ f \in L^q(B) $ for some $q>N$) such that $v$ satisfies the following: $\Delta v(x)=f(x)$ in $B$ with $ v=0$ on $ \partial B$ and $ x \cdot \nabla v(x)=h(x)$ on $ |x|=1$. In the case of $h=C>0$ (constant) one sees they can just explicitly write out a solution. In the case of $h(x)=1 +\epsilon g(x)$ ($g$ fixed $ \epsilon $ small they can also do it). Any comments would be greatly appreciated. REPLY [3 votes]: It is possible to do this. Here is a sketch of the construction: 1) Let $w = (|x|-1)h(x/|x|)$. Then $w$ satisfies the desired boundary conditions, and is smooth away from the origin with $\Delta w = (n-1)h > 0$ on $\partial B_1$. The idea is to find an appropriate extension of $w$ from a neighborhood of the boundary to the interior. 2) Let $v_0 = \max\{w, \, c_0(|x|^2 - 1) - c_0\}$. For $c_0 > 0$ small, $v_0 = w$ near the boundary and $v_0$ is the quadratic on a set $E$ whose boundary is a radial graph just inside of $\partial B_1$. Furthermore, $\Delta v_0 > c_1 > 0$ in the distributional sense. 3) Now let $\eta$ be a smooth cutoff function that is $1$ near $\partial B_1$ and $0$ in a neighborhood of $E$. Take $$v = \eta v_0 + (1-\eta) v_{\epsilon},$$ where $v_{\epsilon}$ is a mollification of $v_0$ (so $\Delta v_{\epsilon} > c_1$). Then $v = w$ near $\partial B_1$. Since $v_0 = w$ is smooth where the derivatives of $\eta$ are supported, it is straightforward to check that $\Delta v$ is smooth and positive in $B_1$ for $\epsilon$ small, completing the construction.<|endoftext|> TITLE: Grothendieck Construction, Categories of Operators and Opposites QUESTION [10 upvotes]: Given a symmetric monoidal category $C$, we can construct its endomorphism operad (or multicategory) $End(C)$ whose objects are the objects of $C$, and for which the multimorphisms from $\{c_1,\ldots,c_n\}$ to $c$ are given by the hom set $Hom_C(c_1\otimes\cdots c_n,c)$. From this symmetric operad we can then produce a category of operators whose objects are finite lists of objects of $C$ and whose morphisms are described (roughly) in the following way: given two lists $\{c_1,\ldots,c_n\}$ and $\{d_1,\ldots,d_m\}$, a morphism between them is a morphism $\phi:\langle n\rangle\to\langle m\rangle$ in finite pointed sets (the category often denoted $\Gamma$ by Segal and others, and $\mathcal{F}in_\ast$ by Lurie) and for each element $i\in\langle m\rangle$ and set $\{j\in\phi^{-1}(i)\}$ a multimorphism of $End(C)$ going from $\{c_j\}$ to $d_i$. Let's denote the application of these procedures to $C$ by $End(C)^\otimes$. Essentially by construction, $End(C)^\otimes$ admits a Grothendieck opfibration to $\Gamma$, and as such, corresponds to a pseudofunctor $F_C:\Gamma\to Cat$. One might even say that this pseudofunctor defines the symmetric monoidal structure on $C$. Given a symmetric monoidal category there is a symmetric monoidal structure on $C^{op}$, so we may also consider $End(C^{op})^\otimes$. Similarly, we obtain a pseudofunctor $F_{C^{op}}:\Gamma\to Cat$. The sort of theorem I'm after is the following: The composition of pseudofunctors $\Gamma\overset{F_{C}}\to Cat\overset{op}\to Cat$ is equivalent to the pseudofunctor $\Gamma\overset{F_{C^{op}}}\to Cat$. Or at the very least, I just want to say that if I apply the Grothendieck construction $\Gamma\overset{F_{C}}\to Cat\overset{op}\to Cat$ I get a symmetric monoidal category which is equivalent, as a symmetric monoidal category, to $C^{op}$ with the opposite symmetric monoidal structure. REPLY [2 votes]: This question is answered in the affirmative in a recent paper of mine with Liang Ze Wong. In fact, we prove it more generally for a (strictly) monoidal simplicially enriched category. As any simplicially enriched monoidal category can be rigidified up to monoidal equivalence to a strict one, there's nothing lost by making the strict assumption. The enriched case of course implies the unenriched statement above. The paper is here: https://arxiv.org/abs/1808.08020<|endoftext|> TITLE: Is there an exponential map on (Hahn) ordered fields? QUESTION [5 upvotes]: If $F$ is an ordered field and $G$ is an ordered abelian group, one can define the Hahn product $F \boxtimes G$ to be the set of formal Laurent series with coefficients in $F$ and exponents in $G$. It is easy to see that this is a ring that derives a linear order from the orders on $F$ and $G$ lexicographically. Via the usual division algorithm it can be shown that $F \boxtimes G$ is also a field. Just as orders on $F$ and $G$ induces an order on $F \boxtimes G$, I would like for it to be true that if $G$ is a field, and $F$ and $G$ both admit exponential maps -- meaning a group homomorphism from the additive group to the multiplicative group -- then there is an induced exponential map on $F \boxtimes G$. This exponential map $\textrm{exp}$ would ideally have the (loosely stated) properties that $\textrm{exp}$ maps the whole field $F \boxtimes G$ bijectively to the positive elements $(F \boxtimes G)_{> 0}$ in some fashion $\textrm{exp}$ respects the individual orders -- perhaps strictly respecting the order on $F$ and reversing the order on $G$ in some fashion $\textrm{exp}$ respects the exponential maps on $F$ and $G$ I've been looking for such a map in $\mathbb{R} \boxtimes \mathbb{R}$ and haven't been able to find one that matches the criteria, nor prove that a map with those properties can't exist. REPLY [6 votes]: There is no such exponential map. This was demonstrated in: F.-V. Kuhlmann, S. Kuhlmann, S. Shelah, Exponentiation in power series fields, Proc. Amer. Math. Soc. 125 (1997) 3177–3183.<|endoftext|> TITLE: Necessity of shapes for coherence results in category theory QUESTION [5 upvotes]: The classic coherence theorems of MacLane (Natural associativity and commutativity, Rice U. studies, 1963) talked about natural transformations between functors. By 1971 (Kelly-MacLane, Coherence in closed categories, JPAA) and 1972 (Kelly, An abstract approach to coherence, LNM 281), functors had been replaced by "shapes"---formal combinations of given functors---with comments to indicate that this replacement is necessary to give correct statements. It is not too hard to build examples that show that the usual coherence statements are false if stated for functors and not shapes. However, I have never seen an example in print. Does anyone know of a reference for an example? For the curious, here is an example. Let $A_n$ be $n$-tuples of integers regarded as sequences of length $n$, and let $A_m\otimes A_n$ be $A_{m+n}$ obtained by concatenating the sequences. This makes $\otimes$ strictly associative. Build an (idiotic) isomorphism $\alpha$ from $(X\otimes Y)\otimes Z$ to $X\otimes(Y\otimes Z)$ by negating the values contributed by $X$ and $Z$ but not $Y$. Note that the two functors connected by $\alpha$ are identical even though the shapes are different. Now check that the usual pentagonal diagram connecting all five shapes on four variables commutes. Thus our mongrel $\alpha$ is coherent on shapes, but not on the functors themselves. All five functors on four variables are the same, but are connected by 16 different isomorphisms, corresponding to all the elements in $\mathbf (Z_2)^4$, that can be built from $\alpha$. REPLY [2 votes]: There is an example in Saunders Mac Lane. Topology and Logic as a Source of Algebra (Retiring Presidential Address), Bulletin of the AMS 82:1, January 1976. See Section 5 ("Coherence and canonical maps"), p.29: Hence, this theorem really amounts to the statement that, in a monoidal category, every diagram of canonical maps $a, b, c$ must commute. It was proved by Mac Lane [73] and independently by Epstein. Mac Lane's then proof of the theorem was quite correct, but his statement was deficient. He did not discuss words at all, so that he considered not the formal word $A\otimes B$ in two letters $A$ and $B$, but only the functor $M\times M \to M$ which it determines (on any monoidal category $M$). As a result, his formulation, carefully examined, confused the word $A\otimes B$ with the word $B\otimes A$ and his theorem would thus assert that the two central maps $c:A\otimes B \to B\otimes A$ and $1: A\otimes B \to A\otimes B$ are equal. Kelly subsequently set this all straight [50]. ([50] is G. M. Kelly. On Mac Lane's conditions for coherence of natural associativities, commutativities, etc., J. Algebra 1 (1964), 397-402. MR 32 #132.)<|endoftext|> TITLE: Looking for a good terminology for permutations having no substring QUESTION [10 upvotes]: What is the good name for permutations of [1,...,n+1] having no substring [k,k+1] http://oeis.org/A000255 ? REPLY [10 votes]: These permutations are called "plus irreducible". See http://arxiv.org/pdf/math/0212163v1.pdf<|endoftext|> TITLE: Geometric quantization of Teichmuller space QUESTION [5 upvotes]: The quantizations of Teichmuller space I have seen are via special coordinates (e.g. the paper of Chekhov and Fock) or conformal blocks. Does one get an equivalent quantization by geometric quantization? For example the Weil-Peterson metric is Kahler. Is it known, one way or the other, if doing Kahler quantization gives the same thing? What about a different polarization? REPLY [2 votes]: One of best ways for quantization of Kahler varieties is to use Kahler reduction since quantization commutes with reduction. so you can write Teichmuller space $\mathcal T(M)$ by symplectic reduction, $$\mu^{-1}(0)/Diff^+M\cong\mathfrak{Met}_{-1}(M)/Diff^+M=\mathcal T(M)$$ which $\mathfrak{Met}_{-1}M$ is the space of reimannian metrics of constant scalar curvature $-1$. Hence has $\omega_{WP}$ Weil-Petersson metric. see page 12 http://arxiv.org/pdf/math/0507076.pdf and also nice paper of Maryam Mirzakhani pages 8-9 http://www.ams.org/journals/jams/2007-20-01/S0894-0347-06-00526-1/S0894-0347-06-00526-1.pdf For example for quantization of coadjoint orbit, you need to quantize $T^*G$, since coadjoint orbit is just symplectic quotient of $T^*G$. See my master project presentation http://fr.slideshare.net/HassanJolany/geometric-quantization-on-coadjoint-orbits<|endoftext|> TITLE: Have finite doubly transitive groups been classified? QUESTION [27 upvotes]: I am trying to determine whether the literature contains a complete proof of the classification of finite 2-transitive groups. This is a fundamental result with important applications in many areas of mathematics, so it seems worthwhile to make sure that it has actually been proved. In short, the issue is that it is common to credit Hering for classifying the nonsolvable affine 2-transitive groups, but it seems that Hering never published a paper containing such a classification; and the one paper which does purport to contain a proof, namely a 1985 paper by Liebeck, begins with a logical leap which I cannot follow. [Added later: thanks to Michael Giudici's answer, I can now follow the first step in Liebeck's proof. I have not yet gone through the rest of the proof, but do not presently have any reason to doubt its validity.] Let me summarize the literature to clarify the issue. By Burnside's classical result, a (finite) 2-transitive group $G$ satisfies one of $L\le G\le\text{Aut}(L)$ for some nonabelian simple group $L$ $G\le\text{AGL}_r(p)$ in the usual action on $\mathbf{F}_p^r$, where in addition $G$ contains all translations by elements of $\mathbf{F}_p^r$. The possibilities in case 1 were determined by Curtis-Kantor-Seitz 1976 in case $L$ is a Chevalley group of ordinary or twisted type (full citations are at the end of this question, in chronological order). It is easy to resolve the case $L=A_n$ (some people cite Maillet 1895 for this, but I haven't read that paper). When $L$ is sporadic, Cameron 1981 and Kantor 1985 stated the result without proof, and Praeger-Soicher 1997 published a proof (summarized in their Table 5.1). This completes the proof in case 1. So the question is case $2$, when $G$ is a group of affine transformations of a vector space with $G$ containing all translations, or in short, $G$ is an "affine group". It seems that the first reference claiming that $2$-transitive affine groups had been classified is Cameron 1981, which does not indicate what the groups actually are, but instead merely states as Theorem 5.3 that "All finite 2-transitive groups are known." More informative statements of the result appear in Huppert-Blackburn 1982, Kantor 1985, Abhyankar 1992, Dixon-Mortimer 1996, and Cameron 1999, all of which include lists of groups. Cameron 1981 says that a paper by Hering (cited as "to appear", but with no title) "examined the known simple groups and determined all such situations in which each could occur as such a composition factor [of an affine 2-transitive group]". I will discuss this unknown paper below. Huppert-Blackburn 1982 gives a proof in case $G$ is solvable (Thm. XII.7.3; this case was first resolved in Huppert 1957), and then states the full result (without proof) as Remark XII.7.5, attributing it to Hering 1974. But Hering 1974 does not contain the stated result; instead it only determines the 2-transitive affine groups having a composition factor isomorphic to $\text{PSL}_2(q)$ or $A_n$ or the smallest of Janko's four sporadic groups. Kantor 1985 says that the result follows from Curtis-Kantor-Seitz 1976, Hering 1973, Hering 1974, Hering 1985, Huppert 1957, and Maillet 1895, in addition to further analysis of sporadic groups which is left to the reader. The only papers in Kantor's list which address the affine case are the three papers by Hering, of which the second was summarized above and the other two will be discussed below. (A minor side remark: the statement of the result in Kantor 1985 is actually tautological; one should add the condition $n>1$ in case (B2) to make it have content.) Abhyankar 1992 (on p.87) cites Cameron 1981, Kantor 1985, Curtis-Kantor-Seitz 1976, and O'Nan 1975. O'Nan 1975 contains wonderful results about 2-transitive groups (impressively, achieved without using the classification of finite simple groups), but does not purport to classify nonsolvable affine 2-transitive groups. Abhyankar does not cite Hering, but also Abhyankar was not a group theorist and did not claim to know the proof, instead saying that the classification of 2-transitive groups "was communicated to me by Cameron". Dixon-Mortimer 1996 (on p.244) cites Huppert 1957, Hering 1974, and Huppert-Blackburn 1982 for the classification of affine 2-transitive groups. Thus, everything they say about the issue is contained in Huppert-Blackburn 1982. Cameron 1999 (on p.110) says that the nonsolvable affine 2-transitive groups were determined by Hering; the only paper of Hering's in the bibliography of Cameron's book is Hering 1974, which (on p.194) is claimed to contain precise conditions for which of the groups listed on p.195 of Cameron 1999 are actually 2-transitive (but Hering 1974 does not contain such conditions). Thus, the group theory community seems united in their belief that Hering has classified the (nonsolvable) affine 2-transitive groups. But I see no evidence that Hering ever published either a statement or a proof of this classification. The only papers of Hering cited by any of the above-mentioned sources are Hering 1973, Hering 1974, and Hering 1985 (as well as the untitled "to appear" paper of Hering's cited in Cameron 1981). I discussed Hering 1974 above: it covers only the case that some composition factor of the group is $\text{PSL}_2(q)$ or $A_n$ or the smallest Janko group. Hering 1985 does not claim to classify nonsolvable affine 2-transitive groups. Instead, the introduction of Hering 1985 says the paper achieves this classification only under the additional assumption that some composition factor of the group is a Chevalley group (of ordinary or twisted type). However, Hering's paper does not contain either a statement or a proof of this promised classification; instead it proves results in this direction and then says "It now is not difficult to check which of the 10 types not investigated above can occur", and performs this check in one case. At the end of Hering 1985 there is a citation to a "to appear" paper by Hering (with a title -- this is the final item in the bibliography below), and the assertion that that paper "contains the corresponding investigation for the known sporadic simple groups". But I checked Mathscinet, and could not find any paper of Hering's with the stated title, or even any paper of Hering's on this topic since his 1985 paper. Of course, the paper may still be "to appear", but perhaps after 31 years the odds are not high. After digging through all of the above and getting increasingly frustrated, I was quite happy when a friendly group theorist pointed me to Liebeck 1987, which is the only reference I've seen which purports to prove the classification of affine 2-transitive groups without merely pointing the reader to one or more papers by Hering. (Minor issue: one must add the condition $a>1$ to case A2 of the result stated in Liebeck 1987, in order to make the result have content.) I am baffled by Liebeck's assertion that "The 2-transitive affine groups have been determined by Hering in [Hering 1985]", since he almost surely would have looked at the paper and noticed that it did not contain the result; but for whatever reason it seems taboo in the group theory community to admit that Hering 1985 (or even Hering 1974) does not even claim to prove this result. But I also cannot follow the very first step of the proof in Liebeck 1987. It begins by letting $G_0$ be the stabilizer of the zero vector, so that $G_0\le\text{GL}_r(p)$, and choosing $a$ to be minimal so that $G_0\le\Gamma\text{L}_a(p^{r/a})$. Write $q:=p^{r/a}$. Cases with $a=1$ are included in the statement of the result, so one may assume $a>1$. The statement of the result also includes cases where $G_0$ normalizes $\text{SL}_a(q)$, so one may assume that this does not hold either. Liebeck then asserts that the main theorem of Aschbacher 1984 implies that one of these holds: $L\le G_0/Z(G_0)\le \text{Aut}(L)$ for some nonabelian simple group $L$ which isn't one of a few specific groups $G_0$ normalizes $\text{Sp}_a(q)$ $G_0$ is contained in the normalizer in $\Gamma\text{L}(\text{F}_q^a)$ of some $\ell$-group $R$, where $\ell$ is a prime. Aschbacher 1984 exclusively addresses groups between a simple group and its automorphism group, so it seems it can only be relevant to ruling out some of the groups stated not to occur in case 1. Does anyone see how to justify Liebeck's assertion? Or alternately, does anyone know any other reference containing a proof of the classification of nonsolvable affine 2-transitive groups? Final remark: the above comments subsume all comments and answers to mathoverflow.32351. ================================================================ Bibliography: Maillet 1895: Sur les isomorphes holoédriques et transitifs des groupes symétriques ou alternés, J. Math. Pures Appl. (5) 1 (1895), 5-34. Huppert 1957: Zweifach transitive, auflösbare Permutationsgruppen, Math. Z. 68 (1957), 126-150. Hering 1973: On linear groups which contain an irreducible subgroup of prime order, Proc. Int. Conf. Proj. Planes, Washington State Univ. Press, Pullman, 1973, pp. 99-105. Hering 1974: Transitive groups and linear groups which contain irreducible subgroups of prime order, Geometriae Dedicata 2 (1974), 425-460. O'Nan 1975: Normal structure of the one-point stabilizer of a doubly transitive permutation group I, II: Trans. Amer. Math. Soc. 214 (1975), 1-74. Curtis-Kantor-Seitz 1976: The 2-transitive permutation representations of the finite Chevalley groups, Trans. Amer. Math. Soc. 218 (1976), 1-59. (For corrigenda, see the paper's Math.Review.) Cameron 1981: Finite permutation groups and finite simple groups, Bull. London Math. Soc. 13 (1981), 1-22. Huppert-Blackburn 1982: Finite Groups III. Springer-Verlag, Berlin, 1982. Aschbacher 1984: On the maximal subgroups of the finite classical groups, Invent. Math. 76 (1984), 469-514. Kantor 1985: Homogeneous designs and geometric lattices, J. Comb. Theory Ser. A (1985) 38, 66-74. Hering 1985: Transitive groups and linear groups which contain irreducible subgroups of prime order, II: J. Algebra 93 (1985), 151-164. Liebeck 1987: The affine permutation groups of rank three, Proc. London Math. Soc. (3) 54 (1987), 477-516. Abhyankar 1992: Galois theory on the line in nonzero characteristic, Bull. Amer. Math. Soc. 27 (1992), 68-133. Praeger-Soicher 1997: Low Rank Representations and Graphs for Sporadic Groups. Cambridge Univ. Press, Cambridge, 1997. Dixon-Mortimer 1996: Permutation Groups. Springer-Verlag, New York, 1996. Cameron 1999: Permutation Groups. Cambridge Univ. Press, Cambridge, 1999. Hering future(?): On representations of Chevalley groups and translations planes, to appear (as of 1985). REPLY [13 votes]: The theorem of Aschbacher that is cited by Liebeck, looks at the maximal subgroups $G_0$ of a group $G$ satisfies $SL(a,p^{r/a})\leqslant G \leqslant \Gamma L(a,p^{(r/a})$. For the application of Liebeck, you take $G=\Gamma L(a,p^{r/a})$. Aschbacher gives 9 classes of such subgroups. They usually preserve some sort of geometric structure on the vector space and so you can easily see that a maximal subgroup of that form is not transitive on the set of nonzero vectors of $V$. This leaves the three classes stated.<|endoftext|> TITLE: Interpolation between $L_1^0$ and $L_2^0$ QUESTION [10 upvotes]: Let $L_p^0$ be the mean zero functions in $L_p(G)$, where, say, $G$ is an infinite compact group endowed with normalized Haar measure. Suppose that $T$ is a bounded linear operator on $L_1$ that maps $L_1^0$ into itself and $L_2$ into $L_2$. Suppose that $\|T\|_{L_1 \to L_1} =1$ and $\|T\|_{L_2^0\to L_2^0} <1$. Q1: If $1 TITLE: Ordinary primes vs supersingular primes QUESTION [7 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Q}$ without CM. As shown by Serre, the set of supersingular primes for $E$ has density zero. Is the analytic rank of $L(E,1)$ determined only by the local factors at the ordinary primes for $E$? REPLY [9 votes]: Well, yes, but no. So the function obtained by removing from the usual $L$-series of $E$ all the supersingular factors is probably not a very nice function. I doubt that it has an analytic continuation and hence I am not sure we should talk about its order of vanishing at $s=1$. But I do not know enough about these things to make that more precise, maybe someone else will correct me on this. On the other hand, the original idea of looking at the product of $N_p/p$ where $N_p$ is the number of solutions modulo $p$ still makes sense. Let $f(x) = \log(\prod_{p\leq x} N_p/p)$, the earliest version of the conjecture of Birch and Swinnerton-Dyer says that $f(x)$ grows like $r\log\log(x)$ where $r$ is the rank. The plot below is $f(x)$ against $\log\log(x)$. I took the first ten curves of rank $0$ (blue) $1$ (red) and $2$ (orange) in Cremona's tables, but now running only over prime $p\leq x$ that are good ordinary. So it looks very much like this conjecture does not care about the few supersingular primes - without much surprise.<|endoftext|> TITLE: When is the category of Gorenstein projective $R$-modules Frobenius? QUESTION [5 upvotes]: Let $R$ be a ring (associative with unit, but not necessarily commutative, and definitely not necessarily Noetherian.) Then the category $\operatorname{GP}(R)$ consists of those $R$-modules having a complete projective resolution, i.e. that are expressible as the image of the map $P_{-1}\to P_0$ in some sequence $$\cdots\to P_{-2}\to P_{-1}\to P_0\to P_1\to P_2\to\cdots$$ of projective $R$-modules, such that this sequence remains exact under $\operatorname{Hom}_R(-,P)$ for any projective $P$. (For extra emphasis: I do not make any finite-generation assumptions on $M$, or on the $P_i$.) My questions are then the following: Is $\operatorname{GP}(R)$ a Frobenius category? Depending on the answer, there are natural follow-up questions: If yes, is there a good reference for this, ideally with a complete argument? If not, can we get this property back by assuming a little more about $R$? As a target, I would like to deal with the case that $R$ is a complete preprojective algebra of an arbitrary finite quiver (which is not typically Noetherian). Henrik Holm has a very nice paper called 'Gorenstein homological dimensions', which proves lots of things about $\operatorname{GP}(R)$ for a general ring $R$, but does not directly address this question. He does show that this category is resolving, which helps a bit (in particular, it is closed under extensions and so inherits an exact structure from $\operatorname{Mod}{R}$). Assuming I haven't made any mistakes, I think the facts that $\operatorname{GP}(R)$ contains all projective $R$-modules, which are also injective in $\operatorname{GP}(R)$, and that every Gorenstein projective admits both an epimorphism from and a monomorphism to such a module are all pretty much clear from the definition. So the only thing that might go wrong (I think!) is that there could be more projectives/injectives that don't agree with each other. There is some extra context which might give a flavour of the kind of sources I would most appreciate (although of course any answer is appreciated!). I am aware that many authors consider the case that $R$ is Iwanaga–Gorenstein, meaning that $R$ is Noetherian and of finite injective dimension as a module on each side, and then consider the Frobenius category $$\operatorname{GP}'(R):=\{X\in\operatorname{mod}{R}:\operatorname{Ext}_R^i(M,R)=0\ \forall\ i>0\}.$$ I am interested in certain (probably) non-Noetherian rings $R$, but that still have finite injective dimension as a module on each side, and in certain (possibly not finitely-generated) $R$-modules $M$ such that $\operatorname{Ext}^i_R(M,P)=0$ for any projective $P$ and any $i>0$. My feeling is that there should be some reasonable Frobenius category of 'Gorenstein projective-like' modules associated to $R$ and containing $M$, by focussing on the homological conditions and forgetting about any finiteness (even something like $\operatorname{GP}'(R)$ as defined above, but with $\operatorname{mod}{R}$ replaced by $\operatorname{Mod}{R}$, and $R$ replaced by an arbitrary projective in the condition – one of the things that Holm proves is that this category is then very close to $\operatorname{GP}(R)$ as defined at the top of the question, but might also include some modules with no finite resolution by Gorenstein projectives). However, I am not very familiar with what can go wrong when you drop finiteness conditions, and am concerned that I may lose the Frobenius property somewhere. If it helps, I may in the end want my category to be Krull–Schmidt (in the strong sense that indecomposables are characterised by having local endomorphism rings) which means I will have to require that $R$ is semi-perfect. This gives a bit more control over $\operatorname{Proj}{R}$, as it means that there are finitely many indecomposable projectives such that every projective is a (possibly infinite) direct sum of these. REPLY [4 votes]: It's always a Frobenius category and its projective-injective objects are the projective modules. Your analysis is essentially right. In fact, this holds very generally. I worked out the following when I read a bit in Enochs-Yenda, but I suppose it's well-known: Given an exact category $(\mathcal{A}, \mathcal{E})$, let $\mathcal{P}$ be the full subcategory of its projective objects. You can induce a new exact structure $\mathcal{E}_{\mathcal{P}}$ on $\mathcal{A}$ which consists of the $\mathcal{E}$-exact sequences $0 \to A' \to A \to A'' \to 0$ such that $$0 \to \hom(A'', P) \to \hom(A, P) \to \hom(A', P) \to 0$$ is exact for all $P \in \mathcal{P}$. Note that the objects in $\mathcal{P}$ are projective and injective with respect to the exact structure $\mathcal{E}_{\mathcal{P}}$ (there may be more $\mathcal{E_P}$-projective and injectives, of course). Define $\mathcal{G_P}$ to be the category of Gorenstein projective objects with respect to $\mathcal{E}_{\mathcal{P}}$: these are the objects $G$ appearing as the image of a differential of a bi-infinite $\mathcal{E}_{\mathcal{P}}$-acyclic complex with components from $\mathcal{P}$. We have $\mathcal{P} \subseteq \mathcal{G_P}$ because we can just take $\cdots \to 0 \to P \xrightarrow{\rm id} P \to 0 \to \cdots$. The key fact is: An $\mathcal{E}$-admissible epic $p\colon A \to G$ with $G \in \mathcal{G_P}$ is necessarily $\mathcal{E_P}$-admissible. Since $G \in \mathcal{G_P}$ there is an $\mathcal{E_P}$-admissible epic $q\colon P \to G$ with $P \in \mathcal{P}$. Since $P$ is projective, there is $f \colon P \to A$ such that $q = pf$. Since $p$ has a kernel, Quillen's "obscure axiom" on exact categories implies that $p$ is an $\mathcal{E_P}$-admissible epic. You can also play around with long exact sequences and $\operatorname{Ext}^i(G,P)$ to see this. It follows from this and the horseshoe lemma that $\mathcal{G_P}$ is closed under extensions in $(\mathcal{A}, \mathcal{E})$. Hence $\mathcal{G_P}$ is also closed under extensions in $(\mathcal{A}, \mathcal{E_P})$. Proposition. The two exact structures on $\mathcal{G_P}$ induced by $\mathcal{E}$ and by $\mathcal{E_P}$ coincide. With respect to this exact structure $\mathcal{G_P}$ is a Frobenius category whose projective-injective objects are precisely the objects from $\mathcal{P}$. That the two induced exact structures coincide on $\mathcal{G_P}$ follows again from the key observation. Since the objects in $\mathcal{P}$ are projective and injective in $(\mathcal{A}, \mathcal{E_P})$, they are also projective and injective in $\mathcal{G_P}$. If $G$ is either projective or injective in $\mathcal{G_P}$, it is a retract of an object of $\mathcal{P}$, hence it belongs to $\mathcal{P}$. To see this, suppose $G$ is projective in $\mathcal{G_P}$. Then there is an $\mathcal{E_P}$-admissible epic $P_{-1} \to G$, which splits since $G$ is projective. So $G$ is a retract of $P_{-1}$. It follows that $G$ is both projective and injective in $\mathcal{G_P}$, and also projective in $(\mathcal{A},\mathcal{E})$. Similarly, if $G$ is injective, it is a retract of $P_0$, hence projective and injective in $\mathcal{G_P}$ and also projective in $(\mathcal{A}, \mathcal{E})$. That there are enough projectives and injectives in $\mathcal{G_P}$ holds by definition.<|endoftext|> TITLE: When do non-exact functors induce morphisms on $K$-theory? QUESTION [10 upvotes]: Let $\mathcal{A}$ and $\mathcal{B}$ be Waldhausen or exact categories, so that we can take the $K$-theory spectrum of $\mathcal{A}$ and $\mathcal{B}$. An exact functor $F: \mathcal{A} \to \mathcal{B}$ induces a morphism of $K$-theory spectra $K(\mathcal{A}) \to K(\mathcal{B})$. Under what conditions does a non-exact functor $\mathcal{A} \to \mathcal{B}$ induce a morphism of $K$-theory spaces $\Omega^\infty K(\mathcal{A}) \to \Omega^\infty K(\mathcal{B})$? To give an example, suppose $\mathcal{A}$ is the category of finite sets and $\mathcal{B}$ is the category of finite $C_p$-sets. Then there is a functor $\mathcal{A} \to \mathcal{B}$ given by multiplicative induction (it sends a set $S$ to $\mathrm{Map}(C_p, S)$) and that should induce a morphism on $K$-theory spaces. This gives a map $\Omega^\infty S^0 \to \Omega^\infty S^0_{C_p}$, which should arise from the Hill-Hopkins-Ravenel "multiplicative norm" construction. (This sort of construction is used in Segal's paper on the Kahn-Priddy theorem in the 1974 "New developments in topology.") Has this been written down somewhere? REPLY [6 votes]: As suggested by Dustin Clausen in his answer, polynomial functors induce maps on $K$-theory. In the setting of stable $\infty$-categories, you proved this in your joint work with Barwick, Glasman, and Nikolaus.<|endoftext|> TITLE: On Johansson's Theorem on homotopy equivalences of 3-manifolds QUESTION [5 upvotes]: Johansson's theorem states the following: Given $f:M_1\rightarrow M_2$ (not a pair map) an homotopy equivalence between 3-manifolds with incompressible boundary. Let $V_i$ be the components of the characteristic submanifolds meeting the boundary then we can homotope $f\simeq g$ so that $g$ is an homotopy equivalence on $V_i$ and is an homeomorphism on $\overline{M_1\setminus V_1}\rightarrow \overline{M_2\setminus V_2}$. I also know about Waldhausen's theorem that if $f$ sends $\partial M_1$ to $\partial M_2$ homeomorphically then $f$ can be homotoped rel $\partial M_1$ to a homeomorphism. My question is that if we know that $f\vert_{\partial M_1}$ is an embedding can we homotope $f:M_1\rightarrow M_2$ rel boundary to be an embedding? REPLY [6 votes]: Yes, this is true, with the appropriate assumption that $M_1$ is irreducible (this is needed for Johansson and Waldhausen's statements), and let's say orientable. One may reduce to Waldhausen's theorem if we know that $f_{|\partial M_1}$ is homotopic to a homeomorphism $f'_{|\partial M_1}:\partial M_1\to \partial M_2$, by homotopy extension. Consider the image surface $f(\partial M_1) \subset M_2$ (assume this embedding is transverse, so doesn't meet $\partial M_2$). Since $\partial M_1$ is homologically trivial in $M_1$, $f(\partial M_1)$ is homologically trivial in $M_2$, hence bounds a submanifold $M \subset M_2$. Then $\partial M =f(\partial M_1)$ is homotopic to $\partial M_2$ iff $M_2-M$ is a product manifold (this follows from a result of Waldhausen, see Corollary 5.5 of Friedhelm Waldhausen, MR 224099 On irreducible $3$-manifolds which are sufficiently large, Ann. of Math. (2) 87 (1968), 56--88.). Take a homotopy inverse $g:M_2\to M_1$ to $f$. Then $g_{|M}$ is $\pi_1$-injective, and we may homotope $g$ so that $g_{|\partial M}\to \partial M_1$ is a homeomorphism. This map is also degree one, and $\pi_1$-injective, hence a homotopy equivalence. But this implies that $\pi_1M\to \pi_1 M_2$ is an isomorphism. In turn, this implies that $M_2-M$ is a product. (I'm leaving out many details here, and I don't mean to give a proof by intimidation, so let me know if you would like to have any details on steps of this argument.) As a reality check, the standard example of a homotopy equivalence which is not a homeomorphism is obtained by permuting the pages of a book of $I$-bundles. But the boundary of one will immerse in the other, and hence will not satisfy your hypothesis.<|endoftext|> TITLE: Centreless semisimple Lie group that is not real algebraic QUESTION [6 upvotes]: Let $G$ be a connected semisimple Lie group with trivial centre and $\mathfrak{g}$ its Lie algebra. The adjoint representation of $G$ defines an isomorphism of $G$ onto the connected component of the identity of $Aut(\mathfrak{g})$. The latter is a real algebraic group, but its identity component (in the Lie group topology) needs not be. However, in many cases $G$ is indeed isomorphic to a real algebraic group. Is there an example of a connected semisimple Lie group $G$ with trivial centre which is not isomorphic (as a Lie group) to a real algebraic group? Here it's really important that $G$ be centreless, otherwise it's easy to find counterexamples (such as the universal cover of $SL_2\mathbb{R}$). The situation is complicated by the following fact. The real algebraic group $SO(n,1)$ is connected in the Zariski topology, while it has two components as a Lie group; thus, its identity component $G_n$ might be a good candidate for a counterexample, since it is centreless and doesn't inherit a real algebraic structure from $SO(n,1)$. However, at least for $n\leq 3$, $G_n$ happens to be isomorphic to a real algebraic group in an unrelated way (namely $\mathbb{R}$, $\mathbb{P}SL_2\mathbb{R}$ and $\mathbb{P}SL_2\mathbb{C}$ for $n=1,2,3,$ respectively). There are several other questions around on the relationship between algebraic and Lie groups, but none of them seems to be relevant in this case... REPLY [7 votes]: You basically answered your own question: The connected component $G$ of $SO(2,1)$ is not real algebraic. As such, $G$ should be isomorphic to (the real points of) $PSL(2,\mathbb R)$. But it isn't. As algebraic groups $PSL(2,\mathbb R)$ and $PGL(2,\mathbb R)$ are isomorphic (in fact, the natural homorphism between them is an isomorphism over $\mathbb C$ and therefore over $\mathbb R$). But $PGL(2,\mathbb R)$ clearly has two components distinguished by the sign of the determinant.<|endoftext|> TITLE: Compatibility conditions for Yetter-Drinfeld modules QUESTION [6 upvotes]: In the paper, page 28, Definition 4.2.1, the compatibility condition for a Yetter-Drinfeld module over $H$ is $$ h_{(1)} v_{(-1)} \otimes h_{(2)}.v_{(0)} = (h_{(1)}.v)_{(-1)}h_{(2)} \otimes (h_{(1)}.v)_{(0)}, v \in V, h \in H. $$ On the other hand, in the article, the compatibility condition for a Yetter-Drinfeld module over $H$ is $$ \delta(h.v) = h_{(1)} v_{(-1)} S(h_{(3)}) \otimes h_{(2)}.v_{(0)}, v \in V, h \in H. $$ Are the two conditions equivalent? Thank you very much. REPLY [3 votes]: Assume that the second equation holds. Then \begin{align*} \delta(h_{1}.v)=(h_{1}. v)_{-1}\otimes(h_{1}.v)_{0}=h_{1,1}v_{-1}Sh_{1,3}\otimes h_{1,2}. v_{0} \end{align*} and therefore $$ (h_{1}.v)_{-1}h_{2}\otimes(h_{1}.v)_{0}=h_{1,1}v_{-1}Sh_{1,3}h_{2}\otimes h_{1,2}.v_{0}=h_{1}v_{-1}\otimes h_{2}.v_{0}. $$ Conversely, assume that the first equation holds. Then \begin{align*} (m\otimes &id)(h_{11}v_{-1}\otimes Sh_2\otimes(h_{12}.v_0) )\\ &=(m\otimes id)\left( (h_{11}.v)_{-1}h_{12}\otimes Sh_2\otimes (h_{11}. v)_0 \right)\\ &=(h_1.v)_{-1}h_2Sh_3\otimes (h_1.v)_0\\ &=(h.v)_{-1}\otimes (h.v)_0. \end{align*}<|endoftext|> TITLE: The "Peano phenomenon" for differential equations QUESTION [7 upvotes]: Consider the following statement: If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function, for the autonomous equation $$x' = f (x)$$ the "Peano phenomenon" can arise only at those values of $\bar x$ for which $f(\bar x) = 0$. (The "Peano phenomenon" = Cauchy problems associated to the above equation can admit more than one solution.) Can someone provide me with a (proof or a) reference for a proof ? According to this paper, this statement is "well known", so a proof of it should be found in textbooks, but I couldn't find any. A paper that contains the proof is mentioned in the linked paper above, but I don't have access to that paper. REPLY [5 votes]: I'm not sure this is "well known" (it wasn't to me). We can do this with a Lyapunov function $V(x,y)=g(y)-g(x)$, with $g(x)=\int_a^x (1/f(u))\, du$. This is well defined near an $a$ with $f(a)\not= 0$. Now if $x(t),y(t)$ both solve $x'=f(x)$, $x(0)=a$, then $W(t)= V(x(t),y(t))$ satisfies $W(0)=0$, $$ W'(t)= g'(y)f(y)-g'(x)f(x)= 0 , $$ by our choice of $g$. It follows that $W\equiv 0$, and this gives uniqueness because $g$ is strictly monotone.<|endoftext|> TITLE: Equivalence between complex and real operator norms QUESTION [8 upvotes]: Consider a real $m\times n$ matrix $A$ and the $p$-norms in $\mathbb{C}^n$ and $\mathbb{C}^m: \|x\|=\left(\sum|x_i|^p\right)^{1/p}$. One defines the real $p$-norm of $A$ as $\|A\|=\sup\frac{\|Ax\|}{\|x\|}$, where the $\sup$ ranges over $0\neq x\in\mathbb{R}^n$. Similarly, one can define the $p$-norm of the complexified operator $A$, that is, $\|A\|_c=\sup\frac{\|Ax\|}{\|x\|}$, where the $\sup$ ranges over $0\neq x\in\mathbb{C}^n$. I'm looking for a simple proof of the fact that $\|A\|=\|A\|_c$, or, in particular, that $\|A\|\geq|\lambda|$, for any (possibly complex) eigenvalue $\lambda$ of $A$ (when $m=n$). Clearly, it holds that $\|A\|_c\geq|\lambda|$. I have seen some results on this issue, but in a more abstract context. My question is about a simple proof of this fact for the finite dimensional case. Also, does this result hold for other norms? REPLY [2 votes]: A discussion of the relation of norms on real and complexified spaces can be found in the paper O. Holtz, M. Karow. Real and complex operator norms. arxiv-paper For the finite-dimensional case, where $A$ maps from $(\mathbb{C}^m,\|\cdot\|_p)$ to $(\mathbb{C}^n,\|\cdot\|_q)$ the paper presents an elementary proof for the equality of real and complex operator norm in the case $1 \leq p \leq q \leq \infty$. This inequality is sharp, i.e. for other values of $p,q$ it does not hold in general. There are some further cases treated in the paper. All this probably follows from the more abstract results you have already found, but maybe the presentation will appeal to you in terms of having a simple proof.<|endoftext|> TITLE: Definability using rudimentary function QUESTION [6 upvotes]: Denote by RUD the set of all rudimentary functions, together with the function that takes any set to its transitive closure. Assume that I know that a binary relation $R$ is definable by some function in RUD. Is there a way to define $R^+$, the transitive closure of the relation $R$, using functions in RUD? Maybe there isn't one, but I strongly feel there should be a way and I cannot seem to find it. I can be even more specific, I want to be able to define the function that given a set $x$ returns the set $\{ y | R^+(x,y) \}$, given that I know there is a function in RUD which defines the function that takes $x$ to $\{ y | R(x,y) \}$. Thanks for your help! REPLY [8 votes]: If I understand you correctly I believe the answer is no. Let $S(x)$ be the rud. function that returns $x \cup \{x\}$. Let $R(x,y) \leftrightarrow y = S(x)$. Then $F(x) = \{y\mid R(x,y)\} = \{S(x)\}$ is rud. But $G(x)=\{y\mid R^+(x,y)\}\supseteq \{y \mid \exists p<\omega\,\, y = SS...SS(x)\, (p \mbox{ times }) \}$. However then there is no $q$ so that $\forall x : rk(G(x))\leq rk(x)+q$. Hence $G$ cannot be rud.<|endoftext|> TITLE: Set of w*-continuous operators closed for the weak* topology or not? QUESTION [6 upvotes]: Let $X$ be a dual Banach space, i.e. $X=(X_*)^*$ for some Banach space $X_*$. Consider the weak* topology of $B(X)$, i.e. the topology of pointwise convergence on $X$ endowed with the $\sigma(X,X_*)$-topology. Consider the set $B_{w^*}(X)$ of $w^*$-continuous bounded operators on $X$. Is it a closed subset of $B(X)$ for the weak* topology of $B(X)$ ? REPLY [5 votes]: The answer is no. I know that for some people here, saying "It's false for $X = \ell^1$" would be a good enough hint, but I also know that this question originated on Math StackExchange, so I've included many of the details, making a rather long answer. The spaces I use for a counterexample are $\newcommand{\R}{\mathbb{R}}X = \ell^1$, and $X_{*} = c_0$, using the usual pairing $\langle\mbox{-},\mbox{-}\rangle_1 : \ell^1 \times c_0 \rightarrow \R$ defined by: $$ \langle \phi, a \rangle_1 = \sum_{n=0}^\infty \phi(n)\cdot a(n). $$ I prefer a different notation from that used in the question, which I'll explain here. I'll use $\ell^1$ for that space considered as a Banach space, and $\ell^1_\sigma$ when equipped with the weak-* topology $\sigma(\ell^1,c_0)$. Then $L(\ell^1) = L(\ell^1,\ell^1)$ is the space of bounded/continuous linear maps from $\ell^1$ to itself in the norm topology, and $L(\ell^1_\sigma)$ the space of continuous linear maps $\ell^1_\sigma \rightarrow \ell^1_\sigma$. The reason the answer is no is that $L(\ell^1_\sigma)$ is dense in $L(\ell^1)$, both in the $\sigma(L(\ell^1),\ell^1 \hat{\otimes} c_0)$-topology and the weak-* operator topology described in the question, but $L(\ell^1_\sigma)$ is a proper subset of $L(\ell^1)$. So $L(\ell^1_\sigma)$ is not closed in either of those topologies. We'll begin with some definitions. For each $\newcommand{\N}{\mathbb{N}}n \in \N$ let's define $\delta_n : \N \rightarrow \R$ such that $\delta_n(m)$ is 1 if $n=m$, and $0$ otherwise. These functions belong to both $\ell^1$ and $c_0$ and are the usual Schauder bases of these spaces. Let's also define $e_n : \ell^1 \rightarrow \R$ by $e_n(\phi) = \phi(n)$. These functionals are weak-* continuous (i.e. $\sigma(\ell^1,c_0)$-continuous), essentially by definition. We also define $s : \ell^1 \rightarrow \R$ by $s(\phi) = \sum_{i=0}^\infty \phi(i)$. This is norm continuous on $\ell^1$ but not weak-* continuous because $\delta_n \to 0$ in the weak-* topology, but $\phi(\delta_n) = 1 \not\to 0$. Therefore the rank 1 operator $T(\phi) = s(\phi) \cdot \delta_0$ is in $L(\ell^1)$ but not $L(\ell^1_\sigma)$. So all that remains is to show that $L(\ell^1_\sigma)$ is dense in $L(\ell^1)$. I'll give two proofs. I came up with the second one first, and it is simpler but requires more background knowledge. The first proof is more "bare hands" and is likely more what the OP was looking for. In both cases, we actually show that the linear span of the rank one operators $T_{n,m}(\phi) = e_n(\phi) \cdot \delta_m \in L(\ell^1_\sigma)$ is dense in $L(\ell^1)$. First proof: The key statement in this proof is: Given $f \in L(\ell^1)$, define: $$ f_n = \sum_{i=0}^n \sum_{j = 0}^n f(\delta_i)(j) \cdot T_{ij}. $$ This is a sequence of finite rank operators in $L(\ell^1_\sigma)$ converging to $f$ in the strong operator topology. We prove this as follows. As part of the proof that $(\delta_n)$ is a Schauder basis for $\ell^1$, we know that for all $\phi \in \ell^1$, $\left(\sum_{i=0}^n \phi(i) \cdot \delta_i\right) \to \phi$. So for each $\epsilon > 0$, there exists an $N \in \N$ such that for all $n \geq N$, $\| \phi - \sum_{i=0}^n \phi(i) \cdot \delta_i \| < \frac{\epsilon}{\| f \|}$. So for all $n \geq N$: $$ \| f(\phi) - f_n(\phi) \| = \sum_{p = 0}^\infty |f(\phi)(p) - f_n(\phi)(p)| $$ $$ = \sum_{p=0}^\infty \left| f(\phi)(p) - \sum_{i=0}^n \sum_{j=0}^n f(\delta_i)(j) \cdot e_i(\phi) \cdot \delta_j(p) \right| $$ $$ = \sum_{p=0}^\infty \left| f(\phi)(p) - \sum_{i=0}^n f(\delta_i)(p) \cdot \phi(i) \right| $$ $$ = \left\| f(\phi) - f\left(\sum_{i=0}^n \phi(i) \cdot \delta_i \right) \right\| \leq \|f\| \cdot \frac{\epsilon}{\|f\|} = \epsilon $$ So $f_n \to f$ pointwise in $\ell^1$, i.e. in the strong operator topology. Since the weak-* topology is coarser than the norm topology, it also converges pointwise for that, so in the weak-* operator topology. To prove that $f_n \to f$ in $\sigma(L(\ell^1),\ell^1 \hat{\otimes}c_0)$, we observe that by comparing the neighbourhood bases to each other, the weak-* operator topology and the topology $\sigma(L(\ell^1),\ell^1 \otimes c_0)$ agree (i.e. for the weak-* topology without completing the tensor product). Now, $\| f_n \| \leq \| f\|$, so we can use the following standard fact: If $E$ is a Banach space, $D \subseteq E$ a dense subspace, then $\sigma(E^{*},E)$ and $\sigma(E^{*},D)$ agree on all norm bounded subsets of $E^{*}$. I don't know a good reference for it in the above formulation, the closest being Schaefer's Topological Vector Spaces III.4.5 (taking $F = \R$ in that theorem). To finish, we apply it with $E^* = L(\ell^1)$, $E = \ell^1 \hat{\otimes} c_0$ and $D = \ell^1 \otimes c_0$, and the norm bounded set $\{ g \in L(\ell^1) \mid \| g \| \leq \| f \| \}$. Second proof: In this case, the key statement is The set of maps $(T_{n,m})_{n,m \in \N}$ separates the points of $\ell^1 \hat{\otimes} c_0$ under the pairing with $L(\ell^1)$. By a standard consequence of the Hahn-Banach theorem (see e.g. Schaefer's Topological Vector Spaces IV.1.3) this shows that the linear span of $(T_{n,m})_{n,m \in \N}$ is $\sigma(L(\ell^1),\ell^1 \hat{\otimes} c_0)$-dense. As the weak-* operator topology on $L(\ell^1)$ is coarser than this topology, this also proves density in that topology. We use a theorem of Grothendieck to identify $\ell^1 \hat{\otimes} c_0$ with $\ell^1[c_0]$, i.e. the space of absolutely summable sequences in $c_0$. Under this isomorphism, the pairing of $L(\ell^1)$ with $\ell^1 \hat{\otimes} c_0$ is mapped to the following, where $f \in L(\ell^1)$ and $\Phi \in \ell^1[c_0]$: $$ \langle f, \Phi \rangle_2 = \sum_{n = 0}^\infty \langle f(\delta_n), \Phi(n)\rangle_1 $$ If $\langle T_{n,m}, \Phi \rangle_2 = 0$ for all $n,m \in \N$, this means $$ 0 = \sum_{i=0}^\infty \langle T_{n,m}(\delta_i), \Phi(i) \rangle_1 = \sum_{i=0}^\infty \sum_{j=0}^\infty T_{n,m}(\delta_i)(j) \cdot \Phi(i)(j) = \sum_{i=0}^\infty \sum_{j=0}^\infty \delta_i(n)\cdot \delta_m(j) \cdot \Phi(i)(j) $$ $$ = \Phi(n)(m) $$ for all $n,m \in \N$, and therefore $\Phi = 0$. As discussed, this shows that the linear span of $(T_{n,m})_{n,m \in \N}$ is $\sigma(L(\ell^1),\ell^1[c_0])$-dense in $L(\ell^1)$ and so $L(\ell^1_\sigma)$ is also dense.<|endoftext|> TITLE: A Galois extension over $\mathbb{Q}$ with Galois group $A_4$ and with cyclic decomposition groups QUESTION [10 upvotes]: Does there exist a Galois extension $L/\mathbb{Q}$ with Galois group $A_4$ (the alternating group on four letters) such that all the decomposition groups are cyclic? This question is motivated by the answer by Kasper Andersen to my question. Namely, a desired example (if exists) would permit one to answer in the negative this hard question. REPLY [15 votes]: Yes. Note that Daniel Loughran's comment to David Speyer's answer to this question states that for any solvable group $G$, there is a Galois extension $L/\mathbb{Q}$ with all decomposition groups cyclic. It's not hard to find a concrete example. If you let $L$ be the splitting field of $x^{4} - x^{3} - 7x^{2} + 2x + 9$, then $L/\mathbb{Q}$ is an $A_{4}$ extension ramified only at $163$. (This is one of the $A_{4}$ fields from Klueners and Malle's excellent database.) There are four prime ideals, $\mathfrak{p}_{1}$, $\mathfrak{p}_{2}$, $\mathfrak{p}_{3}$ and $\mathfrak{p}_{4}$ above $163$ in $L$, and so $ef = 3$ for each such prime ideal. Therefore, the decomposition group for each such prime ideal has order $3$.<|endoftext|> TITLE: Which spaces have enough curves QUESTION [14 upvotes]: Let $\mathbf{Top}$ be the category of topological spaces, and let $I\in\mathbf{Top}$ be the unit interval $I=[0,1]\subset\mathbb{R}$. For any space $X$, let $|X|$ denote the underlying set of points; similarly if $f\colon X\to Y$ is continuous, $|f|$ denotes the underlying function. Define a curve in $X$ to be a continuous map $I\to X$. Given spaces $X$ and $Y$ and a function $f\colon |X|\to |Y|$ on underlying sets, say that $f$ sends curves to curves if composing $f$ with any curve $c\colon I\to X$ yields a (continuous) curve in $Y$. Of course, if $f$ is continuous then it sends curves in $X$ to curves in $Y$, but the converse may not hold. I'll say that $X$ has enough curves if, for any $Y$ and function $f\colon|X|\to|Y|$, we have that $f$ is continuous if and only if it sends curves to curves. An example of a space that does not have enough curves is the "sequence space" $S=\{0\}\cup\{\frac{1}{n+1}\mid n\in\mathbb{N}\}\subset I$. All curves in $S$ are constant, so they cannot detect non-continuous maps out of $S$. Question: Which well-known classes of spaces have enough curves? REPLY [8 votes]: One can prove the following : (1) If $X$ has enough curves, then $X$ is sequential and locally path-connected. (2) If $X$ is first-countable and locally path-connected, then $X$ has enough curves. Note that "first-countable" implies "sequential", but the opposite implication is not true in general. Proof of (1) : As noted by Martin Sleziak in his first comment, if $X$ has enough curves, then it is a quotient of the disjoint sum of copies of $I$. Now, $I$ is sequential and locally path-connected, and these two properties are preserved by disjoint sums and by quotients, hence the result. Proof of (2) : Let $f : X \rightarrow Y$ be a map sending curves to curves. We have to show that $f$ is continuous. Since $X$ is first-countable, it is sequential, hence it suffices to show that for any sequence $(x_n)_{n \geq 1}$ in $X$, converging to some $x \in X$, the sequence $(f(x_n))_{n \geq 1}$ converges to $f(x)$. Let $(U_k)_{k \geq 1}$ be a countable basis of path-connected open neighbourhoods of $x$, such that $U_k \supset U_{k+1}$ for each $k$. Wlog $x_n \in U_1$ for each $n$. For each $k \geq 1$, let $N_k$ be the smallest positive integer such that $x_n \in U_k$ for each $n \geq N_k$. In particular $N_1 = 1$. Since $U_k$ is path-connected, one can find for each $n \in [N_k, N_{k+1} [$ a continuous map $c_n : [\frac{1}{n+1}, \frac{1}{n}] \rightarrow U_k$ with $c_n(\frac{1}{n+1}) = x_{n+1}$ and $c_n(\frac{1}{n}) = x_n$. The collection $(c_n)_{n \geq 1}$ yields a continuous map $c : ]0,1] \rightarrow X$ with $c(]0,\frac{1}{N_k}]) \subseteq U_k$. In particular, setting $c(0) = x$ yields a continuous map $c : I \rightarrow X$. Since $fc$ is continuous, the sequence $f(x_n) = fc(\frac{1}{n})$ converges to $fc(0) = f(x)$. REPLY [7 votes]: A space $X$ whose topology agrees with the final topology with respect to all maps $I\to X$ is often called a delta-generated ($\Delta$-generated) space. The category of $\Delta$-generated spaces is a convenient category of spaces (in the sense of Steenrod) and is used in Diffeology, directed homotopy, and generalized covering spaces theories.<|endoftext|> TITLE: Clarification on Étienne Ghys' "Feuilletages riemanniens sur les variétés simplement connexes" paper QUESTION [5 upvotes]: I apologize for this type of question, but I'm having some trouble to understand remark 3.4(4) on page 212 of this article, that reads The restriction of $\overline{\mathcal{G}}$ (the foliation aproaching $\overline{\mathcal{F}}$ obtained in Theorem 3.3) to the closure of a leaf of $\overline{\mathcal{F}}$ is defined by a fibration over the torus $\mathbb{T}^n$, and the group structure of this torus is well-defined. The author indicates Lemma 3.2 as the justification, but I do not see how it follows, as that lemma only applies to the 1-forms $\alpha_i$ that define $\overline{\mathcal{F}}$, and not (necessarily) to the harmonic 1-forms $u_i$ that he uses, in the proof of Theorem 3.3, to perturbate $\alpha_i$ and get $\overline{\mathcal{G}}$. I wonder if it's perhaps a misprint and the justification is actually Lemma 3.1, the torus appearing as the quotient of $\mathbb{R}^n$ (as he shows that, restricted to a closure $N$ of a leaf, $\overline{\mathcal{F}}$ is a Lie $\mathbb{R}^n$-foliation) by the holonomy representation $H(\pi_1(N))$, but I think in the end this would be equivalent to affirm that Lemma 3.2 also holds for the $u_i$s. Note that this remark is used in the proof of Theorem 3.5. I'd appreciate any insight on that matter. REPLY [3 votes]: I think I've figured it out, so I'm posting it here for future reference. (I use a slightly different notation here: in the paper and in the question the overline in $\overline{\mathcal{F}}$ is used to denote the lifted foliation of Molino's construction (see below), which I denote in the answer by $\mathcal{F}^\#$. The overline notation is now almost aways used to denote the singular riemannian foliation obtained by taking the closures of the leaves) We will assume for simplicity that $\mathcal{F}$ is transversely oriented (otherwise it can be lifted to an orientation covering). Lemme 3.1 of the paper in question shows that the structural algebra of $\mathcal{F}$ is $\mathbb{R}^d$. We consider the transverse orthonormal bundle $M^\#$ of $\mathcal{F}$ (a.k.a. the Molino bundle) with the lifted foliation $\mathcal{F}^\#$. It's known by Molino's theory (see the book "Riemannian Foliations" by P. Molino) that the restriction of $\mathcal{F}^\#$ to a leaf closure $N=\overline{L^\#}$ is developable to a Lie-$\mathbb{R}^d$ foliation. In other words, this restriction is defined by closed $1$-forms $\alpha_1,\dots,\alpha_d$, that is, $$\mathcal{F}^\#|_N=\bigcap_{i=1}^d \ker(\alpha_i).$$ Let's see this in more detail: we have that $\alpha=(\alpha_1,\dots,\alpha_d)$ is a $\mathbb{R}^d$ valued Maurer-Cartan form on $N$ that defines $\mathcal{F}^\#|_N$. Now we aply Darboux construction: consider the connection $\eta$ on $N\times\mathbb{R}^d$ given by $$\eta_{(x,g)}(\xi,v)=\alpha_x(\xi)+v$$ (here we identify $T_g\mathbb{R^d}\equiv\mathbb{R}^d$). It's not difficult to see that $\eta$ is flat and so it determines a foliation $\mathcal{F}_\eta$ on $N\times\mathbb{R}^d$. Choose $x_0\in N$ and take $\hat{N}\in\mathcal{F}_\eta$ the leaf containing $(x_0,0)$. Then $\pi:=\mathrm{pr}_1|_{\hat{N}}:\hat{N}\to N$ is a covering and $f_\alpha:=\mathrm{pr}_2|_{\hat{N}}:\hat{N}\to\mathbb{R}^d$ is a submersion. E. Fedida's theorem (Theorem 4.1 in Molino's book) states that the foliation $\pi^*(\mathcal{F}^\#|_N)$ coincides with the foliation determined by the fibers of $f_\alpha$ (that is, $\mathcal{F}^\#|_N$ is developable over $\mathbb{R}^d$). Now the action of $\pi_1(N,x_0)$ on $\hat{N}$ defines the homomorphism $h_\alpha:\pi_1(N,x_0)\to\mathbb{R}^d$, whose image is $$H_\alpha=\{g\in\mathbb{R}^d\ |\ \hat{N}g=\hat{N}\}$$ (Juxtaposition represents the natural action of $\mathbb{R}^d$ on $N\times\mathbb{R}^d$). By definition, we have the identity $\hat{x}\cdot\gamma=\hat{x}h_\alpha(\gamma)$, for all $\hat{x}\in\hat{N}$ and all classes $[\gamma]\in\pi_1(N,x_0)$, which induces the $h_\alpha$-equivariancy of $f_\alpha$: $$f_\alpha(\hat{x}\cdot\gamma)=f_\alpha(\hat{x})+h_\alpha(\gamma).$$ Denote by $\overline{H_\alpha}$ the closure of $H_\alpha$ in $\mathbb{R}^d$. Then $\mathbb{R}^d/\overline{H_\alpha}$ is a Lie group and, by the $h_\alpha$-equivariancy, $f_\alpha$ induces a submersion $\overline{f_\alpha}:N\to\mathbb{R}^d/\overline{H_\alpha}$ whose fibers are the closures of the leaves of $\mathcal{F}^\#|_N$. (In our case specifically, $\mathcal{F}^\#|_N$ has dense leaves, so $\overline{H_\alpha}=\mathbb{R}^d$ and $\overline{\mathcal{F}^\#|_N}$ is the trivial foliation with $N$ the only leaf) Let $\gamma:[0,1]\to N$ be a smooth path with $\gamma(1)=x_0$ and let $\hat{\gamma}$ be the lifting of $\gamma$ to $\hat{N}$ satisfying $\hat{\gamma}(1)=(x_0,0)$. Then we may write $\hat{\gamma}=(\gamma,\tau)$, $\tau$ being a smooth path in $\mathbb{R}^d$. We have $\tau(1)=0$ and $\tau(0)=f_\alpha(\hat{\gamma}(0))$. As $\hat{\gamma}([0,1])\subset\hat{N}\in\mathcal{F}_\eta$, it follows that $$0=\eta\left(\frac{d\gamma}{dt},\frac{d\tau}{dt}\right)=\alpha\left(\frac{d\gamma}{dt}\right)+\frac{d\tau}{dt},$$ and so $$f_\alpha(\hat{\gamma}(0))=\tau(0)=\int_0^1\alpha\left(\frac{d\gamma}{dt}\right)dt=\int_{[0,1]}\gamma^*\alpha.$$ In particular, for $[\gamma]\in\pi_1(N,x_0)$, we have \begin{equation}\displaystyle \begin{array}{rcl} h_\alpha(\gamma)&=&\displaystyle f_\alpha((x_0,0)\cdot\gamma)-f_\alpha((x_0,0))=f_\alpha(\hat{\gamma}(0))=\int_{[0,1]}\gamma^*\alpha\\ &=&\displaystyle\left(\int_0^1\alpha_1\left(\frac{d\gamma}{dt}\right)d t,\dots,\int_0^1\alpha_d\left(\frac{d\gamma}{dt}\right)d t\right).\end{array} \end{equation} The homomorphism $h_\alpha$ has abelian image, so it factors to $\overline{h_\alpha}:H_1(N,\mathbb{Z})/T\to\mathbb{R}^d$, where $T$ is the torsion subgroup of $H_1(N,\mathbb{Z})$. Denoting $\overline{h_\alpha}=(\overline{h_\alpha^1},\dots,\overline{h_\alpha^d})$, the equation above shows that $\ell_{DR}[\alpha_i]=\overline{h_\alpha^i}$, where $\ell_{DR}:H^1_{DR}(N)\to H^1(N,\mathbb{R})$ is De Rham's isomorfism. Take $[\vartheta_1],\dots,[\vartheta_r]$ a free abelian basis of $H^1(N,\mathbb{Z})\subset H^1(N,\mathbb{R})$. Then, $$\ell_{DR}([\alpha_i])=\sum_{j=1}^r c_i^j[\vartheta_j],$$ $c_i^j\in\mathbb{R}$. If $\sigma_1,\dots,\sigma_r$ are smooth loops representing the basis of $H_1(N,\mathbb{Z})/T$ dual to $\big([\vartheta_j]\big)_{j=1}^r$, it follows that $$H_\alpha=\overline{h_\alpha}\big(H_1(N,\mathbb{Z})/T\big)=\left\langle\big(c_1^j,\dots,c_d^j\big)_{j=1}^r\right\rangle,$$ because $c_i^j=\int_{\sigma_j}\alpha_i$. Now we take $[u_i]\in H^1_{DR}(N)$ such that $[\alpha_i]+[u_i]$ are rational cohomology classes. We can choose $[u_i]$ small enough so that $\alpha_i'=\alpha_i+u_1$ are linearly independent in every point of $N$. Then $\alpha'=(\alpha_1',,\dots,\alpha_d')$ is a new Maurer-Cartan form that defines a foliation $\mathcal{G}^\#$ of $N$. (this is the construction that Ghys does in the proof of Théorème 3.3). Because the new classes are rational, we now have $$H_{\alpha'}=\overline{h_{\alpha'}}=\left\langle\left(\int_{\sigma_j}\alpha_1',\dots,\int_{\sigma_j}\alpha_d'\right)_{j=1}^r\right\rangle,$$ with $\int_{\sigma_j}\alpha_i'\in\mathbb{Q}$ for all $i$ and $j$, so $\overline{H_{\alpha'}}=H_{\alpha'}$ is a lattice of $\mathbb{R}^d$ and so $$\overline{f_{\alpha'}}:N\to\mathbb{R}^d/H_{\alpha'}\cong\mathbb{T}^d$$ is a fibration defining $\overline{\mathcal{G}^\#}$. It also shows that $\mathcal{G}^\#$ is a closed foliation.<|endoftext|> TITLE: Density of convolution QUESTION [5 upvotes]: Let $\{X_i\}$ be i.i.d random variables uniform on a measurable, symmetric set $A$ contained in $[-1,1]$. Let $g_{n}$ be density of $X_1+\ldots + X_n$. Question (general): Is there any non-trivial bound from below on $g_n(0)$ in terms of the Lebesgue measure of $A$. Let us consider an example, say sets with $|A|=1/4$. It is easy to find an example of $A$ such that $g_1(0)=g_3(0)=0$. One checks also that $g_2(0)=1/|A|$. Question (specific): What about $g_4(0)$? REPLY [2 votes]: I'll only discuss even $n$, I don't have any ideas for odd $n$, plus, as your comments indicate, there may not be any lower bounds for odd $n$. Let's write $f(t)=|A|^{-1}\widehat{\chi}_A(t)=(1/|A|)\int_A e^{2\pi itx}\, dx$. Then $$ g_n(0)=\int \widehat{g_n}(x)\, dx = \int f^n(x)\, dx . $$ First attempt: To obtain a lower bound, I'll just focus on the positive contribution coming from $x$ close to zero. Notice that $f(0)=1$, $f'(0)=0$ (because $A$ is symmetric), $|f'(x)|, |f''(x)|\le C$ (the constants could be worked out here). So $h=f^n$ satisfies $h(0)=1$, $h'(0)=0$, $|h''(x)|\lesssim n^2$, thus $$ |f^n(x)|\ge 1-cn^2x^2 . $$ It follows that $$ g_n(0)\ge C/n $$ for even $n$. This was clearly not a very sophisticated estimate and there's probably room for improvement; for example, for $n=2$, we're off by a factor of $|A|$ for small $|A|$. Second attempt: A perhaps more useful bound that tries more seriously to keep track of the dependence on $A$ is obtained if we use Hölder as follows: $$ \frac{1}{|A|}=\int f^2 = \int f^{a+2-a} \le \left(\int f^n\right)^{a/n}\left( \int |f|^{(2-a)n/(n-a)}\right)^{(n-a)/n} $$ If I could take $a=n/(n-1)$ here, this would (formally) give that $$ g_n(0)\ge |A|^{-1/(n-1)} \left( \int |f|\right)^{-(n-2)/(n-1)^2} . $$ That was not correct because $f\notin L^1$, but if $f\in L^{1+\epsilon}$ (this is an assumption on $A$, it's not true automatically), I can take a slightly smaller $a$ to produce a bound of the form $g_n(0)\gtrsim |A|^{-1/(n-1)+\delta}$.<|endoftext|> TITLE: $C^1$-functions on Banach spaces QUESTION [7 upvotes]: For Banach spaces $X,Y$ and an open subset $U$ of $X$ a function $f:U\to Y$ is $C^1$ if $U\to L(X,Y)$, $x\to f'(x)$ is continuous where, by definition, the derivative $f'(x)$ is a continuous linear operator from $X$ to $Y$ satisfying $$\|f(x+h)-f(x)-f'(x)(h)\|_Y = o(\|h\|_X)$$ and $L(X,Y)$ is the Banach space of continuous linear operators endowed with the operator norm. Then $f'$ is again a map between Banach spaces and one can define higher derivatives. For Frechet instead of Banach spaces there are many different topologies on $L(X,Y)$ but none of them is Frechet. To stay in the category one thus needs a different definition. In his exposition of the Nash-Moser theorem Hamilton therefore tells $f$ continuously differentiable if all directional derivatives $Df(x)(v)=\lim\limits_{t\to 0}(f(x+tv)-f(x))/t$ exist in $Y$ and $Df: U\times X \to Y$ is continuous. He then writes: Note well that this definition of a $C^1$ map does not agree with the usual definition for a Banach space. The derivative will of course be the same map (i.e. $Df(x)(v)=f'(x)(v)$), but our continuity requirement is weaker. What is an example (in Banach spaces) showing that the latter definition is strictly weaker than the former? REPLY [4 votes]: Here is a different and very natural example from the forthcoming book Infinite-dimensional Lie Groups. General Theory and Main Examples by H. Glöckner and K.-H. Neeb. I describe it here with the permission of H. Glöckner. The function $f:L^1[0,1]\to L^1[0,1]$ is the superposition $\xi\mapsto \sin\circ\xi=\sin(\xi)$. The directional derivatives are $Df(\xi)(\eta)=\cos(\xi)\eta$ -- which looks very plausible but the proof needs some care. On can show that $Df:L^1[0,1] \times L^1[0,1]\to L^1[0,1]$ is continuous but $Df:L^1[0,1] \to \mathcal L_{op}(L^1[0,1])$ is not: Because of $Df(0)=id$ the inverse mapping theorem would imply that the range $f(L^1[0,1])$ contains a $0$-neighbourhood in $L^1[0,1]$ which is not true because this range is contained in the unit ball of $L^\infty[0,1]$ (which is of first category). If one prefers a Hilbert space example one can replace $L^1$ by $L^2$.<|endoftext|> TITLE: Why is 12 the smallest weight for which a cusp forms exists QUESTION [17 upvotes]: I have already asked this question on MathStackExchange (here) but did not receive an answer, therefore I am trying it here: On wikipedia (here) I have read the following: Twelve is the smallest weight for which a cusp form exists. [...] This fact is related to a constellation of interesting appearances of the number twelve in mathematics ranging from the value of the Riemann zeta function at $−1$ i.e. $\zeta(-1) = -1/12$, the fact that the abelianization of $SL(2,\mathbb{Z})$ has twelve elements, and even the properties of lattice polygons. I know the dimension formula and the weight formula for modular forms as well as the proof of the weight formula via the residue theorem, but I do not understand how this relates to any of the three concepts mentioned in the quote above. Can anybody tell me more? REPLY [8 votes]: These facts are related via the geometric interpretation of cusp forms. Essentially, identifying cusp forms as differential forms on modular curves allows to establish the relation to the ${\rm SL}_2\mathbb{Z}$-action on the upper half-plane. The following is very sketchy, but I hope it properly outlines the main points. The first step is to use the transformation behaviour to interpret cusp forms as certain holomorphic differential forms on a modular curve. Alternatively, cusp forms are viewed as global sections of a suitable line bundle on the compactification of ${\rm SL}_2\mathbb{Z}\backslash\mathbb{H}$. Then one can apply Riemann-Roch to compute the dimension of the space of cusp forms in terms of the Euler characteristic of the modular curve. A formula for dimensions of spaces of cusp forms for congruence subgroups can be found as Theorem 4.9 in Milne's lecture notes. Formulas also applicable to ${\rm SL}_2\mathbb{Z}$ are discussed in the following lecture notes these lecture notes. For ${\rm SL}_2\mathbb{Z}\backslash\mathbb{H}$, the formula for the Euler characteristic gives $-\frac{1}{12}$ because there is one point with isotropy $\mathbb{Z}/4\mathbb{Z}$ and one point with isotropy $\mathbb{Z}/6\mathbb{Z}$. (There is some subtlety here with the isotropy which leads to the Euler characteristic not being integral. One can also deal with this by considering congruence subgroups $\Gamma$ such that $\Gamma\backslash\mathbb{H}$ is a Riemann surface and apply some form of Riemann-Hurwitz to compare to ${\rm SL}_2\mathbb{Z}\backslash\mathbb{H}$.) Now the Euler characteristic can be compared to the order of the abelianization. The fundamental domain for the ${\rm SL}_2\mathbb{Z}$-action retracts equivariantly onto one of its boundary segments, leading to an equivariant retraction of the upper half plane onto a tree. From the action of ${\rm SL}_2\mathbb{Z}$ on the tree we get the well-known amalgam decomposition $\mathbb{Z}/4\mathbb{Z}\ast_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/6\mathbb{Z}$, where the subgroups are the isotropy group from before. The natural map from this amalgamated product to $\mathbb{Z}/12\mathbb{Z}$ is in fact the abelianization -- hence we get a relation between the abelianization and the isotropy of the action on $\mathbb{H}$. Now that we have the Euler characteristic, we can use the Gauss-Bonnet theorem (in a version for finite-volume hyperbolic surfaces) to relate volume and Euler characteristic $$ -{\rm vol}(M)=2\pi\chi(M) $$ This should rather be applied to ${\rm PSL}_2\mathbb{Z}$ because of the ubiquitous complication that the center (of order 2) in ${\rm SL}_2\mathbb{Z}$ fixes all of $\mathbb{H}$ -- the result is a factor 2. In the end, the volume for ${\rm SL}_2\mathbb{Z}\backslash\mathbb{H}$ turns out to be $\pi/3$. Now we need to compare this with another computation of the volume related to the zeta values. There is an equality of volumes $$ {\rm Vol}({\rm SL}_2\mathbb{Z}\backslash\mathbb{H})\times {\rm Vol}({\rm SO}(2,\mathbb{R}))=2{\rm Vol}({\rm SL}(2,\mathbb{R})/{\rm SL}_2\mathbb{Z}). $$ The right-hand side is the Tamagawa volume which is $\zeta(2)$. The volume of ${\rm SO}(2,\mathbb{R})$ is $\pi$. The combination tells us that the volume of ${\rm SL}_2\mathbb{Z}\backslash\mathbb{H}$ is $2\zeta(2)/\pi$. Together with the functional equation for $\zeta$, we see that $-\zeta(-1)$ is the Euler characteristic of ${\rm SL}_2\mathbb{Z}\backslash\mathbb{H}$ which entered in the computation for the cusp forms. For the volume statements, see chapter 11 of MacLachlan-Reid: The arithmetic of hyperbolic 3-manifolds. The appearance of $\zeta(2)$ can be seen directly in the volume computation.<|endoftext|> TITLE: Geometric interpretation of splitting of sequence associated to a homogeneous space QUESTION [5 upvotes]: Let $G$ be a Lie group acting transitively on a smooth manifold $M$. Let $\mathfrak{g}$ be the Lie algebra of $G$ and let $\xi : \mathfrak{g} \to \Gamma(TM)$ be the Lie algebra homomorphism sending $X \in \mathfrak{g}$ to the generating vector field $\xi_X$. Fix a point $p \in M$ and consider the evaluation $\operatorname{ev}_p : \mathfrak{g} \to T_p M$ sending $X \in \mathfrak{g}$ to $\xi_X(p)$. The kernel of this map is the Lie algebra $\mathfrak{h}_p$ of the stabiliser $H_p$ of $p$ in $G$. This gives rise to a short exact sequence $$ 0 \longrightarrow \mathfrak{h}_p \longrightarrow \mathfrak{g} \longrightarrow T_pM \longrightarrow 0 $$ I'm particularly interested in the case when the homogeneous space is not reductive, so that the sequence does not split as $\mathfrak{h}_p$-modules. My question is the following: What is the geometrical interpretation of a (vector space) splitting $T_p M \to \mathfrak{g}$ of the above exact sequence? I suspect that this is standard, but I cannot find much literature in the non-reductive case. REPLY [2 votes]: I am not aware of a general "geometric interpretation" of such a vector space splitting in the non-equivariant case, but there is at least one interesting application, which is along the lines of the comment by @FriedrichKnop. Namely, you get a family of distinguished "normal" coordinates and a family of distinguished curves in $M$. Identifying $M=G/H$, let $\mathfrak m\subset\mathfrak g$ be the image of the splitting, i.e. a linear subspace complementary to $\mathfrak h$. Denoting by $\pi:G\to G/H$ the natural projection you see that $X\mapsto \pi(g\cdot exp(tX))$ restricts to a local diffeomorphism from an open neighbourhood of $0$ in $\mathfrak m$ onto an open neighbourhood of $\pi(g)$ in $G/H$. In each point, this gives you a family of local coordinates parametrized by the fiber over the point. The projections of the exponential curves make sense for all times, so you get a family of distinguished parametrized curves. This is a family of reasonable size, because fixing the initial velocity, there is just a finite dimensional freedom. To see that this is interesting in examples, consider the cases of $S^n$ as a homogeneous space of $SO(n+1,1)$ or of $S^{2n+1}$ as a homogeneous space of $SU(n+1,1)$. In both cases, you get a natural vector space splitting for $\mathfrak g$, which is not $H$-equivariant. In the first case, the resulting curves are the conformal circles on $S^n$. In the second case, you can get the chains for the natural CR structure on $S^{2n+1}$ as distinguished curves of that type (with a slight additional normalization).<|endoftext|> TITLE: Did Euler prove theorems by example? QUESTION [41 upvotes]: In his 2014 book, Giovanni Ferraro writes at beginning of chapter 1, section 1 on page 7: Capitolo I Esempi e metodi dimostrativi Introduzione In The Calculus as Algebraic Analysis, Craig Fraser, riferendosi all'opera di Eulero e Lagrange, osserva: A theorem is often regarded as demonstrated if verified for several examples, the assumption being that the reasoning in question could be adapted to any other example one chose to consider (Fraser [1989, p. 328]). Le parole di Fraser colgono un aspetto poco indagato della matematica dell'illuminismo. I am not fluent in Italian but the last sentence seems to indicate that Ferraro endorses Fraser's position as expressed in the passage cited in the original English without Italian translation. I was rubbing my eyes as I was reading this so I decided to check in Fraser's original, thinking that perhaps the comment is taken out of context. I found the following longer passage on Fraser's page 328 quoted by Ferraro: The calculus of EULER and LAGRANGE differs from later analysis in its assumptions about mathematical existence. The relation of this calculus to geometry or arithmetic is one of correspondence rather than representation. Its objects are formulas constructed from variables and constants using elementary and transcendental operations and the composition of functions. When EULER and LAGRANGE use the term "continuous" function they are referring to a function given by a single analytical expression; "continuity" means continuity of algebraic form. A theorem is often regarded as demonstrated if verified for several examples, the assumption being that the reasoning in question could be adapted to any other example one chose to consider. Let us examine Fraser's hypothesis that in Euler and Lagrange, allegedly "a theorem is often regarded as demonstrated if verified for several examples." I don't see Fraser presenting any evidence for this. Now Wallis sometimes used a principle of "induction" in an informal sense that a formula verified for several values of $n$ should be true for all $n$, but for this he was already criticized by his contemporaries, a century before Euler and Lagrange. Several articles were recently published examining Euler's proof of the infinite product formula for the sine function. The proof may rely on hidden lemmas, but it is a sophisticated argument that is a far cry from anything that could be described as "verification for several examples." It seems to me that this passage from Fraser is symptomatic of an attitude of general disdain for the great masters of the past. Such an attitude unfortunately is found among a number of received historians. For example, we find the following comment: Euler's attempts at explaining the foundations of calculus in terms of differentials, which are and are not zero, are dreadfully weak. (p. 6 in Gray, J. ``A short life of Euler.'' BSHM Bull. 23 (2008), no.1, 1--12). In a similar vein, in a footnote on 18th century notation, Ferraro presents a novel claim that for 18th-century mathematicians, there was no difference between finite and infinite sums. (footnote 8, p. 294 in Ferraro, G. ``Some aspects of Euler's theory of series: inexplicable functions and the Euler-Maclaurin summation formula.'' Historia Mathematica 25, no. 3, 290--317.) Far from being a side comment, the claim is emphasized a decade later in the Preface to his 2008 book: a distinction between finite and infinite sums was lacking, and this gave rise to formal procedures consisting of the infinite extension of finite procedures (p. viii in Ferraro, G. The rise and development of the theory of series up to the early 1820s. Sources and Studies in the History of Mathematics and Physical Sciences. Springer, New York.) Grabiner doesn't hesitate to speak about shaky eighteenth-century arguments (p. 358 in Grabiner, J. ``Is mathematical truth time-dependent?'' Amer. Math. Monthly 81 (1974), 354--365); it is difficult to evaluate her claim since she does not specify the arguments in question. Instead of viewing Fraser's passage as problematic, Ferraro opens his book with it, which is surely a sign of endorsement. The attitude of disdain toward the masters seems to have permeated the field to such an extent that it has acquired the status of a sine qua non of a true specialist. In my study of Euler I have seen sophisticated arguments rather than proofs by example, except for isolated instances such as de Moivre's formula. On the other hand Euler's oeuvre is vast. Question. Can Euler be said to have proved theorems by example in other than a handful of exceptional cases, in any meaningful sense? Note 1. Some editors requested examples of what I described above as a disdainful attitude toward the masters of the past on the part of some historians. I provided a couple of additional ones. Editors are invited to provide examples they have encountered; I believe they are ubiquitous. Note 2. We tried to set the record straight on Euler in this recent article and also here. REPLY [7 votes]: Not sure if this answer adds anything to the ones already given. I write it because It is an example where Euler explicitly writes about the necessity of giving a proof, and more importantly, calls a proof given by himself "Attempt at a proof". The following is his remarks before his "attempt at a proof" of the sum of two squares in 1758, "On numbers which are the sum of two squares". Two years later, he has another paper with the title " Proof of Fermat’s Theorem That Every Prime Number of the Form 4n + 1 is the Sum of Two Squares". In a way, even the titles of these two papers suggest an answer to your question. All prime numbers which are sums of two squares, except 2, form this series: 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, 137, 149, etc. Not only are these contained in the form 4n + 1, but also, however far the series is continued, we find that every prime number of the form 4n+1 occurs. From this, we can conclude by induction 6 that it is likely enough that there is no prime number of the form 4n+ 1 which is not also a sum of two squares. Nevertheless, induction, however extensive, cannot fulfill the role of proof. Even if no one doubts the truth of the statement that all prime numbers of the form 4n+ 1 are sums of two squares, until now mathematics could not add this to its established truths. Even Fermat declared that he had found a proof, but because he did not publish it anywhere, we properly extend confidence toward the assertion of this most profound man, and we believe that property of the numbers, but this recognition of ours rests on pure faith without knowledge. Although I labored much in vain on a proof to be discarded, nevertheless I have discovered another argument to be given for this truth, which, even it if it is not fully rigorous, still appears to be equivalent to induction connected with nearly rigorous proof. The following is from the introduction of the second paper where Euler summarizes the first paper. I next put forth an attempt of the proof from which the validity of this theorem is revealed much more clearly, even if it should be set aside by the standards of rigorous proof.<|endoftext|> TITLE: Are there non-trivial infinite chains of adjoint functors? QUESTION [26 upvotes]: There are self-adjoint functors $A \dashv A$. There are also functors $A$ that are both left- and right-adjoint to another functor $B$. $$A \dashv B \dashv A$$ There are also finite cyclic chains of adjoint functors, that can have the initial functor equal to the final functor in the chain. $$A_0 \dashv A_1 \dashv \dotsb \dashv A_{n-1} \dashv A_0$$Not including those cases, are there nontrivial examples of infinite chains of adjoint functors, going infinitely to the left and infinitely to the right? $$ \dotsb \dashv A_{n-1} \dashv A_n \dashv A_{n+1} \dashv \dotsb $$ Does it make sense to consider such a chain as being an analogue of a long exact sequence, and if so, does it make sense to consider the monads $A_{n+1} \circ A_n$ and comonads $A_{n-1} \circ A_n$formed by composition of adjacent pairs to be at all analogous to homology or cohomology groups generated by long exact sequences? Is this a useful concept? Do these infinite chains appear often enough to warrant such an analogy? REPLY [4 votes]: My paper Non-symmetric $*$-autonomous categories. Theoretical Computer Science, {\bf139} (1995), 115-130, might be thought of as a generalization of Lambek's to Chu categories. It certainly leads to doubly infinite sequences of adjunctions.<|endoftext|> TITLE: The finiteness criterium $F$ under quasi-isometry QUESTION [13 upvotes]: A group $G$ is defined to have $F$ if there exists a finite $K(G,1)$. This property is clearly not invariant under quasi-isometry as one can see from the trivial group and $\mathbb{Z}_2$. My question: Is there also a torsion-free counterexample? Meaning are there two torsion-free, quasi-isometric groups of which one has $F$ and theother one not? REPLY [10 votes]: This is not an answer but a summary of what I know about this issue: The property $F_n$ is QI invariant (the groups here are not assumed to be torsion free). You can find a proof either in section 18.2 of R.Geoghan, "Topological methods in group theory" or in chapter 6 of my book with Cornelia Drutu. It is quite possible that some version of the arguments in these proofs will handle your question, it is worth checking. If a group $G$ has type $FP_n$ and is $n$-dimensional (here and below everything is over ${\mathbb Z}$) then $G$ has type $FP$. In particular, if a group has type $F_\infty$ and has finite cohomological dimension then it has type $FP$. If a finitely presented group $G$ has type $FL$ then there is a finite $K(G,1)$. This is in Brown's book "Cohomology of groups", Ch. 8.8. There are no known examples of finitely presented groups which are of type FP but not of type FL. Putting it all together and assuming that FP=FL, the positive answer to your question would follow from: Suppose that $G$ admits a finite $K(G,1)$ and $H$ is torsion free quasiisometric to $G$. Then $H$ has finite cohomological dimension. This is quite likely to be true, since cohomological dimension over ${\mathbb Q}$ is QI invariant (R.Sauer, "Homological invariants and quasi-isometry", GAFA, 2006).<|endoftext|> TITLE: Question on period map, Gauss-Manin connection and complex coordinates of $\mathcal{H}^1(k)$ QUESTION [6 upvotes]: Let $\mathcal{L}_g$ be the space of abelian differentials on Riemann surfaces of genus $g\ge 2$ and $\mathcal{TH}_g:=\mathcal{L}_g/Diff_0^+(S_g)$ be the Teichmuller space of abelian differentials on Riemann surfaces as it is defined in this article http://arxiv.org/pdf/1311.2758v1.pdf by Giovanni Forni and Carlos Matheus. Let $\mathcal{TH}(k)$ be a stratum of $\mathcal{TH}_g$. At page 20 they state: "It is possible to prove that every $\omega_0\in \mathcal{TH}(k)$ has an open neighborhood $U_0\subset\mathcal{TH}(k)$ such that, after identifying, for all $\omega\in U_0$, the cohomology $H^1(M,\Sigma(\omega),\mathbb{C})$ with the fixed complex vector space $H^1(M,\Sigma(\omega_0),\mathbb{C})$ via the Gauss-Manin connection (i.e., through identification of the integer lattices $H^1(M,\Sigma(\omega),\mathbb{Z}+i\mathbb{Z})$ and $H^1(M,\Sigma(\omega_0),\mathbb{Z}+i\mathbb{Z})$), the period map $\Theta : U_0 \rightarrow H^1(M,\Sigma(\omega_0),\mathbb{C})$ that is, the map defined by the formula $\Theta(\omega) :=( \gamma \mapsto\int_\gamma\omega )\in Hom(H^1(M, \Sigma(\omega), \mathbb{Z}); \mathbb{C}) \simeq H^1(M, \Sigma(\omega), \mathbb{C}) \simeq H^1(M, \Sigma(\omega_0), \mathbb{C})$ , is a local homeomorphism." In fact in the next page they also state that the period map $\Theta$ gives $\mathcal{TH}(k)$ the structure of a complex manifold. My question is really silly, but I'm not sure which are the complex coordinates given by $\Theta$. My guess is this: fix $\omega_0$ and $U_0$. Then choose a basis $\{\tau_1,\dots,\tau_{2g+\sigma-1}\}$ of $H_1(M,\Sigma(\omega_0),\mathbb{Z})$. Then for every $\omega\in U_0$, $\Theta(\omega)=(\int_{\tau_1}\omega,\dots,\int_{\tau_{2g+\sigma-1}}\omega)\in\mathbb{C}^{2g+\sigma-1}$. So the complex coordinates are $(\int_{\tau_1}\omega,\dots,\int_{\tau_{2g+\sigma-1}}\omega)$ and a change of atlas consists in a change of basis of $H_1(M,\Sigma(\omega_0),\mathbb{Z})$. Am I right? REPLY [3 votes]: Yes, $(\int_{\tau_1}\omega,...,\int_{\tau_{2g+\sigma-1}}\omega)$ are the complex coordinates 'given by $\Theta$'. In fact, the authors prove (using the Moser trick) that these integrals are locally injective. To prove that $\Theta$ maps to an open subset in $\mathbb C^{2g+\sigma-1}$ one observes that there exist closed complex-valued $\alpha_1,..,\alpha_{2g+\sigma-1}$ 1-forms such that they vanish identically on a neighborhood of the given set of zeros, and such that $(\int_{\tau_j}\alpha_i)$ is invertible. Coordinate changes are then given by either changing the basis of $H^1(M,\Sigma(\omega_0),\mathbb Z)$ or by changing $\Sigma(\omega_0)$, i.e. the Riemann surface and the abelian differential.<|endoftext|> TITLE: information measure for matrix that is analogous to rank QUESTION [5 upvotes]: Is there a measure for matrix that is analogous to rank of the matrix, but it is continuous on matrix elements? Say, we could say the information in identity matrix $I_n$ is $n$, and when the off-diagonal elements change from 0 to 1, the information contained in the matrix reduces continuously. Example: considering a matrix $A(a)$ \begin{pmatrix} 1&0&0&\\ 0&1&a\\ 0&a&1 \end{pmatrix}. How do I define an information measure $I(A(a))$ that is continuous to $a$ for matrix $A(a)$, so that $I(A(0)) = 3$ and $I(A(1)) = 2$? Rank is not continuous, Shannon information entropy on eigenvalues does not give desired values. REPLY [4 votes]: The notion of stable rank is often used in the low-rank matrix approximation literature to offer a more tractable surrogate for rank. This does not necessarily satisfy all your requirements, but might be still suitable for your purposes.<|endoftext|> TITLE: Are homology spheres stably parallelisable? QUESTION [21 upvotes]: A homology sphere is a closed smooth $n$-dimensional manifold with the same homology groups as $S^n$. Igor Belegradek's answer to a previous question of mine shows that the smoothness hypothesis is not necessary except in dimension four where it is not yet known whether every homology sphere is smoothable. If a homology sphere is simply connected, it follows from (one version of) Whitehead's Theorem that it is in fact homotopy equivalent to $S^n$, i.e. it is a homotopy sphere. It was shown by Kervaire and Milnor in Homotopy Groups of Spheres: I that homotopy spheres have stably trivial tangent bundle. Do homology spheres have stably trivial tangent bundle? If the answer is no, is there a way to see how the fundamental group obstructs stable triviality? REPLY [21 votes]: Yes, they have stably trivial tangent bundles. A remark to this effect can be found on page 70 of M. Kervaire "Smooth Homology Spheres and their Fundamental Groups" but it is a little terse. It is essentially the argument as for homotopy spheres, due to Kervaire--Milnor, with a little elaboration. Let me try to explain it. Let $M^n$ be a $\mathbb{Z}$-homology sphere. It is therefore orientable, so let $\tau : M \to BSO$ classify its stable tangent bundle. As the target is simply-connected, this map factors over the plus-construction $M^+ \simeq S^n$, giving a map $\tau' : S^n \to BSO$. The claim is that $\tau'$ is nullhomotopic, as then $\tau$ is too. We know the homotopy groups of $BSO$ by Bott periodicity: Sometimes $\pi_n(BSO)=0$, in which case we are done. Sometimes $\pi_n(BSO)=\mathbb{Z}$, namely when $n=4k$, in which case homotopy classes are detected by evaluation against the Pontrjagin class $p_k$. But $$\langle \tau', p_k \rangle = \langle [M], p_k(TM) \rangle$$ is a non-zero multiple of the signature of $M$ by Hirzebruch's signature theorem (as all other Pontrjagin classes must vanish on $M$), and $M$ has signature zero as it has no middle-dimensional homology. Thus we are done in this case too. Finally, sometimes $\pi_n(BSO)=\mathbb{Z}/2$. In this case let us reinterpret the element $\tau'$: by the plus-construction discussion it follows that $M$ is stably parallelisable away from a small disc, and it is of course also stably parallelisable on this disc. The difference of these two parallelisations on the boundary of the disc gives a map $\tau'' : S^{n-1} \to SO$, and this corresponds to $\tau'$ under $\pi_n(BSO) \cong \pi_{n-1}(SO)$. Consider the composition $J \circ \tau'' : S^{n-1} \to SO \to G$, where the second map is the $J$-homomorphism. Now $\pi_{n-1}(G) \cong \Omega_{n-1}^{fr}$ framed cobordism, and under this isomorphism $J \circ \tau''$ corresponds to $S^{n-1}$ with the framing which comes from changing the standard framing by the map $\tau''$. But by construction this framing on $S^{n-1}$ bounds, namely it bounds the framing we chose on $M$ minus a small disc: thus $J \circ \tau''=0 \in \pi_{n-1}(G)$. Now one invokes the theorem of Adams that the $J$-homomorphism is injective in these dimensions, so $\tau''$ and hence $\tau'$ and $\tau$ are nullhomotopic too.<|endoftext|> TITLE: What was Gödel's Constitutional Problem? QUESTION [10 upvotes]: It is well known that Kurt Gödel had doubts concerning the US constitution and believed that it somehow was inconsistent and opened up for a dictatorial grab. What was he thinking? REPLY [10 votes]: The history of how Gödel studied the US constitution in preparation for the citizenship exam is well documented by Jeffrey Kegler, as pointed out in the comments. Perhaps more relevant for MO (as opposed to HSM) is what the flaw in mathematical logic might have been that Gödel had noticed in the constitution. A scholarly account of the latter issue is given in Gödel's loophole by Enrique Guerra-Pujol (2012). The mathematician and philosopher Kurt Gödel reportedly discovered a deep logical contradiction in the US Constitution. What was it? Although no one knows with certainty of what Gödel’s loophole or “constitutional contradiction” consists, in view of Gödel's research interests it is reasonable to search for a problem of self-reference. We will refer to a constitutional design defect as “Gödelian” if the flaw is contained in a constitutional statement that is self-referential. . We have identified one particular “design defect” in the Constitution that qualifies as a “Gödelian” design defect. In summary, Gödel’s loophole is that the amendment procedures set forth in Article V self-apply to the constitutional statements in Article V themselves, including the entrenchment clauses in Article V. Furthermore, not only may Article V itself be amended, but it may also be amended in a downward direction (i.e., through an “anti-entrenchment” amendment making it easier to amend the Constitution). Lastly, the Gödelian problem of self-amendment or anti-entrenchment is unsolvable.<|endoftext|> TITLE: Generalizations of the handle trading techniques QUESTION [7 upvotes]: As Theorem 8.1 in "Lectures on the h-cobordism theorem (written by J.Milnor)" show, we can choose a handle decomposition of cobordism (satisfying some connectivity and dimensional assumptions) with no 0,1-handles, which sometimes we call this techniques handle trading (we actually did trading all 1-handles with the same number of 3-handles). Then, my question is the following : Can we generalize handle trading techniques (for instance, trading all 2-handles with 4-handles, 3-handles with 5-handles, …) under some higher-connectivity assumptions? That is all of my question. Thank you for your help. (I have edited the question to make it clear.) REPLY [8 votes]: Yes; see Smale, On the structure of manifolds (Amer. J. Math. 84 1962 387–399) where it's shown that in high dimensions, you can eliminate handles under various connectivity assumptions. The h-cobordism theorem is a special case. For index greater than 1, one usually doesn't do handle-trading, because you can in fact just do handle cancellation, which is simpler. The reason for doing handle trading for 1-handles (and dually, for (n-1)-handles) instead of cancellations is to avoid tricky issues related to presentations of the fundamental group, related to the Andrews-Curtis conjecture. REPLY [6 votes]: You might find the paper by C.T.C Wall: Geometrical connectivity I, J. London Math. Soc. 3 (1971), p. 597-604, interesting. What Wall proves, entirely by handle trading, is that if $W:M_0 \to M_1$ is an $n$-dimensional cobordism and the inclusion $M_0\to W$ is r-connected, then you can built W from M_0 using only handles of index $\geq r+1$, provided that $r \leq n-4$.<|endoftext|> TITLE: On describing a sort of "well-behaved" subgroups of a free abelian group QUESTION [7 upvotes]: I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finitely generated case. Let $M$ be a free abelian group and $N$ a subgroup of $M$. Suppose that $M/N$ is a direct sum of cyclic groups. Can we always find a basis $\{x_i\}_{i\in I}$ of $M$ such that $N=\bigoplus_{i\in I}(N\cap\mathbb Zx_i)$? My attempt: I feel it's true so I tried to prove it. Let $\varphi$ denote the canonical projection $M\to M/N$. To begin with, I'd like to consider a special case that $M/N$ is $p$-primary for a certain prime number $p$. First, define a property $P$ on elements of $M$: for any $x\in M$, $P(x)$ is true iff for any $y\in M$ and $n\in\mathbb Z$, $ny=x$ implies $y\in\ker\varphi$. Using Zorn's Lemma, we can find a maximal submodule $L_1$ of $M$ consisting of elements satisfying $P$. Second, define a property $Q$ on submodules of $M$: for any $L\leqslant M$, $Q(L)$ is true iff $L$ is generated by some $\{a_i\}_{i\in I}$ such that $M/N=\bigoplus_{i\in I}\mathbb Z\varphi(a_i)$. I proved that $\{a_i\}_{i\in I}$ constructed above is always linearly independent (by considering at a time a certain finitely generated subgroup of $M$). I want to show that there is a maximal submodule $L_2$ of $M$ subject to $Q$ and that $M=L_1\oplus L_2$, but I got stuck here. Any thought will be appreciated. REPLY [2 votes]: It is proved in Joel M. Cohen and Herman Gluck, MR 254028 Stacked bases for modules over principal ideal domains, J. Algebra 14 (1970), 493--505, that the answer is yes. They credit Kaplansky for asking the question. The paper also contains an example which seems to show why a Zorn's Lemma argument along the lines of the OP can't work, though I haven't checked that this is exactly what the example shows.<|endoftext|> TITLE: DG categories in algebraic geometry - guide to the literature? QUESTION [29 upvotes]: Although my experience with DG categories is pretty basic I find them to be a very neat tool for organizing (co-)homological techniques in algebraic geometry. For someone who has algebro-geometric application in mind they seem more attractive then stable $(\infty,1)-$categories which seem to carry data in a slightly more convoluted way (which I realize makes for more powerful techniques and generalizations). So far though I've only seen a very limited amount of how actual algebraic geometry looks from a DG point of view and most of the stuff I read about DG categories was either definitions or general theory (papers by Toen for example). Here are several questions I have in mind: What is the "correct" DG category associated to a scheme/algebraic space/stack? Can the different possiblities here be organized as different "DG-stacks" (of certain dg-categories of sheaves) on the relevant site? How can I see the classical "category" of derived categories as some kind of "category" of homotopy categories of dg-categories? (I'm putting category in brackets since i'm not sure that there's such an object, what I really want is to really understand the link between all the classical theory of derived categories and dg-categories). In particular the six functor formalism. I realize that these question might not have a straight yes/no answer and so what I'm looking for is a kind of roadmap to the relevant litrature where the application and formalization of the place of dg-categories in algebraic geometry is discussed. Main question: What are some relevant articles/notes/books which establish and discuss the details of the formalism of DG-categories in the algebro-geometric world? REPLY [21 votes]: Let me try to address the bulleted questions and simultaneously advertise the G-R book everyone has mentioned. Since the main question was about literature, I could also mention Drinfeld's article "DG quotients of DG categories," which nicely summarizes the state of the general theory before $\infty$-categories shook everything up. However, it doesn't contain any algebraic geometry. If $X = \text{Spec } A$ is an affine scheme, it's reasonable to define the category of quasicoherent sheaves $\text{QCoh}(X) := A\text{-mod}$ as the category of $A$-modules. Any other definition (e.g. via Zariski sheaves) must reproduce this answer anyway. If we understand this as the derived category of $A$-modules, then there is a canonical DG model: the homotopically projective complexes in the sense of Drinfeld's article. The next step is to construct $\text{QCoh}(X)$ for $X$ not necessarily affine. So write $X = \cup_i \text{Spec } A_i$ as a union of open affines (say $X$ is separated to simplify things). It would be great if we could just "glue" the categories $A_i\text{-mod}$, the way that we compute global sections of a sheaf as a certain equalizer. Concretely, a complex of sheaves on $X$ should consist of complexes of $A_i$-modules for all $i$, identified on overlaps via isomorphisms satisfying cocycle "conditions" (really extra data). This is the kind of thing that totally fails in the triangulated world: limits of 1-categories just don't do the trick. Even if we work with the DG enhancements, DG categories do not form a DG category, so this doesn't help. As you might have guessed, this is where $\infty$-categories come to the rescue. Let me gloss over details and just say that there is a (stable, $k$-linear) $\infty$-category attached to a DG category such as $A$-mod, called its DG nerve. If we take the aforementioned equalizer in the $\infty$-category of $\infty$-categories, then we do get the correct $\infty$-category $\text{QCoh}(X)$, in the sense that its homotopy category is the usual derived category of quasicoherent sheaves on $X$. (Edit: As Rune Haugseng explains in the comments, it's actually necessary to take the limit of the diagram of $\infty$-categories you get by applying $\text{QCoh}$ to the Cech nerve of the covering. The equalizer is a truncated version of this.) But, you might be thinking, I could have just constructed a DG model for $\text{QCoh}(X)$ using injective complexes of Zariski sheaves or something. That's true, and obviously suffices for tons of applications, but as soon as you want to work with more general objects than schemes you're hosed. True, there are workarounds using DG categories for Artin stacks, but the theory gets very technical very fast. If we instead accept the inevitability of $\infty$-categories, we can make the following bold construction. A prestack is an arbitrary functor from affine schemes to $\infty$-groupoids (i.e. spaces in the sense of homotopy theory). For example, affine schemes are representable prestacks, but prestacks also include arbitrary schemes and Artin stacks. Then for any prestack $\mathscr{X}$ we can define $\text{QCoh}(\mathscr{X})$ to be the limit of the $\infty$-categories $A\text{-mod}$ over the $\infty$-category of affine schemes $\text{Spec } A$ mapping to $\mathscr{X}$. A cofinality argument for Zariski atlases shows this agrees with our previous definition for $\mathscr{X}$ a scheme. For example, if $\mathscr{X} = \text{pt}/G$ is the classifying stack of an algebraic group $G$, then the homotopy category of $\text{QCoh}(\mathscr{X})$ is the derived category of representations of $G$. Even cooler: if $X$ is a scheme the de Rham prestack $X_{\text{dR}}$ is defined by $$\text{Map}(S,X_{\text{dR}}) := \text{Map}(S_{\text{red}},X).$$ Then, at least if $k$ has characteristic zero, our definition of $\text{QCoh}(X_{\text{dR}})$ recovers the derived category of crystals on $X$, which can be identified with $\mathscr{D}$-modules. So we put two different ``flavors" of sheaf theory on an equal footing.<|endoftext|> TITLE: Origin of the name ''momentum map'' QUESTION [19 upvotes]: Why is the momentum map in the differential geometry of symmetries called the ''momentum'' (or ''moment'') map? REPLY [24 votes]: According to §1.3 and §11.2 of Marsden and Ratiu [1994] (see detailed citation given below), the momentum map concept can be traced back to Sophus Lie's 1890 book, and is an English translation of the French words application moment. To quote directly from Marsden and Ratiu: The notion of the momentum map (the English translation of the French words “application moment”) also has roots going back to the work of Lie. Many authors use the words “moment map” for what we call the “momentum map.” In English, unlike French, one does not use the phrases “linear moment” or “angular moment of a particle,” and correspondingly, we prefer to use “momentum map.” We shall give some comments on the history of momentum maps in §11.2. Add For additional clarification, I asked Tudor Ratiu about the origins of the term momentum map. In an email dated Sept 8th 2016, he gave a beautiful response on the evolution of ideas that led to the modern momentum map. At the very end, he made some comments on what Sophus Lie did. Here are the main points from his email, whose contents can mostly be found in the historical notes provided in Marsden and Ratiu [1994]. References are given at the bottom. (1) The momentum map is a generalization of the usual notions of linear and angular momentum from physics, which have a long history. (2) The first person to introduce the modern version of the momentum map was Bertram Kostant at a conference in Japan. He did not give it a name. He used it to prove what is today called Kostant's symplectic covering theorem. (3) Just a few weeks later, Jean-Marie Souriau independently introduced the modern momentum map. He correctly named it in French application moment. He also realized its physical significance, and linked it to linear and angular momentum. This was an enormous breakthrough. (4) Marsden and Weinstein, in their famous 1974 paper on reduction used the term moment map, which was their translation of the French words application moment. This had an immediate reaction from Hans Duistermaat who pointed out that this is a misnomer and physically incorrect. Indeed, even the wiki entry for moment, indicates in the first line that moment should not be confused with momentum. It's analogous to confusing force with linear momentum. (5) For a circle action, the momentum map (without any name) was introduced by Theodore Frankel in an Annals paper in the 1950s. This paper is famous because it links the existence of fixed points of the action to the Hamiltonian character of the action. This bit of history was not known to Marsden and Ratiu when they wrote their book, so it won't be found there. (6) Sophus Lie knew a lot of Poisson geometry and many people now regard him as the founder of Poisson geometry. However, Lie did not introduce symplectic geometry. The link to symplectic geometry, coadjoint orbits, etc. is due to Kostant and Souriau. See the historical notes in Marsden and Ratiu [1994] for details. References Frankel, T. [1959]. Fixed points on Kahler manifolds. Ann. Math. 70, 1–8. Marsden, Jerrold E. and Ratiu, Tudor [1994]. Introduction to Mechanics and Symmetry. Second Edition, 1999. Current 2nd Printing, 2003. New York, Springer-Verlag. Kostant, B. [1966] Orbits, symplectic structures and representation theory. Proc. US– Japan Seminar on Diff. Geom., Kyoto. Nippon Hyronsha, Tokyo 77. Ratiu, Tudor. Personal Communication, Sept 8th, 2016. Souriau, J.-M. [1970]. Structure des systèmes dynamiques, Maîtrises de mathématiques, Dunod, Paris, 1970. ISSN 0750-2435.<|endoftext|> TITLE: A $p$-adic sum of reciprocals of powers QUESTION [6 upvotes]: Let $p$ be a prime number and $k\geq 2$ an even integer. Consider the following $p$-adic integer: $$ S_{p,k} := \lim_{r\to+\infty} \sum_{a=1}^{p^r} \big(\frac{p^r}{a}\big)^k $$ Convergence is easy to see. This can also be written as $$ S_{p,k} = \frac{1}{2} \sum_{0\neq x \in B_p} x^{-k} $$ where $$ \begin{aligned} B_p &:= \big\{\frac{a}{p^r}\; : \; a\in\mathbb{Z},\; r\in\mathbb{N},\; |a|_{\infty} \leq p^r\big\}\\ &=\{x\in\mathbb{Q} : |x|_v \leq 1 \text{ at all places } v\neq p\}\\ \end{aligned} $$ (Here, $|\cdot|_\infty$ is the usual real absolute value, and “places” [of $\mathbb{Q}$] mean either a prime $\ell$ or the symbol $\infty$.) The point of this last presentation is that it suggests the analogy with $B_\infty = \mathbb{Z}$, where $S_{\infty,k} = \sum_{n=1}^{+\infty} n^{-k}$ is the usual Riemann zeta function (at even integers): in a certain sense, the $S_{p,k}$ are a $p$-adic analogue of the $\zeta(k)$ (though probably not the most intelligent or satisfactory analogue). Question(s): Have these $S_{p,k}$ appeared in the literature? Do they have a name? Do they satisfy some known relations? (E.g., can $S_{p,4}$ be expressed in function of $S_{p,2}^2$? Experimentally, the ratio $S_{p,4}/S_{p,2}^2$ does not appear to be rational, but maybe something else can be said.) For what it's worth, Wolstenholme's theorem $\sum_{a=1}^{p-1}\frac{1}{a^2}\equiv 0\pmod{p}$ tells us that $S_{p,2} \equiv 1\pmod{p^4}$. There seem to be many generalizations in various directions (see, e.g., here) involving Bernoulli numbers, Kummer congruences, or the Kubota-Leopoldt $p$-adic zeta function, but I couldn't find a direct connection with the above quantity. REPLY [3 votes]: As far as I know this quantity does not have a name. There is a mention of a general family of $p$-adic limits, of which $S_{k,p}$ is a special case, on p. 31 of [1]. The quantity $S_{k,p}$ can be expressed as an infinite series involving the Kubota-Leopoldt $p$-adic zeta function $\zeta_p(n):=L_p(n,\omega_p^{1-n})$: $$ S_{p,k}=1-\sum_{n\geq 1}{-k\choose n}\left(1-p^{-(n+k)}\right)^{-1}\zeta_p(n+k). $$ This can be derived from the main formula of [2], and something much more general appears in [1]. The even $p$-adic zeta values vanish, and the odd ones are expected to be algebraically independent. We can define a sort of formal version of $S_{k,p}$ by $$ s_{k} := 1-\sum_{\substack{n\geq 1\\n+k\text{ odd}}}{-k\choose n} z_{n+k} \in \mathbb{Q}[[z_3,z_5,\ldots]]. $$ One can check that the elements $s_k$ are algebraically independent in $\mathbb{Q}[[z_3,z_5,\ldots]]$ (since they are homogeneous of degree $1$, it suffices to check they are linearly independent, and this is true because the coefficients grow like polynomials of different degrees). So at least there aren't any "obvious" algebraic relations among the $S_{k,p}$ coming from the expression in terms of $\zeta_p$. My guess is that the $S_{k,p}$ are algebraically independent. [1] David Jarossay, Algebraic relations, Taylor coefficients of hyperlogarithms and images by Frobenius - III. arXiv:1601.01159 [2] Lawrence C. Washington, MR 1611077 $p$-adic $L$-functions and sums of powers, J. Number Theory 69 (1998), no. 1, 50--61.<|endoftext|> TITLE: Is the dimension of $V//G$ always the same as the dimension of $V^*//G$? QUESTION [9 upvotes]: I would like to know whether there is an example of a reductive algebraic group $G$ (say, over the complex numbers $\mathbb{C}$) and a finite dimensional representation $V$ of $G$ such that dim$(V//G)$ is different from dim$(V^*//G)$. Here $V//G=Spec\ \mathbb{C}[V]^G$ is the categorical quotient and $V^*$ is the dual representation of $G$. More generally, I am interested in the relations between the structure of $G$-orbits of $V$ and $V^*$, so if everyone knows a good reference about this topic, please let me know. Thank you in advance. REPLY [15 votes]: If $G$ is connected and $T\subseteq G$ is a maximal torus then there is an involution $\theta:G\to G$ with $\theta(t)=t^{-1}$ for all $t\in T$. It has the property that as a representation $V^*$ is isomorphic to the $\theta$-twisted representation $V$, i.e., to $G\overset\theta\to G\to GL(V)$. To see this just look at the highest weights. Thus there is an isomorphism $\phi:V\overset\sim\to V^*$ with $\phi(gv)=\theta(g)\phi(v)$. This induces $V^*/\!/G=V/\!/\theta(G)=V/\!/G$. Moreover, the $G$-orbit structures of $V^*$ and $V$ are the same. If $G$ is not connected, I am not so sure what to do. If the ground field is $\mathbb C$ then one can argue as follows: Let $K\subseteq G$ be a maximal compact subgroup and choose a $K$-invariant Hermitian scalar product on $V$. Since $K$ is Zariski dense in $G$ one has $\tau(G)=G$ where $\tau(g):=(g^{-1})^\dagger$ and $g^\dagger$ is the adjoint with respect to the scalar product. This induces an isomorphism $\phi:\overline V\to V^*$ with $\phi(gv)=\tau(g)\phi(v)$. So, topologically, $V^*$ and $V$ are equivariantly the same. In particular $V^*/\!/G=\overline{V}/\!/G=V/\!/G$ where the last equality is an antiholomorphic isomorphism. Edit: Here is a general Theorem. Let $k$ be any field, $G$ a reductive $k$-group and $V$ a finite dimensional representation of $G$. Then $\dim V/\!/G=\dim V^*/\!/G$. First reduction: One may assume that $k$ is algebraically closed field since field extension commutes with taking invariants. Second reduction: One may assume assume that $G$ is connected (obvious). Now observe that the involution $\theta$ above exists over $k$. The same argument shows that the assertion holds if $V$ is irreducible or, more generally, if $V$ is completely reducible. Now the theorem follows from Theorem. Same assumptions as above. Let $U_1,\ldots,U_r$ be the composition factors of $V$ and $\overline V:=\bigoplus_{i=1}^rU_i$. Then $\dim V/\!/G=\dim\overline V/\!/G$. This is seen by a deformation argument. Let $0=V_0\subset V_1\subset\ldots\subset V_r=V$ be a composition series. Put $$\tilde V:=\oplus_{i=1}^\infty V_i\otimes_kt^i\subseteq V\otimes_kk[t]$$ where $V_i=V$ for $i\ge r$. Then $\tilde V$ is a $G$-vector bundle over $\mathbf A^1$ whose general fiber is $V$ and whose zero-fiber is $\overline V$. Let $\mathbf V\to\mathbf A^1$ be its geometrical realization such that $\mathbf V_1=V$ and $\mathbf V_0=\overline V$. The quotient $\mathbf V/\!/G$ is faithfully flat over $\mathbf A^1$. In particular, the fiber dimensions are constant: $\dim(\mathbf V/\!/G)_1=\dim(\mathbf V/\!/G)_0$. Geometric reductivity implies that the morphisms $V/\!/G\to(\mathbf V/\!/G)_1$ and $\overline V/\!/G\to(\mathbf V/\!/G)_0$ are completely inseparable. In particular the dimensions are the same. This shows the claim.<|endoftext|> TITLE: Which of these 4 definitions of Galois coverings of integral schemes are equivalent? QUESTION [9 upvotes]: Here are four possible definitions for an etale, finite, surjective map $X\rightarrow Y$ between integral schemes to be considered Galois: There exists a finite group $G$, and an action $\varphi: G\times X\rightarrow X$, so that the induced map $\varphi\times p_2: G\times X\rightarrow X\times_Y X$ is an isomorphism. There exists a finite group $G$, and an action $\varphi: G\times X\rightarrow X$, so that $G$ acts freely an transitively on the geometric fibers of $X\rightarrow Y$. There exists a finite group $G$, and an action $\varphi: G\times X\rightarrow X$, so that the extension of function fields $\kappa(X)/\kappa(Y)$ is Galois with group isomorphic to $G$. The extension of function fields $\kappa(X)/\kappa(Y)$ is Galois. Obviously, 1 implies 2 implies 3 implies 4. Note that the difference between 3 and 4 is that in 4 you are not guaranteed that the action extends from the generic point to $X$. I am inclined to believe that 4 does not imply 3, but I can't think of a counterexample. I think that definitions 1 and 2 are probably equivalent. This is because definition 2 is probably the same as saying that $\varphi\times p_2$ is a bijection on geometric points, which, at least if $X$ and $Y$ are varieties, should imply that it is an isomorphism. I am completely in the dark about whether or not definitions 2 and 3 are equivalent. I'm not sure what I should believe... So: Questions Is 1 equivalent to 2? Is 2 equivalent 3? What is a counterexample for the equivalence of 3 and 4? Is it at all helpful to ask that $X$ and $Y$ be regular for any of these definitions? Any other adjectives that need adding? REPLY [13 votes]: I can prove $(3) \Rightarrow (1)$ when $X$ and $Y$ are irreducible (Lemma 1 below; no integrality assumptions needed. Irreducible is needed for the question to make sense), and I can prove $(4) \Rightarrow (1)$ when $X$ and $Y$ are normal, connected, and Noetherian (in particular integral, see Tag 033M) (Lemma 2 below). I do not know how to approach the general case if $Y$ is not normal; see the remark at the end. Maybe you can weaken it to something like unibranch? The property that I need is that a connected finite étale cover of $Y$ is always integral; I don't know in what generality this is true. Reference. A good introductory reference to some of the stuff I'm using is the chapter Fundamental groups of schemes of the Stacks project. (I make precise references as I go along.) Notation. Let $\eta_X$ (resp. $\eta_Y$) be the generic point of $X$ (resp. $Y$). I will use the following definition. Definition. Let $\phi \colon X \to Y$ be a morphism of schemes, let $G$ be a finite group (viewed as constant étale group scheme), and let $G \times X \to X$ be an action of $Y$-schemes. Then $\phi$ is Galois with group $G$ if the induced map $G \times X \to X \times_Y X$ is an isomorphism. If $F \colon \operatorname{FÉt}_Y \to \operatorname{Set}$ is a fibre functor (notably, $F = F_{\bar y}$ associated to taking the geometric fibre over a geometric point $\{\bar y\} \to Y$), there is the following equivalent notion of being Galois. Auxiliary lemma. Let $\phi \colon X \to Y$ be a finite étale morphism of connected schemes. Then $|\!\operatorname{Aut}_Y(X)| \leq |F(X)|$, and equality holds if and only if $\phi$ is Galois. Proof. Let $G = \operatorname{Aut}_Y(X)$, with the natural action on $X$, and consider the induced map $$\psi \colon G \times X \to X \times_Y X.$$ Applying the fibre functor $F = F_{\bar y}$ gives \begin{align*} F(\psi) \colon G \times F(X) &\to F(X) \times F(X)\\ (\sigma, s) &\mapsto (\sigma(s),s). \end{align*} By Tag 0BN0 (7), this map is injective, and it is bijective if and only if $|\!\operatorname{Aut}_Y(X)| = |F(X)|$. Thus we need to show that $\psi$ is an isomorphism if and only if $F(\psi)$ is a bijection. This follows since $F$ reflects isomorphisms (see Tag 0BMY and Tag 0BNB). $\square$ Lemma 1. Let $\phi \colon X \to Y$ be a finite étale morphism of irreducible schemes. Assume there exists a finite group $G$ and an action $G \times X \to X$ whose base change to $\{\eta_X\}$ makes $\{\eta_X\} \to \{\eta_Y\}$ a Galois extension with group $G$. Then $X \to Y$ is Galois with group $G$. Proof. By the auxiliary lemma, we only have to show that $X$ has enough automorphisms. This is now a simple degree/counting argument (the elements of $G$ providing the automorphisms). $\square$ Lemma 2. Let $\phi \colon X \to Y$ be a finite étale morphism of connected normal Noetherian schemes, and assume that the extension $\{\eta_X\} \to \{\eta_Y\}$ is Galois. Then $\phi$ is Galois. Proof. We want to apply Tag 0BN6 $(3) \Rightarrow (2)$ to the pullback functor \begin{align*} H \colon \operatorname{FÉt}_Y &\to \operatorname{FÉt}_{\eta_Y}\\ Z &\mapsto Z_{\eta_Y}. \end{align*} Assume $Z$ is a connected scheme finite étale over $Y$. Then $Z$ is normal (Tag 025P), connected, and Noetherian; hence integral (Tag 033M). Hence, it only has one generic point $\eta_Z$, and this is the only point lying over $\eta_Y$. Thus, $H(Z)$ is connected, so Tag 0BN6 $(3) \Rightarrow (2)$ implies that $H$ is fully faithful. Note that $H$ also reflects isomorphisms, since $F_{\overline{\eta_Y}}$ does, which factors through $H$. For a fully faithful functor that reflects isomorphisms, we have $$\operatorname{Aut}_Y(X) = \operatorname{Aut}_{H(Y)}(H(X)),$$ so $X$ has enough automorphisms since $H(X)$ does so. $\square$ Remark. We see that $H$ is fully faithful if and only if for every connected étale $Y$-scheme $Z$, the generic fibre $Z_{\eta_Y}$ is just a point. This is certainly the case if $Z$ is integral, and I think the converse holds as well. (See below for an example where $Z$ has multiple components.) Also, we might not need the full strength of $H$ being fully faithful: we only need to get back all automorphisms. Example. Let $A = \mathbb R[x,y]/(x^2+y^2)$. Then $A$ is a domain. It has a finite étale extension $$B = \mathbb C[x,y]/(x^2+y^2) = \mathbb C[x,y]/(x+iy) \times \mathbb C[x,y]/(x-iy),$$ which is connected but not integral. The fibre above the generic point consists of two points. Correspondingly, if $Y = \operatorname{Spec} A$, the functor $$H \colon \operatorname{FÉt}_Y \to \operatorname{FÉt}_{\eta_Y}$$ is not fully faithful, by Tag 0BN6 again. I do not know whether there exists an integral finite étale $A$-algebra $B$ for which $\operatorname{Frac}(A) \to \operatorname{Frac}(B)$ is Galois, but $A \to B$ is not.<|endoftext|> TITLE: Why sum of three squares of real polynomials is a sum of two squares? QUESTION [15 upvotes]: If $f(x),g(x)$ are real polynomials, then $f^2+g^2+1$ is a sum of two squares of polynomials. This easily follows from Fundamental Theorem of Algebra, but is there an argument avoiding it? What are fields for which such a statement holds? REPLY [9 votes]: Theorem 4.2 from this paper says that in general you will need the coefficients of the two squares to be algebraic numbers of exponential, in $\deg (f^2+g^2+1)$, degrees. More precisely, if the Galois group of $f^2+g^2+1\in\mathbb{Q}[x]$ is the symmetric group $S_{d}$, then you will need algebraic numbers of degree exponential in $d$.<|endoftext|> TITLE: Structure of a real 3x3 positive-semidefinite matrix whose eigenvalues verify the triangle inequalities QUESTION [5 upvotes]: It is known that a 3 by 3 real symmetric matrix $A$ has an eigendecomposition $$ A = Q E Q^T $$ where $Q$ is an orthogonal matrix and $E$ is a diagonal matrix whose elements, $E_{11}$, $E_{22}$ and $E_{33}$, are the eigenvalues of $A$. Moreover, if those eigenvalues are non-negative then $A$ is positive-semidefinite. The question is: if those eigenvalues are not only non-negative but also verify the triangle inequalities $$ \begin{aligned} E_{11} + E_{22} &\geq E_{33}\\ E_{11} + E_{33} &\geq E_{22}\\ E_{22} + E_{33} &\geq E_{11} \end{aligned} $$ is there anything special about the structure of $A$ besides the fact that it is positive-semidefinite? Can those extra triangle inequalities constraints be written as functions of the elements of $A$, just like the positive-semidefinite constraints can be written down using the Sylvester's criterion? Can those extra constraints be somehow represented in a semidefinite programming / linear matrix inequalities framework? REPLY [8 votes]: Condition $$E_{11}+E_{22}\ge E_{33}$$ is equivalent to $$E_{11}+E_{22}+E_{33}\ge 2 E_{33}$$ or $$E_{33} \le \frac12 tr\ A$$ since the sum of eigenvalues is equal to the trace. Combining with the other two conditions gives $$\lambda_{max}(A) \le \frac12 tr\ A$$ which is semidefinite representable as $$A \preceq \frac12(A_{11}+A_{22}+A_{33}) I$$<|endoftext|> TITLE: On the sum of uniform independent random variables QUESTION [22 upvotes]: Let $X_1,...,X_n$ be independent uniform random variables in [0,1] and assume $c>1/2$. Is it true that $$\mathbb{P}\left[\sum_{i=1}^n X_i \leq n \cdot c\right]$$ is increasing with respect to $n$? I know that the sum of uniform independent random follows Irwin–Hall distribution, but it seems hard to work with the pdf. Maybe a coupling argument? I want to stress that we want to show the result for all $n$, not only for $n$ sufficiently large. REPLY [5 votes]: The answer is yes. Let $S_n:=\sum_{i=1}^n X_i$, $x\in\mathbb R$, $n=2,3,\dots$, and \begin{equation} G_n(x):=P(S_n/n\le x)=\frac1{n!}\,\sum_j(-1)^j\binom nj (nx-j)_+^n, \end{equation} where $u_+:=0\vee u$; cf. [Irwin--Hall distribution]. The summation here is over all integers $j$, with the usual convention that $\binom nj=0$ if $j\notin\{0,\dots,n\}$. Let \begin{equation} D_n(x):=G_{n+1}(x)-G_n(x). \end{equation} We have to show that $D_n(x)\ge0$ for $x\ge1/2$. Clearly, $D_n$ is $n-1$ times continuously differentiable, $D_n=0$ outside $(0,1)$, and $D_n(1-x)=-D_n(x)$ (symmetry). For small $x>0$, one has $G_n(x)=\frac1{n!}\,n^nx^n>\frac1{(n+1)!}\,(n+1)^{n+1}x^{n+1}=G_{n+1}(x)$, whence $D_n(x)<0$, whence $D_n>0$ in a left neighborhood of $1$, with $D_n(1)=0$. Also, $D_n(1/2)=0$. So, it suffices to show that $D_n$ has no roots in $(1/2,1)$. Suppose the contrary. Then, by the symmetry, $D_n$ has at least $3$ roots in $(0,1)$. Then, by the Rolle theorem, the derivative $G_n^{(n-1)}$ of $G_n$ of order $n-1$ has at least $3+n-1=n+2$ roots in $(0,1)$. This contradicts the fact (proved below) that, for each integer $j$ such that $n-1\ge j\ge\frac{n-1}2$, $D_n$ has exactly one root in interval $$h_{n,j}:=[\tfrac jn,\tfrac{j+1}n).$$ Indeed, take any $x\in[1/2,1]$. It is not hard to see that, for $j_x:=j_{n,x}:=\lfloor nx\rfloor$, \begin{equation} G_n^{(n-1)}(x)=n^{n-1}\sum_{j=0}^{j_x}(-1)^j\binom nj (nx-j) \end{equation} \begin{equation} =\frac{n^{n-1}}{n-1}(-1)^{j_x}\binom n{j_x+1}({j_x}+1) ((n-1)x-{j_x}). \end{equation} Similarly, for $k_x:=j_{n+1,x}$, \begin{equation} G_{n+1}^{(n-1)}(x)=\frac{(n+1)^{n-1}}{2n(n-1)} (-1)^{k_x}\binom{n+1}{k_x+1}(k_x+1)P(n,k_x,x), \end{equation} where \begin{equation} P(n,k,x):=-2 k \left(n^2-1\right) x+k (k n-1)+n \left(n^2-1\right) x^2. \end{equation} Note that $k_x\in\{j_x,j_x+1\}$. Now we have to consider the following two cases. Case 1: $k_x=j_x=j\in[\frac{n-1}2,n-1]$, which is equivalent to $x\in h'_{n,j}:=[\frac jn,\frac{j+1}{n+1})$. It also follows that in this case $j\ge n/2$. In this case, it is not hard to check that $D_n(x)$ equals $(-1)^j P_{n,j,1}(x)$ in sign, where \begin{equation} P_{n,j,1}(x):=2 j n^n (n-j)+x \left(-2 (n-1) n^n (n-j)-2 j (n+1)^n \left(n^2-1\right)\right)+j (n+1)^n (j n-1)+(n+1)^n \left(n^2-1\right) n x^2, \end{equation} which is convex in $x$. Moreover, $P_{n,j,1}(\frac jn)$ and $P_{n,j,1}(\frac{j+1}{n+1})$ each equals $2 n^n - (1 + n)^n<0$ in sign. So, $P_{n,j,1}<0$ and hence $D_n$ has no roots in $h''_{n,j}=[\frac jn,\frac{j+1}{n+1})$. Case 2: $k_x=j+1$ and $j_x=j\in[\frac{n-1}2,n-1]$, which is equivalent to $x\in h''_{n,j}:=[\frac{j+1}{n+1},\frac{j+1}n)$. In this case, it is not hard to check that $D_n(x)$ equals $(-1)^{j+1} P_{n,j,2}(x)$ in sign, where \begin{equation} P_{n,j,2}(x):=(j+1) \left((n+1)^n (j n+n-1)-2 j n^n\right)+2 (j+1) \left((n-1) n^n-(n+1)^n \left(n^2-1\right)\right) x+n \left(n^2-1\right) (n+1)^n x^2, \end{equation} which is convex in $x$. Moreover, $P_{n,j,1}(\frac{j+1}n)$ and $-P_{n,j,1}(\frac{j+1}{n+1})$ each equals $2 n^n - (1 + n)^n<0$ in sign. So, $P_{n,j,2}$ has exactly one root in $h''_{n,j}$ and hence so does $D_n$. Since the interval $h_{n,j}$ is the disjoint union of $h'_{n,j}$ and $h''_{n,j}$, $D_n$ has exactly one root in $h_{n,j}=[\tfrac jn,\tfrac{j+1}n)$, and the proof is complete.<|endoftext|> TITLE: Create a graph with a specified number of spanning trees QUESTION [5 upvotes]: I read that one of the current challenging problems in mathematics is constructing a minimal graph with a specified number of spanning trees (say, $k$). However, is there a quick way to create some graph $G$ (not necessarily minimal) that has $k$ spanning trees ? We can compute the number of spanning trees of a graph $G$ using the pseudo-determinant of the Laplacian matrix of $G$ and the number of labelled vertices (Kirchoff's theorem). Perhaps answering the following question helps in answering the original question: is there a way to generate a Laplacian matrix given its pseudo-determinant value? Thanks in advance! REPLY [12 votes]: A $k$-cycle works if $k>2$. For $k=1$ any tree works. I don't think $k=2$ is possible unless you allow double edges: if the graph is not a tree then it has an $m$-cycle $C$ for some $m>2$; remove edges off $C$ until the graph is connected but has no cycle other than $C$, and then removing an arbitrary edge of $C$ yields at least $m$ spanning trees. REPLY [6 votes]: If $G_1$ and $G_2$ are graphs let $G_1 \vee G_2$ denote their wedge sum. That is, $G_1 \vee G_2$ is obtained by taking the one-point union of $G_1$ and $G_2$. It will not matter what vertices we decide to identify. If $G_1$ and $G_2$ have $k_1$ and $k_2$ spanning trees respectively, then $G_1 \vee G_2$ has $k_1k_2$ spanning trees. As Noam points out a $k$-cycle has $k$ spanning trees for $k \geq 3$. So, we have a graph with $k$ spanning trees on $k$ vertices. When $k$ is composite (with some caveats for the prime $2$) we can quickly do better using the above construction. For example, if we want $k = 9$ we take $K_3 \vee K_3$ which has $5$ vertices instead of $9$.<|endoftext|> TITLE: Interaction of Grothendieck Construction with Coherent Nerve QUESTION [8 upvotes]: There are a number of Grothendieck constructions: one for discrete categories, one for enriched categories (see Tamaki's paper here) and one for quasicategories (see the Unstraightening and Straightening correspondence of Lurie's Higher Topos Theory that goes between certain model categories whose underlying quasicategories are the ones of simplicial sets over a fixed simplicial set, and functors from that simplicial set to the quasicategory of quasicategories). Moreover, there is a construction that takes a simplicially enriched category whose morphism simplicial sets are Kan complexes and produces a quasicategory, namely the homotopy coherent nerve. My question is about how the coherent nerve construction interacts with the enriched Grothendieck construction and the Grothendieck construction on simplicial sets described by Lurie. Now, suppose I have a pseudofunctor from a diagram category to simplicially enriched categories $F:I\to sCat$. The enriched Grothendieck construction gives me a functor $C\to I$, where $C$ is a simplicially enriched category. We can extend this functor along the homotopy coherent nerve $hN:sCat\to sSet$, and by Proposition 3.2.5.18 of Higher Topos Theory, we know that there is a corresponding morphism $X\to N(I)$ where $X$ is a simplicial set. How does this object relate to the coherent nerve applied to $C$, i.e. the morphism of simplicial sets $N(C)\to N(I)$, where $I$ is thought of as a simplicially enriched category by thinking of its morphism sets as 0-dimensional simplicial sets? REPLY [4 votes]: Answering this question took me some time. First of all, Liang Ze Wong and I had to write down a version of the enriched Grothendieck construction that worked for these purposes, as Tamaki's construction, which I linked to above, didn't quite work. That paper can be found here. Next, so long as we assume that $I$ is an ordinary category (it seems to take more work, thought it might be doable, in the case that $I$ is itself a simplicial category) then we can get that the simplicially enriched Grothendieck construction defined in the above cited paper, recovers Lurie's Grothendieck construction after applying the simplicial nerve. So suppose we've got a functor $F:I\to sCat$. Then we can produce the functor $f:I\to sSet$ by composing with the simplicial nerve functor $N:sCat\to sSet.$ Then we can construct the relative nerve of $f$, denoted $N_f(I)$, from Chapter 3 of Lurie's HTT. Lurie shows then that there is a coCartesian fibration $N_f(I)\to N(I)$ and that this is the fibration associated to $f$ by his $\infty$-categorical Grothendieck construction. Then what Wong and myself show in this paper is that this relative nerve, $N_f(I)$ is actually isomorphic to the simplicial nerve of the enriched Grothendieck construction, $N(GrF)$ (this is Theorem 2.3.1 of this second cited paper). Since Lurie already told us that $N_f(I)\simeq Gr_\infty(f)$, we're done!<|endoftext|> TITLE: O-minimal spectrum is a spectral space QUESTION [6 upvotes]: I'm trying to understand a proof on "Sheaves of Continuous Definable Functions" (Pillay, Anand. "Sheaves of continuous definable functions." The Journal of symbolic logic 53.04 (1988): 1165-1169.) Let $\mathcal{N} = (N,<,\ldots)$ be an o-minimal structure and let $X \subset N^m$ be a definable set, the o-minimal spectrum $X^\sim$ of $X$ is the set of complete m-types $S_m(N)$ of the first-order theory $Th_N (\mathcal{N})$ which imply a formula defining X equipped with the topology generated by the basic open sets of the form $$U^\sim = \{p \in X^\sim : U ∈ p\}$$ where U is a definable, relatively open subset of X. We call this topology on $X^\sim$ the spectral topology. My goal is to understand why is $X^\sim$ a spectral space. The proof goes along the lines: Let $F \subset X^\sim$ be a closed, irreducible space and let $\Phi = \{B \subset X : B $ is closed, definable and $B \in p$ for every $p \in F\}$, $\Phi'= \{ X \setminus C: C$ closed, definable and $C \notin \Phi\}$ and $\Phi_1 = \Phi \cup \Phi'$. Since $F$ is irreducible, $\Phi_1$ is consistent and thus determines a type $p$ and clearly $F = \overline{\{p\}}$ I already know that $\Phi$ is consistent (since $\Phi \subset p$ for every $p \in F$) and that for every $\varphi \in \Phi'$ either $\Phi \cup \varphi$ or $\Phi \cup \lnot \varphi$ is consistent. But I can't relate that to $F$ irreducibleness. REPLY [5 votes]: In this answer, I'm going to differ from Pillay's terminology by writing "compact" instead of "quasicompact" and "sober" for the condition that every irreducible closed set is the closure of a unique point (called the generic point). In the statement of Lemma 1.1, Pillay writes "$S^t_n(M)$ is a spectral space, i.e. it has a basis of quasicompact open sets and every irreducible closed set is the closure of a unique point." He has omitted one of the conditions in the definition of a spectral space: A space is spectral iff It is sober. The compact open sets form a base for the topology. The compact open sets are closed under finite intersection (in particular, the whole space - the empty intersection - is compact). The first paragraph of the proof essentially proves both 2. and 3. Pillay explains that the compact open sets in the topology are exactly the definable sets which are open in the o-minimal sense (the latter are a base for the topology by definition, and they are closed under finite intersections). All that remains is to show that $S^t_n(M)$ is sober, and this is what is done in the paragraph quoted in your question. Note first that the intersection of the sets in $\Phi$ is exactly $F$ (any closed set is the intersection of the basic closed sets containing it). Then by compactness, there are finitely many closed sets $C_1,\dots,C_n$ not in $\Phi$ such that $$F\cap (X\setminus C_1)\cap\dots\cap (X\setminus C_n) = F\setminus (C_1\cup\dots\cup C_n) = \emptyset.$$ Now for each $i$, the closed set $C_i' = F\cap C_i$ is a proper subset of $F$, since $C_i\notin \Phi$. So $F = C_1'\cup \dots \cup C_n'$, contradicting irreducibility of $F$. Extending $\Phi_1$ to a complete type $p$, you can now check that $p$ is a generic point of $F$. Again, Pillay does not make this explicit, but the uniqueness of $p$ follows from the key fact (stated in the first paragraph of Section 1) that every definable set is a Boolean combination of open definable sets, so if $p\neq q$, one of $p$ or $q$ is contained in an open set not containing the other (i.e. the topology is $T_0$), so the closures of $p$ and $q$ cannot agree. This completes the proof. Now I'd like to add that I would prefer to prove this theorem a different way. Stone duality for distributive lattices tells us that the category of distributive lattices is dual to the category of spectral spaces and spectral maps (also called coherent spaces and coherent maps) in a way that extends Stone duality for Boolean algebras. So let's say we have a sub-lattice $L$ of a Boolean algebra $B$ ($L$ is distributive, since $B$ is). Then in the inclusion $L\to B$ induces a (continuous) spectral map in the other direction $f\colon \mathrm{Spec}(B)\to \mathrm{Spec}(L)$, given by restricting a prime filter of $B$ to a prime filter of $L$. The map $f$ is always surjective: To see this, let $F$ be a prime filter in $L$, and $I$ its complementary prime ideal. Then check that $F\cup \{\lnot x\mid x\in I\}$ generates a proper filter in $B$ and hence can be extended to an ultrafilter containing $F$ and disjoint from $I$. Now if, additionally, $f$ is injective, then we can identify the points of $\mathrm{Spec}(L)$ and $\mathrm{Spec}(B)$, and $\mathrm{Spec}(B)$ inherits the spectral topology from $\mathrm{Spec}(L)$. It's not hard to see that the following are equivalent: $f$ is injective Whenever $p$ and $q$ are distinct ultrafilters in $B$, there is some $A$ in $L$ such that $A\in p \leftrightarrow A\notin q$. $L$ generates $B$ as a Boolean algebra. Enough generalities now: Suppose $B$ is the Boolean algebra of formulas in some variable context, modulo provable equivalence (equivalently, the Boolean algebra of definable sets of $N^n$ for some $n$). Then $\mathrm{Spec}(B)$ is the usual Stone space of first-order types. If we specify a distinguished subset of these formulas, which are closed under equivalence and finite conjunctions and disjunctions (e.g. the definable open sets in an o-minimal theory), then we get a sub-lattice $L$ of $B$. If, moreover, $L$ generates $B$ as a Boolean algebra, i.e. every formula is equivalent to a Boolean combination of distinguished formulas (in the o-minimal context, this is the key fact referred to above), then the spectral topology on $\mathrm{Spec}(L)$ induces a spectral topology on $\mathrm{Spec}(B)$, where the points are types, and the basic open sets have the form $\{p\mid \varphi\in p\}$ whenever $\varphi$ is one of the distinguished formulas in $L$. In the o-minimal case, this is exactly the o-minimal spectrum as defined by Pillay.<|endoftext|> TITLE: Congruence equation and quadratic residue QUESTION [12 upvotes]: The following observation makes me quite confused when I am trying to count the number of solutions of the equation: $$\sum_{k=0}^{M}{M \choose k}^2x^k=0$$ on finite fields $\mathbb{F}_p$ with the prime number $p>3$ and $M=(p-1)/2$. Observation: I tried the count the number of solutions with Mathematica. To my surprise, the number of solutions coincides with three times the sequence in OEIS. Question: I think what I observed is true for the first 100 primes. Is it true for all prime numbers $p>3$? REPLY [10 votes]: When $p \equiv 1 \bmod 4$, both expressions are $0$ and I understand why. For $p \equiv 1 \bmod 4$, the number $k$ is a QR iff $p-k$ is. Pairing off $k$ and $p-k$, the OEIS expression is $$\frac{1}{p} \left( \frac{p\cdot \# \mbox{QRs}}{2} - \frac{p \cdot \# \mbox{non-QRs}}{2} \right) = 0.$$ Meanwhile, values of $x$ in $\mathbb{F}_p$ are in bijection with elliptic curves $E$ over $\mathbb{F}_p$ and a chosen bijection between the $\mathbb{F}_p$-points of the $2$-torsion $E[2]$ and $(\mathbb{Z}/2)^2$. As js21 explains, we have $\sum \binom{(p-1)/2}{k}^2 x^k = 0$ if and only if $E$ is supersingular. So we need to show that (when $p \equiv 1 \bmod 4$) there are no supersingular elliptic curves $E$ whose $2$-torsion is entirely defined over $\mathbb{F}_p$. An supersingular elliptic curve always has $p+1$ points over $\mathbb{F}_p$, so if $p \equiv 1 \bmod 4$ then $\#E(\mathbb{F}_p) \equiv 2 \bmod 4$ and cannot have $(\mathbb{Z}/2 \mathbb{Z})^2$ as a subgroup. I found the key fact for treating $p \equiv 3 \bmod 4$ in Theorem 2.7 of a paper by Delfs and Galbraith. (See also Galbraith's slides.) If $E$ is a supersingular curve defined over $\mathbb{F}_p$, then Frobenius obeys $\mathrm{Frob}^2 = -p$ and $\mathrm{End}(E) \cap \mathbb{Q}[\mathrm{Frob}]$ is either $\mathbb{Q}\left( \tfrac{1-\sqrt{-p}}{2} \right)$ or $\mathbb{Q}(\sqrt{-p})$. In the former case, Delfs and Galbraith say that the curve is "on the surface", otherwise it is "on the floor". (The metaphor is subterranean, so the floor is below the surface.) As Theorem 2.7 says, curves on the surface always have three $2$-isogenies from them which are defined over $\mathbb{F}_p$, so their $2$-torsion is entirely defined over $\mathbb{F}_p$, while curves on the floor always have one $2$-isogeny, so their $2$-torsion is not entirely defined over $\mathbb{F}_p$. One can see this directly: $\mathrm{End}(E) \cap \mathbb{Q}[\mathrm{Frob}]$ is $\mathbb{Q}\left( \tfrac{1-\sqrt{-p}}{2} \right)$ iff $[2]$ divides $1-\mathrm{Frob}$, if and only if the kernel of $[2]$ (namely $E(2)$) is contained in the kernel of $1-\mathrm{Frob}$ (namely $E(\mathbb{F}_p)$). So roots of your polynomial come from curves on the surface -- generally, $6$ roots per curve, but fewer for $j=0$ or $1728$. By a theorem of Deuring, the number of curves on the surface is the same as the number of ideal classes in $\mathbb{Q}\left( \tfrac{1-\sqrt{-p}}{2} \right)$, namely, $h(-p)$. Combine all of this with Dirichlet's class number formula, and get all the details right, and you should have a proof. Examples: $p=83$. The class number is $3$ (see here) as is the OEIS sequence. Since $p \equiv 3 \bmod 8$, there are $2 \cdot 3 = 6$ $j$-invariants of supersingular curves defined over $\mathbb{F}_{83}$: $0$, $17$, $28$, $50$, $67$, $68\! \equiv\! 1728$. Each of them has two non-isomorphic twists over $\mathbb{F}_{83}$, so $12$ curves over $\mathbb{F}_{83}$ in all. The $2$-isogenies between them are shown on Galbraith's slide 27. Both of the twists of $j=50$, and one of the twists of $j=1728$, have three $2$-isogenies from them, so $E[2]$ defined over $\mathbb{F}_{83}$. The roots of your equation over $\mathbb{F}_{83}$ are $2$, $18$, $24$, $39$, $42$, $45$, $60$, $66$, $82$. Of these roots, $2$, $42$ and $82$ correspond to $j=1728$. The other six correspond to one of the two twists with $j=50$; the other twist is $-y^2 = X(X-1)(X-x)$ where $x$ takes one of these six values. $p=103$: This time, the class number is $5$ and $p \equiv 7 \bmod 8$ so there are $5$ supersingular $j$-invariants in $\mathbb{F}_{103}$: $23$, $24$, $34$, $69$ and $80 \! \equiv \! 1728$. Each of them has two twists. You can see the $2$-isogenies on Slide 33. In particular, the full $2$-torsion is defined over $\mathbb{F}_{103}$ for both twists of $j=23$ and $j=69$, and for one of the twists of $j=1728$. The roots of your polynomial are $2$, $12$, $25$, $28$, $30$, $33$, $43$, $52$, $61$, $71$, $74$, $76$, $79$, $92$, $102$. As $x$ ranges over these values, $y^2 = X(X-1)(X-x)$ covers $j=1728$ and one of each of the twists of $j=23$ and $j=69$, and $-y^2 = X(X-1)(X-x)$ covers the other twist of the latter cases.<|endoftext|> TITLE: A geometric proof of the Gauss-Lucas theorem QUESTION [13 upvotes]: Motivated by a geometric proof of the Fundamental Theorem of Algebra we ask: Is there a geometric proof for the Gauss-Lucas theorem? Since we are working on a half plane, can one imagine a possible proof of the above fact with some methods in hyperbolic geometry? REPLY [4 votes]: Charles R. Johnson and I gave a proof of the Gauss–Lucas theorem that utilizes the field of values (or numerical range) of a matrix, which can be viewed as geometric in nature (the field itself is an uncountably infinite intersection of half-planes). The proof can be found in the article Matricial Proofs of Some Classical Results about Critical Point Location, which recently appeared in The American Mathematical Monthly. Our methods also yield a proof of Marden's theorem and a weaker version of Siebeck's theorem.<|endoftext|> TITLE: A certain mathematical competition in the UK QUESTION [15 upvotes]: There is a foreword, written by professor Snow, to the book A mathematician's apology. In the foreword, it is written some thing like the following: "Hardy was opposed to a certain mathematical competition in the UK because he believed that such competitions destroyed real mathematics in the UK during one century." My question is: What was that competition and why did he believe that the competition destroyed real mathematics for a century? REPLY [18 votes]: Cambridge Mathematical Tripos examination was “a high speed marathon whose like has never been seen before or since” (http://apollonius.math.nthu.edu.tw/d1/disk5/js/history/cambridge.pdf --- "Old Cambridge Days" by Leonard Roth. Another great sources of interesting information about Tripos are the following books: "Mr Hopkins’ men: Cambridge reform and British mathematics in the 19th century" by A.D.D. Craik and "Masters of theory: Cambridge and the rise of mathematical physics" by A. Warwick. Examples below are from these books). Tripos procedure was quite cruel, some kind of Darwinian natural selection at work, as the following examples illustrate. C.T. Simpson, the Second Wrangler in 1842, had worked twenty hours a day for a whole week before the examination and during the Tripos he almost broke down from overexertion. James Wilson, the Senior Wrangler in 1859, experienced a severe mental and physical breakdown immediately after the examination. It took three months him to recover from the illness and after the recovery he found that he had forgotten utterly all the mathematics that he had learned at Cambridge apart from elementary algebra and Euclid. Nevertheless, according to Leonard Roth, Hardy's position about complete abolishment of Tripos examination was too extremist. Various generations of Cambridge men have each shaped the examination according to their light, but the work is never complete and probably never will be. Other intending reformers of the Tripos are even now waiting in the wings; indeed some among them would reform it altogether. Such a notion, startling as it may appear, is by no means novel; it was held more than forty years ago, by Hardy himself, who had backed the 1909 reform as only a first stage of the program. Hardy firmly believed that the Tripos was an unmitigated evil, for which one must blame the inferior performance of British pure mathematician vis-a-vis their European colleagues. So, away with the examination. Now the weakness of this argument resides in its lack of supporting evidence. It would be very difficult to unearth any specific cases of careers which have undoubtedly been ruined or even seriously damaged by the Cambridge mathematical system: on the other hand, the supporters of the status quo can for their part point to a long line of distinguished mathematical physicist, some of whom we have already mentioned, who achieved success either because or in spite of it: looking at their record one could scarcely suppose that they would have done more or better work had they been spared the ordeal of the examination; and, in the past, what an ordeal it was! Abolitionists are such charming people; their motives are so patently pure, and only rarely do they foresee the full consequence of their projected panaceas. All his lifelong Hardy moved in the highest academic circles and tutored the most talented of young men. Had he troubled to consult any lecturer from a provincial university, or even (it may be) a don from a Cambridge college less exalted than Trinity, he could easily have learned a simple but significant truth: if students know beforehand that a particular subject is not to be examined upon, they will, almost to a man - or a woman - altogether decline to study it. Even Forsyth could have told Hardy that much: for he had been a professor at London University.<|endoftext|> TITLE: Smallest solution to $x^2 \equiv x\pmod{n}$ QUESTION [10 upvotes]: Given $n$, is it possible to upper bound the smallest $x > 1$ that satisfies the congruence $x^2 \equiv x\pmod{n}$? Obviously when $n$ is a prime power $x = n$, and we are in the worst situation. However for other $n$ we can make $x \leq \frac{n}{2}$. Perhaps better bounds can be obtained if we know that $n$ has more prime factors. REPLY [6 votes]: We may really get better estimates if $n$ have at least three prime divisors, like, say $0n-1-u_i>0$ as we need). Then $u_1+u_2\in (2n/3,n-1)$ (case $u_1+u_2=n-1$ is impossible since $u_1+u_2$ is divisible by $q_3$), take a solution $n-1-u_1-u_2$. Update. David Speyer shows in his answer that this may be generalized to the estimate $(n-1)/k$ for $n$ having $k$ distinct prime divisors. Let me show that the constant $1/k$ can not be improved for a fixed $k$. Fix $k\geqslant 2$. We need a Lemma. For any $m\geqslant 1$ there exist (arbitrarily large) primes $p_1,\dots,p_m$ congruent to 1 modulo $k$ such that $\frac{p_i-1}k \prod_{j\ne i} p_j\equiv 1\pmod {p_i}$ for all $i=1,2,\dots,m-1$. Proof. Induction in $m$. Base $m=1$ is formally trivial (take any $p_1$ congruent to 1 modulo $k$). Induction step from $m$ to $m+1$. Take $p_{m+1}$ congruent to 1 modulo $kp_1\dots p_{m-1}$ and to $-k/p_1\dots p_{m-1}$ modulo $p_m$. This is possible by Dirichlet theorem (and Chinese Remainders Theorem.) Use this lemma for $m=k$ and choose primes $p_1,\dots,p_k$ satisfying conditions of lemma, set $n=p_1\dots p_k$. Then $u_i=\frac {p_i-1}{k}\cdot \frac{n}{p_i}$ are our $u_i$'s for $i=1,2,\dots,k-1$, $u_k=1-(u_1+\dots+u_{k-1})$. They are all pretty close to $n/k$, thus for any $I\subset \{1,\dots,k\}$, $0<|I| TITLE: Non-isomorphic rings that are localizations of each other QUESTION [17 upvotes]: Do there exist commutative rings $A$ and $B$ and multiplicative subsets $S\subseteq A$, $T\subseteq B$ such that $A\not\simeq B$ but $S^{-1}A \simeq B$ and $T^{-1} B\simeq A$? This question comes from a deleted claim in an answer of Qiaochu Yuan. The original claim says that (the proper class of isomorphism classes of) commutative rings equipped with the relation $A\ge B$ iff $B$ is a localization of $A$ is a (large) poset. The claim is clearly equivalent to the negative answer to the question. However, I do not find any reason that such examples do not exist; neither can I construct an example. REPLY [37 votes]: Example. Let $k$ be a field, and let $K = k(x_1,x_2,\ldots)$ be the fraction field of $k[x_1,x_2,\ldots]$. Let $$A = K[y_1,y_2,\ldots],$$ and $$B = A[y_1^{-1}].$$ Then $B$ is a localisation of $A$. If we further localise at the multiplicative set $S = K[y_1]\setminus\{0\} \subseteq B$, we get a ring that is isomorphic to $A$ by shifting all the indices by $1$: \begin{align*} S^{-1} B &\stackrel\sim\longrightarrow A\\ x_i &\longmapsto x_{i+1},\\ y_i &\longmapsto \left\{\begin{array}{ll}x_1 & \text{if } i = 1,\\ y_{i-1} & \text{if } i > 1.\end{array}\right.\\ \end{align*} To see that $A$ and $B$ are not isomorphic (even as rings), note that the set $A^\times \cup \{0\}$ is closed under addition, and the same is not true in $B$. $\square$ Remark. There are no examples of finite presentation (in a ridiculously general setting): Lemma. Let $R$ be a ring, and assume $A$ and $B$ are $R$-algebras of finite presentation. Assume there exist multiplicative sets $S \subseteq A$ and $T \subseteq B$ and $R$-linear isomorphisms \begin{align*} S^{-1}A \stackrel\sim\longrightarrow B, & & T^{-1}B \stackrel\sim\longrightarrow A. \end{align*} Then $A$ and $B$ are isomorphic (as $R$-algebras). Proof. Consider the composite map $\phi \colon A \to A$ obtained by $$A \to S^{-1} A \stackrel\sim\to B \to T^{-1}B \stackrel\sim\to A.$$ Note that this map is the localisation at the inverse image of $ST \subseteq B$ in $A$. To avoid confusion, denote the codomain of $\phi$ by $A'$. Since the natural map $$A' \otimes_A A' \to A'$$ is an isomorphism (this is always true for localisations), we conclude that the map $$\phi^* \colon \operatorname{Spec} A \to \operatorname{Spec A}$$ induced by $\phi$ is a monomorphism. By EGA IV$_4$, Proposition 17.9.6 (you can think of this as a fancy version of the Ax-Grothendieck theorem), this implies that $\phi^*$ is an automorphism, so $\phi$ is an isomorphism. But a localisation that is an isomorphism is the identity (i.e. $ST$ consists of units of $A$). Then $S$ contains only units of $A$, so $S^{-1}A = A$. $\square$ Remark. We actually only used that one of them was of finite presentation. Remark. One may object that the example above is not $K$-linear (and therefore it violates two of the assumptions of the lemma). However, it is $k$-linear (or $\mathbb Z$-linear...), so in that sense it only violates the finite presentation assumption.<|endoftext|> TITLE: Representability of Weil Cohomology Theories in Stable Motivic Homotopy Theory QUESTION [16 upvotes]: My understanding is that one purpose of stable motivic homotopy theory is to emulate classical stable homotopy theory. In particular, we would like Weil cohomology theories to be representable by motivic spectra. Can someone clarify this and/or give references for the construction of the spectrum representing, say, l-adic cohomology? REPLY [13 votes]: Let me elaborate the comment to be more precise. Question:Can someone clarify if Weil cohomologies are representable and give a reference? Answer: Yes, but it requires a hypothesis: the cohomology has to be defined as a sheaf cohomology. In the paper Mixed Weil cohomologies Cisinski-Déglise (C-D) defined the natural notion of mixed Weil cohomology (which restricted to the pure case is a Weil cohomology). Provided that your Weil cohomology is defined as a sheaf cohomology with the natural properties one may expect (check C-D's notion of mixed Weil theory for the precise definition) C-D constructed a spectrum representing your cohomology. At the end of the paper you can check how do C-D work out the main examples. Question: Can someone give a reference for the construction of the spectrum representing $l$-adic cohomology? Answer: There is no reference for this, to my knowledge [Edit: check Marc Hoyois comment from below. It seems there is such spectrum]. To apply above results the cohomology has to be a sheaf cohomology, as we said above. To my knowledge $l$-adic cohomology is not a sheaf cohomology. Nevertheless, there is something. Let $k$ be a countable perfect fiedl and $X$ be a smooth $k$-scheme. Denote $\bar X=X\otimes \bar k$ for $\bar k$ a separable closure of $k$. Deligne defined a sheaf computing the $\mathbb{Q}_l$-cohomology $H(\bar X,\mathbb{Q}_l)$ of $\bar X$. This cohomology is representable in the stable homotopy category (check the end of C-D paper). Remark: The result of C-D has been extended to apply to other cohomologies with twists. The first result I know is from B. Drew's thesis in 2.1.8 for absolute Hodge cohomology, but the reference requires a lot of investment from the reader. The concrete case of real Deligne cohomology was worked out with analogue ideas to that of C-D by Holmstrom-Scholbach here in section 3 and a general result was then obtained by Déglise-Mazzari here (check the introduction and 1.4.10).<|endoftext|> TITLE: Does the topology induced by the Hausdorff-metric and the quotient topology coincide? QUESTION [8 upvotes]: Assume that $X$ is a metric space, and $\sim$ is an equivalence relation on $X$. Furthermore we assume that the number of elements in each equivalence class is bounded by a positive constant. Does the quotient topology on $X/{\sim}$ and the topology induced by the Hausdorff-metric on $X/{\sim}$ coincide? REPLY [12 votes]: Let $X=[0, \infty)$ with the metric $d(x,y)=|x-y|$. The equivalence relation is $x\sim y$ iff $x=y$ or $xy=1$. In the Hausdorff metric on $X/{\sim}$, the open ball with radius $1/2$ around the equivalence class $\{0\}$ contains only $\{0\}$. However $\{0\}$ is not open in $X$, so $\{\{0\}\}$ is not open in the quotient topology on $X/{\sim}$.<|endoftext|> TITLE: Irreducible representations of Sp(2) QUESTION [5 upvotes]: I am looking for an explicit description of the finite dimensional irreducible representations of the classical Lie group $\text{Sp}(2) = \{A\in M_2(\mathbb{H})\,|\,A\overline{A}^T = I\}$. I can easily find the above info for orthogonal groups and unitary groups (the representations coming from certain spaces of harmonic polynomials) but cannot seem to find such a description for the groups $\text{Sp}(n)$. In particular I am aware that the irreducible representations of $\text{Sp}(2)$ are classified by Young diagram parameters $(a,b)$ with $a\geq b \geq 0$ but I would like to compute with such spaces so need an explicit description if possible. REPLY [6 votes]: Consider the Lie algebra $\frak{sp}(n)$ of the symplectic group ${\rm Sp}(n)={\rm Sp}(n; \mathbb{C})\cap{\rm SO}(4n)$. For $n=1$, let us denote by $V$ the representation of $\frak{sp}(1)$ on $\mathbb{C}^{2}$. Then, any irreducible $\frak{sp}(1)$-representation is isomorphic to some symmetric power ${\rm Sym}^{p}V$ of $V$. For $n>1$ let us denote by $W$ the $2n$-dimensional complex representation of $\frak{sp}(n)$. Then, the symmetric powers ${\rm Sym}^{p}W$ are also irreducible. But not the exterior powers $\Lambda^{p}W$. Let us denote the kernel of the contraction $\Lambda^{p}W\to \Lambda^{p-2}W$ induced by the invariant 2-form on $W$, by $\Lambda_{0}^{p}W$. For $p\leq n$ one can show that this submodule $\Lambda_{0}^{p}W$ is also irreducible. One can moreover define the modules ${\rm Ker}\{\Lambda^{2}({\rm Sym}^{2}W)\to {\rm Sym}^{2}W\}$ and ${\rm Ker}\{\Lambda^{2}({\Lambda}^{2}_{0}W)\to {\rm Sym}^{2}W\}$, but for $n=2$ the second $\frak{sp}(n)$-module is trivial. Some details can be found in the classical book of Simon Salamon, ''Riemannian geometry and holonomy groups''. In particular, for ${\rm Sp}(2)$ see pages 80-84. Finally notice that since $\frak{sp}(2)\cong\frak{so}(5)$, the spin representation of ${\rm SO}(5)$ is isomorphic to the standard representation $W$ of ${\rm Sp}(2)$ and the standard representation of ${\rm SO}(5)$ is isomorphic with $\Lambda_{0}^{2}W$. Of course, using the LiE program you can take a quick view of the complex irreducible representations of ${\rm Sp}(2)$, in terms of highest weights. added The irreducible representation of ${\rm Sp}(2)$ with highest weight (2, 2) is of dimension 81. Since ${\rm Sp}(2)$ is connected and simply-connected, it coincides wth ${\rm Spin}(5)$. Now, this irreducible representation is identified with the isotropy representation of the homogeneous space ${\rm SO}(14)/{\rm SO}(5)$, which is strongly isotropy irreducible. Notice finally that also the coset ${\rm Spin}(10)/{\rm Sp}(2)$ is a strongly isotropy irreducible space (of dimension 35). Here, the isotropy representation has highest weight (2, 1).<|endoftext|> TITLE: Constructing groups of Type E7 with certain Tits Index QUESTION [5 upvotes]: In a new survey on $E_8$, namely Skip Garibaldi - E8 the most exceptional group , the author gives an example (Example 8.4., page 15) on how to construct a group of type E8 with a prescribed Tits-Index. This construction was actually invented by Tits and bears his name.. From the inclusion of $F_4 \subset E_6 \subset E_8$, after applying $H^1(k,-)$, a isotropic group of type $E_8$ with semi simple anisotropic kernel of type $E_6$ arises from an Albert Algebra $A$, which is a $F_4$ torsor (meaning an element of $H^1(k,F_4)$). (Note that this example serves to explain some properties of the Rost invariant of this $E_8$, but we want to focus on the construction itself.) We will remember this for later and call it #Above In the Paper J. Tits - Stronger inner anisotropic forms of simple algebraic groups ,on the last page the author gives an example of a group of type $E_7$ having semi simple anisotropic kernel $D_5 \times A_1$, or in Tits's notation is of index $E^{48}_{7,1}$, meaning the vertex $6$ is circled. He writes: "..[27, Proposition 5] shows that groups with this index are classified by anisotropic quadratic forms $q$ in $10$ variables, whose invariant $c(q)$ is a division quaternion algebra." 1.Does he mean classified as constructed like in #Above ? If he means like in #Above: 1.1.Is a group with that index constructed by choosing such a $q$, looking at $SO(q)$, which is of type $D_5$ and adjoint, noticing that $D_5\subset D_6 \subset E_7$ and applying $H^1(k,-)$? What is confusing to me is that the anisotropic kernel of the resulting $E_7$ is not just $D_5$ but $D_5 \times A_1$. But on the other hand $D_5$ can never occur as anisotropic kernel, so it might all fit together. 2.What is known about the quadratic forms $q$? I ponder they have splitting pattern $(1,2,2)$, after analyzing the more isotropic Tits-Indexes and taking into account Vishik's classification of splitting patterns. REPLY [5 votes]: This might shed some light on relationship between anisotropic quadratic forms in 10 variables and the desired forms of $E_7$, though it uses results more recent than Tits, and doesn't quite answer your questions as stated. Bruce Allison worked out the following results in his paper "Structurable division algebras and relative rank one simple Lie algebras" in Lie Algebras and Related Topics: Proceedings of a Summer Seminar Held June 26-July 6, 1984. More polished and easier to find is his later paper on "Tensor products of composition algebras, Albert Forms and Some Exceptional Simple Lie Algebras" in Transactions of the American Mathematical Society, Vol. 306, No. 2 (Apr., 1988) Assume the base field $k$ does not have characteristic $2$ nor $3$. Take a division quaternion algebra $B$ and a division octonion algebra $C$ over $k$, for which $A = B \otimes C$ is a division algebra too. Equivalently, $B$ and $C$ contain no common quadratic extension of $k$. This tensor product $A = B \otimes C$ is a 32-dimensional algebra with involution -- a special sort of algebra that Allison calls a structurable algebra. Its skew elements form a 10-dimensional subspace $S$, and (with the division algebra condition), the norm form on $S$ is anisotropic. This is the $q$, and $q$ determines $A$ up to isomorphism (as a $k$-algebra with involution) -- see Theorem 5.4 of Allison's later paper. By construction, we can describe the $q$ in more detail. The norm forms on $B$ and $C$ are Pfister forms, say $<1,b_1> \otimes <1,b_2>$ and $<1,c_1> \otimes <1,c_2> \otimes <1,c_3>$, respectively. The form $q$ is the orthogonal sum of the skew subspace of $B$ and (negative) the skew subspace of $C$. So we have $$q = (b_1, b_2, b_1 b_2, -c_1, -c_2, -c_3, -c_1 c_2, -c_2 c_3, - c_3 c_1, -c_1 c_2 c_3).$$ It may be easy to describe the possible $q$, using the classification of quaternion and octonion algebras. Allison uses the structurable algebra $A$ to construct a 5-graded Lie algebra $K_{-2} \oplus K_{-1} \oplus K_0 \oplus K_1 \oplus K_2$ of the desired type. $K_{\pm 1}$ are 32-dimensional, identified with $A$. $K_{\pm 2}$ are 10-dimensional, identified with $S$. Lie brackets on $K_{\pm 1}$ arise from the structurable algebra via $[a_1, a_2] = a_1 \bar a_2 - a_2 \bar a_1$. $K_0$ is harder to describe. Allison proves that all Lie algebras of type $E_{7,1}^{48}$ arise from this construction, and by the end of his latter paper (Transactions 1988) he constructs examples, e.g., over a field $R((T_1, T_2))$ of Laurent series.<|endoftext|> TITLE: Hyperplane sections with chi non-zero QUESTION [7 upvotes]: Let $X$ be a smooth, projective variety over $\mathbb{C}$ for which $\chi(X) = 0$. Here by $\chi$, I mean the topological Euler characteristic of $X(\mathbb{C})$; this number can also be computed as the degree of the top Chern class of the tangent bundle of $X$, and there are various other equivalent definitions. I want to find/construct smooth hyperplane sections $Y$ of $X$ for which $\chi(Y) \neq 0$; I do not care too much about the projective embedding. Under which additional conditions (if any) is doing such a thing possible or reasonable? What are potential obstructions, et cetera? Motivation: I can prove an interesting statement for varieties with $\chi \neq 0$. I believe it to be true for varieties with $\chi = 0$ as well, but alas my arguments do no longer apply. I want to try to reduce it to the case $\chi \neq 0$ using suitably chosen hyperplane sections. REPLY [6 votes]: There is no easy way to express $\chi (Y)$ in terms of $\chi (X)$. Using $\chi (X)= c_n(T_X)$ (with $n:=\dim X$ and the identification $H^{2n}(X,\mathbb{Z})=\mathbb{Z}$), and the exact sequence $0\rightarrow T_Y\rightarrow T_{X|Y}\rightarrow N_{Y/X}\rightarrow 0\ $ you get $\ \ \ \chi (Y)=c_{n-1}\cdot h-c_{n-2}\cdot h^2+\ldots +(-1)^{n-1}h^{n}$, where the $c_i$ are the Chern classes of $X$ and $h\in H^2(X,\mathbb{Z})$ the class of a hyperplane section. It is not easy to decide whether this is $0\,$! On the other hand if you are willing to replace your embedding by a multiple, say $h$ by $kh$, you know that $|\chi (Y)|$ will be $\gg 0$ for $k$ large enough, since $\chi (Y)$ is given by a fixed polynomial in $k$ with leading coefficient $(-1)^{n-1}h^n$.<|endoftext|> TITLE: Question on Weil-Petersson metric on Teichmuller space QUESTION [5 upvotes]: I'm reading Ahlfors' original articles about Weil-Petersson metric: "Some remarks on Teichmüller's space of Riemann surfaces" and "Curvature properties of Teichmüller's space". The tangent space at a point $C$ (here identified with a Riemann surface of genus $g$) of the Teichmuller space can be identified with the space of Beltrami differentials on $C$ and the author defines the Weil-Petersson hermitian form as $\langle \nu,\mu\rangle=\int_C\nu\varphi(\mu)dxdy$, for every Beltrami differentials $\nu,\mu$ on $C$, where $\varphi(\mu)$ is the quadratic differential corresponding to $\mu$. On the Teichmuller space $\mathcal{T}_g$ there are complex coordinates $t_1,\dots,t_n$, because every Beltrami differential $\mu$ can be written $\mu=t_1\nu_1+\dots+t_n\nu_n$, where $\nu_1,\dots,\nu_n$ is a base. Then author then wants to write the Weil-Petersson form in this coordinates as $\sum_{\alpha,\beta}g_{\alpha\overline{\beta}}(\mu)dt_\alpha d\overline{t}_\beta$. What I don't get is why he writes $g_{\alpha\overline{\beta}}(\mu)=\langle L^\mu\nu_\alpha,L^\mu\nu_\beta\rangle$ The definition of $L^\mu\nu$ is the following. Consider $f^\mu:\hat{\mathbb{C}}\rightarrow\hat{\mathbb{C}}$ such that $\partial_{\overline{z}}f^\mu=\mu\partial_zf^\mu$, then, given three Beltrami differentials $\mu,\rho$ and $\lambda$, we write $\rho=\mu|\lambda$ iff $f^\mu=f^\rho\circ f^\lambda$ which is true iff $\rho\circ f^\lambda=\frac{\mu-\lambda}{1-\overline{\lambda}\mu}(\partial_zf^\lambda/|\partial_zf^\lambda|)^2$. Then, by abuse of notation, we can define the function on Beltrami differentials $\rho:\mu\mapsto \mu|\lambda$ and $\frac{\partial}{\partial \nu}\rho(\lambda):=lim_{t\rightarrow 0}\frac{\rho(\lambda+t\nu)-\rho(\lambda)}{t}=L^\lambda\nu=(\nu\frac{(\partial_zf^\lambda)^2}{|\partial_zf^\lambda|^2-|\partial_\overline{z}f^\lambda|^2})\circ(f^\lambda)^{-1}$ Why this definition of $L^\mu\nu$ should imply that, imposing $g_{\alpha\overline{\beta}}(\mu)=\langle L^\mu\nu_\alpha,L^\mu\nu_\beta\rangle$, we get the metric $\int_C\nu\varphi(\mu)dxdy$? REPLY [2 votes]: One way to define Teichmueller space is to fix a Riemann surface $X$, with a fixed complex structure, and define the space of all Beltrami differentials $\mathcal{M}(X)$ on $X$. Notice that in this way we have fixed a base point $X$ of Teichmueller space with a fixed "background" complex structure. By a Beltrami differential I mean a $(-1,1)$ form $\mu(z)\frac{d\bar{z}}{d z}$ such that $\|\mu\|_{\infty} < 1$. By an infinitesimal Beltrami differential I simply mean a $(-1,1)$ form without the latter inequality, all of which form the set $\mathcal{IB}(X)$. Then one way to define Teichmueller space is $\mathcal{T}(X) = \mathcal{M}(X) / \sim$, where two Beltramies $\mu_1$ and $\mu_2 \in \mathcal{M}(X)$ are equivalent, i.e. $\mu_1 \sim \mu_2$, exactly when there is a quasi-conformal homeomorphism $f : X \to X$ homotopic to identity such that $$\mu_1 = f^*\mu_2 = \frac{\bar{\partial} f + (\mu_2\circ f) \overline{\partial f}}{{\partial} f + (\mu_2 \circ f) \partial\bar{f}}. $$ Furthermore, the tangent space $T_X\mathcal{T}(X)$ of $\mathcal{T}(X)$ at the base point $X$ can be thought of as $\mathcal{IB}(X)/\sim$. Now, if by $X^{\mu}$ I denote the Riemann surface with complex structure induced by the Beltrami $\mu$ on $X$, then the tangent space of the Teichmueller sapce at that point is $T_{X^{\mu}}\mathcal{T}(X) = T_{[\mu]}\mathcal{T}(X) = \mathcal{IB}(X^{\mu})/\sim$. The Wiel-Petersson hermitian form, calculated at the point $[\mu]$, is defined as $$\langle\nu_1, \nu_2\rangle_{\mu} = \int_{X^{\mu}} \nu_1 \varphi(\nu_2),$$ where $\nu_1$ and $\nu_2 \in \mathcal{IB}(X^{\mu})$ are infinitesimal Beltramies on $X^{\nu}$ (not on $X$!). This is the reason for which one needs to use the formulas you are asking about. One needs to look at all the infinitesimal Beltramies $\nu = \sum_{\alpha} t_{\alpha}\nu_{\alpha}$ on the base point Riemann surface $X$ from the perspective of the Riemann surface determined by $[\mu] \in \mathcal{T}(X)$, i.e. to look at the infinitesimal Beltriemies $\nu|_{\mu} = \sum_{\alpha} t_{\alpha}\nu_{\alpha}|_{\mu}$ on the surface $X^{\mu},$ i.e. $\nu|_{\mu} \in \mathcal{IB}(X^{\mu})$. In order to do that, first one needs a way to move all Beltramies from $\mathcal{M}(X)$ to $\mathcal{M}(X^{\mu})$ by establishing an appropriate isomorphism. Then, by differentiating that isomorphism at the point $[\mu]$ one would find out how infinitesimal Beltramies are moved from $X$ to $X^{\mu}$ isomorphically, i.e. one would obtain the tangent isomorphism. Consequently, one would have an identification between $\mathcal{IB}(X^{\mu})/\sim$ and $\mathcal{IB}(X)/\sim$. Given a Beltrami $\lambda$ denote by $f^{\lambda} : X \to X^{\lambda}$ the quasi-conformal map (unique up to post-composition with a holomorphic isomorphism) that determines the Beltrami $\lambda = \frac{\bar{\partial} f^{\lambda}}{\partial f^{\lambda}}$ on $X$. Also denote by $f^{\lambda|_{\mu}} : X^{\mu} \to X^{\lambda}$ the quasi-conformal map that determines the (intermidiate) Beltrami $\lambda|_{\mu} = \frac{\bar{\partial} f^{\lambda|_{\mu}}}{\partial f^{\lambda|_{\mu}}}$ on $X^{\mu}$ (not on $X$!). Then, (technically up to holomorphic isomorphism) $f^{\lambda} = f^{\lambda|_{\mu}} \circ f^{\mu} : X \to X^{\lambda},$ where $ f^{\lambda|_{\mu}} : X^{\mu} \to X^{\lambda}.$ After differentiating the latter identity between the three quasi-conformal maps, $df^{\lambda} = df^{\lambda|_{\mu}} \circ df^{\mu}$, and then combining it with the expressions of the type $df(z) = \partial f(z) dz + \bar{\partial} f(z) d\bar{z}$, one obtains again the formula, already featured above, $$\lambda = f^{\mu*}(\lambda|_{\mu}) = \frac{\bar{\partial} f^{\mu} + (\lambda|_{\mu} \circ f^{\mu}) \overline{\partial f^{\mu}}}{{\partial} f^{\mu} + (\lambda|_{\mu} \circ f^{\mu}) \partial\overline{f^{\mu}}}. $$ By solving for $\lambda|_{\mu} \circ f^{\mu}$ and using the fact that $\bar{\partial} f^{\mu} = \mu \partial f^{\mu}$, one obtains exactly the expression $$\lambda|_{\mu} \circ f^{\mu} = \frac{\lambda - \mu} {1 - \bar{\mu}\lambda}\left(\frac{\partial f^{\mu}}{|\partial f^{\mu}|}\right)^2.$$ This is basically the map from $\mathcal{M}(X)$ to $\mathcal{M}(X^{\mu})$ sending Beltramies on $X$ to Beltramies on $X^{\mu}$. To obtain a tangent vector at $[\mu] \in \mathcal{T}(X)$, take an infinitesimal Beltrami $\nu$ on $X$ and form the one parameter family $\mu_t = \mu + t \nu$. On $X^{\mu}$ this family becomes $$\mu_t|_{\mu} \circ f^{\mu} = \frac{\mu + t \nu - \mu} {1 - \bar{\mu}(\mu + t \nu)}\left(\frac{\partial f^{\mu}}{|\partial f^{\mu}|}\right)^2 = t \frac{\nu } {1 - \bar{\mu}(\mu + t \nu)}\left(\frac{\partial f^{\mu}}{|\partial f^{\mu}|}\right)^2.$$ To obtain the tangent vector $\nu|_{\mu}$ at the point $\mu$ simply differentiate the last identity with respect to $t$ and then set $t=0$. The result is $$\nu|_{\mu} \circ f^{\mu} = \nu \frac{(\partial f^{\mu})^2} {|\partial f^{\mu}|^2 - |\bar{\partial} f^{\mu}|^2}.$$ After pre-composing with $(f^{\mu})^{-1}$, one obtains exactly the formula you mention $$L^{\mu}\nu = \nu|_{\mu} = \left(\nu \frac{(\partial f^{\mu})^2} {|\partial f^{\mu}|^2 - |\bar{\partial} f^{\mu}|^2}\right)\circ (f^{\mu})^{-1}.$$ That is why the Weil-Petersson form should look like $$\langle\nu_{\alpha}|_{\mu}, \nu_{\beta}|_{\mu}\rangle_{\mu} = \langle L^{\mu}\nu_{\alpha}, L^{\mu}\nu_{\beta}\rangle_{\mu} = \int_{X^{\mu}} \nu_{\alpha}|_{\mu} \varphi(\nu_{\beta}|_{\mu}) = \int_{X^{\mu}} L^{\mu} \nu_{\alpha} \varphi(L^{\mu} \nu_{\beta}),$$ for any $\nu_{\alpha}$ and $\nu_{\beta}$ from $\mathcal{IB}(X)$ (notice that the base point is $X$!).<|endoftext|> TITLE: Under what condition is a fiber bundle cobordant to the trivial bundle? QUESTION [5 upvotes]: Let $E$ be the total space of a fiber bundle with base $B$ and fiber $F$, where $B$ and $F$ are smooth manifolds. Under what condition is $E$ unoriented cobordant to $B\times F$? And what happens if we take $B$, $F$, and $E$ oriented. Is $E$ oriented cobordant to $B\times F$? Edit : As mentioned in the comment, there are classical counterexamples, due to Atiyah, in the oriented case when the base is not simply connected. Furthermore Dold showed that the unoriented cobrodism ring is a polynomial ring over $\mathbb{Z}/2$ with generator the $\mathbb{R}\mathbb{P}^{i}$ for $i$ even, and some fiber bundle with base $\mathbb{R}\mathbb{P}^{2^r-1}$ and fiber $\mathbb{C}\mathbb{P}^{s2^r}$ with $s,r\geq 1$. see http://www.map.mpim-bonn.mpg.de/Unoriented_bordism or http://link.springer.com/article/10.1007%2FBF01473868. Those bundles have $\mathbb{R}\mathbb{P}^{2^r-1}$ as base spaces which are null-bordant, so they provide counterexample. However, in both those cases, counterexamples (seem to) come from the fact that the base space is not simply connected. In fact, signature is known to be multiplicative in the simply-connected case, see http://www.maths.ed.ac.uk/~aar/papers/chs.pdf. So I will try to reformulate my question with those facts in mind : Given that $B$ is simply connected, is there some reasonable hypothesis on $F$ assuring that $E$ is cobordant to $B\times F$? REPLY [4 votes]: Here are three special cases where it is always true: Sphere bundles of vector bundles. If $E \to B$ is a vector bundle and $S(E) \to B$ is the associated sphere bundle, then $S(E)$ bounds the disc bundle $D(E)$, so it is nullcobordant. On the other hand, $S^{n-1}\times B$ is nullcobordant as $S^{n-1}$ is, so $S(E)$ and $S^{n-1}\times B$ are cobordant. Provided $E$ is an orientable bundle, the same is true in oriented cobordism. Finite covering spaces. If $F \to E \xrightarrow{\pi} B$ is a finite covering space (i.e. $F$ is finite), then $TE \cong \pi^*TB$. It follows that the Stiefel-Whitney numbers of $E$ are just the Stiefel-Whitney numbers of $B$ multiplied by $\deg \pi = |F|$. So $E$ has the same Stiefel-Whitney numbers as the disjoint union of $|F|$ copies of $B$ (Stiefel-Whitney numbers add under disjoint sum), which is nothing but $F\times B$. A similar consideration with Pontryagin numbers shows that if $E$ and $B$ are oriented, then the same conclusion holds in oriented cobordism. Principal bundles. If $G \to E \xrightarrow{\pi} B$ is a principal $G$-bundle, and $\dim G \geq 1$, then $E$ is nullcobordant. To see this, note that $TE \cong V\oplus H$ where $V$ and $H$ are the vertical and horizontal spaces respectively. Now $V\cong E\times\mathfrak{g}$ while $H \cong \pi^*TB$, so $TE \cong (E\times\mathfrak{g})\oplus\pi^*TB$. In particular, $$w(TE) = w((E\times\mathfrak{g})\oplus\pi^*TB) = w(\pi^*TB) = \pi^*w(TB).$$ As $\dim E = \dim B + \dim G > \dim B$, all the Stiefel-Whitney numbers of $E$ vanish, so it is nullcobordant. On the other hand, $G\times B$ is nullcobordant because $G$ is (Lie groups are parallelisable, so all their Stiefel-Whitney numbers are zero). Again, provided that $E$ and $B$ are orientable, the same is true in oriented cobordism (the Pontryagin numbers of $E$ are also zero).<|endoftext|> TITLE: Does $\mathbb{CP}^2$ admit a Riemann surface lamination structure? QUESTION [9 upvotes]: Does $\mathbb{CP}^2$ admit a Riemann surface lamination structure? Every paper or article I looked at, talk only about singular laminations on $\mathbb{CP}^2$. I was wondering why. If you know something about it or you can give some reference, it would be nice. REPLY [11 votes]: It is conjectured that $\mathbb {CP}^2$ contains no embedded compact laminated set (without singularities) apart the smooth algebraic curves. This is a strong form of the "Minimal Exceptional" conjecture, stating that for a singular holomorphic foliation of $\mathbb{CP}^2$, every leaf accumulates in the singular set. Nice references about this subject are the following surveys: É. Ghys - Laminations par surfaces de Riemann. Dynamique et géométrie complexes - Panor. Synthèses, 8, 1999. S. Zakeri - Dynamics of singular holomorphic foliations on the complex projective plane. Laminations and foliations in dynamics, geometry and topology - Contemp. Math., 269, Amer. Math. Soc., 2001. J. E. Fornaess and N. Sibony - Riemann surface laminations with singularities. - J. Geom. Anal. 18 (2008), no. 2, 400–442.<|endoftext|> TITLE: Can you use Chevalley‒Warning to prove existence of a solution? QUESTION [15 upvotes]: Recall the Chevalley‒Warning theorem: Theorem. Let $f_1, \ldots, f_r \in \mathbb F_q[x_1,\ldots,x_n]$ be polynomials of degrees $d_1, \ldots, d_r$. If $$d_1 + \ldots + d_r < n,$$ then the number of common zeroes of $f_1, \ldots, f_r$ is $0$ modulo $p$. In particular, if there exists a zero (e.g. if all the $f_i$ have no constant coefficient), then there exists another one. On the other hand, it can easily happen that the $f_i$ have no common zeroes at all, e.g. if $n = 3$ and $f_1 = x_1$, $f_2 = x_1 + 1$. The problem is that the scheme $$\operatorname{Spec} \mathbb F_q [x_1,x_2,x_3]/(x_1, x_1 + 1)$$ is empty, so you're never going to find any solutions. A more interesting example is when $n = 3$ and $f_1 = x_1^2 - a$ for $a \in \mathbb F_q$ not a square. Then $$\operatorname{Spec} \mathbb F_q[x_1,x_2,x_3]/(x_1^2 -a ) \neq \varnothing,$$ but there are no rational solutions. Question. Let $f_i$ as in the Chevalley‒Warning theorem. Assume that $$X = \operatorname{Spec} \mathbb F_q [x_1,\ldots,x_n]/(f_1, \ldots, f_r) \neq \varnothing.$$ Is there a bound $B$ in terms of the $d_i$ and $n$ such that $X(\mathbb F_{q^k}) \neq \varnothing$ for some $k \leq B$? In particular, I was hoping that there is some trick to reduce this to Chevalley‒Warning (e.g. introduce an extra variable, cf. the example below). However, other methods for attacking this question are welcome as well. This question is motivated by, but not identical to, the finite field case of this question. Example. An example where one can use such a phenomenon is with the reduced norm on central simple algebras over $\mathbb F_q$: Let $A$ be a division algebra over $\mathbb F_{q}$ of dimension $n$. Then the reduced norm is a homogeneous polynomial $P$ of degree $n$ in $n$ variables. For any $a \in \mathbb F_{q}$, consider the homogeneous polynomial $$P(x) - a z^n$$ of degree $n$ in $n + 1$ variables. It has a solution $(0,\ldots,0)$ since it is homogeneous, so by Chevalley‒Warning it has to have a nontrivial solution. But since $A$ is a division algebra, we cannot have $P(x) = 0$ for $x \neq 0$, so the solution has to satisfy $z \neq 0$. This proves that the reduced norm is surjective. (This basically proves that the Brauer group of a finite field (more generally a $C_1$ field) is trivial.) Remark. A standard trick for this type of question is to view $\mathbb F_{q^k}$ as a $k$-dimensional vector space over $\mathbb F_q$. If $\alpha$ is a primitive element, we can introduce new variables $x_{1,1}, \ldots, x_{n,k}$ corresponding to $$x_i = x_{i,1} + \alpha x_{i,2} + \ldots + \alpha^{k-1} x_{i,k}.$$ Then solutions of the $f_i$ over $\mathbb F_{q^k}$ correspond to solutions of suitable polynomials $\tilde{f}_i$ of the same degree $d_i$ over $\mathbb F_q$. This allows us to get rid of the degree assumption in my question: take $k$ such that $d_1 + \ldots + d_{r} < k n$, and do this substitution to reduce to the question I posed above. I guess that this means that my question is equivalent to the finite field case of the question I linked earlier. REPLY [6 votes]: The answer by js21 is a great answer, and it gives an explicit choice of $B$. However, you have not asked for any restrictions on $B$ other than it depends only on $n$, $r$, and $d_1,\dots,d_r$ (I assume you want it to be independent of $q$ and of the coefficients of $f_1,\dots,f_r$). There is, in fact, such a $B$ for every choice of $n$, $r$, and $d_1,\dots,d_r$, even if $r>n$, and this only uses the basic techniques of existence of the Hilbert scheme. Over $\text{Spec}(\mathbb{Z})$, you can form the following parameter space as a product of projective spaces $$\Pi = \mathbb{P}H^0(\mathbb{P}^n,\mathcal{O}(d_1))\times \dots \times \mathbb{P}H^0(\mathbb{P}^n,\mathcal{O}(d_r)).$$ Inside the product $\Pi\times \mathbb{P}^n$, you can form the universal closed subscheme $Z$ that is the simultaneous zero locus of an $r$-tuple of homogeneous polynomials of degrees $d_1,\dots,d_r$. The scheme $\Pi$ is finite type over $\text{Spec}(\mathbb{Z})$, and the projection morphism $\pi:Z\to \Pi$ is projective. Thus, only a finite number of Hilbert polynomials may occur for geometric fibers of $\pi$. More precisely, for every scheme-theoretic fiber of $\pi$ over a point, and for every nonempty open subset of the fiber, we may consider the scheme-theoretic closure of that open subset as a closed subscheme of the fiber. There are only finitely many Hilbert polynomials that can occur for all of these schemes. This follows quicky from the existence of a flattening stratification of $\pi$. So now let $\mathcal{E}=(e_1(t),\dots,e_N(t))$ be a finite set of numerical polynomials. We say that $\mathcal{E}$ is "saturated" if for every field $F$, both (a) for every $e(t)$ in $\mathcal{E}$, the first difference $e(t) - e(t-1)$ is also in $\mathcal{E}$, and (b) for every closed subscheme $Y\subset \mathbb{P}^n_F$ whose Hilbert polynomial is in $\mathcal{E}$, for every nonempty open subset of $Y$ with its reduced structure, the scheme-theoretic closure of that open set also has Hilbert polynomial in $\mathcal{E}$. By the existence of the Hilbert scheme with fixed Hilbert polynomial as a finite type scheme, by the existence of the flattening stratification, etc., every finite $\mathcal{E}$ is contained in a finite $\mathcal{E}$ that is saturated with respect to (b). It is straightforward to saturate with respect to first differences without altering $\mathcal{E}_m$. Thus, by induction on $m$, every $\mathcal{E}$ is contained in a finite $\mathcal{E}$ that is saturated. We can partition $\mathcal{E}$ as $\mathcal{E}_0\sqcup \dots \sqcup \mathcal{E}_m$, where $\mathcal{E}_k \subset \mathcal{E}$ is the subset of numerical polynomials of degree $k$. Fix a hyperplane $H=\mathbb{P}^{n-1}\subset \mathbb{P}^n$, i.e., a "hyperplane at infinity" whose complement is an affine space $\mathbb{A}^n$. The claim is that there exists an integer $d=d(\mathcal{E})$ such that for every field $F$ (e.g., a finite field), for every closed subscheme $Y\subset \mathbb{P}^n_F$ whose Hilbert polynomial is in $\mathcal{E}$ and such that $Y$ is not contained in $\mathbb{P}^{n-1}_F$, then there exists a field extension $F'/F$ of degree $\leq d$ such that $Y(F')$ contains an element that is not in $\mathbb{P}^{n-1}(F')$. Up to replacing $Y$ by one irreducible component $Y_j$ that is not contained in $\mathbb{P}^{n-1}_F$ (this is allowed since $\mathcal{E}$ is saturated), we may assume that $Y$ is integral and $Y\cap \mathbb{P}^{n-1}_F$ is the zero scheme of a regular section of $\mathcal{O}_Y(1)$. In particular, the Hilbert polynomial of this zero scheme is the first difference of the Hilbert polynomial of $Y$, thus it is also in $\mathcal{E}$ since $\mathcal{E}$ is saturated. The claim is proved by induction on $m$. The base case is when $m$ equals $0$. In this case, every $e_i(t)$ is a constant polynomial, and we let $d$ be the maximum of these integers. Since the Hilbert polynomial of $Y$ is constant, $Y$ is zero-dimensional. Then $Y_j$ has total degree $\leq d$. Thus, the associated reduced scheme of $Y_j$ has total degree $\leq d$. This reduced scheme is a singleton set of a closed point that is not contained in $\mathbb{P}^{n-1}_F$. Thus, the residue field of this closed point gives $F'/F$ of degree $\leq d$. This proves the claim when $m$ equals $0$. By way of induction, assume that $m>0$ and assume the result has been proved for smaller $m$. By the previous paragraph, there exists an integer $d$ that works for $\mathcal{E}_0$. Thus, without loss of generality, assume that the Hilbert polynomial of $Y$ has degree $>0$, i.e., $Y$ has dimension $>0$. Since the Hilbert polynomial of $Y\cap \mathbb{P}^{n-1}_F$ is in $\mathcal{E}$, and since $\mathcal{E}$ is saturated, every irreducible component $D_j$ of $Y\cap \mathbb{P}^{n-1}_F$ has Hilbert polynomial in $\mathcal{E}$. The number $\ell$ of such components is bounded by the degree of $Y$, which in turn is bounded by the maximum $d_0$ of the (normalized) leading coefficients of the finitely many elements of $\mathcal{E}$. By Gotzmann's Regularity Theorem, there exists an integer $\rho=\rho(\mathcal{E})$ such that the Castelnuovo-Mumford regularity of $Y$ and every $D_j$ is $\leq \rho$ (actually the existence of such $\rho$ is older, presumably due to Mumford, but Gotzmann gave an explicit upper bound on $\rho$). Thus for the ideal sheaf $\mathcal{I}_Y$ of $Y$ in $\mathbb{P}^n$, $\mathcal{I}_{Y}(\rho)$ is globally generated, and $H^0(\mathbb{P}^n_F,\mathcal{O}(\rho))\to H^0(Y_j,\mathcal{O}_{Y}(\rho))$ is surjective, and the same is true for $D_j$. In particular, for distinct components $D_j$ and $D_k$ of $D$, the image of the restriction map $$H^0(\mathbb{P}^n_F,\mathcal{I}_{D_j}(\rho))\to H^0(D_k,\mathcal{I}_{D_j}\cdot\mathcal{O}_{D_k}(\rho))$$ has nonzero image. This can be checked after base change to an algebraic closure of $F$, and it follows there from the fact that $\mathcal{I}_{D_j}(\rho)$ is globally generated. Thus, there exists $f_{j,k}\in H^0(\mathbb{P}^n_F,\mathcal{O}(\rho)$ such that $f_{j,k}$ vanishes on $D_j$ but not on $D_k$. The element $f_k = \prod_{j\neq k} f_{j,k}$ is an element of $H^0(\mathbb{P}^n_F,\mathcal{O}((\ell-1)\rho))$ that vanishes on every $D_j$ with $j\neq k$, yet does not vanish on $D_k$. Thus, $f=\sum_k f_k$ is an element of $H^0(\mathbb{P}^n_F,\mathcal{O}((\ell - 1)\rho))$ that does not vanish identically on any $D_k$. The same is true for the power $f^s$. Thus, there exists a section of $H^0(\mathbb{P}^n_k,\mathcal{O}((d_0-1)!\rho)))$ whose zero scheme $Z$ on $Y$ does not contain any $D_j$. In particular, $Z$ is not contained in $\mathbb{P}^{n-1}_F$ (there may have been a faster way to arrange that). By the same quasi-compactness arguments as above, the Hilbert polynomials of $Z$'s are bounded and have degree $ TITLE: Differential structures on compact Lie groups QUESTION [5 upvotes]: Given a compact Lie group can there be a differential structure on it with respect to which one cannot define a smooth group operation? REPLY [5 votes]: This is definitely not my expertise, but here is a stab at it. A solution to Hilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group. On the other hand, there are exotic structures on Lie groups (including compact ones). So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness part of the solution to Hilbert's 5th Problem.<|endoftext|> TITLE: Is SO(2n+1)/U(n) a symmetric space? QUESTION [15 upvotes]: I am a physics student with only a rudimentary knowledge of differential geometry, so please feel free to point out if I miss something elementary / trivial. According to https://arxiv.org/abs/1408.2760, $ SO(2n+1)/U(n) $ is not a symmetric space because it does not have the right Cartan decomposition of the Lie algebra. That is, suppose that $ \mathfrak{g} $ is the Lie algebra of $ SO(2n+1)$ and $ \mathfrak{h} $ is the Lie algebra of $ U(n) $. There is a decomposition $$ \mathfrak{g} = \mathfrak{h} + \mathfrak{p} $$ for some $ \mathfrak{p} $ such that $ [\mathfrak{h},\mathfrak{h}] \subset \mathfrak{h}$ and $[\mathfrak{h},\mathfrak{p}] \subset \mathfrak{p}$. But $ [\mathfrak{p},\mathfrak{p}] $ is not in $ \mathfrak{h} $, so we do not have a Cartan involution on this space. I'm wondering if it is that simple. I'll be grateful if someone can clear up my confusion below. I think that $SO(2n+1)/U(n)$ and $SO(2n+2)/U(n+1)$ are diffeomorphic. For instance, this book shows that the two are the same homogeneous spaces by showing that Any element $SO(2n+1)$ can be written as an ordered product of two elements, one in $SO(2n+1)$ and another in $U(n+1)$. They write this as $SO(2n+2)=SO(2n+1)\cdot U(n+1)$ The quotient $X = SO(2n+2)/U(n+1) = SO(2n+1)\cdot U(n+1) / U(n+1)$ can be thought of as $ SO(2n+1)/U(n)$ because the $SO(2n+1)$-action on $X$ is transitive and the stabilizer of the identity $eU(n+1)$ of $X$ are the elements of $SO(2n+1)$ that are also in $U(n+1)$: $SO(2n+1) \cap U(n+1) = U(n)$. An example in low dimensions is $ SO(5)/U(2) = SO(6)/U(3) = \mathbb{C}P^3$, a complex projective space. But $SO(2n+2)/U(n+1)$ is a symmetric space. So why is $SO(2n+1)/U(n)$ not? REPLY [3 votes]: The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a generalized flag manifold of the compact simple Lie group $SO(2n+1)$. It arises from the more general family $$ M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) ) $$ for $p=n$. Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands: $${\frak m}\cong T_{o}M_{n, p}=\frak{m}_{1}\oplus\frak{m}_{2}.$$ This is true, since in this case you paint black in the Dynkin diagram of $G=SO(2n+1)$ a simple root of Dynkin mark equal to 2. It is easy to see that $[\frak{m}, \frak{m}]\neq \frak{k}$, in particular one computes $$ [\frak{m}_{1}, \frak{m}_{1}]\subset\frak{k}\oplus\frak{m}_{2}, \quad [\frak{m}_{1}, \frak{m}_{2}]\subset\frak{m}_{1},\quad [\frak{m}_{2}, \frak{m}_{2}]\subset\frak{k} $$ For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible (in this case, in the Dynkin diagram of $G=SO(2n+1)$ we paint black the first simple root, which has Dynkin mark equal to 1, recall that the highest root of $SO(2n+1)$ is given by $\tilde{\alpha}=\alpha_1+2\alpha_2+\cdots+2\alpha_{n}$ where $\{\alpha_1, \ldots, \alpha_{n}\}$ is a basis of simple roots. The coefficients $\{1, 2, \ldots, 2\}$ are the so called Dynkin marks). In fact, the only generalized flag manifolds which are the same time symmetric spaces are the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which are isotropy irreducible. (see Which Kahler Manifolds are also Einstein Manifolds? for more details) Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$. For example $$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$ However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $$SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).<|endoftext|> TITLE: Are There Mutually Exclusive Large Cardinal Axioms in ZFC? QUESTION [7 upvotes]: The various large cardinal axioms are usually described in terms of some roughly linear hierarchy of varying consistency strengths. However, some cardinal axioms potentially contradict one another. As an example, asserting the existence of unboundedly many strong limit cardinals implies AC. However, asserting the existence of Reinhardt cardinals or Berkeley cardinals implies the negation of AC (all in ZF). This seems to imply different mutually exclusive hierarchies of large cardinal axioms in ZF. Are there other mutually exclusive LCA's in ZFC? If so, what are some of the exact statements that they conflict on? If not, why wouldn't we expect this in ZFC when it seems to be the case in ZF? REPLY [6 votes]: The answer to your question depends on what counts as a large cardinal axiom, and there is no agreed-upon official definition for that term. On the one hand, it is easy to formulate incompatible theories involving large cardinals. If these theories themselves count as large cardinal axioms, then they provide the answer. There is an inaccessible cardinal with CH versus there is an inaccessible cardinal with $\neg\text{CH}$. The least supercompact cardinal is Laver-indestructible versus the least supercompact cardinal $\kappa$ has $V_\kappa\not\subset\text{HOD}$. The least measurable cardinal is strongly compact versus every strongly compact cardinal is supercompact. There is an inaccessible cardinal and V=L versus $0^\sharp$ exists. $V=L[\mu]$ versus $0^\dagger$ exists. Some researchers prefer to adopt a narrow sense of what counts officially as a large cardinal axiom, and on that perspective, these theories are not large cardinal axioms themselves, but theories about large cardinals, and so they do not constitute examples. Other researchers use a broader sense for what counts as a large cardinal axiom, and with that perspective, there are abundant examples. So it may come down to an issue of semantics.<|endoftext|> TITLE: Critical points in $ZF$ without Choice QUESTION [5 upvotes]: Recall the definition of critical point for set theory: A critical point of an elementary embedding of one transitive class into another transitive class is the smallest ordinal not mapped to itself. (This is from the Wikipedia article "Critical point (set theory)", which claims this definition is from Jech's Set Theory (2002 edition)). What theorems about critical points of elementary embeddings can be proven in $ZF$ without recourse to the Axiom of Choice? To be specific, are these theorems enough to prove anything useful regarding critical points of nontrivial elementary embeddings $j$: $V$$\rightarrow$$V$? REPLY [5 votes]: If $j\colon V\to M$ is a nontrivial elementary embedding with $M\subseteq V$ a transitive class, then $j$ has a critical point: Suppose there isn't one, then by induction on rank we prove that $j(x)=x$ for all $x\in V$. First note that $\operatorname{rank}(x)=\operatorname{rank}(j(x))$ for all $x$, since the rank function is absolute and if $\operatorname{rank}(x)<\operatorname{rank}(j(x))$, then we have an ordinal which was moved by $j$. Suppose that $j(y)=y$ for all $y\in x$, then $M\models j(y)\in j(x)$ and therefore $x\subseteq j(x)$; on the other hand, if $z\in j(x)$, then the rank of $z$ is less than the rank of $x$, therefore $j(z)=z$ and therefore $M\models j(z)\in j(x)$, so $V\models z\in x$. Therefore $j(x)\subseteq x$. In particular, $j\colon V\to V$ without a critical point must be the identity. REPLY [5 votes]: In ZF, many of the usual arguments about critical points still go through. For example, every critical point $\kappa$ of an elementary embedding $j:V\to M$ is regular, since if $\kappa$ is the supremum of a short sequence $s$ below $\kappa$, then it is easy to see that $j(s)=s$, which would imply $j(\kappa)=\kappa$, contradicting the assumption that $\kappa$ is the critical point. If $\kappa$ is the critical point of $j:V\to M$, then $\kappa$ is weakly inaccessible. To see this, note that if $\kappa=\delta^+$, then $j(\kappa)=(\delta^+)^M$, which of course is at most $\kappa$, since $P(\delta)^V=P(\delta)^M$, and this contradicts $\kappa\kappa$ we may look at $j\upharpoonright V_\lambda:V_\lambda\to V_{j(\lambda)}$, which is elementary and verifies weak extendibility (weak = we do not insist on $\lambda TITLE: Inverting a function QUESTION [8 upvotes]: I posted this question on crypto.SE but got no answer: Let $w = a_0 \cdot a_1 \cdots a_{n-1} $ be a word from $ \{0,1\}^n $, $|w| = n$ Let $m = \sum_{i=0}^{n-1}{ a_i \cdot 2 ^ {n-1-i} } $ be the corresponding binary number constructed from the word. Let $k= \left \lfloor \frac{n!}{2^n} \right \rfloor \cdot (m+1)$ , then $ 1 \le k \le n! $. Compute the Lehmer-Permutation $\pi_k$ from $k$ on $n$ numbers. ( https://en.wikipedia.org/wiki/Lehmer_code ) Set $ x := \pi_k \cdot w = a_{\pi_k(0)} \cdot a_{\pi_k(1)} \cdots a_{\pi_k(n-1)} $ Then $f(w) := x$. So the function permutes the digits in the word $w$ and the permutation is determined by $w$. Suppose you randomly choose uniformly a word from $\{0,1\}^{1000}$ and then you apply the function. Is it practically possible to invert the constructed word? That is, does somebody have an idea on how to invert the word? More details may be found on: http://orgesleka.blogspot.de/2015/09/candidate-one-way-function.html This picture shows f applied on all words of length 7: After two years, also posted on cs: https://cs.stackexchange.com/questions/110790/inverting-a-function REPLY [7 votes]: Yes, you can use the Lehmer-Permutation to make a function that is suitable for cryptography, whose solution is just as hard as the Diffie-Helman problem. The relevant papers are: (1) Roberto Mantaci, "A permutation representation that knows what "Eulerian" means", Discrete Mathematics and Theoretical Computer Science (4): 101–108, 2001, link. and (2) Iharantsoa Vero Rahrinirina, "Use of Signed Permutations in Cryptography", 2017, arXiv link. In (1), the author shows that the Lehmer code of a permutation is a subexceedant function, and indeed you have a one-to-one correspondence between subexceedant functions and permutations. In (2), the author shows how to use a subexceedant function in cryptography. He discusses the importance of choosing a good base, just like you did by choosing $k$. For such a base, Section 3.1 shows how to do an analogue of the Diffie-Helman public key exchange with a subexceedant function. He points out in 3.4 that it's just as hard as the classic Diffie-Helman problem, i.e. hard enough for cryptography. Lastly, the problem you raised, of choosing $w$ at random, is a special case of the public key setup of (2). Here Alice is the oracle who "knows" the random number chosen, and an attacker is trying to figure out Alice's private key based on her public key. If your choice of $k$ does what you claim on your blog (making the permutation well-distributed), then (1) and (2) show that your proposed $f$ is good for cryptography.<|endoftext|> TITLE: Do you know this Burnside ring module? QUESTION [8 upvotes]: Let $G$ be a finite group and $\Omega(G)$ its Burnside ring. There is a certain $\Omega(G)$-module, let's call it $M(G)$, that appears in something that I am thinking about. As an abelian group $M(G)$ is the direct sum $\oplus_HR(W_GH)$, over conjugacy classes of subgroups $H\subset G$, of the Grothendieck group of complex representations of the Weyl group $W_GH=N_GH/H$. I don't have a lot of experience with $G$-spaces, but I would guess that this module has appeared before. My question is, does anybody know its name, or better yet does anybody know an illuminating way of thinking about it? The module structure is as follows: let $K$ be a subgroup of $G$. Multiplication by the element of $\Omega(G)$ corresponding to the orbit $G/K$ is the composition of two maps $M(G)\to M(K)\to M(G)$. The map $M(G)\to M(K)$ takes the element $V\in R(W_GH)$ to the sum, over conjugacy classes of subgroups $L\subset K$ such that $L$ is conjugate to $H$ in $G$, of the restriction of $V$ along an injection $W_KL\to W_GH$. The map $M(K)\to M(G)$ takes the element $V\in R(W_KL)$ to the sum, over $N_KL$-conjugacy classes of subgroups $H\subset G$ such that $H\cap K=L$, of a representation of $W_GH$ obtained from $V$ by restricting and then inducing along injections $W_KL\leftarrow (N_GH\cap N_KL)/L\to W_GH$. ADDED LATER: This turns out to be one of those cases where asking a question stimulates one to answer it oneself. But I would still be interested in references or further information. REPLY [4 votes]: I've got it. Here's another way to think about $M(G)$. Let $\mathcal G(G)$ be the groupoid whose objects are the finite $G$-sets and whose morphisms are the isomorphisms. Let $Rep(G)$ be the additive category of all functors from $\mathcal G(G)$ to finite-dimensional complex vector spaces that are monoidal in the sense of taking disjoint union to direct sum. Then $M(G)$ is the Grothendieck group of $Rep(G)$. For a homomorphism $i:K\to G$ we get a forgetful functor $\mathcal G(G)\to \mathcal G(K)$ and its left adjoint, both of them monoidal. These induce additive functors from $Rep(K)\to Rep(G)$ and vice versa, hence homomorphisms $i_\ast: M(K)\to M(G)$ and $i^\ast: M(G)\to M(K)$, making $M$ both a covariant and a contravariant functor. For a finite $G$-set $X$, the monoidal functor $S\mapsto S\times X$ from $\mathcal G(G)$ to itself gives a map $M(G)\to M(G)$, and this is how $M(G)$ becomes a Burnside module. In the case when $X$ is $G/K$, this is the same as $i_\ast\circ i^\ast$ for the inclusion $i:K\to G$.<|endoftext|> TITLE: List of invariants that distinguish homotopy equivalent non-homeomorphic spaces QUESTION [6 upvotes]: It is written on wikipedia article (https://en.wikipedia.org/wiki/Analytic_torsion) that the Reidemeister torsion is the first invariant that could distinguish between spaces which are homotopy equivalent but not homeomorphic. Can any one give me examples of other invariants which do this? REPLY [4 votes]: Intersection (co)homology is a (co)homology theory, defined for stratified pseudomanifolds that is not homotopy invariant. Examples of applications to your question have been done by Dirk Schuetz in his papers: "Intersection homology of linkage spaces" and "Intersection homology of linkage spaces in odd dimensional Euclidean space". Using intersection cohomology he is able to distinguish non-homeomorphic spaces. Intersection cohomology can also be used to get the topological classification of weighted projective spaces: "Intersection homology of weighted projective spaces and pseudo-lens spaces." Masato Kuwata<|endoftext|> TITLE: Is the wonderful compactification of a spherical homogeneous variety always projective? QUESTION [16 upvotes]: Let $G/H$ be a spherical homogeneous variety, where $G$ is a complex semisimple group. Assume that the subgroup $H$ is self-normalizing, i.e., $\mathcal{N}_G(H)=H$. Then by results of Brion and Pauer and Knop there exists a wonderful compactification of $G/H$. Is this wonderful compactification always projective, or in some cases it can be complete, but not projective? Feel free to vote to close this elementary question. REPLY [4 votes]: Apart from the general argument mentioned by Friedrich in his answer, in the particular case $N_G(H) = H$ the projectivity of the wonderful compactification $X$ of $G/H$ can be seen directly. Namely, let $\mathfrak g$ and $\mathfrak h$ be the Lie algebras of $G$ and $H$, respectively. Then $X$ is isomorphic to the $G$-orbit closure $\overline{G \cdot [\mathfrak h]}$ in the Grassmannian $\mathrm{Gr}_{\dim \mathfrak h}(\mathfrak g)$. This result was proved by Losev in his paper Demazure embeddings are smooth from 2009. In fact, the projectivity of $X$ is deduced from the weaker fact due to Brion stating that $X$ is isomorphic to the normalization of $\overline{G \cdot [\mathfrak h]}$; see his paper Vers une généralisation des espaces symétriques from 1990.<|endoftext|> TITLE: Showing that $\alpha$ isn't a cardinal in $J_{\alpha+1}^{\vec E}$ for a fine extender sequence $\vec E$ QUESTION [8 upvotes]: In [FSIT] and [OIMT] it is claimed that there is a surjection from $P(\kappa)\cap J^{\vec E}_{\nu(E_\alpha)}\times[\nu(E_\alpha)]^{<\omega}$ onto $\alpha$, and that this surjection lies in $J_{\alpha+1}^{\vec E}$. It is stated as a rather trivial fact, but I'm having trouble with seeing how this map should look like. Here $\nu(E_\alpha)$ is the natural length of the $(\kappa,\alpha)$ pre-extender $E_\alpha$ and $\vec E$ is a fine extender sequence, the definition of which can be found in [OIMT] at page 11. Since $E_\alpha$ is the trivial completion of $E_\alpha\upharpoonright\nu(E_\alpha)$ I can see that $\nu(E_\alpha)$ somehow "carries enough information" to determine $\alpha$, but this vague analogy just doesn't give me anything concrete. Of course, if a given surjection is found and is shown to be definable over $J_\alpha^{\vec E}$, then it's in $J_{\alpha+1}^{\vec E}$. Thanks in advance! References: [FSIT] "Fine structure and iteration trees" by Steel and Mitchell [OIMT] "Outline of inner model theory" in the handbook, by Steel (preprint at https://math.berkeley.edu/~steel/papers/steel1.pdf) REPLY [2 votes]: Here is an alternative answer (it is not true in general that $(E_{\alpha})_{a} \in J_{\alpha}^{E}$, for example when $E_{\alpha}$ is only a measure ) We use the fact that $g \in J_{\alpha+1}^{E} \cap P(J_{\alpha}^{E})$ iff $g$ is definable over $(J_{\alpha}^E,\in,E|\alpha,F)$. Define \begin{gather*} (\xi,\beta,\eta) \in g \\ \longleftrightarrow \\ (J^{E}_{\alpha}, \in , E|\alpha, F) \models \xi < \kappa^{+} \\ \& \\ \exists \gamma_{\xi} \exists f_{\xi} [ (\gamma_{\xi} \ \text{is the least ordinal such that } \ F \cap J^{E}_{\xi} \in J^{E}_{\gamma_{\xi}} ) \\ \& \\ ( f_{\xi} \ \text{is the } \ <_{J^{E}_{\alpha}}- \text{least bijection from} |\gamma_{\xi}| \ \text{onto} \ \gamma_{\xi} \\ \& \\ ( (\beta,\eta) \in f_{\xi} \\ \vee \\ ( \beta < |\nu| \ \& \ \beta \not\in dom(f_{\xi}) \ \& \ \eta = \emptyset ))] \end{gather*} Using lemma 2.9 from OIMT and the fact that there are no cardinals above $\nu$, it follows that $g:\kappa^{+}\times |\nu|^{J_{\alpha}^{E}} \rightarrow \alpha $ is onto.<|endoftext|> TITLE: Does there exist a ``continuous measure'' on a metric space? QUESTION [16 upvotes]: Let $X$ be a separable complete metrizable space. Does there exist a complete metric $d$ and a Borel measure $\mu$ such that (a) $\mu(B_r(x))<\infty$ for every open ball $B_r(x)$ of radius $r>0$ around $x\in X$, (b) for each $r>0$ the map $ x\mapsto \mu(B_r(x)) $ is continuous on $X$, (c) $\mathrm{supp}(\mu)=X$? REPLY [6 votes]: By continuing the line of thought from Nate Eldredge's comment you can handle finitely many separated parts of the space by a distance having values at most one and exactly one between points in different connected components. However, a possible counter-example (I did not yet check the details) should follow by continuing this to countably many separated parts that accumulate. For instance: Take $$X = [-1,0] \cup \{\frac1n\,:\,n \in \mathbb N\}$$ with the induced topology from $\mathbb R$. Each isolated point $\frac1n$ has to have positive measure. Since they accumulate to $0$ you cannot put a distance on $X$ separating them from the interval $[-1,0]$. This should force a discontinuity for a suitable $r>0$ when $x$ travels along $[-1,0]$.<|endoftext|> TITLE: Homotopical interpretation of flatness? QUESTION [9 upvotes]: I have read a discussion (in a less common language) which discussed a homotopical interpretation of flatness, which went something like: A map of commutative algebras is flat if pushing it out along any other morphism is quasi-isomorphic to the derived pushout when the algebras are embedded into dg-algebras. Hence flatness is just another name for quasi-fibrations. Another bloke remarked: Flatness is about giving the correct pullback, but it may also be obtained without flatness. Of course the correct condition is Tor-independence, i.e the tensor product $A\otimes_\Bbbk B$ is correct iff $\operatorname{Tor}^n_\Bbbk(A,B)=0$ for all $n$. Where is this viewpoint written down and/or developed clearly? I'm not at a level where I can make out the picture from a few hints. Does 'derived' means 'homotopy' here? Why do dg-algebras pop up? What's this Tor-independence business all about? Finally, the first bloke also wrote there's a homotopical characterization of étaleness, which I would also like to know more about. REPLY [3 votes]: The derived tensor product is defined as a homotopy pushout of $M\leftarrow R\rightarrow N$ in commutative DGAs. Now, if $R$ is a commutative ring and $M$ and $N$ are $R$-algebras, then the homology groups $H^i(M\otimes^\mathbf{L}_R N)=\operatorname{Tor}^R_i(M,N)$. Therefore $M\otimes_R^\mathbf{L} N$ and $M\otimes_R N$ are quasi-isomorphic for all $N$ iff $M$ is flat. Tor-independence is an isomorphic condition (Definition 55.1 in http://stacks.math.columbia.edu/download/more-algebra.pdf).<|endoftext|> TITLE: Does anyone have Delzell's Thesis on Bad Points of Forms? QUESTION [5 upvotes]: Since a number of papers (e.g. this one) treating denominators in Hilbert's 17th problem point to E.G. Strauss's unpublished letter to G. Kriesel or to Chapter 5 of Delzell's Thesis, which contains an extensive history of "bad points" for real forms, it would be nice if either of these things could be readily found online. It would also be great to have access to the paper: Delzell, C. N., Unavoidable singularities when writing polynomials as sums of squares of real rational functions. Which is cited as in preparation many places... Question: Are there any papers freely available online that treat in detail bad points for homogeneous positive semidefinite forms? If so, where are they? REPLY [4 votes]: Delzell's thesis is available here: http://www.math.ens.fr/~benoist/textes/Delzell.pdf .<|endoftext|> TITLE: What do we call this quantifier ("binder")? QUESTION [5 upvotes]: There's a quantifier ("binder", whatever), call it $\alpha$, defined as follows: $\alpha x.\tau$ is the (usually infinite) expression obtained by applying the substitution $\{x \mapsto \tau\}$ to the expression $x$ an infinite number of times. Its probably a bit unclear what I mean here, so let me give some intuition and a couple of examples. The idea is to read $\alpha x.(x+1)$ as denoting an infinite syntactic expression corresponding to the equation $x=x+1$. In particular, suppose $x=x+1$. Then we can write: $$x = x+1 = (x+1)+1 = ((x+1)+1)+1) = \cdots$$ We think of $\alpha x.(x+1)$ as the 'limit' of this process. So: $$\alpha x.(x+1) = ((...)+1)+1.$$ Notice there's no $x$ in the 'expression' $((...)+1)+1$; once we've passed to infinity, all our variables disappear. So $x$ is 'bound' in the expression $\alpha x.(x+1).$ Its not free. In some sense, we might say: there's no $x$ in $\alpha x.(x+1).$ More generally, we're meant to think of $\alpha x.\tau$ as the infinite expression corresponding to the assumption $x=\tau$. For instance: $$\alpha x.(x+x) = ((...)+(...))+((...)+(...))$$ etc. Question. What do we call this 'quantifier' and where can I learn more about it? I'm also interested in variants and generalizations, so don't be hesitant to post an answer even if it doesn't quite answer the question. REPLY [15 votes]: You are describing the regular tree grammars. Here is the basic idea. It is useful to think of syntactic expressions as abstract syntax trees. In our case we are looking for a tree $\alpha$ which satisfies the equation $$\alpha = \alpha + 1$$ The tree is infinite, but it is also regular (both intuitively and in a precise formal sense): In general you might want to solve a system of such equations, for instance \begin{align*} \alpha &= \beta + \gamma \\ \beta &= 1 + \beta \\ \gamma &= (\gamma + \beta) \end{align*} gives the infinite epression $\alpha$ indicated by $$ (1 + (1 + (1 + \cdots))) + ((\cdots + (1 + (1 + (1 + \cdots)))) + (1 + (1 + (1 + \cdots)))). $$ Regarding the question "What is this binder called?" the answer is a (least) fixed-point operator. It is usually written as $\mu$ or $\mathsf{fix}$ and its defining equation is, unsurprisingly, $$\mu x \,.\, \phi(x) = \phi(\mu x \,.\, \phi(x).$$ Your fixed-point operator works at the level of syntax as it is building an infinite syntactic tree. There are other fixed point operators. For instance, given a monotone map $f : L \to L$ on a complete lattice, $\mu f$ would be the least fixed point of $f$. Such operators are the basis of recursive and inductive definitions in programming languages, and have many other uses as well.<|endoftext|> TITLE: Is there a 2-connected k-regular graph without Hamiltonian path? QUESTION [12 upvotes]: In this paper (Construction 2.6 p860) the authors have built examples of connected $k$-regular graph without Hamiltonian path, but with a cut-vertex (i.e. it is not $2$-connected). Question: Is there a $2$-connected $k$-regular graph without Hamiltonian path? Remark: every $2$-connected $k$-regular graph with at most $3k+3$ vertices admits a Hamiltonian cycle or is one of the two exceptions, which admit a Hamiltonian path. REPLY [4 votes]: The smallest graph coming from the construction of Fedor Petrov's answer can be realized as follows:<|endoftext|> TITLE: Lexicographic distribution of irreducible polynomials QUESTION [8 upvotes]: Let $A = {\mathbb F}_2[X]$, though the following can be adapted to $p \neq 2$ too. Order the elements of $A$ lexicographically. Equivalently, take a polynomial such as $P = X^4 + X + 1$, write its coefficients as binary digits $\mathbf{b}10011$ and find that it is the $19$th polynomial. Let $f$ be the resulting bijection from $A$ to $\mathbb N$, so $f(P) = 19$. The prime number theorem and Riemann hypothesis are known for the ring $A$. In fact, I learned from Paul Pollack's thesis (http://alpha.math.uga.edu/~pollack/thesis/thesis-final.pdf) that it is essentially contained in Gauss's unpublished 8th chapter of the Disquisitiones. But what about in this not-so-natural lexicographic order? In other words, let $\pi(x)$ be the number of irreducible polynomials $P$ in $A$ such that $0 \leq f(P) \leq x$. Is it known or expected that the prime number theorem or Riemann hypothesis is true for $\pi(x)$? Is $\pi(x) \approx li(x)$ with error $O(\sqrt{x} \cdot \log(x))$? REPLY [6 votes]: This is true. By Gauss's theorem (the inclusion-exclusion formula for the number of irreducibles of a given degree), we may restrict to polynomials of a fixed degree $r$. A moment of reflection then shows that what is needed for lexicographical PNT is exactly the following: For every $k \in \mathbb{N}$, and every length $k$ binary vector $v$ with first letter $1$, we have as $r \to \infty$ (for $k$ fixed) that the proportion of degree $r$ irreducibles having $f(P)$ beginning with $v$ is $\sim 1/2^{k-1}$. For $k = 2$ this means that about half the polynomials have $f(P) \in [2^{r-1},2^{r-1}+2^{r-2}]$ and about half have $f(P) \in [2^{r-1}+2^{r-2},2^{r}]$, etc. But, by the involution taking $P$ to its reciprocal, this the same as asking that, asymptotically as $r \to \infty$ for a fixed $k$, the degree $r$ irreducibles equidistribute in the $2^{k-1}$ invertible residue classes of $\mathbb{F}_2[X]/(X^k)$. This is a particular case of the analog of the PNT in arithmetic progressions, well known to hold in any global field: the modulus $X^k$ here can be replaced by any polynomial element. I think the RH statement should hold as well: the lexicographical writings of the degree-$r$ irreducible polynomials are uniformly distributed with square root error in $[2^{r-1},2^r]$. You could try looking at what the above outline gives with Weil's RH in global function fields: not the rational field $\mathbb{F}_2(X)$ itself, but Carlitz's $X$-primary level analogs of the cyclotomic extensions of $\mathbb{Q}$.To put this (very) loosely, just as ERH is known to imply that primes $< x$ begin to equidistribute over $(\mathbb{Z}/q)^{\times}$ as soon as $x \gg q^{2+\epsilon}$, so similarly we should be able to take $r \approx 2k$ in the above, using Weil's theorem. [Added: This is indeed carried out in Pollack's paper provided by So-called friend Don's comment. ]<|endoftext|> TITLE: Intuitive descriptions of some large cardinals QUESTION [6 upvotes]: I was trying to formulate intuitive descriptions of some large cardinals. Roughly something equivalent to "A manifold is an object which looks like patches of $R^n$ glued together". Not perfectly rigorous, but hopefully conveys the basic picture. Here are three descriptions I have in mind: 1) An inaccessible cardinal is a set so large that it can't be reached from smaller infinite sets using unions and power set operations. 2) A measurable cardinal is a set so large that it can't be reached from smaller infinite sets using any set theoretic formula. (I am going for V $\neq$ L) 3) A Reinhardt cardinal is a set so large that all possible properties of the entire set theoretic universe are also true of this set. (If I am correct, this would capture the intuition of why Reinhardt cardinals don't exist. They are simply too ambitious). I am pretty confident that description 1 is correct in essentials, but are 2) and 3) hopelessly off ? If so, how would one appropriately modify the descriptions ? Or are object like measurable cardinals just too abstract to be expressed in anything but technical definitions ? Apologies in advance if this question is too elementary for MathOverflow. REPLY [7 votes]: (1) seems okay, but I'm afraid that most large cardinal properties beyond inaccessibility probably aren't going to admit such simple formulations. I'm not sure I understand the precise meaning of (2), but in any case it doesn't seem right. For one thing, the existence of $0^\sharp$ is weaker than the existence of a measurable cardinal and still implies $V \ne L$. Also if $0^\sharp$ exists then there are many cardinals, namely indiscernibles for $L$, that seem to satisfy (2) but are not measurable. I don't think (3) is right either. To me it sounds like it describes a cardinal $\kappa$ such that $V_\kappa$ is an elementary substructure of $V$. The existence of such a cardinal is much weaker than the existence of a Reinhardt cardinal, and in fact is equiconsistent with $\mathsf{ZFC}$ by a compactness argument.<|endoftext|> TITLE: Does every cocompact lattice admit a homomorphism (with infinite image) into a compact Lie group? QUESTION [6 upvotes]: Let $\Gamma$ be a cocompact arithmetic lattice in a semisimple algebraic group. Does it admit a homomorphism $\Gamma \to K$ with infinite image into a compact real Lie group $K$? REPLY [14 votes]: No, there are cocompact lattices for which no such homomorphism exists. This is Warning 16.4.3 on page 330 of my book Introduction to Arithmetic Groups. Assume $\Gamma$ is irreducible, $G$ has no compact factors, and $\mathrm{rank}_{\mathbb{R}} G \ge 2$. Modulo finite groups, the Margulis Arithmeticity Theorem tells us there is a semisimple algebraic group $S$, defined over $\mathbb{Q}$, such that $\Gamma = S(\mathbb{Z})$. One version of the Margulis Superrigidity Theorem (namely, Corollary 16.4.1 in my book) tells us (modulo finite groups) that any homomorphism from $\Gamma$ to $\mathrm{GL}(n,\mathbb{R})$ must extend to be defined on all of $S(\mathbb{R})$. If $S(\mathbb{R})$ has no compact factors (that is, if $G$ has no compact factors and $\Gamma = G(\mathbb{Z})$), this implies that $\Gamma$ has no infinite-image homomorphism to any compact, connected Lie group $K$. @YCor's construction using norm 1 elements in a division algebra does indeed give an example, since $\Gamma = S(\mathbb{Z})$ and $S(\mathbb{R}) \cong \mathrm{SL}(n, \mathbb{R})$ has no compact factors.<|endoftext|> TITLE: Hilbert modular forms twist-equivalent to their conjugates QUESTION [5 upvotes]: Let $L / K$ be a solvable (or cyclic) Galois extension of totally real fields, and let $f$ be a Hilbert modular newform over $L$. Suppose that, for every $\sigma \in Gal(L / K)$, the conjugate newform $f_\sigma$ is twist-equivalent to $f$, i.e. there exists a Hecke character $\chi_\sigma$ of $L$ such that $a_{\sigma(\mathfrak{p})}(f) = \chi_{\sigma}(\mathfrak{p}) a_\mathfrak{p}(f)$ for all but finitely many primes $\mathfrak{p}$ of $L$, where $a_{\mathfrak{p}}(f)$ is the Hecke eigenvalue at $\mathfrak{p}$. Does it follow that $f$ is twist-equivalent to the base-change to $L$ of a Hilbert modular form over $K$? I'm happy to assume that $f$ is non-CM, and that all weights of $f$ are $\ge 2$, if that helps. (Note that Galois is acting on $L$ here, not on the coefficients of $f$ -- this is not the same setup as the Ribet--Momose theory of "inner twists".) REPLY [2 votes]: I'm going to add an answer to my own question, not because there's anything wrong with user94346's answer, but because I found a paper in the literature treating exactly this question: Erez Lapid and Jonathan Rogawski, On twists of cuspidal representations of GL(2). Forum Math. 10 (1998), no. 2, 175–197. DOI: 10.1515/form.10.2.175<|endoftext|> TITLE: Winding number of a random walk on the square lattice before hitting the origin QUESTION [7 upvotes]: Let us consider a simple random walk on $\mathbb{Z}^2$ started at $(x,0)$ and killed upon hitting the origin. Define the total winding number $w_x$ around the origin to be the (signed) number of complete rotations around 0 (up to the last step before the random walk is killed). What is known about the distribution of $w_x$ as $x\to\infty$? More precisely, does $w/\log(x)$ converge in distribution? So far I have not been able to find a reference addressing this particular question. Indeed, most references on winding numbers of random walks seem to focus on the distribution after a (large) fixed number of steps. One would expect the distribution to be related to the analogue for Brownian motion started at $(x,0)$ and killed upon hitting the unit circle. By going to radial coordinates it is easy to see that this total winding is given by the integer part of a Cauchy random variable with scale parameter $\log(x)/(2\pi)$. REPLY [2 votes]: Ofer Zeitouni's suggestions based on the paper of Shi can most likely be made precise to prove the asymptotic result using Brownian motion. At the risk of self-promotion and self-answering my 2-year-old question, let me advertise an alternative combinatorial solution that gives exact statistics for the total winding angle $\Theta_x~(\approx 2\pi w_x$) of a simple random walk started at $(x,x)$ and killed upon hitting $(0,0)$ for any finite $x\geq 1$. Let $[\Theta_x]\in\pi(\mathbb{Z}+\tfrac{1}{2})$ be this angle rounded to the nearest half-integer multiple of $\pi$. Then it has characteristic function given explicitly by $$ \mathbb{E}[e^{i b[\Theta_x]}] = \frac{1}{2\cos(\tfrac{\pi b}{2})} [z^{2x}] \left(\frac{1-z}{1+z}\right)^{\!1-|b|}\qquad\text{for }b\in[-2,2], $$ where $[z^{2x}]\,\cdot$ means the coefficient of $z^{2x}$ in the series expansion around $z=0$. This is proved in Theorem 6 (together with Proposition 5) of T. Budd, The peeling process on random planar maps coupled to an O(n) loop model (with an appendix by Linxiao Chen), arXiv:1809.02012, where it arises as a byproduct of the study of a certain exploration process of random planar maps. Straightforward singularity analysis then shows that $$ \mathbb{E}[\exp(ib \tfrac{[\Theta_x]}{\log x})] \xrightarrow{x\to\infty} e^{-|b|}\qquad\text{for }b\in\mathbb{R}, $$ which is precisely the characteristic function of the standard Cauchy random variable. Hence we deduce the convergence in distribution of $[\Theta_x]/\log x$ to Cauchy, and the same for $\Theta_x/\log x$ and $2\pi w_x/\log x$.<|endoftext|> TITLE: Why relative consistency results by forcing arguments are provable in finitistic metatheory QUESTION [7 upvotes]: It is claimed in many textbooks that relative consistency results, such as $\text{Con}(\text{ZFC})\rightarrow\text{Con}(\text{ZFC}+2^{\aleph_0}\geq\aleph_2)$, are provable in the finitistic metatheory. It is also claimed by MO users that it is actually provable in PA. Formal proof of Con(ZFC) => Con(ZFC + not CH) in ZFC However, PA may still be too strong for being finitistic. According to Simpson's Subsystems of Second Order Arithmetic, systems like PRA or $I\Sigma_1^0$ may be accepted by some finitists. Question: are those relative consistency results provable in finitistic systems, such as PRA or $I\Sigma_1^0$? If yes, how to see that? If not, which is the weakest natural system we need to produce the proof? I have also notice the answer Formal systems needed to formalize relative independence results. But more details are prefered. ==================================== I am trying to do as much as I can. Fix a forcing notion, say $\mathbb{P}=\text{Fn}(\aleph_2\times\aleph_0,2)$. I think it can be shown that there is a primitive recursive mapping taking each sentence $\sigma$ in the language of set theory (not the forcing language) to the the set theory statement $1\Vdash\sigma$ (or $\forall p\in\mathbb{P}~ p\Vdash\sigma$). Fix $\sigma\in(\text{ZFC}+2^{\aleph_0}\geq\aleph_2)$. I think the Robinson arithmetic $Q$ is sufficient to prove: $\text{ZFC}\vdash (1\Vdash\sigma)$. Now the question is are the following provable in $I\Sigma^0_1$? $\forall x[x\in(\text{ZFC}+2^{\aleph_0}\geq\aleph_2)\rightarrow\text{ZFC}\vdash(1\Vdash x)]$; $\forall x\big[\big((\text{ZFC}+2^{\aleph_0}\geq\aleph_2)\vdash x\big)\rightarrow\text{ZFC}\vdash(1\Vdash x)\big]$. REPLY [8 votes]: @AsafKaragila This is really a comment, not an answer, but the system wouldn't let me enter it as a comment. (My apologies if this is inappropriate.) I did some work on this approach to forcing (the one you heard of from Magidor) many years ago. I'll describe it, but I'm not sure whether this is all commonly-known folklore; I haven't seen it published. The motivation was to extend a nice property of Prikry forcing to forcing in general — namely, that one can actually carry out Prikry forcing, in an appropriate sense, without passing to a generic extension. Let κ be a measurable cardinal with some given normal ultrafilter, and let P be the partial ordering for Prikry forcing (which would make κ cofinal with ω). Then there is an inner model M and an elementary embedding j: V ⧼ M such that you can carry out (in V) Prikry forcing over M: there exists (in V) an M-generic filter over j(P). Because of the elementary embedding, having a Prikry sequence through j(κ) for M is more or less as good as having a Prikry sequence through κ for V. But the nice thing is that we can actually get an M-generic sequence in V. (The proof proceeds by taking iterated ultrapowers of V by the original normal ultrafilter on κ. M is the ωth iterate, the elementary embedding j is jω, and the desired Prikry sequence over M is 〈jn(κ) | n < ω〉.) One can carry out a similar sort of "internal forcing" for any notion of forcing P, by extending the collection of dense open sets of P to an ultrafilter U, and letting j be the elementary embedding mapping V to a submodel M of the ultrapower (one uses a limit ultrapower containing "eventually constant" functions with respect to P). As I recall, names can be used as usual to construct a generic extension M[G] over j(P). Note that this is all taking place in V. The catch, of course, is that M will generally not be well-founded, so it won't be isomorphic to a transitive class with the usual membership relation. Incidentally, one way to think of M[G] is as a limit ultraproduct, a submodel of an ultraproduct Πp∈P M[Gp] / U, where Gp is an M-generic filter (not necessarily in V) on P containing p.<|endoftext|> TITLE: Subsets of $M$ such that $2M \cong M$ QUESTION [5 upvotes]: I have a question concerning cardinal arithmetic in $\sf{ZF}$. Write $ X \cong Y$ if there is a bijection between $X$ and $Y$. Let $M$ be a set such that $2M\cong M$. Can one show, in $\sf{ZF}$, that for any infinite subset $X$ of $M$, we also have $2X \cong X$ ? Thank you :) REPLY [12 votes]: No. Simply because if $A$ is a set of any cardinality, there is $B$ such that: $|A|\leq|B$, and $|B|+|B|=|B|$, and in fact we can require $|B\times B|=|B|$. Just take $B=A^\omega$, all the functions from $\omega$ to $A$. Then we have that $|B|=|A|^{\aleph_0}$ and therefore $|B\times B|=|A|^{\aleph_0+\aleph_0}=|A|^{\aleph_0}=|B|$. Now take $A$ to be some set such that $|A|<2|A|$, e.g. an infinite Dedekind-finite set, and you get a counterexample. (As Joel remarks, taking $B=\omega\times A$ is sufficient for $|B|+|B|=|B|$. However, it might not be enough if we want $|B|^2=|B|$.) Another fun way of proving this is to prove that for every $\alpha$, $V_\alpha$ satisfies that $|V_\alpha|=2|V_\alpha|=|V_\alpha|^2$. So if the additivity would be hereditary (for infinite subsets), it would be provable in $\sf ZF$ that for every infinite cardinal, $|A|+|A|=|A|$. But this is not true, of course.<|endoftext|> TITLE: Can the Hochschild cochain complex be given the structure of a "homotopy BV algebra"? QUESTION [7 upvotes]: In a 1993 letter, Deligne posed the following (paraphrased from a paper of Gerstenhaber and Voronov's): Conjecture (Deligne). The Hochschild cochain complex $CC^*(A)$ of an associative algebra $A$ has a natural structure of an algebra over a chain operad of the little 2-disks operad $E_2$. ("Chain operad of $E_2$" means any operad in $Ch$ whose homology is isomorphic to the homology of $E_2$, i.e. to the Gerstenhaber operad.) Various versions of Deligne's conjecture have been proven, e.g. by Gerstenhaber-Voronov, Tamarkin, Voronov, McClure-Smith (see also 1, 2), and Kontsevich-Soibelman. Now, in some situations, $HH^*(A)$ can be equipped with a BV operator $\Delta$ compatible with the Gerstenhaber algebra structure, so that it becomes a BV algebra. For instance, this was proven by Tradler in the case that $A$ is unital and equipped with a symmetric, invariant, and non-degenerate inner product. And BV algebras are the same thing as algebras over the framed little 2-disks operad $\tilde E_2$. So... Question: With suitable hypotheses on $A$, can $CC^*(A)$ be given the structure of an algebra over a chain operad of $\tilde E_2$? REPLY [5 votes]: Yes, this is the "cyclic Deligne conjecture", which also has several proofs by now. I believe the first one was Kaufmann's, using the cacti operad.<|endoftext|> TITLE: Involutions and Little Adjoint Representations of Simple Algebras QUESTION [6 upvotes]: In what follows I'm going to use $V_{\theta_s}$ for the little adjoint representation af a Lie algebra i.e. the representation associated with the highest short rooth $\theta_s$. Is easy to see that simple algebras of types $B_n$, $C_n$ and $F_4$ can be found as subalgebras of respectively $D_{n+1}$ with $n>3$, $A_{2n-1}$ and $E_6$ looking at the fixed poits algebra of an opportune involution (induced by an automorphism of the Dynkin Diagram) In this article http://arxiv.org/pdf/math/0303222.pdf the autor says that if $\mathfrak{g}= \mathfrak{g}_0\oplus \mathfrak{g}_1$ is the decompoisition induced by the involution as above, then $\mathfrak{g}_1$ is isomorphic to the $\mathfrak{g}_0$-module $V(\theta_s)$. Looking at the article this seems to be very esasy to prove but I have no ideas and no reference. Any help is going to be well accepted. Probably this question is very easy for people on this site, in such a case forgive me, please. REPLY [4 votes]: The involutive compact Lie algebras yield symmetric spaces and the representations in point are the corresponding isotropy representations. The first symmetric space is the sphere $SO_{2n+2}/SO_{2n+1}=S^{2n+1}$, whose isotropy representation is the standard action of $SO_{2n+1}$ on $\mathbb R^{2n+1}$. Its highest weight (after complexifying) is $\epsilon_1$, where we denote the roots of $SO_{2n+1}$ by $\pm\epsilon_i\pm\epsilon_j$ (long) and $\pm\epsilon_i$ (short), hence, the highest short root. The second one is $SU_{2n}/Sp_n$ with isotropy representation $\Lambda^2\mathbb C^{2n}\ominus\mathbb C$ and highest weight $\epsilon_1+\epsilon_2$, where the roots are denoted $\pm\epsilon_i\pm\epsilon_j$ (short) and $\pm2\epsilon_i$ (long), hence again the highest short root. The last one is $E_6/F_4$ with isotropy representation given by the irreducible 26-dimensional representation of $F_4$. The roots of $F_4$ are $\pm\epsilon_i\pm\epsilon_j$ (long; there are 24 of them) and $\pm\epsilon_i$ (short; 8 roots) and $\frac12(\pm\epsilon_1\pm\epsilon_2\pm\epsilon_3\pm\epsilon_4)$ (short; 16 roots). The highest short root is $\epsilon_1$ , and this is also the highest weight of the considered representation. About references: you can find the isotropy representations in Joseph Wolf's book Spaces of constant curvature, ch.8. For $F_4$, you find complementary information in J. F. Adams' Lectures on exceptional Lie groups (Edited by Zafer Mahmud and Mamoru Mimura) and in the planches at the end of Bourbaki's book on Lie groups, ch. IV to VI. There could exist some conceptual argument, but I am not sure the work is worth for three representations only. In any case, I will write again if anything occurs to me.<|endoftext|> TITLE: Geometric and arithmetic Frobenius QUESTION [9 upvotes]: I read in Serre's "Lectures on $N_X(p)$" that when $X$ is a scheme defined over $\mathbb{F}_q$ (a finite field), the geometric Frobenius $F: X \mapsto X$ is defined by fixing every element of the topological space of $X$, and acting as $f \mapsto f^q$ on the structure sheaf. As a particular case, if $X$ is affine (say $Spec(\mathbb{F}_{q^n}[x,y])$), then "$F$ is the "standard Frobenius map" (on points): $(u,v) \mapsto (u^q,v^q)$." Why is that ? Doesn't the action on the structure sheaf implies that every element of $\mathbb{F}_{q^n}[x,y]$ is raised to the $q$-th power ? Also, I have seen other definitions of geometric Frobenius. (I have noticed related questions on MO, but the answers confuse me even more.) Secondly, it is mentioned that the geometric and arithmetic Frobenius act in the same way on $X(\overline{\mathbb{F}_q})$. Can anyone explain this in some detail ? Thanks ! REPLY [8 votes]: One common use of the terminologies geom/arith Frob is as follows: Let $X=Spec R$, $R$ a finite type $F_q$-algebra,then the endomorphism $a\mapsto a^q$ induces an endomorphism $Fr$ on $X$, which is usually referred as absolute Frobenius. Let $Y=X\times Spec F$ be the base change of $X$ to $F$, where $F$ is an algebraic closure of $F_q$. Then one calls the endomorphism $Fr\times Id$ a geometric Frobenius on $Y$. Meanwhile, the Galois conjugation $r\mapsto r^q$ gives an automorphism $fr$ of $Spec F$, and the automorphism $Id\times fr$ on $Y$ is called an arithmetic Frobenius. In particular, the above geometric Frobenius is an endomorphism over $F$, while the arithmetic Frobenius is not defined over $F$. On the rational points $Y(F)$ the geometric Frob is the inverse to the arithmetic Frob. If you define the arithmetic automorphism as the inverse to the automorphism we defined here, then your last assertion holds, so it is a choice of terminology.<|endoftext|> TITLE: Cohomology operations on unoriented cobordism QUESTION [9 upvotes]: In unoriented cobordism there exist stable cohomology operations looking similar to Steenrod squares (they were used by Quillen to compute the unoriented cobordism ring with its formal group law defined by tensor product of bundles). Since their properties are so similar to the properties of Steenrod squares, my questions are: do they become normally Steenrod squares in $N^*(X) \otimes_{N*} \mathbb{Z}_2 $? Do they correspond to $Sq^i \otimes id_{N^*} $ under the isomorphism $N^*(X) \simeq H^*(X) \otimes_{\mathbb{Z}_2} N^*$? REPLY [5 votes]: There is a natural transformation $N^*(X)\to H^*(X;\mathbb{Z}/2)$ given by the universal Thom class, and this maps the Steenrod-tom Dieck operations in cobordism to the Steenrod squares in cohomology. A reference is Section 15 of Tammo tom Dieck, MR 244989 Steenrod-Operationen in Kobordismen-Theorien, Math. Z. 107 (1968), 380--401. This answers your first question, I think. As to your second question, I don't know the answer but I suspect the Steenrod-tom Dieck operations are non-trivial on $N^*(pt)$.<|endoftext|> TITLE: Speed of convergence in Lebesgue's density theorem QUESTION [10 upvotes]: Let $\lambda=\text{unif}([0,1])$ be uniform distribution on $[0,1]$ and $B$ be any Borel set. Lebesgue's density theorem states that for $\lambda$-almost all $x\in[0,1]$ the limit $$\lim_{\epsilon\downarrow o}\frac{\lambda([x-{\epsilon},x+\epsilon]\cap B)}{2\epsilon}$$ exists and is either $0$ or $1$. Im interested in a 'global' approximation. For $n\geq 1$ let $$D_n:=\big\{\frac{0}{2^n},\frac{1}{2^n},\frac{2}{2^n},\dots,\frac{2^n-1}{2^n}\big\}.$$ Now i look at the expression $$a_n:=(1/2)^n\cdot\sum_{x\in D_n}\frac{\lambda\big([x,x+(1/2)^n]\cap B\big)}{(1/2)^n}\cdot\bigg[1-\frac{\lambda\big([x,x+(1/2)^n]\cap B\big)}{(1/2)^n}\bigg].$$ So for large $n$ every summand should be close to Zero. Additionally every summand gets weighted by $(1/2)^n$. The number $a_n$ tells you something about how 'intervall-like' $B$ is. Since you can approximate every $B$ by finite unions of intervalls you can show that $a_n\rightarrow 0$ for each $B$. I want to know: How fast does $a_n$ tend to Zero? Is it true that $\sum_{n\geq 1}a_n<\infty$ for any measurable $B$? If this is not the case, can you provide an example of a set $B$ with $\sum_n a_n=\infty$? Remark: This is a special instance of a more general question I posted on Stackexchange (before I knew that Overflow exists, sorry about that): https://math.stackexchange.com/questions/1835620/switching-independent-experiments-does-i-ax-1-dots-x-n-y-n1-y-n2-dots REPLY [5 votes]: The answer is negative. In fact, let us show that $a_n$ as defined in the question may converge to $0$ however slowly. Indeed, let $(\epsilon_j)$ is any sequence in $(0,1]$ converging to $0$ slowly and regularly enough, in the sense that \begin{equation} \epsilon_{j-1}-\epsilon_j\ge2^{2-j}\tag{1} \end{equation} eventually in $j$ -- that is, for some natural $j_0$ and all natural $j\ge j_0$. For instance, this condition will hold if $\epsilon_j$ is (eventually in $j$) of the form $2^{2-j}$ or $j^{-a}$ or $(\underbrace{\ln\dots\ln}_N j)^{-a}$ for some real $a$ and some natural $N$. In what follows, $j$ is a natural number $\ge j_0$. Let $$k_j:=\lfloor2^j\epsilon_j\rfloor$$ and $$r_j:=k_j/2^j.$$ Then \begin{equation} \epsilon_j-\tfrac1{2^j} TITLE: Polylogarithm sheaves QUESTION [10 upvotes]: In many different places, I could find the notion on ''(poly)logarithm sheaves''. As is indicated in the name of it, I guess that it should have something to do with (poly)logarithm function: $\mathrm{Li}_s(z)$. How are they related to each other? Any reference would be helpful, but it would be better if it requires less preliminaries. REPLY [4 votes]: I would suggest looking at Hain's article (arXiv version here) R. Hain. Classical polylogarithms. Motives Proceedings, vol II, Proc. Symposia Pure Math 55.2, 1994. Very roughly, one writes out a multi-valued function on $\mathbb{P}^1\setminus\{0,1,\infty\}$ which takes values in ${\rm GL}_{n+1}(\mathbb{C})$ where the entries of the matrix are built from (poly)logarithms up to ${\rm Li}_n$. This is the fundamental solution of a linear differential equation. The monodromy of the differential equation gives rise to a local system on $\mathbb{P}^1\setminus\{0,1,\infty\}$. This is the $n$-th polylogarithm local system (which can equivalently be regarded as a sheaf). Actually, doing this for all $n$ gives rise to a pro-local system which you might call the polylogarithm sheaf. Under the correspondence between local systems and representations of the fundamental group, this corresponds to the representation of $\pi_1(\mathbb{P}^1\setminus\{0,1,\infty\})$ on the completion of its group ring w.r.t. the augmentation ideal. This has a Hodge-theoretic and even motivic refinement. These are relevant when trying to express regulators on K-theory in terms of polylogarithms (on the way to formulas for special L-values). You can find all this and more in Hain's article, also lots of references in there. Unfortunately, this is not quite the discussion that would "require less preliminaries"...<|endoftext|> TITLE: Renorming a Banach space to make projections contractive QUESTION [5 upvotes]: Let $X$ be a Banach space and $P$ be a projection in $B(X)$. Then $X$ can be renormed so that $P$ has norm $1$. Can the same be done for a family of projections? That is, given finitely many projections $P_1$,...,$P_n$ in $B(X)$ is there an equivalent norm under which all projections become contractive? Are there any assumptions other than $P_i$ commuting that would ensure this? REPLY [16 votes]: Let $X$ be the Euclidean plane, and consider the two projections $$ P_1=\begin{pmatrix}1&A\\0&0\end{pmatrix}\quad\text{and}\quad P_2=\begin{pmatrix}0&0\\A&1\end{pmatrix}, $$ where $A$ is a big constant (actually any $A>1$ will do). If a norm on $X$ gave both of them norms $\leq1$, then the same would be true for their product, and for powers of their product,which have the form $$ (P_1P_2)^n=\begin{pmatrix}A^{2n}&A^{2n-1}\\0&0\end{pmatrix}. $$ But these powers send the vector $\binom10$ to larger and larger vectors, which cannot all be in any ball.<|endoftext|> TITLE: A question about composition of functions QUESTION [9 upvotes]: Recently, I heard this question: are there two functions $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ such that $f\circ g$ is strictly increasing on $\mathbb{R}$ and $g\circ f$ is strictly decreasing on $\mathbb{R}$? I have the feeling that it is not new. Does anybody know any references? REPLY [13 votes]: Yes. $$f(x)=(-1)^{\lceil x \rceil}\text{abs}(x+\text{sign}(x)),\ g(x)=(-1)^{\lceil x \rceil}\text{abs}(x)$$<|endoftext|> TITLE: Does Littlewood's bound on $\zeta(1+it)$ extend to all the partial sums? QUESTION [15 upvotes]: Littlewood established that $2e^{\gamma} \geq \limsup_{t \to \infty} |\zeta(1+it)| / \log{\log{t}} \geq e^{\gamma}$, the lower bound unconditionally and the upper bound on RH. It now seems to be generally believed that the lower bound represents the truth, and even, in the most optimistic form, that quite possibly there is an absolute constant $C$ such that $|\zeta(1+it)| \leq e^{\gamma}(\log{\log{t}} + \log{\log{\log{t}}}+C)$ for all $t > 10$. Suppose in these questions that we replace $|\zeta(1+it)|$ by $$ B(t) := \sup_k \Big| \sum_{n=1}^k n^{-1-it} \Big|, \quad \textrm{ resp. } \widetilde{B}(t):= \sup_{m,k} \Big| \sum_{n=m}^k n^{-1-it} \Big|, $$ the largest partial sums, respectively the largest subsum over all intervals. Does Littlewood's RH result extend to $B$ or even $\widetilde{B}$? That is: should the exponential sums bound $\widetilde{B}(t) = O(\log{\log{t}})$ be possible to prove on RH, or what kind of bound is available in this uniformity? Should it be reasonable to expect the strongest possible bound $\widetilde{B}(t) \leq e^{\gamma}(\log{\log{t}} + \log{\log{\log{t}}}+C)$, for all $t$? There are evident versions of this question for $L(1,\chi)$, $\frac{\zeta'}{\zeta}(1+it)$ and $\frac{L'}{L}(1,\chi)$. In the last of these, the strongest bound seems to tie well with the belief that the smallest quadratic non-residue mod $q$ is $\ll \log{q}\log{\log{q}}$ -- which it certainly implies, at least when $q$ is prime. REPLY [11 votes]: The short answer is yes, and this is treated explicitly for characters in the work of Granville and Soundararajan (the paper appeared in J. Amer. Math. Soc.). Their Theorem 2 gives that on GRH for $x\le q$ and a primitive character $\chi \pmod q$ one has $$ \Big| \sum_{n\le x} \chi(n) \Big| \ll \Psi(x, (\log q)^2 (\log \log q)^{20}), $$ where $\Psi(x,y)$ denotes the number of integers up to $x$ composed only of prime factors below $y$. From this and partial summation, it follows at once that for $x\le q$ $$ \Big| \sum_{n\le x} \frac{\chi(n)}{n} \Big| \ll \prod_{p\le (\log q)^2(\log \log q)^{20}} \Big(1- \frac{1}{p}\Big)^{-1} \ll \log \log q. $$ In fact their Theorem 2 is a bit more precise, and one should be able to get the Littlewood type constant also. This result will also extend to the $t$-aspect with minor changes, and establish the bound that you want. Alternatively you could look at another paper of Granville and Soundararajan (also in JAMS, and also in the case of characters); the argument given there in Section 5 would give a slightly simpler approach to such results. Here's a quick sketch proof based on the second approach. Let $|t|$ be large, and assume that $x\le |t|$ (for larger $x$ the partial sum simply approximates $\zeta(1+it)$). Let $y$ be a parameter to be chosen, and note that $n\le x$ is either $y$-smooth, or may be written as $n=mp$ where $p\ge y$ is the largest prime factor of $n$ (so that $m$ is $p$-smooth). Thus, with $P(m)$ denoting the largest prime factor of $m$, $$ \sum_{n\le x} \frac{1}{n^{1+it}} = \sum_{\substack{n\le x\\ p|n \implies p\le y}} \frac{1}{n^{1+it}} + \sum_{m\le x/y} \frac{1}{m^{1+it}} \sum_{\max(y, P(m)) \le p \le x/m} \frac{1}{p^{1+it}}. $$ The first term in the RHS is simply bounded by $\prod_{p\le y} (1-1/p)^{-1}$. As for the second term, RH can be used (usual contour shift argument) to show that the sum over $p$ there is $$ \ll \frac{(\log |t|)}{\sqrt{y}}, $$ and bounding the sum over $m$ trivially, the second term is $$ \ll \frac{(\log |t|) \log x}{\sqrt{y}} \ll \frac{(\log |t|)^2}{\sqrt{y}}. $$ Now choosing $y=(\log |t|)^4$ gives a bound of $\ll \log \log |t|$ for your partial sums. With more care the product may be truncated at $y=(\log |t|)^{2+\epsilon}$. One would expect that truncation at $y=(\log |t|)^{1+\epsilon}$ (or even $y=C \log |t| \log \log |t|$) is the truth.<|endoftext|> TITLE: Can a Morse function define a unique structure on a closed manifold? QUESTION [6 upvotes]: I was thinking about the doubt that if $M$ and $N$ are closed manifold and if there exists two Morse function $f$ and $g$ respectively on $M$ and $N$ with the following property that they both have same number of $i$ index critical points for each $i\in \mathbb N$, then whether I can conclude they are homeomorphic or not. It is clear that they are of same dimension since the maximum of $f$ and $g$ are of same index. And we know that if a closed manifold has exactly two critical points then it is homeomorphic with $S^n$. But still I don't expect an positive answer for my doubt. So to find a contradiction I was thinking about handle decomposition of some closed manifold. And now I got stuck with this doubt that suppose I have a handle decomposition of a closed manifold $X$, then can I construct a Morse function $f$ on $X$ such that corresponding to this Morse function I get the same handle representation?? This is kind of converse of Morse handle presentation theorem. So far I could not able to construct such a function. And also I could not able to find any counter example for my initial doubts. It'll be very helpful if someone clarify my these two doubts. REPLY [4 votes]: In answer to your second question, yes, given a handle decomposition it is possible to find a Morse function whose corresponding handle decomposition is the same as the one you started with. I think there's a proof, via "gradient-like vector fields", in Milnor's book on Morse theory (though I don't have a copy in front of me now).<|endoftext|> TITLE: Why is every object cofibrant in an excellent model category? QUESTION [14 upvotes]: In Appendix A.3 of the book higher topos theory appears the notion of an excellent model category (see Definition A.3.2.16). The main feature of this notion is that when $\mathbf{S}$ is an excellent model category one has a model structure on the category of small $\mathbf{S}$-enriched categories which enjoys several good properties. Condition (A2) in the definition of an excellent model category states that every monomorphism is a cofibration and that the collection of cofibrations is closed under taking products. In remark A.3.2.17 which appears right after the definition it is said that condition (A2) implies that every object of $\mathbf{S}$ is cofibrant. This remark is reiterated (and relied upon) in the very nice paper http://arxiv.org/abs/1602.05313 of Tyler Lawson where it showed that the first four conditions in the definition of an excellent model category imply the fifth (which is essentially the only one which is hard to check in practice). I am probably just missing something trivial here, but I just don't see why condition (A2) implies that every object is cofibrant. This would be the case, of course, if the map $\emptyset_{\mathbf{S}} \to X$ from the initial object would always be a monomorphism, but this is not true in general, not even for presentable categories (for example, in the category of rings, the map $\mathbb{Z} \to 0$ from the initial ring to the terminal ring is not a monomorphism). My question is then: how does condition (A2) imply that every object is cofibrant? REPLY [4 votes]: I am now under the impression that it is simply not true that in an excellent model category every object is cofibrant. Let $\mathbf{S}$ be an excellent model category in which the monoidal structure is the Cartesian one (e.g., simplicial sets). For every $X \in \mathbf{S}$ we may consider the coslice category $\mathbf{S}_{X/}$ with its canonical model structure, in which all three classes are created by the projection $p:\mathbf{S}_{X/} \to \mathbf{S}$. We now note that $p$ creates limits and connected colimits (i.e., colimits indexed by connected diagrams), and in particular pushouts. It then follows that the Cartesian product on $\mathbf{S}_{X/}$ is compatible with the model structure. We now claim that $\mathbf{S}_{X/}$ is an excellent model category (with respect to the Cartesian monoidal structure). First note that $\mathbf{S}_{X/}$ is combinatorial. Property (A2) discussed above follows from the same property for $\mathbf{S}$ since $p:\mathbf{S}_{X/} \to \mathbf{S}$ is a right functor and hence preserves products and monomorphisms (while creating cofibrations by definition). Weak equivalences in $\mathbf{S}_{X/}$ are closed under filtered colimits since $p$ preserves filtered colimits and detects weak equivalences. Finally, the invertibility hypothesis follows from Lemma A.3.2.20 of HTT since we have a left Quillen functor $\mathbf{S} \to \mathbf{S}_{X/}$. On the other hand, $\mathbf{S}_{X/}$ will typically not have all its objects cofibrant. Take for example $\mathbf{S}$ to be simplicial sets and $X = \Delta^0 \coprod \Delta^0$, in which case the object $X \to \Delta^0$ is not cofibrant in $\mathbf{S}_{X/}$. Edit: Actually the monoidal product on $\mathbf{S}_{X/}$ cannot not be taken to be the Cartesian one, because the Cartesian product on $\mathbf{S}_{X/}$ doesn't preserve colimits in each variable separately, so this argument doesn't work as is. However, all the rest seems to still be correct, so that $\mathbf{S}_{X/}$ will be an excellent model category as soon as it has some compatible monoidal structure. In particular, $\mathbf{S}_{X/}$ satisfies (A2) but not every object is cofibrant in general. Second Edit: Constructing monoidal structures on $\mathbf{S}_{X/}$ turned out to be quite of a headache. However, there is a natural left action of $\mathbf{S}$ on $\mathbf{S}_{X/}$ which is given by $$ Z \otimes (X \to Y) := X \to X \coprod_{Z \times X} \left[Z \times Y\right] $$ and a natural right action given by $$ (X \to Y) \otimes Z := X \to X \coprod_{X \times Z} \left[Y \times Z\right] .$$ Here the pushouts are performed with respect to the projections $Z \times X \to X$ and $X \times Z \to X$ respectively (these two actions are naturally isomorphic, but it is convenient to have them written on both sides). These actions preserve colimits in each variable separately, and it is not hard to verify that if the Cartesian product on $\mathbf{S}$ satisfies the pushout-product axiom then these actions satisfy the pushout-product axiom as well. We may hence think of $\mathbf{S}_{X/}$ as a bimodule model category over $\mathbf{S}$. In this case we can construct a new (non-Cartesian) monoidal model category, the "square-zero extension" $\mathbf{S}_{X/} \rtimes \mathbf{S}$, whose underlying model category is the product model category $\mathbf{S}_{X/} \times \mathbf{S}$, and such that $$(Z,X \to Y) \otimes (Z',X' \to Y') := (Z \times Z',\left[Z \otimes (X' \to Y')\right] \coprod \left[(X \to Y) \otimes Z'\right])$$ When $\mathbf{S}$ is an excellent Cartesian model category (e.g. simplicial sets) then $\mathbf{S}_{/X}$ satisfies all the axioms of an excellent model category except the existence of a monoidal structure (see above). This implies that the monoidal model category $\mathbf{S}_{X/} \rtimes \mathbf{S}$ is actually an excellent model category. However, in general not every object in $\mathbf{S}_{X/} \rtimes \mathbf{S}$ is cofibrant, e.g. when $\mathbf{S}$ is simplicial sets and $X = \Delta^0 \coprod \Delta^0$. The conclusion of this discussion is not that provocative. Since the condition that every object is cofibrant is required (or at least used) for setting up a model structure on $\mathbf{S}$-enriched categories (see Proposition A.3.2.4 of HTT), all this means is that one should simply add "every object is cofibrant" to the list of axioms defining an excellent model category.<|endoftext|> TITLE: On spatial tensor products of von Neumann algebras QUESTION [14 upvotes]: Let $H$ be a Hilbert space, and let $A_1,A_2,A_3\subset B(H)$ be three commuting von Neumann algebras. We write $\odot$ for the algebraic tensor product, and $\bar\otimes$ for the spatial tensor product of von Neumann algebras. Suppose that for every $i,j\in\{1,2,3\}$, the map $A_i\odot A_j\to B(H)$ extends to a map $A_i\,\bar\otimes\, A_j\to B(H)$. Does it follow that the map $A_1\odot A_2\odot A_3\to B(H)$ extends to a map $A_1\,\bar\otimes\, A_2\,\bar\otimes\, A_3\to B(H)$? REPLY [6 votes]: No. Consider the case where $G$ is an ICC group, $H=\ell_2(G\times G)$, $A_1=(\lambda\otimes\lambda)(G)''$, $A_2=(1\otimes\rho)(G)''$, and $A_3=(\rho\otimes1)(G)''$. By the Fell absorption principle $(g,h)\leftrightarrow (g,gh)$, one has $A_1\vee A_2 \cong (\lambda\otimes1)(G)''\vee(1\otimes\rho)(G)''\cong A_1\mathbin{\bar{\otimes}}A_2$, and similarly for other $A_i \vee A_j$. But since the diagonal $(\lambda\otimes\lambda)(g)\cdot(1\otimes\rho)(g)\cdot(\rho\otimes1)(g)$ fixes $\delta_e\otimes\delta_e$, one has $A_1\vee A_2\vee A_3 \not\cong A_1\mathbin{\bar{\otimes}}A_2\mathbin{\bar{\otimes}}A_3$.<|endoftext|> TITLE: Bijection modeling isomorphism of infinite-dimensional vector spaces QUESTION [7 upvotes]: Let $T : V \to W$ be an isomorphism of vector spaces with bases $B_V$ and $B_W$, which may be of any cardinality. Does there exist a bijection $f : B_V \to B_W$ such that, for each $b_V \in B_V$, the coefficient of $f(b_V)$ in $T(b_V)$ is nonzero? If $V,W$ are finite-dimensional, the answer is yes: $\det(T) \ne 0$, hence some monomial term is nonzero. This exhibits a satisfactory $f$. Alternately, build $f$ inductively by Laplace expansion along a row or column. (Some term $a_{ij} \cdot \text{(complementary minor)}$ is nonzero, and so on.) Does this hold in general? I have tried using Zorn's lemma, but it seems tricky enough that maybe I'm overlooking a straightforward counterexample. I can state what I've attempted if there's interest. (I have tagged this as linear and homological algebra since my desired application is in the latter.) REPLY [5 votes]: Yes. Assume $B_V, B_W$ are infinite. First (merely to simplify notation) reduce to $B_V, B_W$ countable: start with any $v \in B_V$, include the finitely many $w \in B_W$ that are involved in $T(v)$, include the finitely many $v \in B_V$ that are involved in all $T^{-1}(w)$, include the finitely many $w \in B_W$ that are involved in all $T(v)$, etc. Thus $V$ is a direct sum of subspaces each spanned by a countable subset of $B_V$ and mapped isomorphically by $T$ onto a subspace spanned by a countable subset of $B_W$. Let $B_V = \{{v_1,v_2,...\}}, B_W = \{{w_1,w_2,...\}}$. Write $w_1=a_1T(v_{i_1})+...+a_nT(v_{i_n})$ with all $a_i \neq 0$. Some $T(v_{i_j})$ has non-zero $w_1$-component, say $T(v_{i_1})$. Claim: $T : \oplus V_{i:i\neq i_1} \simeq W/w_1$ so define $f(v_{i_1})=w_1$ and continue the process inductively, ping-ponging between $T$ and $T^{-1}$ to ensure that $f$ is bijective. Proof of claim: $T |\oplus V_{i:i\neq i_1}$ is injective because $a_1 \neq 0$; it is onto $w_2$ (for example) because if $w_2=b_1T(v_{k_1})+...+b_mT(v_{k_m})$ and if (say) $k_1=i_1$ then substitute $$b_1T(v_{k_1}) = b_1 a_1^{-1}(w_1-a_2T(v_{i_2})-...-a_nT(v_{i_n})).$$ Note that $f$ has the required property for $T$ and as well $f^{-1}$ has the required property for $T^{-1}$. Edit: $V_i$ means $Kv_i$. Edit: Another (similar) way to see the induced smaller map is an isomorphism uses the snake lemma. Let $C_V = \mathrm{span}(v_i : i \ne i_1)$ and $C_W = W/(w_1) = \mathrm{span}(w_i : i \ne 1)$. We have a diagram $$\begin{array}{ccccccccc}0&& C_V & = & C_V \\&& \downarrow & &\downarrow \tilde{T} \\ (w_1) & \rightarrow & W & \rightarrow & C_W \end{array}$$ which gives from the snake lemma $$0 \to \ker(\tilde{T}) \to (w_1) \to W/T(C_V) \to \mathrm{coker}(\tilde{T}) \to 0.$$ Note that $W/T(C_V)$ is one-dimensional, spanned by $T(v_{i_1})$, so the middle map is an isomorphism or zero. It is not zero because $w_1 \notin T(C_V)$ (because $a_1 \ne 0$). Thus $\ker(\tilde{T}) = \mathrm{coker}(\tilde T) = 0$.<|endoftext|> TITLE: Basic Definition and Notations in RWRE QUESTION [6 upvotes]: From the definition of Zeitouni's lecture notes on RWRE: $(V, E)$ is a special graph, and $N_v:= \{k \in V: (v,k) \in E\}$ is the neighborhood of $v \in V$. $\Omega = \prod_{v \in V} M_1(N_v)$ together with the $\sigma-$algebra $\mathcal{F}$ generated by cylinder functions forms a measurable space $(\Omega, \mathcal{F})$. We define a probability measure $P$ on $(\Omega, \mathcal{F})$ and a time homogeneous Markov chain $\{X_n\}_{n \ge 0}$ with transition probabilities \begin{equation} Q_{\omega} (X_{n+1} = k|X_n=v) =\omega_v (k) \end{equation} $Q_{\omega}$ is called the quenched law of the RW, and the marginal of $\mathbb{P}:=P\otimes Q_{\omega}$ on $V^{\mathbb{N}}$ is the annealed law of the RW. I intentionally used the letters $Q$ and $k$ above. 1) Is this what I should understand from this definition: $\omega$ is simply an element of the set $\Omega$, and $P$ is a (probability) measure on $(\Omega, \mathcal{F})$. Using $\Omega$ as an index set, $Q_{\omega}$ is the probability measure on $(E_{\omega}, \Sigma_{\omega})$ and $X_n=X_n(\omega, z)$ is a $V-$valued measurable function, i.e., $X_n: (E_{\omega}, \Sigma_{\omega}, Q_{\omega}) \to (V,\mathcal{S})$ for each $\omega \in \Omega$. 2) The environment $\omega$ can be viewed as a sequence of random variables. How can I connect these two definitions? Thanks for your time and comments in advance. REPLY [2 votes]: Unfortunately, the probabilistic argot can be sometimes not so easy to understand for outsiders. Also, I think that your notation is unnecessarily complicated. So, what do we have here? The state space is a graph $V$ (no idea, why it has to be ``special''). We consider only the nearest neighbor random walks, i.e., transitions are allowed only to neighbors, and the space of environments $\Omega$ is the set of all such transition operators (or transition matrices). Once you have fixed a starting point (or, more generally, an initial distribution) on $V$, each environment $\omega\in\Omega$ produces the corresponding Markov measure $\mathbf P_\omega$ on the space of nearest neighbor sample paths in $V$ (probabilists like to call the measures $\mathbf P_\omega$ quenched laws). If you now average the measures $\mathbf P_\omega$ with respect to a distribution $P$ on the space of environments $\Omega$, the resulting measure $\mathbf P = \int \mathbf P_\omega\,dP(\omega)$ on the space of nearest neighbor sample paths in $V$ is called the annealed law.<|endoftext|> TITLE: What is the smallest density of a metrizable space without countable separation? QUESTION [11 upvotes]: A Tychonoff space $X$ is defined to have countable separation if some (equivalently, any) compactification $bX$ of $X$ contains a countable family $\mathcal U$ of open sets such that for any points $x\in X$ and $y\in bX\setminus X$ there is a set $U\in\mathcal U$ containing exactly one point of the doubleton $\{x,y\}$. The class of spaces with countable separation is very rich: for every compact Hausdorff space $K$ the family of subspaces of $K$ having countable separation is a $\sigma$-algebra, closed under the Suslin $A$-operation. This notion was indroduced by Kenderov, Kortezov and Moors in 2001 (at least). It can be shown that each metrizable space of density at most continuum has countable separation. On the other hand, by tranfinite induction it is possible to construct a metrizable space of density $\beth_\omega$ without countable separation (this follows from the fact that each Tychonoff space $(X,\tau)$ of cardinality $|X|\ge|\tau^\omega|>\mathfrak c$ contains a subspace $Y$ without countable separation). Let us recall that $\beth_\omega=\sup_{n\in\omega}\beth_n$ where $\beth_0=\omega$ and $\beth_{n+1}=2^{\beth_n}$, so $\beth_\omega$ is rather large. The key property of the cardinal $\kappa=\beth_\omega$ in this context is that $|\kappa^\omega|=2^\kappa>\mathfrak c$. Question: What is the smallest density of a metrizable space without countable separation? Can it be equal to $\mathfrak c^+$? Or it is always $\ge \aleph_\omega$? It can be shown that every (complete) metric space $X$ of cardinality $|X|=\mathfrak c^+$ contains a subset $B\subset X$ such that for every uncountable Polish subspace $P$ of $X$ both sets $P\cap B$ and $P\setminus B$ are not empty. Such set $B$ will be called a Bernstein set in $X$. Question. Can a Bernstein set $B$ in the Hilbert space $\ell_2(\mathfrak c^+)$ of density $\mathfrak c^+$ have countable separation? A negative answer to the last question would follows from the affirmative answer to this problem. REPLY [2 votes]: Since this problem has an affirmative answer, the last question should have a negative answer and then the smallest density of the space in the first question can be $\frak c^+$.<|endoftext|> TITLE: History of the notation for substitution QUESTION [20 upvotes]: One of the very common notations for syntactic substitution is $[\ /\ ]$. However, there seems to be an inconsistency in the literature about its usage. Many write $[t/x]$ for "substitute $t$ for $x$" (Girard, Buss, ...). Others use $[x/t]$ for "replace $x$ with $t$" (van Dalen, Troelstra, Martin-Löf, ...). I am wondering about the history of this notation. In particular, who was the first person to use this notation for syntactic substitution in logic? REPLY [4 votes]: (To the title question, and @AnstenKlev’s answer:) Sarrus (1848) introduced maybe the earliest signe de substitution : $\style{font-family:sans-serif;font-style:italic}1\,^t_x$.<|endoftext|> TITLE: Bernstein sets of large cardinality QUESTION [6 upvotes]: A metrizable space $X$ will be called a generalized Bernstein set if every closed completely metrizable subspace $C$ of $X$ has cardinality $|C|<|X|$. It is well-known that the real line contains a (generalized) Bernstein set of cardinality $\mathfrak c$. Moreover, for every cardinal $\kappa$ with $\kappa^\omega=2^\kappa$ there exist a generalized Bernsten set $X$ of cardinality $|X|=2^\kappa=\kappa^\omega$. In particular, there exists a generalized Bernstein set of cardinality $\beth_{\omega+1}$. Question. What are possible cardinalities of generalized Bernstein sets? For example, is there a generalized Bernstein set $X$ of cardinality $|X|=\mathfrak c^+$? REPLY [2 votes]: After thinking a night on this question and waking up, I realized that the answer is almost trivial: there are restrictions on possible cardinalities of generalized Bernstein set. Any metrizable space $X$ of density $\kappa$ has cardinality $|X|\le\kappa^\omega$ and contains a discrete (and hence completely metrizable) subspace $D$ of cardinality $|D|=\kappa$. If $\kappa=\kappa^\omega$ (which is the case for the cardinal $\mathfrak c^+$), then $X$ cannot be a generalized Bernstein set. On the other hand, there is also a less trivial fact involving (even more) generalized Bernstein sets. Let $\mathfrak c^{+0}=\mathfrak c$ and $\mathfrak c^{+(n+1)}=(\mathfrak c^{+n})^+$ for $n\in\omega$. So, $\mathfrak c^{+n}$ is the $n$th successor cardinal of $\mathfrak c$. Let also $\mathfrak c^{+\omega}=\sup_{n\in\omega}\mathfrak c^{+n}$. By induction it can be shown that each metrizable space $X$ of cardinality $|X|<\mathfrak c^{+\omega}$ contains a subset $B\subset X$ of cardinality $|B|=|X|$ such that for every uncountable Polish subspace $P\subset X$ both sets $P\cap B$ and $P\setminus B$ are not empty. This implies that $B$ contains no uncountable Polish subspaces, so $B$ can be considered as a generalized Bernstein set (in a somewhat different sense).<|endoftext|> TITLE: Can all Local Martingales Be Represented using Only Brownian Motion and Finite Variation Processes? QUESTION [6 upvotes]: This is a cross-post of my unanswered (more than a week) question on Math.SE. Since it covers topics from my graduate-level course on stochastic processes, I thought it might be appropriate to try to ask it here. Prove or Disprove: A process $(X_t)_{t \ge 0}$ is a (continuous) local martingale if and only if it can be represented in the form: $$\int_0^t \xi dB = \large B_{\int_0^t \xi_s^2 ds} $$ where the quadratic variation $\langle X \rangle_t=\int_0^t \xi_s^2 ds$ for some increasing finite variation process $\xi$ (whose paths are therefore absolutely continuous with respect to Lebesgue measure). In particular, a (continuous) local martingale is a deterministic time change of Brownian motion if and only if its quadratic variation is a deterministic function (and absolutely continuous with respect to Lebesgue measure). Basically I want to know how much the theory of stochastic integration/local martingales can be reduced to the study of Brownian motion and finite variation processes only. Is this the same thing as saying thing that all local martingales are Ito processes and vice versa? EDIT: See here for my latest attempt to solve this problem using Stephan Sturm's answer below. Related: this question covers a lot of similar ground, although it does not address how that specific representation can be made to relate to time changes. https://math.stackexchange.com/questions/1457270/representation-theorem-for-local-martingales This question on MathOverflow seems somewhat similar in spirit: The only continuous martingales with stationary increments are Brownian motions Also, and most importantly, these two blog posts by George Lowther (who answered the above linked MO question) are the primary basis for my question, because I think the answer might be contained somewhere within the two posts, but I am having difficulty putting everything together. https://almostsure.wordpress.com/2010/04/20/time-changed-brownian-motion/ https://almostsure.wordpress.com/2010/04/13/levys-characterization-of-brownian-motion/ Motivation: Basically I am trying to devise as simple of an introduction to the concept of local martingale as possible for my study guide for stochastic integration, to be found here. I find the most commonly used characterization, as a localization of a martingale, to be unsatisfactory for several reasons. First, it affords none of the intuition developed from studying martingales in the discrete-time case, since in the discrete time case, local and regular martingales coincide. Hence, for someone learning for the first time, it is not immediately clear why the two concepts shouldn't coincide in continuous time, or why localization of processes should be at all interesting, when it is not at all interesting in discrete time. Second, the commonly used intuition that "a martingale is just an integrable local martingale" is incorrect, as Rick Durrett found out the hard way after publishing the first edition of his textbook on stochastic calculus. In fact the necessary and sufficient condition (Theorem 2.5 on p. 41) is $$\mathbb{E}\left( \underset{0 \le s \le t}{\sup} |X_s| \right) < \infty $$ which is a lot less elegant than would be desired. Third, and along the lines of the second objection, as noted on p. 123 in Revuz and Yor, local martingales are much more general than martingales, not just "a little more", making the name "local martingale" something of a misnomer. Revuz and Yor list at least two integrability classes of stochastic processes, $(D)$ and $(DL)$ on p. 124 which one presumably would think of before thinking of the concept of local martingale. Fourth and finally, local martingales as a concept were introduced in order to generalize the integral with respect to Brownian motion developed by Ito, not to generalize martingales, so I think it would be more appropriate to characterize them in terms of stochastic integration. Both my professor, who mentioned Kunita and Watanabe's work, as well as Revuz and Yor in their remark preceding Theorem 1.8 on p.124, state that local martingales as a concept were introduced in order to advance the theory of stochastic integration. By the Bachteler-Dellacherie theorem, which tells us that the most general class of "good" stochastic integrands is equal to a local martingale up to (plus) a finite variation process, if we were able to characterize local martingales entirely in terms of Brownian motion and finite variation processes, I think it would be an impressive conceptual achievement. Maybe not a practical one, but a conceptually elegant one, which is what interests me at the moment. REPLY [7 votes]: First, a martingale is always only specified with respect to a filtration, and so is thus a local martingale. You do not specify any filtration in your problem, so I assume you mean the natural filtration of the local martingale (i.e., the smallest filtration w.r.t. which $X$ is a local martingale). Second, your prove/disprove statement does not consist of one, but of multiple claims. I will try to disentangle them first before discussing their correctness. I will refer throughout to the the text Continuous Martingales and Brownian Motion which you mention, too. (1) Let $X$ be a continuous local martingale, then there exists a Brownian motion $B$ and a predictable process $\xi$ such that $X_t = B_{\int_0^t \xi_s^2 \, ds}$ for all $t \in [0, \infty)$. (2) Let $X$ be a continuous local martingale, then there exists a Brownian motion $\tilde{B}$ and a predictable process $\tilde{\xi}$ such that $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$ for all $t \in [0, \infty)$. (3) $\xi$ and $\tilde{\xi}$ as well as $B$ and $\tilde{B}$ agree in an appropriate sense, say, are versions of each other. (4) The quadratic variation of $\langle X \rangle_t$ is then given by $\int_0^t \xi^2_s \, ds$ (resp. $\int_0^t \tilde{\xi}^2_s \, ds$). (5) It holds true that $\xi$ is increasing and of finite variation. (6) The statement (3) holds true at least in the case when $\xi$ is is deterministic. Lets now asses the claims in detail: (1) This statement holds true as long as $\langle X \rangle_\infty = \infty$, it is called Dambis -- Dubins-Schwarz theorem (cf. R-Y, Thm. V.1.6). Without the assumption this is not necessarily true in the natural filtration (as it might be not rich enough to support a Brownian motion). However, it is possible to enlarge the probability space resp. the filtration such that this statement is true in the larger space (cf. R-Y, Thm V.1.7). (2) This is the \textit{martingale representation theorem}. It is usually formulated in the context of Brownian filtrations, i.e., it is a priori assumed that $X$ is a local martingale with respect to a filtration generated by a Brownian motion. However, it holds true without this assumption as long as the measure generated by the quadratic variation process, $d\langle X \rangle_t$, is equivalent to the Lebesgue measure $dt$. If there is no equivalence between the measures but at least $d\langle X\rangle_t$ is absolutely continuous with respect to the Lebesgue measure, one can save again the situation by enlarging the probability space (cf. R-Y, V.3.8 and discussion thereafter). Without absolute continuity the statement is false as can be seen on the following example. Let $C_t^i$ be copies of the Cantor function on the unit interval $[0,1)$ and define the extension to the positive halfline by $$ K_t = \sum_{i=0}^\infty i + C_{t-i}^i {{\mathchoice{1\mskip-4mu\mathrm l}{1\mskip-4mu\mathrm l} {1\mskip-4.5mu\mathrm l}{1\mskip-5mu\mathrm l}}}_{[i,i+1)}(t) $$ Let $W$ be a Brownian motion and define $X$ by $X_t = W_{K_t}$. This is clearly a martingale with quadratic variation $\langle X \rangle_t = K_t$. However, if $X$ could be written $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$, then we would have by (4) $\langle X \rangle_t = \int_0^t \tilde{\xi}^2_s \, ds$. However, this is always an absolutely continuous function whereas $K$ is not, therefore they cannot agree. (3) From the restrictions above it is clear that this cannot hold true in general. But you can even find very simple counterexamples, e.g., set $\tilde{\xi} =1$. Then you have $$ \int_0^t (-1) \, d\tilde{B}_s = - \tilde{B}_t = B_t = B_{\int_0^t 1^2 \, ds} = B_{\int_0^t (-1)^2 \, ds}$$ So in this case $\tilde{B} = - B$ whereas $\xi$ is not uniquely determined (e.g., it could be $1$ or $-1$). This can be of course made worse using processes jumping between $1$ and $-1$. (4) This is true and in both cases (direct calculation using properties of Brownian motion and R-Y, Prop. IV.2.7 resp.). (5) Every increasing process is of finite variation by definition. However, the claimed result is clearly not true, just think about an arbitrary integral with deterministic non- monotone integrand, i.e. $X_t = \int_0^t \sin{(s)} \, \, dW_s$ for some Brownian motion $W$. I assume this is simply a typo and you intended to claim that the quadratic variation itself is increasing (or rather: non-decreasing) and thus of bounded variation (which is evidently true, you just add more squares). (6) No, the example from (3) or the Cantor function example from (2) are counterexamples. Let me conclude with two remarks about the context: (a) The idea to place stochastic calculus on change of time instead of stochastic integration has some interesting history. Actually, it predates Ito calculus and was the first form of stochastic calculus, developed by Wolfgang Döblin (Vincent Doblin) in his famous lettre scellée to the French Academy of Science, https://mathoverflow.net/a/100040/20026. (b) I disagree with your statement that local martingale is a misnomer. Yes, a local martingale is much more general than a martingale. But it is intentionally not called generalized martingale but local martingale, as it is locally (i.e. up to an increasing sequence of stopping times) a martingales. Generalizations that resemble locally the original object are very often far more general than the original object. The gap between manifolds and Euclidean spaces is for sure much larger than that between true and local martingales.<|endoftext|> TITLE: Does a continuous map from $\kappa^\omega$ to $[0,1]^\omega$ have a non-scattered fiber? QUESTION [7 upvotes]: Question. Let $\kappa>\mathfrak c$ be a cardinal endowed with the discrete topology and $f:\kappa^\omega\to[0,1]^\omega$ be a continuous map. Is there a point $y\in[0,1]^\omega$ whose preimage $f^{-1}(y)$ is not scattered? Let us recall that a topological space $X$ is scattered if each non-empty subspace of $X$ contains an isolated point. An affirmative answer to this question would give an affirmative answer to this problem. Remark. If $\kappa<\kappa^\omega$, then the answer to the above question is affirmative. Indeed, assuming that every preimage $f^{-1}(y)$, $y\in[0,1]^\omega$, is scattered, we conclude that $|f^{-1}(y)|\le w(f^{-1}(y))\le w(\kappa^\omega)=\kappa$ and hence $|\kappa^\omega|\le\sum_{y\in[0,1]^\omega}|f^{-1}(y)|\le\mathfrak c\cdot \kappa=\kappa<|\kappa^\omega|$, which is a desired contradiction. So, actually, the question concerns cardinals $\kappa=\kappa^\omega>\mathfrak c$. The smallest such cardinal is $\mathfrak c^+$, the successor of the continuum. REPLY [2 votes]: The continuity of $f$ is not needed. Indeed, suppose to the contrary that $\kappa^\omega$ is a union of the family $\{F_\alpha:\alpha<\frak c\}$ of fibers of $f$. Let $\alpha<\frak c$ be any index. Since the fiber $F_\alpha$ is scattered, there is an injective enumeration $F_\alpha=\{f_{\alpha,\beta}:\beta<\beta_\alpha\}$ such that for each $\beta<\beta_\alpha$, $f_{\alpha,\beta}=(f_{\alpha,\beta,n})_{n\in\omega}$ is an isolated point of the space $F_{\alpha,\beta}=F_\alpha\setminus \{f_{\alpha,\gamma}:\gamma<\beta\}$. Therefore there exists a natural number $m=m(f_{\alpha,\beta})>0$ such that $(f_{\alpha,\beta,n})_{n\in m}\ne(f_n)_{n\in m}$ for each $f=(f_n)_{n\in\omega}\in F_{\alpha,\beta}\setminus\{ f_{\alpha,\beta}\}$. It follows that for each natural $m>0$ the projection $\pi_m:\kappa^\omega\to \kappa^m$ is injective on the set $F_{\alpha,m}=\{f\in F_{\alpha}: m(f)=m\}$. Since $\kappa>\frak c$, by induction we can construct a sequence $g=(g_m)_{m\in\omega}$ of elements of $\kappa$ such that $g|m\not\in\pi_m(F_{\alpha,m-1})$ for each $\alpha<\frak c$ and $m>1$. It follows that $g$ does not belong to any $F_{\alpha,m}$, a contradiction.<|endoftext|> TITLE: Euclid vs Eratosthenes QUESTION [8 upvotes]: Very little is known about Euclid's life--much less than about other famous ancient Greek mathematicians, which is puzzling. It is also strange to me that Euclid didn't write about the Eratosthenes sieve. Thus I'd like to ask mathematical historians and everybody else: is there any clear proof that Euclid and Eratosthenes are two different people? I wonder if Euclid's Elements were considered by the top geometers of Archimedes times (before and afterward for awhile) as teaching materials (or even philosophical too) rather than mathematical research. Especially that Elements were not mentioned much in those years, they became popular quite a bit later. For a longer time, even after Archimedes, nobody talked about Euclid of Alexandria. Instead, they talked only about "the author of Elements". Possibly, the geometric part of Elements was a cumulation of the work by several mathematicians, of which Eudoxes was a big part. But the number theoretical part of Elements and the Eratosthenes Sieve were perhaps by the same author, namely Erathostenes. The Elements had no name nor any pseudonym assigned to it. That's why nobody was talking explicitly about the author. Bourbaki used to have meetings together in a single room (perhaps a different room at the different occasions). They have established common notation and conventions, etc. Thus their textbooks had a fairly consistent style. They used to write in that style also monographs by single Bourbakist authors. In the case of the ancient authores such consistency was impossible due to the time and space span. This is why I conjecture that Eratosthenes combined the past results by several authors (especially by Eudoxes) and Eratosthenes rewritten these results in his own hand. This would explain why there was no author mentioned but the style of Elements was consistent. REPLY [5 votes]: C.K. Raju goes to some length to argue that Euclid did not exist at all, in Good-Bye Euclid! He starts from the established fact that, while Euclid was first mentioned by the 5th century philosopher Proclus, our sources for Proclus are a translation from Arabic made in Toledo around the year 1000. Raju has a political agenda for wishing to debunk Euclid, to which I do not subscribe, but some of his arguments are intriguing: Possibly, the name "Euclid" was inspired by a translation error made at Toledo regarding the term uclides which has been rendered by some Arabic authors as ucli (key) + des (direction, space). So, uclides, meaning “the key to geometry”, was possibly misinterpreted as a Greek name Euclides. . In attributing the Elements to an early Greek called Euclid, we are supposing that there was a fixed text which was repeatedly copied out without any significant change by subsequent scribes. But the available papyri on geometry from Alexandria do not correspond to the received text, and do not show any such evidence of the existence of a fixed, early text. Mikhail Katz pointed me, via a MSE thread, to a review by José Ferreirós on Raju's book, since it's behind a paywall I quote the relevant paragraph: In his interest to revise traditional historiography and oppose proof-centred mathematics, Raju devotes a lot of effort to questioning the existence of Euclid and insisting that the text of the Elements originates at the earliest in 370 CE (with Theon) or perhaps even in the tenth century. In my opinion, this is useless and does not help advance the author’s main theses. For historical purposes, what is relevant is that Elements represents a systematisation of a large portion of geometrical knowledge in the Greek-speaking world before the common era. (‘Euclid’ is simply the name of its otherwise unknown author, whose dates—it is true—are dubious; incidentally, an interesting question would be whether philologists find reason to think that the text of Elements was written by different authors.) Raju insists on the idea that Proclus’s views represent the original philosophy of mathematics in the Elements (p. 25), and he overemphasizes the connections between geometrical proof, Platonism, and Christian religion (see below).<|endoftext|> TITLE: Bounds on Matrix Exponential QUESTION [6 upvotes]: Suppose $A$ and $B$ are (non-commuting) hermitian $n\times n$ matrices and $k$ is a large positive number. Suppose we write the product of matrix exponentials as $e^{kA + B} e^{-kA} = e^{C(k)}$ for some matrix $C(k)$. I am interested in how large $C(k)$ is (with respect to some appropriate norm, lets say the operator norm of matrices). In particular is $\| C(k)\| = O(k)$? Some numerical experiments suggest a bound of the form $||C(k)|| \le O(k) ||[ A,B]||$ but I cannot prove this. I know the Baker-Campbell-Hausdorff formula gives us an (explicit) equation for $C(k)$. But bounding this expression term-by-term using the triangle inequality gives a terrible bound. I am hoping somebody knows either another argument (or a trick) that gives a more sensible bound. Thanks a lot! REPLY [4 votes]: $\|C(k)\|$ can be arbitrarily large, since e.g. you can add some large integer multiple of $2\pi i I$ without changing $e^{C(k)}$. If you want to try to avoid this, you might specify that $C(k)$ is the principal branch of the logarithm of $e^{kA+B} e^{-kA}$ (note that although $e^{kA+B} e^{-kA}$ is not hermitian, it has the same eigenvalues as $e^{-kA/2} e^{kA+B} e^{-kA/2}$ which is positive definite). Then the eigenvalues of $C(k)$ will be $O(k)$; unfortunately, since $C(k)$ is not hermitian or normal this does not imply $\|C(k)\| = O(k)$.<|endoftext|> TITLE: Decay of Fourier coefficients for compact Lie groups QUESTION [7 upvotes]: Let $G$ be a compact Lie group, $G^\natural$ the space of conjugacy classes in $G$ with the natural pushforward of $G$'s Haar measure. Let $f\in L^2(G^\natural)$. Then the Peter–Weyl Theorem tells us that there is a unique sequence $\{a_\pi\}_{\pi\in \widehat G}$ such that $$ f = \sum_\pi a_\pi \mathrm{tr}(\pi) $$ in $L^2(G^\natural)$. This doesn't guarantee that $$ f(x) = \sum_\pi a_\pi \mathrm{tr}(\pi(x)) $$ for all $x\in G^\natural$. If $G=S^1$, then $\widehat G=\mathbf{Z}$, and there are nice equivalences of the form ``$f$ is absolutely continuous iff $\widehat f(n) = O(1/n)$." Is there a similar characterization of (absolutely) continuous (/ $C^k$ and/or smooth) functions on general $G^\natural$ in terms of the decay of their $a_\pi$? REPLY [5 votes]: I believe the Sugiura article is, from MathSciNet, (MR0294571 Reviewed) Sugiura, Mitsuo Fourier series of smooth functions on compact Lie groups. Osaka J. Math. 8 (1971), 33–47. Apart from deciding on a summation order, there is the more pointed issue of an upper bound for sup-norm in terms of $L^2$ norm of functions in the irreducibles, obviously far less trivial than the bound $|e^{inx}|=1$ for the circle. But it is not hard: up to a measure constant the bound of sup-norm divided by $L^2$ norm is square root of the dimension of the repn (and the repn need not even be irreducible, in fact). I saw this sort of thing proven in Stein-Weiss ("Fourier analysis on Euclidean spaces") and a suitable form of the argument easily generalizes (e.g., see section 6 of my course notes http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf).<|endoftext|> TITLE: A question on invariant measures QUESTION [6 upvotes]: Let $(X, \mathcal{B}, T)$ be a topological dynamical system and $M(X, T)$ be the set of all invariant measures. I do not know is there some nice functional characterization of the following set $\{f\in C(X): \text{the set}\{\int fd\mu: \mu \in M(X,T)\}\text{is a singleton} \}$. Any comments and remarks will be appreciated. REPLY [7 votes]: Your set consists exactly of weak coboundaries plus constants. A function $f\in C(X)$ is called a coboundary if $f = h \circ T - h$ for some $h\in C(X)$. A function is called a weak coboundary if it is a uniform limit of coboundaries. Proposition A. $f \in C(X)$ is a weak coboundary if and only if $\int f d\mu=0$ for every $\mu \in M(X,T)$. Proof. The "only if" part is trivial, so let's prove the "if" part. Assume $\int f d\mu=0$ for every $\mu \in M(X,T)$. Denote by $f^{(n)} := f + f\circ T + \dots +f\circ T^{n-1}$ the $n$-th Birkhoff sum of $f$. As explained in RW's answer, the sequence of Cesàro averages $\frac{1}{n}f^{(n)}$ converges uniformly to $0$. So the proposition will be a consequence of the following lemma: Lemma B. For every $f \in C(X)$ and every $n$, the function $\frac{1}{n}f^{(n)} - f$ is a coboundary. It's an entertaining exercise to prove the lemma; but if you're in a hurry, here's the solution. Proof. Let $h := \frac{1}{n}\sum_{i=0}^{n-1} f^{(i)}$ (where $f^{(0)}:=0$). Then $$ h \circ T - h = \frac{1}{n}\sum_{i=0}^{n-1} \left(f^{(i)}\circ T-f^{(i)}\right) = \frac{1}{n}\sum_{i=0}^{n-1} \left(f \circ T^i - f\right)=\frac{f^{(n)}}{n} - f. $$ Remarks: The terminology weak coboundary comes from the 2002 Inventiones paper by Bousch and Jenkinson. See their paper for an explicit example of a weak coboundary that is not a coboundary. I think that Proposition A and Lemma B above are kind of folklore. Proposition A is Lemma 3 from Bousch-Jenkinson; for a more complete statement (and a proof using Hahn-Banach), see Proposition 2.13 in Katok-Robinson's survey. Lemma B is used as obvious in Bousch-Jenkinson; I learned it from Andrés Navas. We have used these ideas in non-commutative settings.<|endoftext|> TITLE: What is the intuitive meaning of the coskeleton of a simplicial set? QUESTION [20 upvotes]: Consider the inclusion $i_n: \varDelta_{\leq n} \hookrightarrow \varDelta$. This induces a truncation functor $tr_n:\mathbf{sSet}\to\mathbf{sSet}_{\leq n}$, which has a left and a right adjoint, $tr_n^L$ and $tr_n^R$, obtained via Kan extension. This allows us to define the $n$-(co)skeleton of a simplical set $X$: $$ sk_n X := tr_n^L tr_n X, \quad cosk_n X := tr_n^Rtr_n X. $$ It is easy to see that these functors form an adjunction. $sk_n$ is analogous to forgetting about cells of dimension greater than $n$ in a CW complex, i.e. taking the n-skeleton. What is the dual analogy (if it exists) for the coskeleton of a simplical set $X$? REPLY [7 votes]: For a simplicial set $X$ you can see the $n$-th skeleton to have no $m$-simplices other than the degenerate ones for $m>n$.(As you say) For $m\leq n$ both the $n$-skeleton and the $n$-coskeleton have the same $m$-simplices as $X$. And to answer your question for $m>n$ the $n$-coskeleton has $m$-simplices ALL possible configurations of a $m+1$ number of $(m-1)$-simplices in $\mathrm{cosk}_nX$ that agree in the appropriate $m-2$-faces to form the shape of a standard m-simplex. If I say it in a lower dimension it will sound less tongue-twisty and easier to imagine it geometrically, which I think is what you are looking for: Take $n=2$ For $m=3$ the 3-simplices of the 2-coskeleton are all four possible triangles in $X$ that you can arrange in a tetrahedral shape. For $m=4$ the 4-simplices are all five possible tetrahedra in $\mathrm{cosk}_2X$ that form the shape of a standard 4-simplex (pentachoron). So in this sense, the coskeleton is the largest simplicial set containing X and the skeleton the smallest.<|endoftext|> TITLE: Constructively, are all fibrations cloven? QUESTION [6 upvotes]: A "cloven fibration" is a fibration for which we have an explicit choice of cartesian liftings; this is often phrased as, "We can pick a lifting without using the axiom of choice". Firstly, I'm a bit perplexed by this language, since the intensional character of the formal proof of a statement ought not to factor in the definition / meaning of that statement. Is there a precise definition of "cloven fibration" in a classical setting which does not appeal to the intensional structure of a formal proof (namely, the use of the axiom of choice)? I guess I am looking for an "internal" definition, which could distinguish internally between a (proper) fibration and a cloven fibration. The reason the usual definition makes me uncomfortable is, in a classical metatheory you cannot in fact distinguish between an object which was "constructed" using the axiom of choice and one which is effectively given, so I have trouble understanding what this definition actually means. Secondly, suppose I am working in a constructive metatheory, such as extensional type theory (where the Axiom of Choice holds without restriction—and is not susceptible to Diaconescu's argument in this case). Then are all fibrations cloven, as far as I am concerned? Thanks! And please let me know if there is any way in which I can improve the statement of my question. REPLY [6 votes]: If the axiom of choice holds then every fibration is cloven, and I would not be surprised if the converse holds. So this talk about not using the axiom of choice is best understood as closet-constructivism. Regarding extensional type theory, I would advise being very careful about doing mathematics in it. If your theory is not susceptible to a Diaconescu-style argument then I suppose you are using dependent sums $\Sigma$ instead of existentials $\exists$, and you do not have any kind of quotients or propositional truncations lying around (otherwise the ghost of Diaconescu appears). On the other hand propositional equality satisfies Uniqueness of identity proofs. The thing to pay attention to are statements and definitions which turn from a propostion to a structure when we replace $\exists$ with $\Sigma$. Two examples: The usual definition of "$f$ is a cartesian arrow" is a statement of the form "for every ... there exist unique .... such that some equations hold". If we change this into "$\prod$ ... $\sum$ unique ... such that some equations hold" we still have an h-propostion, thanks to uniqueness and extensional equalty. The usual definition of "$p : E \to B$ is a fibration" is of the form "for every arrow there exists a cartesian arrow such that some equation holds". When we turn this into "$\prod$ arrow $\sum$ cartesian arrow such that equation holds" we changed "$p$ is a fibration" to "$p$ is a cloven fibration". (In fact there is no way to say "$p$ is a non-cloven fibration", is there?) Consequently, any theorem that speaks about equality of cloven fibrations changes its meaning, because we are now comparing fibrations with cleaveges. The upshot of this is that one has to be quite careful about veryfing that everything actually works. Good luck!<|endoftext|> TITLE: Confusion regarding statement of mirror symmetry for elliptic curves QUESTION [5 upvotes]: I am a little bit unsure about the mirror symmetry statement for elliptic curves; specifically, how the flipping of the Kähler and complex moduli works. Perhaps I should say at the outset, the reason I have been thinking about this, is that I am doing a computation involving a torus with parameter $\tau \in \mathbb{H}$, and my answer in invariant under $\tau \to \tau+1$, but not $\tau \to -1/\tau$. So I am thinking that maybe I am only using the Kähler structure, not the complex structure. Of course, the complex structure is given by $\tau \in \mathbb{H}/\rm{PSL}(2, \mathbb{Z})$. I think for the Kähler structure, we choose a class $[\omega] \in H^{2}(X,\mathbb{C})$, which we can parameterize by Kähler parameter $t=t_{1} + i \, t_{2}$: $$t=\frac{1}{2\pi i } \int_{X} [\omega].$$ We identify $t_{2}>0$ with the area of the torus. So unlike the complex structure, Kähler structures related by $\rm{PSL}(2,\mathbb{Z})$, are not necessarily identical, correct? After all, one will have small area, the other large. So I am confused about how the mirror symmetry acts. If mirror symmetry is some mysterious equivalence of the torus under interchange of the complex and Kähler moduli, does that not imply that the space of equivalent Kähler structures is also $\mathbb{H}/\rm{PSL}(2, \mathbb{Z})$. Is this correct? REPLY [4 votes]: This is related to the fundamental question about how to define the Kähler moduli space. The Kähler moduli space is often not as naive as one thinks. A solution for a genus 1 curve is given by Bridgeland in the last section of this article, where the "extended" Kähler moduli is identified with the space of Bridgeland stability conditions modulo autoequivalences of derived category. After taking quotient by the natural $\mathbb{C}^\times$-action, the space is biholomorphic to $\mathbb{H}/\mathrm{PSL}_2(\mathbb{Z})$. The key idea is, under mirror symmetry, the monodromy (symplectomorphism) group corresponds to the group of autoequivalences. The appearance of the $\mathrm{SL}_2(\mathbb{Z})$-action on the derived category of an elliptic curve was first observed by Mukai in his article on the Fourier-Mukai transformations. This action is not geometric as it mixes the rank and degree of objects.<|endoftext|> TITLE: Are there any unitary matrices which satisfy the Yang-Baxter equation which are universal for quantum computation? QUESTION [8 upvotes]: Let $H$ be a finite dimensional hilbert space. Let $L:H\otimes H\rightarrow H\otimes H$ be a unitary transformation. Then the equation $$(L\otimes I)(I\otimes L)(L\otimes I)=(I\otimes L)(L\otimes I)(I\otimes L)$$ where $I:H\rightarrow H$ is the identity mapping is known as the Yang-Baxter equation. We shall call a linear transformation $L:H\otimes H\rightarrow H\otimes H$ a universal quantum gate if there is some $n$ such that the unitary transformations of the form $I^{\otimes i}\otimes L\otimes I^{\otimes (n-i-2)}$ along with the mappings $\lambda\cdot I^{\otimes n}$ where $|\lambda|=1$ generate a dense subgroup of the unitary group $U(2^{n})$. Does there exist a universal quantum gate $L:H\otimes H\rightarrow H\otimes H$ which satisfies the Yang-Baxter equation? If so, then what is the least natural number $n$ so that there is a universal quantum gate $L:\mathbb{C}^{n}\otimes\mathbb{C}^{n}\rightarrow \mathbb{C}^{n}\otimes\mathbb{C}^{n}$ which satisfies the Yang-Baxter equation? $\textbf{Motivation}$ If $L$ is a quantum gate which satisfies the Yang-Baxter equation, then for all $n$, the function $\phi:B_{n}\rightarrow H^{\oplus n}$ defined by $\phi(\sigma_{i})=I^{\oplus(i-1)}\oplus L\oplus I^{\oplus(n-i-1)}$ is a group homomorphism where $B_{n}$ denotes the $n$-strand braid group. If $L$ is also universal, then any quantum algorithm (called algorithm A) can be emulated by a unitary transformation $\phi(w)$ for some braid word $w$. However, if $\sigma_{i_{1}}^{e_{1}}...\sigma_{i_{n}}^{e_{n}}$ is one of the normal forms for the braid for $w$, then the sequence of quantum gates $(\phi(\sigma_{i_{1}})^{e_{1}},...,\phi(\sigma_{i_{n}})^{e_{n}})$ obfuscates the original quantum algorithm so that a person cannot figure out how algorithm A works simply by looking at the braid $\sigma_{i_{1}}^{e_{1}}...\sigma_{i_{n}}^{e_{n}}$. The idea of using braid group representations to obfuscate quantum algorithms has been proposed in this recent paper where they have observed that the Fibonacci representation of braid groups could be used to obfuscate quantum algorithms. However, a universal unitary transformation that satisfies the Yang-Baxter transformation seems to be a more direct way of using braids to obfuscate quantum algorithms. REPLY [5 votes]: Braiding Operators are Universal Quantum Gates, by Louis Kauffman and Samuel Lomonaco (2004) In this paper, we prove that certain solutions to the Yang-Baxter equation together with local unitary two-dimensional operators form a universal set of quantum gates. In particular, we show that a single specific solution of the Yang-Baxter Equation, the Bell basis change matrix, is a universal gate. If we wish to do away with the local unitaries, and only use unitary solutions of the Yang-Baxter equation, then universal quantum computation is not known to be possible, because a classical simulation is available, see Classical simulation of Yang-Baxter gates (2004).<|endoftext|> TITLE: A follow up question to: Number of walks on integer lattice with self-edge at zero QUESTION [5 upvotes]: Let $a(n)$ be the number of lattice paths in ${\mathbb{Z}^2}$ of length $n$ which start at the origin $(0,0)$ and end up at $(n,0)$ and have only up-steps $U:(i,j) \to (i + 1,j + 1)$, down-steps $D:(i,j) \to (i + 1,j - 1)$ and horizontal steps $H:(i,0) \to (i + 1,0)$ on the $x-$axis. Is there a direct combinatorial way (i.e. without use of generating functions) to show that $a(n) = \sum\limits_{k = 0}^{\left\lfloor {\frac{n}{2}} \right\rfloor } {\frac{{{2^k}(n - 1)!!}}{{k!(n - 1 - 2k)!!}}} $ or of the special case $a(2n+1)=5^n?$ REPLY [2 votes]: The idea of Fedor Petrov led me to the following proof of the above formula. In arXiv:1203.5424 R. Duarte and A. G. de Oliveira gave a combinatorial proof of the identity $$S_m(k)=\sum_{i_1+\dots+i_m=k}\binom{2i_1}{i_1}\dots\binom{2i_m}{i_m}=4^k\binom{k+m/2-1}{k}.$$ If we attach a horizontal step $H$ at the beginning then each such path has a uniquely determined decomposition in the form ${L_{{a_1}}}{L_{{a_2}}} \cdots {L_{{a_m}}}$ where ${L_a} = H{M_a}$ and ${M_a}$ is a path of length $2a$ starting and ending on height $0$ with $a$ up-steps and $a$ down-steps. Therefore $\sum\limits_{i = 1}^m {\left( {1 + 2{a_i}} \right) = n + 1} $ or if we set $\sum\limits_{i = 1}^m {{a_i} = k} $ then $m = n + 1 - 2k.$ The number of such paths is ${S_{n + 1 - 2k}}(k) = {4^k}\binom{\frac{{n - 1}}{2}}{k}.$ Summing over all $k$ gives the above result.<|endoftext|> TITLE: Stabilizers of pairs of ternary quadratic forms QUESTION [5 upvotes]: Let $A,B$ be two ternary quadratic forms with real coefficients, given by symmetric matrices $$\displaystyle 2A = \begin{pmatrix} 2a_{11} & a_{12} & a_{13} \\ a_{12} & 2a_{22} & a_{23} \\ a_{13} & a_{23} & 2a_{33} \end{pmatrix}, 2B = \begin{pmatrix} 2b_{11} & b_{12} & b_{13} \\ b_{12} & 2b_{22} & b_{23} \\ b_{13} & b_{23} & 2b_{33} \end{pmatrix}.$$ Let $V_\mathbb{R}$ denote the 12 dimensional real vector space of $(A,B)$ over $\mathbb{R}$, and let $G(\mathbb{R}) = \operatorname{GL}_2(\mathbb{R}) \times \operatorname{SL}_3(\mathbb{R})$. Let $(r,g)$ be an element of $G(\mathbb{R})$, where $$\displaystyle r = \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}.$$ Then $(r,g)$ acts on $(A,B) \in V_\mathbb{R}$ by sending $(A,B)$ to $$(r_1 (gAg^T) + r_2 (gBg^T), r_3 (gAg^T) + r_4 (gBg^T)).$$ Bhargava, in his paper The density of discriminants of quartic rings and fields, stated that the stabilizer in $G(\mathbb{R})$ of an element $(A,B) \in V_\mathbb{R}$ has order 24 if $A,B$ have four common real zeroes over $\mathbb{P}^2$, order 8 if they have exactly one pair of real zeroes over $\mathbb{P}^2$, and order 4 if they have no common real zeroes. He simply said that "one easily checks" that this is the case. Can anyone give an explanation as to why this should be easy to see, and a proof of why it's true? REPLY [4 votes]: It's actually order 8 for no real zeroes and order 4 for two real zeroes, not the other way around. (See the bottom of page 1038 of the paper.) The symmetry groups are taken modulo $\{ \pm 1 \}$, so we work projectively in ${\rm PGL}_2({\bf R}) \times {\rm SL}_3({\bf R})$. We must assume that we're in the generic case that $A,B$ span a two-dimensional space of conics that vanish on four distinct points of ${\bf P}^2$ in general linear position (i.e., no three on a line). Then over ${\bf C}$ the stabilizer is always $S_4$, because it is known (and easy) that ${\bf PGL}_3$ acts simply-transitively on ordered four-tuples in general linear position. But over ${\bf R}$ we must use permutations that commute with complex conjugation. This conjugation acts on the four points as the identity, a simple transposition, or a double transposition according as four, two, or none of them are real. The commutators have orders $24$, $4$, and $8$ respectively, as claimed.<|endoftext|> TITLE: Power sums of p-th roots of unity QUESTION [7 upvotes]: The following question was asked by a colleague of mine. For any prime $p$ consider $$ M_p:=\min_{z_1,\dots,z_p}\max_{j,k}\left|z_1^k+\dots+z_j^k\right|,$$ where $z_1,\dots,z_p$ are the complex $p$-th roots of unity in any order, and $j,k\in\{1,\dots,p-1\}$ are arbitrary. Is $M_p$ bounded? REPLY [7 votes]: This is closely related to the problems surrounding Turán's power sum method. For example see the chapter in Montgomery's Ten Lectures book, or this paper of Gonek. Lemma 1 (attributed to Cassels) there shows that if $b_j >0$, and $|z_j|=1$ (for $j=1$, $\ldots$, $N$) then for $K>N$ $$ \max_{k\le K} \Big| \sum_{j=1}^{N} b_j z_j^k \Big| \ge \frac{\sqrt{K-N}}{\sqrt{K}} \Big(\sum_{j=1}^{N} b_j^2 \Big)^{\frac 12}. $$ Now apply this in your situation with $N=(p-1)/2$ (this is $j$ in your problem) and $K=(p-1)$ to get essentially the result you have. Of course your proof is easy enough in this case, but I thought you might appreciate the general context.<|endoftext|> TITLE: Equivalence of questions regarding restrictions of pure states QUESTION [6 upvotes]: In Davidson and Szarek's article "Local Operator Theory, Random Matrices and Banach Spaces" in the Handbook of the Geometry of Banach Spaces, the authors discuss the (now solved) Kadison-Singer problem of unique extensions of pure states from atomic masas to $\mathcal{B}(H)$. After describing the form of pure states on atomic masas (really, point evaluations on $\beta\mathbb{N}$), they say: This suggests the open question raised in [99, §5] whether every pure state on $\mathcal{L}(\ell_2)$ is an extension of a pure state on some discrete masa, or indeed a given masa. However, it has been shown in [6], cf. [4, §7.3], that the two questions are in fact equivalent. [4] is Anderson, J., Extensions, Restrictions, and Representations of States on C*-Algebras, Trans. AMS, 1979 [6] is Anderson, J., Extreme Points in Sets of Positive Linear Maps on $\mathcal{B}(H)$, J. Funct. Anal., 1979 [99] is Kadison, R., Singer, I., Extensions of Pure States, Amer. J. Math., 1959 I assume that the "two questions" being referred to are (1) whether a given pure state restricts to a pure state (i.e., a homomorphism to $\mathbb{C}$) on some atomic masa, and (2) whether the same is true on a given masa. There is also the related question (1'), usually credited to Kadison and Singer, of whether the same is true on some masa. My question is, are these statements in fact equivalent? Skimming through [6], I didn't find a direct answer to my question, though I may have missed something. Corollary 7.3 in [4] contains the relevant fact that if $h$ is a non-zero homomorphism to $\mathbb{C}$ on a non-atomic masa, then it extends to a pure state on $\mathcal{B}(H)$ which itself restricts to a homomorphism on a particular atomic masa. However, the extension of $h$ need not be unique. Relevant is the paper by Akemann and Weaver (PNAS, 2008), producing a pure state which is not pure on any masa, in the presence of the continuum hypothesis. That paper, and others I've looked at with refinements of their construction, do not seem to mention the equivalence of these two problems, hence my surprise. REPLY [3 votes]: I just ran across this question and apologize for not noticing it earlier. I think the statement is incorrect --- I'm not aware of such an equivalence. As far as I know it is open whether there are models of ZFC in which every pure state is pure on some masa, but some pure state is not pure on any atomic masa.<|endoftext|> TITLE: Extending the topology on a set to the group/vector space it generates QUESTION [6 upvotes]: The multiplicative group $\Bbb Q^+$ can be viewed as a $\Bbb Z$-module. To see this, note that any rational can be decomposed into the form $2^{n_2} \cdot 3^{n_3} \cdot 5^{n_5} \cdot ...$ The tuple of coefficients $(n_2, n_3, n_5, ...)$ is then an element in the module $\Bbb Z^{(\omega)}$, the set of integer sequences with finite support. The dual module $Hom(\Bbb Q^+, \Bbb Z)$ mapping from rationals to integers is $\Bbb Z^\Bbb N$, the set of all integer sequences. For various reasons in musical tuning theory, an important class of these sequences are given by $f_r = (\lfloor r \cdot \log(2) \rceil, \lfloor r \cdot \log(3) \rceil, \lfloor r \cdot \log(5) \rceil, ...)$ for $r \in \Bbb R$, and where $\lfloor x \rceil$ rounds x to the nearest integer. It has sometimes proven useful to put the Euclidean topology on these functionals, because then we can find local maxima with respect to various functions on them. This brings me to my questions: Is it possible, in a natural way, to extend the topology on the set of $f_r$ to the set of all finite $\Bbb Z$-linear combinations of that set? Likewise, is it possible to extend the topology for infinite $\Bbb Z$-linear combinations, when such sums converge? We can also treat the above as a discrete lattice of vectors embedded in a real vector space. Is it possible to extend the topology on the $f_r$ to the $\Bbb R$-linear span of finite weighted combinations? Likewise, is it possible to extend the topology to infinite $\Bbb R$-linear combinations, when they converge? I have a hunch one may be able to use the product topology for this, but am not quite sure. REPLY [2 votes]: The answer to your questions 1 and 3 is that there is a universal way to do it, but I don't know how this works for infinite linear combinations. Unfortunately, I only know how to do it in the category of compactly generated topological spaces (or any cartesian closed category of spaces, really). Hence by a topological ring I mean a ring $R$ equipped with a compactly generated topology making the operations continuous with respect to the compactly generated product (and similarly for topological $R$-modules). My claim is that the forgetful functor from the category of topological $R$-modules to the category of (compactly generated) spaces has a left adjoint. Here is an explicit construction. Let $X$ be a compactly generated space and let $R X$ be the set of all $R$-linear combinations of the points of $X$. We topologize it as follows. For each $m \in \mathbb{N}$ we consider the function $\phi_m \colon R^m \times X^m \to R X$ defined as $\phi_m(a, x) = a_1 x_1 + \ldots + a_m x_m$. We take the finest topology on $R X$ making $\phi_m$ continuous for all $m$. It is quite routine to verify that $R X$ satisfies the universal property of a free topological $R$-module on $X$. The only wrinkle appears when we check the continuity of the addition $R X \times R X \to R X$. Here we need to use the fact that products preserve colimits in order to show that the topology of $R X \times R X$ is generated by the maps $\phi_m \times \phi_n$. If we work with all topological spaces, then the forgetful functor still has a left adjoint as follows by the Adjoint Functor Theorem. However, in this case I don't know how to explicitly describe the resulting topology. The construction above will sometimes fail, e.g. if $X$ is discrete of cardinality $\mathfrak{c}$ then the addition map on $\mathbb{R} X$ is discontinuous with respect to the standard product topology. Note: this answer is based on a discussion from nForum.<|endoftext|> TITLE: Nonabelian $H^2$ and Galois descent QUESTION [9 upvotes]: I would like to know whether the following metatheorem on nonabelian $H^2$ has been ever stated and/or proved. Let $k$ be a perfect field and $k^s$ its fixed separable closure. Let $X^s$ be a variety with additional structure over $k^s$ (I don't want to specify what I mean by additional structure). By a $k$-model of $X^s$ I mean a variety with additional structure $X$ over $k$ together with a $k^s$-isomorphism $$ X\times_k k^s\overset{\sim}{\to} X^s.$$ Metatheorem. Let $k$ be a perfect field and $k^s$ its fixed separable closure. Let $X^s$ be a variety with additional structure over $k^s$. Write $A^s=\mathrm{Aut}(X^s)$, and assume that $A^s$ "is" an algebraic group over $k^s$. Assume that for any $\sigma\in\mathrm{Gal}(k^s/k)$ there exists a $k^s$-isomorphism $$\lambda_\sigma\colon \sigma X^s\to X^s,$$ where $\sigma X^s$ is the variety obtained from $X^s$ by transport of structure. Assume also that $X^s$ admits a $k_1$-model over a finite Galois extension $k_1/k$ contained in $k^s$. Then these data define a $k$-kernel $$\kappa\colon\mathrm{Gal}(k^s/k)\to \mathrm{Out}(A^s)$$ and a cohomology class $\eta\in H^2(k,A^s,\kappa)$. If $\eta$ is not neutral, then $X^s$ has no $k$-model. If $\eta$ is neutral and the variety $X^s$ is quasi-projective, then $X^s$ admits a $k$-model $X$. Moreover, set $A=\mathrm{Aut}(X)$, then there is a canonical bijection between $H^1(k,A)$ and the set of isomorphism classes of $k$-models of $X^s$. Example of application of the metatheorem: If $k=\mathbb{R}$, $k^s=\mathbb{C}$, $A^s$ is a finite abelian group of odd order, then $H^2(\mathbb{R},A)=1$ and $H^1(\mathbb{R},A)=1$ (because $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ is of order 2), hence $X^s$ has a unique model over $\mathbb{R}$. I would be also glad to have references where this metatheorem was proved in special cases. I know that it was proved in the case when $X^s$ is a principal homogeneous space of $G^s$ dominating $Y^s$, where $Y$ is a given homogeneous space (not necessarily principal) of an algebraic group $G$ defined over $k$, see Springer, Nonabelian $H^2$ in Galois cohomology. In: Algebraic Groups and Discontinuous Subgroups (Proc. Sympos. Pure Math., Boulder, Colo., 1965), pages 164--182. Amer. Math. Soc., Providence, R.I., 1966. Borovoi, Abelianization of the second nonabelian Galois cohomology, Duke Math. J. 72(1), 217--239, 1993. Flicker, Scheiderer, Sujatha, Grothendieck's theorem on nonabelian $H^2$ and local-global principles. J. Amer. Math. Soc. 11(3), 731--750, 1998. REPLY [9 votes]: Let me elaborate more on the remark above. Let $k$ be a perfect field. Let $\mathrm{Field}_k$ denote the category of finite extensions of $k$, i.e., the objects of $\mathrm{Field}_k$ are fields $k'$ equipped with an embedding $k \to k'$ such that $k'$ is finite dimensional over $k$. The morphisms are the maps of fields $k' \to k''$ which respect the embedding (here we do not think of all the $k'$ as subfield of a fixed seperable closure). Suppose we have a functor $F:\mathrm{Field}_k \to \mathrm{Grpd}$ to the category of small groupoids. For example, $F$ may be the functor which sends $k'$ to the groupoid whose objects are varieties with a certain structure defined over $k'$ and whose morphisms are structure preserving $k'$-isomorphisms between them. For $k'$ in $\mathrm{Fields}_k$ the inclusion $\iota:k \to k'$ can be considered as a morphism in $\mathrm{Fields}_k$, and we consequently have an associated functor $F(\iota):F(k) \to F(k')$ which we can call the base change functor. If $k'$ is also a Galois extension of $k$ with (finite) Galois group $G$, then the automorphism group of $k'$ in the category $\mathrm{Fields}_k$ is exactly $G$. In particular, $G$ now acts on the groupoid $F(k')$ (via functors). Given an object $X \in F(k')$ let us denote by $X^{\sigma}$ the image of $X$ under the action of $\sigma \in G$ on $F(k')$. Now whenever we have a group $G$ acting on a groupoid $Z$, we have an associated notion of a $G$-equivariant object of $Z$. This is an object $X \in Z$ equipped with a compatible collection of (iso)morphisms $f_{\sigma}: X \to X^{\sigma}$. We may also call this a twisted action of $G$ on $X$. Let us denote by $Z^{hG}$ the groupoid of $G$-equivariant objects in $Z$ (where the notation echoes the fact that we think of $G$-equivariant objects as homotopy fixed points). Now the mere fact that $F$ is a functor implies that if $X$ is an object of $F(k)$ then the object $F(\iota)(X) \in F(k')$ carries a natural twisted action of $G$. We hence obtain a functor $$ T_{k'/k}:F(k) \to F(k')^{hG} .$$ We can now say that $F$ satisfies Galois descent if $T_{k'/k}$ is an equivalence of groupoids for every finite Galois extension $k'/k$. Now the $H^2$ and $H^1$ business is something that has only to do with computing groupoids of equivariant objects, and has nothing to do with, say, algebraic varieties. Let $Z$ be a groupoid equipped with an action of a group $G$. Let $\pi_0(Z)$ denote the set of isomorphism classes of $Z$, so that we have an induced action of $G$ on $\pi_0(Z)$. If $x \in \pi_0(Z)$ is an isomorphism class fixed by $G$, then we have an induced action of $G$ on the connected component $Z_x \subseteq Z$ corresponding to $x$. Let $X \in Z_x$ be any object and let $A = Aut(X)$ be its automorphism group. Since $Z_x$ is a connected groupoid, the group of connected components of self-homotopy equivalences of $Z_x$ is naturally isomorphic to $Out(A)$. We hence obtain a natural map $G \to Out(A)$, i.e., a pseudo-action of $G$ on $A$. Classical obstruction theory now associates to $X$ an obstruction element $o_X \in H^2(G,A)$, which is neutral if and only if $X$ admits a $G$-equivariant structure (i.e., a twisted action of $G$). The object $o_X$ is the one associated with a certain group extension $$ 1 \to A \to G_X \to G \to 1 $$ where $G_X$ is the group whose elements are pairs $(f,\sigma)$ where $\sigma$ is an element of $G$ and $f:X \to X^{\sigma}$ is a morphism (composition of elements is defined in a natural way). If the obstruction element $o_X$ is neutral then we can choose a section $G \to G_X$. Each such section determines a twisted action of $G$ on $X$. Furthermore, two such twisted actions result in isomorphic $G$-equivariant objects if and only if the two sections are conjugate by an element of $A$. This data is now classified by the cohomology group $H^1(G,A)$, and we obtain a bijection between $H^1(G,A)$ and the set of isomorphism classes of $G$-equivariant objects in the component $Z_x$. This is the way to compute groupoids of $G$-equivariant objects. Edit: If $Z,W$ are two groupoids then the functor category ${\rm Fun}(Z,W)$ is a groupoid as well. Two functors $f,g: Z \to W$ are homotopic if they are isomorphic in ${\rm Fun}(Z,W)$, and a functor $f: Z \to W$ is a homotopy equivalence if it has an inverse up to homotopy. For a groupoid $Z$ we have the full subgroupoid ${\rm Equiv}(Z,Z) \subseteq {\rm Fun}(Z,Z)$ spanned by the homotopy equivalences. Then $\pi_0{\rm Equiv}(Z,Z)$ (i.e., the set of isomorphism classes of the groupoid ${\rm Equiv}(Z,Z)$) is naturally a group by composition. This is the "group of connected components of self homotopy equivalences" alluded to in the answer.<|endoftext|> TITLE: On thin $\Sigma^1_2$ equivalence relations QUESTION [8 upvotes]: This question is regarding Hjorth's paper "Some applications of coarse inner model theory", J. Symbolic Logic 62 (1997), no. 2, 337–365. Hjorth claims that if $E$ is a thin $\Sigma^1_2$ equivalence relation, then $E$ is $\Delta^1_2(m)$ for some $\Delta^1_3$ real $m$. One of the key steps is Lemma 2.5. The problem is, I don't see why the "Claim" in the proof of Lemma 2.5 holds. In the "Claim", $M^*$ is a countable model with a Woodin cardinal without a top measure, so I don't understand the sentence "..., all iterates of $M^*$ with respect to the topmost measure must be wellfounded." So I can only see that $E$ is $\Delta^1_2(x)$ for some $x$ coding $M_1|(\delta^+)^{M_1}$, where $\delta$ is the Woodin of $M_1$. Am I missing something? REPLY [5 votes]: Lemma 2.5 of Hjorth's paper is wrong. Here is a counterexample produced from a discussion with Philipp Schlicht. There is a thin $\Sigma^1_2$ equivalence relation which is not $\Pi^1_2(a)$ for any $a \in \mathbb{R}\cap M_1$. Proof: Let $xEy$ iff there is a countable transitive model $M$ of enough fragment of ZFC containing $x,y$ such that $M$ satisfies either neither $x$ nor $y$ is a standard code for a 1-small, $\Pi^1_2$-iterable, sound premouse projecting to $\omega$, or $x,y$ are both standard codes for 1-small, $\Pi^1_2$-iterable, sound premice projecting to $\omega$ and these two mice have a common nondropping iterate. The equivalence classes of $E$ look like this: If $M_1|\alpha$ projects to $\omega$, then the $E$-class of the standard code for $M_1|\alpha$ has only one element. All the other elements are in one $E$-class. $E$ is $\Sigma^1_2$. The reason why $E$ is an equivalence relation is the following: if $P,Q$ have a common nondropping iterate and $P$ is iterable, then $Q$ is iterable. However, $E$ is not $\Pi^1_2(a)$ for any $a \in \mathbb{R} \cap M_1$. Otherwise, let $\alpha$ be the least such that $M_1|\alpha$ projects to $\omega$ and $a \in M_1|\alpha$. Let $$\varphi(v,a)$$ be the statement: $v$ is the standard code for a 1-small, sound premouse $P$ projecting to $\omega$, $P$ looks like the sharp of $Q$ where $Q$ is some proper initial segment of $P$ projecting to $\omega$, $a \in Q$, but for any proper initial segment $Q'$ of $Q$ that projects to $\omega$, $a \notin Q'$. Note that If $v$ codes $(M_1|\alpha)^\#$, i.e. $v$ codes $M_1|\beta$ where $M_1|\beta$ projects to $\omega$ and $\mathbb{R} \cap L(M_1|\alpha) = \mathbb{R} \cap M_1|\beta$, then $\varphi (v,a)$ holds. If $v$ codes $M_1|\beta'$ and $\beta' \neq \beta$ ($\beta$ as in 1) then $\varphi(v,a)$ doesn't hold. There are $v \neq v'$ such that $v E v'$ and $\varphi(v,a)$, $\varphi(v',a)$ both hold. The premice coded in $v,v'$ are not iterable. Therefore, $$v \text{ codes }(M_1|\alpha)^\#$$ iff $\varphi(v,a)$ holds and there are $v' \neq v''$ such that $v' E v'' \wedge \neg(v E v') \wedge \neg(v E v'') \wedge \varphi(v',a) \wedge \varphi(v'',a)$. This means $(M_1|\alpha)^\#$ is a $\Sigma^1_2(a)$ singleton. So $(M_1|\alpha)^\# \in L(M_1|\alpha)$. Contradiction!<|endoftext|> TITLE: Computational complexity of computing simplicial homology QUESTION [11 upvotes]: Is there any literature regarding the fastest known algorithm to compute the homology groups of a simplicial complex (on n vertices)? What about computing the fundamental group? The context is to tell whether a given simplical complex is contractible by showing that the fundamental group and all reduced homology groups vanish, but if there is a faster way to compute whether a simplicial complex is contractible then that would be helpful as well. REPLY [14 votes]: Homology groups can be computed with Smith normal form (see this survey). As for deciding if a simplicial complex is contractible, that is difficult. It is undecidable to tell if a simplicial complex is contractible (see appendix A of this paper). The same paper shows that it is NP-hard to decide if a simplicial complex is collapsible (a condition which implies contractible).<|endoftext|> TITLE: Methods for defining/calculating homotopy limits of quasicategories QUESTION [5 upvotes]: I am working on a project which requires that I calculate homotopy limits of homotopy theories (i.e. $(\infty,1)$-categories). It may be relevant that the homotopy limits which interest me are in the shape of towers; that is, the indexing category looks like $\cdots\rightarrow\cdot\rightarrow\cdot$. Because I am interested in some category-theoretic constructions (e.g. homotopy adjunctions between $(\infty,1)$-categories, Kan extensions, and homotopy (co)limits within $(\infty,1)$-categories), quasicategories seem like a natural choice for modeling $(\infty,1)$-categories, since they have the most well-developed category theory. However, I would be open to using a different model (e.g. Bergner's model structure on simplicial categories or Rezk's complete Segal spaces) if it proved more convenient. Anyway, my question is simply how to calculate these homotopy limits. I am aware of Emily Riehl's proof that there is a model structure on the category of marked simplicial sets which is Quillen equivalent to the Joyal model structure on simplicial sets. This is nice, because the former model category is a simplicial model category, which the latter is not (although it is Cartesian closed). If it comes down to it, I should be able to do everything I want to do in this setting, bringing to bear the techniques developed in Riehl's book Categorical Homotopy Theory for calculating homotopy limits in simplicial model categories. But I'd like to know if there's a more straightforward approach which does not leave the Joyal model structure on simplicial sets. Sticking with Riehl's approach to homotopy limits via weighted limits, is there a particular weight I should be using for calculating homotopy limits in $\mathbf{qCat_\infty}$ as a simplicial category (but again, not a simplicial model category)? I don't know that $N(\mathcal{D}/-):\mathcal{D}\to\mathbf{sSet}_\mathrm{Joyal}$ is cofibrant in $\left(\mathbf{sSet}_\mathrm{Joyal}\right)_\mathrm{proj}^\mathcal{D}$. Does anyone know if it is, or if not, what the cofibrant replacement looks like? Would we take the free groupoid of $\mathcal{D}/-$ before taking the nerve (just a guess)? One last question: the theory developed by Riehl and Verity in their series of papers makes the 2-categorical approach to the study of $(\infty,1)$-categories appealing (i.e. working in the homotopy 2-category of the $(\infty,2)$-category of $(\infty,1)$-categories). Does anyone know if homotopy limits in $\mathbf{qCat_\infty}$ agree with $\mathbf{Cat}$-enriched (conical) limits in $\mathbf{qCat_2}$? That would be a useful result, but I don't think I've seen anything to that effect in Riehl's book. Last note: one of the limits which interests me is of a tower of isofibrations, but I don't know that the morphisms involved in the second limit are inner fibrations (although the diagram is certainly pointwise fibrant). REPLY [6 votes]: When working with quasi-categories, it is often more convenient (and more compatible with existing machinery) not to work with actual strict diagrams of quasi-categories but rather with coCartesian fibrations. In your case this would be a coCartesian fibration of the form $\pi:\mathcal{C} \to N(I)$ where $I$ is your indexing category (e.g., the tower category) and $N(I)$ is its nerve. The limit is then given explicitly as the quasi-category of coCartesian sections $N(I) \to \mathcal{C}$ of $\pi$ (that is, sections which send every edge of $N(I)$ to a $\pi$-coCartesian edge of $\mathcal{C}$). This is a sub-simplicial set of the mapping quasi-category cut out by straightforward degree-wise conditions, so it doesn't get more explicit than that. Of course, you need first to translate your diagram to a coCartesian fibration. In your case the indexing diagram is discrete so you can use the relative nerve construction featured in section 3.2.5 of higher topos theory (here you will again have an explicit formula for the simplices of $\mathcal{C}$, so all in all the end result will be explicit). It could also be that there is a better way to get a coCartesian fibration over $N(I)$ in your particular case. For example, it could be that you constructed your diagram from some initial data, but that you could in fact construct a coCartesian fibration from that initial data just as easily. Concerning the possibility of working with the homotopy 2-category of $(\infty,1)$-categories, the latter most likely does not have (ordinary or $\mathrm{Cat}$-enriched) limits and colimits in general, no more than the the homotopy category of spaces has limits and colimits (though I admit I don't have a concrete counter-example in mind of a diagram in the homotopy 2-category which does not have a limit).<|endoftext|> TITLE: Cofinal monotone maps from $\omega^\omega$ to $\kappa^\kappa$ QUESTION [11 upvotes]: Given a cardinal $\kappa$ consider the set $\kappa^\kappa$ of all functions from $\kappa$ to $\kappa$, endowed with the partial order $f\le g$ iff $f(\alpha)\le g(\alpha)$ for all $\alpha\in\kappa$. A map $f:P\to Q$ between two partially ordered sets is called monotone if for any points $x\le y$ in $P$ we get $f(x)\le f(y)$ in $Q$; cofinal if for any $q\in Q$ there is $p\in P$ with $f(p)\ge q$. Problem 1. Is there an uncountable regular cardinal $\kappa$ admitting a monotone cofinal map $f:\omega^\omega\to\kappa^\kappa$? Remark. It can be shown that such cardinal $\kappa$ belongs to the half-interval $[\mathfrak b,\mathfrak d)$, where $\mathfrak b$ and $\mathfrak d$ are the boundedness and dominating numbers of the poset $\omega^\omega$. Therefore, under $\mathfrak b=\mathfrak d$ the answer to the above problem is "No". Is the same "No" true in ZFC? This question can be reformulated in topological terms as follows. Following Cascales, Orihuela and Tkachuk, we say that a topological space $X$ is strongly $\omega^\omega$-dominated if there exists a family $(K_\alpha)_{\alpha\in\omega^\omega}$ of compact subsets of $X$ such that $K_\alpha\subset K_\beta$ for every $\alpha\le\beta$ in $\omega^\omega$ and for every compact set $K\subset X$ there exists $\alpha\in\omega^\omega$ such that $K\subset K_\alpha$. It is easy to see that Problem 1 is equivalent to the topological Problem 2. Is the space $\kappa^\kappa$ strongly $\omega^\omega$-dominated for some uncountable cardinal $\kappa$? Here $\kappa^\kappa$ carries the Tychonoff product topology and $\kappa$ is endowed with the standard order topology. A Known Related Fact (Cascales-Orihuela-Tkachuk). For any uncountable cardinal $\kappa$ the space $\omega^\kappa$ is not strongly $\omega^\omega$-dominated and hence does not admit a monotone cofinal map $f:\omega^\omega\to\omega^\kappa$. Concerning the consistency strength of the statements in Problems 1,2, let us recall the following result of Cummings and Shelah. For a cardinal $\kappa$ let $\mathfrak b_\kappa$ and $\mathfrak d_\kappa$ be the boundedness and dominating numbers of $\kappa^\kappa$ endowed with the partial preorder $f\le^* g$ iff $|\{\alpha\in\kappa: f(\alpha)>g(\alpha)\}|<\kappa$. Theorem. Let $F$ be a class function assigning to each regular cardinal $\kappa$ a triple of cardinals $(\beta(\kappa),\delta(\kappa),\lambda(\kappa))$ such that $cf(\lambda(\kappa))>\kappa<\beta(\kappa)=cf(\beta(\kappa))\le\delta(\kappa)\le\lambda(\kappa)$ and $\lambda(\kappa)\le\lambda(\kappa')$ for any regular cardinals $\kappa<\kappa'$. Then there exists a class forcing poset $\mathbb P$ preserving all cardinals and cofinalities such that in the generic extension, $\mathfrak b_\kappa=\beta(\kappa)$, $\mathfrak d_\kappa=\delta(\kappa)$ and $|\kappa^\kappa|=\lambda(\kappa)$. This result of Cumming and Shelah implies the consistency of $\mathfrak b=\omega_1$, $\mathfrak d=\omega_2$ and $\mathfrak b_\kappa=\mathfrak d_\kappa>\omega_2$ for all regular uncountable cardinals $\kappa$. This impies that the consistency of the answer "No" to Problems 1,2 is strictly weaker that the consistency of the equality $\mathfrak b=\mathfrak d$. On the other hand, the same result of Cummings and Shelah implies the consistency of $\mathfrak b=\omega_1$ and $\mathfrak b_{\omega_1}=\mathfrak d_{\omega_1}=\mathfrak d=\omega_1^{\omega_1}=\omega_2$, witnessing that a (desirable for me negative answer) to Problem 1 should use hardly the monotonicity (not only cofinality) of the map $f:\omega^\omega\to\kappa^\kappa$. I need this "No" ZFC-answer (very much) for studying free topological groups. Please, help! REPLY [5 votes]: My former doctoral student Lubomyr Zdomskyy has resolved this problem, noticing that adding $\omega_2$ Cohen reals to a model of GCH produces a model in which the cardinal $\omega_1$ admits a monotone cofinal map $\omega^\omega\to(\omega_1)^{\omega_1}$. Alternatively the same result can be derived from the existence of a K-Lusin set of cardinality $\mathfrak c=2^{\omega_1}$ in $\omega^\omega$. A subset $L$ of a Polish space $X$ is called a K-Lusin set in $X$ if for every compact subset $K\subset X$ the intersection $K\cap L$ is countable. Theorem (Zdomskyy). If there exists a $K$-Lusin set $L\subset\omega^\omega$ of cardinality $|L|=\mathfrak c=2^{\omega_1}$, then there exists a mootone cofinal map $f:\omega^\omega\to(\omega_1)^{\omega_1}$. Proof. Let $g:L\to (\omega_1)^{\omega_1}$ be any surjective map. Observe that for every $x\in\omega^\omega$ the set $L_x=\{y\in L:y\le x\}$ is countable. Then the formula $f(x)=\sup(g(L_x))$ defines a required cofinal monotone map $f:\omega^\omega\to(\omega_1)^{\omega_1}$. $\square$ The existence of a K-Lusin set of cardinality $\mathfrak c$ in the Cohen model was established by Bartoszynski and Halbeisen.<|endoftext|> TITLE: Random walk to stay in an interval forever QUESTION [6 upvotes]: Consider a random walk on the real time, starting from $0$. But this time assume that we can decide, for each step $i$, a step size $t_i>0$ to the left or the right with equal probabilities. To formalize this, we have $(X_n)_{n\geq 0}$ such that $X_0=0$, $Pr[X_n=t_n]=0.5$ and $Pr[X_n=-t_n]=0.5$ (hence still $E(X_i)=0$). Let $I=[-1,1]$ and $$p=Pr[\sum_{i=1}^n X_i \in I. \forall n\geq 0]$$ The question is: does there exist a sequence $(t_i)$ such that $t_i>0$ and $\sum_{i=0}^\infty=\infty$ (i.e. $(t_i)$ diverges), to ensure $p>0$. Clearly such a sequence does not exist to let $p=1$, and intuitively it seems to me that neither exists $(t_i)$ to force $p>0$, but I am looking for a rigorous proof or disproof. Moreover, I am not exactly sure whether this kind of questions has been studied, so references would be appreciated. REPLY [10 votes]: Yes. Indeed, if $s = \sum_{i \geq 1} t_i^2 <1$, then $$ \mathbb{P}[ \ \ \forall n, \sum_{i=1}^n X_i \in [-1,1] \ \ ] \geq 1-s > 0. $$ To see this, note that $M_n = |\sum_{i=1}^n X_i|$ is a nonnegative submartingale, so that Doob's martingale inequality yields $$ \mathbb{P}[ \max_{1 \leq j \leq n} M_j > 1 ] \leq \mathbb{E}[M_n^2] = \sum_{i=1}^n t_i^2 \leq s. $$ Letting $n$ tends to infinity, one gets $$ \mathbb{P}[ \sup_{j \geq 1} M_j > 1 ] \leq s, $$ hence the result by taking the complement. Remark : In the other direction, one can show that if $$ \mathbb{P}[ \ \ \forall n, \sum_{i=1}^n X_i \in [-1,1] \ \ ] \geq c > 0, $$ then $$ s = \sum_{i \geq 1} t_i^2 \leq \frac{14}{c^2}. $$ One first note that one must have $|t_i| \leq 2$ for all $i$. One has $|\cos(2 \pi \xi)| \leq e^{-2 \pi^2 \xi^2 }$ whenever $|\xi| \leq 2 \delta$, with $\delta = 0.14$, hence $$ |\mathbb{E}[e^{-2i \pi \xi \sum_{j=1}^n X_j} ]| = \prod_{j=1}^n |\cos(2 \pi \xi t_j)| \leq e^{-2 \pi^2 \xi^2 \sum_{i=1}^n t_i^2} $$ whenever $|\xi| \leq \delta$. Let $\chi$ be the Beurling-Selberg majorant of $\mathbb{1}_{[-1,1]}$ with parameter $\delta$ : one has $\chi \geq \mathbb{1}_{[-1,1]}$, $||\chi||_{L^1}= 2 + \delta^{-1}$, and its Fourier transform $\hat{\chi}$ is supported on $[-\delta,\delta]$. In particular : $$ c \leq \mathbb{E}[\chi(\sum_{i=1}^n X_i)] = \int_{\mathbb{R}} \hat{\chi}(\xi) \mathbb{E}[e^{-2i \pi \xi \sum_{j=1}^n X_j} ] d \xi \leq (2 + \delta^{-1}) \int_{\mathbb{R}} e^{-2 \pi^2 \xi^2 \sum_{i=1}^n t_i^2} d \xi, $$ and thus $$ c \leq \frac{2 + \delta^{-1}}{\sqrt{2 \pi \sum_{i=1}^n t_i^2}}. $$ This implies $$ \sum_{i=1}^n t_i^2 \leq \frac{(2 + \delta^{-1})^2}{2 \pi c^2} \leq \frac{14}{c^2}, $$ hence the result.<|endoftext|> TITLE: Is a polarization on an abelian scheme an open condition? QUESTION [8 upvotes]: Let $A/S$ be an abelian scheme such that the dual abelian scheme $A^{\vee}/S$ exists and let $\lambda : A \to A^{\vee}$ be a morphism of abelian schemes. Is the locus of points in $S$ where $\lambda$ is a polarization open? EDIT: Recall that a polarisation on $A/S$ is a morphism of abelian schemes $\lambda :A \to A^t$ such that, for every geometric point $\overline{s}$ of $S$, $\lambda_{\overline{s}}$ is of the form $a \mapsto t_a^*\mathcal{L}\otimes \mathcal{L}^{-1}$ for some ample line bundle $\mathcal{L}$ over $A_{\overline{s}}$. Here, $t_a$ denotes the translation by a point $a$ in $A_{\overline{s}}$. REPLY [8 votes]: A nice way to understand this is to give a "better" characterization of polarizations that avoids any reliance on structures that only are available on geometric fibers. The claim is that a homomorphism $\lambda:A \rightarrow A^t$ is a polarization if and only if it satisfies the following conditions: (i) $\lambda$ is symmetric with respect to double duality, (ii) the line bundle $(1, \lambda)^{\ast}(\mathscr{P}_A)$ on $A$ is fiberwise ample. This characterization deserves to be more widely known, and it is a bit surprising that (as far as I am aware) it is not mention in any of Mumford's papers or books. [We are assuming that $A$ is projective Zariski-locally on $S$ so that it admits a dual abelian scheme by Grothendieck's work, or one appeals to deeper methods relying on algebraic spaces to make the dual (as a scheme, not just algebraic space) in general.] By rigidity considerations (after passing to the case of a noetherian base, by a small argument) whether or not (i) holds is checkable on geometric fibers, and likewise for (ii), so to prove such an equivalence it suffices to check for abelian varieties over an algebraically closed field. If $\lambda = \phi_L$ for some line bundle $L$ on $A$ then the symmetry of $\lambda$ is part of the classical theory, in which case $(1, \lambda)^{\ast}(\mathscr{P}_A) \in L^{\otimes 2} \cdot A^t(k)$ inside ${\rm{Pic}}(A)$. But ampleness or not of a line bundle on such $A$ only depends on the line bundle class in ${\rm{Pic}}(A)/A^t(k)$, so the condition (ii) is equivalent to ampleness of $L^{\otimes 2}$, which in turn is equivalent to ampleness of $L$ (and then $\phi_L$ is an isogeny by the classical theory too). To complete the proof of the equivalence, it suffices to show conversely that every symmetric homomorphism $\lambda$ arises as $\phi_L$ for some line bundle $L$ on $A$. This is a consequence of Theorem 2 in section 20 and Theorem 3 in section 23 of Mumford's book on abelian varieties, along with consideration of Weil pairings. (Poonen and Stoll gave examples of smooth projective geometrically connected curves over $\mathbf{Q}$ for which the principal polarization of the Jacobian $J$ does not arise as $\phi_L$ for any $L$ on $J$.) Having established the equivalence, it suffices to show that each of (i) and (ii) corresponds to an open condition on the base. More specifically, for (i) it suffices to show that if a homomorphism $f:A \rightarrow B$ between abelian schemes (e.g., $\lambda - \lambda^t$) vanishes on fibers at some $s \in S$ then it vanishes over an open neighborhood of $s \in S$, and for (ii) it suffices to show that if $X \rightarrow S$ is a proper finitely presented map of schemes and $L$ is a line bundle on $X$ then the locus $U$ of $s \in S$ such that $L_s$ is ample on $X_s$ is open in $S$. The former is a consequence of rigidity (after passing to the case when the base is local with $s$ as its closed point, and then reducing to the case when the local base ring is noetherian), and the latter is EGA IV$_3$, 9.6.4 (and by IV$_3$ 9.6.5 the restriction of $L$ to $X_U$ is relative ampleness over $U$ -- i.e., ample over every affine open in $U$, quite remarkable without any flatness hypotheses and absolutely crucial for applications to the use of ampleness for proving effective descent in the context of moduli problems). In fact, being a polarization is a (finitely presented) closed condition on $S$ too. We may reduce to the case of noetherian $S$, and then (i) this is again a consequence of rigidity considerations, whereas for (ii) it lies deeper: by the valuative criterion for properness (applied to open immersion in the base) it suffices to consider the case of abelian schemes over a discrete valuation ring, but one has to do a fair bit more work to handle that; this is discussed early in the book of Faltings and Chai.<|endoftext|> TITLE: Wrong Tits-Index of E7 from Springer 's book QUESTION [5 upvotes]: In the his book Linear algebraic groups, by T.A. Springer, there is a list of possible Tits-Indexes. For the $E_7$ case, there is an index shown, such that vertex $1$ and $7$ are circled (Bourbaki notation). I just realized that yesterday by coincidence. However in Tits's original paper this index is not listed. On top of that neither here http://arxiv.org/pdf/1511.02538.pdf nor here http://arxiv.org/abs/0807.2140 do we find it. What did go wrong in Springers book? REPLY [10 votes]: That is a typo. In index 14 on p. 321 the vertex 6 should be black. In Proposition 17.8.2, this index is correct. It corresponds to $E_{7,3}^{28}$ in Tits' notation.<|endoftext|> TITLE: Coherence theorem for symmetric lax monoidal functors QUESTION [5 upvotes]: Let $V$ and $W$ be symmetric monoidal categories. Let $F:V\to W$ be a lax symmetric monoidal functor with multiplication $\nabla:FA\otimes FB \to F(A\otimes B)$. Consider the following statements: 1) There are lax symmetric monoidal functors $F(-)^{\otimes n}:V\to W$ and $F((-)^{\otimes n}):V\to W$ for each $n\geq 2$, 2) There is a lax monoidal transformation $F(-)^{\otimes n} \Rightarrow F((-)^{\otimes n})$. They seems plausible to me. For example, if $n=2$ these things are a long but doable diagram chase. As $n$ gets bigger, it gets more and more unwieldy to actually verify it by hand. It seems to me this is a sort of "coherence theorem". For example, consider the second statement for $n=3$. There are two reasonable natural transformations to define by suitably tensoring $\nabla$ with the identity, yet they are equal by associativity of $F$. As $n$ grows, the amount of "obvious but provably equal" options for defining the natural transformation grows. Are these statements true, and how might one go about actually proving them? Note: there is a hidden (iterated) diagonal which might be a red herring. We could be considering the functors in 1) to be $V^{\times n}\to W$. REPLY [6 votes]: I think they are true. First of all, let's break out the diagonal as you suggested by writing $(-)^{\otimes n}:V\to V$ as the composite $V \xrightarrow{\Delta} V^n \xrightarrow{\otimes_n} V$. Since the 2-category of symmetric monoidal categories and lax symmetric monoidal functors has finite products, $F$ commutes with the $\Delta$'s (which are strict monoidal), so it suffices to show that $\otimes_n$ is monoidal and that we have a symmetric monoidal transformation (it doesn't make sense for a transformation to be "lax") $F \circ \otimes_n \to \otimes_n \circ F^n$. Now, it's fairly straightforward to show that if $V$ is symmetric monoidal, then $\otimes : V\times V\to V$ is strong monoidal. By taking products with the identity and composing, we find that $\otimes_n : V^n \to V$ is also strong monoidal, and therefore $F\circ \otimes_n$ and $\otimes_n\circ F^n$ are lax monoidal. More generally, if $X$ is a symmetric pseudomonoid in any 2-category with products, then $\otimes :X\times X\to X$ is strong monoidal, and hence so is $\otimes_n:X^n \to X$. But a lax symmetric monoidal functor can be identified with a symmetric pseudomonoid in the 2-category $\mathrm{Oplax}(\mathbf{2},\mathrm{Cat})$ whose objects are the arrows of Cat (functors) regarded as functors from the interval category $\mathbf{2}$ to Cat, and whose morphisms are oplax transformations (which here are just 2-cells fitting in a square). Similarly, a strong symmetric monoidal morphism in that 2-category can be identified with a symmetric monoidal transformation $G\circ F \to F'\circ H$ where $G$ and $H$ are strong symmetric monoidal and $F$ and $F'$ are lax symmetric monoidal. Thus, applying the general result about symmetric pseudomonoids to our $F$, we get the desired transformation.<|endoftext|> TITLE: Question on a paper by Benjamini/Kozma/Wormald about a “well known fact” QUESTION [11 upvotes]: In "The mixing time of the giant component of a random graph" by the aforementioned authors, in the last proof on page 19 it says something along the lines of "It is well known and easy to verify that, since $m=O(n)$ , asymptotically almost surely the maximum degree occurring in $\mathcal{C}_3(n,m)$ is $o(n^{0.02})$" where $\mathcal{C}(n,m)$ denotes a random graph with $n$ vertices, $m$ edges, uniformly sampled among such graphs, and $\mathcal{C}_3(n,m)$ denotes the variable conditioned on minimum degree 3 (which shouldn't influence the above statement). I am wondering where the $n^{0.02}$ comes from, and how this is "well known". Can someone provide an original source for this statement or this kind of statement? REPLY [10 votes]: I have no idea where the figure $n^{0.02}$ comes from. I would usually say it's well known that the maximum degree is $O(\log n)$ (actually even this is an overestimate). It's for example found in an Erd\H{o}s-R\'enyi paper with a title about random matrices, I think; or surely in any of the 'Random Graphs' books. In any case, it is very easy to prove. One (lazy) way is: generate $G(n,p)$ with $p=1000(\log n)/n$. Standard applications of the Chernoff bound give you that a.a.s. this has all vertex degrees between $500\log n$ and $1500\log n$. Take a uniform random subgraph of this with $m$ edges and minimum degree three. Over both levels of randomness this is uniformly distributed, so it's $C_3(n,m)$: done.<|endoftext|> TITLE: Ordinal-definable witnesses to the perfect set property? QUESTION [6 upvotes]: This possibly a very basic descriptive set-theory question; if it is too basic for MO, feel free to migrate. Throughout we work in ZF+AD. My question is: If $A$ is an uncountable OD set of reals, need $A$ have an OD perfect subset? Motivation: The Solovay sequence is given by: $\theta_0$ is the least ordinal onto which $\mathbb{R}$ doesn't surject via an OD function; $\theta_\lambda=\sup\{\theta_\alpha: \alpha<\lambda\}$ for $\lambda$ limit; and $\theta_{\alpha+1}$ is the least ordinal onto which $\mathcal{P}(\theta_\alpha)$ doesn't surject via an OD function. Ignore the later stages of the hierarchy; if the answer to my question is no, then we get a potentially interesting way to compare the "size" of two uncountable sets of reals under AD (which of course have cardinality continuum): let $\theta_0(X)$ be the least ordinal onto which $X$ doesn't surject in an OD way. If the answer to my question is "no," we might have $\theta_0(X)<\theta_0$ for some uncountable $X$, even assuming AD. This looks very interesting to me. . . . which is why I think the answer to my question will be "yes" :P. But oh well. (To keep going and define $\theta_\alpha(X)$, we need $X$ to be "ordinal-independent": that is, we want a "$\kappa$-version" $X_\kappa\subseteq 2^\kappa$ of $X$ for all (or enough) $\kappa$s. Then we can talk about things like the rate at which the Solovay sequence grows, for a given such $X$; but I think I'm getting well ahead of myself, so I'll stop for now.) A note on tags: I used the inner-model-theory tag because of the importance of the Solovay sequence in descriptive inner model theory, and more generally the hope that inner model theory might be relevant to this question. This might be incorrect on my part; if so, feel free to remove that tag. REPLY [12 votes]: Let $A$ be the set of those reals that are not OD. Then $A$ is OD (since I've just defined it), and it's uncountable (since its complement, being a well-orderable set of reals, must be countable under AD). But any perfect OD set $P$ has an element that is OD and thus outside $A$, namely the first element of $P$ (in the usual ordering of the reals, or, if your reals are $2^\omega$, in the lexicographic order), which exists as $P$ is compact. That first element is OD because it's defined from the OD set $P$.<|endoftext|> TITLE: Rational curves on the Fermat quartic surface QUESTION [13 upvotes]: Let $X$ be the Fermat quartic $x^4+y^4+z^4+w^4=0$ in $\mathbb P^3$. It is known that $X$ contains infinitely many $(-2)$-curves, that is, smooth rational curves. (One way to obtain in infinitely many is to use the various elliptic fibrations on $X$, and use translations in the fibers.) Note however that these curves are typically not defined over $\mathbb Q$, even though $X$ is. My question is the following: Is there a finite extension $K$ of $\mathbb Q$ such that all of the smooth rational curves are defined over $K$ and have a $K$-rational point? I suspect that the answer is no, but I don't see how to prove it. REPLY [13 votes]: I am posting my comments as an answer. I am concerned that I misunderstand the OP, so let me state first the result. There exists a finite field extension $K/\mathbb{Q}$ such that for every closed immersion $\mathbb{P}^1_\mathbb{C} \hookrightarrow X\otimes_{\mathbb{Q}}\mathbb{C}$, there exists a closed immersion $\mathbb{P}^1_K \hookrightarrow X \otimes_{\mathbb{Q}} K$ with the same image after base change to $X\otimes_{\mathbb{Q}}\mathbb{C}$. To prove this, first observe that the geometric Picard group $\text{Pic}(X\otimes_{\mathbb{Q}}\mathbb{C})$ is finitely generated. Thus, there exists a finite field extension after which a finite generating set of the geometric Picard group is defined over that field. Second, the geometric automorphism group of the complex scheme, $\text{Aut}_{\mathbb{C}}(X\otimes_{\mathbb{Q}}\mathbb{C})$, is a discrete group that is finitely generated by Corollary 2.4 of Chapter 15, p. 315 of the following; result due to H. Sterk -- Finiteness results for algebraic K3 surfaces. Math. Z., 189(4):507–513, 1985. Daniel Huybrechts Lectures on K3 Surfaces Part of Cambridge Studies in Advanced Mathematics September 2016 ISBN: 9781107153042 http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf Thus, after a further finite field extension, there exists a finite subset of $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ whose images generate the geometric automorphism group. Thus, $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ is already the full geometric automorphism group. Third, by Corollary 4.6 of Chapter 8, p. 161 -- result again proved by Sterk in the article cited above -- the action of the geometric automorphism group on the set of smooth rational curves has only finitely many orbits. For each of those finitely many orbits, after a degree $2$ field extension, every curve in that orbit acquires a rational point. Thus, after a further finite field extension, every smooth, genus $0$ curve on $X$ acquires a rational point. Please note, all of this is valid for every K3 surface over every characteristic $0$ field, not only for the Fermat quartic surface over $\mathbb{Q}$. Using extensions of Sterk's results by Lieblich and Maulik, this should also be true in positive characteristic.<|endoftext|> TITLE: Infinite dimensional version of a simple fact on certain singular matrices QUESTION [7 upvotes]: We consider the following simple fact about matrices. Then we try to generalize it in the context of smooth manifolds; Let $L$ be the collection of all $n \times n$ real matrices $A=(a_{ij})$ with the following property: $$\sum_{i=1}^{n} a_{ij}=0$$ for every fixed $j$. Obviousely $L$ is a Lie algebra.(As I have already learned from Qiaochu Yuan in another MO post) Moreover the linear map $X \mapsto AX$ has non trivial kernel. These simple facts can be modelized in an infinite dimensional manner.(note that a vector $X\in\mathbb{R}^{n}$ can be considered as a function on a finite ($n$ pointed)set $M$ equiped with discrete counting measure and a matrix is a function on $M\times M$. Now the product $AX$ has an integral representation if we replace $\sum$ by the integral sign. That is we read the expersion $\sum a_{ij}x_{j}$ in the integral form $\int_{M} a_{ij}x_{j}$ where the integration is based on the normalized counting measure. Now we state our questions as generalization of the above simple fact about matrices. Assume that $M$ is a compact orientable manifold or a Lie or topological group. So $M$ has a natural measure, correspond to volum form or the invariant metric or Haar measure. Assume that $g: M \times M \to \mathbb{R}$ is a smooth function which satisfies $$\int_{M} g(x,y)dx=0\;\;\;\;\;(1) $$ for all $y \in M$. Does the linear map $A$ on $C^{\infty} (M)$ has nontrivial kernel? $$A(f)(x)=\int_{M} g(x,y)f(y)dy$$ Note that for topological groups we consider continuous functions, since smoothness is meaningles. For our next question, we assume that $M$ is a symplectic manifold, so $M \times M$ has a natural symplectic structure. Let $L$ be the space of all smooth functions on $M\times M$ which satisfy the equation (1). Is $L$ closed under Poisson bracket? REPLY [4 votes]: Willie Wong answered the general case, and I'd like to give a counterexample for the symplectic case: Let $M=S^2$, the standard 2-sphere embedded in $\mathbb{R}^3$ as the unit sphere, with the standard symplectic form corresponding to the Euclidean metric on $\mathbb{R}^3$. Take on $(M \times M, \Omega = \omega \oplus - \omega)$ the functions \begin{align} f(z,w) &= z_3\cdot F(w), \\ g(z,w) &= z_3 \cdot G(w) \end{align} both of which are in $L$. Now, $\{ f, g \} = \Omega(\mathrm{sgrad}(g), \mathrm{sgrad}(f)) =: \Omega'(\mathrm{d}f, \mathrm{d}g)$. Then, \begin{align} \{f,g\} &= (\omega' \oplus (-\omega')) \left( F(w)\mathrm{d} z_3 + z_3 \mathrm{d} F(w), G(w)\mathrm{d} z_3 + z_3 \mathrm{d} G(w) \right) \\ &= F(w)G(w) \omega'(\mathrm{d}z_3,\mathrm{d}z_3) - z_3^2 \omega'(\mathrm{d} F(w),\mathrm{d} G(w)) \\ &= z_3^2 \{G,F\}(w) =:h(z,w) \end{align} Now, unless $\{G,F\} \equiv 0$, $h$ will not be in $L$.<|endoftext|> TITLE: Semisimplicity of Frobenius on *integral* Tate module QUESTION [8 upvotes]: Let $K$ be a number field and $A/K$ an Abelian variety; let $l$ be a (rational) prime. Do there exist infinitely many primes $\mathfrak{p}$ of $K$ such that the Frobenius at $\mathfrak{p}$ acts semisimply on the integral Tate module $$T_l(A)\otimes \overline{\mathbb{Z}_\ell}?$$ By this I mean that there is a $\overline{\mathbb{Z}_\ell}$-basis of eigenvectors for the Frobenius action. The answer is well-known to be yes if the integral Tate module is replaced with the rational Tate module (this goes back to Weil). If this is true, it should probably boil down to some Chebotarev argument, but I'm blanking at the moment. Some remarks: As znt notes, the answer is "yes" if the image of the representation $G_K\to GL(A[\ell])$ contains a matrix with distinct eigenvalues, by Chebotarev. Moreover, one can show that for any prime $\mathfrak{p}$ at which $A$ has good reduction, the Frobenius at $\mathfrak{p}$ acts semisimply on the $\ell$-adic Tate module for almost all $\ell$ (this follows from the theory of Frobenius tori). If the Mumford-Tate conjecture holds for $A$, I think it's enough to find any torus with good reduction at $\ell$ in the Mumford-Tate group of $A$, though I haven't written out the details. In any case, to find a counterexample it should be enough to find an Abelian variety over $K$ whose Mumford-Tate group contains no torus with good reduction at $\ell$. One may also show the result holds if the image of the map $G_k\to GL(T_\ell(A))$ contains any element with distinct eigenvalues which is semi-simple over $\overline{\mathbb{Z}_\ell}$. REPLY [5 votes]: Probably not. Let $E_1$ and $E_2$ be two distinct elliptic curves with identical $\ell$-torsion representations, and let $A$ be $E_1 \times E_2$ mod the diagonal $\ell$-torsion representation. Then at any prime $\mathfrak p$ where Frobenius acts semisimply on $A[\ell]$, $E_1$ and $E_2$ are isogenous. Indeed, if the eigenvalues of Frobenius of $E_1$ and $E_2$ are distinct, an eigenbasis must consist two elements of $T_\ell (E_1)$ and two elements of $T_\ell(E_2)$, but there is no basis of $T_\ell(A)$, the lattice of elements of $T_\ell(E_1)+ T_\ell(E_2)$ where both parts are congruent mod $\ell$, consisting only of elements of $T_\ell(E_1)$ and $T_\ell(E_2)$. If the eigenvalues of Frobenius of $E_1$ and $E_2$ are equal, then they are isogenous. Of course this is no problem as there are infinitely many such primes (https://arxiv.org/abs/1411.2914). However, one can take a product of many such abelian surfaces for many different pairs of elliptic curves. The product is only semisimple if each pair of curves is isogenous. Standard heuristics suggest there should be only finitely many primes where all these pairs are isogenous (already 3 independent pairs should be sufficient).<|endoftext|> TITLE: Corollaries of the Yoneda Lemma in Analysis? QUESTION [26 upvotes]: This is a cross-post of my ~2 weeks (canonically) unanswered question on Math.SE: https://math.stackexchange.com/questions/1830287/corollaries-of-the-yoneda-lemma-in-analysis. I am looking for some simple examples of how the Yoneda Lemma can be applied in analysis and probability theory and related fields. One candidate example is the Fourier transform. See this question on Math.SE: https://math.stackexchange.com/questions/1667473/is-the-fourier-transform-a-special-case-of-this-version-of-the-yoneda-lemma. Another candidate example would be the Riesz Representation Theorem (see this unanswered question on Math Overflow Can the similarity between the Riesz representation theorem and the Yoneda embedding lemma be given a formal undergirding? or one of the responses to my original question on Math.SE). There are also many coordinate-free representations of objects in analysis, and all seem to be characterized by the interactions of the objects with all possible coordinate systems. (Although this seems to be the principle of equivalence rather than the Yoneda Lemma per se, see the comment by David Roberts below.) Motivation/Context: For example, in Qiaochu Yuan's answer to a similar question, the Yoneda Lemma shows us that every element in a poset can be determined by either the set of all elements it is greater than or equal to, or the set of all elements which is less than or equal to, essentially each element is equivalent to a Dedekind cut. If we consider the partial order of equivalence classes of Cauchy convergent sequences of rational numbers, then this reasoning seems to justify the construction of the real numbers from Dedekind cuts (although I am not entirely sure). A simple candidate example that I can think of and somewhat understand is any linear transformation on a finite-dimensional vector space. These can be represented by a matrix, which determines the action of the transformation on a certain basis of the vector space. But seemingly we can characterize the "inherent/intrinsic" linear transformation in a coordinate-free way by specifying its action on every possible set of bases. (This is basically the idea that tensors are actually coordinate-free objects.) Why do I think that the Yoneda Lemma might be related to coordinate-free representations of vectors? Because the automorphism group on a vector space (which consists of changes of basis and which is a monoidal category with one object), can be lifted to the category whose objects are the various representations of a single vector in different coordinate systems, using the group action. Whether or not this lifting constitutes a functor, I don't know. In any case, this lifting from a category with one object to the category of all possible coordinate representations of a vector suggests intuitively that we can consider all of those coordinate representations to be a single object. My question is whether or not the Yoneda lemma formalizes that intuition. In another answer on MathOverflow, which was mentioned towards the end of Tom LaGatta's talk about category theory at the NYC Lisp meetup, an analogy to particle physics was made. Basically the intuition behind Yoneda's Lemma is supposed to be that one can characterize an object (up to equivalence I guess) by probing it via its interactions (i.e. morphisms) with all other objects. In the above example, we would be smashing a vector against all possible changes of basis in order to understand it completely. Despite being an analyst/probabilist, Tom LaGatta did not go further into examples besides this particle physics metaphor. (He did say something along the lines of "You will find it meaningful in your own context, I guarantee it" around 1:34:00. However, I am curious, because this analogy suggests to me coordinate-free representations of objects. EDIT: Related post on MO Is there an introduction to probability theory from a structuralist/categorical perspective? REPLY [5 votes]: Thanks William for reaching out (and thanks David Roberts for the hat tip to my talk). Let me give an intentionally fuzzy, high-level answer. Generally speaking, the Yoneda Lemma allows you to make an identification between an "object" and the "maps into (or out of) the object". Because of the structural depth of this result, just about every result in analysis that exploits such an identification could be proven using the Yoneda Lemma. This may or may not add value depending on the audience of such a work: analysts are mostly uninterested in categorical formulations of their models (unless the work proves a novel analytical result). I'm sure category theorists would be interested in an analytical survey using CT. The basic application is of course Cayley's theorem: every group can be admitted as a group action on some space. For a more modern example, I believe the Kolmogorov-Centsov theorem can be formulated in this fashion: every stochastic process can be characterized using its finite-dimensional distributions (uniqueness), and all consistent families of finite-dimensional distributions generate a stochastic process (existence). The uniqueness result should follow immediately using Yoneda, and existence should follow from some "existence of limits" axiom of the category. I suggest trying to write this up as an exercise. Moving beyond that, every analytical or probability category should admit some Yoneda consequence, since the result holds for every category. In most categories, this will be uninteresting. In some categories, this will correspond to some interesting classical theorem. Those are the ones to latch onto, because now every other classical CT result might correspond to some interesting theorem. Many will be known (so analysts won't care), but you have a chance of proving new, cutting-edge results too. My suspicion is that if you're very good at both analysis and CT, then this can lead to a profitable research career: Use analytical results (e.g. Central Limit Theorem) to motivate the construction of suitable concrete categories, Grab general CT theorems and apply them to those categories, Prove new analytical results by unwrapping the abstract formalism into concrete, quantitative statements. Hope this helps!<|endoftext|> TITLE: Bounding the union of conjugates of a maximal subgroup of the Symplectic group over a finite field QUESTION [5 upvotes]: Let $g \geq 1$ be a positive integer, and let $p$ be a prime. Consider the symplectic group $G := \operatorname{Sp}_{2g}(\mathbb{F}_p)$ of symplectic matrices with entries in $\mathbb{F}_p$. Let $M \subset G$ be a maximal subgroup and let $S = \bigcup_{h \in G} hMh^{-1}$. Question: Is it true that $|S| = \alpha \cdot |G| + O(1/p)$, where $\alpha \in (0,1)$ is a constant, bounded away from $1$ and independent of $p$? It's not too hard to see the answer is yes in the case where g = 1; see http://www.math.cornell.edu/~zywina/papers/Quantitative-HIT.pdf, Lemma 7.3. For higher g, I'm aware that there is a fairly nice classification of the maximal subgroups, such as that given in Kleidman and Liebeck, table 3.5.C. I think my question can be affirmatively answered for many of the geometric subgroup classes, by noting that the maximal subgroups are normalizers of certain finite index "classical subgroups" (as termed by Kleidman and Liebeck) and the complements of the classical subgroups in the maximal subgroups have trace 0, but I'm having difficulty showing this for all maximal subgroups, particularly for subgroups of Lie type. REPLY [5 votes]: Let $\Omega$ be the set of cosets of $M$, and consider the natural action of $G$ on $\Omega$. The set $S^C$ (the complement of $S$) is the set of derangements in this action. So the upper bound you seek is equivalent to a lower bound on the proportion of derangements in the action of $G$ on $\Omega$. The lower bound you need is given by a much more general result of Fulman and Guralnick. To state it, we need some notation: Let $H$ be a group acting on a set $X$. We write $\delta(H,X)$ for the proportion of elements in $H$ that are derangements in the action on $X$. Theorem There exists a $\delta>0$ such that for any almost simple group $G$ and any set $\Omega$ on which $G$ acts transitively, $\delta(G,\Omega)>\delta$. For a reference see the following paper and its sequels: Derangments in simple and primitive groups by Fulman and Guralnick. In Groups and Geometries, Durham, 2001.<|endoftext|> TITLE: L1 analog of Bernstein's inequality QUESTION [7 upvotes]: Let $p(x)$ be a degree $n$ polynomial over $[-1, 1]$, and let $q(x) = p'(x) \sqrt{1-x^2}$. Is it true that $$ \|q\|_1 \leq O(n) \|p\|_1 $$ where we define $\|f\|_p := \left(\int_{-1}^1 |f(x)|^pdx\right)^{1/p}$? For reference, Bernstein's inequality shows that $$ \|q\|_\infty \leq n\|p\|_\infty $$ with equality at the $n$th Chebyshev polynomial, and looking at Legendre polynomials shows that $$ \|q\|_2 \leq \sqrt{n(n+1)}\|p\|_2 $$ with equality at the $n$th Legendre polynomial. REPLY [4 votes]: Appendix A4 of the book P. Borwein, T. Erdelyi, Polynomials and Polynomial inequalities, Graduate Texts in Mathematics 161, Springer should be a good source for your question. In particular, (A.4.22) gives $$\|P'\|_p\leq cn^2\|P\|_p,$$ for every polynomial $P$ of degree $n$ and $0 TITLE: Is $\mathfrak b_a$ a new cardinal characteristic of the continuum? QUESTION [6 upvotes]: By a partial function from $\omega$ to $\omega$ we understand a function $f:dom(f)\to\omega$ defined on an infinite subset of $\omega$. A family $\mathfrak F$ of partial functions from $\omega$ to $\omega$ is called domain almost disjoint (briefly, DAD) if the family $(dom(f))_{f\in\mathcal F}$ is almost disjoint in the sense that for any distinct partial functions $f,g\in\mathcal F$ the intersection $dom(f)\cap dom(g)$ is finite. Let $\mathfrak b_a$ be the smallest cardinality of a DAD family $\mathfrak F$ of partial functions from $\omega$ to $\omega$ such that for every function $g:\omega\to\omega$ there exists a partial function $f\in\mathcal F$ such that $f(n)\not\le g(n)$ for infinitely many numbers $n\in dom(f)$. It is easy to see that $\mathfrak b\le\mathfrak b_a\le \mathfrak d$. Question. Is the strict inequality $\mathfrak b<\mathfrak b_a$ consistent? Remark. The affirmative answer to this question implies a positive answer to this problem. Namely, the strict inequality $\mathfrak b<\mathfrak b_a$ implies the exitence of a monotone cofinal map $\omega^\omega\to\mathfrak b^{\mathfrak b}$. REPLY [3 votes]: After thinking some time I realized that $\mathfrak b_a=\mathfrak b$, so this question has answer "No" and this "No" does not help to solve the original question. To show that $\mathfrak b_a=\mathfrak b$, consider the binary tree $2^{<\omega}$ and take a family $\mathcal B$ of functions from $2^{<\omega}$ to $\omega$ of cardinality $|\mathcal F|=\mathfrak b$, which is not upper bounded in the set $\omega^{2^{<\omega}}$ endowed with the partial order $\le^*$ of almost dominance. Replacing each function $f\in\mathcal B$ by a larger function, we can assume that $f$ is constant on each level $2^n\subset 2^{<\omega}$, $n\in\omega$, of the tree $2^{<\omega}$. Let $\mathcal A$ be the (almost disjoint) set of branches of the binary tree $2^{<\omega}$. Choose any injective map $dom:\mathcal B\to\mathcal A$ and consider the DAD family $\mathcal F=\{f|dom(f)\}_{f\in\mathcal B}$ of partial functions from $2^{<\omega}$ to $\omega$. We claim that the family $\mathcal F$ witnesses that $\mathfrak b_a=\mathfrak b$. Indeed, take any function $g:2^{<\omega}\to \omega$ and find a function $\hat g:2^{<\omega}\to\omega$ such that $\hat g\ge g$ and $\hat g$ is constant on each level $2^n$ of the tree $2^{<\omega}$. By the choice of the unbounded family $\mathcal B$, there exists a function $f\in\mathcal B$ such that $f\not\le^* \hat g$. Since $f$ and $\hat g$ are constant on each level and $dom(f)$ meets each level $2^n$ of the tree $2^{<\omega}$, the relation $f\not\le^* \hat g$ implies $f|dom(f)\not\le^* \hat g|dom(f)$ and hence $f|dom(f)\not\le^* g|dom(f)$.<|endoftext|> TITLE: Isomorphism in cyclic cohomology vs isomorphism in Hochschild cohomology QUESTION [8 upvotes]: Let $A$ be a unital algebra over a field $K$, $C^n(A)$ a space of all $n+1$ linear maps into scalar field $k$ (I'm interested in case $k=\mathbb{C}$) and $$(bf)(a_0,...,a_{n+1})=\sum_{i=0}^n(-1)^if(a_0,...,a_ia_{i+1},...,a_{n+1})+(-1)^{n+1}f(a_{n+1}a_0,a_1,...,a_n)$$ for $f \in C^n(A)$. One checks that $b^2=0$ so we get a cochain complex $$C^0(A) \stackrel{b}{\longrightarrow} C^1(A) \stackrel{b}{\longrightarrow} C^2(A) \stackrel{b}{\longrightarrow} ... $$ The cohomology of this complex is the Hochschild cohomology. We further introduce the cyclic operator $\lambda:C^n(A) \to C^n(A)$ where $\lambda f (a_0,a...,a_n)=(-1)^nf(a_n,a_0,a_1,...,a_{n-1})$. Then we put $C^n_{\lambda}(A):=ker(1-\lambda)$. Remarkably we have $bC^n_{\lambda}(A) \subset C^{n+1}_{\lambda}(A)$ hence we have a subcomplex. The cohomology of this complex is the cyclic cohomology. Suppose that $f: (C^*(B),b) \to (C^*(A),b)$ be a map of Hochschild complexes commuting with the cyclic operator. Then it induces two morphisms: at the level of Hochschild cohomology and at the level of cyclic cohomology. My question is: Why is it true, that if the induced maps for Hochschild cohomology are isomorphisms then all induced maps in cyclic cohomology are isomorphisms as well? The idea is to use so called $IBS$ long exact sequence of the form: $$... \to HC^n(A) \stackrel{I}{\rightarrow} HH^n(A) \stackrel{B}{\rightarrow} HC^{n-1}(A) \stackrel{S}{\rightarrow} HC^{n+1}(A) \to ... .$$ (Here $HC$ is cyclic cohomology, $HH$ is Hochschild cohomology, $I$ comes from the inclusion of complexes, there are also exact formulas for $B$ and $S$ but I don't know whether we have to use those explicit formulas)). I wonder is it enough to use some general cohomological lemma (something like five lemma) in order to conclude that the maps for cyclic cohomology are isomorphisms. The problem is that if we write another IBS-sequence for an algebra $B$ then one vertical arrow is isomorphism, after that we have two arrows for which we don't know whether are they isomorphisms and then another isomorphism: so I don't see how to apply five lemma. How to handle this problem? REPLY [9 votes]: This was something that used to puzzle me when I was first learning this stuff. Refreshing my memory just now, I think that the trick is to use the fact that the Connes–Tsygan sequence has some zero entries in low degrees. It starts $$ 0 \to HC^0(A) \to HH^0(A) \to 0 \to HC^1(A) \to HH^1(A) \to HC^0(A) \to HC^2(A) \to \dots $$ So, in the context of the original question, if the induced maps $HH^n(B)\to HH^n(A)$ are isomorphisms in all degrees, the maps $HC^0(B)\to HC^0(A)$ and $HC^1(B)\to HC^1(A)$ are iso. Now we can set up an inductive argument: suppose we know that $HC^{k-1}(B)\to HC^{k-1}(A)$ and $HC^k(B)\to HC^k(A)$ are both iso for some $k\geq 1$. Then by considering $$ HH^k(B)\to HC^{k-1}(B) \to HC^{k+1}(B) \to HH^{k+1}(B)\to HC^k(B) $$ $$ HH^k(A) \to HC^{k-1}(A) \to HC^{k+1}(A) \to HH^{k+1}(A)\to HC^k(A) $$ and applying the five lemma, we deduce that $HC^{k+1}(B)\to HC^{k+1}(A)$ is iso. So by induction, the induced maps between cyclic cohomology groups are iso in all degrees.<|endoftext|> TITLE: Comparing the sizes of uncountable sets of reals under AD QUESTION [5 upvotes]: Working in ZF+AD, let $$\theta_0(X)=\min\{\alpha\in ON: \not\exists f: X\rightarrow \alpha\mbox{ surjective and OD}\}$$ be the least ordinal onto which $X$ does not surject in an OD way, for $X\subseteq \mathbb{R}$ uncountable and OD. Clearly if $X$ has an OD perfect subset, then $\theta_0(X)=\theta_0$ (the first term in the Solovay sequence); howevr, this need not occur, as Andreas Blass' answer to Ordinal-definable witnesses to the perfect set property? shows (consider the set of non-OD reals). My question is: is it consistent with ZF+AD that there is an uncountable OD set of reals $X$ with $\theta_0(X)<\theta_0$? If so, what can we say about the possible values of the $\theta_0$-map? My motivation is, ultimately, the question Cardinal characteristics without choice; I'm especially interested in cardinal characteristics which have natural "$\kappa$-versions" for arbitrary $\kappa$ (e.g. the idea of the dominating number makes sense on $\kappa^\kappa$, not just $\omega^\omega$), and the possibility of comparing such cardinal characteristics by looking at the Solovay-type sequences their "large" sets induce; but that's jumping the gun by a substantial amount. I mention this here only in case this motivation is helpful re: finding references, etc. REPLY [3 votes]: The following is due to John Steel - who is not on mathoverflow - following a suggestion (see Ordinal-definable witnesses to the perfect set property?) of Vladimir Kanovei; since none of this is my work, and indeed I don't currently understand the crucial theorem used, I've made this community wiki. I will leave this up for a while before accepting it, for two reasons: first, this uses enough inner model theory that I am not comfortable claiming correctness, and I want to see if the inner model theorists around here agree with the argument below. Second, it leaves open the question of whether the opposite answer is consistent, and I would love information on that, too. Extremely surprisingly, at least to me, the answer to the question is consistently yes! We can indeed have "small $\theta_0$". This raises the question of what the possible values of $\theta_0(X)$ are, but I think that's a job for a later question. CLAIM. Assume $AD^++V=L(\mathcal{P}(\mathbb{R}))$. Then the set $X$ of reals which are Cohen over HOD satisfies $\theta_0(X)=\omega_1$. To prove this claim, we use the following theorem (John says it follows from http://www.math.uci.edu/~ntrang/AD+reflection.pdf, but I don't immediately see how): Theorem. Assume $AD^++V=L(\mathcal{P}(\mathbb{R}))$. Then there are definable maps $A$ and $W$ from reals such that $A(x)=\mathbb{P}_x$ is a forcing notion, $W(x)=W_x$ is a $\mathbb{P}_x$-name for a symmetric submodel of the generic extension of $HOD[x]$ by $\mathbb{P}_x$, and for every real $x$, $V=W_x^G$ for some $\mathbb{P}_x$-generic $G$. OK, let's take that for granted for a moment; so what? Well, let $f:\mathbb{R}\rightarrow ON$ be OD via $\varphi$ and ordinal parameters $\overline{\gamma}$. Then we have $$\mbox{$f(x)=\alpha$ $\quad$ iff $\quad$ $HOD[x]\models$"$\Vdash_{\mathbb{P}_x}$"$W_x\models \varphi(x, \overline{\gamma}, \alpha)."$"}$$ But then the set of possible values for $f(x)$, for $x$ Cohen over $HOD$, is countable in $HOD$! Why? Well, if $f(x)=\alpha$ for $x\in X$, then this was forced by some $p\prec x$. But there are only countably many Cohen conditions, so we're done. Now, I believe that this generalizes to the following (this is not in John's email, so perhaps there is a subtlety of Cohen-ness I'm missing): SECOND CLAIM: Assume $$\mbox{$AD^++V=L(\mathcal{P}(\mathbb{R}))+$"There are uncountably many $\mathbb{Q}$-generic reals over $HOD$"}$$ (for $\mathbb{Q}$ some forcing notion). Then the set $Y_\mathbb{Q}$ of reals which are $\mathbb{Q}$-generic over $HOD$ has $\theta_0(Y_\mathbb{Q})\le \vert\mathbb{Q}\vert^+$. Based on this, I suspect it is in fact consistent with $AD^++V=L(\mathcal{P}(\mathbb{R}))$ that - for each $\omega<\kappa<\Theta$ - there is an OD set $Z_\kappa$ such that $\theta_0(Z_\kappa)=\kappa$. But I am galactically far from proving anything like this! Heck, I don't even understand everything I've written above. :P The crucial point - even assuming everything above is correct - is establishing the lower bound. In the case of the original $X$, the lower bound was trivial: any uncountable OD set $A$ surjects onto $\omega$ in an OD way, since we can (OD-ly) find a sequence $I_n$ ($n\in\omega$) of disjoint rational intervals such that $I_n\cap A\not=\emptyset$ for all $n$. But this breaks down above $\omega$, so I'm completely lost!<|endoftext|> TITLE: Infinite families in stable homotopy groups QUESTION [9 upvotes]: I will be very grateful for any advise or reference on the following. 1- How much is known about infinite families in ${_2\pi_*^s}$, the $2$-component of the stable homotopy ring? 2- How much is known about possible geometric construction of any of these families?! I like to know about known constructions of such families by geometric methods. The geometric constructions that I am interested in are (i) factoring an element of $f\in{_2\pi_*^s}$ through some finite dimensional complexes. The $\eta_i$ elements of Mahowald family are constructed in such a way. Also representing $f$ in terms of triple or higher Toda bracket will lead to such a factorisation, not necessarily unique. ADDED I think Joel Cohen's result on representing elements of ${_2\pi_*^s}$ by (higher) Toda brackets, despite the question about the indeterminacy, means that any stable map $S^n\to S^0$ admits a factorisation through finite number of finite dimensional (stable) CW-complexes. For instance, the $\mu_i$ element in ${_2\pi_{8i+1}^s}$ coming from ${_2\pi_*}J$ with $J$ being the fibre of $\psi^3-1:BSO\to BSO$, is represented by a triple Toda bracket and by the construction of Adams, it factors through a $2$-cell complex. (ii) constructing elements using homotopy operations arising as described by Bruner. For instance, Bruner's $\tau_i$ family is constructed using $\cup_1$ operation as described by Bruner. I doubt if there is any structural result on the existence of such families; I presume whether or not if there exist finite number of such families is not known?! and if anything known would be a collection of latest results, something like what we find in Ravenel's Green book (do not know if a more updated reference exists!). ADDED In particular, I like to know of any geometric construction of infinite families detected in the Adams or Adams-Novikov spectral sequences, such as those coming from Greek letter constructions? I must say that I dno't know much about the Greek Letter elements, so these might be very well documented somewhere in the literature. I am happy even to know about any conjectural construction or those which are folklore and believed to be true?! REPLY [7 votes]: As you probably know, the existence of Greek letter elements relies on the existence of (generalized) Smith-Toda complexes -- the best introduction to those is probably still Section 1.3 of Ravenel's Green book, but Lee Nave's paper http://annals.math.princeton.edu/wp-content/uploads/annals-v171-n1-p10-p.pdf is certainly also work a look. A generalized Smith-Toda complex is a finite complex whose $BP$-homology is $BP_*/(p^{i_0}, v_1^{i_1},\dots, v_n^{i_n})$ and it is often denoted by $M(i_0,\dots, i_n)$. They arise inductively by defining a self-map of $M(i_0, \dots, i_{n-1})$ that induces multiplication by $v_n^{i_n}$ on $BP$-homology and taking its cone. In general, the existence of such a $v_n$-self map $f$ is only known for some value of $i_n$ (by the periodicity theorem by Hopkins and Smith). The determination of a minimal $i_n$ is very difficult in general. The basic procedure how these things give rise to periodic families in the stable homotopy groups of spheres is the following. Let's start with an element $x \in \pi_k M(i_0, \dots, i_{n-1})$. Then we can look at the composite $$S^k \to M(i_0,\dots, i_{n-1}) \xrightarrow{f^m} \Sigma^?M(i_0, \dots, i_{n-1}) \to S^?,$$ where the last map is "projection to the top cell". The value of this class in the Adams-Novikov spectral sequence can (usually) be explicitly computed although it can be a major issue to decide whether it is zero or not. For example, $M(1)$ (i.e. the mod $p$ Moore spectrum) admits a $v_1^1$-self map for $p>2$, producing the $\alpha$-family; $M(1,1)$ admits a $v_2^1$-self map for $p>3$, producing the $\beta$-family; $M(1,1,1)$ admits a $v_3^1$-self map for $p>5$, producing the $\gamma$-family. Showing that the $\gamma$-family is indeed nonzero in the the Adams-Novikov spectral sequence was accomplished in the Miller-Ravenel-Wilson article Periodic Phenomena in the Adams-Novikov Spectral Sequence. Note that none of this directly works for $p=2$. The mod-$2$ Moore spectrum $M(1)$ only admits a $v_1^4$-self map (see Adams's $J(X)$ IV). The resulting $M(1,4)$ admits only a $v_2^{32}$-self map. The latter result is proven in the Behrens-Hill-Hopkins-Mahowald paper https://www3.nd.edu/~mbehren1/papers/v2_32.pdf. The consequences for periodic families are discussed in Section 11 of the earlier preprint http://hopf.math.purdue.edu/Hopkins-Mahowald/eo2homotopy.pdf by Hopkins and Mahowald (although the account is rather condensed and I have never worked through it). For further background see theTMF-book and Homology of $tmf$ by Mathew. There is a also later work on $v_2$-self maps at $p=2$ by Bhattacharya, Egger and Mahowald: http://arxiv.org/abs/1406.3297 -- I know nothing though about its consequences for periodic families.<|endoftext|> TITLE: Is there a way to embed Clifford algebras into the corresponding tensor algebra? QUESTION [12 upvotes]: $\newcommand{\talg}{\mathcal{T}(V)}$$\newcommand{\clalg}{\mathcal{Cl}_q(V)}$$\newcommand{\qalg}{\mathcal{I}_q(V)}$Is there a way to embed Clifford algebras into the corresponding tensor algebra? There are simple and straightforward embeddings of the underlying vector space $V$ into its corresponding tensor algebra $\talg$ and any of its corresponding Clifford algebras $\clalg$ (where $q$ denotes the quadratic form defining the Clifford algebra). This fact is what makes both tensor analysis and geometric (Clifford) algebra compatible with ordinary vector algebra or calculus. However, even though any $\clalg$ can be formed as a quotient of $\talg$, or perhaps because of that fact, no $\clalg$ seems to be "compatible" with $\talg$ in a simple way, at least in the sense that there does not appear to exist any "simple" embedding of $\clalg$ into $\talg$, such that we could use all of the structure and geometric intuition afforded by the $\clalg$ framework while still working in the most general possible space, $\talg$. This seems like a big problem to me, because even if one isn't interested in tensors of rank $>2$, rank 2 tensors (also known as matrices) are ubiquitous, and I have the impression that the greater compatibility of vector algebra with matrix algebra is the main obstacle to more widespread implementation of geometric algebra. (I.e. because unlike vector algebra, geometric algebra doesn't "play nice" with matrix algebra or tensor algebra in general.) This seems like a major pedagogical problem, since it seems like many subjects could be much more easily understood through the prism of geometric algebra, but the (seeming) incompatibility of any $\clalg$ with $\talg$ vastly (and arguably rightly) dampens any enthusiasm to pursue such an approach. Even if there can not exist any simple embedding, what about a simple way to switch between the two systems? I am also doubtful in this regard, since the definition of exterior algebra and outer products in terms of tensor products is very unwieldy (a sum over the symmetric group). And the exterior algebra is the most degenerate type of Clifford algebra possible. Here is a similar question on Math.SE, corroborating my claim that every multivector corresponds to a tensor but not vice versa: What is the relationship of tensor and multivector. These two documents explain how to represent some multivectors as tensors for the special case that $V=\mathbb{R}^3$ and $q=I$ i.e. just the identity, so that the inner product is just the dot product: (1) (2). The best I could find so far is the following (see page 7 of this document): Given $\qalg:= \{ \sum_k A_k \otimes (v\otimes v -q(v))\otimes B_k: v \in V, A_k, B_k \in \talg \} $ We have for $A,B \in \clalg$: $AB = A \otimes B + \qalg$, where $AB$ denotes the Clifford (geometric) product and $\otimes$ is of course the tensor product. EDIT: Thinnking about Oscar Cunningham's comments below, I think we can write $$\talg = \clalg \oplus \qalg$$ (or something that's actually mathematically correct but similar "in spirit"). $\clalg$ are exactly those tensors which have a unique embedding as an element of the Clifford algebra, and $\qalg$ consists of exactly those tensors which can not be represented in the Clifford algebra, and hence are mapped to 0 by the quotient map. Thus the problem reduces to: Given any arbitrary tensor $t \in \talg$, determine its (unique since direct sum?) representation as $$t= C + \tau,$$ where $C \in \clalg$ is a member of the Clifford algebra, and $\tau \in \qalg$ is a member of the ideal $\qalg=\{ \sum_k A_k \otimes (v\otimes v -q(v))\otimes B_k: v \in V, A_k, B_k \in \talg \} $. Since we have this explicit representation for $\qalg$, I think the problem might be much easier than I originally anticipated. For example, if $t$ is a rank 4 tensor in some four dimensional vector space $V$, we could write something like $$t=\sum_{\sigma \in S_4} \left[e_{\sigma(1)}\otimes(\langle e_{\sigma(2)},e_{\sigma(3)}\rangle_q)\otimes e_{\sigma(4)}\right] + \sum_{\sigma\in S_4} \left[e_{\sigma(1)}\otimes(e_{\sigma(2)}\otimes e_{\sigma(3)} - \langle e_{\sigma(2)}, e_{\sigma(3)} \rangle_q )\otimes e_{\sigma(4)}\right],$$ where $e_1, e_2, e_3, e_4$ are basis vectors and $\langle \cdot, \cdot\rangle_q$ is the inner product formed from the quadratic form $q$ via polarization. Of course, I am going somewhat out on a limb here in assuming, i.e. I do not know how to prove that $$\sum_{\sigma \in S_4} \left[e_{\sigma(1)}\otimes(\langle e_{\sigma(2)},e_{\sigma(3)}\rangle_q)\otimes e_{\sigma(4)}\right]=C \in \clalg$$ or that $$\sum_{\sigma\in S_4} \left[e_{\sigma(1)}\otimes(e_{\sigma(2)}\otimes e_{\sigma(3)} - \langle e_{\sigma(2)}, e_{\sigma(3)} \rangle_q )\otimes e_{\sigma(4)}\right]=\tau\in \qalg.$$ The other strategy I was thinking of was to represent tensors of rank 2 or greater only via their isomorphisms with multilinear maps, since then that would come down to representing them as functions of (contravariant) vectors, a function of $k$ vectors, a $k$-linear map, for a $k$ tensor. This would work insofar as all contravariant vectors have canonical embeddings in both the Clifford and tensor algebras, but I think we still have as unresolved the problem of how to represent covariant vectors (row vectors) as well as general covariant/mixed variance tensors. Some possible problems: The 0 element of a Clifford/geometric algebra has all grades (i.e. for any $k$ it is a $k$-blade/multivector), whereas I don't think this is true in the tensor algebra (at least thinking in terms of representing tensors via k-dimensional arrays -- you can have the scalar 0, but also 0 column vectors, 0 row vectors, 0 matrices, etc.). Intuitively, one wants to think of a geometric/Clifford algebra as "symmetric algebra + exterior algebra" (commutative inner product $\frac{1}{2}(v\otimes w + w \otimes v)$ and anticommutative outer product $\frac{1}{2}(v\otimes w - w \otimes v)$), especially since (at least for vector spaces over fields of characteristic zero, which is what I personally am interested in) both the symmetric and exterior algebras can be identified with certain subclasses of tensors (see: Wikipedia - symmetric algebra vs. symmetric tensors). But how do we do this for a general Clifford algebra? Maybe $$vw = v \otimes w + \qalg \\= \frac{1}{2}(\langle v, w \rangle_q + v \wedge w)= \frac{1}{2}(v\otimes w + w \otimes v + \qalg)+\frac{1}{2}(v \otimes w - w \otimes v)?$$ It's clear that $v \wedge w$ can always be identified as a subalgebra of the Clifford algebra, and this sort of definition somewhat explains that, but then is it true that the inner product generated by polarization of $q$ must equal $\frac{1}{2}(v\otimes w + w \otimes v + \qalg)$? And even if it does, does that mean that the inner product of a Clifford algebra is commutative if and only if $q$ is a symmetric quadratic form? Since in general the product derived from a quadratic form via polarization need only be bilinear, but not necessarily symmetric (i.e. commutative) unless $q$ is symmetric too. Related: Which concepts in differential geometry cannot be represented using geometric algebra? (The answer is essentially: tensors.) How would one express the result of a tensor product (of two vectors) in the geometric algebra? Is geometric algebra isomorphic to tensor algebra? (No. Most tensors do not admit of a unique representation in geometric algebra due to the quotient structure.) What is the hierarchy of algebraic objects meant to capture geometric intuition? (Tensor algebras are essentially the most general possible, even though geometric algebras are in general less unwieldy.) REPLY [5 votes]: $\newcommand{\qalg}{\mathcal{I}_q(V)}$As K. Conrad points out, this question has actually been answered already on MathOverflow by user MTS: see these answers here and here. The essential idea is this: despite the fact that Clifford algebras have non-zero quadratic form in general, we can still "piggyback" on the embedding of the exterior algebra in the tensor algebra to represent the Clifford algebra, regardless of the choice of quadratic form. Note: The embedding of the exterior algebra into the tensor algebra can be given by defining the exterior=wedge product in terms of the tensor product. Discussion of two canonical ways of doing this are given here on MathOverflow (the top answer being again by MTS) as well as on Math.SE. Beautiful and rigorous proofs of this fact are given in the two answers (1) (2) linked to above by user MTS; in what follows I will merely try to motivate this result for novices like me. 1. First, we might expect this result intuitively by considering the decomposition of the Clifford product for vectors into an "inner" and "outer" product. The former is the inner product generated by polarization of the quadratic form $q$ and the latter is "equivalent" in some sense to the exterior/wedge product from the exterior algebra. Since the inner product is grade-reducing, while the outer product is grade-increasing, we might expect that the inner product would be "incapable of producing new basis elements", hence the only basis elements of the Clifford algebra we should expect are those created by the corresponding exterior algebra, i.e. there is some reason to expect a priori that a given Clifford algebra is "no more rich than" making the corresponding exterior algebra an inner product space, i.e. by defining a quadratic form on the exterior algebra. 2. MTS's answers make this suspicion rigorous by showing that the ideals for the Clifford algebra and the exterior algebra are sufficiently similar so that the quotients of the tensor algebra by them can be related in a straightforward manner. Let us motivate this fact by considering $\mathcal{I}_q(V)$ explicitly. Remember that $$\mathcal{I}_q(V):= \left\{ \sum_k A_k \otimes (v \otimes v - q(v)) \otimes B_k : v \in V, A_k, B_k \in \mathcal{T}(V) \right\}$$ Using the distributivity of the tensor product (e.g. here) I rewrite this as $$\qalg = \left\{ \sum_k \left[A_k \otimes v \otimes v \otimes B_k - A_k \otimes q(V) \otimes B_k \right] : v \in V, A_k, B_k \in \mathcal{T}(V)\right\} \\ \overset{?}{=} \left\{ \sum_k \left[A_k \otimes v \otimes v \otimes B_k - q(V) \cdot( A_k \otimes B_k) \right] : v \in V, A_k, B_k \in \mathcal{T}(V)\right\}$$ In any case, the motivating idea here is that the right term arising from the quadratic form, always being of strictly lower grade than the left term, should in some sense be "negligible" compared to the left term (MTS's second answer makes this idea explicit and rigorous). In other words: $$\qalg \approx \left\{ \sum_k A_k \otimes v \otimes v \otimes B_k: v \in V, A_k, B_k \in \mathcal{T}(V) \right\}$$ where the right hand side is obviously $\mathcal{I}_0(V)$, the ideal which forms the exterior algebra. 3. In the very simple case of geometric algebra over $\mathbb{R}^n$, this fact is used all the time in creating a basis for the geometric algebra. Namely, it is stated that $\mathbb{G}^n \cong \mathbb{R}^{2^n}$ (vector space isomorphism), with the basis being given by: $$\{e_1, \dots, e_n, e_{\sigma_2(1)}\wedge e_{\sigma_2(2)}, e_{\sigma_3(1)}\wedge e_{\sigma_3(2)} \wedge e_{\sigma_3(3)}, \dots, e_1 \wedge \dots \wedge e_n : \sigma_i \in A_i \}$$ where $A_i$ denotes the set of all even permutations of $i$ elements selected from $\{1, \dots, n\}$ and $\{e_1,\dots,e_n\}$ is an orthonormal basis of $\mathbb{R}^n$, orthogonal with respect to the inner product of $\mathbb{G}^n$. It should be evident that this basis is isomorphic to the basis of the exterior algebra over $\mathbb{R}^n$. Since in any Clifford algebra $vw = v \wedge w$ ($vw$ denoting the Clifford/geometric product) if and only if $\langle v, w \rangle_q=0$ i.e. if and only if $v$ and $w$ are orthogonal with respect to the inner product induced by $q$ via polarization, the possible choices of orthonormal basis will in general depend on the choice of quadratic form for the geometric algebra. Nevertheless, the one-to-one correspondence of the bases with the basis for the exterior algebra $\mathbb{R}^n$ will still remain and will be sufficient to generate a vector space isomorphism, since vector space isomorphisms only depend on the linear independence and dimensions of the bases, and not their orthogonality under an inner product. In other words, we can always use the vector space basis of the exterior algebra over $\mathbb{R}^n$ as a vector space basis for the geometric algebra $\mathbb{G}^n$ regardless of our choice of quadratic form, although this basis will not always be orthogonal. Hence by identifying the geometric algebra with the exterior algebra + inner product structure, we get a linear embedding of the geometric algebra into the tensor algebra "for free" by using the exterior algebra's embedding. (See p.4 of this document for discussion of the canonical basis of a geometric algebra generated by an orthonormal basis of $\mathbb{R}^n$.)<|endoftext|> TITLE: Negative probabilities - what are two ordinary pgfs that correspond to the gf of a half-coin? QUESTION [5 upvotes]: In Half of a Coin: Negative Probabilities, author considers pgf of a fair coin represented by random variable, $X = 1_H$: $$G_X(z) = E[z^X] = \sum_{x=0,1} z^xP(X=x) = (z^0)(1/2) + (z^1)(1/2) = \frac{z+1}{2}$$ Author defines gf of a half-coin (a coin with sides $n=0,1,2,...$ with some sides having negative probability): $$G_{X_{0.5}}(z) = \sqrt{\frac{z+1}{2}} = \frac{1}{\sqrt{2}}\sum_{k=0}^{\infty} \binom{1/2}{k} z^k$$ That is, $$P(X = k) = \frac{1}{\sqrt{2}} \binom{1/2}{k}$$ If we flip two independent half-coins, their sum is 0 or 1 with probability 1/2. So it's like flipping one fair coin. This is based on pgf of sums of independent random variables being the products of the pgfs of the random variables. Later on, there's this fundamental theorem saying that for any gf $f$, there exists two pgfs $g, h$ such that $$fg = h$$ How is that related to the half-coin? What are possible $g$ and $h$ that correspond to $$f(z) = \sqrt{\frac{z+1}{2}}$$ ? I was expecting something like two gfs for one pgf as was done with the fair coin and the two half coins (product of two gfs for the half coins is the pgf for the fair coin). REPLY [4 votes]: My previous answer, in much simplified and more explicit form: take $g(z):=\frac{1-\sqrt{1-z}}{z}$ and $$h(z):=f(z)g(z)=\frac{1-\sqrt{1-z}}{z}\,\sqrt{\frac{1+z}{2}} =\frac{\sqrt{1+z}-\sqrt{1-z^2}}{z\sqrt2}.$$ Then $g$ and $h$ are pgf's. Indeed, as before, let \begin{equation} c_j:=\frac1{j2^{2j-1}} \binom{2 j-2}{j-1}>0. \end{equation} Then \begin{equation} \sqrt{1+z}=1+\sum_{j=1}^\infty(-1)^{j-1}c_j z^j,\quad \sqrt{1-z^2}=1-\sum_{i=1}^\infty c_i z^{2i}, \end{equation} \begin{equation} g(z)=\frac{1-\sqrt{1-z}}{z}=\sum_{j=1}^\infty c_j z^{j-1}, \end{equation} \begin{equation} h(z)\sqrt2=\frac{\sqrt{1+z}-\sqrt{1-z^2}}{z} =\sum_{i=0}^\infty c_{2i+1} z^{2i}+\sum_{i=1}^\infty (c_i-c_{2i}) z^{2i-1}. \end{equation} It remains to check that $c_i\ge c_{2i}$ for $i\ge1$. Let $r_i:=c_{2i}/c_i$. Then $r_{i+1}/r_i=\frac{16i^2-1}{16i^2-4}>1$ for $i\ge1$, so that $r_i$ is increasing in $i\ge1$ to $\lim_{i\to\infty}r_i=\frac1{2\sqrt2}<1$. So, $r_i<1$ for $i\ge1$, which confirms that $c_i\ge c_{2i}$. This completes the proof. I am retaining the previous answer, because it shows some of the process by which the second answer was obtained.<|endoftext|> TITLE: On the use of Weisfeiler-Leman refinement in Babai's GI proof QUESTION [9 upvotes]: This question is for those familiar with the methods behind Babai's recent proof that graph isomorphism can be decided in quasipolynomial time. I am a newcomer to the GI problem, so I apologize if my question is overly naive. 1. My first question is on some of the background Babai assumes about the Weisfeiler-Leman canonical refinement process. On several occasions he remarks that certain regularity properties his proofs exploit come 'for free' as a result of WL refinement. See for example the proof of the basic Design Lemma (especially on page 41 in the linked paper). The intuition goes as "$k$-dimensional WL refinement creates regularity on the scale of $k$ vertices", and if you examine each of his regularity conditions, you will find that they are indeed "local" w.r.t. the dimension of refinement. This is further supported by this paper by Immerman and Lander, where they prove among other things that any two graphs with the same $k$-dimensional canonical WL refinement agree on all first-order with counting formulas on $k$ variables (where the variables can take values in the vertices of each graph). However, I haven't been able to see a nice, uniform way in which the regularity results Babai needs follow from this property of $k$-WL. Am I missing something (maybe relevant previous work?)? 2. My second question is a sort of continuation: Babai outlines a way to get the same regularity properties without WL refinement (see the footnote on page 41). I think I got his idea to formally work, as described in section 3.3. of this essay; however it requires tiny ad-hoc arguments for each kind of regularity you want to use. So my question is, if there is no uniform way to prove such regularity statements using WL (i.e. if the answer to my first question is 'no'), then both the WL and non-WL methods require us to look into the details of the current situation, so what's the point of using WL at all? Dispensing of it would make for a more self-contained, first-principles paper. I guess one reason would be that WL has become a common tool in the study of the GI problem, so the people who have worked in that field are well aware of its magic, and to a large extent it guides the intuition behind many of the proofs? REPLY [6 votes]: Not to toot my own tooting of other people's horns, but you'll find an explanation of k-ary Weisfeiler-Leman (complete with pseudocode) in section 2.5 of my Bourbaki talk on Babai's work (and Luks's, and Weisfeiler-Leman's...), now in the arxiv (in french). Note: this is what I reconstructed based on what other people said the method did. It works, but experts may find it odd (and then they should tell me).<|endoftext|> TITLE: Kunneth formula for de Rham cohomology twisted by flat vector bundle QUESTION [6 upvotes]: It is well known that for two manifolds $M$ and $N$ (let's say they are compact), the Kunneth formula says that $H(M\times N)=H(M)\otimes H(N)$, where $H$ denotes de Rham cohomology with complex coefficients. We also know that one can twist de Rham cohomology by a complex flat vector bundle $E\to M$ with a flat connection $\nabla^E$. Then the differential of the chain complex $\Omega(M; E)$ is coupled with $\nabla^E$. Denote by $H(M, E)$ the corresponding de Rham cohomology. Let $E\to M$ and $F\to N$ be complex flat vector bundles with flat connections $\nabla^E$ and $\nabla^F$. Write $p_M:M\times N\to M$ be the standard projection map and similarly for $p_N$. Then $p_M^*E\otimes p_N^*F\to M\times N$ is a complex flat vector bundle with flat connection $p_M^*\nabla^E\otimes p_N^*\nabla^F$. I wonder there should be a Kunneth formula for de Rham cohomology twisted by complex flat vector bundles, that is, $$H(M\times N, p_M^*E\otimes p_N^*F)\cong H(M, E)\otimes H(N, F).$$ Is this assertion obviously true (or obviously wrong)? If the former case is true, is there a reference for a proof of this? REPLY [7 votes]: Yes, the Künneth formula holds in this framework. Notice that the De Rham cohomology of a flat vector bundle on a manifold is simply the cohomology of the corresponding locally constant sheaf (= local system). So what you are asking for is a Künneth formula for cohomology with local coefficients; this is worked out thoroughly in this paper.<|endoftext|> TITLE: Linear algebra in terms of abstract nonsense? QUESTION [39 upvotes]: The categories of vector spaces and finite dimensional vector spaces are pretty much as nice as can be, I think. I was wondering what portions of basic linear algebra (first couple of courses) fall out by saying "big"(er) words, and also what standard facts admit a clarifying categorical phrasing. What are some interesting examples of facts about vector spaces and linear maps that admit a nice categorical formulation? Edit. I'm not looking for (completely) elementary things like definitions of universal constructions by universal properties instead of concrete ad hoc realizations, though I can't think of any "nonelementary" things. If you write an elementary property of the category of vectors spaces, e.g a property of any abelian category, please give some nice examples of where it lends its power. REPLY [2 votes]: Here is an example (more along the lines of "linear algebra ~ $K$-theory of a point"): The categorical characterization of dimension: Let $\chi: \operatorname{Obj}(\operatorname{fVct}_{\mathbb{K}})\to G$ be a function from f.d. $\mathbb{K}$-vector spaces to an abelian group $G$ satisfying $V\cong W \implies \chi(V)=\chi(W)$, If $0\to U \to V \to W \to 0$ is exact, then $\chi(U)-\chi(V)+\chi(W)=0$. Then $\chi=\chi(\mathbb{K}^1)\, \dim_\mathbb{K}$. This is from Kostrikin & Manin's Linear Algebra and Geometry (p.89); it includes other basic categorical statements as well.<|endoftext|> TITLE: Preservation of Woodinness when it overlaps the active extender QUESTION [5 upvotes]: I'm trying to show that if a premouse $\mathcal M$ is 1-small then it's also tame. Definition. $\mathcal M$ is 1-small if for every extender $E$ on the $\mathcal M$-sequence, $\mathcal J^{\mathcal M}_{\text{crit }E}\models\text{ there is no Woodin cardinal}$. Definition. $\mathcal M$ is tame if for every extender $E$ on the $\mathcal M$-sequence, $\mathcal J^{\mathcal M}_{\text{lh }E}\models\forall\delta\geq\text{crit } E:\delta\text{ is not Woodin}$. The argument as I see it is that, assuming $\mathcal M$ isn't tame with $E$ and $\delta$ witnessing it, we form the $E$-ultrapower, note that $\delta$ is still a Woodin cardinal in the ultrapower, so that the ultrapower thinks that there is a Woodin cardinal between $0$ and $i_E(\text{crit }E)$, so $\mathcal M$ thinks that there is a Woodin between $0$ and $\text{crit }E$, so $\mathcal J^{\mathcal M}_{\text{crit }E}\models\text{there is a Woodin}$, thereby showing that $\mathcal M$ isn't 1-small. So, my main question is why $\delta$ is still Woodin in the ultrapower. Couldn't we wind up with subsets of $\delta$ that can't be reflected by any $\kappa<\delta$ in the ultrapower? This argument is from Proposition 2.31 in Löwe-Steel's "An introduction to core model theory", but it seems like they assume that the model thinking that there is a Woodin between some $\alpha$ and $\beta$ is equivalent to the $\beta$-initial segment thinking that there is a Woodin above $\alpha$. Is this true? Thank you in advance! REPLY [3 votes]: The Woodiness of $\delta$ in $\mathcal{J}^{\mathcal{M}}_{lh(E)}$ is witnessed by a bunch of extenders, which are either on the $\vec{E}$ sequence of $\mathcal{M}$ or are definable from elements of $\vec{E}$. As $\mathcal{M}$ is a premouse we know that its extender sequence $\vec{E}$ satisfies the coherence condition, i.e. if $i: \mathcal{J}^{\mathcal{M}}_{lh(E)} \rightarrow Ult(\mathcal{J}^{\mathcal{M}}_{lh(E)}, E)$ denotes the ultrapower embedding for the extender $E$, and let $\kappa$ be its critical point, then $i(\vec{E} \upharpoonright \kappa) \upharpoonright \alpha = \vec{E} \upharpoonright \alpha$, which has as a consequence that also the ultrapower $Ult(\mathcal{J}^{\mathcal{M}}_{lh(E)}, E)$ has all the necessary extenders to witness the Woodiness of $\delta$ in it.<|endoftext|> TITLE: Are accessible $\infty$-categories closed under accessible localizations? QUESTION [7 upvotes]: The defining problem of homotopy theory is that often when one localizes a nice category at a reasonable class of morphisms, the result is a very bad category. Does passing to the $\infty$ world fix this (so that homotopy theory becomes no different from ($\infty$)-category theory)? A "nice" category is an accessible one, a reasonable class of morphisms is one which is accessible and accessibly embedded in the arrow category, and the localization is a category with the appropriate universal property. This all transports into the $\infty$ setting. To make this precise, one might consider the strict 2-category (actually, the link only talks about the strict 2-category of all quasicategories, we pass to a locally full sub-2-category) $\mathbf{Acc}$ of accessible $\infty$-categories and accessible $\infty$-functors (which I guess has PIE limits as in the ordinary case, including cotensors to define the arrow category), and ask whether this 2-category has co-inverters. This fails in the non-$\infty$-case -- weak equivalences of simplicial sets are accessible and accessibly embedded, but the co-inserter is the homotopy category, which is not accessible. Or maybe there's a more $\infty$-way to put this, talking about codescent diagrams or something. More generally, one could ask whether the 2-category of accessible $\infty$-categories has all co-PIE colimits. Of course, all of this may be asking too much -- after all, the 2-category of $\lambda$-accessible categories and $\lambda$-accessible functors should be a 2-category of free algebras for the free $\lambda$-filtered cocompletion 2-monad $\mathrm{Ind_\lambda}$ -- and one generally expects that free algebras won't be closed under colimit-like constructions like quotients (or localizations). I suppose that if the $\infty$-category $\mathcal{C}$ being localized is locally presentable, then presenting $\mathcal{C}$ as a combinatorial model category, we should be able to use Jeff Smith's theorem to see that the localization is again presented by a combinatorial model structure. At least, this is so long as the localizing subcategory satisfies 2-out-of-3. But it would be nice to remove this technical hypothesis. REPLY [3 votes]: Just a few points: first, if you localize at a subcategory $W\subset\mathcal{M}$ that doesn't satisfy 2-out-of-3, it doesn't really matter, in the sense that you'll really be localizing at the smallest supercategory $W\subset \overline{W}\subset\mathcal{M}$ which does satisfy 2-out-of-3 (called the saturation of $W$). After all, if we localize at $g$ and $gf$, we get $f^{-1}=(gf)^{-1}g$ (note that we really do need to localize at $g$ as well for this equation to hold). Second, as long as you're talking about presentable $(\infty,1)$-categories and localizing at small classes of morphisms, the $\infty$-version of localization is really just reflecting into the full subcategory of $\overline{W}$-local objects -- i.e. it is a left adjoint, which means that it preserves colimits (so if they existed before, they still exist). This is Proposition 5.5.4.15 of Higher Topos Theory. Also, note that $\mathrm{Hom}(-,A\times B)\cong\mathrm{Hom}(-,A)\times\mathrm{Hom}(-,B)$, which implies that the product of local objects is local, so the localization functor preserves products (this is true of infinite products as well), and the equalizer of $A\rightrightarrows B$ is $\overline{W}$-local as long as $A$ is, so we have all small 1-limits. Limits of diagrams indexed by non-discrete $(\infty,1)$-categories probably work out as well, but I don't know a proof off the top of my head. The cautionary point is that all of this discussion takes place in the context of $\mathbf{qCat_\infty}$, not $\mathbf{qCat_2}$. And also, localizations of $(\infty,1)$-categories that are not presentable are not necessarily so well-behaved.<|endoftext|> TITLE: Consistent sentences with no arithmetically definable models QUESTION [5 upvotes]: I've seen a construction of a sentence of first order logic that is consistent, but has no models with underlying set $\mathbb{N}$ and recursive functions and relations. Do there also exist consistent sentences with no model on $\mathbb{N}$ with arithmetically definable functions and relations? If no, at which levels of the arithmetical hierarchy are there consistent sentences with no model at that level? If yes, what about the same question for analytically definable models? REPLY [6 votes]: In this edit Propositions A and B are fine-tuned (thanks to input from Alex Mennen and David Marker). This note complements the answer by Joel Hamkins. The question was posed for consistent sentences and asked for the arithmetical definability of the open diagram of some model of the given sentence, but as pointed out by Joel, a much stronger result is true, namely: every consistent arithmetically definable theory has a model whose elementary diagram is arithmetically definable. In contrast, there are extensions $T$ of PA (Peano arithmetic) with the property that the open diagram of no model of $T$ is arithmetically definable; indeed something much stronger is true: Proposition A. For every $X \subseteq \omega$ there is a consistent theory $T$ in the language of arithmetic extending PA such that $X$ is Turing reducible to the open diagram of every model of $T$ whose universe is $\omega$. Proposition A can be readily established as an immediate corollary of the following classical result: Theorem B (Scott). Every countable Scott set can be realized as the standard system of some pointwise definable model of PA. Proof sketch of the proposition. It is easy to see that any $X\subseteq\omega$ is included in a countable Scott set (for example the Scott set can be chosen as all subsets of $\omega$ that are arithmetical in $X$), so by Theorem B there is a pointwise definable model $\mathcal{M}_0$ of PA whose standard system $\mathrm{SSy}(\mathcal{M}_0)$ includes $X$. Let $T$ be the first order theory of $\mathcal{M}_0$, and note that $X$ is in the standard system of every model of $T$. On the other hand, it is well-known that if $\mathcal{M}$ is a model of PA whose universe is $\omega$, then every member of $\mathrm{SSy} (\mathcal{M}$) is Turing reducible to the graph of both the addition and the multiplication operation of $\mathcal{M}$ [this fact plays a key role in the proof of Tennenbaum's theorem that states that there is no nonstandard model of Peano arithmetic whose open diagram is computable]. This implies that if $\mathcal{M}$ is additionally a model of $T$, then every member of $\mathrm{SSy}(\mathcal{M}$), and in particular $X$, is Turing reducible to the graph of the addition operation of $\mathcal{M}$.<|endoftext|> TITLE: An efficient isomorphism between finite fields QUESTION [12 upvotes]: Let $p$ be a prime number. Let $f$ and $g$ be irreducible polynomials over $\mathbb{F}_p$, both of degree $n$. We know that factor-rings $\mathbb{F}_p[x]/(f)$ and $\mathbb{F}_p[x]/(g)$ are isomorphic (both of them is isomorphic to $\mathbb{F}_{p^n}$). My question is: Is it possible to make this isomorphism efficient? I.e. can we find a matrix that makes an isomorphism between $\mathbb{F}_p[x]/(f)$ and $\mathbb{F}_p[x]/(g)$ for poly(n) operations in $\mathbb{F}_p$? REPLY [13 votes]: Google provides an answer to this question. The first deterministic polynomial time algorithm for this is due to H. W. Lenstra, Jr., in his paper "Finding isomorphisms between finite fields" (Mathematics of Computation, v. 56 (1991), 329-347). Another algorithm appears in the paper by B. Allombert, "Explicit computation of isomorphisms between finite fields" (Finite Fields and their Applications, v. 8 (2002), 332-342). However, the latter paper does not describe the running time of its algorithm, beyond saying that it is fast. An excellent survey of algorithms in number theory is H. W. Lenstra, Jr., "Algorithms in algebraic number theory" (Bulletin of the American Mathematical Society, v. 26 (1992), 211-244). Although it was written 25 years ago, still this is well worth reading.<|endoftext|> TITLE: On the history of the Artin Reciprocity Law QUESTION [24 upvotes]: At the beginning of Milne's notes on class field theory, he has a quote by Emil Artin (as recalled by Mattuck in Recountings: Conversations with MIT mathematicians): I will tell you a story about the Reciprocity Law. After my thesis, I had the idea to define L-series for non-abelian extensions. But for them to agree with the L-series for abelian extensions, a certain isomorphism had to be true. I could show it implied all the standard reciprocity laws. So I called it the General Reciprocity Law and tried to prove it but couldn't, even after many tries. Then I showed it to the other number theorists, but they all laughed at it, and I remember Hasse in particular telling me it couldn't possibly be true... I assume he is referring to the definition of the Artin map as sending an unramified prime ideal $\mathfrak p$ to the Frobenius element $(\mathfrak p, L/K) \in \operatorname{Gal}(L/K)$ for an abelian extension $L/K$. Questions: Why was it so hard for other number theorists to believe it? Artin probably talked to other people around 1924-27 (he says he took 3 years to prove the theorem in the same quote later) and Chebotarev had just recently proven his density theorem. Surely the density theorem provided strong evidence of the importance of the Frobenius element. It is also something that one can explicitly verify at least in small cases and as Artin says, one can show that it also implies the other reciprocity laws proved by then. It seems very strange to outright dismiss the idea. Were there other candidates for what the map should be or any reason to suspect that there should be no such canonical map? Surely there must have been strong reasons for such a strong rejection. REPLY [17 votes]: As GH has already remarked, the same thing happened a lot later after Taniyama and Shimura asked whether elliptic curves defined over the rationals are modular. To begin with your last question, there were no other candidates for the Artin isomorphism; reciprocity laws at the time were intimately connected to power residue symbols. Actually Euler had formulated the quadratic reciprocity law in the correct way (namely that the symbol $(\Delta/p)$ only depends on the residue class of $p$ modulo $\Delta$), but Legendre's formulation prevailed. I'm pretty sure that Hasse would not have laughed had Artin shown him right away that the general reciprocity law is equivalent to the known power reciprocity laws. Hasse did not have to wait for Chebotarev's density law to be convinced that Artin was right - by the time Chebotarev's ideas appeared it was clear that Artin must have been right. And, by the way, Chebotarev's article did a lot more than prove the importance of the Frobenius element - it provided Artin with the key idea for the proof of his reciprocity law, namely Hilbert's technique of abelian crossings. Let me also add that in 1904, Felix Bernstein conjectured a reciprocity law in 1904 that is more or less equivalent to Artin's law in the special case of unramified abelian extensions. Its technical nature shows that it was not at all easy to guess a simple law such as Artin's from the known power reciprocity laws.<|endoftext|> TITLE: Completeness of 2-category of Monoidal Categories QUESTION [8 upvotes]: Is the 2-category of monoidal categories complete? If not, can any conditions be imposed to satisfy completeness? REPLY [11 votes]: It depends on what you mean by "the 2-category of monoidal categories" and also what you mean by "complete". The 2-category of monoidal categories and strict monoidal functors is complete as a Cat-enriched category in the sense of enriched category theory, hence also complete as a bicategory. The 2-category of monoidal categories and strong monoidal functors is not complete as a Cat-enriched category, but it has all PIE-limits and hence all pseudo-limits, and therefore is complete as a bicategory (has all "bilimits", i.e. bicategorical limits). The 2-category of monoidal categories and lax (resp. colax) monoidal functors is not even complete as a bicategory, but it does have an interesting class of limits, including all colax (resp. lax) limits.<|endoftext|> TITLE: Can a Riemannian metric be analytic in non-analytically different coordinates? QUESTION [14 upvotes]: Suppose I have two coordinates on the same (subset of a) Riemannian manifold. If the metric tensor is analytic in both coordinates, is the change of variables between them necessarily analytic? In other (and perhaps clearer) words, is this conjecture true? Let $U,V\subset\mathbb R^n$ be domains. Let $g$ be a real-analytic Riemannian metric on $U$ and let $\phi\colon V\to U$ be a smooth diffeomorphism. If $\phi^*g$ is a real-analytic metric on $V$, then $\phi$ is real analytic. I have not been able to prove this or find a counterexample. The difficulty in using the formula for the metric pulled back over $\phi$, $$ \phi^*g_{ij}(x) = \partial_i\phi^k(x)g_{kl}(\phi(x))\partial_j\phi^l(x), $$ is that $\phi$ appears several times and in different ways in the right-hand side. Taking determinants yields $\det(\phi^*g(x))=\det(g(\phi(x)))\det(D\phi(x))^2$, which might help if one could change coordinates analytically on $U$ so that $\det(g(y))$ is constant and show that analyticity of $\det(D\phi)$ implies analyticity of $\phi$. (Compare this with the one-dimensional proof below.) The claim is true if $n=1$, but I do not see how to generalize this argument: A Riemannian metric on $\mathbb R$ is just a positive scalar function. After a real-analytic change of coordinates on $U$ one can assume that $g\equiv1$ (Euclidean metric). Then $\phi^*g(x)=\phi'(x)^2$. Since this is analytic and $\phi'$ never vanishes, the function $\phi$ is analytic. REPLY [21 votes]: The answer is yes — the change of variables between is necessarily analytic. The harmonic functions have to be analytic in both of your coordinates. So you can pass from one of your coordinates to some harmonic coordinates and then pass from the harmonic coordinates to the other of your coordinates. P.S. This idea is due to Shefel, and it was rediscovered independently by DeTurck and Kazdan. See correspondingly Шефель, С. З. "Конформное соответствие метрик и гладкость изометрических погружений." Сибирский математический журнал 20.2 (1979): 397—401. Dennis DeTurck, Jerry Kazdan Some regularity theorems in Riemannian geometry. Ann. Sci. École Norm. Sup. (4) 14 (1981), no. 3, 249–260.<|endoftext|> TITLE: The Teichmüller space $T_g$ of a closed riemann surface $S_g$ of genus $g \geq 2$ can't be parametrized by $6g−6$ geodesic length functions QUESTION [17 upvotes]: I asked this question almost a month ago on Math SE. After waiting three weeks for an answer or a comment, I opened a bounty on the question in hope that it might get an answer this way. The bounty expires in a day, and sadly I didn't get any answer or hint on it until now. However, someone said in a comment that this question might be better suited for MathOverflow, and this is why I decided to post it now on MO. I apologize if this question isn't appropriate for MathOverflow. For better readability, I decided not to fully copy-paste my question and only write the important parts. For the research and attempt I've done on the question and for further references, I kindly refer to the original question found here: The original question on Math SE Setting: It is well known that the Teichmüller space $T_{g,b}$ of an oriented Riemann surface $S_{g,b}$ of genus $g \geq 2$ with $b \geq 1$ geodesic boundary components (satisfying $2g + b \geq 3$) can be globally parametrized by $6g - 6 + 3b$ geodesic length functions, a number which equals its dimension (see Schmutz [1]). However, in the case of a closed surface $S_{g,0}$ (i.e. without boundary components), the minimal number of geodesic length functions needed for a parametrization of the corresponding Teichmüller space $T_{g,0}$ is $6g-5$ (first proven by Schmutz [1]), which is greater than its dimension (which is $6g-6$). The fact that $6g-5$ is minimal is not proven in Schmutz' paper. Note that Schmutz' given parametrization is to be understood as embedding of $T_{g,0}$ into $\mathbb{R}^{6g-5}$ via the length functions of $6g-5$ geodesics, and not as homeomorphism between $T_{g,0}$ and $\mathbb{R}^{6g-5}$. Question: The question is why a global parametrization of the Teichmüller space $T_{g,0}$ (of a closed surface $S_{g,0}$) by $6g-6$ geodesic length functions is not possible, i.e. why the number $6g-5$ is minimal. What I am looking for: A proof to the question or a reference where I can find a proof. Also hints that may lead to a proof would be kindly appreciated. I know from several references (see original question on Math SE) that it has something to do with Scott Wolpert's studies on the convexity of the geodesic length functions and that the result might be stated in terms of trace functions for marked Fuchsian groups (these references also state that Wolpert first found the result). Overall, I'd also be interested in the reason why the parametrization can be done by a number of geodesic length functions that equals the dimension of the Teichmüller space in the case of surfaces with boundary components, but not in the case of closed surfaces. Edit: Thanks to a comment of Mr. Agol I noticed a mistake in my initial statement of Hamenstädt's result. What is shown in [2] is that for surfaces with $n \geq 1$ punctures, $6g - 5 + 2n$ geodesic length functions provide coordinates on $T_{g,n}$, and not $6g - 6 + 2n$ as I stated. So in case of surfaces with punctures, we have the same situation as for closed surfaces. References: [1] P. Schmutz, Die Parametrisierung des Teichmüllerraumes durch geodätische Längenfunktionen, Comment. Math. Helv. 68, 1993, no. 2, 278-288 (found here in german or here in french, sadly not available in english) [2] U. Hamenstädt, Length functions and parametrizations of Teichmüller space for surfaces with cusps, Ann. Acad. Sci. Fenn. Math. Vol. 28, 2003, 75 - 88 (found here) REPLY [18 votes]: I think Scott's argument is that the lengths of $6g-6$ curves can't form coordinates for Teichmuller space. If one has $6g-6$ geodesics which parameterize, then they must be filling (they meet every simple closed curve). But the length of a filling (immersed) curve is proper in Teichmuller space, and hence the infimum of the length is achieved (in fact, at a unique point). Hence the lengths of $6g-6$ geodesics can't form local coordinates near the global (and hence local) infimum of the sums of the lengths. Now we apply invariance of domain to conclude that we cannot even have a parameterization. The reason that the geodesics must be filling is that otherwise, there is a disjoint simple closed curve. For any choice of lengths of the curves, we may twist about the disjoint simple closed geodesic, showing that the length of the curve does not uniquely determine the metric. The reason that the length of the (geodesic representative of) a filling curve is proper is that if we consider the part of Teichmuller space in which the curve has bounded length, then any given (simple closed) curve will have bounded length as well, since it has bounded intersection with the filling curve, and hence can be cut and pasted out of boundedly many pieces of the filling curve. Moreover, there is also a lower bound on the length of simple closed curves by the collar lemma (if a simple closed curve were too short, then it would have a large collar, and force a filling curve to have length greater than the width of the collar). If we take a union of two pants decompositions of the surface which fill, then it is not hard to show that the space of metrics in which the lengths of the cuffs of the pants remain bounded (above and below) lies in a compact part of Teichmuller space (one may use Fenchel-Nielsen coordinates with respect to one pants decomposition, and use the boundedness of the lengths of the other pants decomposition to bound the twist parameters).<|endoftext|> TITLE: Topological invariance of Stiefel-Whitney classes for open smooth manifolds QUESTION [7 upvotes]: It is well known that Stiefel-Whitney classes are homotopy invariant for closed smooth manifolds. But in the case of open manifolds even $w_1$ is not a homotopy invariant (take just open cylinder and open Mobius strip). Therefore, the following question naturally arises. Conjecture 1. Suppose that $f\colon M^n \to N^n$ is a purely continuous homeomorphism of two connected open manifolds homotopy equivalent to a finite polyhedron. Then $f^*w_k(N^n) = w_k(M^n)$ for all $1\le k\le n.$ The conjecture is trivially true for $w_1$ (a loop preserve or change the orientation), and for $w_n=0$. Due to the fact that the direct product with euclidian space preserve S-W classes, we can assume that the dimension $n$ is as big as we want. Also, the prominent theorem of Thom states that any mod 2 homology class of an arbitrary connected topological space can be realized by the image of the fundamental class of some connected closed smooth (possibly unorientable) manifold. Fix any $k\ge 2$. Suppose $g\colon L^k \to M^n$ is a continuous mapping ($L^k$ is connected, closed and smooth). Then $w_k(g_*[L^k])$ depends only on the behavior of the manifold $M^n$ around the image $g(L^k)\subset M^n$. Therefore, it is easy to see that the above conjecture is equivalent to the following Conjecture 2. Fix any $k\ge 2.$ Then for some $N_0\gg 1$ and any $n\ge N_0$ the following statement holds true. Suppose $\overline{M}^n$ is a smooth connected $n$-manifold with a connected boundary. Suppose also, that $N^n$ is a smooth open connected $n$-manifold, which is homeomorphic to the interior $M^n := \mathrm{int}\overline{M}^n$. Then for any purely continuous homeomorphism $f\colon M^n \to N^n$ one has $f^*w_k(N^n) = w_k(M^n)$. I can prove the Conjecture 2 for the case when the Whitehead torsion $\mathrm{Wh}(\pi_1 \partial \overline{M}^n)$ is zero. REPLY [4 votes]: The topological invariance of Stiefel-Whitney classes of smoothable manifolds is just a classic theorem of Thom, written in one of his most famous papers. It was pointed out to me by Alexander A. Gaifullin, and I am very grateful to him. Rene Thom, Espaces fibres en spheres et carres de Steenrod, Annales Scientifiques de l'E.N.S. 3e serie, tome 69 (1952), p. 109-182. Theorem III.8. Suppose $M^n$ is a smoothable connected (paracompact) $n$-manifold (it could even be of an infinite homology type). Suppose also that $\Sigma$ is a smooth structure on $M^n$. Then Stiefel-Whitney classes of a tangent bundle of $M^n$ does not depend on $\Sigma$. This theorem is written on the page 151. The proof is just as in my previous answer. Also, on the pages 151-152 of the paper Thom gives the Wu formula for S-W classes of a closed manifold (this proves the homotopy invariance of S-W classes in this case).<|endoftext|> TITLE: Is the image of an $S$-arithmetic subgroup under a surjective $k$-morphism $S$-arithmetic? QUESTION [9 upvotes]: Let $k$ be a global field and let $S$ be a non-empty set of places containing all archimedean ones. Suppose $f:G\to H$ is a surjective $k$-morphism of $k$-groups and let $\Gamma\leq G(k)$ be an $S$-arithmetic subgroup. Is $f(\Gamma)$ an $S$-arithmetic subgroup? If $k=\mathbb{Q}$ and $\Gamma=G(\mathbb{Z})$ this is Theorem 4.1 in "Algebraic Groups and Number Theory" by Platonov and Rapinchuk. I assume that this can be extended to $k$ any number field and $S$ any finite set of places. What about the positive characteristic case? Is there a similar theory? Is this result still true? What is an adequate reference for this topic when $k$ is of positive characteristic? REPLY [9 votes]: Let $k$ be a global field, and $S$ a non-empty finite set of places of $k$ containing the archimedean places. It is certain that you meant to assume $G$ and $H$ are smooth and affine (hence the surjection between them is flat); I will assume both of these in what follows. The answer is typically negative. For example, the Frobenius endomorphism of any ${\rm{SL}}_n$ (or any Chevalley group, for that matter) over $\mathbf{F}_p$ provides a counterexample. Indeed, there is an evident $S$-integral model for the situation (even one arising over $\mathbf{F}_p$ or $\mathbf{Z}$), and by basic facts about the well-posedness of $S$-arithmeticity it is equivalent to check if the coset space for the map between $S$-integral points is finite. To contradict this it is enough to do so with an $\mathscr{O}_{k,S}$-subgroup $U$ provided that $U(\mathscr{O}_{k,S})$ is the full preimage of itself under the induced map on $\mathscr{O}_{k,S}$-points. Any root group $U \simeq \mathbf{G}_a$ does the job, which is to say the infinitude is forced by that of the quotient of $\mathscr{O}_{k,S}$ modulo its subgroup of $p$th powers. But the answer is affirmative whenever the $k$-group scheme $K := \ker f$ satisfies two properties: (i) the finite etale component group of $K/K^0$ has order not divisible by ${\rm{char}}(k)$ and (ii) $K^0$ is an extension of a smooth group by one of multiplicative type. (Note that (i) and (ii) always hold when ${\rm{char}}(k)=0$.) This comes down to the finiteness of the degree-1 fppf-cohomology set for any affine flat $\mathscr{O}_{k,S}$-group $\mathscr{K}$ whose generic fiber $\mathscr{K}_k$ satisfies (i) and (ii). That finiteness rests on reduction theory for semisimple groups over global fields (in the guise of its consequences for Galois cohomology of such groups) and on the structure theory of pseudo-reductive groups (since in (ii) we impose no reductivity hypotheses on the smooth group). To carry out the reduction to that H$^1$-finiteness one has to prove that any quotient map between affine $k$-group schemes of finite type extends to a flat surjection (not just homomorphism!) between affine flat $\mathscr{O}_{k,S}$-groups of finite type. The point is that with such a flat surjective homomorphism over $\mathscr{O}_{k,S}$ in hand, the coset space for the induced map on $\mathscr{O}_{k,S}$-points (the finiteness of which is equivalent to the question posed about $S$-arithmeticity of an image) is a subset of such an H$^1$. We'll come back to the construction of such an $\mathscr{O}_{k,S}$-map at the end. As for the H$^1$-finiteness theorem over $\mathscr{O}_{k,S}$ that we have mentioned, in the number field case it is Prop. 5.1 in the paper "Actions algebriques de groupes arithmetiques" by Gille and Moret-Bailly, a preprint version of which is available at http://math.univ-lyon1.fr/homes-www/gille//publis/action.pdf. In the function field case, to prove the same H$^1$-finiteness result (under the additional restrictions on the $k$-group that we have mentioned), first one shows that it is permissible to enlarge $S$ due to the argument given near the end of the proof of that Prop. 5.1 which deduces it from the "finiteness of $S$-class numbers" of arbitrary affine group schemes of finite type. In the function field case, this "class number" finiteness for such group schemes is Theorem 1.3.1 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio 148. By enlarging $S$, the general finiteness of ${\rm{H}}^1(\mathscr{O}_{k,S}, \mathscr{K})$ (assuming $\mathscr{K}_k$ satisfies the conditions mentioned above) reduces to three cases: (i) $\mathscr{K}$ is finite etale with order not divisible by ${\rm{char}}(k)$, (ii) $\mathscr{K}$ is of multiplicative type, (iii) $\mathscr{K}_k$ is smooth and connected. By using inclusion into Weil restriction from a big enough connected finite etale cover of ${\rm{Spec}}(\mathscr{O}_{k,S})$, and using that short exact sequences of $k$-tori split up to isogeny, for (i) and (ii) we respectively reduce to the constant case and the split case. The constant case of (i) is a consequence of the finiteness of the number of finite Galois extensions of $k$ with bounded degree not divisible by ${\rm{char}}(k)$ and with ramification contained within a fixed finite set of places. The split case of (ii) is a consequence of the $S$-unit theorem for global function fields. The hard case is (iii), and by the finiteness of degree-1 Tate-Shafarevich sets of affine $k$-group schemes of finite type (which is Theorem 1.3.3(i) in the Compositio 148 paper mentioned above) one can push through the proof of Gille and Morey-Bailly's Prop. 5.1 (see Theorem 7.2.1 in that Compositio paper, which has a typo of $\mu_p$ that should be $\alpha_p$ in the line just below its statement). Now we come to the loose end which is crucial for reducing the original problem to that of finiteness of an fppf H$^1$: Theorem. Let $f:G \rightarrow H$ be a surjective flat homomorphism between affine $k$-group schemes of finite type. There exists a surjective flat homomorphism $F: \mathscr{G} \rightarrow \mathscr{H}$ between affine flat $\mathscr{O}_{k,S}$-group schemes of finite type such that $F_k = f$. Proof: Let $\mathscr{G}$ be a flat affine $\mathscr{O}_{k,S}$-group scheme with generic fiber $G$ (e.g., make this via schematic closure of $G$ in ${\rm{GL}}_{n, \mathscr{O}_{k,S}}$ relative to a choice of closed $k$-subgroup inclusion of $G$ into some ${\rm{GL}}_n$), and let $\mathscr{K}$ be the schematic closure in $\mathscr{G}$ of $\ker f$. By $\mathscr{O}_{k,S}$-flatness, $\mathscr{K}$ is normal in $\mathscr{G}$. By work of Artin, the group-sheaf quotient $\mathscr{G}/\mathscr{K}$ for the fppf topology is an algebraic space that is separated, flat, and of finite type over $\mathscr{O}_{k,S}$. It suffices to show that this quotient is represented by an affine scheme. A reference for that affineness (as well as for good counterexamples if the Dedekind base is replaced with the affine plane over a field) is given in the answer to Is the category of affine fppf groups closed under normal quotients? QED<|endoftext|> TITLE: Proof of Laurent Phenomenon for Cluster Algebras QUESTION [6 upvotes]: I went through the proof of the Laurent phenomenon for Cluster Algebras in Fomin and Zelevinsky's initial paper: Cluster Algebras I: Foundations. I am stuck at their claim that the gcd of two exchange polynomials is 1. I am working in a setup, where all coefficients are set to 1, thus the coefficient group is just the group of integers (therefore the cluster algebras are of geometric type). But why do the exchange polynomials have to be different? In my opinion their gcd could be a binomial in the cluster variables. This would not be a unit in the ring of Laurent polynomials. What am I missing? Cheers REPLY [4 votes]: Exchange polynomials can be the same. What is proven is not that the exchange polynomials are coprime in the ring of Laurent polynomials with whatever coefficients you are using, but rather that they are coprime in the ring of Laurent polynomials with "universal coefficients." See the discussion on page 11 within the proof of Theorem 3.1 for the universal coefficients set up. Essentially what is done is a polynomial ring with an indeterminate for each for each coefficient in the exchange polynomials. That is $\mathbb P$ is the free abelian group generated by a certain collection of coefficients, and then we look at $\mathcal{L}_0$ which is a Laurent polynomials ring over $\mathbb Z \mathbb P$. It is only in the ring of Laurent polynomials with these coefficients the coprime condition happens. Note with this set up exchange polynomials which were the same before become distinct. What I have said isn't precise; so, see the article for the exact definitions. The important thing is you must go the universal coefficients because it may not be true in the original coefficients for exactly the reason you point out.<|endoftext|> TITLE: Global elements in categories with no terminal object? QUESTION [16 upvotes]: Let $\mathcal{C}$ be a category. Suppose $\mathcal{C}$ contains a terminal object, which I will denote by $\boldsymbol{1}$. Then for any object $B$ in $\mathcal{C}$, a global element of $B$ is a morphism $\boldsymbol{1}\longrightarrow B$. (In many concrete categories, the terminal object is a singleton set, so this definition picks out something close to what we would normally call an "element" of the object $B$.) Given another object $A$ in $\mathcal{C}$, we can say that a morphism $f:A\longrightarrow B$ is constant if there is some global element $\epsilon:\boldsymbol{1}\longrightarrow B$ such that $f = \epsilon \circ \alpha$, where $\alpha:A\longrightarrow \boldsymbol{1}$ is the (unique) morphism from $A$ to the terminal object $\boldsymbol{1}$. So far, these definitions are totally standard and well-known. But they require the existence of a terminal object. So my questions are the following: Suppose $\mathcal{C}$ is a category which does not necessarily have a terminal object. Is there a suitable definition of "global elements" for the objects of $\mathcal{C}$? Is there a suitable definition of "constant morphisms"? Here, by a "suitable" definition, I mean a definition which reduces to the definitions I gave above in the case when $\mathcal{C}$ has a terminal object. Also a suitable definition of "constant morphism" should have the following property: composing any morphism with a constant morphism yields a constant. (Note that the definition via terminal objects has this property.) REPLY [2 votes]: A Paragraph Proof Say an arrow in a category is “constant” if it's post-composition with any two arbitrary compatible arrows is indistinguishable. In the case that there's a terminal object , an arrow $f : a ⟶ b$ is constant precisely when it factors through the unique arrow $!ₐ : a ⟶ $. This is not terribly difficult to see provided that “$a$ is non-empty”, i.e., it has a “point”, call it $¡ : ⟶ a$. Indeed, if f is constant then the factorization is witness by f ∘ ¡ since $(f ∘ ¡) ∘ !ₐ = f ∘ (¡ ∘ !ₐ) = f ∘ id = f$, where the second equality holds since f is constant. Conversely, if $f$ factorizes through the unique arrow $!ₐ : a ⟶ $ by some arrow $ε$ then, for any composable maps $x,y$ we have $f ∘ x = ε ∘ !ₐ ∘ x = ε ∘ !ₐ ∘ y = f ∘ y$ and so f is constant; where the second equality holds since $!ₐ ∘ x$ and $!ₐ ∘ y$ have the same type going to but such arrows are unique. (Challenge: is the factorisation of constants $a ⟶ b$ via the unique arrow $a ⟶ $ itself unique?) A Calculational Proof Below is the originally posted answer, but with the requested notational preliminaries. • The Booleans are denoted and have values {true, false}. • Functions, in type theory or set theory, of the form X → are called “predicates”. • Unlike any other equality in mathematics, the equality on is associative! Due to the special statues of this equality, it has a few notations: ≡ and ⇔ being the most popular, with = receiving less usage in this context. ⟪ I do not use the associativity of boolean equality anywhere in this post; an nifty example usage can be found in The associativity of equivalence and the Towers of Hanoi problem ⟫ • Rather than write, for example, $∑_{p ∈ ℕ, p \text{ prime}} f(p)$ where the range of the quantification is made a second-class citizen in the usual 2-dimensional notation, I use linear notation of Z and write $(∑ p : ℕ ❙ prime.p • f.p)$. • More generally, we can define this quantification notation for any monoid $(M, ⊕, u)$ and write $(⊕ x : X ❙ r.x • f.x)$ for the ⊕ of the terms f.x where x ∈ X satisfies predicate $r$. “Empty range”: (⊕ x : X ❙ false • f.x) = u “Finite range”: (⊕ i : ℕ ❙ 0 ≤ i ≤ n • f.i) = f.0 ⊕ f.1 ⊕ ⋯ ⊕ f.n “Abbreviation”: (⊕ x : X • f.x) = (⊕ x : X ❙ true • f.x) In particular, for the booleans we have the conjunction ∧, “and”, is associative with unit being “true” and so this notation applies to (, ∧, true). It is then conventional to define a synonym: $(∀ x : X ❙ r.x • f.x) ≔ (∧ x : X ❙ r.x • f.x)$. Likewise for the disjunctive monoid (, ∨, false) and the ∃xistential quantifier. • Finally, a proof of an equality can be rendered by a sequence of steps as is done in elementary school with the justification of each step annotated between the transitions. Informal “P = Q = R where the first equality follows because of reason₁ and the second follows because of reason₂” can be rendered in a simpler style as P =⟨ reason₁; i.e, a hint explaining why P = Q ⟩ Q =⟨ reason₂; i.e, a hint explaining why Q = R ⟩ R The conclusion $P = R$ follows by transitivity of equality. • This approach to proofs is very popular among Functional Programmers and computer science category theorists; e.g., A Gentle Introduction to Category Theory --- the calculational approach. It is also used in the popular proof-assistant Agda; where it is not built-in but is in-fact a user-defined construct! This is possible since Agda allows mixfix unicode lexemes as identifiers. I am not using any proof assistant for this post. Below is my original answer. Let's start with the second question, then follow up with the first. General definition Definition 0. for any category , we define the predicate constant : Arr → constant.f ≡ (∀ x, y • f ∘ x = f ∘ y) (Where the bullet “•” serves to sepeate the quantifer dummies from the quantifer body). Definition reduction We now show that the above definition reduces to that of OP's when terminal objects exist. Theorem 1. In any category with terminal object , constant.(f : a ⟶ b) ≡ (∃ ε : ⟶ b • f = ε ∘ !ₐ) , provided ∃ ¡ₐ : ⟶ a where !ₓ denotes the unique morphism to the terminal object: [! characterisation] ∀ x : Obj • ∀ f : Arr • f = !ₓ ≡ f : x ⟶ Proof ∷ [⇒] We have $f : a ⟶ b$ and we need to define $ε : ⟶ b$ and this can be accomplished if we only had some element $ ⟶ a$. In set theoretical terms, this is tantamount to $a$ being non-empty. ─Challenge: in a category with and , is it the case that $x ≠ ≡ ( ⟶ x) ≠ ∅$?─ Anyhow, we can use our proviso here and so define $ε ≔ f ∘ ¡ₐ$. It remains to show that we have the property OP uses in his/her definition. ε ∘ !ₐ =⟨ definition of ε ⟩ f ∘ ¡ₐ ∘ !ₐ =⟨ constant.f ⟩ f ∘ id₁ =⟨ identity ⟩ f [⇐] Assuming the existence, let $x, y : p ⟶ a$ then we prove $f ∘ x = f ∘ y$: f ∘ x =⟨ assumption ⟩ ε ∘ !ₐ ∘ x =⟨ !-characterisation since !ₐ ∘ x : p ⟶ ⟩ ε ∘ !ₚ =⟨ !-characterisation since !ₐ ∘ y : p ⟶ ⟩ ε ∘ !ₐ ∘ y =⟨ assumption ⟩ f ∘ y Constant closure Finally, we prove the property that OP is interested in, namely: ∀ f,g • constant.g ⇒ constant.(f ∘ g) Indeed, given arrows $x$ and $y$, we have f ∘ g ∘ x =⟨ constant.g ⟩ f ∘ g ∘ y Notice that this is much simpler than a proof using your more particular definition ─less complexity since we avoid existential quantifiers. It is interesting to note that constant.g ≡ (∀ f • constant.(f ∘ g)) Global elements Let us say that a global element of an object $b$ is any constant map with target $b$, let's denote such elements by a new predicate: e ⟨∈⟩ b ≡ constant.e ∧ tgt.e = b where f : a ⟶ b ≡ src.f = a ∧ tgt.f = b Let us show that this reduces to the definition of global elements as we know them when terminals exist. In particular, let's show that there's a correspondence between the two notions. [⇒] given $e ⟨∈⟩ b$, we have some global element $ε : ⟶ b$ by the reduction theorem earlier ─of course this relies on us having the same proviso! [⇐] conversely, given any global element $ε : ⟶ b$, we know it is constant with target $b$ and so $ε ⟨∈⟩ b$. Neato! Hope this helps :-)<|endoftext|> TITLE: explicitly embedding a simplicial $d$-complex into $\mathbb{R}^{2d+1}$, or algorithms for doing so QUESTION [9 upvotes]: A classical result in topology for which I can't find a reference for is that a simplicial complex $K$ of dimension $d$ with $n$ vertices can be linearly embedded into $\mathbb{R}^{k}$ when $k=2d+1$. Does anyone know if an example of such a linear embedding can be given explicitly (or determined efficiently)? If not, what is the smallest $k\leq n$ for which a general explicit construction is known (keeping $n$ and $d$ fixed)? REPLY [22 votes]: Consider the curve $C$ given by $(t,t^2, \ldots, t^{2d+1})$ in ${\mathbb R}^{2d+1}$. Any collection of distinct $2d+2$ points on this curve is in general position. In particular, the two $d$-dimensional affine subspaces of ${\mathbb R}^{2d+1}$ determined by any two distinct collections of $d+1$ points on this curve are disjoint. Given a simplicial $d$-complex $K$, map its vertices to distinct points on the curve $C$, and extend linearly to each simplex. By the general position property of the curve $C$, this is an embedding of $K$ into ${\mathbb R}^{2d+1}$.<|endoftext|> TITLE: An elliptic curve for Ramanujan-type cubic identities? QUESTION [8 upvotes]: Given the roots $x_i$ of the depressed cubic, $$x^3+px+q=0$$ with rational coefficients. It can be shown that, in general, one can find rational $u,v$ such that, $$(u-x_1)^{1/3}+ (u-x_2)^{1/3}+ (u-x_3)^{1/3} = {v}^{1/3}\tag1$$ Solution 1: $$u = \frac{-p^6q+27q^5}{p(p^6+9p^3q^2+27q^4)}$$ $$v = \frac{27p^2q^3}{p^6+9p^3q^2+27q^4}$$ Solution 2: $$u=\frac{p^{12} - 135 p^6 q^4 - 729 p^3 q^6 - 729 q^8}{9 p q (p^3 + 6 q^2) (p^6 + 9 p^3 q^2 + 27 q^4)}$$ $$v =\frac{3 (p^8 + 12 p^5 q^2 + 36 p^2 q^4)}{q (p^6 + 9 p^3 q^2 + 27 q^4)}$$ and so on, with the $u,v$ becoming increasingly long. (One $u$ was a rational $30$-deg poly in $p$.) Note: The $u$ satisfy the rational Diophantine equation, $$u^3+pu+q=w^3\tag2$$ Eq. $(2)$ can be transformed to an elliptic curve. However, not all of its rational $u$ will yield rational $v$. Question: How do we show that the two $u,v$ above, in fact, are just the first pairs of infinitely many rational solutions? REPLY [6 votes]: The following statement can be viewed as an addendum to Zieve's answer. Proposition. Let $p$ and $q$ be rational numbers such that $$ p \ne 0,\quad q \ne 0 , \quad p^3+6q^2 \ne 0,\quad p^3+9q^2 \ne 0 . $$ Then the curve $E : y^2=x^3+27p^3+\frac{729}4q^2 $ has infinitely many rational points. Proof. Let us consider two cases. Case 1: $27p^3+\frac{729}4q^2 = 0$; then one can obtain infinitely many rational points on $E : y^2=x^3 $ via a rational parametrisation. Case 2: $27p^3+\frac{729}4q^2 \ne 0$; then $E$ is a nonsingular cubic with a point $P:= (-3p : \frac{27}{2} q : 1) $, that is, an elliptic curve (we use homogeneous coordinates here). It is clear that $P\ne O$, where $O:=(0:1:0)$ is the point at infinity. Futhermore, the order of $P$ can be $2$, $3$ or $6$ (see [1, p.134, Theorem 5.3] ). Using standart computer software one can obtain $$ 2P=( pq(p^3 + 6q^2) : -(p^6 + 9p^3q^2 + \frac{27}2q^4) : q^3 ),$$ $$ 3P=( -3p(p^3 + 9q^2) (p^9 - 81p^3q^4 - 243q^6) : \frac{81}2 q (p^3 + 6q^2) (p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6) : p^3(p^3+9q^2)^3 ),$$ $$ 6P=( \frac19 qp(p^3 + 6q^2)(p^3 + 9q^2)(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)(p^9-...)(p^{27}+...) : -\frac1{27}(p^6+...)(p^{12}+...)(p^{36}+...) : q^3p^3(p^3 + 6q^2)^3(p^3 + 9q^2)^3(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)^3 ).$$ Let us prove that in fact the point $P$ is of infinite order in the group $E(\mathbb Q)$. Suppose $2P=O$; this means that $$ ( pq(p^3 + 6q^2) : -(p^6 + 9p^3q^2 + \frac{27}2q^4) : q^3 ) \sim (0:1:0) $$ or equivalently, $$ \left\{ \begin{aligned} &q=0 \\ &p^6 + 9p^3q^2 + \frac{27}2q^4 \in \mathbb Q^{\times}\\ \end{aligned} \right. $$ But $q \ne 0$ by hypothesis, hence the system above has no solutions. Suppose $3P=O$; reasoning as above, we get $$ \left\{ \begin{aligned} &p(p^3+9q^2)=0 \\ &q (p^3 + 6q^2) (p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6) \in \mathbb Q^{\times}\\ \end{aligned} \right. $$ The constraints $p \ne 0$ and $p^3+9q^2 \ne 0$ ensure that this system has no solutions. Suppose $6P=O$; similary we have $$ \left\{ \begin{aligned} &qp(p^3 + 6q^2)(p^3 + 9q^2)(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)=0 \\ &(p^6+...)(p^{12}+...)(p^{36}+...) \in \mathbb Q^{\times} \\ \end{aligned} \right. $$ To complete the proof of the proposition it suffices to show that the equation $$ p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6 =0 $$ has no solutions over $\mathbb Q$. Assume that there exists such a rational solution $(p,q)$. Then $(\tilde p,\tilde q):=(p^3+3q^2 , -3q^2)$ is a rational solution of $\tilde p^3=2 \tilde q^3$, which is a contradiction. Aside: We omitted the coordinates of $6P$ for reasons of space. The following Magma code can be useful. F := FunctionField(Rationals(),2); E:=EllipticCurve([0, 27*p^3+729/4*q^2]); P:=E![-3*p,27/2*q,1]; "the coordinates of 6P:"; 6*P; a:=(6*P)[1]; b:=(6*P)[2]; "the numerator of the first coordinate of 6P:"; Factorisation(Numerator(a)); "the denominator of the first coordinate of 6P:"; Factorisation(Denominator(a)); "the numerator of the second coordinate of 6P:"; Factorisation(Numerator(b)); "the denominator of the second coordinate of 6P:"; Factorisation(Denominator(b)); References: [1] Knapp, A. W. (1992). Elliptic curves (Vol. 40). Princeton University Press.<|endoftext|> TITLE: Finite-index iff positive density? QUESTION [9 upvotes]: Let $G$ be a finitely generated group and $S$ a symmetric generating set. Define density (lower density, say) with respect to the sequence of balls $S^n$. Is it true that a subgroup of $G$ has positive density iff it has finite index? Low-hanging fruit: This is certainly true if $G$ has subexponential growth, as then $(S^n)$ is a Folner sequence. In general, finite index implies positive lower density: Suppose $G = \bigcup_{i=1}^m g_i H$ where $m = [G:H]$. Then for every $n$ there is some $i$ such that $|g_iH\cap S^n| \geq |S^n|/m$. But $|g_i H\cap S^n| = |H \cap g_i^{-1} S^n| \leq |H \cap S^{n+r}|$, where $r$ is chosen so that $g_1,\dots,g_m \in S^r$. Thus $|H\cap S^{n+r}| \geq |S^n|/m \geq |S^{n+r}|/(m|S^r|)$, so $H$ has lower density at least $1/(m|S^r|)$. There are a couple of results in the special case in which $G$ is a free group and $S$ is the standard set of generators. Woess [1] proves this when $H$ is normal in $F_r$, while [2, Theorem 2.5] extends this to non-normal $H$ (at the level of "private communication"). Judging from the speciality of these results one would guess that the answer is no in general, i.e., that at least in some groups with respect to some generating sets there are infinite-index subgroups of positive density. Is that true? Lamplighter groups $\mathbf{Z}/2\mathbf{Z} \wr \mathbf{Z}^k$ are an interesting source of examples, but not quite counterexamples as far as I can tell. [1] Wolfgang Woess, MR 731608 Cogrowth of groups and simple random walks, Arch. Math. (Basel) 41 (1983), no. 4, 363--370. [2] Alexandre V. Borovik, Alexei G. Myasnikov, and Vladimir N. Remeslennikov, MR 2028100 Multiplicative measures on free groups, Internat. J. Algebra Comput. 13 (2003), no. 6, 705--731. REPLY [3 votes]: It's not true. Let $\def\Z{\mathbf{Z}} G = F_2 \times \Z$, let $H = F_2$, and let $S$ be the generating set consisting of the identity, $x,x^{-1},y,y^{-1}\in F_2$ and $1,-1\in\Z$. Then for all $n\geq 0$ and $k\in\Z$ we have $$ |S^n\cap(F_2 + k)| = \begin{cases} 2\cdot 3^{n-|k|} - 1 & \text{if}~ n\geq |k|, \\0 & \text{if}~n < |k|.\end{cases}$$ The reason is that the standard ball in $F_2$ of radius $n$ has size $2\cdot 3^n - 1$ (as there are $4\cdot 3^{m-1}$ words of length $m$ for each $m\geq 1$, and one of length $0$), but it first takes at least $|k|$ steps to get to $F_2 + k$ in the first place. Thus $|S^n| = 4\cdot 3^n - 2n - 3$ and $|S^n\cap F_2| = 2 \cdot 3^n - 1$, so $F_2$ has lower density $1/2$, but infinite index. Moreover if you look at $G = F_r \times \Z$ and do the same calculation you find that $F_r$ has lower density $1-1/r$, so it is possible to have lower density arbitrarily close to $1$ but still have infinite index. On the other hand upper density equal to $1$ forces $H=G$: see Generic set that is a proper subgroup.<|endoftext|> TITLE: Geometric invariant theory and normalizers of stabilizers QUESTION [8 upvotes]: For simplicity, work over an algebraically closed field of characteristic $0$. Let $$\begin{aligned} X &= \text{a smooth projective variety,} \\ G &= \text{a reductive group acting linearly on $X$,} \\ H &= \text{a finite subgroup of $G$,}\\ N(H) &= \text{the normalizer of $H$ in $G$.}\\ \end{aligned}$$ For $x \in X$, let $$\begin{aligned} \text{Stab}(x) &= \{ g \in G : g \cdot x = x \},\\ X_H &= \{ x \in X : \text{Stab$(x)$ contains $H$ }\} \end{aligned}$$ Then $G$ doesn't necessarily act on $X_H$, but $N(H)$ does, so we can look at the $N(H)$-stable and semi-stable points of $X_H$ in the sense of GIT. Questions (1) Are the $N(H)$-semi-stable points of $X_H$ automatically $G$-semi-stable points of $X$? Ditto for stable points. (This seems unlikely, but I don't know a counterexample.) (2) If not, is there some simple criterion for determining the points $x\in X_H$ such that $x$ is $G$-stable when viewed as a point of $X$? If it helps, the specific situation I have is: $$\begin{aligned} X &= \mathbb P^N \\ G &= \text{PGL}_k~\text{or SL}_k \\ N(H) &= \text{group of diagonal matrices in $G$.}\\ \end{aligned}$$ In particular, $N(H)$ is a maximal torus. REPLY [5 votes]: The answer to (1) is affirmative for semistable points. This is basically a result of Luna in his paper "Adhérences d'orbite et invariants. Invent. Math. 29 (1975), 231–238". More precisely: We only assume that $H$ is reductive (not necessarily finite). Let $\tilde X$ be the affine cone over $X$. For any $x\in X^H$ ($=X_H$ in the Op's notation) let $\tilde x\in\tilde X$ be a lift. Then, by definition, $x$ is semistable for $G$ (or $N$) if the orbit $G\tilde x$ (or $N\tilde x$) does not contain the vertex $0$ in its closure. This already shows that $\tilde x$ is $N$-semistable if it is $G$-semistable. Conversely assume that $x$ is $N$-semistable. By replacing $\tilde x$ by a point in the (unique) closed $N$-orbit of $\overline{N\tilde x}$ we may assume that $N\tilde x$ closed and $\ne\{0\}$. The group $H$ acts on the line $k\tilde x$ by a character $\chi$. If $\chi$ is of infinite order then $x$ is not $N$-semistable. Thus, $S:=\chi(H)\subseteq\mathbf G_m$ is a finite group. Put $\tilde G:=G\times S$ and $\tilde H:=\Delta(H)$ where $\Delta(h):=(h,\chi(h)^{-1})$. The point is that $\tilde G$ acts on $\tilde X$ and $\tilde x$ is an $\tilde H$-fixed point. Observe that the normalizer $\tilde N$ of $\tilde H$ in $\tilde G$ is of finite index in $N\times S$. In particular, $\tilde N\tilde x$ is closed and $\ne\{0\}$. Now we invoke Luna's result, Cor. 1 in op.cit., which says that "$\tilde N\tilde x$ closed" implies "$\tilde G\tilde x$ closed". In particular $0\not\in \overline{G\tilde x}$, i.e. $\tilde x$ is $G$-semistable. qed. Preservation of stability cannot be expected, in general. For example assume assume $H$ is contained in an infinite reductive subgroup $F$ of $G$ such that $N_F(H)$ is finite. Then $x=eF\in X=G/F$ has a finite $N$-stabilizer but the $G$-stablizer is $F$. On the positive side, Luna proves something different which might be useful: the morphism $X^H/\!/N\to X/\!/G$ is always finite.<|endoftext|> TITLE: Is the $L^1$-space dual to a Banach space QUESTION [8 upvotes]: Let $(\Omega,\mu)$ be a measure space. It is well known that for $1 TITLE: How many edges can be added to two circles before the graph becomes Hamiltonian? QUESTION [11 upvotes]: Start with two $n$-circles $(v_1\cdots v_n)$ and $(w_1\cdots w_n)$ of vertice sets $V$ and $W$, where $n\ge 5$. Add a number of vertex-disjoint edges between $V$ and $W$ (thus no chords) in a way that the resulting graph is still non-Hamiltonian. What is the maximum number $f(n)$ of such edges (let us call them links) that can be added? Equivalently: How many edges can a non-Hamiltonian graph on $2n$ vertices with two induced $n$-cycles and maximal degree 3 have? Obviously such a graph cannot contain a $C_4$. Thus, if in a construction there is an edge $(v_i,w_j)$, the four edges $(v_{i\pm1},w_{j\pm1})$ are excluded. It is clear that $f(n)$ grows linearly with $n$: a trivial lower bound is $f(n)\ge [\frac n2]$, realized by adding the links $(v_{2i},w_{2i})$ for $i=1,\dots,[\frac n2]$. What can be said about $\lim\limits_{n\to\infty}\dfrac {f(n)}n$? Or at least about $\liminf\limits_{n\to\infty}\dfrac {f(n)}n$ and $\limsup\limits_{n\to\infty}\dfrac {f(n)}n$? The Petersen graph shows that $f(5)=5$, and it is not hard to see that $f(6)<6$, thus $f(6)=5$ by the left displayed graph where the circles are $abcde\!f$ and $123456$. Further $f(8)\ge6$ by the right graph. To see that it is not Hamiltonian, note that the two violet edges cannot belong to a H-circle. I have also found $f(9)\ge 7$ and generally $f(2k-5)\ge k$ for $k\ge7$ by "splitting" two vertices of the Petersen graph, but as $k$ grows, this is no big improvement on the trivial lower bound. Yet, there should be much better lower bounds. As for upper bounds, I suppose they are notoriously hard to find. Is it possible at least to prove that $f(n) TITLE: why is paraproduct or paradifferential calculus important in PDE theory? QUESTION [5 upvotes]: In the article https://www.baidu.com/link?url=W1BjGmDoZM8QkrV_Qd_26vzNhCJGWyfH79q5cn7q0QQxomVLtH7Fw_mApElkfCZUWiDcYjNhoLhMrGFEXtf4O_&wd=&eqid=a93906890002f93700000003577cbb98, it says that "...paraproducts have played a central role in analysis and PDEs..." In what sense and how does the development of the paradifferential calculus influence the modern analysis of PDEs? And how does the development of analysis of PDEs influence the paradifferential calculus theory? I know one example, Danchin's work on the existence of smooth solutions for Navier-Stokes equations in critical spaces (Danchin R. Global existence in critical spaces for compressible Navier-Stokes equations[J]. Inventiones Mathematicae, 2000, 141(3): 579-614.). But I am not familiar with the other examples. REPLY [6 votes]: How does the development of paradifferential calculus influence the modern analysis of PDEs? You can get a quick glance by looking at Bony's 1981 paper with its 476 citations on MathSciNet. (Also, the few sentences after those which you quoted in the article you linked to gave already several broad applications that are extremely useful in PDEs.) One item not on that list, however, is Hormander's 1990 paper on Nash-Moser Implicit Function Theorem, from which we can quote: The Nash-Moser method contains constructions which very much resemble the dyadic decompositions which are central in the paradifferential calculus of Bony. In fact, we show that the Nash-Moser technique can often be replaced by elementary nonlinear functional analysis combined with the paradifferential calculus of Bony. In terms of paraproducts, I think Tao's lecture notes give a fairly accurate account of how many PDE people actually think about paraproducts: as a particular way of exposing structures arising from interactions of different frequencies in a product; these structures can then be coupled with knowledge of the linearized evolution to get sharper estimates.<|endoftext|> TITLE: Does the Lebesgue measure induce a finitely additive measure on the Boolean algebra of regular open subsets of (0,1)? QUESTION [6 upvotes]: Let $(0,1)$ the unit interval. An open subset $\mathcal{R}\subseteq(0,1)$ is regular if it is the interior of its own closure. The intersection of two regular open sets is regular. Unfortunately, the union of two regular open sets is generally not regular. (For example, if $0 TITLE: Does an expander remain an expander after removing few vertices and edges? QUESTION [7 upvotes]: Consider a sequence of expander graphs ($G_n$); say $G_n$ has $n$ vertices. Remove $o(n)$ vertices (and the edges emanating from these vertices) and cut $o(n)$ edges. Call $G'_n$ the largest connected component of the resulting graph. Are the ($G'_n$) still expanders ? REPLY [4 votes]: No. Assume that $G_n$ has bounded degree (this is probably an assumption of yours). By removing $0$ vertices and $O(\log n)$ edges, you can make sure that $G'_n$ has $\geq n/2$ vertives and contains a "segment of length $\geq c\log n$" (that is a sequence $x_1,\dots,x_{k}$ of vertices in $G'_n$ such that $k \geq c \log n$ and for all $i TITLE: Most papers ever "recalled" due to a flawed result? QUESTION [15 upvotes]: Prompted by this bit of news, http://www.wired.co.uk/article/fmri-bug-brain-scans-results where a bug in MRI software has the potential to nullify up to 40,000 published papers. Has anything analogous happened in mathematics? Obviously not on this scale, but was that ever a bug which, once discovered, caused a large number of follow-up results to become falsified as well? REPLY [3 votes]: One I have heard of is the Italian school of algebraic geometry, 1885 to 1935. REPLY [2 votes]: I would risk to state that in mathematics such things (on large scale) do not happen:-) There are indeed some published and "accepted" false proofs, and false statements. The link given in the comment of Wojowu gives plenty of examples. However, I am sure that the list of results which depend on these incorrect results and thus are false will be very small. The reason is that most mathematicians (at least most of those who prove significant results:-) usually take care to check what their proofs rely on. This is in the nature of mathematics: if you use a fact that you do not really understand, you do not understand what you are doing yourself. I would state this as a bit ambitious principle: "You have to know the proofs of all results that you use". Of course this is not always so in practice, but this is an ideal, a goal. The result is that more some theorem is used, more times it is checked by various people. And the theorems which are really used much are checked so well, that one can be reasonably sure that they are correct. I mean that none of the examples listed in the answers to Wojowu's question produced a chain of false results. By the way, this is one reason why "computer-assisted proofs" are considered less satisfactory by many mathematicians than real "human" proofs.<|endoftext|> TITLE: How are eigenvalues and eigenvectors affected by adding the all-ones matrix? QUESTION [17 upvotes]: Given an $n \times n$ matrix $A$ and the $n\times n$ all-ones matrix $J = (1)_{ij}$, I'm interested in the relation between the eigenvalues and eigenvectors of the matrices $A$ and $A+J$, or more generally $A_t := A + tJ$. Is there a nice description of the eigenvalues or eigenvectors of $A_t$ in terms of those of $A$? If not, what about for $t$ small? It would be great to have an answer for general coefficient fields, but I would also be interested in the case with $A \in M_n(\mathbb{C})$ or $A \in M_n(\mathbb{R})$ with all nonnegative or positive entries, if they have nicer answers. Thank you. REPLY [12 votes]: A cute fact that is trivial to prove is this: define the characteristic polynomial of a matrix $M$ by $\phi_M(x) = |xI-M|$. Then for any $A$ and any $s$, $$\phi_{A+sJ}(x) = (1-s)\phi_A(x)+s\phi_{A+J}(x).$$ The proof only needs the fact that a determinant doesn't change when a multiple of one row is added to a different row, so I don't see why it wouldn't be true for all fields.<|endoftext|> TITLE: Fundamental group of the space of maps into a classifying space QUESTION [7 upvotes]: Let $P : E \to X$ be a principal $G$-bundle, where $G$ is a connected topological group. $P$ is classified by a map $f: X \to BG$. The group of gauge transformations $\mathcal{G}$ of $P$ is defined to be the group of $G$-equivariant homeomorphisms $f : E \to E$ over $X$. With the usual notations, in homotopy we have: $B\mathcal{G} \simeq Map_{f}(X,BG)$ I am wondering, what is known about the homotopy groups $\pi_{k}(B\mathcal{G})$ when $X$ has dimension bigger than four and $G=SU(n)$? In particular, it is known if $B\mathcal{G}$ is simply connected? Thanks. REPLY [7 votes]: Very little is know about these spaces. Most of the literature focuses on studying the homotopy of the gauge groups of principal $G$-bundles with $G$ a simply connected, compact Lie group and $X$ a $4$-sphere or other simply connected $4$-manifold. And these alone are very delicate problems. Very little is known about these gauge groups when $X$ is of larger dimension. For an idea of the complexity of the problem see "Homotopy type of gauge groups of quaternionic line bundles over spheres" by Claudio and Spreafico. They look at $SU(2)$-gauge groups over spheres of various dimensions - and even here it is a case-by-case problem! You would do well to keep track of the work of Stephen Theriault and his students, whom I know are interested in working with more intricate base spaces $X$. I know there is work being done with $X$ a 2-connected $7$-manifold. There is one solution, whose application and finer details I will leave you to decide. For the classical groups $G=SO(n),SU(n),Sp(n),Spin(n)$ and so on you have the infinite dimensional groups $SO(\infty),SU(\infty)$, and so on. Let us assume for simplicity that $G=SU=SU(\infty)$. The same reasoning applies to all such groups. Then the gauge groups $\mathcal{G}(X,SU)$ of these infinite dimensional groups are all trivial. This follows since one may use the $H$-structure of $BSU$ to constuct a section to the evaluation fibration $ev:B\mathcal{G}\simeq Map_f(X,BSU)\rightarrow BSU$. Similarly we may use the induced $H$-structure to show that all the components of $Map(X,BSU)$ have the same homotopy type. Thus $B\mathcal{G}\simeq BSU\times Map_*^0(X,BSU)$, where $Map_*^0(X,BSU)$ is the component of the constant map in the space of pointed maps $X\rightarrow BSU$. Thus $\pi_kB\mathcal{G}=\pi_kBSU\oplus K^k(X)$. Now for each $n$ the map classifying the inclusion $Bj_n:BSU(n)\rightarrow$ is $(2n+1)$-connected. One sees that $\pi_kB\mathcal{G}(X,SU(n))$ stabilizes as $n$ grows (analogous to Bott periodicity). Using the above construction then, for a given $n$ one readily obtains information about $\pi_kB\mathcal{G}(X,SU(n))$ in a range $0 TITLE: Inequality for hook numbers in Young diagrams QUESTION [27 upvotes]: Consider a Young diagram $\lambda = (\lambda_1,\ldots,\lambda_\ell)$. For a square $(i,j) \in \lambda$, define hook numbers $h_{ij} = \lambda_i + \lambda_j' -i - j +1$ and complementary hook numbers $q_{ij} = i + j -1$. Let $$H(\lambda) = \prod_{(i,j) \in \lambda} h_{ij} \,, \qquad Q(\lambda) = \prod_{(i,j) \in \lambda} q_{ij}\,. $$ Question: Is there an elementary proof of the following inequality: $$H(\lambda) \le Q(\lambda), $$ where the inequality becomes the equality only for rectangular shapes. For example, when $\lambda = (3,2,1)$ we have $$H(\lambda)=5\cdot 3 \cdot 3 \cdot 1\cdot 1 \cdot 1 = 45, \qquad Q(\lambda)=1\cdot 2 \cdot 2 \cdot 3\cdot 3 \cdot 3 = 108. $$ Let me mention that $$\sum_{(i,j) \in \lambda} h_{ij} = \sum_{(i,j) \in \lambda} q_{ij}, $$ so somehow this says that $q_{ij}$ are more evenly distributed than $h_{ij}$. Note: this inequality is a corollary of the results in our recent paper. The proof of the main result is algebraic and quite involved. P.S. Originally posted on MSE since I thought this might be an easy exercise. Now I don't. UPDATE (July 8, 2015): Petrov's elegant proof gives a stronger result. In particular, it proves what I suggested above, that the variance of complementary hooks $(q_{ij})$ is smaller than that of the usual hooks $(h_{ij})$. To see this, take $\varphi = x^2$ and use $Var(X) = E[X^2] - E[X]^2$. Note also, as explained in the comments, the proof shows the hook numbers majorize the complementary hooks, when both are ordered from largest to smallest. For the example above: $5 \ge 3,$ $5+3\ge 3+3$, etc. This is quite remarkable and perhaps even counterintuitive. SECOND UPDATE: We just wrote a paper on the subject with a different proof. REPLY [4 votes]: I was recently thinking about this question again and I came up with another way to explain the result $$\mathcal H_{\lambda}=\left\{ h(i,j)\right\}_{(i,j)\in \lambda}\succcurlyeq \left\{ h^{*}(i,j)\right\}_{(i,j)\in \lambda}=\mathcal H^{*}_{\lambda} \tag{1}$$ in a way that (i) combinatorially describes the sequence of Robin Hood moves to get from $\mathcal H_{\lambda}$ to $\mathcal H^{*}_{\lambda}$ and (ii) gives bijective proofs of the various inequalities that follow. In particular there is a bijective explanation for the original inequality: $$\prod_{(i,j)\in \lambda} h(i,j)\le \prod_{(i,j)\in \lambda} h^{*}(i,j) \tag{2}$$ Majorization and RH moves: Suppose that $\mu, \nu$ are two integer partitions of $n$. We say that $\mu\succcurlyeq \nu$ if $$\sum_{i=1}^{r}\mu_i\geq \sum_{i=1}^r \nu_i $$ for all $r\geq 1$ (if the partitions do not have the same number of parts, we can add extra zero entries at the end). A Robin Hood (RH) move on $\mu$ is to pick to parts $\mu_i> \mu_j$ and replace them with $\mu_i -1, \mu_j+1$ and reorder the terms. This can be pictured on the associated Young diagram as the move which removes a single square at the end of a row and attaches it at the end of some row of shorter length while preserving the property of being a partition. These and their inverse also more frequently known as lowering/raising operators, I just like the metaphor of giving to the less fortunate parts of the partition. A fundamental fact is that RH moves describe the covering relation in the majorization poset of partitions. In other words, $\mu\succcurlyeq \nu$ is equivalent to being able to obtain $\nu$ from $\mu$ by successively applying RH moves. When working with majorization in combinatorial settings this description is often more desirable than the analytic description (such as when working with symmetric functions, like in Macdonald's book, for example). Partitions as directed graphs: Given a partition $\lambda=(\lambda_1,\dots,\lambda_k)$, its Young diagram can also be viewed as a directed graph $G(\lambda)$ by taking the vertices to be the lattice points $(i,j)$ with $1\le i\le \lambda_j$ and $1\le j\le k$, and edges to be $(i_1,j)\to (i_2,j)$ when $i_2\geq i_1$ and $(i,j_1)\to (i,j_2)$ when $j_2\geq j_1$. Notice that equality is allowed, so there is a self loop at every vertex. The graph $G^{*}(\lambda)$ is defined the same as $G(\lambda)$ but with all the edges reversed. The set of outdegrees of $G(\lambda)$ is precisely $\mathcal H(\lambda)$ and similarly the outdegrees of $G^{*}(\lambda)$ for the set $\mathcal H^{*}(\lambda)$. We will think of the multisets $\mathcal H(\lambda),\mathcal H^{*}(\lambda)$ as partitions of $|\lambda|+N$ where $$N= \sum_{i=1}^{k}\binom{\lambda_i}{2}+\sum_{i=1}^{r}\binom{\lambda'_i}{2}.$$ Here $\lambda'$ is the conjugate partition of $\lambda$. We can put a total order on the edges of $G(\lambda)$ that aren't loops by making all horizontal edges come before vertical edges and the edge $(i_1,j_1)\to (i_2,j_1)$ comes before $(i_3,j_2)\to (i_4,j_2)$ if $j_1\deg(v)$ in $G(t)$, but this follows by checking that for every edge $v\to w$ we have and edge $u\to w$ together with the fact that there is no edge $v\to u$. Corollary (1): We have $\mathcal H_{\lambda}\succcurlyeq \mathcal H^{*}_{\lambda}$. Moreover $H(N_0)$ for $N_0=\sum_{1}^k\binom{\lambda_i}{2}$ recovers the multiset of semi-hooks from the Pak-Petrov-Sokolov paper, and we have $$H(0)\succcurlyeq H(N_0)\succcurlyeq H(N).$$ Bijective proof of (2): The quantity $\prod_{(i,j)\in \lambda} h(i,j)$ counts the number of subgraphs of $G(\lambda)$ of outdegree 1 at every vertex.The inequality follows if we can describe an injection from the set of such subgraphs to the subgraphs of outdegree 1 of $G^{*}(\lambda)$. Start with a subgraph of outdegree 1 of $G(1)$ and start reversing the edges one by one according to the order defined previously producing an outdegree 1 subgraph of $G(t)$ for every $t$. When we reverse the edge $u\to v$ we also replace the outgoing edge $v\to w$ by $u\to z$ where $z$ is such that $v+z=u+w$ as vectors (in other words, $z,w,v,u$ are the vertices of a (possibly degenerate) rectangle). This ensures that the subgraph remains of outdegree 1. Notice that this is an injection because it can be performed backwards in a unique way whenever it is possible. When the partition is not a rectangle and we try to run the process backwards, the solution to $v+z=u+w$ may take us to a square outside of $\lambda$ so we conclude that our map will be a strict injection in that case. Note: The injection defined above may seem a bit arbitrary but it is simply combining the RH moves with the bijective proof of $$(a-1)(b+1)\geq ab$$ for some integers $a>b$.<|endoftext|> TITLE: A sufficient condition for a morphism to be a closed immersion? QUESTION [5 upvotes]: Let $R$ be an integral $\bar{\mathbb{F}}_p$-algebra of finite type, let $V$ be an $R$-algebra. Consider a morphism $f \colon \mathrm{Spec}(V) \rightarrow \mathbb{A}^n_R$ that has the following properties: every fibre of $f$ is a closed immersion $f_K\colon \mathrm{Spec}(V)_K \hookrightarrow \mathbb{A}^n_K$ the induced morphism of functors $\mathrm{Hom}_R(-, \mathrm{Spec}(V)) \rightarrow \mathrm{Hom}_R(-, \mathbb{A}^n_R)$ is injective $f$ is universally closed $V$ is at most countably generated as an $R$-algebra $\mathrm{Spec}(V)$ has a $\mathbb{G}_m$-action and $f$ is compatible with the $\mathbb{G}_m$-actions Is it true that then $f$ is always a closed immersion? This should be clear if $f$ is of finite type, but in the particular example I am interested in, this is (a priori) not the case. If not, are there any easy to understand counterexamples? Some of the conditions above may be irrelevant (like $R$ living over $\bar{\mathbb{F}}_p$ or the group scheme action), but I listed them anyways on the off-chance that they could be useful. REPLY [5 votes]: Yes, it is true. More precisely any universally closed monomorphism of schemes (your second and third condition) with locally Noetherian target is a closed immersion. This is proven in Proposition 3.8 of Ferrand: Monomorphismes de schémas noethérian (http://www.numdam.org/article/SAC_1967-1968__2__A7_0.pdf).<|endoftext|> TITLE: 2-torsion in class groups of cubic fields QUESTION [12 upvotes]: I was wondering if there are good bounds for the $p$-parts of the class group of a number field $F$ in terms of its discriminant $D_F$. More precisely, the bound for the order of the full class group of $F$ is of order $\sqrt {D_F}$ and I was wondering whether for a fixed prime $p$ there is a bound for the $p$-torsion of order $D_F^\delta$ for small $\delta > 0$ (ideally arbitrarily small). I am especially interested in $p=2$, but I think it is interesting to ask the more general question. Here is an example: if $F$ is q quadratic field, then the 2-torsion in the class group $C_F$ is generated by primes dividing the discriminant (see for example the answer to this question). In particular the order of the 2-torsion is an $O(D_F^\delta)$ for any $\delta > 0$. I would be interested in similar bounds for fields of higher degree, in particular for non-real cubic fields. (In this case there seems to be a relation between the 2-torsion of the class group and elliptic curves but I am not competent to exploit it). There are also much better bound "on average" for the 2-torsion of class groups of cubic fields: Manjul Bhargava has proven that when cubic fields are ordered by discriminant the mean order for the 2-torsion tends to a constant which is equal to 1.25 in the case of non-real fields. (I read about that and more in this preprint.) REPLY [15 votes]: I'm happy to announce a new result of the shape you ask for: if $F$ is a cubic field (of any signature) then the size of the 2-torsion in its class group is bounded above by $O(D_F^{0.2785})$. The same is also true if $F$ is a number field of any degree $n$, but in this case one has to replace $0.2785$ with a function of $n$ that rapidly tends to $1/2$. This is joint work of Manjul Bhargava, Arul Shankar, Takashi Taniguchi, Jacob Tsimerman, Yongqiang Zhao, and myself. (We did this at an AIM mini-workshop; I highly recommend these to anyone who has the chance!) I regret that the paper is in a rather rough state of preparation, and so I don't have a preprint to share. Once we're done, I'd be happy to e-mail a copy to you or to anyone else who leaves their contact information in the comments.<|endoftext|> TITLE: Convergence issues with infinite product of formal series QUESTION [6 upvotes]: Question first: Show that if $s_1 < s_2 < \dots$ is an increasing sequence of positive integers and $P(x)$ is a nonzero polynomial then we cannot have $$ P(x) \equiv \prod_{j=1}^\infty (1 - x^{s_j}) $$ as formal series. The right-hand side really means $\lim_{N \to \infty} \prod_{j=1}^N (1 - x^{s_j})$ where the notion of convergence is as described in this math.SE answer by Bill. The intuition is that if we take $x \to 1^-$ then the left-hand side tends to zero at a slower rate than the right-hand side, since $P$ has the root $1$ with only finite multiplicity, while every factor on the RHS tends to zero. However, since formal and functional power series aren't actually the same thing (see link above), I'm not sure how to make this precise, even after formally inverting the $1-x$ factors on the left-hand side (by multiplying both sides by $(1+x+x^2+\dots+)^k$). Does anyone know the correct incantations? (For context, this came up in a solution to a recent IMO proposal. Since it was for high school students I swept the convergence issues under the rug, but myself I'd like to know exactly what the right thing to do is.) REPLY [4 votes]: Equivalently, we'll show that we cannot have $$\frac{1}{P(x)} = \frac{1}{\prod_{j=1}^{\infty} (1 - x^{s_j})}$$ as formal power series. The idea is that the LHS has a pole of finite order at $x = 1$ while the RHS has an essential singularity at $x = 1$. Precisely, the coefficients on the LHS have asymptotic growth a polynomial times an exponential. On the other hand, the coefficients of the RHS can be shown to have growth both strictly larger than any polynomial (by truncating the product) and strictly smaller than any exponential (by comparing to the growth rate in the case where $s_j = j$, which is known). So the two rates of growth can't match.<|endoftext|> TITLE: Complexity for solving linear equations? QUESTION [5 upvotes]: What is the best known complexity for finding a vector $x \in \mathbb{R}^n$ to minimize $||Ax - b||^2$ and/or to solve (when possible) the system of linear equations $Ax=b$? I am interested in approximation methods that produce $\epsilon$-approximations for arbitrarily small $\epsilon>0$. Specifically, if $A$ is a real-valued $m \times n$ matrix and $b$ a column vector of size $m$, if I can get an answer within $\epsilon$ of optimality via an algorithm that has total complexity $O(mn \log(mn/\epsilon))$, is that good? Or known? Any related thoughts or references are appreciated. Some basic comparisons: i) $O(mn \log(mn/\epsilon))$ seems to be better than Gaussian elimination, which can have higher complexity and has “divide by small number” issues. ii) The least-square problem of minimizing $||Ax-b||^2$ is the same as finding a vector $x$ to solve the orthogonality equations $A^TAx = A^Tb$, and these orthogonality equations always have (possibly many) solutions $x \in \mathbb{R}^n$. Yet, computing $A^TA$ requires $mn^2$ computations, which is already larger than $O(mn\log(mn/\epsilon))$ (assuming we overlook the log() factors). iii) Gauss-Seidel iterative methods seem nice but I think (I am not sure) these methods need matrices to be square and to have special structure. REPLY [4 votes]: There is a meaningful oracle model where you can obtain a provably optimal method when searching for approximate solutions: this is the "matrix-vector multiplication oracle", where you want to solve a linear system $Ax=b$ with the minimum number of products of the form $Ax$ or $A^Ty$ (that's what the oracle provides). In this model, one can prove that the conjugate gradient method is the best you can do in many (if not all) situations. For example, when you have an upper bound on the eigenvalues of $A^TA$, say $L$, then the number of iterations is $O(\sqrt{LR^2/\varepsilon})$, where $R>0$ is the distance to the solution from your starting point. Also, if you have upper and lower bounds on the spectrum you can get linear convergence (that is, logarithmic dependence on $\varepsilon$) for conjugate gradients, and the speed of this linear convergence is optimal (this is the square root of the condition number). For more on this, check Chapter 12 on Nemirovski's book (http://www2.isye.gatech.edu/~nemirovs/Lec_EMCO.pdf), or for a short version Vishnoi has these nice lecture notes (http://theory.epfl.ch/vishnoi/CargeseLectures.pdf).<|endoftext|> TITLE: Decomposition of pullback metric QUESTION [5 upvotes]: Let $(M^3,g)$ be a complete riemannian manifold and $\Sigma ^2\subset M^3$ a embedded minimal compact surface. Consider the immersion $\phi: \Sigma \times [0,\varepsilon)\to M$ given by $$\phi(p,t)=\exp_p(tN(p)),$$ when $N$ is a unit normal vector field along to $\Sigma$. I would like to show that if we take the pullback metric $\phi^*g$ in $\Sigma\times [0,\varepsilon)$, then $\phi^*g=d\sigma_t^2+dt^2$, where $d\sigma_t^2$ is a smooth family of metrics in $\Sigma$. I saw already argument like that in many papers, but I failed in prove it. Anyone has a little help? Thanks so much. REPLY [9 votes]: This has nothing to do with minimality of $\Sigma$. By definition of the exponential map, the mapping $\gamma_p(t) = \phi(p,t)$, as a map $[0,\epsilon)\to M$ is a geodesic ray, with unit speed, initial position $p\in \Sigma$, and initial velocity $N_p$. Let $V$ be a vector in $T_p \Sigma$ extended to $\Sigma \times [0,\epsilon)$ along $\gamma_p(t)$ by the product structure. Equivalently, you have that $V$ is Lie-transported along $\gamma_p(t)$, or that $[\dot{\gamma_p}, V] = 0$. (Vector field commutators are independent of metric!) The pushforward of $V$ is a vector field that satisfies $V(t) = 0$. To get your splitting it suffices to show that $\langle V, \dot{\gamma_p}\rangle_g = 0$. But this follow because at time $t = 0$ the equality is true, and $$ \nabla_{\dot{\gamma_p}} \langle V, \dot{\gamma_p}\rangle_g = \langle \nabla_V \dot{\gamma_p}, \dot{\gamma_p}\rangle_g = 0 $$ by construction and geodesy.<|endoftext|> TITLE: What's an example of a subcategory whose closure under colimits takes a lot of steps to form by iteration? QUESTION [17 upvotes]: Let $\mathcal{C}$ be a cocomplete category and $\mathcal{S} \subseteq \mathcal{C}$ be a full subcategory. The colimit completion $\mathrm{Colim}^\mathcal{C}(\mathcal{S})$of $\mathcal{S}$ in $\mathcal{C}$ is the smallest full subcategory $\mathcal{S} \subseteq \mathrm{Colim}^\mathcal{C}(\mathcal{S}) \subseteq \mathcal{C}$ which is closed under the formation of small colimits in $\mathcal{C}$. Naturally, it can be attained via a transifinite induction: set $\mathrm{Colim}^\mathcal{C}_0(\mathcal{S}) = \mathcal{S}$; for successor ordinals define $\mathrm{Colim}^\mathcal{C}_{\alpha+1}(\mathcal{S})$ to be the full subcategory of $\mathcal{C}$ consisting of objects which are the colimits of small diagrams in $\mathrm{Colim}^\mathcal{C}_\alpha(\mathcal{S})$; and for limit ordinals take unions. If we iterate through all ordinals, we get $\mathrm{Colim}^\mathcal{C}(\mathcal{S})$ at the end. In category theory, one is often interested in cases where this process stabilizes after one step. For instance if $\mathcal{S}$ is a dense generator in $\mathcal{C}$ (or even just colimit-dense), then the process stabilizes after one step. If $\mathcal{S}$ is a regular generator in $\mathcal{C}$, then the process stabilizes after at most two steps. And the canonical example of a category lacking a strong generator is $\mathsf{Top}$ where one can argue as for $k$-spaces that the process always stops after one step. One is also interested in cases where we close under not all colimits, but just those of certain shapes; if the class of diagrams is saturated or almost so, then in the free case where $\mathcal{C}$ is a presheaf category and $\mathcal{S}$ is the representables, the process stabilizes in one step. But I'd like to see some examples where things go bad. Question: What's an example of $\mathcal{S} \subseteq \mathcal{C}$ where $\mathrm{Colim}^\mathcal{C}(\mathcal{S})$ takes more than two steps to attain? Infinitely many steps? A proper class of steps? I'm really interested in seeing a proper class of steps, but I'd be happy to at least see something easier. And I'm not at all averse to taking advantage of duality / doing everything with limits instead of colimits. And on the flip side, If $\mathcal{S}$ is small and $\mathcal{C}$ is locally presentable, must the process stop after a small number of steps? REPLY [3 votes]: I wish I could give a reference but in 1965 John Isbell claimed that the limit completion of the 2 element group (considered as a one object category) requires a proper class of steps, but I didn't understand his argument and I don't believe he ever published it.<|endoftext|> TITLE: Three homothetic centers are collinear QUESTION [8 upvotes]: I am looking a proof for the problem as follows: Let a convex hexagon, such that its principal diagonals are concurrent. For each side of the hexagon, extend the adjacent sides to their intersection, forming a triangle exterior to the given side. Then show that: Three external (or internal) homothetic centers of three pair circumcircle of opposite triangles are collinear. Please see the applet in Geogebra PS: The line through the three external homothetic centers are perpendicular to the line through three internal homothetic centers. Another applet: Three homothetic centers are collinear associated a circumscribed conic hexagon REPLY [2 votes]: At least for the external homothetic centers, this follows from Monge's theorem, which actually tells you that "for any three circles in a plane, none of which is completely inside one of the others, the intersection points of each of the three pairs of external tangent lines are collinear".<|endoftext|> TITLE: Why is the Berkovich spectrum of a C*-Algebra the same as the Gelfand spectrum? QUESTION [9 upvotes]: Let $A = \mathcal{C}(X)$ be a commutative (unital) C*-Algebra. Let $Spec(A)$ denote its Gelfand spectrum $$ Spec(A) = \{A \rightarrow \mathbb{C} : \text{non-zero *-homomorphism} \} \simeq X. $$ Now view $A$ as a normed algebra over $\mathbb{C}$ with respect to the supremum norm $\| a\|_{sup} = \sup_{x \in X} (\|a(x)\|)$. Its Berkovich spectrum $Sp^B(A)$ is then given by $$Sp^B (A) = \{|\cdot|: A \rightarrow \mathbb{R}_+ \text{ multiplicative seminorm bounded by }\| \|_{sup} \}.$$ Why do we have $Spec(A) = Sp^B(A)$? (as for example stated in the wikipedia article on Berkovich spectra https://en.wikipedia.org/wiki/Berkovich_space#Berkovich_spectrum) It is clear that every homomorphism $A \rightarrow \mathbb{C}$ composed with the absolute value gives a seminorm of the required form. However I struggle to see the converse. Given a multiplicative seminorm $|\cdot| : A \rightarrow \mathbb{R}_+$ bounded by $\| \|_{sup}$, why must there be an $x \in X$ such that for all $a \in A:$ $|a| = \| a(x) \|$? Using the boundedness and multiplicativness of $|\cdot|$, I can show that for each $a \in A$ there must be an $x$ such that $|a| = \| a(x) \|$. But a priori these $x$ can be different for different $a$ - I simply fail to see why this cannot happen. REPLY [4 votes]: Observe that every commutative C*-algebra $A \not\simeq \mathbb{C}$ has a non-trivial zero-divisor. To see this view $A$ as continuous functions over its spectrum which contains at least two points (and is Hausdorff) and use Urysohn lemma to construct two continuous functions supported at separating open sets of these two points. It follows that every closed$^1$ prime ideal is the kernel of a homomorphism to $\mathbb{C}$. Now, given a bounded multiplicative seminorm $s$ on $A$, observe that its kernel $p$ is a closed prime ideal and it defines a multiplicative norm on $A/p\simeq \mathbb{C}$. This norm must coincide with the usual absolute-value $|\cdot|$. Denoting the map $A\to\mathbb{C}$ by $a\mapsto a(x)$ we conclude that for every $a\in A$, $s(a)=|a(x)|$. Note that the commutativity assumption was inessential in all of the above. 1: Thanks to Robert Furber for correcting my omission here.<|endoftext|> TITLE: Why is there a factor $p$ in the definition of $T_p$ via Hecke correspondences on modular curves? QUESTION [16 upvotes]: Fix $N\ge4$. Let $Y_1(N)$ and $X_1(N)$ be the usual modular curves. I want to view them as schemes over $\mathbb Z$ representing the moduli functors of (usual or generalized) elliptic curves with (Drinfeld) $\Gamma_1(N)$-structures. That they exist in this form is shown in Brian Conrad's paper "Arithmetic moduli of generalized elliptic curves". Let $f\colon E_1(N)\rightarrow Y_1(N)$ be the universal elliptic curve. Let further $Y_1(N,p)$ and $X_1(N,p)$ denote the moduli spaces for $\Gamma_1(N,p)$ structures used in the definition of Hecke correspondences, and denote the universal elliptic curve over $Y_1(N,p)$ by $E_1(N,p)$. Consider the diagram $$\begin{array}{ccccccccc} & & E_1(N,p) & & \xrightarrow{\phi} & & E_1(N,p)/C & & \\ & \swarrow && \searrow & & \swarrow & & \searrow &\\ E_1(N) & & & & Y_1(N,p) & & & & E_1(N)\\ & \searrow & & \swarrow{\pi_1} & & \searrow{\pi_2} & & \swarrow & \\ & & Y_1(N) & & & & Y_1(N) & & \end{array} $$ where $\pi_1$,$\pi_2$ are the usual degeneracy maps, $\phi$ is the universal $p$-isogeny, and the two squares are cartesian. This diagram can be used to describe Hecke operators on (Eichler-Shimura type) cohomology groups. More precisely, if we denote analytifications by a superscript "an", then the Hecke operator $T_p$ on the cohomology group $H^1(Y_1(N)^{\mathrm{an}},\operatorname{Sym}^k R^1f_*\mathbb Z)$ (or similar groups) equals $$T_p=\operatorname{tr}_{\pi_1}\circ\phi^*\circ\pi_2^*. $$ This is written e. g. in Delignes paper "Formes modulaires et représentations l-adiques" (Prop. 3.18 (i)). One has a similar relation also in étale cohomology. In the above-mentioned article "Arithmetic moduli of generalized elliptic curves" (§4.5), Brian Conrad does a similar thing on modular forms themselves, so he considers the cohomology group $H^0(X_1(N),\omega^{\otimes k})$. However, in his case he gets $$pT_p=\operatorname{tr}_{\pi_1}\circ\phi^*\circ\pi_2^*. $$ I am surprised that one has to divide by $p$ in the second case to get the classical Hecke operator from the composition $\operatorname{tr}_{\pi_1}\circ\phi^*\circ\pi_2^*$, while in the first case one doesn't. What is the reason behind this? Any other references on this topic are welcome. REPLY [14 votes]: This question is somehow a "characteristic 0" question, so let me treat $Y = Y_1(N)$ and $X = X_1(N)$ as $\mathbf{Q}$-varieties rather than doing anything complicated with integral models. There's an isomorphism of sheaves on $X$, the Kodaira-Spencer map, $$\omega^2 \to \Omega^1_{X / \mathbf{Q}}(C)$$ where $C = X - Y$ is the divisor of cusps. So we have an isomorphism $$ H^0(X, \omega^k) = H^0(X, \omega^{k-2} \otimes \Omega^1(C)).$$ However, this isomorphism is not Hecke equivariant for the obvious actions of Hecke operators on both sides: you pick up a factor of $p$ in the Hecke operator $T_p$. Concretely, a section of $\omega^k$ looks like $f(\tau) (\mathrm{d}z)^k$, and a section of $\omega^{k-2} \otimes \Omega^1$ looks like $f(\tau) (\mathrm{d}z)^{k-2} \mathrm{d}\tau$, where $\tau$ is a coordinate on the upper half-plane, and $z$ is a coordinate on the universal elliptic curve; the action of $T_p$ involves pullback via a map which looks locally like $(\tau, z) \mapsto (\tau/p, z)$, and hence there's an extra factor of $p$ coming from $\mathrm{d}\tau$ which $\omega^k$ doesn't see. So this is why you have to modify the definition of $T_p$ on $H^0(X, \omega^k)$: it is in order to get it to match with the natural action on $H^0(X, \omega^{k-2} \otimes \Omega^1)$. The latter is somehow the "correct" action, since $H^0(X, \omega^{k-2} \otimes \Omega^1(C))$ is canonically a subspace of $H^1_{\mathrm{dR}}(Y_1(N), \operatorname{Sym}^{k-2} R^1f_*\mathbf{Q})$ and will compare well with etale, Betti, etc cohomology.<|endoftext|> TITLE: Uncle of Witt algebra QUESTION [11 upvotes]: A Witt algebra W is an infinite-dimensional Lie-algebra defined by the generator relations: W: $[L_{j},L_{k}]:=(j-k)\cdot L_{j+k}$ And my first thought was: What about the analogous algebra defined by W': $[L_{j},L_{k}]:=j\cdot{L_k}-k\cdot{L_j}$ (Yes, I checked, the Jacobi relation is fulfilled.) There are about two possibilities: a) W' is uninteresting, question can be closed. b) W' is interesting and someone had the idea before me. Inventor's name suffices then. :-) (Unfortunately, there is no way to google such a formula.) REPLY [14 votes]: Interesting/uninteresting is a very subjective thing, so let me try to just say several things that I see immediately. 0) This algebra, unlike the Witt algebra, does not have any [obvious] grading, and this makes it in a sense less interesting for some purposes. 1) This algebra, unlike the Witt algebra, comes from antisymmetrising an associative product $L_j\circ L_k=j L_k$ (in the case of the Witt algebra, it is not associative, but only satisfied the so called [left] pre-Lie identity $(a\circ b)\circ c-a\circ(b\circ c)=(b\circ a)\circ c-b\circ(a\circ c)$. So in a sense it is more unusual: Lie algebras do not frequently arise that way. 2) The associative product above is not only associative but it does in fact satisfy the so called [left] permutative identity $a\circ b\circ c=b\circ a\circ c$ (so this algebra is very close to commutative; it is only non-commutative because there is no unit element). The operad of permutative algebras is the Koszul dual of the operad of pre-Lie algebras, so it is kind of an amusing coincidence. 3) This Lie algebra contains a one-dimensional ideal spanned by $L_0$, and the quotient by this ideal, after a linear change of basis $M_j=\frac{L_j}{j}$, has simpler commutation relations $[M_j,M_k]=M_k-M_j$ ($j,k\ne 0$).<|endoftext|> TITLE: When can a finite subgroup of $GL(2n,\mathbb{R})$ be viewed as a subgroup of $GL(n,\mathbb{C})$? QUESTION [17 upvotes]: A finite group acting on a complex vector space of dimension $n$ can be seen as acting on a real vector space of dimension $2n$ just by forgetting the complex structure of the space. My question is, if I am handed a real vector space $V$ of dimension $2n$, and a group $G$ acting on it, is there a test I can perform to determine if the action arose from a complex action in this way? Sometimes it is easy to rule out: for example, if $G$ contains anything orientation-reversing, then clearly it doesn't arise in this way. Or if one knows enough about $G$ (abstractly as a group) to know it doesn't have any faithful $n$-dimensional representation. But I would like an if-and-only-if criterion: Is there a test I can perform on the pair $G,V$ to determine whether the action of $G$ can be obtained by beginning with a complex $n$-dimensional representation and forgetting the complex structure of the vector space? To make a little more precise what I mean by "can be obtained": if there is an element $J\in GL(V)$ that commutes with the action of $G$ and satisfies $J^2 = - I$, then the action of $G$ "can be obtained from a complex $n$-dimensional action by forgetting the complex structure", since one can regard $V$ as a complex vector space via the action of $J$. So the question is, if I am handed $G$ and $V$, is there a test for the existence of such a $J$? REPLY [18 votes]: It's cleaner to ask about an arbitrary finite-dimensional real representation $V$ of a finite group $G$; the hypothesis that $V$ is faithful isn't particularly helpful. $V$ has a decomposition $\bigoplus_i n_i V_i$ into irreducible components with multiplicities, and so its endomorphism algebra takes the form $$\text{End}(V) \cong \prod_i M_{n_i}(D_i)$$ where $D_i = \text{End}(V_i)$ are division algebras over $\mathbb{R}$ by Schur's lemma, so either $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$. The question is when there is a morphism (necessarily a monomorphism) $\mathbb{C} \to \text{End}(V)$ of $\mathbb{R}$-algebras, and the answer is iff there is such a morphism into each $M_{n_i}(D_i)$, hence for each $i$ either $D_i = \mathbb{R}$ and $n_i$ is even, or $D_i = \mathbb{C}$ or $\mathbb{H}$. We can test for this as follows. If $W$ is an irreducible real representation, then $\text{End}(W \otimes \mathbb{C}) \cong \text{End}(W) \otimes \mathbb{C}$ (all tensor products here and below taken over $\mathbb{R}$), and so exactly one of three things happens: $\text{End}(W) \cong \mathbb{R}$, so $\text{End}(W \otimes \mathbb{C}) \cong \mathbb{C}$, meaning that $W \otimes \mathbb{C}$ remains irreducible. $\text{End}(W) \cong \mathbb{C}$, so $\text{End}(W \otimes \mathbb{C}) \cong \mathbb{C} \otimes \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$, meaning that $W \otimes \mathbb{C}$ is a direct sum of two complex conjugate and nonisomorphic irreducibles. $\text{End}(W) \cong \mathbb{H}$, so $\text{End}(W \otimes \mathbb{C}) \cong \mathbb{H} \otimes \mathbb{C} \cong M_2(\mathbb{C})$, meaning that $W \otimes \mathbb{C}$ is a direct sum of two isomorphic irreducibles. These three cases can be distinguished by the value of $$\langle W, W \rangle = \frac{1}{|G|} \sum_{g \in G} \chi_W(g)^2$$ as Claudio says; it takes the values $1, 2, 4$ in the above three cases. With this modification to the orthogonality relations you can try to figure out the decomposition of $V$ into real irreducible representations and then compute the $n_i$ and the $D_i$ using the above test. See also the Frobenius-Schur indicator for some discussion of how to classify the real irreducible representations given knowledge of the complex irreducible representations.<|endoftext|> TITLE: Bounds for number of edges of a graph, given girth and number of vertices QUESTION [7 upvotes]: In reading a paper, I came across an affirmation "a graph of girth $g$ and $q$ vertices has at most $q^{1+(O(1)/g)}$ edges" In a previous question I asked in this site about it, I was reffered to a theorem by Bondy and Simonovits, that states: $ex(q,C_{2k}) < (k - 1 + o(1)) q^{1+1/k}$ This is a bound that does the job, assuming a fixed girth. However, in my case, I have a girth that grows with $n$ (a variable independent of $q$). So, now I ask a more generic question: what kind of bounds are known, to limit a graph's number of edges, given its girth and its number of vertices? REPLY [9 votes]: The bound you quote is actually a little inefficient, because it is harder to find a cycle of fixed length $t$ than to find a cycle of length at most $t$. I recommend that you look at Section 4.1 of the excellent survey by Füredi and Simonovits: https://arxiv.org/pdf/1306.5167.pdf In particular, Theorem 4.1 of that survey, saying that $$\textrm{ex}(n, \{C_3, C_4, \dots, C_{2k}\}) < \frac{1}{2} n^{1 + 1/k} + \frac{1}{2} n,$$ gives a fairly satisfactory answer to your question for odd girth.<|endoftext|> TITLE: Posets obtained from a semigroup by the definition $x \leq y \iff x \cdot y = x$ QUESTION [7 upvotes]: A po-groupoid is a groupoid $\langle A,\cdot\rangle $ such that the relation defined by $$ x \leq y \text{ if and only if } x \cdot y = x $$ is a partial order on $A$, the order related to $\langle A,\cdot\rangle $. For every poset $\langle A,\leq\rangle $ one can define a po-groupoid operation $*$ on $A$ setting $$ x * y := \begin{cases} x & \text{if } x\leq y \\ y & \text{if } x\not\leq y. \end{cases} $$ such that $\leq$ is the order related to $\langle A,*\rangle $. Po-groupoids are obviously idempotent, but need not be associative nor commutative. In spite of this, I think that every down-directed poset is related to a commutative po-groupoid. An important example is given by semilattices: Commutative idempotent semigroup; these are exactly the po-groupoids related to posets $\langle A,\leq\rangle $ such that for every $x,y\in A$, $\inf\{x,y\}$ exists (where the product is given by the infimum). I think I have an example of a poset that has no associative po-groupoid (a po-semigroup) related to it. My question is, Is there any characterization in the literature of posets related to po-semigroups? An important piece of information is that po-semigroups form a variety axiomatized by the following identities: \begin{align*} (x \cdot y) \cdot z &\approx x \cdot (y \cdot z) \\ x \cdot x &\approx x \\ x \cdot y \cdot x &\approx y \cdot x. \end{align*} REPLY [4 votes]: This is far from a complete answer, but you might be interested in taking a look at Section 3.5 of Lawrence Valby's 2015 PhD thesis, Some Case Studies in Algebra Motivated by Abstract Problems of Language. It is available on ProQuest, but I can email you a copy if you don't have access to ProQuest. Lawrence gives a simple (second-order) characterization of a class of posets he calls $\downarrow$-posets and proves that this class of posets is not first-order axiomatizable. Here are the definitions: An action (of $S$ on $C$ on the right) is an algebra in the signature consisting of two sorts, $C$ and $S$, and a function $C\times S\to C$. Any action $(C,S,\cdot)$ gives rise to a poset $(C,\leq)$, defined by $c\leq d$ if and only if there is a finite sequence $s_1,\dots,s_n$ (possibly the empty sequence) of elements of $S$ such that $ds_1\dots s_n = c$. Instead of $S$, we can look at the set $S'$ of all functions $C\to C$ induced by the action of finite sequences of elements from $S$ on $C$. Then $S'$ is a monoid (with identity given by the empty sequence and multiplication given by composition), and we can equivalently define the poset $(C,\leq)$ by $c\leq d$ if and only if there exists $s\in S'$ such that $ds = c$. So if we're interested in the induced posets, we might as well be looking at monoid actions. If $C$ and $S$ are collections of sets, we say that $S$ acts on $C$ by intersection if whenever $A\in C$ and $B\in S$, then $A\cap B\in C$. A $\downarrow$-action is an (abstract) action which is isomorphic to an action by intersection. It turns out that these are exactly the idempotent commutative actions (satisfying $css = cs$ and $cst = cts$, for all $c\in C$ and $s,t\in S$). If we replace $S$ by $S'$, we get exactly the semilattice actions. A $\downarrow$-poset is a poset arising from a $\downarrow$-action. So $\downarrow$-posets are simpler than yours in that the algebras are commutative, but they're more general in that they come from actions on arbitrary sets, not necessarily the action of the algebra on itself. And they're not just the posets with finite meets! Anyway, the ideas in Lawrence's thesis might be relevant to your case.<|endoftext|> TITLE: Find the best constant to this bounded inequality QUESTION [5 upvotes]: Let $n$ be postive integer number, and $x_{i}\ge 0$, such $$x_{i}x_{j}\le 4^{-|i-j|},1\le i,j\le n$$ then I have prove $$x_{1}+x_{2}+\cdots+x_{n}<\dfrac{5}{3}$$ Edit Add Proof:since $x^2_{i}\le 1,0\le x_{i}\le 1$,Let $S_{j}=\sum_{i=1}^{j}x_{i},S=\sum_{i=1}^{n}x_{i}$,then we have $$0=S_{0}\le S_{1}\le S_{2}\le\cdots\le S$$so there exist $k$,such $S_{k}\le\dfrac{S}{2}\le S_{k+1}$, if we let $$T_{k}=S-S_{k},T_{k+1}=S-S_{k+1}$$ then we have $$|S_{k}-T_{k}|+|S_{k+1}-T_{k+1}|=|2S_{k}-S|+|2S_{k+1}-S|=2x_{k+1}\le 2$$ then for $l\in\{k,k+1\}$,we have $$|S_{l}-T_{l}|\le 1\tag{1}$$ and we have $$S_{l}T_{l}=\sum_{i=1}^{l}\sum_{j=l+1}^{n}x_{i}x_{j}\le\sum_{i=1}^{l}\sum_{j=l+1}^{n}4^{-|i-j|}\le\sum_{i=1}^{l}\dfrac{1}{4^{l-i}}\sum_{j=l+1}^{n}\dfrac{1}{4^{j-l}}<\dfrac{4}{3}\cdot\dfrac{1}{3}\tag{2}$$ use $(1)$ and $(2)$ we have $$x_{1}+x_{2}+\cdots+x_{n}=S_{l}+T_{l}=\sqrt{(S_{l}-T_{l})^2+4S_{l}T_{l}}\le\dfrac{5}{3}$$ Question : Let $n$ be postive integer number, and $x_{i}\ge 0$, such $$x_{i}x_{j}x_{k}\le 4^{-|i-j-k|},1\le i,j,k\le n$$ then I have prove $$x_{1}+x_{2}+\cdots+x_{n} TITLE: The naive approach to deriving profunctors - What's wrong with it? QUESTION [7 upvotes]: Let $(\mathcal{C,W})$ be relative category (equipped with a wide subcategory of weak equivalences satisfying 2 out of 3 property). Consider a profunctor $F: \mathcal{C \times D^{op}} \to \mathsf{Set}$ where $\mathcal{D}$ is an ordinary category. We can define right "derived" functor of $F$ by the following procedure (the left derivation is completely analogous): The right derived functor of $F$ is defined by $RF(X)= \lim_{{X \to Z} \in \mathcal{W}} F(Z)$ for $X\in \mathcal{C}$ where the limit is taken in the presheaf category $[\mathcal{D}^{op},\mathsf{Set}]$. If we're lucky and $RF(X)$ is isomorphic to a representable presheaf for every $X \in \mathcal{C}$ then we have the actual derived functor. Otherwise we can still define it as a profunctor by the above procedure. I assume that to actually compute this guy we need a more structured situation for the localization (on both the target and the source categories) to be tractable enough (e.g. model categories / triangulated categories). But as a definition I don't see anything wrong with it. What's wrong with this naive approach? Why haven't I found it anywhere in the literature? (Is this the same as a Kan extension and I'm just being dumb?) REPLY [5 votes]: The problem with this definition is that the formula $\mathrm{colim}_{X \to Z \in \mathcal{W}} F(Z)$ does not, in general, depend functorially on $X$. For it to depend functorially on $X$ you need a way to push forward weak equivalences along arbitrary maps. This would work, for example, if $\mathcal{C}$ is a cofibration category and you take the colimit indexed only by trivial cofibrations out of $X$. Of course, taking the actual colimit may be too strict, but if you take the homotopy colimit $$\displaystyle\mathop{\mathrm{hocolim}}_{X \to Z \in \mathcal{W} \cap \mathcal{C}of} F(Z)$$ (computed, say inside the simplicial presheaf category $\mathrm{Fun}(\mathcal{D}^{op}, \mathrm{Set}_{\Delta})$), then you do get something sensible. For example, if $\mathcal{C}$ is a cofibration category and $X,Y$ are cofibrant objects then $$\displaystyle\mathop{\mathrm{hocolim}}_{X \to Z \in \mathcal{W} \cap \mathcal{C}of} \mathrm{Hom}(Y,Z)$$ is a model for the derived mapping space $\mathrm{Map}(Y,X) = \mathbb{R}\mathrm{Hom}(Y,X)$.<|endoftext|> TITLE: Lifting symmetries to the universal cover QUESTION [6 upvotes]: If $X$ is a connected topological space with universal cover $p: \tilde{X} \to X$, I believe any homeomorphism $f : X \to X$ can be 'lifted' to a homeomorphism $\tilde{f} : \tilde{X} \to \tilde{X}$. (I'm using the word 'lift' in a nonstandard way here. What I mean is that we can find a map $\tilde{f} : \tilde{X} \to \tilde{X}$ such that $p \circ \tilde{f} = f \circ p$. I believe, but haven't checked, that any choice of such $\tilde{f}$ is a homeomorphism.) However, there is in general no canonical way to choose $\tilde{f}$ given $f$, so if we have a discrete group $G$ acting as homeomorphisms of $X$ this action may not give rise to an action of $G$ on $\tilde{X}$. Is there a functorial way to build a group $G_X$ such that any action $\alpha$ of $G$ on $X$ gives rise to an action $\tilde{\alpha}$ of $G_X$ on $\tilde{X}$? That question is a bit vague; to make it precise I think I want an onto homomorphism $\rho: G_X \to G$ such that for all $g \in G_X$, $\tilde{\alpha}(g) : \tilde{X} \to \tilde{X}$ 'lifts' $\alpha(\rho(g)) : X \to X$ in the above sense. If $G$ acts on $X$ in a way that preserves a basepoint $\ast$ it would act on $\pi_1(X,\ast)$, allowing us to define the semidirect product of $G$ and $\pi_1(X,\ast)$, and I would try using this as $G_X$. However, I'm mainly interested in the case where the action of $G$ does not preserve any point of $X$. REPLY [3 votes]: Assuming that $G$ is discrete, the homotopy quotient $X/G$ fits into a fiber sequence $$X \to X/G \to BG$$ and hence, by the long exact sequence in homotopy, its fundamental group $\widetilde{G} = \pi_1(X/G)$ fits into a short exact sequence $$1 \to \pi_1(X) \to \widetilde{G} \to G \to 1$$ and moreover $X$ and $X/G$ have the same universal cover, hence $\widetilde{G}$ acts on $\widetilde{X}$ and this action extends the action of $G$ on $X$ as expected. (All of this is in a homotopical sense; I haven't thought about what it would take to get a point-set action.) One way to describe the classification of group extensions is that extensions of $G$ by $N$ correspond to actions of $G$ on the groupoid $BN$, and the extension above comes in this way from the induced action of $G$ on the fundamental groupoid $\Pi_1(X)$. This extension splits as a semidirect product iff the corresponding action has a homotopy fixed point, which is weaker than the condition that some particular point-set model of it has an actual fixed point but still need not be satisfied in general. REPLY [3 votes]: Inspired by Qiaochu Yuan's answer, I'll note that we can solve my problem rather tautologously in the universal example where $G$ is the group of all homeomorphisms of $X$ (treated as a discrete group, for simplicity) and $\alpha$ is the identity. In this case we can let $G_X$ be the group of all homeomorphisms $f : \tilde{X} \to \tilde{X}$ that 'lift' homeomorphisms $f : X \to X$, in the sense that $p \circ \tilde{f} = f \circ p$ . For any such $\tilde{f}$ there is a unique $f$ that makes this equation hold, so there is a homomorphism $\rho: G_X \to G$ with $$ \rho(\tilde{f}) = f .$$ The kernel of $\rho$ is group of deck transformations of $\tilde{X}$, so we have a short exact sequence $$1 \to \pi_1(X) \to G_X \stackrel{\rho}{\longrightarrow} G \to 1$$ so $G_X$ is an extension of $G$ by $\pi_1(X)$. When $G$ fixes a point $\ast \in X$, then $G_X$ is a semidirect product of $G$ and $\pi_1(X,\ast)$. REPLY [3 votes]: A way to solve this is to consider for every action $\alpha$ of $G$ on $X$ the group of gauge transformations $G(\alpha)$. An element of $G(\alpha)$ is an automorphism $h$ of the universal cover $\tilde X$ such that there exists $g\in G$ such that $p\circ h=\alpha(g)\circ p$. Then you can naturally defines a notion of (iso)morphism between representations $\alpha$. I assume the isomorphim class of these representations is a set $A$. Write $G_X=\Pi_{\alpha\in A}G(\alpha)$. Any action of $G$ on $X$ isomorphic to $\alpha$ yields to the action of $G_X$ on $\tilde X$ defined by the projection $G_X\rightarrow G(\alpha)$.<|endoftext|> TITLE: A proof of Theorem 9.2.12. in the Gompf-Stipsicz QUESTION [6 upvotes]: I'm seeking for a proof of Theorem 9.2.12. in the Gompf-Stipsicz "4-Manifolds and Kirby Calculus" (for the statement, see the following image). But the textbook omits any proofs and only gives a weaker version (Theorem 9.2.13.). My question is: How can we find a proof of the theorem? Any references will be fine. Thank you for your help. REPLY [5 votes]: The weaker version is a result due to Wall: C.T.C. Wall, On simply connected 4-manifolds, J. London Math. Soc., 39 (1964), 141–149 A stronger statement has been proven by Kreck (who also gives some references), but I don't know about the original reference Gompf and Stipsicz were thinking about. Matthias Kreck, h-cobordisms between 1-connected 4-manifolds Geom. Topol., 5 (2001), 1-6<|endoftext|> TITLE: How to construct an abelian variety with CM by a given CM field? QUESTION [5 upvotes]: Let $F$ be a totally real number field, and let $K$ be a quadratic extension of $F$ which cannot be embedded into $\mathbb{R}$. Then $K$ is a so called CM field. For instance, take $F = \mathbb{Q}(\sqrt2)$, and $K = \mathbb{Q}(\zeta_8) = F(i)$ where $\zeta_8$ is a primitive complex $8$-th root of unity. Denote by $\mathcal{O}_K$ the ring of integers of $K$. My question is: How to construct a number field $L$ and an abelian variety $A/L$ such that $\mathrm{End}(A) \cong \mathcal{O}_K ?$ For an example with elliptic curves, take $F = \mathbb{Q}$, and $K = \mathbb{Q}(\sqrt{-14})$ so that $\mathcal{O}_K = \mathbb{Z}[\sqrt{-14}]$. Evidently, $\mathcal{O}_K$ is a lattice in $\mathbb{C}$ so $E_0 = \mathbb{C}/\mathcal{O}_K$ is an elliptic curve with $\mathcal{O}_K \subseteq \mathrm{End}(E_0)$. Since $\mathrm{End}(E_0)$ is an order in an imaginary quadratic number field, we have $\mathrm{End}(E_0) \cong \mathcal{O}_K$ as $\mathcal{O}_K$ is the maximal order. Furthermore, $j(E_0)$ is an algebraic number (in fact, an algebraic integer) which generates a field isomorphic to $L = \mathbb{Q}(\sqrt{2\sqrt{2}-1})$ over $\mathbb{Q}$. It follows that we can define an elliptic curve $E_1/L$ such that $j(E_1) = j(E_0)$ so that $E_1 \cong_{\mathbb{C}} E_0$ and in particular $\mathrm{End}(E_1) \cong \mathrm{End}(E_0) \cong \mathcal{O}_K$ as required. Can I tell a similar story (with abelian vaieties instead of elliptic curves) for any CM field? It is only clear to me that the answer is positive for imaginary quadratic number fields. REPLY [3 votes]: A reasonably good answer to your question of finding a number field $L$ is given in the comments to Is there an excplicit number field of definition for an Abelian Variety $A/\mathbb{C}$ with CM? . However, the answer is roughly "no, there's no easy way to find $L$". But first you need to understand the difference between the field of moduli and fields of definition. For elliptic curves, the field of moduli is generated by the $j$-invariant (and if you also want the CM maps defined, you need to adjoin $j$ to the CM field, at least if $\text{End}(A)$ is the full ring of integers). For abelian varieties, the field of moduli is the field generated by the point in moduli space, or alternatively the fixed field of $$ \{\sigma\in\text{Gal}(\overline{\mathbb Q}/\mathbb Q) : \sigma(A)\cong A\}.$$ But there need not be a model for $A$ that is defined over this field. There's a brief discussion of CM abelian varieties in Shimura's Introduction to the Arithmetic Theory of Automorphic Forms (Section 5.5). More detailed information is in books on CM abelian varieties such as those by Shimura-Taniyama and Lang.<|endoftext|> TITLE: A property of minimal prime ideals in commutative reduced ring QUESTION [5 upvotes]: Let $\{\mathfrak{p}_i\}_{i\in I}$ ($I$ is an infinite set) be a family of minimal prime ideals in a commutative reduced ring $R$ with identity, and let $a, b \in R$. If the ideal $\langle a, b\rangle$ generated by $a$ and $b$ is contained in the union $\bigcup_{i\in I}\mathfrak{p}_i$, can we deduce that $\langle a, b\rangle \subseteq \mathfrak{p}_j $ for some $j\in I$? REPLY [5 votes]: This is an interesting question. Quentel's Example provides an example. Let $R$ be a reduced ring, $Min(R)$ its space of minimal prime ideals, and $q(R)$ its classical quotient ring. Theorem The following are equivalent. $q(R)$ is von Neumann regular. $R$ satisfies Property $A$ and $Min(R)$ is compact. $Min (R)$ is compact and whenever a f.g. ideal is contained in the union of minimal prime ideals, then it is contained in one them. Consequently, Quentel's Example is a reduced ring $R$ with $Min(R)$ compact but $q(R)=R$ not von Neumann regular. Therefore, it must possess a f.g. ideal which is contained in a union of minimal prime ideals but contained in any single one. Glaz, Sarah Controlling the zero divisors of a commutative ring. Commutative ring theory and applications (Fez, 2001), 191–212, Lecture Notes in Pure and Appl. Math., 231, Dekker, New York, 2003. J. A. Huckaba, Commutative Rings With Zero Divisors, Marcel Dekker, Inc. (1988)<|endoftext|> TITLE: Determining a function is harmonic from mean value property for just three(?) radii QUESTION [5 upvotes]: A couple days ago I posted this on MSE (here) but in retrospect it might be more appropriate for this site. This theorem is well-known (maybe it can be called Morera's theorem): A continuous function satisfying the mean value property on balls is harmonic. I was recently surprised to hear in a talk that the conclusion still holds if you only check the mean value property on three (I think) radii. Does anyone have a reference or name for this result? I would enjoy seeing the details and a proof. REPLY [8 votes]: Yes, in fact you only need two radii. More precisely, a theorem on page 167 of this Monthly paper of Zalcman says any two radii $r_1$ and $r_2$ will work unless the quotient $r_1/r_2$ is a quotient of zeros of a certain explicit function. The author says this result "was discovered by Jean Delsarte as far back as 1957." REPLY [3 votes]: I guess this is the theorem : Let $f$ be an infinitely differentiable function defined in $\mathbb R^{n}$, and $u(x,r)$ the mean value of $f$ taken over the sphere with center at $x$ and radius $r$, and let $a$ and $b$ denote two fixed positive numbers. If $u(x,a)=u(x,b)=f(x)$ in $\mathbb R^{n}$, then $f$ is harmonic. When $n>3$, exception must be made of a finite number of ratios $a/b$ which are independent of the function $f$. The theorem is proved in : J. Delsarte, J.-L. Lions, Moyennes généralisées, Comment. Math. Helv. 33 1959 59–69. also in the lecture notes (see in particular Chapter III) : J. Delsarte, Lectures on topics in mean periodic functions and the two-radius theo- rem, Tata Institute, Bombay, 1961. http://www.math.tifr.res.in/~publ/ln/tifr22.pdf<|endoftext|> TITLE: Dropping "generic" from the definition of forcing QUESTION [7 upvotes]: Back when I was first learning about forcing and trying to understand the need to consider generic filters, I came up with the following question. Suppose we have a countable transitive model $M$. Let's say that "$p$ pseudoforces $\phi$" if for every filter (not necessarily generic) $G\in P$, $p\in G$ implies that $\phi$ is true in $M[G]$. Is pseudoforcing definable in $M$? I'll allow a little wiggle room about what $M[G]$ means when $G$ is not generic, but I suspect that the answer is no, regardless. Is that correct? The reason I was led to ask this was that one intuitive justification for why forcing (as standardly defined) is definable in $M$ is that one doesn't need to know anything "specific" about any particular $G$ to decide whether $p$ forces $\phi$. But this level of handwaving would seem to apply to pseudoforcing as well, so I think would be illuminating to understand exactly how genericity comes into play here. EDIT in response to Joel David Hamkins's request that I clarify what I mean by $M[G]$ when $G$ is not generic: I'm going to forget about Boolean-valued models and follow the approach in Kunen's textbook. We have an arbitrary poset $P$ in our countable transitive model $M$ of ZFC. We take the usual definition of a $P$-name: $\tau$ is a $P$-name if and only if $\tau$ is a relation and for all $\langle \sigma,p\rangle \in \tau$, $\sigma$ is a $P$-name and $p\in P$. Next we have define how to evaluate $\tau$ at $G\subseteq P$, but again we can just use the standard definition: $$\tau_G = \{ \sigma_G \mid \exists p\in G : \langle\sigma,p\rangle \in \tau\}.$$ Then $M[G]$ is defined to be the set of all $\tau_G$ as $\tau$ ranges over all $P$-names in $M$. This is already a non-trivial construction because Kunen shows that if $G$ is any nonempty filter then $M[G]$ satisfies Extensionality, Foundation, Pairing, and Union. Now it seems to me that I can define $p \mathrel{?\mathord{\vdash}} \phi$ (read "$p$ pseudoforces $\phi$") analogously to $p\Vdash\phi$ simply by dropping the word "generic" from the definition—instead of "for all generic filters $G$" we say "for all filters $G$." What goes wrong? My guess, based on something Andreas Blass once told me, is that we run into trouble when we try to prove the definability of pseudoforcing. A related question is this. Kunen proves two crucial facts about forcing; (1) it's definable, and (2) every $\phi$ that is true in $M[G]$ is forced by some $p\in G$. Suppose I formulate the conjecture that these two facts are also true of pseudoforcing. Could I then deduce that $M[G]$ satisfies ZFC from this conjecture? Since the conjecture is false (see Goldstern's comment about part (2)), the answer to this question is yes for trivial reasons, but what I'm trying to get at is whether the genericity of $G$ is primarily needed in order to prove these two crucial facts, and that the rest of the proof that $M[G]$ satisfies ZFC follows "formally" from them. REPLY [2 votes]: Your setting is not completely clear to me, but I claim that the truth lemma fails. Here is the argument: Let $H$ be a filter which is not generic and let $D$ be a dense set in $M$ not intersecting $H$. Let $a$ be the canonical name for $D$, and let $c$ be the canonical name usually given for the (generic) filter. Then, $M[H]\models \neg\exists x; (x\in D\cap H)$. So, assuming the truth lemma, $\exists p\in H; p\Vdash \neg\exists x; (x\in a\cap c)$. Fix such a $p\in H$. Since the definition of forcing a negation can be easily recovered, we have that $\forall q\leq p; q\not\Vdash\exists x;(x\in a\cap c)$. Since $D$ is dense, there is a $q\leq p$ such that $q\in D$. Fix such a $q$. Now, $q\in D$, so $\forall F\ni q; M[F]\models q\in D\cap F$. Therefore, $\forall F\ni q; M[F]\models \exists x; (x\in D\cap F)$ and $q\Vdash \exists x; x\in a\cap c$, a contradiction.<|endoftext|> TITLE: Bimodules of fractions QUESTION [5 upvotes]: This question is about extending modules of fractions to bimodules of fractions. I would not be surprised if the result is known, I have tried looking in Goodearl and Warfield, but may have missed the point as I am not an expert in algebra. The motivation lies in looking at invertible bimodules in non commutative algebraic geometry. Given sufficient conditions, for a ring $R$ and a multiplicative set of regular elements $X$ we can form the ring of fractions $RX^{-1}$. Now for an right $R$ module $E$, we can form the right $RX^{-1}$ module $E \otimes_R RX^{-1}$. However suppose that $E$ is actually a $R$-bimodule. When is $E \otimes_R RX^{-1}$ then a $RX^{-1}$ bimodule? As I see it, this question can be phrased in the same sort of terms as the Ore condition on the ring, that is given $x\in X$ and $e\in E$ is there $y\in X$ and $f\in E$ so that $ey=xf$. Is this the right approach? If so, are there general conditions saying when this construction works? Where can I find out about it? Apologies for my lack of knowledge here, I hope that this does not sound too trivial to the experts. The idea is to apply it to localisation of bimodules over quantum groups and their homogenous spaces. Edit: I should say that the reason why I want $E \otimes_R RX^{-1}$ rather than something potentially bigger to be the bimodule is that otherwise taking fractions will not commute with tensor product, or rather we will likely not get a monoiodal functor between bimodule categories. REPLY [3 votes]: Suppose $R,S,T$ are rings. Let $_RM_S$ and $_SN_T$ be bimodules. The tensor product $M\otimes_S N$ is naturally an $R$-$T$-bimodule. The "middle" $S$-structure is gone. In your case, $E$ is an $R$-$R$-bimodule, and you are viewing $RX^{-1}$ as an $R$-$RX^{-1}$-bimodule. That makes $E\otimes_R RX^{-1}$ naturally into an $R$-$RX^{-1}$-bimodule. The left $R$-module structure is unaffected, and there is no natural way to make $E\otimes_R RX^{-1}$ have a left $RX^{-1}$ structure. Of course, the double tensor $RX^{-1}\otimes_R E\otimes_R RX^{-1}$ does have such a structure. To justify my "no natural" claim, take $R=E=\mathbb{Z}[x]$. Define a right $R$-module structure on $E$ in the obvious way. Define a left $R$-module structure on $E$ by letting $\mathbb{Z}$ act normally but left multiplication by $x$ acts like multiplying by $0$. This makes $E$ an $R$-$R$-bimodule. Now taking $X$ to be the set of all regular elements of $R$, we have $S:=RX^{-1}=\mathbb{Q}(x)$ and $E\otimes_R S\cong R\otimes_R S\cong S=\mathbb{Q}(x)$. This is now an $R$-$S$-bimodule, with $x$ acting as $0$ from the left. If we try to invert the $0$-multiplication, we get the zero module. There is no way to make the non-zero module $E\otimes_R S$ have a left $RX^{-1}$ structure, which extends its given left $R$-module structure.<|endoftext|> TITLE: A generalization of holomorphic functions QUESTION [6 upvotes]: Assume that $U$ is an open set in the complex plane $\mathbb{C}$ and $A$ is a real $2\times 2$ matrix. We define $$\mathcal{S}_{A}=\{f:U\to \mathbb{C}\mid Df.A=A.Df \}$$ where $Df$ is the $2\times 2$ Jacobian matrix of smooth function $f:U\to \mathbb{R}^{2}\simeq \mathbb{C}$. Every function with this property is called an $A$-holomorphic function. Example: When $J=\begin{pmatrix} 0&-1\\1&0\end{pmatrix}$, $\mathcal{S}_{J}$ is the space of (usual) holomorphic functions. Assume that this vector space $\mathcal{S}_{A}$ is closed under "uniform convergence on compact subsets of $U$". That is: Assume that $f_{n}$ is a sequence in $\mathcal{S}_{A}$ and $f_{n}$ converges to $f$, uniformly on compact subsets of $U$. Then $f$ must be in $\mathcal{S}_{A}$. Does this imply that $A$ is in the form $$A=\begin{pmatrix} a&-b\\b&a \end{pmatrix}$$ The question is motivated by the fact that the uniform limit of a sequence of holomorphic functions is a holomorphic function. REPLY [4 votes]: Simple counterexample: Suppose $f(x,y) = u(x,y) + i v(x,y)$ is a holomorphic function. Then it satisfies the Cauchy-Riemann equations \begin{align} u_x = v_y\\ v_x = -u_y \end{align} Then the function $g(x,y) = \tilde{u}(x,y) + i \tilde{v}(x,y)$ given by \begin{align} \tilde{u}(x,y) = \frac12 u(2x,y) \\ \tilde{v}(x,y) = v(2x,y) \end{align} satisfies \begin{align} \tilde{u}_x = \tilde{v}_y \\ \tilde{v}_x = -4 \tilde{u}_y \end{align} which is the same as $ A.D g = D g.A $ for $$ A = \begin{pmatrix} 0 & -1/4 \\ 1 & 0 \end{pmatrix} $$ This example is obtained geometrically by doing a linear change of variables $(x,y) \mapsto (2x,y)$ on the domain and $(u,v) \mapsto (\frac12 u,v)$ on the co-domain and so all convergence and closure of holomorphic functions carry over unchanged. More generally, the same procedure shows that for any $A$ of the form $$ A = c P J P^{-1} $$ where $P\in GL(2)$ and $c\in \mathbb{R}_+$ the closure property asked for in the question holds. [Edit to answer this question]: The correct way to think about the question is that the two copies of $A$ in the equation $A (Df) = (Df) A$ are not the same. One is the complex structure on the domain and the other is the complex structure on the codomain. If you conjugate the complex structure on the domain side ONLY what you get is $$ P A P^{-1} (Df) = (Df) A$$ where the second $A$ is unchanged because it is the complex structure on the codomain. To make the equation work as written, we then do the reverse conjugation on the complex structure of the codomain which changes the equation to $$ PA P^{-1} (Df) = (Df) P A P^{-1}$$ The action on the codomain side is the reverse (given by the inverse matrix) simply because covariance versus contravariance in the equation.<|endoftext|> TITLE: Seeking very regular $\mathbb Q$-acyclic complexes QUESTION [10 upvotes]: This question was raised from a project with Nati Linial and Yuval Peled We are seeking a $3$-dimensional simplicial complex $K$ on $12$ vertices with the following properties a) $K$ has a complete $2$-dimensional skeleton (namely, every triple of vertices form a 2-face of $K$) and it has $165$ $3$-simplices. b) Every $2$-face belongs to three $3$-faces c) Every $1$-face belong to fifteen $3$-faces d) Every vertex belongs to fifty five $3$-faces. e) $H_3(K,\mathbb Z)=0$. While all these properties are little negotiable we can ask even that $K$ would admit further interesting symmetries and regularity properties. By property 2, the list of $3$-faces of $K$ is a $3$$-$$(12,4,3)$ design. So perhaps such an example can be found among the known designs. We can hope, more geneally, for $\mathbb Q$-acyclic $k$-dimensional simplicial complexes $K$ on $n=k(k+1)$ vertices with complete $(k-1)$-skeleta and with similar regularity properties. Namely, $K$ must have ${{n-1}\choose {k-1}}$ $k$-faces, and we can demand that every $(k-1)$-face belongs to $k$ $k$-faces, and more generally that every $i$-face $i TITLE: How often does a pair of linear maps generate a Zariski-dense subgroup of $GL(d,\mathbb{R})$? QUESTION [6 upvotes]: I am an analyst working on a number of problems which in some way relate to random matrix products. In this context I frequently find that the analytic properties I am interested in depend in some way on an algebraic equation not being satisfied by the elements of a finitely-generated matrix semigroup. My knowledge of algebraic geometry is very basic, so I am not even sure whether or not the following question is difficult, but I would find an answer extremely useful: How large is the set of pairs $(A,B) \in GL(d,\mathbb{R})^2$ such that the group generated by $A$ and $B$ is Zariski dense in $GL(d,\mathbb{R})$? Specifically, does it include an open dense set in the standard topology? Is it residual? Does it have full Lebesgue measure? All else being equal, I would especially appreciate a "crisp" reference which is easily understood by readers without specific expertise in algebraic geometry. Thanks in advance! REPLY [10 votes]: The collection $A,B$ which generate a Zriski dense subgroup contains a Zariski open set (I should add that a Zariski oepn set is of full measure in $GL(n,\mathbb{R})\times GL(n,\mathbb{R})$). The set of elements $A$ of $G=GL(n,\mathbb {R})$ which have distinct complex eigenvalues $a_i$ such that for different pairs of indices $(i,j)$ and $(k,l)$ the values $a_ia_j^{-1}$ and $a_ka_l^{-1}$ are different form a Zariski open set (regular semi-simple elements form a Zariski open set, and these extra conditions ensure that under the adjoint reepresentation, $Ad(A)$ has as many distinct eigenvalues as are allowed. We assume that $A$ and $B$ have infinite order, since an element of $SO(n)$ will have eigenvalues which are inverses of each other, and generic elements of $GL(n)$ can be chosen to avoid this (assume $n\geq 3$). Hence the connected component of the group gen by $A$ (or by $B$) is not identity and its lie algebra $\mathfrak{T}$ will be non-central in $\mathfrak{gl_n}$. Suppose that $G$ is the Zariski closure of the group generated by $A$ and $B$. Fix a maximal torus of $GL(n,\mathbb{R})$ (up to conjugacy, there are only finitely many). We view $B$ as an element $gCg^{-1}$ for some $g\in GL(n)$ and some $C \in T$. The Lie algebra of $G$ splits into eigenspaces for $Ad(A)$; the condition that $all$ non-trivial eigenspaces are non-zero gives an open condition on the set of $B$'s: the projection of $g\mathfrak{T}g^{-1}$ onto these non-trivial eigenspaces is nonzero for generic $g$. and hence the group generated by $A,B$ will have Lie algebra $\mathfrak{g}$ which contains all the $E_{ij}$ and hence contains $\mathfrak{sl_n}$. Therefore, $G=GL_n$ (if we assume that $(det A)^2$ is not equal to one).<|endoftext|> TITLE: Forcing PFA with ccc forcing QUESTION [14 upvotes]: Is it consistent (from suitable large cardinals) that there is a ccc poset which forces PFA? This seems quite implausible to me. If we could force PFA via ccc forcing, the ground model would have to be quite close to a model of PFA (having the correct continuum, no squares, SCH holding etc.). However, the ground model cannot be a model of full PFA (or even BPFA), since it follows from a result of Caicedo and Veličković that any ccc forcing over such models destroys BPFA. The reference for the Caicedo-Veličković result is Andrés Eduardo Caicedo and Boban Veličković, The bounded proper forcing axiom and well orderings of the reals, Math. Res. Lett. 13 (2006), no. 3, 393--408. (link) They show that if $V$ and an inner model $M$ agree on $\omega_2$ and both satisfy BPFA then they have the same subsets of $\omega_1$. The conclusion above follows since any ccc forcing adds a subset of $\omega_1$. REPLY [13 votes]: I think a negative answer can be derived from the following observations. One. A nontrivial c.c.c. forcing adds a subset of $\omega_{1}$ (consider the least cardinal $\kappa$ for which it adds a subset of $\kappa$, and the tree of possible initial segments for this subset; the splitnodes in the tree give rise to a name for a new subset of $\omega_{1}$). Two. A c.c.c. partial order $P$ has the property that for any ordinal $\delta$, if in a $P$-extension one has a continuous $\subseteq$-increasing chain $\bar{N} = \langle N_{\alpha} : \alpha < \omega_{1} \rangle$ of countable subsets of $\delta$ with union $\delta$, then for some club $C \subseteq \omega_{1}$ in the ground model, $\langle N_{\alpha} : \alpha \in C \rangle$ exists already in the ground model. Now if we look at Justin Moore's MRP coding for subsets of $\omega_{1}$ as in Section 4 of this paper (http://arxiv.org/pdf/math/0501526v1.pdf) we see that every subset of $\omega_{1}$ coded in a $P$-extension by such a sequence $\bar{N}$ would be coded already by restriction of $\bar{N}$ in the ground model.<|endoftext|> TITLE: Is $S^1\vee S^1$ an Eilenberg-Mac Lane Space to a Homotopy Purist? QUESTION [7 upvotes]: I believe that $S^1\vee S^1$ is the Eilenberg-Mac Lane space $K(\mathbb{Z}\ast\mathbb{Z},1)$. One can prove this by constructing its universal cover and observing that it is contractible. My question is this: Is there a simple homotopy-theoretic proof of this result that does not use connected covers? I have never seen one appear in any text. Repeated attemps by myself to use the usual tools (Hilton-Milnor Theorem, examine the fibre of the inclusion $S^1\vee S^1\hookrightarrow S^1\times S^1$, etc...) to prove this elementary result have met with connectivity issues or similar problems. REPLY [6 votes]: I think the following can be turned into a proof, but I haven't checked the details. By a result of Milnor, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $F(S^0 \vee S^0)$, the free group functor on the pointed set $S^0 \vee S^0$. By a version of the Hilton-Milnor theorem, the latter coincides up to homotopy with $F(S^0) \ast F(S^0)$, the latter being the free product with amalgamation of simplicial free groups. But $F(S^0)$ coincides with $\Bbb Z$ up to homotopy. Hence, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $\Bbb Z \ast \Bbb Z$ (as a space). In particular, $\Omega (S^1 \vee S^1)$ is homotopically discrete, which gives the result you are after (in conjunction with the van Kampen theorem).<|endoftext|> TITLE: What ccc forcings add a Suslin tree? QUESTION [14 upvotes]: In a comment to Miha's question in Forcing PFA with ccc forcing, I suggested that if such situation is even possible, it might be achieved by screwing with PFA by some ccc forcing (e.g. adding a Cohen real), and then "fixing all the problems" via a ccc forcing. And that raises the question: Apart of adding a Cohen real, what sort of ccc forcing adds a Suslin tree on $\omega_1$? Of course, take any ccc forcing $\Bbb P$, and consider $\Bbb P\times\operatorname{Add}(\omega,1)$ is an answer. So to avoid this triviality, we only consider forcings which do not add a Cohen real themselves. So we can rephrase the question: Suppose that $\Bbb P$ is a ccc forcing which adds a Suslin tree on $\omega_1$. Does $\Bbb P$ add a Cohen real? If the answer is negative (namely it is possible to add a new Suslin without adding a Cohen), does the answer change if we assume the ground model has no Suslin trees? REPLY [9 votes]: It is consistent that the answer is no. If we start with $L$ as our ground model then whenever $T$ is a Suslin tree, the forcing $\mathbb{P}_T$ which shoots a branch through $T$ will always introduce new Suslin trees. Thus in $L$, $\mathbb{P}_T$ is an example of a forcing which has the ccc, adds Suslin trees and does not add a Cohen real. To see that $\mathbb{P}_T$ adds a new Suslin tree, note first that whenever $G$ is $L$-generic for $\mathbb{P}_T$, $\diamondsuit$ still holds in the extension $L[G]$: if $(a_{\alpha} \, : \, \alpha < \omega_1)$ is $\diamondsuit$-sequence in $L$ then $(a_{\alpha}^G \, : \, \alpha < \omega_1)$ will be a $\diamondsuit$-sequence in $L[G]$ as can be seen using the ccc of $\mathbb{P}_T$ and the fact that $\mathbb{P}_T$ has size $\aleph_1$. Next observe that $\mathbb{P}_T$ will add a new subset of $\omega_1$ and hence a new stationary subset of $\omega_1$ by a theorem of Solovay, saying that each stationary subset of $\omega_1$ can be partitioned into $\omega_1$-many stationary sets. Finally fix a new stationary subset $B \subset (\omega_1 \cap Lim)$ and, working in $L[G]$, use the restricted diamond sequence $\diamondsuit_B := (a_{\alpha}^G \, : \, \alpha \in B)$ to define a new Suslin tree $S$: Build the tree via induction on its height. If $\alpha$ is a limit ordinal then let $S_{\alpha}:= \bigcup_{\beta< \alpha} S_{\beta}$. If $\alpha$ is a double successor ordinal let $S_{\alpha+1}$ be the extension of $S_{\alpha}$ which adjoins to every top level $x \in S_{\alpha}$ infinitely many immediate successors. If $\alpha$ is limit we distinguish two cases in the definition of $S_{\alpha+1}$. If $\alpha \notin B$ let $S_{\alpha+1}$ be the $<_L$-least normal extension of $S_{\alpha}$ (note here that we can use the $L$-wellorder as $\mathbb{P}_T$ is $\omega$-distributive, hence no new countable sets are added by $\mathbb{P}_T$). If $\alpha \in B$ and $a_{\alpha}^G$ is a maximal antichain in $S_{\alpha}$ let $S_{\alpha+1}$ be the $<_L$-least normal extension of $S_{\alpha}$ such that $a_{\alpha}^G$ remains maximal in $S_{\alpha+1}$. And finally if $\alpha \in B$ but $a_{\alpha}^G$ is not a maximal antichain in $S_{\alpha}$, let $S_{\alpha+1}$ be a normal extension of $S_{\alpha}$ which is not the $<_L$-least. The standard argument shows that $S:= \bigcup_{\alpha < \omega_1} S_{\alpha}$ is a Suslin tree. However the way we defined $S$ ensures that $S$ can not be in $L$.Indeed if we assume that $S \in L$ then one could define $B \subset \omega_1$ via just looking at the stages of $S$ where $S_{\alpha+1}$ is not the $<_L$-least normal extension of $S_{\alpha}$. Thus if $S \in L$, $B$ would be in $L$ as well which is a contradiction.<|endoftext|> TITLE: Nuances Regarding Naturality QUESTION [20 upvotes]: It's frequently said, informally, that a natural isomorphism is one that doesn't depend on arbitrary choices. But the phrase "arbitrary choices" lends itself to different interpretations. Consider the following examples, all concerning a finite-dimensional vector space $V$ over a field $k$: A. The natural isomorphism $V\rightarrow V\otimes_kk$ (by which I mean the obvious natural transformation from the identity functor to the functor $ -\otimes_kk$) is defined by a basis-free formula $v\mapsto v \otimes 1$, and is easily proved to be an isomorphism without ever invoking the existence of a basis. B. The natural isomorphism $V\rightarrow V^{**}$ is defined by a basis-free formula $v\mapsto \Big(\phi\mapsto \phi(v)\Big)$, but to prove this is an isomorphism, it seems I have to choose a basis. Likewise for the natural isomorphism $V^*\otimes_k V\mapsto Hom_k(V,V)$. C. The natural isomorphism $Hom_k(V,V)\mapsto V^*\otimes_kV$ cannot, I think, even be defined by any basis-free formula. To define the map, one first chooses a basis and then proves that the definition is independent of that choice. Is there a more formal way to distinguish such cases? I'm looking for a precise definition of what it means to be "definable by a basis-free formula" or "provably an isomorphism without ever mentioning a basis" --- precise enough so that it's possible to prove (not just suggest, as above) that a given isomorphism satisfies one, the other, both, or neither of these conditions. (I suppose that the business about "provably an isomorphism without ever mentioning a basis" is perhaps no more interesting than "provably an isomorphism without ever using $X$", where $X$ is some other known property of vector spaces. But the business about being "definable by a basis-free formula" seems to me to be more interesting.) Edited to add: I appreciate the responses, but I fear the main thing I've learned from them is that I failed to make the question clear. This wasn't intended as a question about vector spaces; the vector spaces were meant to illustrate something more general. I quite understand the reasons why the map $V^{**}\rightarrow V$ can't be constructed without reference to a basis (most notably, this map doesn't exist in the infinite dimensional case, so there's no hope to construct it without using the finite dimensionality, which means referring to a basis). The question is not why this is true but how to state it in a rigorous way. How, formally, and in more general contexts than vector spaces, does one formalize the notion of "not constructible without making some choices that turn out to be irrelevant at the end"? REPLY [3 votes]: Here are two examples based on sets and one more. Again a long comment. 1) For a set $S$ let $\mathcal{L}$ be the set of linear orders and $\mathcal{P}$ be the set of permutations. There is an easy proof that there is a bijection between these two things, but it depends on choosing an arbitrary but fixed distinguished linear order $\ell$ and composing the obvious bijections $\mathcal{L} \rightarrow \ell \times \mathcal{L} \rightarrow \ell \times \mathcal{P} \rightarrow \mathcal{P}$. This seems unavoidable as there is one, but only one, natural distinguished permutation (once there are three elements) but no distinguished linear order. Alternately, the orbit structure of the natural action of $\operatorname{SYM}_S$ on $\mathcal{L}$ and on $\mathcal{P}$ is not the same. I stated it that way to highlight the natural bijection between $\mathcal{L} \times \mathcal{L}$ and $\mathcal{L} \times \mathcal{P}$ which is a projection on the first component and, for each fixed first component, a bijection between $\mathcal{L}$ and $\mathcal{P}.$ That seems like a candidate for a proof that the set of bijections is non-empty. Of course it is a little problematic if $S$ might or might not be $\emptyset.$ 2) Consider the claim that there is a bijection between the even and odd cardinality finite subsets of a set $S.$ The obvious proof depends on choosing an arbitrary but fixed element $s$ and using it to partition the finite subsets into pairs $\{{A,A \oplus \{{s\}}\}}$. Since the claim is true precisely when $S \ne \emptyset,$ this seems unavoidable. 3) This one I am less sure of. The point is that something like a Hamel basis or an everywhere dense non-measurable set seems needed, but the result feels more discrete and concrete than Vitali Sets. Call a real triple of the form $T_x=\{{x,x+1,x+\sqrt{2}\}}$ an ell. I want to claim that $\mathbb{R}$ can be partitioned into disjoint ells (call this a $1$-cover). Alternately, but perhaps not equivalently, there is a subset $X \subset \mathbb{R}$ such that $X,X+1$ and $X+\sqrt{2}$ constitutes a partition of $\mathbb{R}$ into three disjoint congruent sets. The set of all ells is a $3$-fold cover of $\mathbb{R}.$ I will give a natural partition of $\mathbb{R}$ into sets called boards such that there is an explicit and unique set of three $1$-covers for each board. Then a $1$-cover of $\mathbb{R}$ is precisely a selection of one $1$-cover (from the possible three) for each board. It seems pretty natural that the following are equivalence relations in $\mathbb{R}$. $x \sim y$ when $x-y \in \{{a+b\sqrt{2} \mid a,b \in \mathbb{Z}\}}$ $x \approx y$ when $x-y \in \{{a+b\sqrt{2} \mid a,b \in \mathbb{Z} \text{ and }a\equiv b \bmod 3\}}.$ This does not require extending $\{{1,\sqrt{2}\}}$ to an explicit basis of $\mathbb{R}$ over $\mathbb{Z}.$ Note that an equivalence class is everywhere dense. Call an $\sim$-equivalence class a board. Also, consider the equivalence relations on the set of ells defined by $T_x \sim T_y$ and $T_x \approx T_y$ when $x \sim y$ and $x \approx y.$ Each ell lies entirely in one board and its $\approx$-equivalence class is the unique $1$-cover of that board including that ell. A $\sim$-equivalence class constitutes a $3$-cover of its board and it resolves into three $\approx$-equivalence classes which are the only $1$-covers of that board.<|endoftext|> TITLE: Some elementary questions about deformation quantization QUESTION [5 upvotes]: I am interested in deformations of affine Poisson algebras, and so this is the setting in which I shall write out the elementary definitions involved. All algebras and vector spaces shall be over $\mathbb{C}$ although most of my questions make sense over any field. Any answers to any part of my question will be very welcome indeed, thanks in advance! An affine Poisson algebra $A$ is a finitely generated, reduced, commutative algebra with a bracket $\{ \cdot, \cdot\} : A \otimes A \rightarrow A$ which is skew-symmetric, satisfies the Jacobi identity and such that for all $f \in A$ the map $\{f, \cdot\} : A \rightarrow A$ is a derivation. I am aware of two broad notions of deformation. Filtered deformations arise when $A$ is also graded $A = \bigoplus_{i \geq 0} A^i$ and the Poisson bracket is homogeneous of some fixed degree $d$, meaning $\{ \cdot, \cdot\} : A^i \otimes A^j \rightarrow A^{i + j - d}$ for all $i, j$. If $B$ is a non-commutative, filtered algebra with $B = \bigcup_{i \geq 0} B_i$ such that $[B_i, B_j] \subseteq B_{i + j-d}$ then $\operatorname{gr} B$ inherits a Poisson structure by setting $$\{ f, g\} := [\overline{f}, \overline{g}] + B_{\deg f + \deg g - d - 1}$$ for homogeneous elements $f, g \in \operatorname{gr} B$ where $\overline{f}, \overline{g}$ are lifts of $f,g$ to $B$. If $\operatorname{gr} B \cong A$ as Poisson algebras then we say that $B$ is a filtered deformation of $A$. A closely related notion of deformation involves star products. Suppose that $A$ is a Poisson algebra as above. Then a $\ast$-product is an associative $\mathbb{C}[[\hbar]]$-linear product on $A[[\hbar]]$ which is written $$f \ast g = fg + \hbar B_1(f, g) + \hbar^2 B_2(f,g) + \cdots$$ where $f, g \in A$ and $\ast$ satisfies $f \ast g - g \ast f = \hbar\{f, g\} + \mathcal{O}(\hbar^2)$. The algebra $(A[[\hbar]], \ast)$ is then referred to as a deformation quantisation of the Poisson algebra $(A, \{ \cdot, \cdot\})$. Kontsevich famously defined a $\ast$-product on the $\mathcal{C}^\infty(M)$ where $M$ is a Poisson manifold, thus showing that deformation quantisations always exist in this case. Smooth affine Poisson varieties are special cases of these and I believe that the ring of regular functions can be endowed with a $\ast$-product à la Kontsevich. I hope everything that I've written thus far is clear and correct. Question 1: what is the precise connection between these two definitions of quantisation? Subquestion 1a: When $B_i(f,g) = 0$ for $i \gg 0$ we can define $f \ast g|_{\hbar = 1}$ and if this holds for all $(f,g) \in A\times A$ we can define a product $\ast_{\hbar = 1} : A\otimes A \rightarrow A$. Does Kontsevich's $\ast$-product have this vanishing property (when $\operatorname{Spec} A$ is smooth)? Subquestion 1b: If $(A, \{\cdot,\cdot\})$ is graded then is there automatically a filtration on $(A, \ast_{\hbar = 1})$ such that the associated graded is $(A, \{\cdot, \cdot\})$? Subquestion 1c: If the Poisson structure on $A$ is not graded then how is the $\hbar = 1$ product on $A$ related to the Poisson structure? Is there something more general that a filtered deformation coming from this construction and can it be axiomatised? Kontsevich's $\ast$-product is a consequence of his formality theorem, and so the choice of $\ast$-product is only well-defined up to so-called gauge equivalence, but gauge equivalent products subtend isomorphic algebras. Question 2: Is isomorphisms a strictly finer invariant than gauge equivalence? This can be asked in two settings: Subquestion 2a: If $(A[[\hbar]], \ast) \cong (A[[\hbar]], \ast')$ then does it follow that the structures are gauge equivalent? Subquestion 2b: If $(A, \ast_{\hbar = 1}) \cong (A, \ast'_{\hbar = 1})$ then does it follow that the structures are gauge equivalent? Finally I would like to know something about quantising Poisson subvarieties. First comes the general question: Question 3: Is Kontsevich's quantisation functorial in any sense? Now comes the more specific version of this question. Suppose that $I \unlhd A$ is a Poisson ideal, so that $A/ I$ is a Poisson algebra and the projection $A \rightarrow A/I$ is a Poisson map. Kontsevich's quantisation gives products $(A[[\hbar]], \ast)$ and $(A/I[[\hbar]], \ast')$. Subquestion 3a: Do we have a surjection $(A[[\hbar]], \ast) \twoheadrightarrow (A/I[\hbar]], \ast')$? And finally the $\hbar = 1$ version: Subquestion 3b: Do we have a surjection $(A, \ast_{\hbar = 1}) \twoheadrightarrow (A/I, \ast'_{\hbar=1})$? Again, thanks in advance for any comments, answers, advice... anything at all really. REPLY [8 votes]: a lot of questions, let me try on some of them :) The bad news is that in most of the interesting situations the higher order terms of the star product, the $B_i$ will not vanish. Heuristically this can be understood as follows (and can be made precise in many cases): $B_1$ is a bidifferential operator of order at least one, since it has to produce the Poisson bracket as antisymmetric term. Now the associativity failure if you only have $B_1$ is a composition of two $B_1$'s hence generically a tridifferential operator of order $2$. Thus to compensate this you need the $B_2$ (since you want something associative!) which then enters the associativity condition together with $B_0$, i.e. the multiplication. But $B_0$ is bidifferential of order zero, so you need a bidifferential operator $B_2$ in order to produce terms with $2$ derivatives, in order to compensate the $B_1$ terms. This continues by induction. So whenever the bidifferential operators you have already found up to $B_k$ are of order at least $k$, then $B_{k+1}$ tends to be of order $k+1$. "Tends to be" can be made precise, say in the case of symplectic Poisson structures, where the bidifferential operators are sufficiently nondegenerate. The zero Poisson bracket $B_1 = 0$ of course needs no higher order to fix associativity :) But you see that you will typically need infinitely many contributions to get associativity. This makes 1b and 1c sort of obsolete. Of course there might be situations where you only have finitely many $B_i$ different from zero, but I have not yet seen examples (being, of course, only a proof of my limited experience). However: one can of course require honest convergence of the formal series. This is an ongoing huge project of several people (including myself) and by far not well-understood. One enters the realm of strict deformation quantization, rewarding, challenging, but also quite trickey... For 2: the difference between gauge transformations (or equivalence transformations) and isomorphisms is that an isomorphism is of the form \begin{equation} \Phi = \Phi_0 + \hbar \Phi_1 + \hbar^2\Phi_2 + \cdots \end{equation} with linear maps $\Phi_k\colon A \longrightarrow A$, extended $\hbar$-linearly. In this case, $\Phi_0$ is easily shown to be an isomorphism of the underformed algebra $A$. For an equivalence transformation you require $\Phi_0 = id$, thus having slightly less freedom. In case of, say, symplectic star products, one can furthermore show that $\Phi_0$ is the pullback with a symplectic diffeo. But if this diffeo is large enough, it can act nontrivially on the second deRham cohomology. If your characteristic class of the star product is now not invariant under $\Phi_0$ then such an isomorphism really changes the class with respect to equivalence transformations. For 3: this is a long long story. In short: no. In geometric situation the question is probably not even that relevant since there are not too many Poisson ideals. They would correspond to Poisson submanifolds and your functoriality would sort of imply that the big star product is tangent to the submanifold. This is known to be false in most cases, say for linear Poisson structures (see a paper by Cahen, Gutt, Rawnsley?) More important is the question of a Poisson subalgebra which is a associative ideal, i.e. a coisotropic submanifold. They are used to phase space reduction and produce reduced Poisson algebras. Now it is a quite nontrivial question to characterize those coisotropic submanifolds for which you can perform reduction. In general: no, but there are many favorable positive situations. Here papers pf Bordemann or Cattaneo-Felder might give you some inspiration. In the symplectic case there is also a very old approach by Bordemann, Herbig and myself using some BRST like construction. Again 3b is beyond hope.<|endoftext|> TITLE: Balanced Tensor Product of Module Categories QUESTION [6 upvotes]: (Moved from MSE) Let $\mathcal{C}$ be a $k$-linear ($\operatorname{Vect}_k$-enriched) monoidal category and consider the 2-category $\operatorname{Mod}_\mathcal{C}$ of $k$-linear $(\mathcal{C}, \mathcal{C})$-bimodule categories in the sense of Ostrik (https://arxiv.org/abs/math/0111139). Roughly speaking, this construction can be thought of as a categorification of a module over a ring. Let $\otimes$ denote the tensor product of $k$-linear $(\mathcal{C}, \mathcal{C})$-bimodule categories (as defined in the work of Douglas, Schommer-Pries, and N. Snyder in https://arxiv.org/abs/1406.4204). Does $\otimes$ endow $\operatorname{Mod}_\mathcal{C}$ with the structure of a monoidal 2-category? REPLY [6 votes]: Yes, but I don't believe there is a paper on arXiv containing all details (I would love to be corrected!), and if I were refereeing a paper claiming it to be so, I would give the author a hard time. Edit: Per Matthew Titsworth's comment below, Gregor Schaumann's thesis contains these details. Original post: The balanced tensor product is defined by universal property, and this makes everything work out. Let $\mathcal M_n$ denote the bicategory whose objects are $n$-simplices with the following decorations: On each edge (with endpoints $i TITLE: Where am I suppose to actually learn how to compute hypercohomology? QUESTION [16 upvotes]: I'm reading about algebraic de Rham cohomology over characteristic zero which is constructed using hypercohomology. Already, constructing injective resolutions is difficult, and coupling this with finding an injective resolution of the de Rham complex makes this extraordinarily difficult to figure out how to compute an example. Where can I learn about computational techniques? For example, how can I compute the algebraic de rham cohomology of the cubic surface $$ \textbf{Proj}\left(\frac{\mathbb{C}[x,y,z,w]}{(x^3 + y^3 + z^3 + w^3)}\right) $$ ? REPLY [22 votes]: Since you are asking about computing algebraic de Rham, don't use injective resolutions, since they are not really constructive. You can use a Cech complex: If $\{U_i\}$ is an affine open cover of your variety $X$, form a double complex $C^{\bullet\bullet}=C^\bullet(\{U_i\}, \Omega_X^\bullet)$ with Cech coboundary in one direction, and exterior derivative in the other direction. Then $$H^i_{dR}(X) \cong H^i(Tot(C^{\bullet\bullet}))$$ Depending on the example, this may not be a bad way to proceed. If you want a quick and dirty approach for your specific example: use Grothendieck's algebraic de Rham theorem to rewrite it in terms of singular cohomology. Then you have access to all the standard exact sequences. You can also look at Hartshorne's paper in IHES on de Rham cohomology, where he reproduces most of these standard results by algebraic arguments. Using the fact that a smooth cubic surface is the blow up of $\mathbb{P}^2$ at $6$ points, the Betti numbers are 1,0,7,0,1,0...<|endoftext|> TITLE: Legendre Polynomial Integral QUESTION [7 upvotes]: How can I evaluate $$ \int_{-1}^1 P_n(x)P_l(x)x^k dx $$ when $k$ is even? Or what might be a source where I could find integrals like this? REPLY [7 votes]: The integral $$\int\limits_0^1 x^k P_m(x)P_n(x)dx$$ is evaluated in terms of the hypergeometric function $_3F_2$ in http://link.springer.com/article/10.1007/BF01650571 (Some integrals containing products of legendre polynomials, by L. Carlitz). The evaluation is based on $$P_m(x)P_n(x)=\sum\limits_{r=0}^{\mathrm{min}(m,n)}\frac{A_rA_{m-r}A_{n-r}}{A_{m+n-r}}\,\frac{2m+2n-4r+1}{2m+2n-2r+1}P_{m+n-2r}(x),$$ with $$A_r=\frac{(2r-1)!!}{2^rr!},$$ and $$\int\limits_0^1 x^k P_n(x)dx=\frac{k(k-1)\cdots(k-n+2)}{(k+n+1)(k+n-1)\cdots(k-n+3)}.$$ The following formulas $$\int\limits_{-1}^1 x P_m(x)P_n(x)dx=\left\{\begin{array}{cc}\frac{2(m+1)}{(2m+1)(2m+3)}, & n=m+1, \\ \frac{2m}{(2m-1)(2m+1)}, & n=m-1,\end{array}\right . $$ and $$\int\limits_{-1}^1 x^2 P_m(x)P_n(x)dx=\left\{\begin{array}{cc}\frac{2(m+1)(m+2)}{(2m+1)(2m+3)(2m+5)}, & n=m+2, \\ \frac{2(2m^2+2m-1)}{(2m-1)(2m+1)(2m+3)}, & n=m, \\ \frac{2m(m-1)}{(2m-3)(2m-1)(2m+1)}, & n=m-2,\end{array}\right . $$ are given as an exercise 12.9.2 (p. 806) in the book Mathematical Methods for Physicists, 6th Edition, by G.B Arfken and H.J. Weber: https://www.amazon.com/Mathematical-Methods-Physicists-George-Arfken/dp/0120598760<|endoftext|> TITLE: Connection between PageRank and Fiedler vector QUESTION [6 upvotes]: This question is on graph clustering. In its simplest form, the eigenvector corresponding to the second smallest eigenvalue of the normalized Laplacian of a graph provides a relaxed solution to the problem of partioning the graph into two sets of vertices so that there are few connecting links between the two sets(mincut problem). However there are also works that use the personalized pagerank of a graph starting from a certain vertex and finds a cut based on the largest components of the pagerank whose conductance is sufficiently close to the minimum possible. The work of Local Graph Partitioning using PageRank Vectors, by Reid Andersen et. al., is one such work where they derive a Cheeger inequality for the conductance of a set derived this way. I understand at least the intuition behind why the Fiedler vector works for partitioning - when the graph has two disconnected components, the second eigenvector is in the space spanned by $1_S, 1_{S^c}$ and is perpendicular to the all one vector, so it makes sense to consider this vector. The PageRank is however the left eigenvector of the largest eigenvalue of a modified Markov matrix. Is there a way to make a link between the pagerank and the Fiedler vector? As an optimisation problem, pagerank-like result arises as the solution to the Fiedler eigenvector with an additional constraint that the vector is highly correlated with a vector v, which represents prior knowledge(semi-supervised clustering) REPLY [5 votes]: The PageRank vector with seed $s$ and jumping constant $\alpha$ is by definition the solution to the equation $$pr(s,\alpha) = \alpha s + (1-\alpha)pr(s,\alpha)W$$ where $W$ is the random walk matrix $D^{-1}A$ ($D$ is the degree matrix, $A$ the adjacency matrix). Note that I'm right multiplying by $W$ here, as is conventional in the literature on this stuff. A bit of analysis shows that $pr(s,\alpha) = s R_\alpha$ where: $$R_\alpha = \alpha \sum_{i=0}^\infty (1-\alpha)^i W^i$$ I think that the best way to get intuition about the isoperimetric properties of PageRank vectors is to stare at this sum for awhile. Assume that $s$ is the indicator function of a single vertex $v$ (so that the corresponding PageRank vector is "personalized" to $v$). Then the vector $s W^i$ is a probability distribution whose entries correspond to the probabilities that an $i$-step random walk starting at $v$ will end up at other vertices in the graph. So the entries of $sR_\alpha$ can be interpreted as probabilities that "exponentially short" random walks starting at $v$ will visit other vertices. Here "exponentially short" means that we account for random walks of all possible lengths (the index $i$ ranges from $0$ to $\infty$) but we exponentially punish the long walks via the factor $(1-\alpha)^i$. Now take a graph $G$ which is connected but "almost disconnected" meaning there is a set $S$ of vertices whose isoperimetric ratio (conductance) is very small. This means that if an edge has one vertex in $S$ then its other vertex is much more likely to be in $S$ than not, and thus a short random walk which starts in $S$ is unlikely to reach $S^c$. So for many vertices in $S$ we expect that the values of $pr(s,\alpha)$ will drop off considerably (more than you would expect from looking at degrees) on vertices not in $S$. This is the basic principle that Andersen-Chung-Lang use to (nearly) recover $S$ from $pr(s,\alpha)$. Note that the underlying idea of using exponentially short random walks to locally cluster large graphs is really due to Spielman-Teng, who worked directly with random walks instead of PageRank vectors. Chung has also exploited this idea using heat kernels instead of PageRank vectors. All formulations of the idea lead to relatively similar performance and theoretical guarantees, so I think it mostly comes down to preference.<|endoftext|> TITLE: Why do we need a transitive model in forcing arguments? QUESTION [11 upvotes]: One major approach to the theory of forcing is to assume that ZFC has a countable transitive model $M \in V$ (where $V$ is the "real" universe). In this approach, one takes a poset $\mathbb{P} \in M$, uses the fact that $M$ is countable to prove that there exists a generic set $G \in V$, then defines $M[G]$ as an actual set inside $V$ and proves it is a model of ZFC. The downside to this approach is that a countable transitive model may not exist. For example, it is possible that $V = L$ and $V$ is a minimal model of ZFC, so that any smaller model $M \in V$ is non-standard. However, if we only want a countable model of ZFC, there is no problem. First, Gödel's completeness theorem shows that (assuming ZFC is consistent, of course!) there is some model $M_0 \in V$ of ZFC. Then, the Löwenheim-Skolem theorem guarantees that there is a elementary substructure $M \subseteq M_0$ which is countable in $V$. So $M$ is a countable model of ZFC, and therefore, there is a generic filter $G \in V$. Can we continue the proof of forcing along these lines? Of course, transitivity is convenient for many reasons (such as showing that various formulas are absolute), but is it possible to go without it? Perhaps we would need to modify the construction of the $\mathbb{P}$-names and $M[G]$ by only considering elements that are actually in $M$. EDIT To be a bit more clear, I believe that forcing can be done without a countable model at all, using either the syntactic approach or an approach via Boolean-valued models. My question is more humble. The arguments of forcing are very intuitive when $M$ is a countable transitive model; why don't the same (up to relativizing formulas to $M$) arguments work when $M$ is just countable? REPLY [8 votes]: This is really a comment on Hamkins answer, but I'm not permitted to make comments so I'll write it as an answer. Using the standard Schoenfield machinery for forcing, there is no need for a well founded model. The theory of forcing is developed entirely inside the model $M$, including the definition of the class of names and the forcing relation. This uses the axiom of foundation inside $M$, but not external well foundedness. Given a externally defined generic set $G$, then, $M[G]$ is just the set of equivalence classes of names under the relation $\dot x \equiv \dot y \iff \exists p\in G ( M\vDash p\vdash \dot x = \dot y)$. This is, of course, essentially equivalent to Hamkins' answer.<|endoftext|> TITLE: A mysterious quasi-isomorphism in Kashiwara-Schapira's proof of HKR QUESTION [6 upvotes]: On p. 127 of Kashiwara-Schapira's paper "Deformation Quantization Modules", there is the following situation: $X$ is a smooth complex (quasi?)projective variety and $\delta\colon X\to X\times X$ is its diagonal embedding. The goal is to prove the Hochschild-Kostant-Rosenberg theorem in this context, but I'm having trouble understanding what's going on at one particular step. So, Kashiwara-Schapira define a complex of $\mathcal O_{X\times X}$-modules by $$ P_k = \delta_*\Omega_X^k\oplus\delta_*\Omega_X^{k+1} $$ for $k\ge 0$ and $P_k=0$ for $k<0$ and with differentials being given by a composition of a projection and inclusion of factors in a direct sum. They assert that this is an exact complex away from the $0$-th term and so the canonical map $P_k\to\delta_*\mathcal O_X$ into the complex lying in degree $0$ is a quasi-isomorphism. So far so good. But then they claim that by applying functors $\delta^*$ and $H^0,$ one gets $$ \delta^*\delta_*\mathcal O_X \to H^0(\delta^*)(P_\bullet) \simeq \bigoplus\Omega_X^i[i]. $$ This is the point where I get lost. In paritcular, I don't understand where this last isomorphism comes from. I mean, if I look at $$ \delta^*P_k=\delta^*(\delta_*\Omega_X^k\oplus\delta_*\Omega_X^{k+1}), $$ I see no immediate way to relate this with $\Omega_X^k.$ I would maybe sooner expect to see $\delta^*\delta_*\Omega_X^k$ in its place? But really I'm just confused. Seeing how it isn't even remarked upon in the paper, I'm probably missing a very simple point. If somebody could spell it out for me, that would be great. REPLY [4 votes]: First, $P_k$ is not a direct sum. It is, in fact, an extension (non-trivial!) corresponding to the Atiyah class. Of course, $\delta^*P_k$ is not isomorphic to $\Omega^k$, but there is a canonical map $\delta^*P_k \to \Omega^k$ (induced by the projection $P_k \to \delta_*\Omega^k$ and the adjunction for $\delta$). Now the proof splits into two steps: a global statement, and a local statement. The global statement is the fact that the morphisms $\delta^*P_k \to \Omega^k$ combine into a morphism in the derived category $\delta^*P_\bullet \to \oplus \Omega^k[k]$. It follows from the fact that the differentials in $H^0(\delta^*)P_\bullet$ are zero. Alternatively, one can use the $\delta$-adjunction and construct a morphism $P_\bullet \to \oplus \delta_*\Omega^k[k]$. The local statement is that the morphism of complexes $\delta^*P_\bullet \to \oplus \Omega^k[k]$ is an isomorphism. It is proved by a choice of another resolution of the diagonal (that exists only locally), relating it to $P_\bullet$, and using it to compute the map. Alternatively, one could use here the spectral sequence computing the derived pullback of a complex in terms of derived pullbacks of its terms.<|endoftext|> TITLE: Higgs fields whose determinant have only simple zeros QUESTION [5 upvotes]: Is the following property true for every stable holomorphic bundle of rank 2 with trivial determinant on a compact Riemann surface: The space of trace-free Higgs fields, whose determinant have only simple zeros, is open and dense in the space of all trace-free Higgs fields. I would like to see a proof, a reference or a counter-example. Motivation: This property is certainly true for a generic stable bundle, as one might see by investigating the Hitchin fibration. REPLY [7 votes]: The statement is easily checked to be true for very stable bundles, so when you look for a counter example you will need to look for a wobbly (i.e. stable but not very stable) bundle. Suppose that $E$ is a rank two bundle with $E$ stable and $\det E = \mathcal{O}$. We are trying to decide if it can happen that for all traceless $\theta : E \to E\otimes K$ it follows that $\det (\theta)$ has a double zero. This condition looks quadratic but it is in fact linear. Given a point $x \in C$, the traceless condition tells you that $\det (\theta)$ will vanish at $x$ if and only if $\theta(x)$ is a nilpotent $K_{x}$-valued endomorphism of $E_{x}$. Furthermore $\det (\theta)$ will vanish to second order at $x$ if and only if $\theta_{x}$ is the zero endomorphism. Correction: As the OP points out this statement is incorrect. We can have a multiple zero of the determinant with $\theta_{x}$ still being regular nilpotent. A typical example is a Higgs field which locally looks like $$ \theta = \begin{pmatrix} 0 & 1 \\ z^{2} & 0 \end{pmatrix} dz $$ with $z$ being a local coordinate centered at $x$. Because of this the approach I suggested previously will not work. The condition that $\det(\theta)$ has only simple zeroes is equivalent to saying that the spectral curve for $\theta$ is smooth. In other words we want to show that the image of $$ \tag{1} \det : H^{0}(C,\text{End}_{0}(E)\otimes K) \to H^{0}(C,K^{\otimes 2}) $$ is not contained in the discriminant divisor parametrizing singular spectral curves. If $E$ is very stable, then the map (1) is surjective (and in fact is a finite morphism of degree $2^{3g-3}$). So the question is: what happens is when $E$ wobbly? When $E$ is wobbly we have a positive dimensional vector subspace $N \subset H^{0}(C,\text{End}_{0}(E)\otimes K)$ consisting of nilpotent endomorphisms. The map $\det$ induces a rational map $$ f : \mathbb{P}(H^{0}(C,\text{End}_{0}(E)\otimes K) \dashrightarrow \mathbb{P}(H^{0}(C,K^{\otimes 2})) $$ which is given by a $3g-3$ dimensional linear system of quadrics with base locus $\mathbb{P}(N)$. Using the classification of wobbly bundles you can analyze these maps. Most likely blowing up $\mathbb{P}(N)$ will resolve $f$ into a morphism and you may be able to check directly that this morphism is surjective. In fact, now that I have thought more about this, I think there is a round about way to check the statement by studying the map from a general spectral Prym to the moduli $M$ of semistable vector bundles. If you choose a smooth spectral curve $X$, then the map $\pi : \text{Prym}(X/C) \dashrightarrow M$ is dominant and finite of degree $2^{3g -3}$ over the very stable locus. The locus where $\pi$ is not defined consists of nested Brill-Noether loci parametrizing line bundles with unstable push-forward and you can resolve $\pi$ to a morphism by blowing-up these loci in a sequence starting with the smallest one (which consists of isolated points) and continuing up the nested sequence. At the end you get a morphism $\hat{\pi}$ from the blown-up Prym which is finite and of degree $2^{3g -3}$ onto $M$. The exceptional divisors map onto the wobbly locus and each exceptional divisor is contained in the ramification. You can now check that away from the semistable locus $\hat{\pi}$ is not totally ramified at any of the exceptional divisors which says that the original map $\pi$ is surjective on the stable locus. This seems to prove your statement. However this is too complicated and convoluted and I think that analyzing the Hitchin map as above may be much easier. It will take too long to write the details of the blow up of the Prym here, so if you want to know more about that send me an email and I will be happy to explain.<|endoftext|> TITLE: The set of subgroups of $F_2$ QUESTION [10 upvotes]: This question came up in our algebraic topology class and our Professor didn't know the answer. I also couldn't find an answer so far. What is the cardinality of the set of subgroups of $F_2$? Here $F_2 = \mathbb Z * \mathbb Z$ denotes the free group on two generators. The cardinality of the set of subgroups is clearly bounded below by $\aleph_0$ (as $F_2$ contains subgroups of all countable ranks) and above by $2^{\aleph_0}$. REPLY [21 votes]: It is clear that there are $2^{\aleph_0}$ subgroups of the free group $F_\infty$ on countably many generators (because each subset of a free generating set generates a different subgroup). In addition, it is well known that $F_2$ contains a subgroup that is not finitely generated. Since all subgroups of a free group are free, this subgroup is isomorphic to $F_\infty$. Then every subgroup of this $F_\infty$ is also a subgroup of $F_2$, which means that $F_2$ has $2^{\aleph_0}$ subgroups. (However, this simple argument does not prove Andy Putman's stronger statement that there are $2^{\aleph_0}$ normal subgroups.)<|endoftext|> TITLE: Numbers divisible only by primes of the form 4k+1 QUESTION [6 upvotes]: Let $A(N)$ denote the number of positive integers $n\le N$ composed of prime numbers $p\equiv 1\pmod 4$ only. Is there an asymptotic formula for $A(N)$ (as $N$ tends to infinity)? REPLY [10 votes]: Yes, $A(X) = cX/\sqrt{\log(X)} + O(X/\log^{3/2}(X))$ for a positive real number $c$ which I think is 1 (edit: This remark on the constant was just a vague recollection which is wrongly remembered as it turns out. See the comments below.). This result holds in much much more generality, e.g. for primes in congruence classes or even for positive-density subsets of primes defined by modular forms. See Theorem 2.8 of Serre's "Divisibilit\'e de certaines fonctions arithmetiques" from 1976 in L'Enseignement Mathematique. You'd replace $\sqrt{\log(X)}$ with $\log^{1-\delta}(X)$ for a more general set of primes of density $\delta$, but the primes congruent to 1 mod 4 and 3 mod 4 respectively make up density $1/2$ subsets of primes. You can also get secondary, tertiary, or as many error terms as you wish.<|endoftext|> TITLE: "Nice" basis for highest-weight irreducible module of a simple Lie algebra QUESTION [6 upvotes]: Let $\mathfrak{g}$ be a simple complex Lie algebra, $\mathfrak{h}$ a Cartan subalgebra, $\Phi \subset \mathfrak{h}^*$ the associated root system, $\Sigma = \{\sigma_i : i\in I\}$ a basis of simple roots. Further, let $\lambda$ be a dominant weight (maybe even a fundamental weight if this simplifies anything), and let $V_\lambda$ be the irreducible $\mathfrak{g}$-module with heighest weight $\lambda$. Vaguely stated, my question is: Is there a "nice" basis for $V_\lambda$, which makes it algorithmically convenient to deal with? Now I have to admit that I don't know exactly what I want of that basis, except at the very least that it has to be compatible with the weight decomposition (i.e., partitioned into sets of equal weight), that the action of $\mathfrak{g}$ on $V$ should be convenient to write (when $\mathfrak{g}$ itself is given a "nice" basis, see below), and that the symmetry of the Weyl group should be "as apparent as possible". The standard construction of $V_\lambda$ as a quotient of the corresponding Verma module does not seem to give a nice basis (the Verma module itself is reasonably transparent, but when it comes to finding a basis for the quotient, while this is computable, I can't see how to do it in a nicely symmetric way). However, for some particular $\lambda$, a nice basis is known: If $\lambda$ is the highest root, corresponding to $V_\lambda$ being the adjoint representation, then this note by Meinolf Geck describes a basis for $V_\lambda = \mathfrak{g}$ which is due to Lusztig and earlier works of Chevalley and Tits (I don't fully understand how these contributions relate, but that doesn't really matter here). A Chevalley basis consists of elements $h_i$ for $i\in I$ (all having weight $0$) and $e_\alpha$ for $\alpha\in\Phi$ (having the corresponding weights). The adjoint action is given by $[h_i,h_j] = 0$ and $[h_i,e_\alpha] = (\alpha,\sigma_i^\vee)\, e_\alpha$ and $[e_{-\alpha},e_\alpha] = \sum_{i\in I} c_i h_i$ where $\alpha^\vee = \sum_{i\in I} c_i \sigma_i^\vee$ and lastly $[e_\alpha,e_\beta] = \pm (m+1) e_{\alpha+\beta}$ if $\alpha+\beta\in\Phi$ where $m = \max\{p\in\mathbb{N}: \beta-p\alpha\in\Phi\}$ — for some choice of signs (Geck explains how it can be made canonical). Further, the symmetry of the Weyl group is made apparent by the fact that for each $i\in I$ there is an automorphism taking $h_j$ to $h_j - (\sigma_i,\sigma_j^\vee)\, h_i$ and $e_\alpha$ to $\pm e_{s_i(\alpha)}$ (where $s_i(\alpha)$ is $\alpha-(\alpha,\sigma_i^\vee)\, \sigma_i$) — again for some choice of signs —, and these generate an extension of the Weyl group by an elementary abelian $2$-group. In a related note, the same author provides a basis and construction for $V_\lambda$ when $\lambda$ is minuscule, which is very similar but even simpler. So this suggests the question of what can be said for arbitrary $\lambda$. I suppose that nothing quite so canonical or explicit is known, but maybe if we lower the bar to "nice and algorithmically convenient", something can be said? (Certainly multiple weights are problematic; but after all, tensor powers of representations with a "nice" basis themselves have a "nice" basis, so not all hope is lost.) REPLY [2 votes]: I am not an expert, but it seems that Geck's constructions specialize the Lusztig-Kashiwara "canonical" or "crystal" bases of highest weight modules. These also have a Littelmann "path model", related to the earlier Hodge-Young "standard monomial" bases (for SL$_n$).<|endoftext|> TITLE: Are there an infinite number of integers $n$ such that $n, n+1$, and $n+2$ have the same number of divisors? QUESTION [5 upvotes]: Is the set $S:=\{n\in\mathbb{N} \mid \text{$n$, $n+1$ and $n+2$ have the same number of divisors}\}$ infinite? Example: $33\in S$. REPLY [7 votes]: I will convert my comment into an answer, since I suspect it is still the state of the art. The version of your question for two consecutive integers was proved in Heath-Brown, D. R. (1984). The divisor function at consecutive integers. Mathematika, 31(01), 141-149. ISO 690 See this paper of Hildebrand (the Heath-Brown paper is behind a paywall) for an improvement of Heath-Brown's result. As Fedor Petrov mentions, your conjecture is very likely true, but is still an open problem (as far as I know). Indeed, your conjecture appears explicitly in Richard Guy's Unsolved Problems in Number Theory (Section 18B). This section also contains a lot of other related conjectures and results. For example, Erdős conjectured that for any $k$, there exist $k$ consecutive integers with the same number of divisors.<|endoftext|> TITLE: Generalisations of the Kullback-Leibler divergence for more than two distributions QUESTION [13 upvotes]: A very fundamental quantity in information theory is the Kullback-Leibler divergence between two probability distributions over the same random variable, $D_{KL}(Q\|P) = \sum_i q_i \log\frac{q_i}{p_i}$. This can be interpreted as the amount of information gained when changing from a prior $P$ to a posterior $Q$. Many if not all of the other important quantities in information theory can be seen as special cases of the KL divergence. I'm interested in generalisations of the KL divergence to cases where we have three or more distributions rather than just two. In particular, I'm interested in fully understanding the case where one moves from a prior $P$ to a posterior $Q$, but then receives some additional information to arrive at a second posterior $R$. Is there literature on extending or generalising the KL divergence to deal with this three variable case? Of course one can consider $D_{KL}(R\|Q)$ and $D_{KL}(R\|P)$ as well as $D_{KL}(Q\|P)$, and one can also consider something of the form $\sum_i r_i \log\frac{q_i}{p_i}$, which is really $D_{KL}(R\|P)-D_{KL}(Q\|P)$. This latter quantity is interesting because it can be negative as well as positive, and when it's negative it indicates that the information leading to $Q$ was misleading, or contradictory with the information that led to $R$. However, these quantities seem not to tell the whole story by themselves. Intuitively, the information leading to $Q$ could be misleading in some ways but truthful in others, with $\sum_i r_i \log\frac{q_i}{p_i}$ only giving the total amount of misleadingness rather than the full decomposition. So it seems to me that there should be two nonnegative quantities (both functions of the probabilities $p_i$, $q_i$ and $r_i$) that measure the amount of misleading information $M$ and the amount of consistent information $C$, with $\sum_i r_i \log\frac{q_i}{p_i} = C-M$. It is not obvious to me what form these functions should have. The above intuition might not be clear without a motivating example, so I'll include one below. But really my question is just whether anything along the lines of the above has been written about before, or indeed, whether anything has been written about three-distribution analogues of the KL divergence at all. Appendix: a motivating example The above question is self-contained, but an example will make it clearer why I'm asking. I'll start with an example of misleading information and an example of truthful information, and then I'll combine them into a single example. In the combined system the quantity $\sum_i r_i \log\frac{q_i}{p_i} = 0$, but intuitively this is because there is one bit of misleading information and one bit of truthful information, which sum to zero. To start with let's consider a variable $X$ that can take the value 0 or 1. Suppose we have $$ p(X=0)=1/2,\qquad p(X=1)=1/2,\\ q(X=0)=3/4,\qquad q(X=1)=1/4,\\ r(X=0)=0,\qquad r(X=1)=1. $$ On the first piece of information (taking us from $P$ to $Q$), we're led to believe that $X$ is more likely to be zero, but on the second piece of information (taking us from $Q$ to $R$) we find that $X=1$. So intuitively the first piece of information is misleading. Quantitatively, $\sum_i r_i \log\frac{q_i}{p_i} = \log_2\frac{1/4}{1/2} = -1\,\text{bit}$, suggesting that in some sense the information in $Q$ took us one bit "in the wrong direction," away from the "true" posterior $R$. As a second example, consider another Boolean variable $Y$ with the probabilities $$ p(Y=0)=1/2,\qquad p(Y=1)=1/2,\\ q(Y=0)=0,\qquad q(Y=1)=1,\\ r(Y=0)=0,\qquad r(Y=1)=1. $$ Now intuition tells us that all the information about $Y$ was already in $Q$, with $R$ telling us nothing extra. We find that $\sum_i r_i \log\frac{q_i}{p_i} = 1\,\text{bit}$, indicating that $Q$ took us one bit towards the correct posterior. Let us now create a new random variable, $Z$, from the combined states of $X$ and $Y$, assuming that $X$ and $Y$ are independent. This gives us $$ p(Z=00)=1/4,\qquad p(Z=01)=1/4,\qquad p(Z=10)=1/4,\qquad p(Z=11)=1/4,\\ q(Z=00)=0,\qquad q(Z=01)=3/4,\qquad q(Z=10)=0,\qquad q(Z=11)=1/4,\\ r(Z=00)=0,\qquad r(Z=01)=0,\qquad r(Z=10)=0,\qquad r(Z=11)=1.\\ $$ We can now calculate that for $Z$ we have $\sum_i r_i \log\frac{q_i}{p_i} = 0$. But intuitively, this is because we received one misleading bit about $X$ and one truthful bit about $Y$, with zero being the sum of these two independent pieces of information. The information in $Q$ correctly told us that $Z$ was neither equal to $00$ nor $10$, but at the same time it incorrectly indicated that $Z=01$ was more likely than $Z=11$, so it contained some truth and some falsehood simultaneously. My question is about whether this last intuition can be captured formally via a function that looks only at the probabilities listed above for the values of $Z$. This function would return the value 1 bit for the probabilities above, indicating that $Q$ provides one bit of correct information about $Z$, and allowing us to deduce that it also contains one bit of misleading information. I am interested both in whether such a function has explicitly been defined and studied, and in any related concepts that might help me to formulate it myself. REPLY [12 votes]: The Kullback-Leibler divergence $D_{\rm KL}(Q||P)$ of two distributions $Q,P$ has been generalized to multiple distributions in various ways: [1] information radius: $R(P_1,\ldots P_k)=\frac{1}{k}\sum_{i=1}^k D_{\rm KL}(P_i||k^{-1}\sum_i P_i)$ [2] average divergence: $K(P_1,\ldots P_k)=\frac{1}{k(k-1)}\sum_{i,j=1}^k D_{\rm KL}(P_i||P_j)$ [3,4] dissimilarity: the weighted arithmetic mean of the KL distances between each of the $P_i$’s and the barycenter of all the $P_i$’s References [1] Robin Sibson, Information radius. Probability Theory and Related Fields, 14, 149–160 (1969). [2] Andrea Sgarro, Informational divergence and the dissimilarity of probability distributions. Calcolo, 18, 293–302 (1981). [3] Michèlle Basseville, Divergence measures for statistical data processing (2010). [4] Darío García-García and Robert C. Williamson, Divergences and Risks for Multiclass Experiments (2012).<|endoftext|> TITLE: Uncountable models of Kelley-Morse set theory with only a countable number of sets QUESTION [7 upvotes]: The Kelley-Morse set theory can be thought as the "full-secondorderification of $\sf ZFC$", where we switch from sets to classes and allow the comprehension schema to include quantifiers on class variables. As in the usual case of class-set theory, sets are exactly those classes which are elements of other classes. So we can consider this as a one-sort theory and have a definable predicate $\mathrm{Set}(x)\iff\exists y(x\in y)$. The "standard" models of Kelley-Morse set theory are $V_{\kappa+1}$ where $\kappa$ is a strongly inaccessible cardinal, and then $\mathrm{Set}$ is interpreted exactly as $V_\kappa$. But we can use the Lowenheim-Skolem theorem to obtain a countable model of Kelley-Morse. The trick here is that we keep "enough" classes to satisfy the impredicative comprehension schema, but the model is still countable. My question is as follows: Let $M$ be a model of Kelley-Morse What sort of limitations do we have on the cardinality of $M$ and $\mathrm{Set}^M$? Specifically, can we have a model with countably many sets and uncountably many classes? Does the answer change if we assume that $\mathrm{Set}^M$ is transitive? REPLY [11 votes]: The first relevant theorem is the following classical result: Theorem A. (Mostowski, Keisler) If $M$ is a countable model of Kelley-Morse + Choice Scheme, then there is an elementary extension $M^{*}$ of $M$ such that $\mathrm{Set}^{M}= \mathrm{Set}^{M^{*}}$ and $M^{*}$ has cardinality $\aleph_1$. In the above, the Choice Scheme consists of universal generalizations of statements of the following form, where the parameters in $\phi$ are suppressed: $\forall x \exists Y \phi (x,Y) \rightarrow \exists Z \forall x \phi(x,(Z)_x)$, where $x$ range over sets, capital letters range over classes, and $(Z)_x$ [read as: the $x$-th slice of $Z$] denotes $\{t: (x,t)\in Z\}$). The theorem holds for models of Second Order Arithmetic, with the same proof that works for Kelley-Morse. The proof uses omitting types and elementary chains; see, e.g., Chapter 28 of Keisler's monograph Model Theory for Infinitary Logic. It is well-known that Kelley-Morse plus the Choice Scheme is bi-interpretable with with the extension of $ZFC^-$ ($ZFC$ without powerset, where $ZF$ is formulated using the collection and separation schemes) obtained by adding "the last cardinal exists, and it is inaccessible". On the other hand; Theorem B. (Jensen's Gap-1 Theorem). Assuming $V=L$, every first order theory that admits a two-cardinal model of the form $(\aleph_{1}, \aleph_{0})$ admits a two-cardinal model of the form $(\kappa^{+},\kappa)$ for any prescribed infinite cardinal $\kappa$. By putting Theorems A & B together with the Loewenheim-Skolem theorem, we obtain the following corollary; in what follows, the "two-cardinal type" of a model $M$ of Kelley-Morse, is the ordered pair of cardinals $(|M|, |\mathrm{Set}^{M}|)$. Corollary. It is consistent with $ZFC$ that the following holds: for every consistent extension $T$ of Kelley-Morse + Choice Scheme (in the same language), the "two-cardinal" types of models of $T$ are precisely of the form $(\kappa, \kappa)$ or $(2^{\kappa}, \kappa)$, where $\kappa$ ranges over infinite cardinals. Postscript. I suspect that, using forcing, one can build a model of Kelley-Morse with countably many sets and continuumly many proper classes outright in $ZFC$. For models of Goedel-Bernays theory of classes, this was shown by Matt Kaufmann in this paper.<|endoftext|> TITLE: maybe this conjecture also hold to this complex inequality QUESTION [5 upvotes]: I have solve this following Question: Complex numbers ${x_i},{y_i}$ satisfy $\left| {{x_i}} \right| = \left| {{y_i}} \right| = 1$ for $i=1,2,\ldots ,n$. Let $x=\frac{1}{n}\sum\limits_{i=1}^n{{x_i}}$, $y=\frac{1}{n}\sum\limits_{i=1}^n{{y_i}}$ and $z_i=x{y_i}+y{x_i}-{x_i}{y_i}$. Prove that $$\sum\limits_{i=1}^n{\left| {{z_i}}\right|}\leqslant n$$. Proof: Note that $|z_i| = |(x-x_i)(y-y_i) - xy| \leq |x-x_i| |y-y_i| + |x| |y|$. Hence by Cauchy-Schwarz Inequality, \begin{align*} \sum |z_i| &\leq \sum |x-x_i| |y-y_i| + n|x| |y| \\ &\leq \sqrt{\sum |x-x_i|^2}\sqrt{\sum |y-y_i|^2} + n|x| |y|.\end{align*}Now we have \begin{align*}\sum |x-x_i|^2 &= \sum \left(|x|^2 + |x_i|^2 + 2{\rm Re}(x\overline{x_i})\right)\\ &= n|x|^2 + n - 2{\rm Re}(x\sum\overline{x_i})\\ &= n|x|^2 + n - 2n{\rm Re}(x\overline{x})\\ &= n(1-|x|^2), \end{align*}the last equality following from ${\rm Re} (x\overline{x}) = x\overline{x} = |x|^2$. Thus $$\sqrt{\sum |x-x_i|^2}\sqrt{\sum |y-y_i|^2} + n|x| |y| = n\left(\sqrt{1-|x|^2}\sqrt{1-|y|^2} + |x| |y|\right) \leq n,$$ where the last inequality is also proven by Cauchy-Schwarz, and we are done. Now I conjecture and try prove(Now I can't prove it) Question Complex numbers ${x_i},{y_i},z_{i}$ satisfy $\left| {{x_i}} \right| = \left| {{y_i}} \right| = \left|z_{i}\right|=1$ for $i=1,2,\ldots ,n$. Let $x=\frac{1}{n}\sum\limits_{i=1}^n{{x_i}}$, $y=\frac{1}{n}\sum\limits_{i=1}^n{{y_i}},z=\dfrac{1}{n}\sum_{i=1}^{n}z_{i}$ and $w_i=xy{z_i}+yz{x_i}+xzy_{i}-2{x_i}{y_i}z_{i}$. Prove that $$\sum\limits_{i=1}^n{\left| {{w_i}}\right|}\leqslant n$$. REPLY [5 votes]: The following seems to be a counterexample: Let $n=2$, $x_1 = 1 = - y_1$, $x_2= -1 = - y_2$, $z_1=z_2 =1$. Then $x=y=0$, $z=1$. So $w_i = -2 x_i y_i z_i = 2$ for $i=1,2$. Finally, $w_1 + w_2 = 4$.<|endoftext|> TITLE: Moduli of hyperelliptic curves: odd vs even genus QUESTION [5 upvotes]: I'm stumped by Exercise 2.3 in Harris-Morrison, which says: "Show that there does not exist a universal family of curves of genus 2 over any open subset $U \subset M_2$. In general, if $H_g \subset M_g$ is the locus of hyperelliptic curves, for which $g$ does there exist a universal family over some open subset $U \subset H_g$? Answer: for $g$ odd." It's not 100% clear what is meant by "universal family"; possibly it just means some family each of whose (closed) fiber is ``the right curve''. Question: what precisely is the question, and then what is the answer? In any case the "exercise" seems to suggest a fundamental difference between hyperelliptic curves with odd vs even genus. I'd be happy to see an explanation of what this difference is. REPLY [2 votes]: I think that the answer to your question is in http://arxiv.org/pdf/0802.0635.pdf Proposition 4.7. There they also point out that "universal family" should in fact be replaced with "tautological family". (The notion of a tautological family is I believe the one defined in Lemma 3.89 by Harris and Morrison)<|endoftext|> TITLE: Spectral triples which are not $\theta$-summable QUESTION [9 upvotes]: I would like to see an example of a spectral triple $(A,H,D)$ such that the underlying algebra $A$ is commutative but this spectral triple is not $\theta$-summable in the sense that $e^{-tD^2}$ is never traceclass operator. EDIT: Spectral triple is defined as a unital $*$-algebra $A$ represented faithfully on some Hilbert space $H$ together with the (usual unbounded) self adjoint operator $D$ with the property that $D$ has compact resolvent and such that for all $a$ in $A$ the commutator $[D,a]$ is (extends to) a bounded operator REPLY [2 votes]: One can produce an operator in a less artificial way. Consider the Sturm-Liouville operator $Du=-u''+V(x)u$, where $V(x)$ grows at infinity as $\log\log|x|$. The eigenvalues grow very slowly, also as $\log\log$, and therefore the exponent of the operator is not trace class (the details on the asymptotics of eigenvalues can be found, eg, in the book by S.Levendorskii, 'Asymptotic distribution of eigenvalues of differential operators. Kluwer Academic Publishers Group, Dordrecht, 1990.)<|endoftext|> TITLE: Density of smooth functions on Hölder spaces QUESTION [7 upvotes]: The following result is often cited without reference in the context of PDEs: Let $\varOmega \subset\mathbb R^n$ be a bounded open set with smooth boundary. If $0<\beta<\alpha<1$ then $C^\infty(\overline \varOmega)$ is dense in $C^{0,\alpha}(\overline \varOmega)$ in relation to the $C^{0,\beta}$ norm. Question: Where (book or paper) can a proof be found? Remark: simple examples show that the above statement is false for $\beta=\alpha$, see e.g. Density of smooth functions under "Hölder metric" REPLY [9 votes]: It helps to have the concept of a "little Lipschitz function". A Lipschitz function $f: X \to \mathbb{R}$ is little Lipschitz if $\frac{|f(p) - f(q)|}{\rho(p,q)} \to 0$ as $p,q \to 0$. Notation: ${\rm Lip}(X)$ is the set of real-valued Lipschitz functions on $X$ and ${\rm lip}(X)$ is the set of real-valued little Lipschitz functions on $X$. The usual norm is $\|f\|_L = \max(L(f), \|f\|_\infty)$ where $L(f)$ is the Lipschitz number of $f$. Also, for any metric space $X$ and $0 < \alpha < 1$ let $X^\alpha$ be the same set equipped with the metric $\rho^\alpha(p,q) = (\rho(p,q))^\alpha$. Then $C^{0,a}(\overline{\Omega}) = {\rm Lip}(\overline{\Omega}^\alpha)$. The facts you want here are that (1) ${\rm Lip}(X) \subset {\rm lip}(X^\alpha)$ for any compact $X$ and any $0 < \alpha < 1$ and (2) $C^\infty(\overline{\Omega})$ is dense in ${\rm lip}(\overline{\Omega}^\alpha)$. Since $X^\beta = (X^\alpha)^{\beta/\alpha}$, this yields $C^\infty(\overline{\Omega}) \subset {\rm Lip}(\overline{\Omega}^\alpha) \subset {\rm lip}(\overline{\Omega}^\beta)$, with the first space being dense in the last, so the first is dense in the second which is what you asked about. For those two facts see Theorems 2.52 and 8.27 of my book Lipschitz Algebras (second edition). Density follows from Theorem 8.27 because $C^\infty(\overline{\Omega})$ separates points "uniformly"; this can be seen by checking that for any $x,y \in \overline{\Omega}$ the function $z \mapsto a^{\alpha-1}\cdot\max(a - |x - z|, 0)$ with $a = |x - y|$ has Lipschitz number $1$ in ${\rm Lip}(\overline{\Omega}^\alpha)$ and separates $x$ and $y$ to their full distance; smooth this to get a $C^\infty$ function which has Lipschitz number $\leq 1$ and separates $x$ and $y$ to nearly their full distance.<|endoftext|> TITLE: Corepresentations of Tensor Products of Hopf Algebras QUESTION [6 upvotes]: Given two cosemisimple Hopf algebras $H,G$ over ${\mathbb C}$, denote their usual (not braided) tensor product by $G \otimes H$. What conditions do we need to impose on the Hopf algebras to ensure that the corepresentations of $G \otimes H$ are direct sums of corepresentations of the form $V \otimes W$, where $V$ is a corepresentation of $G$ and $W$ is a corepresentation of $H$? I have heard a rumour that this is true when $G$ and $H$ can be completed to compact quantum groups. However, I can't find a reference, and I believe that it should be true for a larger class of Hopf algebras. REPLY [6 votes]: Let $A$ be a coalgebra over a field $\mathbb K$ and $\mathcal A = \mathrm{Comod}^A$ its category of comodules. A well-known result of Sweedler writes $A = \mathrm{colim}_i A_i$ where $A_i$ are finite-dimensional subcoalgebras of $A$. Let $A_i^*$ denote the linear dual to $A_i$ and $\mathcal A_i = \mathrm{Mod}_{A_i^*} = \mathrm{Comod}^{A_i}$ the category of $A_i^*$-modules, equivalently the category of $A_i$-comodules. As observed in my paper with Brandenburg and Chirvasitu, $\mathcal A = \mathrm{colim} \mathcal A_i$ is the colimit (over the same diagram) in the bicategory of $\mathbb K$-linear locally presentable categories. Here and throughout I sloppily write "$=$" for "there is a canonical equivalence". Tensor products of coalgebras distribute over colimits of coalgebras, and tensor products of categories distribute over colimits of categories. Moreover, it is well-known that if $R$ and $S$ are $\mathbb K$-algebras, then $\mathrm{Mod}_{R \otimes S} = \mathrm{Mod}_R\boxtimes \mathrm{Mod}_S$. (This can be checked directly from the universal property of the tensor product.) Suppose that $B = \mathrm{colim}_j B_j$ is another coalgebra over $\mathbb K$ with $B_j$ its finite-dimensional subcoalgebras, and $\mathcal B = \mathrm{Comod}^B$ and $\mathcal B_j = \mathrm{Mod}_{B_j^*}$ as above. We have $$ \begin{aligned}\mathcal A \boxtimes \mathcal B & = (\mathrm{colim}_{i} \mathcal A_i) \boxtimes (\mathrm{colim}_j \mathcal B_j) \\ & = \mathrm{colim}_{i,j} (\mathcal A_i \boxtimes \mathcal B_j) \\ & = \mathrm{colim}_{i,j} (\mathrm{Mod}_{A_i}^* \boxtimes \mathrm{Mod}_{B_j^*}) \\ & = \mathrm{colim}_{i,j} (\mathrm{Mod}_{A_i^* \otimes B_j^*}) \\ & = \mathrm{colim}_{i,j} (\mathrm{Comod}^{A_i \otimes B_j}) \\ & = \mathrm{Comod}^{\mathrm{colim}_{i,j} A_i \otimes B_j} \\ & = \mathrm{Comod}^{A\otimes B}. \end{aligned}$$ On the other hand, the tensor product of categories can be presented as follows. For arbitrary $\mathbb K$-linear locally presentable categories $\mathcal A$, $\mathcal B$, objects of $\mathcal A \boxtimes \mathcal B$ are colimits over objects of the form $V \boxtimes W$ subject to the relation that $\mathrm{colim}_m (V_m \boxtimes W) = (\mathrm{colim}_m V_m) \boxtimes W$, and similarly in the $W$-variable, where on the LHS the colimit is computed in $\mathcal A \boxtimes \mathcal B$ and on the RHS the colimit is computed in $\mathcal A$. It follows from this presentation that if $\mathcal A$ and $\mathcal B$ are both semisimple categories, then so is $\mathcal A\boxtimes \mathcal B$. Indeed, suppose that $\mathcal A$ is semisimple and let $\mathcal I = \{I_1,\dots\}$ be a complete set of simples. By "semisimple" I mean that every object of $\mathcal A$ is a direct sum of objects from $\mathcal I$. Let $\mathbb K_m = \mathrm{End}(I_m)$; by Schur's lemma, it is a division ring. Then $\mathcal A = \bigoplus_m \mathrm{Mod}_{\mathbb K_m}$. If $\mathcal B$ is also semisimple with simples $\{J_1,\dots\}$ and $\mathrm{End}(J_n) = \mathbb L_n$, then $$ \mathcal A \boxtimes \mathcal B = \bigoplus_{m,n} \mathrm{Mod}_{\mathbb K_m} \boxtimes \mathrm{Mod}_{\mathbb L_n} = \bigoplus_{m,n} \mathrm{Mod}_{\mathbb K_m \boxtimes \mathbb L_n}.$$ Suppose now that $\mathbb K$ is algebraically closed and $\mathcal A = \mathrm{Comod}^A$ and $\mathcal B = \mathrm{Comod}^B$. Then $\mathbb K_m = \mathbb L_n = \mathbb K$ for all $m,n$. It follows that every object of $\mathcal A \boxtimes \mathcal B$ is a direct sum of objects of the form $I_m \boxtimes J_n$, as you wanted. (Under the equivalence $\mathcal A \boxtimes \mathcal B = \mathrm{Comod}^{A\otimes B}$, $I \boxtimes J$ corresponds to the $A\otimes B$-comodule $I \otimes J$.) If $\mathbb K$ is not algebraically closed, then one can find many examples where $\mathbb K_m \boxtimes \mathbb L_n$ is not a division ring. It is always, however, a semisimple algebra, and so its module theory is semisimple with simples given by direct summands of the regular module. Thus in this case every object of $\mathrm{Comod}^{A\otimes B}$ is a direct sum of direct summands of objects of the form $I_m \boxtimes J_n$. So as you can see, the answer to your question correctly interpreted is "yes", and it has nothing to do with Hopf algebras or quantum groups.<|endoftext|> TITLE: Is there a compact generated triangulated category which does not have a compact generator? QUESTION [5 upvotes]: Let $\mathcal{T}$ be a triangulated category which has arbitraty direct sums. An object $E\in \mathcal{T}$ is called compact if the functor Hom$(E,-)$ commutes with arbitrary direct sums. A triangulated category $\mathcal{T}$ is called compactly generated if there is a set $\mathcal{S}$ of objects of $\mathcal{T}$ which consists of compact objects and satisfies $$ \text{Hom}(G[n],X)=0, ~\forall G\in \mathcal{S}, \forall n\in \mathbb{Z} \text{ implies } X=0. $$ In particular, an object $E$ is a compact generator of $\mathcal{T}$ if $\{E\}$could play the role of $\mathcal{S}$ in the above definition. It is well-known that for a quasi-compact, separated scheme $X$, the derived category of complexes of quasi-coherent $\mathcal{O}_X$-modules $D(X)$ has a compact generator. My question is: is there a triangulated category $\mathcal{T}$ which is compactly generated but does not have a compact generator? REPLY [7 votes]: Possibly the stack example in pbelman's answer is of this form, but an elementary way to construct examples is by taking infinite products. Let $\{\mathcal{C}_i\}_{i\in I}$ be an infinite collection of nonzero compactly generated triangulated categories, and $\mathcal{C}=\prod_{i\in I}\mathcal{C}_i$. It is easy to see that an object $(X_i)_{i\in I}$ of $\mathcal{C}$ is compact if and only if $X_i$ is compact in $\mathcal{C}_i$ for every $i$ and $X_i=0$ for all but finitely many $i$. Such objects clearly generate $\mathcal{C}$, but a generating set must involve generators of every $\mathcal{C}_i$, so no single compact object of $\mathcal{C}$ is a generator.<|endoftext|> TITLE: Geometric Construct for Integrating Symmetric Tensors? QUESTION [7 upvotes]: I'm interested in finding the appropriate geometric construct for the integration of symmetric tensors, analogous to the way differential forms can be integrated over manifolds. The motivation comes from gauge theories, commonly used in theoretical physics. The simplest example is a vector field $A_i$ defined on a Euclidean space $R^m$, where the vector field is only defined up to addition of a gradient, $A_i \sim A_i + \partial_i \alpha$, for some scalar $\alpha$. The meaningful quantities are the integrals of $A_i$ around closed curves, which eliminates any contribution from the gradient term. More generally, allow $A$ to be an $n$-form defined on $R^m$ ($m>n$), defined only up to addition of an exact form, $A\sim A+d\alpha$. The meaningful quantities are the integrals of $A$ over closed $n$-dimensional submanifolds of $R^m$, which kill the contribution from the exact form. Now, to the main question. Allow the gauge field to be a symmetric tensor $A_{ij}$ defined on the Euclidean space, but only up to a second derivative, $A_{ij}\sim A_{ij}+\partial_i\partial_j \alpha$ for arbitrary scalar $\alpha$. I'm wondering what the appropriate geometric construct is for integrating symmetric tensors which would allow me to obtain meaningful quantities, independent of $\alpha$. I should mention that I have indeed been able to find appropriate "objects" of integration when the theory is defined on a lattice approximation to $R^m$, so I believe the concept to be well-defined. I just need a better way to mathematically characterize these objects in order to make meaningful statements in the continuum limit. Based on the lattice analysis, it seems that the "objects" of integration have well-defined dimension (based on notions of self-similarity, as in fractals), but do not appear to be conventional manifolds. Furthermore, the "dimension" of the objects can actually exceed $m$ in certain cases, leading to a picture in which higher-dimensional objects have been embedded in a lower-dimensional Euclidean space. But these are merely conjectures at this point. Does anyone have any ideas as to what mathematical construct I am looking for? REPLY [6 votes]: Here is one way to construct all "local" gauge-invariant quantities out of a symmetric tensor $A_{ij}$. It would be up to you to decide how it meshes with the intuition you gained from your investigations on the lattice. First, define $R[A]_{ij:kl} = \partial_i\partial_k A_{jl} - \partial_j\partial_k A_{il} - \partial_i\partial_l A_{jk} + \partial_j\partial_l A_{ik}$ and $C[A]_{ij:k} = \partial_i A_{jk} - \partial_j A_{ik}$. You might notice that $R[A]$ is the linearized Riemann tensor, applied to a perturbation $\eta_{ij} \mapsto \eta_{ij} + A_{ij}$ of the flat Euclidean metric (or of whatever signature you like), while $C[A]$ is not exactly the linearized Christoffel connection, but a certain projection of it. It is a straight forward exercise to show that a pure gauge field configuration $A_{ij} = \partial_i \partial_j \alpha$ is annihilated by both $R[A]$ and $C[A]$. But it is also true that any $A_{ij}$ annihilated by both $R[A]$ and $C[A]$ must be of that form, at least locally. First, $R[A] = 0$ implies that $A_{ij} = \partial_i v_j + \partial_j v_i$ for some $v_i$ (at least locally), which is essentially an infinitesimal restatement of the well-known fact that any metric that has zero Riemann curvature admits coordinates in which it takes the standard Euclidean form. Next, for $A_{jk} = \partial_j v_k + \partial_k v_j$ we have $C[A]_{ij:k} = \partial_k (\partial_i v_j - \partial_j v_i)$. By applying the Poincaré lemma, $C[A] = 0$ means that (locally) $v_i = \partial_i \alpha + B_{ij} x^j$, where $B_{ij} = B_{[ij]}$ is constant and the $x^j$ are Euclidean coordinates. But that last term is irrelevant, since we have found that our original $A_{ij} = \partial_i v_j + \partial_j v_i = \partial_i \partial_j (2\alpha)$ (at least locally). Now consider any integral of the form \begin{equation} F(A) = \int ( f^{ijkl}(x) R[A]_{ij:kl} + g^{ijk}(x) C[A]_{ij:k} ) \, dx . \end{equation} For example, the tensors $f$ and $g$ could be arbitrary test functions (smooth and compactly supported), but could also be distributions supported on lower dimensional submanifolds, whatever makes sense in your context. The point is that all (reasonable) linear gauge-invariant functionals of the $A_{ij}$ field can be written in the above form.<|endoftext|> TITLE: Good analytic spaces over a field into locally ringed spaces is fully faithful QUESTION [8 upvotes]: Let $k$ be a field which is complete with respect to a non-trivial non-archimedean rank-1 valuation, and let $X$ be scheme which is locally of finite type over $k$. In section of 3.5 of Berkovich's book, he defines the analytification of $X$ to be the closed $k$-analytic space $X^{\textrm{an}}$ representing the following functor: let $\textrm{An}_k$ be the category of good analytic spaces over $k$, and let $F_X \colon \textrm{An}_k \to (\textrm{Sets})$ be the functor which sends a good analytic space $Z$ over $k$ to the set of morphisms $Z \to X$ of locally $k$-ringed spaces. In order for Berkovich's proof to work, it seems that we need the functor from $\textrm{An}_k$ to the category of locally ringed spaces to be fully faithful; that is, every morphism of locally ringed spaces between two good analytic spaces must in fact be a morphism of analytic spaces over $k$. This fact is stated without proof on page 16 of Berkovich's Trieste notes, but how could one prove this? However, there is also Remark 2.1.13 of Berkovich's book, which gives an example of a morphism (of locally ringed spaces) between $k$-affinoid spaces which is not a morphism of $k$-affinoid spaces. As $k$-affinoid spaces are examples of good analytic spaces, this seems to contradict the above assertion of fully faithfulness (granted, this particular example is over a trivially-valued field). This issue also does not seem to be addressed in Berkovich's IHES paper. REPLY [5 votes]: First, I would like to say that I do not understand why you need the full faithfullness of the analytification functor. It seems to me that the main point is to prove that giving a morphism from an analytic space $X$ to the affine analytic space of dimension $n$ is equivalent to giving $n$ global section of $X$. Second, as you rightly say, the functor is not fully faithful over trivially valued fields nor for non-strict spaces (see Remark 2.1.13 of Berkovich's book), so let us assume that the valuation is non-trivial and that the spaces are strict. We can reduce to the case of a morphism between strictly $k$-affinoid spaces, say $f : X \to Y$. By taking global sections, you find an induced morphism between affinoid algebras $A_Y \to A_X$ and [BGR, Theorem 6.3.1/1] tells you that such a morphism is always continuous. Arguing as for schemes, you then prove that the induced morphism $X = \mathcal{M}(A_X) \to \mathcal{M}(A_Y) = Y$ coincides with $f$ for rigid points. By density of rigid points (which holds over a non-trivially valued field), it coincides with $f$ everywhere.<|endoftext|> TITLE: A model structure on semi-simplicial algebraic Kan complexes? QUESTION [8 upvotes]: By an algebraic semi-simplicial kan complex I mean a semi-simplicial set (i.e. a presheaf on the category of finite ordered sets and injective order preserving maps), which is a Kan complex (in the sense that any map from a semi-simplicial horn to it can be filled into a cell) and in which a filling has been chosen for each such Horn, and with the morphism between them being the semi-simplicial map which preserves those chosen fillers. My question is: Is there a model structure on this category of semi-simplicial algebraic Kan complexes, in which: The weak equivalences are the maps that induce bijections on all the $\pi_n$. The fibrations are the maps which are "Kan fibrations" as map of semi-simplicial sets. the trivial fibrations are the maps which are trivial kan fibration of semi-simplicial sets. In particular, not all objects would be cofibrant: the cofibrant objects are those which are obtained by freely adding cell gradually. This would be very similar to the model structure of algebraic kan complex constructed by T.Nikolaus in this paper, with the exception that it can not come from a model structure on semi-simplicial sets. The reason I'm asking this, is that I have recently observed that there is what Spitzweck called in his thesis a J-semi model structure that fits this description. I.e. I have been able to prove all the axiom of a model structure except that my "trivial cofibrations" (left lifting property with respect to fibrations) are the same as "cofibration and weak equivalences" only if they have a cofibrant domain. to put it more simply: The only thing I don't know is whether maps with a non cofibrant domain can be factored as as acyclic cofibration followed by a fibration, or even more simply whether pushout of Horn inclusion are always weak equivalences when the domain is not cofibrant. Moreover I know that this semi-model structure is Quillen equivalent to the model structure on spaces. I have obtain this as a corollary of a more general results that I was working on, but my approach does not say anything about non cofibrant objects and so I cannot answer the question. It would help a lot to know whether this semi-model structure is actually a model structure. So I was hoping that someone with a different point of view might answer this... REPLY [2 votes]: I went back to this question a few days ago and found the solution: it is indeed a true model structure. I have two (related) approaches to this, but anyway the key point is the semi-simplicial approximation theorem: Theorem: If $A \hookrightarrow B$ is a cofibration (=an inclusion) of semi-simplicial sets, $C$ is a semi-simplicial Kan complex, $f:A \rightarrow C$ is a semi-simplicial map and one has $g: |B| \rightarrow |C|$ a continuous map on the geometric realization that extends $|f|$ then there is a semi-simplicial map $g' : B \rightarrow C$ which extends $f$ and such that $|g'|$ is homotopy equivalent to $g$ relative to $|A|$. This can be found in the first paper of Rourke and Sanderson on semi-simplicial sets as theorem 5.3 From this one can deduce a rather direct proof that the trivial cofibrations of the model structure mentioned in the question are weak equivalences: Lemma: A map of algebraic semi-simplicial Kan complexes is an equivalence if and only if it is an equivalence on the geometric realization of the underlying semi-simplicial sets. Indeed, because of the semi-simplicial approximation mentioned above, the $\pi_n$ of a Kan complex and of its semi-simplicial geometric realization are the same. Now if $A$ is a semi-simplicial algebraic Kan complex, then gluing a horn inclusion to $A$ is done as follows: first, one glues the horn inclusion on the underlying semi-simplicial set and then one iteratively glues all the horn inclusions that do not already have a chosen filling (and one never needs to collapse anything, since the gluing of horn inclusions are monomorphism). In particular, the final Kan complex $A'$ is obtained from $A$ only by gluing horn inclusions and hence $A \rightarrow A'$ is clearly a weak equivalence on the semi-simplicial geometric realization, which proves the claim. But one can also see the whole model structure mentioned in the question in a nicer way: Theorem: There is a 'right' semi-model structure on semi-simplicial sets, whose fibrations and trivial fibrations are the Kan fibrations and Kan trivial fibrations and weak equivalences are the maps that induce weak equivalences on the geometric realizations (cofibrations are the monomorphisms). I am not sure of the utilization of the word "right" here, what I mean is the dual of the notion mentioned in the question, i.e. all trivial cofibrations are weak equivalences, but only the trivial fibrations of fibrant codomain are weak equivalences. From this theorem, one concludes by observing that Nikolaus' construction of model categories of algebraically fibrant objects also works for a semi-model category and produces a semi-model category. But if one applies it to a right semi-model category one gets a right semi-model category in which every object is fibrant and this is an ordinary model category. Proof: One has two weak factorization systems by the small object argument, and weak equivalences obviously satisfies $2$-out-of-$3$ and even $2$-out-of-$6$. Trivial cofibrations are obviously weak equivalences. Using the approximation theorem above, weak equivalences between fibrant objects can be characterized as bijections on the semi-simplicial $\pi_n$, from which one easily deduces that the trivial fibrations between fibrant objects are weak equivalences. We conclude by proving that fibrations between fibrant objects which are also weak equivalences are trivial fibrations: Let $f :X \rightarrow Y$ be a fibration and a weak equivalence, let $a : \partial \Delta_n \rightarrow X$ be a boundary of $X$ and $b : \Delta_n \rightarrow Y$ be a filling of $f(a)$ in $Y$. As $f$ is an equivalence on the geometric realizations, $a$ has a filling on the geometric realization. Morevoer, as $Y$ is fibrant, there is a map $a' : \Delta_n \rightarrow X$ such that $f(a')$ is homotopy equivalent to $b$ relative to the boundary. The point is then that with a bit of work on simplicial combinatorics one can construct a semi-simplicial set $I$ such that: $I$ contains $\Delta_n \coprod_{\partial \Delta_n} \Delta_n$ as a sub-complex, the map $\Delta_n \hookrightarrow I$ is a trivial cofibration and the geometric realization of $I$ is (at least up to homotopy) $\Delta_n \coprod_{\partial \Delta_n} \Delta_n$ with a filled interior. At this point, one has a map from $I$ to $Y$ at the level of geometric realizations that extends our already existing map from $\Delta_n \coprod_{\partial \Delta_n} \Delta_n$. Hence by the approximation theorem one has such a map as a semi-simplicial map, the inclusion of $\Delta_n$ into $I$ is a trivial cofibration and therefore $I$ can be lifted to $X$ and the other inclusion of $\Delta_n$ into $I$ provides our lifting. This concludes the proof of the theorem.<|endoftext|> TITLE: 2-adic valuation of odd harmonic sums QUESTION [5 upvotes]: (This question is cross-posted on math.stackexchange) I'm playing with p-adic valuations, and find that the odd harmonic sums, $\tilde{H}_k=\sum_{i=1}^{k}\frac{1}{2i-1}$, has 2-adic valuation $||k^2||_2=2||k||_2$. E.g.) $\tilde{H}_4=\frac{176}{85}$ has $||\tilde{H}_4||_2=||176||_2=4=2||4||_2$. $\tilde{H}_6=\frac{6508}{3465}$ has $||\tilde{H}_6||_2=2$. $||\tilde{H}_8||_2=||\frac{91072}{45045}||_2=6$. $\cdots$ How can I prove this? (Some simple observation I have tried: As $||H_k||_2 =-r$ for $2^r\leq k <2^{r+1}$, $||H_k||_2=||H_{2k}||_2+1=||\frac{1}{2}H_k||_2$. Thus ultrametric ineq for $\tilde{H}_k=H_{2k}-\frac{1}{2}H_k$ doesn't help at all, yielding trivial result $\tilde{H}_k\in \mathbb{Z}_2$.) (That $||\tilde{H}_k||_2=2||k||_2$ for odd $k$ is almost trivial, so the problem can be reduced to show $||\tilde{H}_{2k}||_2=||\tilde{H}_k||_2+2$, still have no idea) (I've checked this holds up to $\sim 1000$ using Mathematica, and Mathematica spits the result almost immediately for $n\sim 1000$ and takes about a minute for $n\sim 10000$. ) For anyone interested : I've used the following Mathematica code, try it yourself a = Table[ IntegerExponent[ Numerator[HarmonicNumber[2 n] - HarmonicNumber[n]/2], 2]/2, {n, 10000}]; b = Table [IntegerExponent[n, 2], {n, 10000}]; NonNegative[Min[a - b]] REPLY [8 votes]: It is sufficient to prove this result for the sum where $i$ ranges from $m 2^n + 1$ to $(m+1)2^n$. That is because $\tilde{H}_k$ is a sum of an odd number of terms of this form with $n= ||k||_2$, so if they all have $2$-adic valuation $2||k||_2$ then $\tilde{H}_k$ does as well. To that end, consider what happens to the sum if we decrease each $i$ by $2^{n-1}$. That is the same as decreasing the denominator $1/(2i-1)$ by $2^n$. $1/x - 1/(x-2^n) = 2^n/( x^2)+2^{2n}/( x^3) + \dots$ The terms with a factor of $2^{3n}$ or more are irrelevant, as is the $2^{2n}$ term because the sum over all odd numbers in an interval of length $2^n$ of $1/x$ is clearly even. Now in $\sum_{i=r}^{r+2^n-1} \frac{2^n}{ (2i-1)^2}$, $1/(2i-1)$ ranges over all odd residue classes modulo $2^{n+1}$, so its square ranges over all residue classes modulo $2^{n+2}$ that are congruent to $1$ mod $8$, hitting each one twice. The sum of all those residue classes is $2^n + 16 \binom{2^{n-1}}{2} \equiv 2^n$ mod $2^{n+1}$. So the sum is $2^{2n}$ mod $2^{2n+1}$. After performing this decrease an odd number of times, we are left with $\sum_{i=-2^{n-1}+1}^{i=2^{n-1}} \frac{1}{2i-1}=0$ because the positive and negative terms cancel. Hence the original sum is $2^{2n}$ mod $2^{2n+1}$. REPLY [6 votes]: Here is a more direct approach. We split the sum into $$ -\tilde{H}_k = \sum_{i=1}^{k} \frac{1}{1-2i} = \sum_{i=1}^{k} \frac{1}{1-4i^2} + \sum_{i=1}^{k} \frac{2i}{1-4i^2}. $$ The first sum can be computed as $$ \sum_{i=1}^{k} \frac{1}{1-4i^2} = \frac{1}{2} \sum_{i=1}^{k} \frac{1}{1-2i} - \frac{1}{-1-2i} = \frac{1}{-2} - \frac{1}{-2-4k} = -\frac{k}{2k+1}. $$ To compute the second sum, we expand the fraction in $\mathbb{Z}_{2}$: $$\sum_{i=1}^{k} \frac{2i}{1-4i^2} = 2 \sum_{i=1}^{k} i + 2^3 \sum_{i=1}^{k} i^3 + 2^5 \sum_{i=1}^{k} i^5 + 2^7 \sum_{i=1}^{k} i^7 + \cdots. $$ Because all the Faulhaber polynomials have odd denominator, for each odd $p \ge 3$ the sum $\sum_{i=1}^{k} i^p$ has $2$-adic valuation at least $\nu_2(k^2) - 2$. Hence all terms except for the first one has valuation strictly greater than $k^2$. Hence $$ -\tilde{H}_k = -\frac{k}{2k+1} + k^2 + k = \frac{2k^3 + 3k^2}{2k+1} + (\text{even multiple of } k^2) $$ has $2$-adic valuation exactly $\nu_2(k^2)$.<|endoftext|> TITLE: History: What was the Lemma? (Grothendieck Harvard Lectures; Mumford) QUESTION [56 upvotes]: In an article about the life of Grothendieck, available here: http://www.ams.org/notices/200409/fea-grothendieck-part1.pdf Allyn Jackson writes about how Mumford was profoundly impressed: Mumford found the leaps into abstraction to be breathtaking. Once he asked Grothendieck how to prove a certain lemma and got in reply a highly abstract argument. Mumford did not at first believe that such an abstract argument could prove so concrete a lemma. “Then I went away and thought about it for a couple of days, and I realized it was exactly right,” Mumford recalled. What were the lemma and proof that so impressed Mumford? (I have tried asking algebraic geometers and category theorists; the tags attached to this question are speculative.) REPLY [26 votes]: The concrete lemma was "The Theorem of The Cube", (Mumford, Abelian Varieties [AV], Section 6) as indicated by Tabes Bridges. The abstract argument was the second theorem in Section 5 of AV (p.46). In Mumford's collected works, Volume 2, p. 689 there is a letter of Grothendieck to Mumford in which the abstract argument is discussed, and footnote 2 indicates that it was published by Mumford in AV as the second theorem in Chapter 5. All of this was confirmed in an email correspondence with Mumford. I am obliged Mumford (of course), to Tabes Bridges for giving the right answer and to Benjamin Dickman for a reminder to post. I should look at MO more frequently!<|endoftext|> TITLE: Subgroup schemes of $\mathbb{A}^n$ QUESTION [5 upvotes]: Let $R$ be an integral $\bar{k}$-algebra of finite type. Let $V(I) \subseteq \mathbb{A}_R^n$ be a reduced (closed) subgroup scheme such that $V(I)\backslash \{0\} \neq \emptyset$ and the $\mathbb{G}_m^R$-action on $\mathbb{A}_R^n \backslash \{0\}$ restricts to a free action on $V(I)\backslash \{0\}$. Is it true that $I$ can generically be generated by linear polynomials? (If $R=k$ is any field, and $V(I)$ as above, can $I$ be generated by linear polynomials?) If $R=\bar{k}$ this is clear, because we can just check it on $\bar{k}$-points (should not even need the $\mathbb{G}_m^R$-action). In general, intuitively this seems good to me (I would just think of 'generalised' subvector spaces), but I know too few about group schemes to verify it. I guess if there was a counterexample, then it would be in the non-perfect world. Edit: As Jason pointed out there are indeed counterexamples. What if we additionally assume that $V(I)$ has reduced (closed) fibers? REPLY [2 votes]: I am posting my comments above as an answer. If one does not impose a condition on fibers, then there are counterexamples such as when $R=\overline{k}[t]$ and the ideal $I$ in $R[x,y]=\Gamma(\mathbb{A}^2_R,\mathcal{O})$ equals $\langle x^p-ty^p\rangle$. On the other hand, if the fiber over every $\overline{k}$-point of $\text{Spec}(R)$ is reduced, then the ideal $I\subset R[x_1,\dots,x_n]$ is generated by $I_1=I\cap R[x_1,\dots,x_n]_1$. The homogeneity condition on $V(I)$ implies that $I$ is homogeneous, i.e., $I$ equals the direct sum over every integer $d$ of $I_d=I\cap R[x_1,\dots,x_n]_d$. In particular, $I_1$ is compatible with arbitrary base change of $R$, so that also the cokernel $I/I_1 R[x_1,\dots,x_n]$ is compatible with arbitrary base change. If the support of this module is nonempty, then this closed subset of $\mathbb{A}^n_R$ contains a $\overline{k}$-point. This maps to a $\overline{k}$-point of $\text{Spec}(R)$. Thus, the base change of $I$ over this $\overline{k}$-point of $R$ is also not generated by linear polynomials. That implies that the fiber is nonreduced.<|endoftext|> TITLE: Does every smooth manifold admit a metric with bounded geometry and uniform growth? QUESTION [5 upvotes]: Let $M$ be a smooth manifold, $g_M$ a Riemannian metric, and consider for $x\in M$ the volume growth function, $gr_x$ that maps $r>0$ to the volume $vol_{g_M}(B(x,r))$. My interest is to see whether or not any differentiable manifold can be endowed with a metric structure such that the family of growth functions $\{gr_x, x\in M\}$ has bounded variation. The answers posted in Example of a manifold which is not a homogeneous space of any Lie group make clear that some kind of variation must be allowed. A Riemannian manifold is said to be of bounded geometry if the injectivity radius is bounded away from 0 and the norms of the covariant derivatives of the curvature tensor $|\nabla^iR|$ are bounded with bounds independent of the point $x\in M$ (but they may depend on $i$ of course). It is well known that every differentiable manifold admits a metric of bounded geometry. This was proved by R.E. Greene in his paper Complete metrics of bounded curvature on noncompact manifolds. We say that the growth of $M$ is uniformly superlinear or uniformly subexponential if the limits $$\lim \inf_{s\to \infty} \frac{vol(B(x,s))}{s}=\infty,\quad \lim \sup_{s\to \infty} \frac{\ln[ vol(B(x,s))]}{s}=0$$ converge uniformly on $x\in M$ (that is, in the first case for example, given $\epsilon >0$, we can choose $\delta(\epsilon)$ independent of $x\in M$). Question: Can we endow every differentiable manifold $M$ with a metric tensor such that $M$ is of bounded geometry and exhibits either uniform superlinear or uniform subexponential growth? REPLY [4 votes]: The answer is "yes". Read page 96 in "Volume and bounded cohomology" by Gromov or "Manifolds with quadratic curvature decay and slow volume growth" by Lott and Shen.<|endoftext|> TITLE: Is a C*-algebra with an isomorphic predual a von Neumann algebra? QUESTION [13 upvotes]: It is well-known that a C*-algebra $A$ is a von Neumann algebra if and only if it has an isometric predual, that is, if and only if there exists a Banach space $X$ such that $A$ is isometrically isomorphic to $X^*$. Does the same hold for isomorphic preduals? Let $A$ be a C*-algebra such that $A$ is (not necessarily isometrically) isomorphic to $X^*$ for some Banach space $X$. Is $A$ a von Neumann algebra? REPLY [7 votes]: Now that Yemon has given a counterexample, I thought it worth giving a stronger hypthoesis under which we have a positive answer. I looked at something similar here: http://arxiv.org/abs/math/0604372 see Section 4. Theorem: Let $M$ be a commutative von Neumann algebra, and let $N$ be a dual Banach algebra. Any bounded algebra isomorphism between $M$ and $N$ is automatically weak$^*$-continuous. With Le Pham and White we remove the commutative condition in http://arxiv.org/abs/0804.3764 With ten years' of hindsight, I find that I can now improve this result and give an easy proof... A dual Banach algebra is a Banach algebra which is a dual space, in such a way that the multiplication is separately weak$^*$-continuous. Thus the above result says that a von Neumann algebra has a unique weak$^*$-topology making the multiplication map weak$^*$-continuous. If $E$ is a Banach space, then whenever $E$ is isomorphic to $F^*$ for some Banach space $F$, say under $\theta: E \rightarrow F^*$ we can look at the image of $F$ under the adjoint $\theta^*$. This gives a closed subspace of $E^*$, say $X$, such that $E$ is canonically isomorphic to $X^* = E^{**} / X^\perp$, where $X^\perp$ is the annihilator of $X$ in $E^{**}$. This means that: For each $0\not=x\in E$ there is $\mu\in X$ with $\mu(x)\not=0$; For each $\Phi\in E^{**}$ there is $x\in E$ with $\mu(x) = \Phi(\mu)$ for all $\mu\in X$. I call $X$ a "concrete predual" of $E$. The $x\in E$ given by (2) is unique, because of condition (1). Let $\alpha : E^{**} \rightarrow E$ be the map so given. It is a projection of $E^{**}$ onto $E$, once we canonically identify $E$ with its image in $E^{**}$. Theorem: Let $A$ be a C$^*$-algebra which is isomorphic to a dual space $F^*$, with the additional assumption that $F$ is a Banach $A$-bimodule, and the isomorphism $A \cong F^*$ is a bimodule map. Then $A$ is a von Neumann algebra (and the weak$^*$-topology is the canonical one). To prove this, argue as above to reduce to a concrete predual $X\subseteq A^*$. The bimodule condition becomes that $X$ is $A$-invariant for the usual action of $A$ on $A^*$. One can check that $X$ being $A$-invariant is actually equivalent to the map $\alpha:A^{**}\rightarrow A$ being an algebra homomorphism. Thus $\ker\alpha$ is a closed 2-sided ideal in $A^{**}$ and so is $*$-closed. This in turn is equivalent to $X$ being $*$-closed, which in turn implies that $\alpha$ is actually a $*$-homomorphism. Then $\alpha$ is a contractive projection which implies that $A$ is isometrically isomorphic $X^*$, hence a von Neumann algebra. (This proof is obviously influenced by Tomiyama's proof that W$^*$-algebras and von Neumann algebras are the same thing.)<|endoftext|> TITLE: Is there any formula to find number of Pythagorean triplets between two integers 2 and j, j>2? QUESTION [9 upvotes]: Given $j \geq 5$, is there a formula for the number of Pythagorean triplets $(a, b, c)$ satisfying the constraint that $a, b, c \leq j$? There exists at least one Pythagorean triplet for $j\geq5$; the question is how to find the exact number of Pythagorean triplets for large $j$. REPLY [7 votes]: The question here was studied (in a slightly more generalized version) by Sierpinski (1906) in Polish; asymptotics were found by Fricker (1977, 1982) and Fischer (1979) both in German; and an unconditional upper bound was established by Stronina (1969) in Russian. As for English, you can find these references (and others) in, for example, the paper: Nowak, W. G., & Recknagel, W. (1989). The distribution of Pythagorean triples and a three-dimensional divisor problem. Math. J. Okayama Univ, 31, 213-220. Link (no paywall). Here is an excerpt from the first page: The reference here comes from backtracking; first: Benito, M., & Varona, J. L. (2002). Pythagorean triangles with legs less than n. Journal of computational and applied mathematics, 143(1), 117-126. Link (no paywall). In there, the authors point to an earlier paper (p. 118): Kühleitner, M. (1993, December). An omega theorem on Pythagorean triples. In Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg (Vol. 63, No. 1, pp. 105-113). Springer Berlin/Heidelberg. Link. The last paper is, unfortunately, only visible as a preview; but it is written in English (despite the German title) and contains a reference to the Nowak and Recknagel paper excerpted above.<|endoftext|> TITLE: Criterion for being reflexive via Ext QUESTION [7 upvotes]: In this question it was claimed that if a module $M$ over a noetherian domain $R$ satisfies $\rm{Ext}^i(M,R)=0$ for $i=1,2$, then $M$ is reflexive. Is this true? Does someone know a reference or a proof for this? If it is not correct, is there another criterion for being reflexive via some $\rm Ext$-goups vanishing? REPLY [5 votes]: About the last question, I should point out that there is indeed a Ext-vanishing criterion for reflexivity. Namely, $M$ is reflexive iff $Ext^{1,2}(Tr(M),R)=0$ where $Tr(M)$ is the Auslander-Bridger transpose of $M$. For details see for instance Lemma 2.2 in this paper: https://arxiv.org/pdf/0809.1958v3.pdf<|endoftext|> TITLE: $x_1 = 2$, $x_{n + 1} = {{x_n(x_n + 1)}\over2}$, what can we say about $x_n \text{ mod }2$? QUESTION [22 upvotes]: This question was asked on MathStackexchange here, but there was no answer, so I am asking it here. Let$$x_1 = 2, \quad x_{n + 1} = {{x_n(x_n + 1)}\over2}.$$What can we say about the behavior of $x_n \text{ mod }2$? Is there an exact formula for $x_n \text{ mod }2$? REPLY [23 votes]: As remarked by Joe Silverman, a natural way to look at this question is by phrasing it in terms of the map $f(x):=\frac{1}{2}x(x+1)$ on the $2$-adic integers $\mathbb{Z}_2$. We are then asking about the behaviour of the orbit $f^n(2)$ with respect to the partition of $\mathbb{Z}_2$ into two clopen sets $U_1:=2\mathbb{Z}_2$ and $U_2:=1+2\mathbb{Z}_2$. Most questions of this form are very hard, for reasons which I will attempt to explain. Short answer: dynamically, there is no obvious reason why this should be easier than the Collatz problem or the normality of $\sqrt{2}$. Suppose that we are given a continuous transformation $f$ of a compact metric space $X$ and a partition of $X$ into finitely many sets $U_1,\ldots,U_N$ of reasonable regularity (e.g. such that the boundary of each $U_i$ has empty interior). We are given a special point $x_0$ and we want to know how often, and in what manner, the sequence $(f^n(x_0))_{n=1}^\infty$ visits each of the sets $U_i$. We might believe the following result to be useful: if $\mu$ is a Borel probability measure on $X$ such that $\mu(f^{-1}A)=\mu(A)$ for every Borel measurable set $A\subset X$, and if additionally every $f$-invariant measurable set $A\subseteq X$ satisfies $\mu(A)\in \{0,1\}$, then $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\phi(f^i(x)) =\int \phi\,d\mu$$ for $\mu$-almost-every initial point $x$ and every $\mu$-integrable function $\phi \colon X \to \mathbb{R}$. This is the Birkhoff ergodic theorem or pointwise ergodic theorem. In particular we could take $\phi$ to be the characteristic function of a partition element $U_i$, so the above measures the average time spent in $U_i$ by the sequence $f^n(x)$. Or we could take $\phi$ to be the characteristic function of a set such as $U_1\cap f^{-1}U_2 \cap f^{-2}U_1$, which would tell us how often the orbit of $x$ follows the path $U_1 \to U_2 \to U_1$, and so on. The Birkhoff theorem is great if we are content to study points $x$ which are generic with respect to a particular invariant measure. If we want to study a specific $x$ then we have no way to apply the theorem. If multiple different invariant measures $\mu$ exist then the integrals which they assign to the characterstic functions $\chi_{U_i}$ will in general be different and we will get different answers, and the Birkhoff averages along orbits will have fundamentally different behaviours depending on which measure the starting point is generic with respect to (if any). In a broad sense this is an aspect of the phenomenon called "chaos". Here is a nice example: let $X=\mathbb{R}/\mathbb{Z}$, $f(x):=2x$, let $U_1=[0,\frac{1}{2})+\mathbb{Z}$ and $U_2=X\setminus U_1$. What can we say about the manner in which the trajectory $f^n(\sqrt{2})$ visits $U_1$ and $U_2$? For example, does the trajectory spend an equal amount of time in both sets? Put another way, is $\sqrt{2}$ normal to base $2$? Dynamical methods can tell us that Lebesgue a.e. number is normal to base $2$, but they struggle badly for specific numbers. The difficulty of this problem is tied quite closely to the existence of many invariant measures for this map: Lebesgue measure is invariant, but so is the Dirac measure at $0$; there are many invariant measures supported on closed $f$-invariant proper subsets corresponding to restricted digit sets (e.g. numbers where "11" is not allowed in the binary expansion) and also many fully supported invariant measures which are singular with respect to Lebesgue measure (and indeed with respect to one another). Each invariant measure contributes a set of points with its own particular, distinct dynamical behaviour, and given an arbitrary point we have no obvious way of knowing which of these different dynamical behaviours prevails. There is only one situation in which this dynamical problem becomes easy: when a unique $f$-invariant Borel probability measure exists. In this case the convergence in the Birkhoff theorem is uniform[*] over all $x$, and the problem of distinguishing which invariant measure (if any) characterises the behaviour of the trajectory $f^n(x)$ vanishes completely. An example would be the leading digit of $2^n$: take $X=\mathbb{R}/\mathbb{Z}$, $U_k=[\log_{10}k,\log_{10}(k+1))+\mathbb{Z}$ for $k=1,\ldots,9$, $f(x):=x+\log_{10}2$, then the leading digit of $2^n$ is $k$ if and only if $f^n(0) \in U_k$. The irrationality of $\log_{10} 2$ may be applied to show that Lebesgue measure is the only $f$-invariant Borel probability measure, and using the uniform ergodic theorem we can simply read off the average time spent in each $U_k$ as the Lebesgue measure of that interval. So how does this affect $f(x):=x(x+1)/2$ on $\mathbb{Z}_2$? Well, $x=1$ and $x=0$ are both fixed points and carry invariant Borel probability measures $\delta_1$ and $\delta_0$. So the dynamical system is not uniquely ergodic, and the problem of determining which of the (presumably many) invariant measures characterises the trajectory of the initial point $2$ has no obvious solution. [*] to obtain uniform convergence in this case $\phi$ must be a continuous function. By upper/lower approximation we can deduce pointwise convergence to $\int \phi\,d\mu$ for all $x\in X$ in the case where $\phi$ is upper or lower semi-continuous, for example where $\phi$ is the characteristic function of a closed or open set.<|endoftext|> TITLE: How to sample a uniform random polyomino? QUESTION [11 upvotes]: A polyomino is formed by joining finitely many unit squares edge to edge. It may be regarded as a finite subset of the regular square tiling with a connected interior. In particular, for us, polyominos may have holes. (Update: these are sometimes also called two-dimensional lattice animals. For another model of random lattice animal, see this earlier MO question and discussion and references there.) I would like to be able to sample a uniform random polyomino with $n$ squares, where $n$ is of the order $100$ or $1000$. I am thinking of fixed polyominos, where two polyominos are considered equivalent only if they agree after a translation, but where rotations and reflections are in general distinct. But I might also be okay with sampling one-sided or fixed polyominos if that is for some reason easier or more natural. The point is that I don't want to have to enumerate all the polyominos of this size. For $n=100$, enumerating all the polyominoes would already be far beyond the reach of modern computing. I am hoping that there is a random construction, or else a Markov chain that converges to the uniform distribution (i.e. every polyomino having equal probability). I would be interested in such a Markov chain, if it could be proved to converge to the uniform distribution, even without guarantees on the mixing time. My first attempt at a random construction was to apply a randomized version of Redelmeier's method, or at least Wikipedia's description of this algorithm. But I am not sure if Wikipedia's description is correct. I copy it here, for the sake of completeness. The basic idea is that we begin with a single square, and from there, recursively add squares. Depending on the details, it may count each n-omino n times, once from starting from each of its n squares, or may be arranged to count each once only. The simplest implementation involves adding one square at a time. Beginning with an initial square, number the adjacent squares, clockwise from the top, 1, 2, 3, and 4. Now pick a number between 1 and 4, and add a square at that location. Number the unnumbered adjacent squares, starting with 5. Then, pick a number larger than the previously picked number, and add that square. Continue picking a number larger than the number of the current square, adding that square, and then numbering the new adjacent squares. When n squares have been created, an n-omino has been created. This method ensures that each fixed polyomino is counted exactly n times, once for each starting square. I was hoping that by making a uniform random legal choice at each step, I could generate a large random polyomino, but unfortunately the description of the algorithm above seems to sometimes get stuck. See, for example, the figure below. We have consructed a polyomino with 16 tiles, but no further moves are possible. REPLY [5 votes]: Here is an approach to estimating parameters of uniformly random polyominoes of $n$ squares without generating random ones uniformly. First define a rooted tree whose root is the single square and whose nodes at distance $m$ from the root are the $m$-square polyominoes, with no nodes further than $n$ from the root. This is fairly easy: the parent of polyomino $P$ can be $P-s$ where $s$ is the most-northern most-western square such that $P-s$ is a polyomino. Now you need a procedure that, given a polyomino $P$ tells you how many children it has and chooses a random child. If nothing better comes to mind, just try adding a new square in each possible place and reject it if $P$ is not the parent. Now start at the root and follow a random path down until you come to a leaf $L$ (with $m$ squares, say). At each node, choose a child uniformly at random. If $d_0,\ldots,d_{m-1}$ are the degrees (numbers of children) of the nodes along that path (excluding $L$), then $p(L)=(d_0\cdots d_{m-1})^{-1}$ is the probability that a random path ends at $L$. Now, if $\chi(L)$ is the indicator function of "$L$ has $n$ squares", then the expectation of $\chi(L)/p(L)$ is exactly equal to the number of $n$-square polyominoes, so you can estimate that number by sampling repeatedly. In fact, for any function $f$, the sum of $f(P)$ over all $n$-vertex polyominoes is exactly the expectation of $f(P)\chi(L)/p(L)$ for a random path. This lets you estimate the expectation of $f$ by repeated sampling. I've considered the simplest case, where all the children of a node are chosen with equal probability as a random path is traced. The theory is valid for any distribution on the children. To get good estimates you would like the probability of choosing a child to be roughly proportional to the number of $n$-square polyominoes descended from that child. Then each $n$-vertex polyomino comes out with similar probability. If that is far from the truth, the estimates may have large variance even though they have the correct expectation. This is an example of "sequential importance sampling". Searching on that phrase will bring up some papers; look for Diaconis in particular. I've seen this work extremely well and extremely badly on different problems.<|endoftext|> TITLE: When can the Cayley graph of the symmetries of an object have those symmetries? QUESTION [7 upvotes]: Let $P$ be an object in $\mathbb{R}^n$ with symmetry group $G$. Let $C$ be the a Cayley graph of $G$. When can $C$ be embedded in $\mathbb{R}^m$ so that the embedded graph has the same symmetry group $G$ as the original object $P$? For example, the standard Cayley graph for $D_4$, the symmetry group of the square, can be embedded in the plane with the same symmetry:                     (Image from J.Gross.) Let us view the Cayley graph as a directed, unlabeled graph. I would be interested in both positive and negative examples: $P$ such that $C$ can share the same symmetries, and $P$ such that none of its $C$ can realize the same symmetries. REPLY [3 votes]: (Answer rewritten in view of better understanding and interesting discussion in comments:) I first want to show that for "most" finite groups, all Cayley graphs of the group can be embedded into some $\mathbb{R}^n$ such that the symmetry group of the embedded Cayley graph is just $G$ (whether the Cayley graph is directed and labeled or not), and such that the natural action of $G$ on the embedded Cayley graph and the "natural" action on the Cayley graph coincide. (The last point is not true for the example in the question: $D_4$ has two orbits on the vertices of the embedded Cayley graph.) Babai has shown the following: If the finite group $G$ is neither abelian of exponent $>2$ nor generalized dicyclic, then there is some $v$ in some $\mathbb{R}^n$ with a representation $G\to \mathbf{O}(\mathbb{R}^n)$, such that $|G|=|Gv|$ and such that the symmetry group of the point set $Gv$ is just $G$. (Except for elementary abelian $2$-groups, Babai uses the regular representation $G\to \mathbf{O}(\mathbb{R}^{G})$ in his proof.) If we add line segments between $gv$ and $gsv$ corresponding to elements $s\in S$ of some generating set $S$ of $G$, then the resulting embedded Cayley graph has also symmetry group $G$, whether we consider the arcs as directed and labeled or not. (At least when we use the regular representation, these line segments will not intersect.) This argument shows that the question concerning geometric embeddings is in fact easier than the question of graphical regular representations. If, on the other hand, $G$ is abelian of exponent $>2$ or generalized dicyclic, then $G$ has an automorphism sending each group element to its inverse or itself, and which yields a symmetry of all such point orbits $Gv\subseteq \mathbb{R}^n$. This is also a symmetry of the Cayley graphs obtained in the above way, if we forget directions of arcs. On the other hand, in these groups, each generating set must contain elements of order $>2$, so it is not a symmetry of the directed and labeled Cayley graph. But this does not completely clarify the situation for these groups, since there may be other additional symmetries (depending on the representation). The original version of this answer contained the following special case: Suppose that the object $P\subseteq \mathbb{R}^n$ is a convex polytope such that its symmetry group $G$ acts transitively on the vertices of $P$ (and thus $P$ is the convex hull of an orbit of a point, called an orbit polytope). Then there is some point $v\in \mathbb{R}^n$ such that $|Gv|=|G|$ (only the identity of $G$ fixes $v$) and such that the symmetry group of $Gv$ is only $G$. (See Lemma 1 in Babai's paper or Corollary 5.4 in our paper. In fact, "almost all" points have these properties.) Let $Q$ be the convex hull of $Gv$, which is another orbit polytope. The $1$-skeleton of this polytope is in a natural way (an undirected version of) a Cayley graph of $G$, where the generating set $S\subseteq G$ is the set of elements $s\in G$ such that $v$ and $sv$ are connected by an edge (a $1$-dim. face) of $Q$. (Also revelant in this context is a paper of Ellis, Harris and Sköldberg (MR2270569 (2008g:20117)).) REPLY [2 votes]: [Wiki, as already in the comments] For the easy negatives, consider the Cayley graph with respect to the generating set $S=G$. The graph obtained is then a complete graph. It can be embedded in $\mathbb{R}^{G} = \{ (a_g)_{g \in G} \mid a_g \in \mathbb{R} \}$ as follows: $g \mapsto \delta_{g\cdot}$ where $\delta_{gh} = 1$ if $h=g$ and $0$ else. The edges between two vertices all have length $\sqrt{2}$. The permutation matrices are linear isometris of $\mathbb{R}^G$ and will permute the vertices as desired. So the symmetries of the Cayley graph contain $S_{|G|}$ which is strictly larger than $G$ as soon as $|G| \geq 2$. More generally, as Nick Gill pointed out, this embedding shows that (without bound on $m$, or in fact, as soon as $m \geq |G|$) there is no difference between the "abstract" automorphisms and the "geometric" automorphisms. Positive answers for "abstract" are found in this answer of C. Godsil (again, pointed out by N. Gill) https://math.stackexchange.com/questions/1098115/when-is-the-automorphism-group-of-the-cayley-graph-of-g-just-g<|endoftext|> TITLE: Does complex multiplication for higher dimensional abelian varieties give some generalization of class field theory? QUESTION [6 upvotes]: I am currently learning some aspects of the theory of complex multiplication for elliptic curves, and the relationship with class field theory. As I understand it, there is a very special class of elliptic curves $E$ whose automorphism rings are ideals $\mathfrak{a}$ of $\mathcal{O_K}$, the ring of integers of an imaginary quadratic field $K$. And conversely, for every ideal $\mathfrak{a}$ in an imaginary quadratic field $K$, there is an elliptic curve whose endomorphism ring is $\mathfrak{a}$. The correspondence goes by viewing $E$ as $\mathbb{C}/\Lambda$ for some lattice $\Lambda$, and the endomorphism ring as the set of complex numbers $z$ such that multiplication by $z$ maps $\Lambda$ into itself. These special curves give an explicit version of class field theory for $K$: the Hilbert class field is generated by the $j$-invariant of a $E$, and the rest of the maximal abelian extension is generated by numbers determined by the $x$-coordinates of torsion points on $E$. How does this generalize to the world of abelian varieties? I am dimly aware that the idea of complex multiplication is well-defined for an abelian variety $A$. Loosely, this corresponds to the case when the endomorphism ring of $A$ is larger than "expected", and this is related to some algebraic number field in a somewhat similar way to the above relationship. What does this look like on the class field theory side? Does it give an explicit class field theory for more general types of fields than imaginary quadratic fields? Or does it correspond to something more exotic - some Langlands-y analogue of class field theory for non-abelian extensions? I'm also vaguely aware that "explicit class field theory" in a form analogous to the case of complex multiplication has been developed for CM-fields, which are imaginary quadratic extensions of totally real number fields. Is this related to higher-dimensional abelian varieties? REPLY [2 votes]: I think this is due to Shimura: "Abelian Varieties with Complex Multiplication and Modular Functions", http://press.princeton.edu/titles/6242.html Edit: See also the link mentioned by nfdc23 in the comments to this response, http://math.stanford.edu/~conrad/vigregroup/vigre04/mainthm.pdf and http://jmilne.org/math/CourseNotes/cm.html.<|endoftext|> TITLE: linear independence of $\sin(k \pi / m)$ QUESTION [12 upvotes]: I have tried searching the literature for a result like the following, but have not found anything. For a positive integer $m$, is it known that $$\{ \sin (k \pi / m): 1 \leq k \leq m/2, (k,m)=1 \}$$ is linearly independent over the rationals? References or a proof would be greatly appreciated. REPLY [12 votes]: We have $$\sin\frac{\pi}{9}+\sin\frac{2\pi}9-\sin\frac{4\pi}9=\sin\frac{2\pi}{18}+\sin\frac{4\pi}{18}-\sin\frac{8\pi}{18}=\sin\frac{2\pi}{18}-\sin\frac{8\pi}{18}+\sin\frac{14\pi}{18},$$ and the latter, denoting $\xi_{18}=\exp\frac{2\pi i}{18}$, is the imaginary part of $$\xi_{18}-\xi_{18}^4+\xi_{18}^7=\xi_{18}(1-\xi_{18}^3+\xi_{18}^6)=0.$$ Thus, your conjecture is wrong.<|endoftext|> TITLE: Some examples of $\mathbb Q$-Gorenstein smoothing QUESTION [9 upvotes]: I am trying to understand $\mathbb Q$-Gorenstein smoothings, and especially the third condition in the following definition. Definition. For a normal projective surface $X$ with quotient singularities, a $\mathbb Q$-Gorenstein smoothing is a one-parameter flat family of projective surfaces $\psi \colon \mathcal X \rightarrow \Delta$ over a small disk $\Delta$, which satisfies the following three conditions: the general fibre $X_t$ is a smooth projective surface, the central fibre $X_0$ is isomorphic to $X$, the canonical divisor $K_{\mathcal X / \Delta}$ is $\mathbb Q$-Cartier. We say that $X'$ is a $\mathbb Q$-Gorenstein smoothing of $X$ if there exists such a family $\psi \colon \mathcal X \to \Delta $ as above such that $X'=\psi^{-1}(t)$ for some $t\in \Delta$. For example, I am trying to understand the following. Question 1. Assume that there exists a fixed integer $m$ such $mK_{X_t}$ is Cartier for every $t$. Is there any example where 3rd condition is not satisfied? Question 2. If the total space of the family is $\mathbb Q$-Gorenstein, is it true that the 3rd condition is satisfied? References with examples would be appreciated. REPLY [5 votes]: Just a note on the construction of the example Francesco mentioned, since in my opinion it is very instructive. Let $X$ be either the Veronese surface or a rational quartic scroll in $\mathbb P^5$. In the first case, it is $\mathbb P^2$ embedded via the global sections of $\mathscr O_{\mathbb P^2}(2)$ and in the second case it is $\mathbb P^1\times \mathbb P^1$ embedded via the global sections of $\mathscr O_{\mathbb P^1\times \mathbb P^1}(2,1)$. In both cases you get a degree $4$ surface in $\mathbb P^5$ and hence a general hyperplane section of both gives you a rational quartic curve in $\mathbb P^4$, let's call it $C$. So, the projective cone over $X$ in $\mathbb P^5$ has two different significant hyperplane sections: If the hyperplane misses the vertex, then the section is a copy of $X$ and if it goes through the vertex and is general among those, then it is a projective cone over $C$, let's call that $Y$. It is easy to see that the different hyperplane sections are deformations of each other, so this way you get a family with general fiber $X$ and special fiber $Y$. Clearly, for the two different choices of $X$ the two corresponding families can't both be $\mathbb Q$-Gorenstein deformations, since $$9=K_{\mathbb P^2}^2 \neq K_{\mathbb P^1\times \mathbb P^1}^2=8,$$ and as Francesco explained the cone over the quartic has $K^2=9$, so it is the quartic scroll who is the culprit and the reason for this issue is that the canonical of the scroll and the hyperplane class are incomparable, and hence the canonical of the total space is not going to be a torsion in the local class group which is what makes the (relative) canoncial not $\mathbb Q$-Cartier at the vertex. The importance of this example is that it explains an interesting phenomenon in the moduli theory of canonically polarized varieties of dimension at least $2$ (as opposed to curves). While stable curves are Gorenstein and hence any family of stable curves is automatically Gorenstein (not just $\mathbb Q$-Gorenstein), the same is not true for surfaces and higher dimensional varieties. This means that a stable family in dimensions at least $2$ is not simply a family of stable objects, but has an additional condition about the compatibility of the canonical polarizations. There are actually more intriguing issues that come up, but let me not write a book here, so if you want to know more about this check out the following references. The last paper by Altmann and Kollár is actually a major (very recent) breakthrough in this question, resolving an open problem that has baffled many people in the last 25 years. In particular, from their work it follows that even just asking that a stable family be $\mathbb Q$-Gorenstein, one needs to assume something known as "Kollár's condition". Well, Altmann and Kollár call it "qG-deformation", but everyone else calls it "Kollár's condition". You can read about the difference between Viehweg's functor and Kollár's functor in the first two references, and see that in both of them the status is reported as unknown whether those two functors are really different in characteristic $0$, but now due the [AK16] we know they are. Hacon, Christopher D.; Kovács, Sándor J. Classification of higher dimensional algebraic varieties. Oberwolfach Seminars, 41. Birkhäuser Verlag, Basel, 2010. x+208 pp. ISBN: 978-3-0346-0289-1 Kovács, Sándor J. Young person's guide to moduli of higher dimensional varieties. Algebraic geometry—Seattle 2005. Part 2, 711–743, Proc. Sympos. Pure Math., 80, Part 2, Amer. Math. Soc., Providence, RI, 2009. Klaus Altmann, János Kollár; The dualizing sheaf on first-order deformations of toric surface singularities, arXiv:1601.07805 [math.AG]<|endoftext|> TITLE: Small set such that $\{1 , \ldots , n\} \cdot A = \mathbb{Z} / p \mathbb{Z}$ QUESTION [12 upvotes]: Let $p$ be a large prime and $n < p$. What is the smallest size of a set $A \subset \mathbb{Z} / p \mathbb{Z}$ such that $A \cdot \{1 , \ldots , n\} = \mathbb{Z} / p \mathbb{Z}$? Here $\cdot$ denotes the product set $X\cdot Y = \{xy : x \in X , y \in Y\}$. Trivial bounds are $$p/n \leq |A| \leq p,$$ and using the probabilistic method one can improve the upper bound to $|A| \ll (p/n) \log p$, just a logarithm away from the lower bound. REPLY [8 votes]: I believe that the answer to this question can be found in this paper of Chen, Shparlinski and Winterhof. See Theorem 2 on page 6. It seems they give a contruction of a set with size $|A| \leq \frac{2p}{n}$ such that $A \cdot \{1,2,\dots,n\}=\mathbb Z_p$.<|endoftext|> TITLE: When is a functor a right derived functor? QUESTION [10 upvotes]: Suppose we have Grothendieck abelian categories $\mathcal{A}, \mathcal{B}$. Suppose also we have given an exact functor of triangulated categories $$ F \colon D(\mathcal{A}) \to D(\mathcal{B}) $$ where $D(\mathcal{A})$ and $D\mathcal({B})$ denote appropriate derived categories of complexes (possibly bounded below or both ways). Suppose that for any injective $I \in \mathcal{A}$ we have a functorial maps which are quasi-isomorphisms $F(I) \cong h^0F(I)$, and $F(D^{\geq 0}(\mathcal{A})) \subseteq D^{\geq 0}(\mathcal{B})$ ($F$ is $t$-left exact for the standard $t$-structure). Is it then true that $F$ is the right derived functor of is zeroth cohomology? I.e. $F \cong R(h^0F)$. If not, are there known counterexamples? Or which approporiate additional assumptions are needed so that such a statement holds? I believe that the above setup yields that the $i$th cohomology of $F$ is canonically isomorphic to the $i$th right derived functor of $h^0F$. However, I don't see how to extend the given quasi-isomorphism $F(I) \cong h^0F(I)$ for injectives in $\mathcal{A}$ to a natural transformation of functors $F \circ Q \to h^0F$ where $Q$ denotes the natural functor from the homotopy category $K(\mathcal{A})$ to $D(\mathcal{A})$. Once one has this, the universal porperty of the right derived functor should give the result. Some Background: In the situation I am interested in the functor $F$ arises as a composition of a left and right derived functor in a much bigger ambient category, whose restriction to $D(\mathcal{A})$ happens to be left exact and satisfy the condition $F(I) \cong h^0F(I)$ for injective objects of $\mathcal{A}$. I could imagine that this is not an uncommon situation... The result I.Proposition 7.4 in Residues and Duality (Hartshorne) is a statement of the type I am looking for. it says that under similar assumptions as above a right derived functor is (a shift of) the left derived functor of its highest non-vanishing cohomology... REPLY [16 votes]: Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories with enough injective objects. Let me use the notation $D^+(\mathcal{A})$ and $D^+(\mathcal{B})$ to denote the stable $\infty$-categories whose homotopy categories are the (cohomologically bounded below) derived categories of $\mathcal{A}$ and $\mathcal{B}$, respectively (you can also consider unbounded derived categories, but the situation is a bit more subtle). Let $\mathcal{C} \subseteq \mathrm{Fun}( D^{+}( \mathcal{A} ), D^{+}( \mathcal{B}) )$ be the full subcategory spanned by those functors which are exact, left t-exact, and carry injective objects of $\mathcal{A}$ into the heart of $D^{+}( \mathcal{B} )$. Then the construction $$F \in \mathcal{C} \mapsto h^0 F|_{ \mathcal{A} }$$ determines an equivalence from $\mathcal{C}$ to the category of left exact functors from $\mathcal{A}$ to $\mathcal{B}$. The inverse of this equivalence is "taking the right derived functor". Consequently, one can answer your question as follows: given a functor of triangulated categories $G: hD^{+}(\mathcal{A}) \rightarrow hD^{+}(\mathcal{B})$, it arises as a right derived functor (of a left exact functor of abelian categories) if and only if a) The functor $G$ lifts to an exact functor of stable $\infty$-categories $D^{+}(\mathcal{A}) \rightarrow D^{+}(\mathcal{B})$ (anything that you build by composing derived functors will have this property). b) The functor $G$ is left t-exact and carries injective objects of $\mathcal{A}$ into the heart of $hD^{+}(\mathcal{B})$.<|endoftext|> TITLE: k-flats in homogeneous spaces QUESTION [9 upvotes]: In a symmetric space of rank $k$ (and I'll take $k > 1$) every geodesic is contained in a $k$-flat: a totally geodesic, flat, connected, and closed submanifold of dimension $k$. Question. Are there non-symmetric homogeneous spaces that share this property? In this paper the author shows that if we also require that the isometry group act transitively on the set of pairs $(p, \Sigma)$, where $\Sigma$ is a flat and $p$ is a point in it, then the space is symmetric. My main interest is having many examples, homogeneous or not, of compact Riemannian manifolds for which every geodesic is contained in a totally geodesic, flat torus of dimension $k > 1$. REPLY [7 votes]: If the k-flats are compact, then the space must be symmetric (Molina-Olmos, J. Differential geometry 45 (1997) 575-592; see also Proc. Amer. Math. Soc. 129 (2001), 3701-3709). Homogeneous spaces (non-symmetric and irreducible) with the property that every geodesic is contained in a k-flat ($k\geq 2$) can be constructed as follows (due to Ernst Heintze): choose a simple compact Lie group $G$ and a compact subgroup $H$ such that $\mathrm{rank} (G)\geq \mathrm{dim} (H) +2$. Then $G/H$, with the normal hogeneous metric is a desired example (see also Spatzier-Strake, Comment. Math. Helv. 65 (1990) 299-317). Observe that the projection from $G$, with the bi-invariant metric, onto $G/H$, with the normal homogeneous metric, is a Riemannian submersion.} (The known fact that $G/H$ is irreducible, as a Riemannian manifold, can be found in Olmos-Reggiani-Tamaru, Math.Z. 277 (2014), 611-628).<|endoftext|> TITLE: Restriction of irreducible unitary representation to normal subgroup of finite index QUESTION [10 upvotes]: Let $G$ be a Lie group (or more generally a locally compact group), let $N$ be a closed and normal subgroup of $G$ of finite index. Let $H$ be an infinite dimensional complex Hilbert space, and let $\pi$ be an irreducible unitary representation of $G$ on $H$. Now consider the restriction $\pi|_N$ of $\pi$ to $N$. Is it true that $\pi|_N$ is the finite sum of irreducible unitary representations? Perhaps someone could provide an indication of the proof or a reference? ADDED COMMENT: Or how about a more basic question: Must $(\pi|_N, H)$ have a (discretely occuring) irreducible subrepresentation? Or even just an irreducible quotient (as would exist for a finitely generated algebraic representation)? REPLY [8 votes]: I like this question! Restricted to the finite index subgroup $N$, the representation $\pi$ splits into a direct sum of irreducible representations. I could not see an easy proof of this, but the proof goes along the following lines. Suppose $I$ is a totally ordered indexing set and for each $i\in I$, $W_i$ is an $N$ invariant non-zero (closed) subspace of $H$ such that $i TITLE: Artin conjecture on L-functions QUESTION [11 upvotes]: Artin conjecture on Artin $L$-functions asserts that the Artin $L$-function $L(\rho,s)$ of a non-trivial irreducible representation $\rho$ of the Galois group $\Gamma$ of a number field admits analytic continuation to the whole complex plane. It is known for $1$-dimensional and induced representations, plus a few other special cases. What is the status towards a proof? References would be very much appreciated. Langlands, R. P. (1970), "Problems in the theory of automorphic forms", Lectures in modern analysis and applications, III, Lecture Notes in Math 170, Berlin, New York: Springer-Verlag, pp. 18–61. REPLY [8 votes]: This is the status as far as I know. For dimension $\leq 2$ it is up to date. For higher dimensional representations I'm sure it is very incomplete, so feel free to edit or comment. Dimension 1. Known by "Artin-Hecke". Dimension 2. Only open case is even $A_5$ representations. References for the known cases are: $C_n$ "Artin-Hecke" $D_n$ "Artin-Hecke" (see here) $A_4$ Langlands, "Base change for GL(2)" (1980) $S_4$ Tunnell, "Artin's conjecture for representations of octahedral type" (1981) $A_5$ (odd $\rho$ over $\mathbb{Q}$) Khare-Wintenberger, "Serre's modularity conjecture (I)" (2009) $A_5$ (odd $\rho$ over totally real fields) Pilloni-Stroh, "Surconvergence, ramification et modularité" (2013) Dimension 3. Mostly wide open. ($\rho$ induced) Jacquet, Piatetski-Shapiro, Shalika, "Relèvement cubique non normal" (1981) ($\rho$ twist of a symmetric square) Gelbart, Jacquet, "A relation between automorphic representations of GL(2) and GL(3)" (1978) Dimension 4. Only open solvable cases are $E_{2^4}\cdot D_{10}$ and $E_{2^4}\cdot F_{20}$. There are known non-solvable cases, but in general it's wide open. ($\rho$ solvable) $\mathrm{GO}_4$ Ramakrishnan, "Modularity of solvable Artin representations of GO(4)-type" (2001) ($\rho$ solvable) $E_{2^4}\cdot C_5$ Martin, "A symplectic case of Artin's conjecture" (2003) Dimension $\geq 5$ Again there are some known cases, but mostly wide open. The consensus is that a solution of the complete Artin conjecture is only accesible from general functoriality results such as base change or induction that would imply the strong Artin conjecture, and therefore Artin's holomorphy conjecture. This means that the only way we have to prove that an Artin L-function $L(\rho, s)$ is holomorphic is to prove that $\rho$ is modular. An alternative, as pointed out by Julien in the comments, would be to solve Selberg's orthogonality conjecture. That this implies the Artin conjecture is proved in "Selberg's Conjectures and Artin L-functions" (1994).<|endoftext|> TITLE: Geometric intuition for the condition of Galois descent QUESTION [6 upvotes]: Continuing in my attempts to understand bits and pieces of Borceux and Janelidze's Galois Theories, I've just realized that I don't have any geometric intuition for the most convenient characterization of Galois descent for covering morphisms. The general setting is a complete category of the form $\mathsf{Fam}(\mathsf A)$, or alternatively, a complete category with a functorial choice of connected components for each object. The geometric intuition (I hope) comes from looking at locally connected spaces and I want it to understand Galois field extensions geometrically. The idea is that an arrow $p:E\rightarrow B$ is a morphism of Galois descent if and only if the following two conditions hold (see Galois Theories between Prop 6.6.6 and 6.6.7). $p$ is an effective descent morhpsim $p$ is trivialized by itself, i.e $p^\ast p$ is a trivial covering morphism (trivial fiber bundle with discrete fiber). Of course the first condition makes sense, but the second one? What's the idea there? Why would I want $p$ to trivialize itself? For fields all arrows are effective descent morphisms so being of Galois descent amounts to asking for the square below to be a pullback, where the top left corner is a finite coproduct of copies of $\operatorname{Spec} E$: $$\require{AMScd} \begin{CD} \operatorname{Spec} E^{\amalg{n}} @>>> \operatorname{Spec}E \\ @VVV @VV{p}V\\ \operatorname{Spec}E @>>{p}> \operatorname{Spec} B \end{CD}$$ i.e as usual an $E$-algebra isomorphism $E\otimes_BE\cong E^n$. Added. I dug out this paper by Schauenburg on Hopf-Galois and Bi-Galois extensions, and in the long paragraph above Lemma 2.4.2 he says a lot of seemingly nice things which are way over my head. The words 'principal bundle' appear often enough to warrant some hope. In particular, it's written: This is the algebro-geometric version of a principal fiber bundle with structure group $G$, or a $G$-torsor. Now I do not know almost any algebraic geometry beyond the very basics, and I am looking for topological/geometric intuition of the original condition I asked about, so if this addition is irrelevant to the question, just ignore it :) REPLY [6 votes]: Let $G \to E \to B$ be a principal bundle. It's classified by a map $f : B \to BG$ in the sense that $E$ is the homotopy fiber of this map. This means that $E$ has a certain universal property: namely, it is the universal map to $B$ such that the composition with $f$ is equipped with a nullhomotopy. This says precisely that $E$ is the universal space over $B$ equipped with a trivialization of the pullback of the principal bundle to $E$: in other words, $E \to B$ is the universal thing that trivializes itself. The relationship to Galois extensions is that $K \to L$ is a finite Galois extension with Galois group $G$ iff $\text{Spec } L \to \text{Spec } K$ is a principal $G$-bundle in a suitable sense ("$G$-torsor").<|endoftext|> TITLE: Fields generated by torsion points of CM elliptic curves QUESTION [6 upvotes]: I'm using the same setup as Corollary 1.7 on p. 44 of de Shalit manuscript (Iwasawa theory of elliptic curves with complex multiplication). I think there is a mistake in his Corollary 1.7 and I'm wondering if it is possible to fix it. So let $E/F$ be an elliptic curve defined over a number field $F$ such that (1) $E$ has complex multiplication by $\mathcal{O}_K$, (2) $F(E_{tor})$ is abelian over $K$ (this is a strong assumption, since a priori we only know that $F(E_{tor})$ is abelian over $F$). In particular, it follows from (2) that $K(1)\subseteq F$, where $K(1)$ stands for the Hilbert class field of $K$. Let $\mathfrak{f}\subseteq\mathcal{O}_F$ be the conductor of the Groessencharacter associated to $E/F$ (which depends just on the $F$-isogeny class of $E$). For an integral ideal $\mathfrak{m}\subseteq\mathcal{O}_K$, we let (a) $F[\mathfrak{m}]=F(E[\mathfrak{m}])$ (b) $F(\mathfrak{m})=F(x(P):P\in E[\mathfrak{m}])$ It is easy to see that $[L[\mathfrak{m}]:L(\mathfrak{m})]\leq 2$ and that we have an injection $\theta_{\mathfrak{m}}:Gal(F[\mathfrak{m}]/F)\hookrightarrow (\mathcal{O}_K/\mathfrak{m})^{\times}$. [added: Most interesting example: If $F=K(1)$ and if $\mathfrak{m}$ is such that $1\in\mu_K$ is the only root of unity $\zeta\in\mu_K$ such that $\zeta\equiv 1\pmod{\mathfrak{m}}$, then $Gal(F(\mathfrak{m})/F)\simeq (\mathcal{O}_K/\mathfrak{m})^{\times}/\mu_K$. In particular, if $\mu_K=\{\pm 1\}$ then $im(\theta_{\mathfrak{m}})$ has at most index $2$.] Let $\mathfrak{g}\subseteq\mathcal{O}_K$ be another ideal and let us assume that $(\mathfrak{f}\cap\mathcal{O}_K) |\mathfrak{g}$ and that $(\mathfrak{m},\mathfrak{g})=1$. ---------(added on 6th of April 2017: in fact de Shalit's assumptions on the ideals $\mathfrak{g}$ and $\mathfrak{m}$ are more restrictive than what I initially wrote above)--------. Then de Shalit claims the following: (c) $\theta_{\mathfrak{m}}$ is surjective (d) $F[\mathfrak{m}]$ is linearly disjoint from $F[\mathfrak{g}]$ over $F$. In general, I don't see why $\theta_{\mathfrak{m}}$ is surjective, since in one of his computations, he forgets the contribution coming from the roots of unity of $K$. Moreover, if we take into account this omission, it seems to me that we only obtain that $F(\mathfrak{m})$ and $F(\mathfrak{g})$ are linearly disjoint over $F$. Q1 Are (c) and (d) still true ? There seems also to be something wrong with his proposition 1.6 (his proof of Corollary 1.7 is partly based on it), since it seems to me that one could assume from the outset that $F$ contains the coordinates of the points $E[\mathfrak{m}]$ without changing the assumptions (1) and (2) above (but may be his proposition 1.6 is correct if $F=K(1)$ which is the most interesting case). Here is one related question to the previous paragraph (still assuming that (1) and (2) hold true) which I don't know the answer: Q2 Let us assume that $F=K(1)$. Then we know from CM theory that $F(\mathfrak{m})$ corresponds to the ray class field of $K$ of modulus $\mathfrak{m}$. Let us assume that $[F[\mathfrak{m}]:F(\mathfrak{m})]=2$ (which I expect to be the "generic" case). What is the conductor of the abelian extension $F[\mathfrak{m}]/K$ (note in particular that this conductor must be divisible by $\mathfrak{m}$) ? REPLY [3 votes]: The proof of Corollary 1.7 is fine. I had misunderstood his proof. His proof uses in a crucial way his assumption (ii) which appears on the top of p. 41. As is explained on p. 41, this assumption implies that the Groessencharacter $\psi$ associated to E/F (a Groessencharacter on F) comes from a Groessencharacter $\varphi$ on K of type $(1,0)$. Le me just repeat de Shalit's argument in a hopefully slightly more detailed way: (1) We have $[F[E[{\frak{m}}]]:F]\leq \# (\mathcal{O}_{K}/\frak{m})^{\times}$, $K({\frak{g}})=F(E[{\frak{g}}])$ (from his Proposition 1.6) and $K({\frak{m}\frak{g}})=F[E[{{\frak{m}}}]]\cdot K({\frak{g}})=K(E[{\frak{m}}{\frak{g}}])$ (again from his Proposition 1.6) (2) The key observation now is that the set of roots of unity of $F[\frak{g}]$ which are congruent to $1$ modulo $\frak{g}$ is reduced to $\{1\}$. This comes from our assumption that the conductor of $\varphi$ divides $\frak{g}$ and that $\varphi$ has type $(1,0)$ (I had missed the type $(1,0)$ assumption initially). From (2), it follows that (3) $[ K( {\frak{m}} {\frak{g}} ) : K ({\frak{g}}) ]=\# (\mathcal{O}_{K}/\frak{m})^{\times}$. The result now follows by incorporating (1) and (3) into the Hasse diagram which appears below his Corollary 1.7. Note that (a) the linear disjointness of $F(E[\frak{g}])$ and $F(E[\frak{m}])$ over $F$ and the equality (b) $[F[E[{\frak{m}}]]:F]=\# (\mathcal{O}_{K}/\frak{m})^{\times}$ are proved simultaneously!<|endoftext|> TITLE: Stable unions without stable images QUESTION [6 upvotes]: A regular category is one with finite limits and pullback-stable images (i.e. (regular epi, mono) factorizations). A coherent category is a regular category that also has pullback-stable finite unions of subobjects (monomorphisms). These two classes of categories correspond to fragments of first-order logic containing $\wedge,\top,\exists$ and $\wedge,\top,\vee,\bot,\exists$ respectively. But what about categories corresponding to a fragment containing only $\wedge,\top,\vee,\bot$? Are there naturally-ocurring examples of categories having finite limits and pullback-stable unions of subobjects but not pullback-stable images? REPLY [7 votes]: Here's another example, inspired by aws's but not involving any computability. Let $A$ be a distributive lattice and let $\mathrm{Fam}(A)$ be the free coproduct-completion of $A$, whose objects are families $\{a_i\}_{i\in I}$ of elements of $A$ and whose morphisms $\{a_i\}_{i\in I} \to \{b_j\}_{j\in J}$ are functions $f:I\to J$ such that $a_i \le b_{f(i)}$. Then $\mathrm{Fam}(A)$ has finite limits (given by limits in $\mathrm{Set}$ and meets in $A$) and pullback-stable finite unions (given by unions in $\mathrm{Set}$ and joins in $A$; distributivity of $A$ is needed to make these pullback-stable). Its monomorphisms are arbitrary morphisms whose underlying set-function is injective. But it doesn't have images unless $A$ is a complete lattice; the image of the unique map from any family $\{a_i\}_{i\in I}$ to the terminal object would have to be the one-element family on a join of all the $a_i$.<|endoftext|> TITLE: For which Ramsey type results density versions are wrong? QUESTION [8 upvotes]: I look for examples of Ramsey-type statements, for which the density counterparts do not hold. Example: usual Ramsey theorem. If all edges of a complete graph $K_n$ are colored in $c$ colors, there is a monochromatic, say, triangle if $n>n_0(c)$ is large enough. But if we choose more than $\frac1c \binom{n}2$ edges, it may appear that there is no triangle formed by the chosen edges. Another (related) example (Schur theorem): if we color $\{1,\dots,n\}$ in $c$ colors, there is a monochromatic solution of $x+y=z$. It is not true that if we choose a half of numbers, than there exists a solution of above equation with $x,y,z$ chosen. Say, we could choose only odd numbers. On there other side, there are very important examples, when denisty versions are true (Szemeredi theorem, density Hales-Jewett and many others). My question is to 1) give less trivial examples; 2) give some theorems or conjectures on when density versions hold and when fail. REPLY [5 votes]: Here are a few examples from graph-Ramsey theory. In the first pair of examples, the Ramsey version and density version are essentially as far apart as one can get. In the last two pairs of examples, the two versions coincide. Now I wonder if there is an example from graph-Ramsey theory where the bound from the density version is strictly stronger than the Ramsey version? But in general, it seems to me that your question could be narrowed down by simply asking for results in which the density version implies the Ramsey version, since those seem to be more rare. 1R) In every 2-coloring of $K_n$, there is a monochromatic connected subgraph on $n$ vertices. (Folklore) 1D) Every graph on $n$ vertices with at least $\binom{n-1}{2}+1$ edges is connected. (Folklore) 2R) In every 2-coloring of $K_n$, there is a monochromatic path on at least $2n/3$ vertices. (Gerencsér, Gyárfás) 2D) Every graph one $n$ vertices with at least $\frac{2}{3}\binom{n}{2}$ edges has a path on at least $2n/3$ vertices. (Erdős, Gallai) 3R) In every 2-coloring of $K_n$, there is a monochromatic matching covering at least $2n/3$ vertices. (Cockayne, Lorimer) 3D) Every graph on $n$ vertices with at least $\frac{5}{9}\binom{n}{2}$ edges has a matching covering at least $2n/3$ vertices. (Erdős, Gallai) 4R) In every 2-coloring of $K_n$, there is a monochromatic copy of every tree $T$ with at most $n/2+1$ vertices. Furthermore, there are trees with more than $n/2+1$ vertices for which this is not true. (Burr-Erdős conjecture, solved for large $n$ by Zhao) 4D) Every graph on $n$ vertices with at least $\frac{1}{2}\binom{n}{2}$ edges contains every tree with at most $n/2+1$ vertices. (Erdős, Sós conjecture) 5R) In every $r$-coloring of $K_{n,n}$, there is a monochromatic connected subgraph on at least $2n/r$ vertices and this is essentially best possible. (Gyárfás) 5D) Every balanced bipartite graph on $2n$ vertices with at least $n^2/r$ edges has a connected subgraph on at least $2n/r$ vertices and this is also best possible. (Gyárfás; Mubayi; Liu, Morris, Prince)<|endoftext|> TITLE: Mapping a group to a finite group s.t. the image of each generator is nontrivial QUESTION [5 upvotes]: Recall that a group $G$ is called residually finite if for any nontrivial element $g\in G$ there exists a finite group $H$ and a homomorphism $f$ from $G$ to $H$ such that $f(g)\neq1$. My question is for which kind of finitely presented group $G=\{x_1, \cdots, x_n|r_1, \cdots, r_m\}$, there exists a finite group $H$ and a homomorphism $f$ from $G$ to $H$, such that $f(x_i)\neq 1$ $(1\leq i\leq n)$. Finite groups and free groups are naive examples. REPLY [13 votes]: I'm a bit unclear on what you're asking, but I'll assume that you mean that given $G$, is there a presentation $G=\{x_1, \ldots, x_n|r_1, \ldots, r_m\}$ for which the property holds (note that your question only concerns the generators $x_1,\ldots x_n$, so the relators $r_j$ are irrelevant)? Otherwise, if the presentation is given, then the property will hold iff none of the $x_i$ lie in the kernel of the map $G\to \hat{G}$, the profinite completion. Given this interpretation, then the answer is that as long has $G$ has a non-trivial homomorphism to a finite group $H$, then you can always find a generating set with this property. Starting with some generators $G=\langle x_1, \ldots, x_n\rangle$, and a non-trivial homomorphism $f:G\to H$, then $f$ must be non-trivial on some generator, say $f(x_1)\neq 1$. Now, consider all the generators $x_i, \ldots, x_n$ (by relabeling) such that $f(x_j)=1$, $i\leq j\leq n$. We obtain a different generating set $G=\langle x_1,\ldots, x_{i-1}, x_ix_1^{-1}, \ldots, x_nx_1^{-1}\rangle$. Clearly now $f$ is non-trivial on each one of these generators.<|endoftext|> TITLE: Random Cantor sets on the unit interval QUESTION [6 upvotes]: Denote $A=\{0\}, B=\{0,1\}$. Then any subset of $\Omega:=\{A,B\}^{\mathbb N}$ is a continuum provided the number of $B$'s is infinite. We treat these as binary expansions of numbers in $[0,1]$. For instance, $(AB)^\omega$ is the set $\left\{\sum_{n=1}^\infty a_n4^{-n}\mid a_n\in\{0,1\}\right\}$, which is a Cantor set of Hausdorff dimension $\frac12$. Similarly, it is easy to show that $(AB^N)^\infty$ has Hausdorff dimension $\frac{N}{N+1}$. Now, suppose the sequence of $A$'s and $B$'s is random: i.i.d. with distribution $\bigl(\frac1{N+1}, \frac{N}{N+1}\bigr)$. It is obvious that a generic sequence will be a Cantor set. Question. Is it true that a subset of $\Omega$ has Hausdorff dimension $\frac{N}{N+1}$ almost surely? A weaker version: is this a null set almost surely? REPLY [4 votes]: Like the classical Cantor set, your set has a natural representation $C=\bigcap C_n$, where $C_1\supseteq C_2\supseteq C_3\supseteq\ldots$, and each $C_n$ is a disjoint union of $2^{n-X_n}$ intervals of length $2^{-n}$, with $X_n$ denoting the number of $A$'s in your sequence up to position $n$. So, covering by intervals of that length, we have that $H_{2^{-n}}^d(C_n)=2^{n(1-d)-X_n}$. If we write $p=P(A)$, then $X_n/n\to p$ almost surely, so $\dim C\le 1-p$ almost surely. For the lower bound, we observe that $C$ will contain at least one point from each of the $2^{n-X_n}$ intervals of length $2^{-n}$ whose union is $C_n$. This means that we cannot find a $2^{-n}$ covering of $C$ that decreases the number of intervals needed by more than a fixed factor, compared to the one of $C_n$ from above. So $\dim C=1-p$ almost surely, as claimed.<|endoftext|> TITLE: Stochastic Analogue of Stokes Theorem QUESTION [6 upvotes]: Dynkin's formula can be thought of as the stochastic version of the Fundamental Theorem of calculus, $$E^x[f(X_{\tau})]=f(x)+E^x\left[\int_0^{\tau}Af(X_u)du\right],$$ where $\tau$ is a first exit time and $A$ is the generator of the process $X_t$. I'm wondering what the stochastic version of the Stokes theorem, say on a surface with boundary, should be? REPLY [4 votes]: See, e.g., "Integral of differential forms along the path of diffusion processes" by Ikeda and Manabe. Perhaps the "Stokes theorem" in Malliavin's book is also of interest.<|endoftext|> TITLE: Inverse galois problem and étale homotopy QUESTION [8 upvotes]: Is there any relation between étale homotopy theory (Grothendieck-Galois theory) and the inverse Galois problem?...I mean...in classical homotopy theory, every finite group $G$ realizes as a "Galois group"...that is...one has the universal covering $p_G:EG\rightarrow BG$ for which $$Gal(p_G):=Deck(p_G)\simeq\pi_1(BG)\simeq G$$ Hence, my question would be: Is the classifying algebraic stack $\mathscr{B}G$ a sort of Eilenberg-Maclane space $K(G,1)$ in étale homotopy theory?...can this say something about the inverse Galois problem? REPLY [12 votes]: The inverse Galois problem is a statement about the étale fundamental group of a specific scheme. So constructing a space whose fundamental group is $G$ does not tell you whether $G$ is a quotient of the fundamental group of $\operatorname{Spec} \mathbb Q$. An analogous problem in topology would be. Q: Does every finite simple group $G$ occur as the deck transformation group of a covering of the wedge of two circles? A: Yes, because every finite simple group can be generated by two elements (by an explicit calculation with CFSG), hence is a quotient of the free group on two generators, which is the fundamental group of the wedge of two circles. You can see that the linchpin of the argument is explicit calculations involving the specific space, and involving the specific group, and the characterization in terms of $BG$ does not help much.<|endoftext|> TITLE: What did Yu Jianchun discover about Carmichael numbers? QUESTION [38 upvotes]: There's a news story going around (see for example [1]; other accounts are even more breathless) about an amateur mathematician, Yu Jianchun, finding an "alternative method to verify Carmichael numbers". I'm curious as to the method; does anyone know about this? I don't think this is smoke and mirrors, as was the claimed proof of the Riemann Hypothesis by Opeyemi Enoch of the Federal University Oye-Ekiti, Nigeria. [2] Cai Tianxin of Zhejiang University has apparently read the proof and invited Yu Jianchun to give a talk at a graduate seminar, and CNN apparently ran the story by William Banks who apparently compared the discovery to the proof of the infinitude of Carmichael numbers (though from what appears in the article it seems he did not have access to the work itself). But the article (appropriately, I suppose, for its nontechnical audience) doesn't give any real information on the discovery itself. The closest it comes is a picture of a letter (in Chinese) from Yu Jianchun with most of a formula: $$ \frac{\left(\frac{N^{P_1}-N}{N} - \frac{N^{P_1-P_2+1}-N}{P_1-P_2}\right)(P_1-P_2)}{P_2}(N,P_1,P_2 $$ (note that the rightmost part is cut off). There is also what might be a worked example, starting $$ \begin{align} =&\frac183^{3n}(3^n+1)^3(3^n+2)^3(3^{2n-1}+2\cdot3^{n-1}+1)^3\\ =&\left[\frac123^n(3^n+1)(3^n+2)(3^{2n-1}+2\cdot3^{n-1}+1)\right]^3 \end{align} $$ but it's hard to make sense of this without understanding the surrounding text (in Chinese). [1] China's 'Good Will Hunting?' Migrant worker solves complex math problem [2] Riemann Hypothesis not proved REPLY [4 votes]: On page 136 of Cai Tianxin, The Book of Numbers, it says, In 2015 the Chinese amateur, Jianchun Yu, who was then a packer in a bookstore warehouse in Hangzhou found that all the numbers like $(6k+1)(12k+1)(24k^2+6k+1)$ are Carmichael numbers if these three factors are all primes. The proof is left as an exercise on page 142. I found no other mention of Jianchun's work in the book.<|endoftext|> TITLE: Penrose transform and general wave equations QUESTION [14 upvotes]: In the late 1960's Penrose developed twistor theory, which (amongst other things) led to an exceptional description for solutions to the wave equation on Minkowski space via the so-called Penrose transform; If \begin{equation} u(x,y,z,t) = \frac {1} {2 \pi i} \oint_{\Gamma \subset \mathbb{C} \mathbb{P}^{1}} f(-(x+iy) + \lambda (t-z), (t+z) + \lambda (-x + i y), \lambda ) d \lambda, \,\,\,\,\,\,\,\,\,\, (1) \end{equation} where $\Gamma \subset \mathbb{C} \mathbb{P}^{1}$ is a closed contour and $f$ is holomorphic on $\mathbb{C} \mathbb{P}^{1}$ except at some number of poles, then $u$ satisfies the Minkowski wave (Laplace-Beltrami) equation $\square_{\eta} u = 0$. I am aware that there are a number of works in the literature describing twistor theory on curved manifolds, but have not seen explicit constructions along the lines of (1) such that the function $u$ satisfies a wave equation of the form $\square_{g} u = 0$ for (Lorentzian) metric $\boldsymbol{g}$. Is it known how to explicitly construct contour integrals similar to $(1)$ for some class of metrics $\boldsymbol{g}$? What about when $\boldsymbol{g}$ is Einstein (e.g. Schwarzschild), in particular? Are there topological obstructions in spacetimes $I \times \Sigma$? What about de-Sitter space? REPLY [3 votes]: I believe you can express the Penrose transform using a contour integral by expressing Serre duality in terms of Dolbeault cohomology. For example, IIRC, on a half-conformally-flat 4-manifold $M$ with twistor space $Z$, functions solving the Laplacian equation correspond to elements of the sheaf cohomology group $H^1(Z, \mathcal{O}(-2))$ and the evaluation at a point $x \in M$, with corresponding twistor line $P_x \subset Z$, is to restrict to: $$ H^1(P_x, \mathcal{O}(-2)) \simeq H^0(P_x, \mathcal{O})^* \simeq \mathbb{C}. $$ If we represent our element of $H^1(P_x, \mathcal{O}(-2))$ by a form in $H^{0,1}(P_x, \mathcal{O}(-2))$, then choosing a contour that does not meet the singularities from the section $\mathcal{O}(-2)$, the Serre duality map above yields an contour integral by Stokes's theorem.<|endoftext|> TITLE: On a result of Frankl and Wilson QUESTION [8 upvotes]: In the paper 'Intersection theorems with geometric consequences' (Combinatorica 1981) P. Frankl and R. M. Wilson consider families $\mathcal{F}$ of $k$-subsets of $\{1,\dots,n\}$ with the restriction on mutual intersections: $|A\cap B|-k$ is not divisible by $q$, where $q$ is a fixed prime power, whenever $A\ne B$ are from $\mathcal{F}$. They claim that $|\mathcal{F}|\leqslant \binom{n}{q-1}$, it is cited also by R. L. Graham here. This bound looks very strange: what if $n=q-1>k$, for example? Or $q-1>n$? I think that from their proof the following weaker estimate may be obtained: $|\mathcal{F}| \leqslant \sum_{i\leqslant q-1}\binom{n}{i}$. Now the questions: was there errata, or maybe everything is correct and I miss something? What are current best lower and upper bounds in this question? REPLY [2 votes]: The form of the Frankl-Wilson that I know is the following. Frankl-Wilson Theorem. Let $q$ be a prime power and let $\mathcal{F}$ be a family of $k$-sets of an $n$-set such that $|A \cap B| \equiv \lambda_1, \dots,$ or $\lambda_s$ for all distinct $A,B \in \mathcal{F}$ and where $\lambda_i \not\equiv k$ for all $i$. Then $|\mathcal{F}| \leq \binom{n}{s}$. I believe this form of the theorem also appears in the original paper of Frankl and Wilson. It seems to me that there is an implicit assumption that each $\lambda_i$ actually does occur as the intersection number of two sets in $\mathcal{F}$. Otherwise, as you have said, it does not look like the theorem is technically correct. However, with the additional assumption, I think everything is fine. For example, with the additional assumption, we have that $s$ is clearly at most $k$, since each set in $\mathcal{F}$ is a $k$-set. Therefore, in the situation that $s=q-1$ (by taking the set of $\lambda_i$ to be all elements of $\mathbb{Z} / q\mathbb{Z}$ except $k$), we must also have $k \geq q-1$. So the situation you describe $n=q-1>k$ is impossible for $s=q-1$. If we are considering the non-uniform version of the Frankl-Wilson Theorem (where the sets are allowed to have different sizes), then the bound that you prove is indeed the correct one. For example, see Theorem 1.3 of this paper of Alon, Babai and Suzuki.<|endoftext|> TITLE: Rank 2 vector bundles over $\mathbb CP^2$ QUESTION [7 upvotes]: Is there any classification of the rank 2 complex vector bundles over $\mathbb CP^2$ up to diffeomorphism? REPLY [3 votes]: As additional comment, We have Cartan-Serre theorem for construction of rank 2 vector bundles on projetive varieties. Let $X$ be a complex manifold, $L_1, L_2 ∈ Pic(X)$ are line bundles, $Z ⊂ X$ with $codim_X(Z) = 2$. Under some cohomological conditions, the sheaf $E$ sitting in $$0 \to L_1 \to E \to L_2 \otimes J_Z → o$$ is locally free. Moreover, If $X$ is projective, any rank-2 holomorphic vector bundle can be constructed this way<|endoftext|> TITLE: Action of longest element of Weyl group on zero weight space QUESTION [6 upvotes]: Let: $G$ be a real semisimple Lie group; $\rho$ be an irreducible representation of $G$ on a finite-dimensional real vector space; $A$ be a "Cartan subspace" of $G$ (a Lie subalgebra which is abelian, composed of hyperbolic elements, and maximal); $L$ be the centralizer of $A$ in $G$ (this group is often called "$MA$"); $W = N_G(A)/Z_G(A)$ be the restricted Weyl group of $G$, and $w_0$ its longest element. I would like to classify the representations that satisfy the following two conditions: the space $V^L$ of points of $V$ fixed by every element of $L$ is nonzero; the action of $w_0$ on $V^L$ is nontrivial. An important particular case (which it would already be nice to have) is when $G$ is split. In that case, $V^L$ is simply the zero weight space, which is nonzero whenever the highest weight is an integer combination of roots. So we just need to find the action of $w_0$ on the zero weight space. This looks like it should have already been done somewhere, but I have been unable to find any references so far. Any pointers would be appreciated! REPLY [3 votes]: The action of $w_0$ can be computed by Allen Knutson's suggestion (whether or not $-1\in W$). As Allen says, suppose $g\in N(T)$ is a representative of $w_0$. Then $\text{trace}(\pi(w_0)) \text{ on } {V_0}=\text{trace}(\pi(g))\text{ on } V$ (since the only element of the root lattice fixed by $w_0$ is $0$). It is known that $g$ is conjugate to $\exp(\pi i\rho^\vee)$. Note that $g^2=\exp(2\pi i\rho^\vee)\in Z(G)$, and $\lambda(g^2)=e^{2\pi i\langle\lambda,\rho^\vee\rangle}=1$. Write $\Delta(V)$ for the set of weights of $V$. These are in the root lattice; write $\text{ht}(\mu)$ for the height (writing $\mu$ as $\pm$ a sum of simple roots). Then $$\text{trace}(\pi(w_0))=|\{\mu\in\Delta(V)\mid \text{ht}(\mu)\text{ even}\}|-|\{\mu\in \Delta(V)\mid ht(\mu)\text{ odd}\}|$$ Since $\pi(w_0)$ on $V_0$ is diagonalizable with $\pm1$ on the diagonal this computes the number of $1/-1$ eigenvalues.<|endoftext|> TITLE: uniform one-sided van der Corput inequality QUESTION [5 upvotes]: Is the following true (and if yes, where the best proof is written?)? For any $c>0$ for large enough positive integers $N$ we have $\sum_{k=0}^{N-1} \cos(k^2t)\geqslant -cN$ for all real $t$? Hm, if true, it should be hard: it allows to get signs of certain Gauss type sums, for example. REPLY [5 votes]: The answer is negative. As you said if you take $\frac{t}{2\pi}$ to be a rational number, say $\frac{p}{q}$ then it converges to the Gaus-sums: $$ \lim_{N \to \infty} \frac{1}{N} \sum_{k=0}^{N-1} \cos(k^{2} t) = \Re \frac{1}{q} \sum_{k=0}^{q-1} \exp\left( 2 \pi i \cdot \frac{p}{q} k^{2}\right) $$ Now if you take $t=\frac{4 \pi}{5}$ (i.e., $p=2, q=5$) I think it should no be difficult to show that the latter sum is a negative number $< - 0.4$ (even without taking the real part).<|endoftext|> TITLE: Understanding the purely formal part of the sheaf theoretic (cohomological) framework for representation theory QUESTION [10 upvotes]: By now I have the impression that many statements in representation theory can be phrased a lot more elegantly using cohomological language. Therefore I'm trying to understand a bit better the sheaf theoretic framework for representation theory. To be specific I'd like to understand, for example, the purely formal parts (no hard "real" theorems) of the following ideas: Induction and Coinduction as different pushforwards of sheaves Equivariant sheaves as sheaves on quotient stacks and relation to the (Co-)induction functors. I have some background in algebraic geometry and homological algebra (i'm even fine with some moderate stacky language) so I think I have the nessasary tools to understand this yet unfortunately I'm having a hard time finding references for statements appearing, for example, in the answers to the following question. Is there a source that tells the overall cohomological story of representation theory with some sketches of proofs for the obviously formal propositions? (for example that thing with induction and pushforward which looks entirely formal to me). REPLY [5 votes]: Let $G$ be an affine group scheme (e.g. a finite group). There is a stack $BG$ which is more or less determined by either of the following two properties: Maps to $BG$ are the same thing as $G$-torsors. More formally, there's a functor from commutative rings to groupoids sending a commutative ring $R$ to the groupoid of isomorphism classes of $G$-torsors over $\text{Spec } R$; this functor is represented by a stack called $BG$. Quasicoherent sheaves on $BG$ are the same thing as representations of $G$. If $f : H \to G$ is a map of affine group schemes, it induces a map $Bf : BH \to BG$ of stacks, which further induces a pullback functor $Bf^{\ast} : QC(BG)) \to QC(BH)$; as you might imagine, this is restriction of representations. Pullback has, as usual, a right adjoint, namely pushforward $Bf_{\ast} : QC(BH) \to QC(BG)$; this is one version of induction of representations (I guess it might be "coinduction," but I'm happy to just call it "induction" because this adjoint will, I think, always exist, so it ought to have the more fundamental name). To get a better sense of what pushforward is doing, you can think of $BG$ as being the quotient stack $\text{pt}/G$, and $BH$ as being the quotient stack $(G/H)/G$; then you can think of representations of $G$ as $G$-equivariant sheaves on a point, and representations of $H$ as $G$-equivariant sheaves on $G/H$. This lets you interpret push/pull along the map $BH \to BG$ as $G$-equivariant push/pull along $G/H \to \text{pt}$. In the nicest examples, where $G$ is reductive and $H$ is the Borel subgroup, $G/H$ will be a smooth projective variety (the flag variety of $G$); the significance of this is that $G$-equivariant pushforward from $G/H$ to a point (that is, taking global sections) now sends finite-dimensional representations of $H$ to finite-dimensional representations of $G$. What representations you get is described by the Borel-Weil-Bott theorem, and for the nicest statements you should take the derived pushforward. It's unclear to me what part of this story is "purely formal." You can write down adjoints all day, but this feature that sometimes induction sends finite-dimensional things to finite-dimensional things even though $H$ isn't finite index in $G$ doesn't seem formal to me. Also, I don't know a reference for any of this.<|endoftext|> TITLE: Short and elegant definition of the $C^1$ topology QUESTION [7 upvotes]: A friend told me that the $\mathbf{C^1}$-topology on the set $C^\infty(M,N)$ of smooth functions between two smooth manifolds $M$ and $N$ can be defined as the coarsest topology making the map $$ C^\infty(M,N) \to C(TM,TN), \qquad f \mapsto df $$ a homeomorphism onto its image, where $C(TM,TN)$ is the space of continuous functions between the tangent bundles $TM$ and $TN$ equipped with the compact-open topology. Is this definition correct, i.e. equivalent to the usual one by jet spaces? Wikipedia and the other references I found have very long definitions for the C1-topology. REPLY [6 votes]: The answer is almost yes if you mean in your post the following: $C^1$-topology should mean the compact open $C^1$-topology and you should replace the $d$ by the mapping $$T \colon C^1 (M,N) \rightarrow C(M,N) \times C (TM,TN), f \mapsto (f,df)$$ where the factors on the right are endowed with the compact open topology. Then the compact open $C^1$ topology can be defined as the initial topology with respect to $T$ (note this is equivalent to stating that $T$ is a topological embedding onto its closed image (which is what you asked)). Details of this construction can be found in Wockels script https://www.math.uni-hamburg.de/home/wockel/teaching/data/HigherStructures2013/hs.pdf in Section 5 (note that he deals immediately with the compact open $C^\infty$ topology and all proofs deal with this more complicated situation, however all the proofs adapt to switching from $k=\infty$ to $k=1$. Some comments: For many desirable properties of the compact open topology one would like $M$ to be compact (see Wockel's notes). If $M$ is non-compact there are also different function space topologies, often called Whitney topologies which are often considered and which coincide with the compact open $C^1$-topology if $M$ is compact.<|endoftext|> TITLE: Can something finite over $\mathbb{C}(q)$ be a modular form? QUESTION [19 upvotes]: If $f\in\mathbf{C}[[q]]$ is non-constant, and algebraic over $\mathbf{C}[q]$ (in the sense that it is a root of a polynomial with coefficients in in $\mathbf{C}[q]$) then can $f$ be the $q$-expansion of a modular form (for some congruence subgroup of $SL(2,\mathbf{Z})$)? I ask for the following reason. There are geometers in my department who occasionally come up with $q$-expansions (probably from counting things in geometry) and ask if these things are likely to be modular forms. Sometimes they are, sometimes they aren't, sometimes I don't know. But one that came up today I noticed was non-constant and algebraic over $\mathbf{C}[q]$ and so I instantly said that this should rule it out, and then I realised I could not immediately point to a proof of this. Katz proved many years ago that a non-constant polynomial in $q$ can't be the $q$-expansion of a modular form (by which I mean a form which has no poles, even at cusps), because if we have a non-constant polynomial modular form of some weight and level, we can consider all modular forms of that weight and level which are polynomials in $q$, and then it's not hard to check that this space is Hecke stable, but Hecke operators tend to increase the degree of a polynomial modular form if it has positive degree and it's not hard to finish the job now. See Katz Antwerp III, p94 (p26 of the article). Now I've found the time to look at the article, I realise that probably modular forms algebraic over $\mathbf{C}[q]$ might also form a Hecke-stable subspace, although now one can't use the degree trick to finish. I was half-expecting the result to be false in characteristic $p$, but now I'm not so sure. I know that the $\Delta$ function is $\sum_{n\geq1,n\ \mathrm{odd}}q^{n^2}$ in characteristic 2 but presumably this isn't algebraic over $\mathbf{F}_2[q]$; I realise now that I have no good test for this in my head. REPLY [22 votes]: A non-constant modular form has a natural boundary on the real line. A power series that's algebraic in $q = e^{2\pi i \tau}$ can't. REPLY [14 votes]: In characteristic $p>0$ the quotient field ($K$ say) of the ring generated by $E_4,E_6$ (i.e. modular forms of level one) is of transcendence degree $1$. But $q$ (in the sense of the Tate curve) is transcendental over $K$ (I proved this in J. Number Theory 58 (1996) 55-59). So any non constant modular form or modular function is transcendental over $\mathbb{F}_p(q)$.<|endoftext|> TITLE: Wonderful compactification QUESTION [7 upvotes]: Suppose $G$ is a semi-simple group of adjoint type over an algebraic closed field, and $X$ its wonderful compactification a la De Concini and Procesi. Let $P=MU$ be a parabolic subgroup in $G$, and let $Y$ be the wonderful compactification of $M$. Question. Can you describe the closure of $P$ in $X$? In particular, is it true that the closure of $P$ fibers over $Y$? REPLY [6 votes]: I am just writing my comment above as an answer, just in the special case that the parabolic is a Borel subgroup $B$. In this case, the reductive part $M$ is a maximal torus $T$ in $G$. Denote by $W\subset N_G(T)$ a Weyl group. There is a conjugation action of $G$ on itself, and there is an induced conjugation action of $W$ on $T$. Denote by $q_T:T\to T/W$ the quotient of this finite group action. It is a "classical fact" that there is a conjugation-invariant morphism $$q_G: G \to T/W,$$ that extends $q_T$ and that is a geometric quotient of the conjugation action of $G$ on itself, cf. Section 6 of the following. MR0180554 (31 #4788) Reviewed Steinberg, Robert Regular elements of semisimple algebraic groups. Inst. Hautes Études Sci. Publ. Math. No. 25 1965 49–80. 14.50 (20.75) http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1965__25_/PMIHES_1965__25__49_0/PMIHES_1965__25__49_0.pdf In particular, compare the restriction of $q_G$ to $B$ with the composite morphism $$ B = T\times U \xrightarrow{\text{pr}_T} T \xrightarrow{q_T} T/W.$$ By restricting to the open dense subset of regular elements, it is clear that these two morphisms are equal. Why is this relevant? The following is derived from an article with Xuhua He. MR2833468 Reviewed He, Xuhua(PRC-HKST); Starr, Jason(1-SUNYS) Semistable locus of a group compactification. (English summary) Represent. Theory 15 (2011), 574–583. 14L30 (14M27) https://arxiv.org/abs/0907.0281 The Zariski closure $\overline{T}$ in $\widehat{G}$ is projective and normal. There is a maximal open subscheme $\widehat{G}^{ss}\subset \widehat{G}$ containing $G$ that is explicitly described in that article on which the morphism $q_G$ extends to a regular morphism, $$\widehat{G} \supseteq \widehat{G}^{ss} \xrightarrow{\overline{q}_G} \overline{T}/W.$$ If I am reading the article correctly, for the opposite Borel $B'$ to $B$, for the associated left-right action of $B'\times B$ on $\widehat{G}$, and for the conjugation action of a diagonal copy $G_\Delta$ in $G\times G$, the semistable locus is a disjoint union of $G_\Delta\cdot (B'\times B)$-orbits indexed by subsets of the set of simple roots (presumably one for each stratum in the boundary stratification of $\widehat{G}$). Denote by $\overline{B}$ the normalization of the closure of $B$ (I believe the closure is already normal, but I am not completely certain). Edit. Friedrich Knop points out that Springer proved that the closure of $B$ is normal. Denote by $\overline{B}^{ss} \subset \overline{B}$ the fiber product of $\overline{B}$ and $\widehat{G}^{ss}$ over $\widehat{G}$. Denote by $\overline{B}^o$ the maximal open subscheme of $\overline{B}$ on which the following rational transformation extends to a regular morphism, $$\overline{B} \supseteq B \xrightarrow{\text{pr}_T} T \subseteq \overline{T}.$$ By Zariski's Main Theorem, $\overline{B}^o$ is also the maximal open on which this extends to a finite-to-one correspondence. In particular, since $\overline{T}\to \overline{T}/W$ is finite, $\overline{B}^o$ equals the maximal regular domain of the rational transformation, $$ \overline{B} \supseteq B \xrightarrow{q_T} T/W \subseteq \overline{T}/W.$$ Thus $\overline{B}^o$ contains $\overline{B}^{ss}$ (presumably these are equal open subsets). Anyway, the deepest $G\times G$-orbit of $\widehat{G}$ is isomorphic to $G/B\times G/B'$. The intersection with $\overline{B}$ appears to equal $\{B/B\}\times G/B'$ union $G/B\times \{B'/B'\}$. It looks to me like the intersection with $\overline{B}^o$ equals the complement of the point $(B/B,B'/B')$. Certainly in the special case that $G=\textbf{PGL}_2$, all of this is straightforward to compute.<|endoftext|> TITLE: Intuition for the Lefschetz motive (Tate motive)? QUESTION [7 upvotes]: Yo! Maybe this question is too dumb for mathoverflow, but I believe no one will pay attention to it in math.stackexchange, so I will post it here. If this question is not suitable, just delete it. I will denote $W$ the weak equivalences regarding contractibility of $\mathbb{A}^1$ and $(-)^{\#}$ the pseudo-abelian hull. I have been trying to understand better the Tate motive and its realizations in some cohomologies. However I always fail to understand why the Lefschetz motive is not something homotopically trivial. First, let me clarify my intuition. As I understand the category $\mathbf{Spc} (k)$ (let's focus on pure effective geometric motives for now) is the smallest category containing schemes and its (non-existent in the category of schemes) cellular decompositions. Now the localization by weak equivalences $\mathbf{Spc} (k) [W^{-1}]$ leads to homotopy types of cellular decompositions of schemes and the Hurewicz map linearize everything and lands in $\mathbf{DM}^{eff}(k)$, which is analogous to $\mathbb{Z} Sing : \mathbf{Top} \rightarrow Ch(\mathbf{Ab})$ (actually it's analogous applying the Hurewicz map before the localization to be more precise). If my intuition is wrong, please, correct me. Now, in the category of effective geometric motives $\mathbf{DM}^{eff}_{gm}= (K^{b} (Cor_{fin}(k))[\{\text{Mayer Vietoris, W}\}^{-1}])^{\#} \hookrightarrow \mathbf{DM}^{eff}(k) =D^{-} (Sh_{Nis})[W^{-1}]$, the Lefzchetz motive $$\mathbb{L} = \widetilde{[\mathbb{P}^1]} = Cone (x_{*})$$ on LHS and $$\mathbb{L} = C_{*} \mathbb{Z}_{tr} (\mathbb{G}_m, 1) [1]$$ on the RHS, where $x \in \mathbb{P}^1 (k)$. Now in Virtual Lefschetz motive and http://arxiv.org/pdf/0907.4046v2.pdf for instance, it's used the Grothendieck ring of varieties $K_0 (Var_k)$ as a prototype to pure motives and there $[\mathbb{P}^1] = [Spec (k)] + \mathbb{L} = [Spec (k)] + [\mathbb{A}^1]$, hence $\mathbb{L} \cong [\mathbb{A}^1]$. And thinking intuitively on the complex points $\mathbb{P}^1 (\mathbb{C}) = \{x\} \cup \mathbb{C}$. Another interpretation that I see is as an orientation class (angle form) of the sphere by noticing that $\mathbb{L} = \mathbb{Z} (1)[2]$ and $H^2 (\mathbb{P}^1)$ have one generator. However in étale cohomology it behaves as the orientation class of the circle (and maybe this is my main confusion: Why $H^2 (\mathbb{P}^1) \cong H^1 (\mathbb{G}_m)$ as Galois representations? Maybe seeing the Hodge structures clarify it?) So my questions are. 1) Is there a map from or to $K_0 (Var_k)$ to $\mathbf{DM}_{gm}^{eff}(k)$ and in this map what makes $\mathbb{L}$ fails to be contractible? 2) Why $\mathbb{L} = \widetilde{[\mathbb{G}_m]}[1]$ in $\mathbf{DM}_{gm}^{eff}(k)$? In other words why the LHS and RHS above are equal under the embedding of geometric motives into motives? This will explain my confusion with the étale realization of the Lefschetz motive. 3)I would like to read additional remarks or point of views about the Tate and Lefschetz motive which clarify your intuitive understanding. Thanks in advance. REPLY [4 votes]: 1) No, there is no such map. You can only send both of these classes into the group $K_0(DM_{gm}^{eff})\cong K_0(Chow^{eff})$. The main point (and, possibly, the source of your problems) is that if you want these two maps to be compatible then you should send variety into its motif with compact support! Thus $[A^1]=[P^1]-[pt]\neq 0$! So, the motif of an affine line is "trivial", whereas its motif with compact support is not. Just look at the Mayer-Vietoris triangle $M(G_m)\to M(A^1)\bigoplus M(A^1)\to M(P^1)\to M(G_m)[1]$. Since motives of affine lines are "trivial", the latter morphism yields the isomorphism in question. If you want to understand the Euler characteristics of (the motif with compact support of) a variety $V$ then just count points in $V(F_q)$. If you want to study "ordinary motives", look at Hodge structures, yes.<|endoftext|> TITLE: Why did Gödel name his constructible universe $L$? QUESTION [11 upvotes]: It seems like Gödel didn't use the letter $L$ for his model before his book "The Consistency of the Axiom of Choice and of the Generalized Continuum-Hypothesis with the Axioms of Set Theory", which is probably the first place it got used. Do anyone of you know why he used the letter $L$? It does seem like a bit ad hoc in the book, where he names some functions $J_i$, then some other functions $K_i$ and then ends up defining $L$. But why $J_i$ then? (I'm sorry if this is not suited for MO) REPLY [13 votes]: So far as I know, there are four interpretations why Gödel uses $\mathrm{L}$ to denote the constructible universe. By Paul Bernays (see page 64 in Halbeisen 2017), Gödel originally used the old German script to denote the constructible universe, where is a capital $\mathrm{C}$. Although the appearance of is very like the German script $\mathfrak{L}$ for $\mathrm{L}$, it's not the German script $\mathfrak{L}$ for $\mathrm{L}$, and furthermore Bernay's words don't interpret why Gödel uses $\mathrm{L}$ to denote constructible universe in 1940 (in fact he had introduced the constructible universe in 1938 which is a proceeding paper). So I don't think Bernays' interpretation is suitable. By Kai Hauser, the letter $\mathrm{L}$ comes from "law" since the model is constructed using some laws. But in 1940, Gödel didn't mention any law and the word "law" at all. Furthermore, even so, any other model should be named $\mathrm{L}$ since they are also constructed using some other laws. So I don't think Hauser's interpretation is suitable either. By someone (I don't remember who said this first), since von Neumann's universe is the first model of $\mathsf{ZFC}$ which is denoted by $\mathrm{V}$ and the constructible universe may be the second model of $\mathsf{ZFC}$, to distinguish from and show relations to $\mathrm{V}$, Gödel extends the angle of $\mathrm{V}$ to 90，rotates it down to the horizontal line, and then makes it become $\mathrm{L}$. Of course, it's a joke, and can't be suitable either. In my opinion, the only suitable interpretation for Gödel's using $\mathrm{L}$ to denote the constructible universe is that $\mathrm{L}$ is after $\mathrm{J,K}$ in order since Gödel defines the constructible universe $\mathrm{L}$ after defining some functions $\mathrm{J}_\mathrm{i}$ and $\mathrm{K}_\mathrm{i}$ for which beginning with $\mathrm{J}$ is possibly because almost all the letters before $\mathrm{J}$ have been used before introducing the constructible universe in 1940 (and in particular he uses $\mathfrak{Cls}(\mathrm{A})$ related to $\mathfrak{C}$ for $\mathrm{C}$ to denote that $\mathrm{A}$ is a class and $\mathrm{Y}\,\mathfrak{Con}\,\mathrm{X}$ to denote that $\mathrm{Y}$ is connex in $\mathrm{X}$). In fact in 1940, Gödel's usages for many symbols are very random, for example he uses $\mathfrak{A}$ to denote some operation and $\mathfrak{B}$ to denote some notion without any sufficient considerations. By the way, why at first people use $\mathrm{V}$ to denote the cumulative universe of all sets or the class of all sets is also mysterious. There are two interpretations for it. $\mathrm{V}$ is the v in von Neumann. This may be because the cumulative universe is also called von Neumann universe which was first claimed by Gregory H. Moore in 1982 because existence and uniqueness of the general transfinite recursive definition of sets was demonstrated by von Neumann for both Zermelo-Fraenkel set theory in 1928 and his own set theory in 1928a. While in fact the first publication of the von Neumann universe was by Ernst Zermelo in 1930. Furthermore, the $\mathrm{V}$ notation was not used by von Neumann in his 1920s papers. $\mathrm{V}$ is the shape of the picture of the cumulative universe. In fact, Peano, and Whitehead and Rusell had use $\mathrm{V}$ notation to denote the class of all individuals on page XI in 1889 in Latin, and the class of all sets on page 229 in 1910 in English respectively. But at those times they clearly didn't know the two points talked about as above. So both of the two interpretations as above are not quite suitable. In fact, why Peano, and Whitehead and Rusell made such uses are hard to understand to some extent. Of course, you can say that, at some time after 1920s, people began to accept the use of $\mathrm{V}$ to denote the von Neumann universe because they realized the two points as above.<|endoftext|> TITLE: Reference for push-pull formula in cohomology QUESTION [9 upvotes]: I would like a precise reference for the following fact. Assume that $$ \begin{array}{ccc} M\times_B N & \stackrel{f'}{\to} & N \newline \quad\downarrow g' & & \quad\downarrow g \newline M & \stackrel{f}{\to} & B \end{array} $$ is a Cartesian square of manifolds (meaning $f$ and $g$ are transverse, and $M\times_B N$ is their pullback, a submanifold of $M\times N$). Assume all manifolds are closed and oriented (although minimally we would require that $g$ is proper and oriented and $g'$ is endowed with the induced orientation). Then for any integral cohomology class $x\in H(N;\mathbb{Z})$ we have $$ f^*g_!(x) = g'_! f'^*(x)\in H(M;\mathbb{Z}). $$ Here $g_!:H^*(N;\mathbb{Z})\to H^{*+\dim(B)-\dim(N)}(B;\mathbb{Z})$ is the Umkehr map induced by $g$. This is stated by Quillen for complex cobordism in his paper "Elementary proofs of some results in cobordism theory using Steenrod operations", and in fact it is immediate from his geometric interpretation of cobordism classes. It can also easily be deduced for integral cohomology, once one has some geometric interpretation of cohomology classes (such as Kreck's cobordism of stratifolds, or Buoncristiano-Rourke-Sandersons mock bundles), but I'm hoping it may be classical and appear in some older algebraic topology text. REPLY [2 votes]: I think if you can consider the vertical maps in your diagram as fibre bundles, then there exists a commutative square with the same horizontal maps, but the vertical ones replaced by transfers maps (or Umkehr maps). Taking homology gives the formula. For the property mentioned above, you may look at Mann, Benjamin M.; Miller, Edward Y.; Miller, Haynes R. $S^1$-equivariant function spaces and characteristic classes. Trans. Amer. Math. Soc. 295 (1986), no. 1, 233–256.<|endoftext|> TITLE: Classifying map for a surface bundle QUESTION [5 upvotes]: Let $E\longrightarrow X$ be a surface (with holes) bundle. The structure group is then $M_{g, s}$, the mapping class group of the fiber. It follows from the famous work of Penner that the classifying space of $M_{g, s}$ is homotopy equivalent to the geometrical realization of the category of fatgraphs, i.e., $$BM_{g,s} \simeq |\mathit{Fat}_{g,s}|.$$ My question is as follows: how can one construct the classifying map $X\longrightarrow |\mathit{Fat}_{g,s}|$ explicitly given the initial bundle in these terms? This seems pretty intuitive, but I can not get to the formal description. My only thought was to try to find something similar the natural map to the Milnor join construction of the classifying space. Edit: the natural restrictions for the genus and number of holes are $s\geqslant 1$ and $s+2g \geqslant 3$. REPLY [2 votes]: As I said in the comments, the more natural thing would be to construct classifying maps to the moduli space of Riemann surfaces. One then has to choose an identification of this with the complex of fatgraphs. For simplicity, I'm going to work with closed surfaces and with fiber bundles whose bases are smooth manifolds. Consider a fiber bundle $\pi:E \rightarrow X$ whose fibers are closed genus $g$ surfaces $\Sigma_g$ and whose base $X$ is a smooth manifold (not necessarily compact). Our goal is to construct a classifying map $\phi:X \rightarrow \mathcal{M}_g$, where $\mathcal{M}_g$ is the moduli space of Riemann surfaces. I'm going to think of $\mathcal{M}_g$ as the space of isomorphism classes of conformal structures on $\Sigma_g$. We can find a smooth embedding $f:E \rightarrow \mathbb{R}^k$ for some large $k$. For each $x \in X$, the image $A_x:=f(\pi^{-1}(x))$ is a smooth submanifold of $\mathbb{R}^k$ which is diffeomorphic to $\Sigma_g$. Restrict the usual Riemannian metric on $\mathbb{R}^k$ to $A_x$ to make $A_x$ into a Riemannian manifold. The image $\phi(x) \in \mathcal{M}_g$ is then the conformal structure on $A_x$ given by this Riemannian metric. Of course, this depends on the choice of embedding $f$. However, stabilizing $f$ by embedding $\mathbb{R}^k$ into $\mathbb{R}^{k+1}$ changes nothing. By making $k$ large enough, we can ensure that any two choices of embeddings are isotopic, which gives us a homotopy between the two classifying maps.<|endoftext|> TITLE: Uniqueness of sums of roots of unity QUESTION [7 upvotes]: Let $\zeta:=e^{\frac{2\pi i}{n}}$, with $n\geq4$, and let $2\leq k\leq n-2$. Let us suppose that the prime factorization of $n$ is $n=p_1^{\alpha_1}\cdot\dots\cdot p_s^{\alpha_s}$, with $\alpha_i>0$ and $ p_1 TITLE: q-analog of a combinatorial identity involving binomial coefficients QUESTION [8 upvotes]: Using, e.g., properties of iterated finite differences it is easy to show that for any pair of integers $n$ and $m$ with $n>\!>m$ one has the identity $$ \sum_{k=0}^m(-1)^{k-m} {n-k\choose m}{m\choose k}=(-1)^{m}. $$ Motivated by Colored sl(N) link homology via matrix factorizations, by Hao Wu, I am interested in a certain $q$-analog of the above formula. Defining the $q$-integer $[n]_q$ as $$ [n]_q=\frac{q^{n}-q^{-n}}{q-q^{-1}} $$ and consequently the $q$-factorial as $$ [n]_q!=[n]_q[n-1]_q\cdots [1]_q $$ and the $q$-binomial $$ {n \brack m}_q=\frac{[n]_q!}{[m]_q![n-m]_q!} $$ (notice that this convention, which seems to be the standard one among quantum group theorists differ by a power of $q$ and by a change of variable $q\leftrightarrow q^2$ from the maybe more standard convention $[n]_q=(1-q^n)/(1-q)$) An extensive Sage computation shows that the following combinatorial identity holds: $$ \sum_{k=0}^m(-q)^{k-m} {n-k\brack m}_q{m\brack k}_q=(-1)^{m} q^{-(n+1-m)m}. $$ in perfect agreement with the "shifting factor" one finds on page 9 of Hao Wu's paper. My problem is that I have not been able to work out a formal proof of the above identity involving $q$-binomials: any suggestion would be appreciated. (it must also be said that combinatorics is "Here be dragons" to me, and that even working out the above identity involving ordinary binomials, although quite simple in the end, took me some time; so I see the question I am asking can possibly be actually trivial, and if this is the case I apologize for this) REPLY [8 votes]: You can see this as an instance of the q-Vandermonde identity. The q-binomial theorem tells us that the coefficient of $t^k$ in $\prod_{i=0}^{m-1}(1+q^{-2i}t)$ is $q^{-k(m-1)}{m \brack k}_q$ and the coefficient of $t^{n-m-k}$ in $\prod_{i=0}^{m}\frac{1}{1+q^{-2i}t}$ is $(-1)^{n-m-k}q^{-m(n-k-m)}{n-k \brack m}_q$. By taking the product and finding the coefficient of $t^{n-m}$ in two ways, we see that $$\sum_{k=0}^m (-1)^{n-m-k}q^{k-m(n-m)}{n-k\brack m}_q{m\brack k}_q=(-1)^{n-m}q^{2m(m-n)}.$$ From here, just multiply both sides by $(-1)^n q^{m(n-m-1)}$ and obtain your identity.<|endoftext|> TITLE: Exponential approximation for 3F2 hypergeometric function with repeated indices QUESTION [7 upvotes]: In my research I have run across the hypergeometric function $${}_3F_2(d,d,d;d+1,d+1;z)$$ where d is a positive integer and 0≤z≤1. When I plot this as a function of d on a semilog plot, it appears to be exponential (or close to it) for large d, but I haven't been able to prove it or find a formula for the exponential as a function of z. I've poked around on the web, but I'm a physicist rather than a mathematician and I'm a bit overwhelmed by the hypergeometric literature. Can anyone help? Thanks. REPLY [6 votes]: Indeed the hypergeometric series is unimodal (the terms for fixed $z$ and $d$ first grow with $k$ then decrease again). The hypergeometric series is given by $$ {}_{3}F_{2}(d,d,d;d+1,d+1;z)=\sum_{k=0}^{\infty} \frac{(d)_{k}(d)_{k}(d)_{k}}{(d+1)_{k}(d+1)_{k} k!} z^{k}=\sum_{k=0}^{\infty} b(k,d) z^{k}, $$ with the Pochhammer symbol $(a)_{k}=\frac{\Gamma(a+k)}{\Gamma(a)}$. Therefore the ratio of two consecutive terms gives the nicely compact $$ \frac{b(k+1,d) z}{b(k,d)}=\frac{(d+k)^{3}}{(d+k+1)^{2}(k+1)} z $$ which should be equal to 1 for $k=k_{\max}$. Expanding the ratio for large $d$ we find for the highest order in $d$ $$ k_{\max}=\frac{z}{1-z} \ d + O(1). $$ This expansion is valid for $d\ z$ and $d \ (1-z)$ well larger than 1, as can be seen by the following terms in the expansion. Now we do as Noam Elkies suggested in his comment: make the index $k$ continuous and expand the terms of the hypergeometric sum to second order around $k_{\max}$. Actually, it is the logarithm of the terms that gets expanded to afterwards have a nice Gaussian integral, when we switch from summation over $k$ to integral. We get $$ \sum_{k=0}^{\infty} b(k,d) z^{k}\approx \int_{-\infty}^{\infty} \exp\{\ln(b(k_{\max} + \tau,d) z^{k_{\max}+\tau})\}\ d\tau. $$ The exponent is expanded to second order in $\tau$ giving a Gaussian integral that can be calculated easily. (The factor in front of the $\tau^2$ term is indeed negative, the integral is converging.) The extension of the lower integration limit to $-\infty$ is okay, since at this stage the integrand is rapidly vanishing in that region as function of $\tau$. All this can be done nicely with your favourite computer algebra system. After tweaking the result a bit and expanding it for large $d$, the final result is $$ {}_{3}F_{2}(d,d,d;d+1,d+1;z)\approx\exp\left\{\frac{(5 z +1)^2}{8 \ d \ z}\right\}\ (1-z)^{-d+2}. $$ This approximation becomes bad if $d \ z \approx 1$ and smaller (i.e., for very small $z$). But in this regime the even simpler $$ {}_{3}F_{2}(d,d,d;d+1,d+1;z)\approx (1-z)^{-d+2} $$ can be used. For $d\ (z-1)$ smaller than one we get also invalid results. As a byproduct I mention below the asymptotic expansion for hypergeometric functions with higher indices $$ {}_{p+1}F_{p}(\overbrace{d,\dots,d}^{p+1};\underbrace{d+1,\dots,d+1}_{p};z)\approx\exp\left\{\frac{(z (2 p +1) +1)^2}{8 \ d \ z}\right\}\ (1-z)^{-d+p}. $$ Again for small $d\ z$ the exponential prefactor should be avoided for a good approximation. For $z$ approaching 1 the approximation breaks down. All computations were conducted with Mathematica 10.<|endoftext|> TITLE: Groups whose word problem can be solved in constant time QUESTION [10 upvotes]: Given a finitely generated group $G$, define an encoding of $G$ to be a one-to-one function $\Phi:G\to \bigcup_n \{0,1\}^n$ that sends each group element to a unique finite word. For $a,b\in G$, assuming that $\Phi(ab)$ can be computed from $\Phi(a)$ and $\Phi(b)$, my question is: is there a classification for groups where there exists an encoding $\Phi$ such that $\Phi(e_ia)$ can be computed in constant time by a multi-tape Turing machine from $\Phi(e_i)$ and $\Phi(a)$ where $e_1,\ldots,e_n$ is some choice of generators and $a\in G$. For example, $\mathbb Z$ has such an encoding. Expanding the encoding to allow strings with letters $\{0,1,*\}$, we can represent an element in $\mathbb Z$ as $\cdots 000*11\cdots 1\underline{1}000\cdots$ with the appropriate number of $1$s to the right or the left of $*$ for positive or negative values and $\underline{\phantom{a}}$ indicating the position of the head of the Turing machine. Then we just add or remove a $1$. Similarly, using multiple tapes, we can encode $\mathbb Z^n$. Using symbols $\{a,b,a^{-1},b^{-1}\}$ we can represent the free group on $\{a,b\}$ by just writing down its usual representation as words. We could also, for instance, encode the group $\langle a,b \mid aaa=e\rangle$. (For each of these examples, the head of the Turing machine should be left at the end of each encoding, otherwise the algorithm may not be constant time.) However, I'm not sure of more complicated examples. I believe that operating on all encodings rather than directly on the encoding produced by representations in terms of generators makes this question distinct from ``groups where a multi-tape Turing machine can compute the application of a generator $e_i$ to a word $w=w_1\cdots w_n$ when given the input $\cdots 000.w_1\cdots w_n00\cdots$,'' but I don't have an example of a group where there exists a more computationally efficient encoding than just writing down the word in terms of generators. I am also curious about the more general context of simultaneously constant-time operations on sets. For instance, on $\mathbb Z$, the function $f(x)=x+1$ can be accomplished in constant time by encoding $x$ as a string of $x$ $1$s. Similarly, the operation $g(x)=2x$ can be accomplished in constant time by encoding $x$ in binary. In binary $f(x)$ cannot be computed in constant time, and when written as $x$ $1$s, $g(x)$ cannot be computed in constant time. Does there exist an encoding of $\mathbb Z$ such that both $f$ and $g$ can be computed in constant time? Is there a name for such collections of operations? REPLY [8 votes]: This is a long comment rather than an answer. Using the ideas in the examples that you gave, you could show that any group that is virtually a direct product of finitely many free groups of finite rank has your property: if $N \unlhd G$ with $G/N$ finite and $N$ a direct product of free groups, then you can use one tape for each free factor of $N$, and use a constant amount of memory to keep track of which coset of $N$ in $G$ you are in. You could conjecture that all examples have this form, although that could be hard to prove. I find it interesting that groups with your property have real-time word problem, which means that you can decide whether a word $x_1x_2 \cdots x_n$ is equal to the identity in $G$ by reading the letters of the word at constant speed - i.e. you are only allowed constant time between reading letters, and you must halt in constant time after reading the final letter. There is no classification of groups with real-time word problem, but there are various examples known, and I would guess that this is a larger class of groups than those with your property. In fact the configuration at any time of the real-time Turing machine determines the group element represented by the subword that you have read so far. But you could have many different configurations describing the same group element, which is perhaps the main difference from your problem, where you require a unique binary string for each group element. I did some work on this problem myself a few years ado, and two references are D.F. Holt and S. Rees. Solving the word problem in real time. Journal of the London Mathematical Society (2001) , Vol.63 (No.3). pp. 623-639. D. F. Holt and C. E. Röver. On Real-Time Word Problems J. London Math. Soc. (2003) 67 (2): 289-301. Examples include finitely generated nilpotent groups, hyperbolic groups, groups hyperbolic relastive to virtually abelian subgroups,and the solvable Baumslag-Solitar groups $B(1,r)$. The class is closed under direct and free products. There is another conjecture around that the groups whose word problem is an intersection of finitely many context-free languages are groups that are virtually a direct product of free groups i.e. the same class as mentioned above in connection with your problem, but I cannot see any direct link between the two problems.<|endoftext|> TITLE: Chevalley restriction theorem for non-split Cartan QUESTION [7 upvotes]: Let $G$ be a reductive group over a field $k$ with maximal torus $H$. Let $\mathfrak{g}$ and $\mathfrak{h}$ denote the corresponding Lie algebra. If $k$ is algebraically closed, we have a theorem of Chevalley which says that $k[\mathfrak{g}]^G\simeq k[\mathfrak{h}]^W$. A theorem of Chevalley–Shephard–Todd then states that this is a polynomial algebra. I'm wondering to what extent these theorems remain true if $k$ is not algebraically closed. The case I'm actually interested in is when $k=\mathbb{C}((t))$ in which case, the conjugacy class of Cartan subalgebras are in bijection with conjugacy classes in the absolute Weyl group (i.e., the Weyl group over the algebraic closure). Most references I know assume $k$ is algebraically closed or that $\mathfrak{h}$ is split. References dealing with non-split Cartan would be appreciated. REPLY [4 votes]: Taking invariants commutes with flat base change, so, in particular, with extensions of the base field. This implies that Chevalley's restriction theorem works over any field of characteristic zero. Also Shephard-Todd works over any field since a polynomial ring with a positive grading has only trivial forms. To see this take any $k$-basis of $\mathfrak m/\mathfrak m^2$ (with $\mathfrak m$ being the maximal homogeneous ideal) and lift it to homogeneous ring elements which are defined over $k$. These will freely generate the ring.<|endoftext|> TITLE: Is the assignment of a root system to a complex semisimple Lie algebra functorial? QUESTION [13 upvotes]: As described here, we have a category of root systems, where a morphism from a root system $\Phi$ in a Euclidean space $E$ to a root system $\Phi'$ in $E'$ is given by a linear map $f: E \to E'$ such that $f(\Phi) \subseteq \Phi'$ and $f(\alpha) \dashv f(\beta) = \alpha \dashv \beta$ for all $\alpha, \beta \in \Phi$, where $\alpha \dashv \beta \mathrel{:=} 2 \frac{\langle \alpha | \beta \rangle}{\langle \beta | \beta \rangle} \in \mathbb{Z}$ and $\langle ~ |~ \rangle$ denotes the inner product on $E$. For isomorphisms of root systems, this last condition is in fact redundant, and follows automatically from the first two. This seems to be the "correct" notion of morphism for root systems: we have finite coproducts of root systems (which are oddly denoted by a product in most texts) and an initial object (the empty root system). Moreover, the Weyl group construction is a covariant functor on root systems. Now, given a complex semisimple Lie algebra $\mathfrak{g}$, we can assign to it a root system $\Phi$, given by the set of nonzero weights of the representation of a Cartan subalgebra $\mathfrak{h} \leq \mathfrak{g}$ on $\mathfrak{g}$ via the adjoint action. This assignment of a root system to a complex semisimple Lie algebra is an invariant, in the sense that isomorphic Lie algebras have isomorphic root systems (and conversely). But is this construction functorial? That is given a morphism of semisimple Lie algebras $f: \mathfrak{g} \to \mathfrak{g}'$, do we obtain a morphism of their root systems? If not a covariant functor, might this be a contravariant functor? Cross-post math.stackexchange One point that was raised by Mariano Suárez-Alvarez in the comments is that the choice of Cartan subalgebra is not functorial. Nevertheless, in the end the construction of a root system does not depend on the choice of Cartan subalgebra, which leaves me hope that this construction might still be functorial. If it is indeed functorial, then the classification of semisimple complex Lie algebras would have an elegant description in terms of this "root system" functor: it is strong monoidal, conservative, and reflects simple objects (irreducible root systems being the simple objects in the category of root systems). Furthermore, since the Weyl group of a root system is functorial, then the assignment of the Weyl group to a Lie algebra would simply be the composite of the two functors. REPLY [19 votes]: For everything you're requesting, it seems reasonable only to consider this question using maps that are isomorphisms, since: (i) the zero map between semisimple Lie algebras has no reasonable "associated" map between root systems, (ii) inclusions such as ${\rm{Sp}}_{2n} \rightarrow {\rm{SL}}_{2n}$ have no reasonable associated map either way between the root systems (the latter is simply laced of rank $2n-1$ and the former has two root lengths and rank $n$). After one restricts the morphisms under considerations to be isomorphisms (which is where all of the substance lies anyway), the answer is "yes". It comes down to a trick, but there is some real content underlying the trick. I have no idea what "strongly monoidal" means, but if one thinks about how mathematicians use semisimple Lie algebras, semisimple Lie groups, and semisimple algebraic groups (after they're done being classified) then such a classification that is functorial with respect to isomorphisms is not only elegant but also useful to do important things. More to the point, the need for such a version of the classification theorem (really in a form with split connected semisimple groups over general fields in place of semisimple Lie algebras over $\mathbf{C}$) already arose back in the 1960's and especially 1970's, as part of the structure theory of connected semisimple groups over general fields (to classify Galois-twisted forms, especially over fields of arithmetic interest and over $\mathbf{R}$) and in the definition of the Langlands dual group beyond the split case. It is sometimes attributed to Kottwitz, but the main idea can already be found in SGA3 in the discussion of the "scheme of Dynkin diagrams" (see Exp. XXIV, 3.1--3.4) and it has probably been rediscovered multiple times (e.g., by Tits for his notion of $\ast$-action of Galois groups on diagrams when formulating classification theorems over general fields without split hypotheses). So finally the question comes down to this: is there a way to associate a root system $\Phi(\mathfrak{g})$ to a complex semisimple Lie algebra $\mathfrak{g}$ in a manner that is functorial with respect to isomorphisms? Sure! Let $G = G(\mathfrak{g}) = {\rm{Aut}}_{\mathfrak{g}/\mathbf{C}}^0$ be the identity component of the automorphism scheme of $\mathfrak{g}$. It is a basic fact from the structure theory of connected semisimple groups and semisimple Lie algebras over $\mathbf{C}$ that the groups $G(\mathfrak{g})$ are exactly the connected semisimple algebraic groups with trivial center, and the natural map $${\rm{Lie}}({\rm{Ad}}_G): {\rm{Lie}}(G) \rightarrow {\rm{Lie}}({\rm{GL}}(\mathfrak{g})) = {\rm{End}}(\mathfrak{g})$$ is an isomorphism onto $\mathfrak{g}$ (embedded via ${\rm{ad}}_{\mathfrak{g}}$). It is also a general fact that the natural action of $G(\mathbf{C})$ on $\mathfrak{g}$ is transitive on the set of pairs $(\mathfrak{h}, \mathfrak{b})$ consisting of Cartan subalgebras $\mathfrak{h} \subset \mathfrak{g}$ and Borel subalgebras $\mathfrak{b} \subset \mathfrak{g}$ containing $\mathfrak{h}$, and that the stabilizer in $G(\mathbf{C})$ of such a pair is the subgroup $T(\mathbf{C})$ where $T \subset G$ is the unique maximal torus whose Lie algebra is $\mathfrak{h}$ (inside $\mathfrak{g}$). For such a pair, let $(\Phi, \Delta)$ be the associated "based root datum" consisting of the root system for $(\mathfrak{g}, \mathfrak{h})$ and the root basis attached to $\mathfrak{b}$ (i.e., the simple roots in the positive system of roots consisting of those whose root lines are contained in $\mathfrak{b}$). Note that the adjoint action on $T(\mathbf{C})$ on $\mathfrak{h}$ is trivial. Hence, if $(\mathfrak{h}', \mathfrak{b}')$ is another such pair in $\mathfrak{g}$, with $(\Phi', \Delta')$ the associated based root datum, then there is a canonical isomorphism $(\Phi, \Delta) \simeq (\Phi', \Delta')$: the adjoint action on $\mathfrak{g}$ arising from any element $g \in G(\mathbf{C})$ that carries $\mathfrak{h}$ onto $\mathfrak{h}'$ and carries $\mathfrak{b}$ onto $\mathfrak{b}'$ (such $g$ is unique modulo $T(\mathbf{C})$, so this isomorphism between based root data is independent of the choice of such $g$). We define the canonical based root datum $(\Phi(\mathfrak{g}), \Delta(\mathfrak{g}))$ to be the inverse limit of all such $(\Phi, \Delta)$'s along the canonical isomorphisms just specified. More concretely, elements of $\Phi(\mathfrak{g})$ are exactly the compatible systems of roots with respect to those isomorphisms, and likewise for $\Delta(\mathfrak{g})$. It is clear from the construction that this pair is functorial with respect to isomorphisms among such $\mathfrak{g}$'s. Now compose this functorial construction (functorial with respect to isomorphisms) with the "forgetful" functor that drops the data of the root basis!<|endoftext|> TITLE: Curves embedding in plane QUESTION [8 upvotes]: Given two closed simple(no self-intersection point) curves $C_1,C_2$ in the plane $\mathbb R^2$, is there a good way to judge whether one curve can be embedded inside the other one, here embedding inside means by translation and rotation one curve can be put inside the area circled by the other one. Ps: You could give some advice for such computation. Thank you. REPLY [6 votes]: Long ago Chazelle studied the polygonal version of your question, and obtained a polynomial-time algorithm, polynomial in the number of vertices: Bernard Chazelle, "The polygon containment problem," in Advances in Computing Research, Vol I: Computational Geometry, (F.P. Preparata, ed.), JAI Press, Greenwich, Connecticut (1983), 1–33. If both polygons have $n$ vertices and neither can be assumed convex, the polynomial is $O(n^7 \log n).$ If both are convex, then the complexity is reduced to $O(n^3)$. Subsequently his results have been improved in various ways. For example, the time complexity in the general case was reduced by Avnaim & Boissonnat by a factor of $n$.1 Google scholar lists more than 100 papers that cite the original. One application is cartography: placing labels inside map regions:             Fig.1(b) from Aonuma, H., Imai, H., Imai, K., & Tokuyama, T. "Maximin location of convex objects in a polygon and related dynamic Voronoi diagrams." In Proc. 6th Symposium Computational Geometry, 1990, pp. 225-234. ACM link. 1Avnaim, Francis, and Jean Daniel Boissonnat. "Polygon placement under translation and rotation." Annual Symposium on Theoretical Aspects of Computer Science. Springer Berlin Heidelberg, 1988.<|endoftext|> TITLE: Simple proof for $\sum_{i=1}^n a^{\gcd(i,n)} $ is divisible by $n$ QUESTION [10 upvotes]: Burnside's Lemma Deduce That: $$\sum_{i=1}^{n} a^{\gcd(i,n)} $$ is divisible by $n$ it's a beautiful result. but i want to prove it without any abstract algebraic tools such as Burnside's Lemma... is there any simple Number-Theoretic proof for it? REPLY [10 votes]: Our sum equals $$S_n(a):=\sum_{d|n} \varphi\left(\frac{n}d\right)a^d.$$ Fix a prime power $p^k$ which divides $n$ (but $p^{k+1}$ does not divide $n$, we write this as $\nu_p(n)=k$), we have to prove that $p^k$ divides $S_n(a)$. If $p$ divides $a$, then $p^k$ divides each summand. Indeed, $\nu_p(\varphi(n/d))\geqslant \nu_p(n/d)-1=\nu_p(n)-\nu_p(d)-1\geqslant \nu_p(n)-d$ since $d TITLE: Stopping times for Brownian motion QUESTION [9 upvotes]: Let $B_t, t\geq 0$ be standard Brownian motion. Let $\big(\mathcal{G}_t, t\geq 0\big)$ be the natural filtration, defined by $\mathcal{G}_t=\sigma(B_s, 0\leq s\leq t)$. Define also a filtration $\big(\mathcal{F}_t, t\geq 0\big)$ by $\mathcal{F}_t=\bigcap_{\epsilon>0} \mathcal{G}_{t+\epsilon}$. Let $\tau$ be a stopping time with respect to the filtration $(\mathcal{F}_t)$. Does there always exist $\tau'$ which is a stopping time with respect to the filtration $(\mathcal{G}_t)$ such that $\tau=\tau'$ with probability 1? Related: Blumenthal's 0-1 law says that, for fixed $t$, for any event $A\in\mathcal{F}_t$, there is an event $\tilde{A}\in\mathcal{G}_t$ such that the symmetric difference of $A$ and $\tilde{A}$ has probability 0. However, this on its own is not enough. For example, let $U$ be a uniform random variable on $[0,1]$, and define a process $C_t, t\geq 0$ by $C_t=0$ for $t\leq U$ and $C_t=t-U$ for $t\geq U$. Then a similar 0-1 law holds, and $U$ itself is a stopping time for the filtration $\mathcal{F}_t=\bigcap_{\epsilon>0}\sigma(C_s, 0\leq s\leq t+\epsilon)$, but there is no stopping time $V$ for the filtration $\mathcal{G}_t=\sigma(C_s, 0\leq s\leq t)$ such that $U=V$ with probability 1. REPLY [7 votes]: There does, but it's not such an obvious fact. Let $\tau$ be an $(\mathcal F_t)$ stopping time. Then $\tau$ is also a stopping time of the augmented filtration $(\tilde {\mathcal F}_t)$, obtained by adding all null sets in the completion of $\mathcal F_\infty$ to each $\mathcal F_t$. As noted already, it is a consequence of the Blumenthal zero-one law that $(\tilde{ \mathcal F}_t)$ coincides with the analogous augmentation $(\tilde{\mathcal G}_t)$ of $(\mathcal G_t)$. Also, it is known (see section 5 of Chapter IV of the book of Revuz and Yor, for example) that each $(\tilde{ \mathcal F}_t)$ stopping time is predictable. In particular, $\tau$ is $(\tilde{ \mathcal F}_t)$-predictable. Thus the process $Z_t:=1_{\{\tau=t\}}$ is $(\tilde{ \mathcal G}_t)$-predictable. By the "de-completion" result stated as a Lemma in section 6 of Appendix I of Volume B of Probabilités et Potentiel by Dellacherie and Meyer, $Z$ is indistinguishable from a $(\mathcal G_t)$-predictable process $Z'$. The $(\mathcal G_t)$ stopping time $\tau':=\inf\{t:Z'_t=1\}$ is a.s. equal to $\tau$.<|endoftext|> TITLE: How to compute $[CP^2, G/PL]$? QUESTION [8 upvotes]: Let $E^4$ be the two stage Postnikov space appearing in the homotopy type of the classifying space $G/PL$. One of its properties is that it only has two nontrivial homotopy groups $\pi_2(E)=Z/2Z$ and $\pi_4(E)=Z$. Here $Z$ denotes the integers. In Lopez de Medrano's book on involutions, it is stated without proof that the homotopy classes of maps $[CP^2, G/PL]$ is isomorphic to $Z$. Anyone know a proof? He gives a passing hint that it has to do with the k-invariant used to glue the Postnikov pieces together. Thanks. REPLY [7 votes]: This is the second half of an answer completing the first half given by @Qiaochu Yuan. The information we need is that the $k$-invariant is $\beta Sq^2$, and the space $E$ has the $H$-space structure of the fiber of $\beta Sq^2: K(\mathbb Z/2,2)\to K(\mathbb Z,5)$. (Madsen-Milgram Theorem 4.34, but they do not prove this, only refer to Sullivan.) Since all maps between the EM-spaces are $H$-maps, the short exact sequence one obtains is a short exact sequence of groups. Let $F$ be the fiber of $\beta: K(\mathbb Z/2,4)\to K(\mathbb Z,5)$, in fact $F=K(\mathbb Z/2,4)$. The fiber sequence $K(\mathbb Z,4)\to F \to K(\mathbb Z/2,4)$ gives rise to another short exact sequence when applying $[\mathbb C P^2,-]$ to it, namely $0\to \mathbb Z \to \mathbb Z \to \mathbb Z/2\to 0$. Now $Sq^2$ induces a map of the short exact sequences which is the identity on the $\mathbb Z$-terms on the left and also the identity on the $\mathbb Z/2$ terms on the right, since $Sq^2:H^2(\mathbb C P^2,\mathbb Z/2)\to H^4(\mathbb C P^2,\mathbb Z/2)$ is an isomorphism. It follows that the group we are interested in is isomorphic to $\mathbb Z$.<|endoftext|> TITLE: The minimum codimension of Lie subalgebra of $\chi^{\infty}(M)$ QUESTION [8 upvotes]: Assume that $M$ is an arbitrary manifold. Is there a Lie subalgebra of $\chi^{\infty}(M)$, the space of smooth vector fields on $M$, whose codimension is equal to one? If not, what is a counter example? In particular what is the answer to this question for $M=\mathbb{R}^{2}$ or $M=S^{2}$? The question is a particular case of the following general question: Question: For a $n$ dimensional manifold $M$, is it true to say that the minimal codimension of Lie sub algebras of $\chi^{\infty}(M)$ is equal to $n$? The above question serachs for obstruction for finite codimensionality of Lie subalgebras (codimension equal to one) of $\chi^{\infty}(M)$ while the following post concerns diversity and variation of finite dimensional Lie subalgebras. A Manifold for which $\chi^{\infty}(M)$ is rich REPLY [2 votes]: Let $L$ be a sub-algebra of $\mathrm{Vect}(M)$. I think one might be able to prove $\mathrm{codim}\ L \geq \dim M$ by using a recent result of Hurtado. Here is a sketch of the proposed proof, every step of which is difficult: (1) Let $G \subset \mathrm{Diff}(M)$ be the group generated by $\exp(L)$. Problem: The exponential need not converge. (2) Let $N = \mathrm{Diff}(M)/G$. One hopes that $N$ is a manifold of dimension $\mathrm{codim}(L)$. Problem It is not clear how to put a manifold structure on $N$. We then get a homomorphism $\mathrm{Diff}(M) \to \mathrm{Diff}(N)$ by the action of $\mathrm{Diff}(M)$ on the quotient $N$. Hurtado shows that such maps only exist if $\dim M \leq \dim N$. We can, at least momentarily, dodge discussions of quotients and exponentials of infinite dimensional Lie groups by defining $N$ to be be the set of subalgebras of $\mathrm{Vect}(M)$ conjugate to $L$ by an element of $\mathrm{Diff}(M)$ (or perhaps $\mathrm{Diff}_0(M)$.) Hopefully, the set of all codimension $n$ subspaces of $\mathrm{Vect}(M)$ is some sort of manifold by a Grassmannian like construction, and then $N$ can be a closed submanifold of this. If we are lucky, then the tangent space to $N$ at $[L]$ is $\mathrm{Vect}(M)/L$. We clearly have a map $\mathrm{Diff}(M) \to \mathrm{Bijections}(N)$ and, presumably, once we have built the smooth structure on $N$ this will be a map $\mathrm{Diff}(M) \to \mathrm{Diff}(N)$ and Hurtado's result will tell us that $\dim M \leq \dim N = \dim(\mathrm{Vect}(M)/L)$. All of this needs to come with the disclaimer that I am a simple minded algebraic geometer, who doesn't like these infinite dimensional objects.<|endoftext|> TITLE: Isoperimetric inequality via Crofton's formula QUESTION [10 upvotes]: I have seen various assertions that one can derive the isoperimetric inequality in the plane from Crofton's formula in geometric probability. Unfortunately, I have not managed to figure out such a proof (or find a reference containing said proof). Can anyone point me in the right direction? REPLY [4 votes]: A proof of the isoperimetric inequality using Crofton's formula is contained in these notes by Treibergs: http://www.math.utah.edu/~treiberg/isoperim/isop.pdf See also the proof in Integral Geometry and Geometric Probability by Santalo.<|endoftext|> TITLE: Applications of arithmetic topology to number theory QUESTION [35 upvotes]: There is a well-known analogy between 3-manifolds and number fields, with knots corresponding to prime ideals. Are there any results in number theory that have been proven using topology through this analogy? I would also be interested in any result in number theory that had been motivated by an analogous result in topology. Edit: In light of Dan Petersen's comment, I realize that this is to some extent a historical question. After all, if knot theory had been very developed in 1700, it is possible that quadratic reciprocity would be one example. I have not read all of the Morishita book, but it seems to be mainly concerned with proving topological analogues to some famous theorems in number theory. REPLY [17 votes]: Le and Murakami (HERE and HERE) discovered several previously unknown relations between multiple zeta values through the study of quantum invariants of knots. Further relations were later discovered through knot theory by Takamuki, and by Ihara and Takamuki. These relations stem from the fact that the Kontsevich invariant extends to an invariant not of tangles, but of bracketed tangles, also known as q-tangles or non-associative tangles (alternatively, to knotted trivalent graphs, e.g. HERE). An "associator" relates different bracketings, and the relations satisfied by the associator induce the relations between the multiple zeta values, upon choosing a weight system. Now that these relations have been discovered, the scaffolding can be removed and they can be derived directly from the defining relations of the associator as shown by Furusho (see there for further references). The associator is an algebraic object rather than a topological one, so now the number theorist "no longer need the knot theorists" as far as the known relations between multiple zeta values are concerned.<|endoftext|> TITLE: The current status of the conjecture on algebraic matroids QUESTION [10 upvotes]: Can anyone point out some articles for the conjecture: the dual of an algebraic matroid is algebraic? Thank you! REPLY [9 votes]: I believe this is still open. In Matroid Theory Second Edition by Oxley which was published in 2011 this is listed as an open problem. This problem is addressed twice in the book. First on page 219 Oxley writes: "Probably most basic unsolved problem in the study of algebraic matroids is the following..." He then lists exactly this question on the dual of an algebraic matroid. It then says Lindstrom (1985) shows that the question of whether the dual of an algebraic matroid is algebraic can be reduced to the following question: If a matroid is algebraic over a field of a certain characteristic, then is its dual also algebraic over a field of this characteristic? Later on page 585 Oxley says that the intractability of the problem in the original post "prompted Lindstrom (1988) to proposed the following question." If a matroid is algebraic over a field of a certain characteristic, is its dual also algebraic over a field of the same characteristic? These portions of the book appear to be unchanged from the First Edition published in 1992. So, it seems no major progress has been made. I'll list two papers of Lindstrom referenced below. Lindstrom (1985): On algebraic representations of dual matroids. Department of Math., Univ. of Stockholm, Reports, No. 5. Lindstrom (1988): Matroids, algebraic and non-algebraic. In Algebraic, extremal and metric combinatorics 1986. I cannot find Lindstrom (1985), but Lindstrom (1988) can be found on MathSciNet.<|endoftext|> TITLE: A generalization of Erdős–Mordell inequality QUESTION [5 upvotes]: I proposed my conjecture generalization of Erdős–Mordell inequality as following: Let $A_1A_2....A_n$ be a polygon in a plane, $P$ be the point in $A_1A_2....A_n$. Let $d_i$ be the distances from $P$ to $A_iA_{i+1}$, for $i=1,...,n$ and $A_{i+1}=A_1$. Then: $$\sum_{1}^{n}{ PA_i} \ge \frac{1}{cos{\frac{\pi}{n}}}\sum_{1}^{n}{d_i} $$ Equality holds when $A_1A_2....A_n$ be the regular polygon, and $P$ is its center. REPLY [7 votes]: This has been proved here for convex polygons: Lenhard, H. C., Verallgemeinerung und Verschärfung der Erdös-Mordellschen Ungleichung für Polygone, Arch. Math. 12(1961), 311-314.<|endoftext|> TITLE: Group bundles for topological spaces without universal cover QUESTION [6 upvotes]: I‘m currently writing my Bachelor Thesis on (Co-)Homology with local coefficients. Let me first describe the situation: There are two approaches in defining Homology with local coefficients of a topological space $X$ (see [1, p. 328 – 331]). The first one is via modules (see [1, p. 328]) while the second is via bundles of groups (see [1, p. 330]). However, the first approach only works if $X$ admits a universal cover. So my question is: Are there examples of topological spaces that admit a non-trivial bundle of groups but not a universal cover? I have tried to use a covering of the Hawaiian earrings as described in [1, Section 1.3 Exercise 6 p. 79]. However, if I understand correctly, this is no group bundle since the lifts of closed loops are no group homomorphisms of the fiber. I‘d be very thankful if you helped me! Kathy References: [1] Hatcher, Allen: Algebraic Topology, 2002, https://www.math.cornell.edu/~hatcher/AT/AT.pdf, p. 79 and p. 327 - 331 REPLY [3 votes]: You are correct that the covering space of the Hawaiian earrings in the picture on p.79 of Hatcher's book cannot be made into a group bundle. There are other coverings of the Hawaiian earrings which can, however. So the Hawaiian earrings are an example of the spaces you are looking for. The total space of a group bundle always contains one copy of the base space, corresponding to all trivial elements in all fibers. So try to construct a non-trivial group bundle as follows: Let $X$ be the Hawaiian earrings, then construct first a non-trivial double cover $f:Y\to X$. Now take the covering $id \cup f: X\cup Y\to X$. Finally you have to equip this with a group bundle structure.<|endoftext|> TITLE: Is a one-dimensional compact complex analytic space necessarily projective? QUESTION [15 upvotes]: Let $X$ be a compact complex analytic space with singular locus $X^{\mathrm{sing}}$. Suppose that $X\setminus X^{\mathrm{sing}}$ is a Riemann surface. If $X^{\mathrm{sing}} = \emptyset$, then $X$ is a compact Riemann surface and hence projective (i.e. $X$ admits a holomorphic embedding into some complex projective space). In the general case where $X^{\mathrm{sing}}$ is not necessarily empty, does there exist an injective analytic mapping $j : X \to \mathbb{CP}^{N}$ to some projective space, such that $X$ is isomorphic to $j(X)$? REPLY [22 votes]: Yes, every proper 1-dimensional complex-analytic space $X$ admits a closed immersion (in the sense of locally ringed spaces over $\mathbf{C}$) into an analytic projective space and more specifically is the analytification of a 1-dimensional projective $\mathbf{C}$-scheme (uniquely determined up to unique isomorphism by the GAGA theorems). It also holds in the non-archimedean case in complete generality too. Note that $X$ could be everywhere non-reduced, in which case $X^{\rm{sing}} = X$ (so $X - X^{\rm{sing}}$ is empty), so there is no need to mention $X^{\rm{sing}}$ at all. Up to invoking GAGA cohomological comparison isomorphisms and GAGA algebraization of coherent analytic sheaves at appropriate steps, as well as Oka's coherence results for radicals and normalizations, the method is the same proof bootstrapping from the smooth case (a case that I assume you take as known, but see Remark 1 below) as for the proof of projectivity of arbitrary 1-dimensional proper schemes over a field. In particular, we do use some honest analytic input beyond the tools from the smooth case (no surprise, given the context for the question). The beautiful book "Coherent Analytic Sheaves" provides complete proofs of all analytic tools below (aside from the GAGA theorems, which are in Serre's paper). In the argument below, I avoid all appeal to the analytic theory of ampleness, limiting ampleness considerations to the algebraic side. But finite-dimensionality of the space of global sections of coherent sheaves on $X$ is used all over the place (known by Cartan-Serre, and massively generalized by Grauert as proved near the end of the book "Coherent Analytic Sheaves", though in the present case can be deduced from the smooth case by standard coherence arguments due to finiteness of the normalization map for $X_{\rm{red}}$). Remark 1: In Chapter VII, sections 1--2 of the book "Theory of Stein Spaces" one finds an elegant purely cohomological proof (using the "divisor exact sequence") that a connected compact Riemann surface $X$ admits (many) non-constant meromorphic functions. This argument uses as serious analytic input the finite-dimensionality of coherent sheaf cohomology. Thus, for such $X$ that provides a non-constant proper map $f:X \rightarrow \mathbf{CP}^1$, and such a map must have finite fibers and so is finite in the analytic sense. It therefore corresponds to a coherent sheaf of algebras on $\mathbf{CP}^1$, and that coherent sheaf of algebras is algebraic by GAGA, so that gives a proof of the projectivity of such smooth $X$ (since a scheme finite over a projective space is projective). Thus, the smooth case that we are taking as known can be proved in that way. The merit of the argument in Remark 1 is that it applies essentially verbatim (up to minor adjustments to deal with non-rational points) to prove the projectivity in the non-archimedean case for proper rigid-analytic curves that are regular. Regularity is weaker than smoothness when the non-archimedean ground field $k$ is not perfect, but handling regularity is a crucial feature in practice since normalization of a reduced rigid-analytic curve is always regular but typically not smooth for imperfect ground fields $k$. Beware that for affinoid curves over $k$, geometric reducedness generally cannot be attained by killing nilpotents after a finite extension of $k$ when $p = {\rm{char}}(k)>0$ and $[k:k^p]$ is infinite, as can happen even for discretely-valued $k$ with awful residue field. Remark 2: The method below is written so that it applies essentially verbatim (with only minor technical adjustments to deal with non-rational points, and replacing "smooth" with "regular") in the non-archimedean case (where analogues of all of the analytic inputs are known, largely by work of Kiehl) to reduce the projectivity (really even algebracity, by rigid-analytic GAGA) of proper rigid-analytic spaces of dimension 1 to the case of such spaces that are regular, which we have noted are handled by the method in Remark 1. In other words, the method we discuss also settles the non-archimedean case in complete generality. The desire to provide a method that applies also in the non-archimedean case partially governs the expository choices made below (for those who might care). Now we finally take up the main task, to deduce projectivity for general $X$ from the smooth case. As we know from Oka's work, the radical $J$ of $\mathscr{O}_X$ is a coherent sheaf of ideals, so $J^n = 0$ for large $n$. Granting that the underlying reduced space $X_{\rm{red}} = V(J)$ is projective, to get the same for $X$ we consider $X_i = V(J^i)$ for $i = 1, 2, \dots, n$. Note that the closed immersion $X_{i+1} \hookrightarrow X_i$ is defined by a square-zero coherent ideal sheaf (i.e., $J^i/J^{i+1}$ is square-zero inside $O_X/J^{i+1}$) for $1 \le i < n$. Assuming $X_{\rm{red}} = V(J)$ is algebraic, let's deduce the same for $X_i$ for every $1 \le i \le n$ by induction (so the general problem is brought down to the reduced case). If $n=1$ there is nothing to do, so assume $n>1$. Suppose $1 \le i < n$ with $X_i$ algebraic, and we shall deduce $X_{i+1}$ is algebraic. Since $J^i/J^{i+1}$ is a coherent sheaf on $V(J) = X_{\rm{red}}$, by the GAGA cohomological comparison for the algebraic $X_{\rm{red}}$ we have ${\rm{H}}^2(|X|, J^i/J^{i+1}) = 0$ by the algebraic theory. Hence, ${\rm{Pic}}(X_{i+1}) \rightarrow {\rm{Pic}}(X_i)$ is surjective due to arising from the exact sequence of sheaves $$0 \rightarrow J^i/J^{i+1} \rightarrow O_{X_{i+1}}^{\times} \rightarrow O_{X_i}^{\times} \rightarrow 1$$ (where the first map is $s \mapsto 1+s$). Thus, composing this surjections, the natural map ${\rm{Pic}}(X) \rightarrow {\rm{Pic}}(X_{\rm{red}})$ is surjective. Letting $L_0$ be an ample line bundle on $X_{\rm{red}}$ (by our temporary hypothesis of projectivity in the reduced case), we can lift this to a line bundle $L$ on $X$. Coherent sheaf cohomology of $X$ vanishes beyond some finite degree (due to the existence of a finite Stein covering, as for any compact Hausdorff complex-analytic space, or by bootstrapping from the algebraicity of $X_{\rm{red}}$). By the same induction arguments through powers of the coherent radical which show that a line bundle on a proper noetherian scheme (over an affine base) is ample if its pullback to the underlying reduced scheme is ample, $L$ satisfies the cohomological criterion for ampleness (i.e., for any coherent sheaf $F$ on $X$ and all $m \gg_F 0$, $F \otimes L^{\otimes m}$ has vanishing higher cohomology and is generated by global sections). In particular, for any sufficiently large $N$, $L^{\otimes N}$ is generated by global sections and $\Gamma(X, L^{\otimes N}) \rightarrow \Gamma(X_{\rm{red}}, L_0^{\otimes N})$ is surjective. Fix such an $N$, so the natural map $$f:X \rightarrow {\mathbf{P}}(\Gamma(X, L^{\otimes N})^{\ast})^{\rm{an}}$$ lifts the analogous map $$f_0:X_{\rm{red}} \rightarrow {\mathbf{P}}(\Gamma(X_{\rm{red}}, L_0^{\otimes N})^{\ast})^{\rm{an}}$$ between closed subspaces, and by GAGA the map $f_0$ is a closed immersion for large $N$ since $L_0$ is ample. But then $f$ is injective on underlying sets and hence is a proper morphism with finite fibers, so it is a finite morphism of complex-analytic spaces; i.e., it is classified by a coherent sheaf of algebras on the projective space target. By GAGA we can algebraize that coherent sheaf of algebras, so $X$ would be algebraic and hence projective; this just expresses that an analytic space finite (in the analytic sense) over a projective analytic space is itself projective (since a scheme finite over a projective scheme is projective). That slog finally brings us down to the case when $X$ is reduced. We can ignore any isolated points on $X$, so we can assume all analytic irreducible components $Y_i$ of $X$ are 1-dimensional. Let $Y'_i \rightarrow Y_i$ be the finite analytic normalization map (thanks to Oka, or more classical methods, though 1-dimensional singularities can be incredibly nasty), so each $Y'_i$ is algebraic by the known smooth case. Let $Y' := \coprod Y'_i$, so $q:Y' \rightarrow X$ is the finite normalization and by reducedness of $X$ (and the analytic Nullstellensatz) the natural map $O_X \rightarrow q_{\ast}(O_{Y'})$ is an inclusion of coherent sheaves with coherent cokernel having finite support. Let $x_i \in Y_i$ be a smooth point not on any other $Y_j$, so $x_i$ is a smooth point on $X$. Let $L = \otimes O(x_i)$, so $q^{\ast}(L)$ is ample on $Y'$ (by the algebraicity on $Y'$ and choice of the $x_i$'s). For any coherent $G$ on $X$, the natural map $G \rightarrow q_{\ast}(q^{\ast}(G))$ has coherent kernel and cokernel with finite support (isomorphism over $X^{\rm{sm}}$ that is the complement of a finite subset of $X$), so the natural map ${\rm{H}}^i(X, G) \rightarrow {\rm{H}}^i(Y', q^{\ast}(G))$ is an isomorphism for $i>0$. Hence, for any coherent $F$ on $X$, $F \otimes L^{\otimes m}$ has vanishing higher cohomology for large $m$ since the analogue holds on $Y'$ (where $L$ has ample pullback). Thus, for any point $x \in X$, if $I_x$ denote the coherent ideal sheaf of $x$ we see that $\Gamma(X, L^{\otimes m}) \rightarrow L^{\otimes m}(x)$ is surjective for all large $m$ since ${\rm{H}}^1(X, I_x \otimes L^{\otimes m})=0$ for all large $m$ (depending on $x$). Applying to $x=x_i$ for each $i$, for all large enough $m$ we see that $\Gamma(X, L^{\otimes m})$ generates the fiber of $L^{\otimes m}$ at every $x_i$, so by coherence considerations if we pick a large enough $M$ then $\Gamma(X, L^{\otimes M})$ generates the stalks of $L^{\otimes M}$ away from an analytic set disjoint from all $x_i$'s. Such an analytic set must be finite (as every irreducible component contains some $x_i$ and is 1-dimensional), so for a big enough $M' \in M \mathbf{Z}$ we see that $L' := L^{\otimes M'}$ is generated by global sections. It follows that we have a natural map $f:X \rightarrow {\mathbf{P}}(\Gamma(X, L')^{\ast})^{\rm{an}}$ that is injective on tangent spaces at each of the points $x_i \in X^{\rm{sm}}$, so $f$ has no positive-dimensional fibers (as such a fiber would have to contain some $Y_i$ for analyticity and dimension reasons, a contradiction by the tangential injectivity property at each $x_i$). Thus, $f$ has finite fibers, so by properness $f$ is a finite map in the analytic sense. Once again arguing with GAGA for coherent sheaves of algebras on projective space as we did above, $X$ is algebraic. Voila.<|endoftext|> TITLE: Where can I find basic "computations" of equivariant stable homotopy groups? QUESTION [6 upvotes]: I am new to this subject; so please correct me if I will say something wrong or if you don't like my notation. In particular, I don't know whether it is reasonable to consider an infinite group $G$ (should it actually be a compact Lie group?) in my question. Also, for a group $G$ the corresponding stable homotopy category $SH_G$ (should I use another notation?) also depends on the choice of a universe; does one get a canonical category taking a complete universe? As far as I understand equivariant stable homotopy theory, in $SH_G$ one should take spectra $\Sigma_G (G/H)_+$ corresponding to all homogenious spaces $G/H$ to obtain a family of compact generators. Now, I want to know something on morphisms between these generators. In particular, is the group $Mor_{SH_G} (\Sigma_G (G/H_1)_+, \Sigma_G (G/H_2)_+[i])$ zero for any $i>0$ and all subgroups of $G$? This question seems to be related to A heart for stable equivariant homotopy theory Is there any text where I can find results of this sort? Possibly, one can say something about these morphism groups (say) using duality and apply some well-known results after that? Any hints would be very welcome! REPLY [2 votes]: Since Denis gave the right reference, namely http://www.math.uchicago.edu/~may/BOOKS/equi.pdf, I did not follow up and answer this question. We can work with any compact Lie group and any complete universe. Working in the equivariant stable category always, with $S = S^0$, $S^{-n}$ a negative sphere $G$-spectrum and $S^0$ and $S^n$ sphere spaces with trivial $G$-action, $$ \pi_{-n}(S) = [S^{-n},S]_G = [S,\Sigma^{\infty} S^n]_G = [S^0, QS^n]_G = 0 $$ where $Q^G$ is the functor $(colim_V \Omega^V \Sigma^V(-))^G$, with $V$ running through representations of $G$; $Q^G(X)$ is an $(n-1)$-connected $G$-space if $X$ is so. Replacing $S$ by $\Sigma^{\infty} G/H_+$, the same argument works, using change of groups so that one is now looking at $[-,-]_H$. More generally, the negative homotopy groups of the suspension $G$-spectrum of any $G$-space are zero, and the non-negative groups are calculated by the tom Dieck splitting theorem.<|endoftext|> TITLE: Minimum separation among $m$ random points on an $n$-dimensional unit sphere QUESTION [10 upvotes]: Consider $m$ points $v_1, \ldots, v_m \in R^{n}$, which are uniformly distributed on the $n$-dimensional unit sphere $S^{n-1} = \{v:\|v\|_2 = 1\}$. Let the minimum separation be $$ \rho = \min_{i,j\in{\{1,\ldots,m\}}} \|v_i - v_j\|_2. $$ Question: What is the expectation of $\rho$? How fast does $\rho$ converge to its expectation as $m,n\to\infty$? Here are some closely related questions: Mean minimum distance for N random points on a one-dimensional line Mean minimum distance for N random points on a unit square (plane) Mean minimum distance for K random points on a N-dimensional (hyper-)cube However, the case for n-dimensional sphere seems less clear. REPLY [5 votes]: The preprint "Random Point Sets on the Sphere --- Hole Radii, Covering, and Separation" by Johann S. Brauchart, Edward B. Saff, Ian H. Sloan, Yu Guang Wang, and Robert S. Womersley gives the following result in Corollary 3.4: $\mathbb{E}[N^{2/d}\Theta_\text{min}]\to C_d = (\kappa_d/2)^{-1/d}\Gamma(1+\tfrac{1}{d})$ as $N\to\infty$ It also gives a bound on the variance.<|endoftext|> TITLE: Lie subalgebras of $\chi^{\infty}(M)$ of codimension $n = \dim M$ QUESTION [11 upvotes]: For a connected $n$-manifold $M$, the Lie algebra of all smooth vector fields is denoted by $\chi^{\infty}(M)$. For a point $p\in M$ we define $L_{p}=\{X\in \chi^{\infty}(M)\mid X(p)=0\}$. Of course $L_{p}$ is a Lie subalgebra of $\chi^{\infty}(M)$ whose codimension is equal to $n$. Is it true that every codimension-$n$ Lie subalgebra of $\chi^{\infty}(M)$ is necessarily in the form of (or isomorphic to) $L_{p}$ for some $p\in M$? Here is a related post. REPLY [3 votes]: Correction: The argument I gave initially is wrong. I treated $\mathfrak X(M)'$ like the space of differential forms. Only operations on $\mathfrak X(M)$ go over to the dual as (negative) adjoint operations, so $\mathcal L_X$ makes sense but $i_X$ and $d$ do not. Since it created some interest I leave the old answer. Corrected argument: Let $\alpha$ be a 1-form on $M$ which is non-zero and closed near a point $p$ in $M$. Then consider the current $\alpha. \delta_p \in \mathfrak X(M)'$ where $\delta_p$ is the Dirac delta. Consider the the space $$ L^{\alpha}_p = \{X\in \mathfrak X(M): \mathcal L_X(\alpha. \delta_p)=0\} $$ Let us compute this space. The question is local, so we assume that we are in $\mathbb R^n$ and $p=0$. Since $\mathcal L_X$ acts as the negative adjoint, for an arbitrary field $Y$ we have $$ 0=\langle Y, \mathcal L_X(du^1.\delta_p)\rangle = -\langle \mathcal L_XY, du^1.\delta_p\rangle = -[X,Y]^1(0) = \big(-(\partial_iY^1(0))X^i(0) + (\partial_iX^1(0))Y^i(0)\big)\partial_1 $$ Running $Y$ through a basis of $T_0M$ with $\partial_i Y(0)=0$ for all $i$ implies that $\partial_i X^1(0)=0$ $ \forall i$. Choosing $Y$ with $Y(0)=0$ in such a way that $dY^1$ runs through a basis of $T_0^*M$, implies $X^i(p)=0$. The converse is also true, thus $$ L^{\alpha}_0 = \{X\in \mathfrak X(\mathbb R^n): dX^1(0)=0, X(0)=0 \} $$ which has codimension $2n$, SIGH. Remark, and proof of the statement in the question: In a related question it was hinted that the determination of all the maximal ideals of $\mathfrak X(M)$ would be of interest. These are all of infinite codimension and are of the form: For a point $p$ in $M$ consider all vector fields $X$ which vanish at $p$ of infinite order. Here $M$ should be compact or $\mathfrak X(M)$ should be replaced by the space of vector fields with compact support. This is proved by Purcell and Shanks: For a related result and references see MR0516602 Grabowski, J. Isomorphisms and ideals of the Lie algebras of vector fields. Invent. Math. 50 (1978/79), no. 1, 13–33. In fact, this paper contains a proof of your question: Let $M$ be compact or replace $\mathfrak X(M)$ by the Lie algebra $\mathfrak X_c(M)$ of vector fields with compact support. So let $A=C^\infty_c(M)$ and let $\mathcal L = \mathfrak X_c(M)$. By Proposition 3.6, they satisfy the assumtions of the following theorem. Theorem 5.1. Let $A$ be an $I$-algebra and let $\mathcal L$ be an admissible $A$-Lie module. Then for each maximal-prime finite-codimensional ideal $J$ of $A$ the Lie subalgebra $\mathcal L_J$ of $\mathcal L$ is maximal finite-codimensional and the mapping $\mathfrak M_A \ni J \mapsto \mathcal L_J\in \mathfrak M_{\mathcal L}$ is a bijection. Since $\mathfrak M_A = \{A_p: p\in M\}$ and $\mathcal L_J =\{X\in \mathcal L: X(A)\subset J\}$, the result follows. For notation see Grabowski's paper. Old, wrong answer: Here is a counterexample: Let $\alpha$ be a 1-form on $M$ which is non-zero and closed near a point $p$ in $M$. Then consider the current $\alpha\otimes \delta_p \in \mathfrak X(M)'$ where $\delta_p$ is the Dirac delta. Consider the space of all $X\in\mathfrak X(M)$ with $i_X(\alpha\otimes\delta_p) = 0$ and $\mathcal L_X(\alpha\otimes\delta_p) = 0$. Since $i_{[X,Y]} = [i_X,\mathcal L_Y]$ and $\mathcal L_{[X,Y]} =[\mathcal L_X,\mathcal L_Y]$, this space is a Lie algebra. Its codimension is 2. More detail: Choose a Riemannian metric $g$ on $M$ and and consider a chart $(U,u)$ centered at $p$ such that $\alpha|_U = du^1$. Then $0 = i_X(du^1\otimes\delta_p) = du^1(X)(p)= X^1(p)$ and $0=\langle Y, \mathcal L_X(du^1\otimes\delta_p)\rangle = \langle Y, (i_Xd + di_X) (du^1\otimes\delta_p)\rangle = -\langle Y, i_X(du^1\wedge d\delta_p)\rangle = -X^1(p) \text{div}(Y)(p) + Y^1(p) \text{div}(X)(p). $ Since $X^1(p)=0$ and $Y$ is arbitrary, we see that the Lie subalgebra is given by $\{X: X^1(p)=0, \text{div}(X)(p)=0\}$ and thus has thus has codimension $2$. The divergence is with respect to the density of $g$. One can also use $\text{div}(f.X) = f\text{div}(X) + g(\text{grad}^g(f),X)$ to make a more local computation.<|endoftext|> TITLE: Limit space of a sequence of Riemannian manifolds with uniformly bounded below Ricci curvature QUESTION [5 upvotes]: Let $\{M^n_i\}_{i=1}^\infty$ be a sequence of closed smooth Riemannian $n$-dimensional manifolds with uniformly bounded below Ricci curvature and uniformly bounded above diameter. The Gromov compactness theorem says that there exists a subsequence which converges in the Gromov-Hausdorff sense to a compact metric space $X$. Questions. (1) Is the Hausdorff dimension of $X$ necessarily integer? (2) Is it at most $n$? Remark. If one assumes a stronger condition that the sectional (rather than Ricci) curvature of $M_i$ is uniformly bounded below then the answers to both questions are positive as it is shown in the theory of Alexandrov spaces. REPLY [3 votes]: I cannot say anything about question (1), but the answer to question (2) is yes: By the work of Cheeger-Colding, we can assume that $(M_i)$ endowed with the normalized volume measure converges in the measured Gromov-Hausdorff sense. Thus, the limit is a $\mathrm{CD}(K,n)$-space in the sense of Lott-Sturm-Villani (where $K$ is the uniform lower bound on the Ricci curvatures of $M_i$). These spaces have Hausdorff dimension $\leq n$. You should find these results (or references to the original sources) in K.-T. Sturm. On the geometry of metric measure spaces I. and II.<|endoftext|> TITLE: A two-point inequality QUESTION [5 upvotes]: Let $M(p,q) = (2p-\sqrt{p^{2}+q^{2}})\sqrt{p+\sqrt{p^{2}+q^{2}}}$ and set $B(t) = M(x+t, \sqrt{t^{2}+(y+bt)^{2}})$. Given any real $x,y,b$ is it true that $\varphi(t) = B(t)+B(-t)$ is decreasing in $t$ for $t \geq 0$. Motivation: Consider the hamming cube $\{-1,1\}^{N}$. Let $f :\{-1,1\}^{N} \to \mathbb{R}$. Set $f_{k} = \mathbb{E} (f| \mathcal{F}_{k})$ to be a martingale $k=0,..,N$ which takes the average of the function with respect to the variables $(x_{k+1},...,x_{N})$. So $f_{0} = \mathbb{E} f = \frac{1}{2^{N}} \sum_{x \in \{-1,1\}^{N}} f(x)$, $f_{N} := f$. So $f_{k}$ lives on $\{-1,1\}^{k}$. For example $$ f_{N-1}(x) = \frac{1}{2}\left(f(x_{1},\ldots, x_{N-1},1)+f(x_{1},\ldots, x_{N-1},-1) \right) $$ Define $\nabla_{i} f := \frac{1}{2}\left(f(x_{1},x_{2},\ldots, 1,\ldots, x_{N}) -f(x_{1},x_{2},\ldots, -1,\ldots, x_{N})\right)$. And let $|\nabla f|^{2} = \sum_{i=1}^{N}|\nabla_{i} f|^{2}$. Now let $T_{\rho}$ be the Ornstein-Uhlenbeck semigroup on $\{-1,1\}^{N}$ i.e., $$ T_{\rho} f = \sum_{S \in 2^{N}} \rho^{|S|} \hat{f}(S) W_{S}(x) $$ Where $W_{S}(x)$ is the Walsh system , and $\hat{f}(S)$ are Fourier coefficients with respect to this system. My question becomes the claim that the following map $$ \rho \to \mathbb{E}M(T_{\rho}f_{k}, |\nabla T_{\rho} f_{k}|) $$ is monotone for $\rho \in [0,1]$ and any $k \geq 1$. This in particular makes the process $M(f_{k}, |\nabla f_{k}|)$ supermartingale. And then central limit theorem $N \to \infty$ gives some interesting inequalities. REPLY [2 votes]: The following may be the starting point of an answer. Introduce the new variables \begin{equation} u:=\sqrt{(y-b t)^2+t^2+(t-x)^2},\quad v:=\sqrt{(b t+y)^2+t^2+(t+x)^2}, \end{equation} so that $b=\frac{-4 t x-u^2+v^2}{4 t y}$. Further let \begin{equation} r:=\sqrt{-t+u+x},\quad s:=\sqrt{t+v+x}, \end{equation} so that \begin{equation} u = r^2 + t - x, \quad v = s^2 - t - x. \end{equation} Then \begin{equation} \varphi'(t) \frac{32}3\, r s t y^2 =-\left(r^2+t-x\right)^2 \left(-2 (r+s) \left(-s^2+t+x\right)^2+8 t x (r+s)+4 y^2 (s-r)\right) \end{equation} \begin{equation} -16 s t y^2 \left(r^2+t-x\right)-(r+s) \left(r^2+t-x\right)^4-16 r t y^2 \left(-s^2+t+x\right) \end{equation} \begin{equation} +4 \left(-s^2+t+x\right)^2 \left(2 t x (r+s)+y^2 (s-r)\right)-(r+s) \left(-s^2+t+x\right)^4 \end{equation} \begin{equation} -16 t x \left(t x (r+s)+2 y^2 (s-r)\right), \end{equation} which is a polynomial in $x,y,r,s,t$. We need to show that this polynomial is nonnegative on an appropriate set (say $S$) of 5-tuples $(x,y,r,s,t)$. In principle, this can be done by using standard methods of real algebraic geometry. However, I have not yet succeeded in obtaining a tractable description of such a set $S$.<|endoftext|> TITLE: An identity related to partitions into $n$ parts and Schur polynomials QUESTION [8 upvotes]: While working with Schur polynomials I found what seems like a nice identity, and I wonder if it has a simple proof. Notation: Suppose $d,n\in\mathbb{N}$, and $\lambda =(\lambda_1,\dots,\lambda_n)$ is an ordered partition of $d$, that is $\lambda_1 \ge \dots \ge \lambda_n \ge 0$, and $\lambda_1 + \dots + \lambda_n = d$. Let $\Lambda(d,n)$ be the set of all such partitions. For each $\lambda \in \Lambda(d,n)$ define: $$ N(\lambda) = \prod_{1\le i < j \le n} \frac{\lambda_i - \lambda_j + j - i}{j - i}, $$ and $$ W(\lambda) = \prod_{1 \le i \le n} (\lambda_i + n - i)!. $$ Remark: $N(\lambda)=s_\lambda(1,\dots,1)$, where $s_\lambda$ is the Schur polynomial associated with the partition $\lambda$. Now, define $$ A(d,n) = \sum_{\lambda \in \Lambda(d,n)} \frac{N(\lambda)^2}{W(\lambda)}. $$ It seems the following identity holds: $$ A(d,n) = \left( \prod_{k=0}^{n-1} k! \right)^{-1} \frac{n^d}{d!}.$$ Is there a simple proof\explanation? REPLY [10 votes]: We can use the fact that $N(\lambda)=\left|\text{SSYT}(\lambda)\right|$, the number of semistandard Young tableaux of shape $\lambda$, and that $d!\cdot\left(\frac{\prod_{1\le i < j\le n}(\lambda_i - \lambda_j + j - i)}{\prod_{1 \le i \le n} (\lambda_i + n - i)!}\right)=\left|\text{SYT}(\lambda)\right|$, the number of standard Young tableaux of shape $\lambda$. By some simple rearranging your identity becomes $$\sum_{|\lambda|=d}\left|\text{SSYT}(\lambda)\right|\left|\text{SYT}(\lambda)\right|=n^d.$$ Now, if you track the left hand side through the Robinson-Schensted-Knuth correspondence, you'll find that it corresponds to $d\times n$ matrices with row sums all $1$. So there are exactly $n^d$ of these. REPLY [6 votes]: I found a proof which I don't really like, but I'll share it. For two (real) diagonal matrices $A,B$, the Harish-Chandra-Itzykson-Zuber (HCIZ) integral is $$ I(A,B) = \int_{U(n)} e^{\rm{tr}(U^* A U B)} \, \rm{d} U = c_n \frac{\det\left([e^{a_j b_k}]_{j,k=1}^n\right)}{\Delta(a)\Delta(b)}, $$ where $\Delta(a) = \prod_{j TITLE: On certain solutions of a quadratic form equation QUESTION [6 upvotes]: This is a continuation of this question: A class of quadratic equations Let $f(x,y) = ax^2 + bxy + cy^2$ be an irreducible and indefinite binary quadratic form. Consider the equation $$\displaystyle f(x,y) = a,$$ where $a$ is the $x^2$ coefficient of $f$. Plainly, this equation is soluble in the integers since $(x,y) = (1,0)$ is a solution. Since $f$ is indefinite and irreducible, an appropriate action by the unit group of the quadratic field generated by $f$ will give us infinitely many solutions. I am interested in the case when one can find a solution with $y$ co-prime to $a$. This is not always possible. For example, the equation $$\displaystyle 5x^2 - 7y^2 = 5$$ has fundamental solution $(6,5)$ and all solutions are of the form $$\displaystyle \begin{pmatrix} 6 & 7 \\ 5 & 6 \end{pmatrix}^n \binom{\pm 6}{\pm 5}.$$ The issue appears to be that $a$ divides $b$, the $xy$ coefficient of $f$. However, this divisibility property is not preserved under substitution action by $\operatorname{GL}_2(\mathbb{Z})$, so this cannot be an intrinsic obstruction. Here is the ultimate question I am interested in. Let $O_f(\mathbb{R})$ be the maximal subgroup of $\operatorname{GL}_2(\mathbb{R})$ which fixes $f$ by substitution. It is conjugate to the split orthogonal group $O(1,1)$. Let $O_f^-$ denote the subset of $O_f$ consisting of elements of negative determinant. This part is conjugate to $$\displaystyle \left \{ \pm \begin{pmatrix} \cosh t & \sinh t \\ -\sinh t & -\cosh t \end{pmatrix}, t \in \mathbb{R} \right \}.$$ My question is the following: for $f$ an irreducible and indefinite binary quadratic form with integer coefficients, does $O_f^-$ always contain an element in $\operatorname{GL}_2(\mathbb{Z})$? In other words, is $O_f^- \cap \operatorname{GL}_2(\mathbb{Z})$ always non-empty? One can show that the $O_f^-$ contains elements of the shape $$\displaystyle \begin{pmatrix} m & \dfrac{bm + cn}{a} \\ n & - m \end{pmatrix},$$ where $(m,n)$ satisfies $f(m,n) = a$. If $n$ is co-prime to $a$, then since $am^2 + n(bm+cn) = a$, it follows that $n | bm + cn$ and the above is in $\operatorname{GL}_2(\mathbb{Z})$. If $a | n$ and $a | b$, then again it is in $\operatorname{GL}_2(\mathbb{Z})$. I've done quite a few experiments and it seems that $O_f^- \cap \operatorname{GL}_2(\mathbb{Z})$ is always non-empty. Any help would be appreciated! REPLY [3 votes]: Your ultimate question was answered by Gauss: $O_f^- \cap \operatorname{GL}_2(\mathbb{Z})$ is nonempty if and only if the class of $f$ is ambiguous (i.e. its square is the trivial class). Indeed, $f(x,y)$ is improperly equivalent to itself if and only if $f(x,y)$ is properly equivalent to $f(y,x)$. As the classes of $f(x,y)$ and $f(y,x)$ are inverses to each other, the claim follows. For more details see Theorem 2.1 in Chapter 14 of Cassels: Rational quadratic forms. Sections 4 and 6 of the mentioned chapter contain further information on ambiguous classes, e.g. each such class contains a form of type $[A,0,C]$ or $[A,A,C]$, and vice versa.<|endoftext|> TITLE: What should I cite for the Poincaré conjecture? QUESTION [23 upvotes]: I'm writing a paper that, rather unexpectedly, needs the Poincaré conjecture for one of the results. (The paper has almost nothing to do with differential geometry!) The conjecture was famously proved at the beginning of the century by Perelman, in a series of three papers. Unfortunately, I'm not a differential geometer, and I fear that if I read his papers I won't understand anything or be able to pinpoint in which one the conjecture is solved. I'm sure that if I just write "the Poincaré conjecture, due to Perelman" in my paper everybody will understand what I'm talking about. But it still feels "normal" to cite something. So: what should I cite? Nothing? One of the three papers? All three of them? Something else? REPLY [21 votes]: I think it is customary to cite at least the first two papers ("The entropy formula for the Ricci flow and its geometric applications" and "Ricci flow with surgery on three-manifolds"). See this for an example. Together they imply the full geometrization conjecture. The shortest proof of "just Poincare" involves the third paper, but I guess that's beyod the point.<|endoftext|> TITLE: Do character tables determine association schemes up to isomorphism? QUESTION [5 upvotes]: I am interested in commutative, but not-necessarily-symmetric association schemes. I noticed that the conjugacy class scheme of the dihedral group of order 8 has the same character table as the one for the quaternion group, and the schemes are also isomorphic. Are there examples of non-isomorphic association schemes that have the same character table? I am not married to association schemes arising from groups. REPLY [2 votes]: IMHO the isomorphism of association schemes is understood as isomorphism of underlying coloured digraphs. The smallest example of non-isomorphic (in this sense) association schemes with the same character table comes from a pair of nonisomorphic degree 6 strongly regular graphs on 16 vertices. One of these graphs is Shrikhande graph, the other is $4\times 4$ rook graph. In the example in the question one has association schemes of conjugacy classes of groups $D_8$ and $Q_8$, and they happen to be isomorphic in the sense of coloured digraph isomorphism. It is example number 9 in tables by Izumi Miyamoto and Akihide Hanaki. In fact if isomorphism of association schemes was possible to decide from their character tables then the graph isomorphism problem would be easy to solve.<|endoftext|> TITLE: Elementary symmetric functions of reciprocals of monic polynomials in function fields QUESTION [5 upvotes]: Let $q$ be a prime power and $\mathbb{F}_q$ the field of cardinality $q$. Let $A = \mathbb{F}_q[T]$ and let $A_+ \subset A$ be the monic polynomials. Choose any ordering $<$ of $A_+$ and let $k$ be a positive integer. Set $$e(k) = \sum_{\begin{matrix} a_1, a_2, \ldots, a_k \in A_+ \\ a_1 \cdots > a_k}} \frac{1}{a_1^{s_1} \cdots a_k^{s_k}} \in \mathbb{F}_q((T^{-1})), $$ where the ordering is an arbitrarily chosen lexicographic ordering of $A_+$ and $s_1, \ldots, s_k$ are positive integers. It seems then that your value $e(k)$ is $\zeta_l(1, \ldots, 1)$ of depth $k$. He also defines $$ \zeta_d(s_1, \ldots, s_k) = \sum_{\substack{a_1, \ldots, a_k \in A_+ \\ \deg a_1 > \cdots > \deg a_k}} \frac{1}{a_1^{s_1} \cdots a_k^{s_k}}, $$ which is the type of MZV you mentioned studied by Chang. He also defines a couple of other variants. To address one of your questions, from what Thakur says in Remark 5.10.16, it would appear that the $e(k)$'s are probably not algebraically related to the MZV's $\zeta_d(s_1, \dots, s_k)$. Anderson and Thakur ("Multizeta values for $\mathbb{F}_q[t]$, their period interpretation, and relations between them", IMRN, 2009) showed that the MZV's $\zeta_d(s_1,\dots, s_k)$ defined using degrees can be realized as periods of certain higher dimensional Drinfeld modules (Anderson $t$-modules) that are iterated extensions of tensor powers of the Carlitz module. They then can also be expressed as linear combinations of values of Carlitz multiple polylogarithm functions. Using this period interpretation various bits of transcendence machinery can be applied, which is the subject of the paper you mention by Chang (Compositio, 2014). Thakur mentions in that same remark 5.10.16 that the lexicographic MZV's $\zeta_l(s_1, \dots, s_k)$ do not have natural relations with periods of Drinfeld modules, so at least on the surface they seem to represent a different class of numbers than the $\zeta_d(s_1, \dots, s_k)$'s. Thakur follows this up with numerical calculations when $q=2$ and shows that they do not appear to be rational multiples of Carlitz zeta values. Of course this does not answer your question definitively, but I think it would be safe to say that it is not known whether there are any general relations. Also, one minor correction: when you remark that it was shown that the Carlitz zeta values $\zeta(m)$ are algebraically independent over $\mathbb{F}_q(T)$ (aside from the known relations involving Bernoulli-Carlitz numbers and $p$-th powers), this result is actually due to Chang and Yu ("Determination of algebraic relations among special zeta values in positive characteristic," Adv. Math. 216 (2007), 321-345). In the 1991 paper of Yu, he proves that the values are all transcendental over $\mathbb{F}_q(T)$, but the algebraic independence questions were answered later.<|endoftext|> TITLE: Positive semidefinite ordering for covariance matrices QUESTION [6 upvotes]: Suppose that X and Z are matrices with the same number of rows. Let $$ D = \left[\begin{array}{cc} X' X & X'Z \\ Z'X & Z'Z \end{array} \right]^{-1} - \left[\begin{array}{cc} (X' X)^{-1} & 0 \\ 0 & 0 \end{array} \right],$$ where all inverses are assumed to exist and the zeros represent zero matrices of suitable dimensions. How can we prove that $D$ is positive semidefinite? REPLY [6 votes]: Let $$A:=X' X,\quad B:=X'Z,$$ \begin{equation} \left[\begin{array}{cc} U & V \\ V' & T \end{array} \right]:= \left[\begin{array}{cc} X' X & X'Z \\ Z'X & Z'Z \end{array} \right]^{-1}. \end{equation} Then $AU+BV'=I$, $AV+BT=0$, whence \begin{equation} V=-A^{-1}BT,\quad U=A^{-1}+A^{-1}BTB'A^{-1}, \end{equation} \begin{equation} D=\left[\begin{array}{cc} A^{-1}BTB'A^{-1} & -A^{-1}BT \\ -(A^{-1}BT)' & T \end{array} \right]. \end{equation} Also, $T$ is positive definite, as a diagonal block of the positive definite matrix $\left[\begin{array}{cc} X' X & X'Z \\ Z'X & Z'Z \end{array} \right]^{-1}$. Therefore, for any column matrix $\left[\begin{array}{cc} x \\ z \end{array} \right]$ with sub-columns $x$ and $z$ of appropriate heights, straightforward calculations yield \begin{equation} \left[\begin{array}{cc} x \\ z \end{array} \right]'D\left[\begin{array}{cc} x \\ z \end{array} \right] =(w-z)'T(w-z)\ge0, \end{equation} where $w:=B'A^{-1}x$. Thus, $D$ is indeed nonnegative definite.<|endoftext|> TITLE: A $n$-gon is isospectral to a regular $n$-gon (Isospectral $\implies$ isometry ?) QUESTION [9 upvotes]: If an $n$-gon $P$ is isospectral to a regular $n$-gon $Q$, what could we say about the shape of the $P$. Otherwise, what could we say about $Q$? In fact, some hints or simply some ideas would be appreciated. Clarification : I talk about the spectrum of the Laplacian on the interior of the polygon, acting on the space of functions vanishing on the boundary. REPLY [11 votes]: Rowlett is hosting The Sound of Symmetry here. The proof of Theorem 4 is exactly as Noam Elkies suggests: Via the Dirichlet heat trace's asymptotic expansion, both area and perimeter are determined by the spectrum, and so for any $n$-gon $\Omega$ the isoperimetric ratio $|\Omega|/|\partial\Omega|^2$ is determined by the spectrum. The content of the proof is that this ratio is globally maximized among $n$-gons at the regular $n$-gon. This determines the regular $n$-gon and implies that any $n$-gon isospectral to the regular $n$-gon is in fact isometric to it. Unless there's a point in the proof where you're confused, I think this settles the question. (I have a suspicion that the third coefficient of the heat trace in convex polygons $$ \frac{1}{24}\sum_{\mbox{angles }\alpha_i} \bigg(\frac{\pi}{\alpha_i} - \frac{\alpha_i}{\pi}\bigg) $$ is also extremal at the regular $n$-gon. This may be another way to show a version of the desired result for convex $n$-gons.)<|endoftext|> TITLE: Fermionic Wick Theorem QUESTION [7 upvotes]: Assume we are given are fermionic many-particle system with creation operators $c_1^\dagger,c_2^\dagger,...$ acting on some Hilbert space $\mathcal{H}$. That is, the $c_i^\dagger$ and their corresponding annihilation operators $c_i=(c_i^\dagger)^\dagger$ act as bounded operators on $\mathcal{H}$ and satisfy the CAR relations $$\{c_i,c_j\}=\{c^\dagger_i,c_j^\dagger\}=0\quad\text{and}\quad\{c_i^\dagger,c_j\}=\delta_{ij}\quad\forall i,j$$ In the bosonic case, the Wick theorem can be elegantly written in the form $$\prod_{j=1}^nc_{i_j}^{\sigma_j}=\sum_{Q\subset\mathbb{N}_n}\langle\prod_{j\in Q}^nc_{i_j}^{\sigma_j}\rangle:\prod_{j\not\in Q}^nc_{i_j}^{\sigma_j}:$$ for any $i_1,...,i_n\in\mathbb{N}$ and $\sigma_1,...,\sigma_n\in\{-1,1\}$, where $c^-_i:=c_i$, $c^+_i:=c^\dagger_i$ for any $i\in\mathbb{N}$, $\mathbb{N}_n:=\{1,\dots,n\}$, $\langle A\rangle:=\langle\Omega\mid A\Omega\rangle$ tenotes the vacuum expectation value and $:\cdots:$ denotes the Normal ordering. However, for fermions this formula has to be modified by attaching an appropriate sign to every of the summands. What is this sign? REPLY [8 votes]: Here's a trick to recover the fermionic signs, knowing the bosonic formulas. First, introduce a sufficient number of algebraically independent "c-numbers" $\epsilon_k$, that are only required to anti-commute among themselves and with the $c_i$, $c_i^\dagger$ generators. Then, according to the usual sign rules, the products $\epsilon c_i$ and $\epsilon c_j^\dagger$, where $\epsilon$ is any one of the new generators, will obey the bosonic canonical commutation relations with extra anti-commuting factors: $$ [\epsilon c_i, \eta c_j^\dagger] = -\epsilon \eta \{c_i, c_j^\dagger\} = -\epsilon \eta \delta_{ij} = -\epsilon \eta \langle c_i c_j^\dagger \rangle = \langle (\epsilon c_i) (\eta c_j^\dagger) \rangle, $$ for example. The extra intermediate $-$ sign appeared because the $c_i$ and $\eta$ exchanged positions while moving $\eta$ out of the commutator or moving it inside the expectation value. Note that $\epsilon$ and $\eta$ need to be chosen independently, to avoid writing down tautologies like $0=0$ due to the nilpotency $\epsilon^2 = 0$ of the new generators. So, the bosonic Wick formula still works in the following way $$ \prod_{j=1}^n \epsilon_{k_j} c_{i_j}^{\sigma_j}=\sum_{Q\subset\mathbb{N}_n}\langle\prod_{j\in Q}^n \epsilon_{k_j} c_{i_j}^{\sigma_j}\rangle:\prod_{j\not\in Q}^n \epsilon_{k_j} c_{i_j}^{\sigma_j}: , $$ where the $\epsilon_{k_j}$ multipliers are all distinct for distinct $j$. I'll leave it as an exercise to obtain the exact sign factor that you were looking for, which you can get by factoring out $\prod_{j=1}^n \epsilon_{k_j}$ from both sides of the above equation.<|endoftext|> TITLE: Algebraic structure on homotopy groups of spheres QUESTION [9 upvotes]: It is about a "conjecture" I heard (when I was student). There would exist an algebraic structure on the homotopy groups of spheres such that this algebraic structure would be the free algebraic structure generated by one point. In the wikipedia page "Field with one element", it is written that the algebraic $K$-theory of the field with one element is related (can be identified ?) with the stable homotopy groups of spheres. Is it the answer ? REPLY [13 votes]: I think the answer to the last question is no. As far as I understand, the "general linear groups over the field with one element" are supposed to be the symmetric groups. Therefore the statement that the algebraic K-theory of the field with one element can be identified with the stable homotopy groups spheres is essentially the Barratt-Priddy-Quillen theorem, which says that the group completion of $\coprod_n B\Sigma_n$ is the sphere spectrum. As far as I can tell, the Barratt-Priddy-Quillen theorem does not yield that much explicit information about stable homotopy groups of spheres. On the positive side, I would like to mention that Smirnov apparently gave an explicit presentation of the $E_\infty$ page of the Adams spectral sequence as an $A_\infty$ algebra: "Description of stable homotopy groups of spheres in the language of A∞-algebras." (Russian) Uspekhi Mat. Nauk 51 (1996), no. 1(307), 171--172; translation in Russian Math. Surveys 51 (1996), no. 1, 171–172. But I don't know the details of this work.<|endoftext|> TITLE: Hodge standard conjecture for étale cohomology QUESTION [17 upvotes]: It is known that Hodge standard conjecture is true for étale cohomology for a field $k$ of characteristic zero. It means that the following pairing $$ (x,y)\mapsto (-1)^{i}\langle L^{r-2i}(x),y\rangle $$ is positive definite, where $\langle -,-\rangle$ is the cup product, $L$ is the Lefschetz operator for this cohomology and $x,y\in C^i(X)\cap P^{2i}(X)$. In above formula $X$ is a smooth projective scheme over $k$, $C^i(X)$ is a free abelian group generated by closed irreducible subschemes of $X$ of codimension $i$ and $P^i(X)$ is a kernel of $L^{r-i-1}$. My question is : what is the main problem with proving the similar theorem for fields of arbitrary characteristic? REPLY [8 votes]: Another way to phrase the main obstacle in positive characteristic is the following: Although we can formulate the Hodge standard conjecture purely cycle-theoretically, in practice the only way we get a grip on it is through a cohomological representation. For example, in characteristic $0$ we use Hodge cohomology to prove the positivity statement of the pairing on primitive cohomology. However, in positive characteristic, there isn't even a cohomology theory taking values in a field that admits a notion of positivity! The only known cohomology theories are over fields like $\mathbb Q_\ell$ and $W(k)$, neither of which admits the structure of an ordered field (because they have too many roots of unity). Thus, one either has to come up with a cycle-theoretic proof that doesn't make reference to any cohomology theory (as far as I can tell, this would be new even in characteristic $0$), or one has to come up with a Weil cohomology theory taking values in an ordered field (like $\mathbb Q$ or $\mathbb R$, although already over $\mathbb F_{p^2}$ cohomology theories with values in one of these two fields provably do not exist by a result of Serre). Edit: As suggested by Will Sawin below (Serre's original argument is the case $K = \mathbb R$): Claim. There is no Weil cohomology theory with values in any totally ordered field $K$. Indeed, there exists a supersingular elliptic curve $E$ over $\mathbb F_{p^2}$ such that $D = \operatorname{End}(E) \otimes_\mathbb Z \mathbb Q$ is a quaternion division algebra. It is non-split at $\infty$ (e.g. because it is nontrivial but split at all $v \neq p, \infty$), so it is given by $D = (a,b)$ for $a, b \in \mathbb Q_{<0}$. Such a division algebra never splits over an ordered field (since $a$ and $b$ have to remain negative in $K$, so $b$ can never be a norm of $K(\sqrt{a})/K$). $\square$ (In fact, using the Brauer–Hasse–Noether sequence from class field theory and explicit computation of the local invariant maps, one sees that $D$ can be represented as $$D = \left\{\begin{array}{ll}(-1,-1) & p = 2,\\(-1,-p) & p \equiv 3 \pmod{4},\\(-q,-p) & p \equiv 1 \pmod{4},\end{array}\right.$$ where in the third case, $q \equiv 3 \pmod{4}$ is a prime whose residue class mod $p$ is a non-square. However, we only needed the negativity of $a$ and $b$.)<|endoftext|> TITLE: Whether a given algebra is the algebra of endomorphisms for a vector space QUESTION [7 upvotes]: Let $\mathbb{F}$ be a field and let $A$ be an associative unital $\mathbb{F}$-algebra. Is there a criterion to let me know if $A$ is isomorphic to the algebra $\mbox{End}(\mathbf{V})$ of endomorphisms for some $\mathbb{F}$-vector space $\mathbf{V}$? Generalizations to $A$ being an associative unital ring and $\mathbf{V}$ an Abelian group or similar are welcome. Answers for particular cases $\mathbb{F}\in\{\mathbb{R},\mathbb{C}\}$ are also appreciated. Thank you. This is a repetition of the following question I posed on MathSE days ago which has no answers so far. As I am no expert in algebra, so I thought maybe it is a research level question. REPLY [3 votes]: If $V$ is a vector space, the two-sided ideals of $E:=End(V)$ form a chain. The nonzero ideals are exactly those of the form $I_{\alpha}=\{\varphi\in E : {\rm dim}({\rm im}(\varphi))<\alpha \}$, where $\alpha$ is an infinite cardinal. Thus, for instance, if $V$ is a vector space of countable dimension, then there are three ideals in $E$. Viewing $E$ as the ring of column-finite matrices, these ideals are exactly the zero ideal, the ideal consisting of matrices with only finitely many nonzero crows, and $E$ itself. The one-sided ideals are much more complicated. In particular, the Jacobson radical is the zero ideal. The ring $E$ is von Neumann regular; meaning that for every $x\in E$ there exists some $y\in E$ with $xyx=x$. The ring $E$ is clean, meaning that for every $x\in E$ there exists some unit $u$ and idempotent $e$ such that $x=e+u$. There are many other purely ring theoretic properties that such endomorphism rings have. A good source of information is Lam's "A First Course in Noncommutative Rings". I know of no simple criterion for being an endomorphism ring of a vector space over a field. (That's not to say there isn't one.)<|endoftext|> TITLE: Decomposition of an induced representation of $GL(2, q)$ QUESTION [5 upvotes]: Denote $G = GL(2, q) = GL_2(\mathbb{F}_q)$, $B$ its Borel subgroup of upper triangular matrices, $T$ its splitting torus of diagonal matrices. The object I am interested in is $Ind_B^G\rho$, where $\rho$ is a $(q - 1)$-dimensional irreducible representation of $B$. $$[G: B] = \frac{q(q + 1)(q - 1)^2}{q(q - 1)^2} = q + 1 \Longrightarrow \dim Ind_B^G\rho = (q + 1)(q - 1).$$ We know that $B$ have $(q - 1)^2$ many $1$-dimensional irreducible representations (determinant composed by two characters of $\mathbb{F}_q^*$), and $(q - 1)$ many $(q - 1)$-dimensional irreducible representations $\rho$. We can think about them as part of $Ind_T^G\tau$ ($\tau$ an irreducible representation of $T \cong \mathbb{F}_q^* \times \mathbb{F}_q^*$), which decomposes to the direct sum of two irreducible representations of dimension $1$ and $(q - 1)$. While representations of $G$ induced from $1$-dimensional representations of $B$ give principal series representations of $G$, representations of $G$ induced from $(q - 1)$-dimensional representations of $B$ contains the cuspidal representations of $G$. Question. How does $Ind_B^G\rho$ ($\dim \rho = q - 1$ irreducible) decompose into irreducible representations of $G$? REPLY [6 votes]: If $\chi$ is a nontrivial character of the unipotent radical $U$ of $B$ then $\text{Ind}_U^G(\chi)$ is the sum of all irreducible non one-dimensional representations of G with multiplicity one. See See Piatetski-Shapiro, Complex representations of GL(2,K) for finite fields K, Contemporary Mathematics, 1983 (MR0696772), page 2.Therefore $\text{Ind}_B^G(\rho)$ is the sum of all irreducible non one-dimensional representations, having a given central character, with multiplicity one.<|endoftext|> TITLE: For which finite groups $G$ does the Wedderburn decomposition of $\mathbb{Q}[G]$ consist only of fields and division algebras? QUESTION [8 upvotes]: Let $G$ be a finite group. Then the rational group algebra $\mathbb{Q}[G]$ has a wedderburn decomposition of the form $\prod_i M_{n_i}(D_i)$ where each $D_i$ is a division algebra. My question is: for which $G$ do we have $n_i=1$ for all $i$? In other words, for which finite groups $G$ does the Wedderburn decomposition of $\mathbb{Q}[G]$ consist only of fields and division algebras? Clearly, this is true when $G$ is abelian. Another example is $G=Q_8$, the quaternion group of order $8$. Are there other examples? Is there a complete classification of such groups? If so, can you provide a reference? REPLY [11 votes]: The groups you are looking for are precisely those for which the group algebra $\mathbb Q[G]$ does not contain nonzero nilpotent elements. These groups have been classified by Sehgal in here. The finite groups which have this property are the abelian ones and the Hamiltonian groups of order $2^mt$ (where $t$ is odd), such that $2$ has odd order modulo $t$.<|endoftext|> TITLE: $\kappa$-homogeneous topological spaces QUESTION [19 upvotes]: Let $\kappa>0$ be a cardinal and let $(X,\tau)$ be a topological space. We say that $X$ is $\kappa$-homogeneous if $|X| \geq \kappa$, and whenever $A,B\subseteq X$ are subsets with $|A|=|B|=\kappa$ and $\psi:A\to B$ is a bijective map, then there is a homeomorphism $\varphi: X\to X$ such that $\varphi|_A = \psi$. Questions: Is it true that for $0<\alpha < \beta$ there is a space $X$ such that $|X|\geq \beta$, and $X$ is $\alpha$-homogeneous, but not $\beta$-homogeneous? Is there even such a space that is $T_2$? Also it would be nice to see an example for $\alpha=1, \beta=2$. And I was wondering whether there is a standard name for $\kappa$-homogeneous spaces. (Not all of these questions have to be answered for acceptance of answer.) REPLY [7 votes]: The sort of space you describe is usually called strongly $\kappa$-homogeneous. If you google that phrase you will find some interesting results about these kinds of spaces (mostly concerning how this property relates to other homogeneity properties). The earliest reference I could find to strongly $n$-homogeneous spaces (only finite values of $n$ are considered) is in a 1953 paper by C. E. Burgess (available here). Despite the fact that these kinds of spaces have been in the literature for well over half a century, it is unknown whether there is a topological space that is strongly $4$-homogeneous but not strongly $5$-homogeneous. (This is stated explicitly in this paper by Ancel and Bellamy from last year -- see the second paragraph of the second page.) As far as I can tell (although I don't have an authoritative reference), it is also unknown whether there is a strongly $n$-homogeneous space that is not also strongly $(n+1)$-homogeneous for any finite $n \geq 4$. Therefore Joel's answer (along with some of Andreas's comments below it) constitutes the state-of-the-art knowledge on this question.<|endoftext|> TITLE: For which finite groups $G$ is every character a virtual permutation character? QUESTION [16 upvotes]: Let $G$ be a finite group. A (complex) character $\chi$ of $G$ is said to be a virtual permutation character if it can be expressed as a $\mathbb{Z}$-linear combination of characters induced from the trivial characters of subgroups of $G$, i.e., $\chi = \sum_{H \leq G} n_H \mathrm{ind}^{G}_{H} 1_H$ for some $n_H \in \mathbb{Z}$. My question is: which finite groups have the property that every character is a virtual permutation character? Is there a classification of such groups? If so, can you provide a reference? One well-known example is the symmetric group $S_n$. Moreover, it is clear that the class of groups we're looking for is closed under direct products. But are there other examples? Note that every virtual permutation character is rational-valued, but the converse is false. For example, the every character of the quaternion group of order $8$ is rational-valued, but the unique non-linear character is not a virtual permutation character. Moreover, non-trivial Schur induces are not the only obstacle: even if the representation attached to $\chi$ takes values in matrices over $\mathbb{Q}$, the character $\chi$ need not be a virtual permutation character. These obstacles are the topic of the article "Rational representations and permutation representations of finite groups" Math. Ann. 364, Issue 1 (2016), 539-558 by Alex Bartel and Tim Dokchitser (see http://arxiv.org/abs/1405.6616). REPLY [9 votes]: No classification of such groups is known. As you say, for every character to be a virtual permutation character, necessary conditions are that all irreducible characters are $\mathbb{Q}$-valued; equivalently, every element $g$ of the group is conjugate to $g^i$ for all $i$ coprime to the order of $g$; and all Schur indices are trivial. Also, as you say, these conditions are not sufficient, in general. Incidentally, we also do not have a classification of the finite groups satisfying the first condition. Regarding examples: all of the Weyl groups satisfy the first condition above. Of these, in addition to the symmetric groups (Weyl groups of type $A_n$), the Weyl groups of types $B_n$ (isomorphic to $C_2\wr S_n$) and $D_n$ have the property that every character is a virtual permutation character. As for the exceptional types: the Weyl group of type $F_4$ also has that property, while the groups of types $E_6$, $E_7$, and $E_8$ do not. You can find these facts, with proofs, and many more in the book by Dennis Kletzing, Structure and Representations of $\mathbb{Q}$-Groups, Lecture Notes in Mathematics 1084. Miscellaneous remarks: in the chain of inclusions {permutation characters} $\subseteq$ {characters of rational representations} $\subseteq$ {$\mathbb{Q}$-valued characters} $\subseteq$ {all complex characters}, many of the obstructions to these inclusions being equalities can be analysed "locally" at the level of normalisers of cyclic subgroups of your group. For example, the "equivalently..." in condition 1 above already does that for the last inclusion in the chain. The obstruction for the middle inclusion is the theory of Schur indices, and these, too, can be analysed locally, although that analysis is fairly involved – see [1,2]. The obstruction for the left-most inclusion is even more intricate, but one can still draw some conclusion from a local analysis. For example the index of that inclusion is $1$ for the given group $G$ if and only if it is $1$ for all the maximal quasi-elementary subgroups of $G$. This easily follows from Solomon's induction theorem, but one can make much finer statements, as is discussed in my paper with Tim that you have mentioned in the question. [1] U. Riese and P. Schmid, Schur Indices and Schur Groups, II, J. Algebra 182. [2] P. Schmid, Schur indices and Schur groups, J. Algebra 169.<|endoftext|> TITLE: Can a projective solvable group be transitive? QUESTION [6 upvotes]: Let $p > 3$ be a prime number, and let $G \leq \mathrm{PGL}_2(\mathbb{F}_p)$ be a solvable subgroup. Is it possible that the action of $G$ on $\mathbb{P}^1(\mathbb{F}_p)$ is transitive? REPLY [10 votes]: Yes. Take any generator of the multiplicative group of $ \newcommand{\GF}[1]{\mathbb{F}_{#1}} \GF{p^2}$ and make it into an element $g$ of $ \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\PGL}{PGL} \GL_2( \GF{p} )$. Then $g$ is diagonalizable over $\GF{p^2}$ with two Galois conjugate eigenvalues. Thus when $g^k v = av$ for some $v\in \GF{p}^2$ and $a\in \GF{p}$, then necessarily $g^k = aI$. Thus the image of $\langle g \rangle $ in $\PGL_2(\GF{p})$ is a regular, cyclic group of order $p+1$. Added later: As Nick Gill mentions, the $g$ constructed above (or $\langle g \rangle$) is called a Singer cycle in the literature. More generally, a Singer cycle $g\in \GL_n(\GF{q})$, constructed analogously from a generator of $ (\GF{ q^n } )^*$, induces one long cycle on the elements of $ (\GF{q})^n \setminus \{0\}$. (Thus the name Singer cycle.) The corresponding subgroup of $\PGL_n( \GF{q} )$ permutes transitively the points of $\mathbb{P}^{n-1}(\GF{q})$. Bonus: Let's just assume that $p$ is a prime power. If the action of a solvable group $G\leq \PGL_2( \GF{p} )$ on $\mathbb{P}^1(\GF{p} )$ is primitive, then $p=3$ or $p=2^{2^k}$ and $p+1$ is a Fermat prime. Proof. Take a minimal normal subgroup $N$ of $G$. Then $N$ is elementary abelian and transitive, and thus regular, so $p+1$ must be a prime power, too. Thus either $p$ is a Mersenne prime, or $p = 8$ and $p+1 = 9$ or $p+1$ is a Fermat prime. The last case is of course possible. The Sylow $3$-subgroup of $\PGL_2( \GF{8} )$ has order $9$, and its normalizer does not act primitively, so the case $p=8$ is ruled out. If $p=2^m-1$ is Mersenne, then $N$ is elementary $2$-abelian of order $2^m$. It is not too dificult to see that the Sylow $2$-subgroup of $\PGL_2( \GF{p} )$ is a dihedral group. It follows that $|N| \leq 2^2$ and thus $p=3$ as claimed.<|endoftext|> TITLE: Twisted derived Morita theory of schemes QUESTION [5 upvotes]: It has been proved by Toën and Lunts-Schnürer that the dg category $\mathrm{L}_{qcoh}(X\times Y)$ of quasi-coherent sheaves over the product of two quasi-compact, quasi separated (and flat over a commutative ring $k$) schemes is quasi-equivalent to the dg-category of continuous dg functors $\mathbb{R}\underline{\mathrm{Hom}}_c(\mathrm{L}_{qcoh}(X),\mathrm{L}_{qcoh}(Y))$, thus providing a kernel for every triangulated functor between the derived categories of $X$ and $Y$ which is eligible for being of Fourier-Mukai type (i.e. commuting with direct sums). This equivalence also descends to the perfect dg subcategories and the internal (derived) Hom for dg categories above is defined by Toën in his paper https://arxiv.org/abs/math/0408337 by means of some dg bi-modules over the two dg categories. What is known about whether this equivalence still holds when we consider the dg categories of complexes of (perfect or not) twisted sheaves? The paper https://arxiv.org/abs/1002.2599 by Toën proves that that the twisted categories have a generator, is that sufficient to claim some result à la Bondal-Van den Bergh saying that the box product of the two generators is the generator of the dg category of the product? To summarize in one line: is it plausible to have a result like $$ \mathbb{R}\underline{\mathrm{Hom}}_c(\mathrm{L}_{qcoh}(X,\alpha),\mathrm{L}_{qcoh}(Y,\beta)) \simeq \mathrm{L}_{qcoh}(X\times Y, \alpha^{-1}\boxtimes \beta) $$ eventually for perfect complexes? and if it is so plausible that someone already wrote it, where could I read it? Thank you! REPLY [3 votes]: I believe the answer is yes. Let attempt a sketch and give the appropriate references. The general strategy is to prove the main theorems of Ben-Zvi, Francis, Nadler in a twisted fashion. The main content of that paper is that the so-called "perfect stacks" are those algebro-geometric gadgets whose $L_{qcoh}$'s are exactly compactly generated and the dualizables and compacts coincide. As a result, we can keep track of these compact generators and prove that $L_{perf}(-)^c$ takes fiber products to tensor products (of dg-categories or what not). The desired result then follows from the fact that $L_{perf}(X)$ for $X$ again being a "perfect" stack (so that $L_{perf}(X) \otimes_{L_{perf}(k)} L_{perf}(Y) \simeq Fun_{L_{perf}(k)}( L_{perf}(X), L_{perf}(Y))$. Here goes the sketch: In general we can twist by $\alpha: X \rightarrow Pr$, a linear category with descent [Antieau-Gepner, 6.2], but we restrict to the case where factors through $\alpha: X \rightarrow Alg$. For concreteness, this corresponds classically to twisting by an Azumaya algebra. The main point is: [Antieau-Gepner, 6.8], we want $L_{perf}^{\alpha}(U)$ for $U$ affine to be generated by a compact/dualizable/perfect object. Now, as you set it up, suppose $X, Y$ are qcqs schemes flat over a base $S$ (not sure if we need flat), let $\alpha, \beta$ be the Azumayas you want to twist by. In this case, we see that $L_{perf}^{\alpha/\beta}(-)$ for appropriate $(-)= X, Y, S$ are indeed compactly generated. We use the fact that this is true for affines and then use [Antieau-Gepner, 6.11] to see that we can glue the compact generators together in the twisted setting Next, we employ the argument of Ben-Zvi, Francis, Nadler, 4.7, but do it in a twisted way. One sees that all one needs is Ben-Zvi, Francis, Nadler, 4.6 where, in the twisted setting, one needs to prove that if $L_{perf}^{\alpha}(X) \otimes L_{perf}^{\beta}(Y) \simeq L_{perf}^{\alpha \otimes \beta}(X \times Y)$. Doing that only requires Ben-Zvi, Francis, Nadler, 3.24 to know that external products take compact generators to compact generators and that $L_{perf}^{\alpha \otimes \beta}(X \times Y)$ is generated by external products. As you suspect this is an argument ala Bondal-Van Den Bergh and you need only use the compact generation statements above Lastly, you need Ben-Zvi, Francis, Nadler, 4.8 for self-duality, actually it seems that the dual of $L_{perf}^{\alpha}(X)$ is $L_{perf}^{\alpha^{-1}}(X)$, which explains that inversion in your formula. Anyway, the twisted categories are dualizable since they have a compact generator, and performing the diagram chase in that reference gives you the appropriate duality statement. Admittedly, I haven't kept track of the twisting, and I might do so if I have more time to edit this answer<|endoftext|> TITLE: Special values of real analytic Eisenstein series QUESTION [5 upvotes]: Given $\tau$ in the upper half plane, define the normalized real-analytic Eisenstein series by $$ E(\tau, s) = \frac{1}{2} \sum_{(m,n)}' \frac{y^s}{|m\tau + n|^{2s}} $$ It is initially defined for $\text{Re} (s) > 1$, but then analytically continued to the whole plane, except for a simple pole at $s=1$. What are some known special values of this function? The only known values I have seen on the internet are $E(i, s) = 4 \zeta(s) \beta(s)$. I am especially interested in $E(\tau, 0)$ for any other values of $\tau$. REPLY [6 votes]: Combining paul garret's comments (see also his wonderful notes, Standard compact periods for Eisenstein series) with the class number formula, the functional equation of the Dedekind zeta function and the fact that $\zeta(0)=-1/2$, we get: $$E(\tau,0)=\frac{\zeta^*_K(0)}{\zeta(0)}=\frac{2h_KR_K}{w_K}$$ where $h_K$ is the class number, $R_K$ the regulator and $w_K$ the order of the group of roots of unity of $K=\mathbb{Q(\tau)}$. From here it should be very tractable computation for any reasonable $\tau$.<|endoftext|> TITLE: How to compute modular forms of weight one on Shimura curves? QUESTION [11 upvotes]: Classical modular forms of weight $k\geq 1$ can be computed explicitly by exhibiting their $q$-expansion. When the weight is at least two, the most standard method uses modular symbols. When the weight is one, we can not use modular symbols directly because such modular forms are not cohomological, but one can still compute their $q$-expansion by using ideas of Buhler, Frey, Serre, Buzzard and others. Say now that we want to compute explicitly modular forms of weight $k\geq 1$ on a Shimura curve associated to an order in an indefinite quaternion algebra over $\mathbb{Q}$. I can think of at least two ways of exhibiting such modular forms: either by (1) computing their Taylor expansion about a CM point or by (2) invoking Cerednik-Drinfeld theory and computing the harmonic cocycle on the Bruhat-Tits tree associated to it. Method (1) is in principle available for all weights $k\geq 1$ but I'm not sure how well it works systematically in practice, as there is no canonical choice of CM point. Method (2) seems more canonical and robust to me, but it has the drawback that it can only be used directly for $k\geq 2$. My question is: how would you compute in practice quaternionic modular forms of weight $1$? I seek an answer which should allow us to play around with these modular forms and do basic operations like for instance multiplying two forms of weight one and recognize the output as an explicit linear combination of eigenforms of weight two. This is beyond what one can gather from Jacquet-Langlands, I would say. REPLY [2 votes]: Great question! You have a few options. You could use the method of power series expansions (https://math.dartmouth.edu/~jvoight/articles/ModFormPaper-090613.pdf). We replace $q$-expansions with a normalized power series expansions at a point (CM points of class number 1, when they exist, give the nicest expansions); we solve for the undetermined coefficients by writing down linear equations implied by being modular of weight 1 (and, to get a good set of equations, it is best to the Cauchy integral formula). You can then also compute the action of Hecke operators and multiply to your heart's delight. This method is analytic/experimental, exponentially convergent (roughly speaking, each Arnoldi iteration gives you an extra digit of precision in your answer); depending on your application, you may need an extra step to ensure that whatever you need from the output (e.g. multiplication table) is rigorous. For this algorithm, working over a totally real field is just as easy. Paul Nelson gives an algorithm to evaluate the Shimizu lift explicitly (https://arxiv.org/pdf/1210.1243v1.pdf), providing an exponentially convergent series expansion of a Jacquet-Langlands transfer. Paul only looks at even weight, but I don't think this is a serious restriction. Forms in this method are recorded by their values at (sufficiently many) points, and using this you can also multiply. For Shimura curves that arise from congruence subgroups of triangle groups, one can do everything explicitly in terms of the ${}_2F_1$-hypergeometric function (https://math.dartmouth.edu/~jvoight/articles/shimcm-fixed-errata.pdf). Roughly speaking, there is Hauptmodul $t$ for the full triangle group, a modular equation relating $t(z)$ and $t(Nz)$ for level $N$, and canonical generators for the ring of fractional weight modular forms with character to play with. Some of this is work in progress (with Baba-Granath-Nelson-V-Yang), but I would be happy to discuss with you offline if it turns out this is what you need.<|endoftext|> TITLE: What was achieved on IUT summit, RIMS workshop? QUESTION [29 upvotes]: I would like to know what was achieved in the workshop towards the verification of abc conjecture's proof and the advance of understanding of IUT in general. A comment from a participant: C Vincent (via Twitter) Omg, I think the speaker is actually proving ABC right now. There's an inequality that looks suspiciously like what we want. #IUTsummit Inter-universal Teichmüller Theory Summit 2016 (RIMS workshop, July 18-27 2016) Organized by Fesenko, Mochizuki and Taguchi. REPLY [8 votes]: In my opinion this is a "non-question" in MO speak, but there has been an article and blogs about the "summit" by now, either of which may give you (or point to) more information. As indicated in the latter source, one should expect another round of such inquiries in early September (2016).<|endoftext|> TITLE: Mapping Class Group (MCG) of connected sum of 3-torus and $S^2\times S^1$ QUESTION [6 upvotes]: I know that the MCG (isotopy-classes of orientation preserving homeomorphisms) of 3-torus $(S^1\times S^1 \times S^1)$ is $SL(3,Z)$, since it is an Eilenberg–MacLane space, giving $ MCG(T^3)=Out(\pi_1(T^3))=Out(Z^3)=SL(3,Z)$. But $S^2\times S^1$ is not an Eilenberg–MacLane space. I want to know what $MCG(T^3\# S^2\times S^1)$ is? And, what relation does it have to $\pi_1((T^3\# S^2\times S^1)) $? Is there a general way of finding MCG of connected sum of 3-manifolds? Thanks REPLY [16 votes]: The mapping class groups of all compact orientable 3-manifolds are essentially known. A fairly detailed summary of the results, focusing on the nonprime case and with references to proofs in the literature, can be found in Section 2 of a paper of mine with Nathalie Wahl: "Stabilization for mapping class groups of 3-manifolds", Duke Math. J. 155 (2010), 205-269, arXiv:0709.2173. (Section 2 can be read independently of the rest of the paper.) The results are stated in terms of the map $\Phi:MCG(M)\to Out(\pi_1M)$. For $M$ closed and orientable the kernel of $\Phi$ is a direct sum of finitely many copies of ${\mathbb Z}_2$ generated by Dehn twists along 2-spheres. Also, $\Phi$ is surjective if this is true for each prime connected summand of $M$ and a certain other natural condition holds involving permuting different summands with isomorphic fundamental groups. In the case of $T^3 \# (S^1\times S^2)$ the kernel of $\Phi$ is a single ${\mathbb Z}_2$ generated by a Dehn twist along the $S^2$ factor of $S^1\times S^2$. (Interestingly, the twist along the $S^2$ separating the two connected summands is isotopic to the identity.) The image of $\Phi$ is an index two subgroup of $Out({\mathbb Z}^3*{\mathbb Z})$ since one can only realize $SL(3,{\mathbb Z})$, not all of $GL(3,{\mathbb Z})$, by orientation-preserving homeomorphisms of $T^3$. The structure of the automorphism group of a free product seems to be well understood in terms of the automorphism groups of the factors. For $M=T^3 \# (S^1\times S^2)$ this means its mapping class group is generated by the MCG's $SL(3,{\mathbb Z})$ and ${\mathbb Z}_2 \times {\mathbb Z}_2$ of the two summands, along with homeomorphisms obtained by regarding $M$ as obtained from $T^3$ by attaching a "1-handle" and dragging either end of this handle around loops in $T^3$. $\bf Added\ later\colon$ Generators for the mapping class group of $T^3 \# (S^1\times S^2)$ can be described explicitly, using the algebraic fact that the automorphism group of a free product of groups $G_1,\cdots,G_n$ is generated by four types of automorphisms: (1) Automorphisms of the individual factors $G_i$, (2) Permutations of isomorphic factors. (3) Partial conjugations, in which one $G_i$ is conjugated by an element $x_j\in G_j$ and the other factors $G_k$ are fixed. (If $i=j$ this gives an automorphism of type (1), an inner automorphism of $G_i$.) (4) If some $G_i$ is infinite cyclic generated by $g_i$, say, and $x_j$ is an arbitrary element of some $G_j$ with $j\neq i$, there is an automorphism sending $g_i$ to $g_ix_j$ or $x_jg_i$ and fixing the other factors $G_k$, $k\neq i$. For $T^3 \# (S^1\times S^2)$ the type (1) automorphisms for the ${\mathbb Z}^3$ factor form a copy of $SL(3,{\mathbb Z})$ which is the mapping class group of $T^3$ (a result originally due to Waldhausen I think). Since one can isotope an orientation-preserving homeomorphism of $T^3$ to fix a ball pointwise, this homeomorphism of $T^3$ gives rise to a homeomorphism of the connected sum which is the identity on the other summand. For the $S^1\times S^2$ summand the automorphism group of $\pi_1$ is ${\mathbb Z}_2$ with a generator realized by a homeomorphism of $S^1\times S^2$ that reflects both factors simultaneously. This too can be isotoped to extend to the connected sum via the identity on the complement. Note that $SL(3,{\mathbb Z})$ is generated by elementary matrices that differ from the identity matrix in a single off-diagonal entry, which is 1, and these generating automorphisms of ${\mathbb Z}^3$ are realized by Dehn twists along tori in $T^3$. There are no type (2) automorphisms of ${\mathbb Z}^3 * {\mathbb Z}$. For type (3) automorphisms, conjugation of the ${\mathbb Z}^3$ factor by an element of the ${\mathbb Z}$ factor is actually an inner automorphism of the whole group since the ${\mathbb Z}$ factor is abelian. We are interested in $Out({\mathbb Z}^3 * {\mathbb Z})$ so inner automorphisms are factored out. On the other hand, conjugating the ${\mathbb Z}$ factor by an element of ${\mathbb Z}^3$ is realizable by composing two type (4) automorphisms. Type (4) automorphisms are realized by the homeomorphisms that I described earlier as dragging one end of the "1-handle" $S^1\times S^2$ around a loop $\gamma$ in the $T^3$ summand. Concretely, this homeomorphism is a Dehn twist along a certain torus $T_\gamma\subset T^3 \# (S^1\times S^2)$ defined as follows. First take a solid torus neighborhood of the loop $\gamma$. We can view connected sum with $S^1\times S^2$ as removing two disjoint open balls in $T^3$ and identifying the resulting two boundary spheres. Choose one of these two balls in the interior of the solid torus neighborhood of $\gamma$ and choose the other ball in the exterior of this solid torus. The boundary torus of the solid torus then gives the $T_\gamma$ we want. The homeomorphism $h_\gamma$ we are looking for is a twist in the longitudinal direction along $T_\gamma$. The automorphism of $\pi_1={\mathbb Z}^3*{\mathbb Z}$ induced by $h_\gamma$ is the identity on the factor $\pi_1(T^3)$ since generators for this ${\mathbb Z}^3$ can be chosen disjoint from the solid torus. If $x\in \pi_1(T^3)$ is the element represented by $\gamma$ and $g$ is a generator of $\pi_1(S^1\times S^2)$, then it's not hard to check that the automorphism induced by $h_\gamma$ sends $g$ to $xg$ or $gx$. (The two automorphisms given by $xg$ and $gx$ differ by an inner automorphism in the special case we are considering.) Thus a type (4) generator is realized by a twist along a torus. In summary, the mapping class group of $T^3\#(S^1\times S^2)$ is generated by twists along tori and 2-spheres, and also the homeomorphism reflecting the two factors of $S^1\times S^2$.<|endoftext|> TITLE: Proof that the length function $\ell: \operatorname{Teich}(S) \to \mathbb{R}^\mathcal{S}$ is injective without the $9g-9$ theorem QUESTION [5 upvotes]: In Chapter 10 about Teichmüller spaces of Farb and Margalit's "A Primer to Mapping Class Groups", the length function $$\ell: \operatorname{Teich}(S) \to \mathbb{R}^\mathcal{S}$$ is described, where $S$ is a compact surface with finitely many punctures and $\chi(S) < 0$ and $\mathcal{S}$ denotes the set of isotopy classes of essential simple closed curves in $S$. It is already impressive that the map $\ell$ is injective (which they also remark), however they prove the much stronger $9g-9$ theorem. Question: Is there a (simpler) proof that $\ell$ is injective without showing that the length function is actually determined by its value on only $9g-9$ simple closed curves? I've searched for a bit now, and came across e.g. these lecture notes, where Theorem 3.3 claims that the map is injective. The reference given is "Travaux de Thurston sur les surfaces", in which Theorem 7.9 states that $\ell$ is injective. However, it is not proven independently from the $9g-9$ theorem after all. REPLY [2 votes]: There is a representation-theoretic viewpoint, where one can consider traces of $SL_2(\mathbb{R})$ representations (determined by lengths of geodesic up to sign). Then one sees that the trace variety is determined by traces of simple closed curves, modulo certain relations. See papers of Feng Luo, for example. He shows that the modulus of the surface is determined by the moduli of all of the 1-punctured tori and 4-punctured sphere subsurfaces. In turn, it is not hard to show that the modulus of a once punctured torus is determined by the lengths of 3 curves intersecting pairwise once (the complementary regions are two triangles with vertices on the 3 Weierstrass points, and an annulus: the triangle has sidelengths half the geodesic lengths, and then one can see that the lengths determines the geometry of the triangles and annulus; there is a similar picture for 4-punctured spheres). See this question for more references on parameterizing Teichmuller space by geodesic lengths.<|endoftext|> TITLE: Expected distance between points drawn from different distributions QUESTION [5 upvotes]: Let $X$ and $Y$ be independent random variables, with $X_1,X_2$ being independent copies of $X$ and $Y_1,Y_2$ being independent copies of $Y$. Then (is it true that) $2\mathbb{E}|X-Y|\geq\mathbb{E}|X_1-X_2|+\mathbb{E}|Y_1-Y_2|$? This is saying that the expected distance between two points drawn from different distributions is greater than the expected distance between points drawn from one or the other. My attempts so far include: Triangle inequality gives $4\mathbb{E}|X-Y|\geq\mathbb{E}|X_1-X_2|+\mathbb{E}|Y_1-Y_2|$ when we write $|X_1-X_2|=|(X_1-Y_1)+(Y_1-X_2)|$; Squared distances work, i.e. $2\mathbb{E}(X-Y)^2\geq\mathbb{E}(X_1-X_2)^2+\mathbb{E}(Y_1-Y_2)^2$ holds immediately after expanding the squares; Counterexamples? I have tried generating $X$, $Y$ from different distributions, e.g. normal, exponential, simple Bernoulli, etc. and haven't found anything. However, the inequality very much doesn't hold for $X$ and $Y$ dependent - a trivial example is $Y=X$, when the LHS is 0 while the RHS can be positive. Discrete case doesn't seem to shed much more light on it. The inequality reduces to $\sum_{i,j} (p_i-q_i)(p_j-q_j)|i-j|\leq 0$, where $p_i=\mathbb{P}(X=i)$ and $q_i=\mathbb{P}(Y=i)$. I have also thought of Jensen for $|x|, \sqrt{x},x^2$, Holder ($||fg||_1\leq ||f||_p||g||_q$ for $1/p+1/q=1$), and Minkowski ($||f+g||_p\leq||f||_p+||g||_p$ for $p\geq 1$). I would appreciate any ideas/counterexamples/references for this. Thanks! REPLY [2 votes]: This is the result of me trying to prove the identity in Brendan McKay's answer. Consider, a bit more generally, any nonnegative independent r.v.'s $X$ and $Y$, still with $X_1,X_2$ being independent copies of $X$ and $Y_1,Y_2$ being independent copies of $Y$. Then $P(X\wedge Y>x)=F(x)G(x)$ for all real $x$, where $F(x):=P(X>x)$ and $G(x):=P(Y>x)$. So, \begin{equation} E(X\wedge Y)=\int_0^\infty dx\,P(X\wedge Y>x) =\int_0^\infty dx\,F(x)G(x). \end{equation} Similarly, $E(X_1\wedge X_2)=\int_0^\infty dx\,F(x)^2$ and $E(Y_1\wedge Y_2)=\int_0^\infty dx\,G(x)^2$. Note also that $|x-y|=x+y-2(x\wedge y)$ for real $x,y$. It follows that \begin{equation} 2E|X-Y|-(E|X_1-X_2|+E|Y_1-Y_2|) \end{equation} \begin{equation} =-2\int_0^\infty dx\,[2F(x)G(x)-(F(x)^2+G(x)^2)] \end{equation} \begin{equation} =2\int_0^\infty dx\,[F(x)-G(x)]^2\ge0. \end{equation}<|endoftext|> TITLE: Compute the kernel of multiplication of algebraic numbers QUESTION [6 upvotes]: Let $\lambda_1, \dots, \lambda_n$ be the roots of a polynomial $g(x)$ of $n$-degree with rational coefficients and such that $g(0) \neq 0$. (Hence obviously they are non-zero algebraic numbers.) Consider the function $f: \mathbb{Z}^n\rightarrow A^{\ast}$ defined by $f(z_1, z_2, \dots,z_n)=\lambda_1^{z_1}\lambda_2^{z_2}\cdots\lambda_n^{z_n}$, where $A$ is the field of algebraic numbers. I am interested in computing the kernel of $f$. Specifically, I would like to exhibit generators of this kernel. Is there an efficient algorithm? Moreover, is there any estimate of the cardinality of the generators? REPLY [2 votes]: This is basically a re-writing of the answer of Luc Guyot. I would like to present not only the calculation of the rank, but also the algorithm for calculating a basis. Let $K$ be a number field, and let $\lambda_1, \cdots, \lambda_n$ be elements in $K^\times$. Let $f$ be the morphism $\mathbb{Z}^n \rightarrow K^\times$ sending $(z_1, \cdots, z_n)$ to $\lambda_1^{z_1} \cdots \lambda_n^{z_n}$. We want to establish an algorithm for calculating a basis of the kernel $\ker(f)$. Let $\mathcal{O}$ be the ring of integers of $K$. Write $\mathcal{I}$ for the group of fractional ideals, which is a free abelian group generated by prime ideals of $\mathcal{O}$. Denote by $\iota$ the canonical map $K^\times \rightarrow \mathcal{I}$ sending $\lambda$ to the principle ideal $\lambda\mathcal{O}$. The map $f$ induces a map $\tilde{f}: \ker(\iota \circ f) \rightarrow \ker(\iota) = \mathcal{O}^\times$, and it is clear that $\ker(f) = \ker(\tilde{f})$. The algorithm then goes in two steps: Step 1. Compute a basis of $\ker(\iota \circ f)$, i.e. an isomorphism $\tau:\mathbb{Z}^d \rightarrow \ker(\iota \circ f)$. This is done by decomposing every $\lambda_i$ into prime ideals, hence reducing to a morphism between $\mathbb{Z}^n$ and $\mathbb{Z}^m$, where $m$ is the number of prime ideals involved. Step 2. Compute a basis of the kernel of the composition $\tilde{f} \circ \tau:\mathbb{Z}^d \rightarrow \mathcal{O}^\times$, so that $\ker(\tilde{f})$ can be determined via $\ker(\tilde{f}) = \tau(\ker(\tilde{f} \circ \tau))$. Similarly, this only requires a basis of $\mathcal{O}^\times$. Note that the group $\mathcal{O}^\times$ also has a torsion part, which should be taken into account. If we assume that the field extension $K/\mathbb{Q}$ is "small", then there are efficient algorithms for both the ideal decomposition and the unit group computation. In the case of the original problem, one should take $K$ to be the splitting field of the polynomial $g(x)$, which (when $g(x)$ is irreducible) typically has degree $n!$ over $\mathbb{Q}$, hence not quite "small". But it should work well in some special cases (e.g. $g(x)$ only have roots in $\mathbb{Q}$).<|endoftext|> TITLE: Is the Bott-Chern/Aeppli cohomology determined by the de Rham and Dolbeault cohomologies? QUESTION [7 upvotes]: Is there an example of two different complex manifolds that have the same de Rham cohomology and Dolbeault cohomology but different Bott-Chern/Aeppli cohomology? REPLY [9 votes]: As for compact complex surfaces, the Bott-Chern/Aeppli cohomologies are determined by de Rham/Dolbeault cohomologies:this is contained in Lemma 2.3 in: Teleman, A.: The pseudo-effective cone of a non-Kählerian surface and applications. Math. Ann. 335(4), 965–989 (2006), arXiv:0704.2948.<|endoftext|> TITLE: Mochizuki's Gaussian Integral Analogy QUESTION [11 upvotes]: In his latest 115-page overview, Mochizuki spends some time explaining "alien copies" by the analogue of evaluating the Gaussian integral by squaring it and introducing a second variable/dimension. In sections 1.1 to 1.7 (pages 5-10) he goes through this, and then in section 3.8 (pages 89-94) he revisits the analogy. I have two questions about this. Firstly, one gets that $I^2=(\int_{-\infty}^\infty e^{-x^2}\,dx)^2=\pi$, and by positivity one finds $I=\sqrt\pi$. What is the analogue in the IUT setting of "positivity" here? Do the choices up to a unit not matter somehow? Secondly, I wonder about the scope of alien copies. At the end of the day, if I read 3.8 correctly in #10 and #11 in the table on page 92 and the computations in 3.7(iv) (page 89), the main result of the analogy is to evaluate (to first order) the left side of the inequality at the top of page 87. So my question is: are alien copies "only" useful for this Gaussian-like evaluation, or do they play a role throughout? (NB: the word "alien" does not appear in the text from pages 22-90.) Alternatively, do they also enlighten the derivation of the inequality (see 3.7(ii) starting on page 85) in the first place, or the estimation of its right side? Since no one has answered yet, maybe I'll add a third question. In 2.2 on page 11, a distinction (quite important I gather) between étale-like and Frobenius-like structures is made, the former being "zero mass" objects, such as (1.4) the differentials "$dx\,dy$" in the Gaussian evaluation. Are there Frobenius-like structures with the Gaussian analogue, and if so, what are they? REPLY [6 votes]: For all the similarities, there's a significant difference between both parts of the analogy. In the IUT side of the analogy, we are trying to compare the "mutually alien copies" at each end of the $\Theta$-link $$\{\underline{\underline{q^{j^2}}}\} \longleftrightarrow \underline{\underline{q}}\tag{1}$$ But in the case of the Gaussian integral, we are introducing the second (alien) copy in order to evaluate the first, not to compare them. $$\bigg(\int_{-\infty}^\infty e^{-x^2}\,dx\bigg)\bigg(\int_{-\infty}^\infty e^{-x^2}\,dx\bigg)\longleftrightarrow \quad?$$ So at some point we need to reverse the process. That's not case in IUT. To be more precise, the alien copies in question are $\Theta^{\pm\text{ell}}\text{NF}$-Hodge theaters (denoted $\bullet$ in the diagram below) in the domain/codomain of the $\Theta$-link, and the analogy is between a pair of such a Hodge theaters and one evaluation of the Gaussian integral. But there is a infinite lattice of such a pairs at work in IUT, the log-theta lattice. This process doesn't have to be undone in any way, in order to evaluate individual elements of the lattice, the way we need to go from $I^2$ to $I$. The "Gaussian-like evaluation" is basically the main result of IUT, so I would say that it appear throughout and at the same time it is "only" useful for this. A different instance of the use of alien copies would be Mochizuki's theory of mono-anabelian reconstruction. In that case we have Galois theaters and $\mathfrak{log}$-Frobenius functors between them, $$\mathfrak{log}:\mathfrak{Th}_\mathbb{T}^\bullet [Z] \longrightarrow \mathfrak{Th}_\mathbb{T}^\bullet [Z]$$ The derivation of inequality 3.7(ii) follows from $(1)$ and the Kummer isomorphisms [IUTechIII Theorem A, (ii)], which yield $$-|\log (\underline{\underline{\Theta}})| \leq -|\log (\underline{\underline{q}})|$$ All of this involves of course the log-theta lattice (i.e. the "alien copies"). I think the analogy is supposed to be something like this: $$\text{zero mass} \longleftrightarrow \text{étale-like} \longleftrightarrow dx\,dy$$ $$\text{positive/net mass} \longleftrightarrow \text{Frobenius-like} \longleftrightarrow \lim \sum dx\,dy$$<|endoftext|> TITLE: Do homotopy categories of finitely (co)complete quasicategories determine categorical equivalences? QUESTION [6 upvotes]: Let $F : C \to D$ be an exact functor between (co)fibration categories such that $Ho(F) : Ho(C) \to Ho(D)$ is an equivalence of homotopy categories. Cisinski proved that in this case $F$ is an equivalence. It was shown by Szumiło that (co)fibration categories and finitely (co)complete quasicategories are equivalent (as fibration categories). So it seems that analogous theorem might be true for quasicategories. Let $F : C \to D$ be an exact functor (that is, $F$ preserves finite (co)limits) between finitely (co)complete quasicategories such that $Ho(F) : Ho(C) \to Ho(D)$ is an equivalence of homotopy categories. Does this imply that $F$ is a categorical equivalence of quasicategories? Denis-Charles Cisinski, MR 2746284 Invariance de la $K$-théorie par équivalences dérivées, J. K-Theory 6 (2010), no. 3, 505--546. Karol Szumiło, Two Models for the Homotopy Theory of Cocomplete Homotopy Theories. REPLY [6 votes]: Yes, this follows from the proof of my result that you quoted, essentially because in my argument weak equivalences of cofibration categories are defined as exact functors inducing equivalences on homotopy categories while equivalences of quasicategories are standard categorical equivalences (and these homotopy theories turn out to be equivalent). More precisely, let $\mathsf{CofCat}$ and $\mathsf{QCat}_\omega$ be categories of cofibration categories and of finitely cocomplete quasicategories. I construct functors $N_f \colon \mathsf{CofCat} \to \mathsf{QCat}_\omega$ and $\mathrm{Dg} \colon \mathsf{QCat}_\omega \to \mathsf{CofCat}$ that are inverse to each other (in a rather weak but sufficient sense). The proof of Lemma 4.10 in my thesis shows that there is a natural weak equivalence between $\mathrm{Ho}N_f\mathcal{C}$ and $\mathrm{Ho}\mathcal{C}$ for any cofibration category $\mathcal{C}$. Combined with Proposition 4.7 this implies that there is also an equivalence between $\mathrm{Ho Dg}\mathcal{C}$ and $\mathrm{Ho}\mathcal{C}$ for any quasicategory $\mathcal{C}$. Now, let $\mathcal{C} \to \mathcal{D}$ in $\mathsf{QCat}_\omega$ be a functor that induces an equivalence on homotopy categories. It follows that $\mathrm{Dg}\mathcal{C} \to \mathrm{Dg}\mathcal{D}$ is a weak equivalence of cofibration categories and thus $N_f\mathrm{Dg}\mathcal{C} \to N_f\mathrm{Dg}\mathcal{D}$ is a categorical equivalence of quasicategories (since $N_f$ is an exact functor, Theorem 3.3). This last functor is equivalent to $\mathcal{C} \to \mathcal{D}$ in $\mathsf{QCat}_\omega$ (by Proposition 4.7 again) which is therefore also a categorical equivalence.<|endoftext|> TITLE: Hadwiger-Nelson problem for $\ell^\infty$ QUESTION [5 upvotes]: Let $G=(V, E)$ be the following graph: $V=\ell^\infty = $ set of bounded real sequences, with the norm $$\|x\|_\infty = \sup_{n\in\mathbb{N}}|x_n|,$$ $E = \big\{\{x,y\}: x,y\in \ell^\infty \text{ and }\|x-y\|_\infty = 1\big\}$. The unit vectors $e_i$ (defined by $e_i(n) = 1$ for $i=n$ and $e_i(n) = 0$ otherwise) form a clique, so $\chi(G) \geq \aleph_0$. But do we have $\chi(G)=\aleph_0$? REPLY [10 votes]: No. The set of all $\{0,1\}$-sequences is also a clique in $G$. Thus, $\chi(G) \geq 2^{\aleph_0}$. On the other hand, the set of all bounded real sequences has size $2^{\aleph_0}$, so $\chi(G)=2^{\aleph_0}$.<|endoftext|> TITLE: Real vs complex norm for polynomials QUESTION [7 upvotes]: Let $d \in \mathbb{N}$. We denote by $C_d$ the best (smallest) constant satisfying that $$ \sup{\{ |P(z)| \colon z \in \mathbb{D} \}} \leq C_{d} \, \sup{\{ |P(x)| \colon x \in [-1,1] \}} $$ for every polynomial $P$ of degree $\leq d$ with real coefficients. (Notation: $\mathbb{D}$ is the open unit disk of $\mathbb{C}$.) I would like to know whether these constants are known, or the best known estimations for them (with references). Right now, I am only aware of the paper of Erdös: "Some remarks on polynomials" (1947), which I found in the paper of Aron, Beauzamy and Enflo: "Polynomials in Many Variables: Real vs Complex Norms" (1993). REPLY [6 votes]: Let $K=[-1,1]$ and let $\Omega$ be its complement in $\mathbb C$. By the Bernstein's lemma (or Bernstein-Walsh lemma), see Theorem 5.5.7 in T. Ransford, Potential theory in the complex plane, Cambridge, 1995, $$|P_{d}(z)|\leq e^{dg_{\Omega}(z)}\|P_{d}\|_{K},\qquad z\in \Omega,$$ where $g_{\Omega}(z)$ denotes the Green function of $\Omega$ with a logarithmic singularity at infinity, whose explicit expression is $$g_{\Omega}(z)=\log|z+\sqrt{z^{2}-1}|.$$ It is easy to check that $$\max_{|z|=1} g_{\Omega}(z)=g_{\Omega}(i)=\log(1+\sqrt{2})$$ thus $$\|P_{d}\|_{\mathbb D}\leq (1+\sqrt{2})^{d}\|P_{d}\|_{K}.$$ Concerning sharpness, the theorem referenced above also states that there exists a polynomial $Q_{d}$ (a so-called Fekete polynomial) of degree $d$ such that $$ a_{d}^{cd}(1+\sqrt{2})^{d}\|Q_{d}\|_{K}\leq\|Q_{d}\|_{\mathbb D},$$ where $a_{d}<1$ tends to 1 as $d\to\infty$ and $c$ is some constant that could be explicitly computed ($a_{d}$ involves the $d$-th diameter of $K$ and the constant $c$ is given in terms of the Harnack distance for $\Omega$, see the theorem for details).<|endoftext|> TITLE: Existence of range of numbers containing a coprime to given n QUESTION [11 upvotes]: I suspect the following statement is true and I can use it in my work if it is true. However I am not a number theorist and I could not prove it myself. I was wondering if this is known to number theorists. Statement: For each positive real number $\alpha$ there exist a natural number $N$ such that, for every $n \geq N$ each of the intervals $[n^{\alpha},2n^{\alpha}) , [2n^{\alpha},3n^{\alpha}), ...$ contains at least one natural number that has no common factor with $n$. REPLY [17 votes]: This is tightly related to the Jacobsthal function defined to be the smallest integer $j(n)$ such that any segment of $j(n)$ consecutive integers contains an integer co-prime with $n$. Iwaniec ("On the problem of Jacobsthal", Demonstratio Math. 11, 225–231, 1978) proved that $j(n)=O(\log^2(n))$; this implies your statement.<|endoftext|> TITLE: Which algebraic theories are co-sites? QUESTION [10 upvotes]: Given a category $C$, I'll say that a set $J$ of families $\{f_i\colon A\to B_i\mid i\in I\}\;$ is a co-coverage if their opposites $\{f_i^{op}\colon B_i\to A\mid i\in I\}\;$ form a coverage on $C^{op}$. In this case, each family in $J$ will be called a co-covering family and the pair $(C,J)$ a co-site. Let $T$ be a (possibly multisorted) algebraic theory. For each sort $t\in T$ and $n\in\mathbb{N}$, we have a product cone of projections $\pi_i\colon t^n\to t$, which we declare should be preserved by co-presheaves. Thus we "want" them to be co-covering families. One could attempt to declare that such families $\{\pi_i\colon t^n\to t\mid 1\leq i\leq n\}\;$ are the co-covering families (one family for each $t,n$), but this choice rarely seems to form a co-coverage. Question: What are some sufficient conditions under which the evident "product sketch" for an algebraic theory forms a co-coverage? Question: What are some sufficient conditions under which the category $T-\mathbf{Alg}$ is a topos? REPLY [15 votes]: In the following I'll speak of coverages on $T^\mathrm{op}$ rather than co-coverages on $T$. Let me take the liberty of loosening your question. It seems a little funny to me to to ask that the finite coproduct cocones in $T^\mathrm{op}$ form a coverage on the nose. After all, the point of a coverage is to present a notion of sheaf, or equivalently a Grothendieck topology. And every set of cocones generates a Grothendieck topology by just throwing in all the additional sieves that the axioms of a Grothendieck topology tell you have to be in there. So let's just let $J$ be the Grothendieck topology on $T^\mathrm{op}$ generated by the finite coproduct cocones. In general there's no reason for this site to be subcanonical. But what I think you're really driving at is the question: When do the $J$-sheaves $T \to \mathsf{Set}$ coincide with the $T$-algebras (= the finite-product-preserving functors $T \to \mathsf{Set}$)? Note that the $J$-sheaves are always contained in the $T$-algebras. If they coincide, $T\mathrm{-Alg}$ is a Grothendieck topos on the site $(T^\mathrm{op},J)$ (and in particular $J$ is subcanonical). The converse is also true: to see this, note that the proof of Giraud's theorem actually shows that any set of generators for a topos can be taken as a subcanonical site of definition (and the Grothendieck topology is uniquely determined). So we can reformulate this question as: When is $T\mathrm{-Alg}$ a (necessarily Grothendieck) topos? In the single-sorted case, Johnstone has given a syntactic characterization. That article is behind a paywall, but the jist of it is this. All of the hypotheses of Giraud's theorem are satisfied by $T\mathrm{-Alg}$ for any $T$, except for disjointness and pullback-stability of coproducts; Johnstone translates these conditions into the following: Theorem (Johnstone): If $T$ is single-sorted, then $T\mathrm{-Alg}$ is a topos if and only if $T$ is degenerate (i.e. the terminal category) or else the following criteria are satisfied: $T$ has no pseudo-constants. Every operation of $T$ is sufficiently unary. A pseudo-constant is a unary operation $u$ such that $u(x) = u(y)$ for ally $x,y$. The definition of "sufficiently unary" is distributed throughout the paper in such a way that I can't easily extract it. But the idea is that $T\mathrm{-Alg}$ is the topos of $M$-sets when $T$ is the free category with finite products on a monoid $M$ -- i.e. if all the operations of $T$ are unary. Johnstone's perspective seems to be that you can't stray too far from this case and still have a topos. I'm sure these conditions could all be formulated more categorically, and you could probably do it if you took some time with Johnstone's paper. Or maybe it would become clear just thinking about what disjoint and universal coproducts mean. I don't know how easy it would be to extend the results to the many-sorted case, although the basic observation -- that if $T$ is the free theory on a category then $T\mathrm{-Alg}$ is a topos -- remains true. When $T$ is freely generated by a category $C$ with a sketched terminal object $1$, you will get a topos if the terminal object is strict, because in this case $T$ is just freely generated under finite products by $C \setminus \{1\}$. But if 1 is not strict, then if we assume that $T\mathrm{-Alg}$ is a topos and in particular $J$ is subcanonical, then by the Yoneda lemma the initial object of $T\mathrm{-Alg}$ is not strict, a contradiction (the initial object in any topos is strict). Johnstone gives the example where $T$ is the theory of a Jónsson-Tarski algebra, i.e. a set equipped $X$ with a bijection with $X \times X$. The Jónsson-Tarski algebras form a topos, which is kind of cool. Added: An example which has gained relevance of late with the advent of condensed / pyknotic mathematics is the (large) site $\mathcal C$ of free compact Hausdorff spaces. This is the Kleisli category for the ultrafilter monad on $Set$. So it is a full subcategory of compact Hausdorff spaces, where the objects are the Stone-Cech compactifications of discrete spaces. Sheaves on $\mathcal C$ are called condensed or pyknotic sets (depending on how one deals set-theoretically with the largeness of the site $\mathcal C$). They form a topos (modulo set-theoretic details). And a presheaf $F: \mathcal C^{op} \to Set$ is a sheaf precisely if $F$ preserves finite products. That is, $\mathcal C$ is a Lawvere theory whose algebras form a topos.<|endoftext|> TITLE: Geometric interpretation of horizontal and vertical lift of vector field QUESTION [10 upvotes]: In many References such as D.E. Blair, Riemannian Geometry of Contact and Symplectic Manifolds chapter 9, and Differential Geometric Structures By Walter A. Poor Page 54; the horizontal and vertical lift(space) of a vector field on $M$, $X\in\Gamma(TM)$, are defined as follows: If $X$ is a vector field on $M$, its vertical lift $X^V$ on $TM$ is the vector field defined by $X^V\omega = \omega(X)\circ \pi$, where $\omega$ is a 1-form on $M$, which on the left side of this equation is regarded as a function on $TM$. For an affine connection $\nabla$ on $M$, the horizontal lift $X^H$ of $X$ is defined by $X^H\omega = \nabla_X\omega$. The span of the horizontal lifts at $t ∈ TM$ is called the horizontal subspace of $T_tTM$. The other approach is as follows: It is well-known that the tangent space to $TM$ at $(x, u)$ splits into the direct sum of the vertical subspace $VTM_{(x,u)}=ker\pi_*|_{(x,u)}$ and the horizontal subspace $HTM_{(x,u)}$ with respect to $\nabla$ $$TTM=HTM\oplus VTM.$$ For $X\in T_xM$, there exists a unique vector $X^h$ at the point $(x, u)\in TM$ such that $X^h\in HTM_{(x,u)}$ and $\pi_*(H^h) = X$. $X^h$ is called the horizontal lift of $X$ to $(x, u)$. There is also a unique vector $X^v$ at the point $(x, u)$ such that $X^v\in VTM_{(x,u)}$ and $X.(df) = Xf$ for all functions $f$ on $M$. $X^v$ is called the vertical lift of $X$ to $(x, u)$. My problems are: How can I split every vector field in $TTM$ into horizontal and vertical part? What is the geometric interpretation of horizontal and vertical spaces? why the tangent sphere bundle and tangent bundle is different in the sense of horizontal and vertical part? It seems that after solving the question I can to prove the following identities: $$[X^v,Y^v]=0,\quad dX(Y)=Y^h+(\nabla_YX)^v\quad X,Y\in\Gamma(TM).$$ Thanks. REPLY [18 votes]: I find the following viewpoint helpful to translate between the different incarnation of a connection. To every vector bundle $\pi: E \to M$ (in your case $E = TM$) we have an associated exact sequence of vector bundles (sometimes called the Atiyah sequence, at least in the principal bundle case): $$ 0 \to V E \to TE \to \pi^* TM \to 0 $$ Here $VE$ denotes the bundle of vertical tangent bundles. Now there are three ways to split an exact sequence: Write down an isomorphism between the middle term and the sum of the terms on the right- and left-hand side. In our case, this corresponds to the decomposition $TE = VE \oplus HE$. Split on the left, i.e. give a map $TE \to VE$. This is the connection form $\omega$. Split on the right, i.e. specify a map $\pi^* TM \to TE$. This corresponds to lifting a tangent vector from $M$ to $E$. So now it should be pretty clear how to translate between the different viewpoints (modulo some natural isomorphisms). For example, every tangent vector $X \in T_e E$ can be written as a sum $X^v + X^h$, where $X^v = \omega(X)$ and $X^h$ is the horizontal lift of some vector $Y \in T_{\pi(e)}M$.<|endoftext|> TITLE: Existence of imaginary quadratic fields of class numbers coprime to $p$ with prescribed splitting behaviour of $p$ QUESTION [7 upvotes]: Let $x\in\{\text{totally ramified, inert, totally split}\}.$ If $p\geq 5$ is a prime, are there infinitely many imaginary quadratic fields $K=\mathbb{Q}(\sqrt{-d})$ of class number coprime to $p$ so that $p$ has ramification behaviour $x$ in $K/\mathbb{Q}$? REPLY [10 votes]: A more general version of this statement was shown by Kimura in Acta Arith. (2003). His corollary gives $\gg \sqrt{X}/\log X$ such fields ${\Bbb Q}(\sqrt{-d})$ with $d\le X$, and also allows you to add further splitting conditions. There is an extensive literature on divisibility and indivisibility of class numbers and Kimura's paper has many other relevant references. Update There is also an erratum to Kimura's paper -- it seems that he needs the existence of one such field to get a lower bound for the number of such fields; for large $p$ this is guaranteed by work of Horie. In the meantime a paper of Wiles has addressed the general version of this question, and there is a quantification of Wiles's result due to Beckwith.<|endoftext|> TITLE: What is the interface between functional analysis and algebraic geometry? QUESTION [13 upvotes]: This is a very open ended curiosity of mine and I would be grateful to hear any comments in this direction. In particular I am interested in functional analysis/algebraic geometry books/papers references which show this bridge from functional analysis into algebraic geometry. I am not sure if its related but what are the good references for ``functional analysis on manifolds"? Like how do we characterize the function space based on the domain manifold properties or for specific manifolds like say spheres. (the related things I see are courses like, http://www.math.uiuc.edu/~palbin/Math524.Spring2012/LectureNotesMay1.pdf or http://www.math.harvard.edu/~canzani/math253.html but these seem more about understanding specific differential operators on manifolds rather than the space of functions on a manifold) Like is there any meaning to wondering, "What is the Hilbert space of functions on a sphere?" REPLY [7 votes]: Have a look at how the Hirzebruch-Riemann-Roch can be deduced as a special case of the Atiyah-Singer index theorem. The idea is to consider the hodge operator $\overline{\partial} + \overline{\partial}^\ast$ on the bundle of differential forms with values in a given holomorphic vector bundle $V$ over a complex manifold $X$. This is an elliptic differential operator, and the index theorem says that its Fredholm index is the integral of the product of the Chern class of $V$ and the Todd class of $X$. On the other hand the Fredholm index is the holomorphic Euler characteristic of $V$ by Hodge theory, yielding HRR. The proof of the index theorem invariably uses a lot of functional analysis - either analysis of heat kernels, pseudo-differential operator theory, or operator algebras, depending on your preferences. This provides a foundation for a number of other interactions between algebraic geometry and functional analysis, like generalizations to holomorphic non-commutative spaces or analytic counterparts of Grothendieck-Riemann-Roch.<|endoftext|> TITLE: (Non-trivial) local coefficient system which is not a bundle of groups QUESTION [6 upvotes]: Let me first give the definition of a local coefficient system (see also [2, p. 257] and [3, p. 35]): Let $X$ be a topological space. A local coefficient system is a functor from the category $\Pi_1(X)$ (= the fundamental groupoid) to the category AbGrp of abelian groups. Such a functor assigns to each $x \in X$ an abelian group $G(x)$ and to each homotopy class $\xi \in \pi_1(X; x_1, x_2)$ (the set of all endpoint-preserving homotopy classes of paths from $x_1$ to $x_2$) a homomorphism $G(\xi): G(x_2) \to G(x_1)$; these are required to satisfy (i) if $\xi \in \pi_1(X, x) = \pi_1(X; x, x)$ is the identity, then $G(\xi): G(x) \to G(x)$ is the identity; (ii) if $\xi \in \pi_1(X; x_1, x_2)$, $\eta \in \pi_1(X; x_2, x_3)$, then $G(\xi \eta) = G(\xi) \circ G(\eta): G(x_3) \to G(x_1)$ So my question is: It is easy to see that every bundle of groups (defined in [1, p. 330]) is a local coefficient system, but I think the converse is not true (as stated without proof in [3, p. 35]), so I am looking for a (non-trivial) local coefficient system which is not a bundle of groups. Please note: A good choice for the space $X$ in the above definition of a local coefficient system could for example be the Hawaiian earrings. Thank you in advance! References: [1] Hatcher, Allen: Algebraic Topology, 2002, https://www.math.cornell.edu/~hatcher/AT/AT.pdf, p. 330 [2] Whitehead, George: Elements of Homotopy Theory, New York: Springer, 1978 (Graduate Texts in Mathematics Vol. 61), p. 257 [3] Hutchings, Michael: Introduction to higher homotopy groups and obstruction theory, 2011, https://math.berkeley.edu/~hutching/teach/215b-2011/homotopy.pdf, p. 35 REPLY [13 votes]: Strip away the group structure and you get the simpler question: Does every functor from the fundamental groupoid of $X$ to Set correspond to a covering space (a bundle of sets)? As a special case this includes the question, does every subgroup of the fundamental group of a path-connected space come from a connected covering space.The answer to all of these is no. A standard example is when $X$ is the "Hawaiian earring" a.k.a. "clamshell space" a.k.a. "shrinking wedge of circles" (union of infinite sequence of circles in the plane with diameters tending to zero and one point in common). I like to organize these ideas like this: There is a functor, obviously faithful, from the category $Cov(X)$ of covering spaces of $X$ to the category $Fun(\Pi_1(X),Set)$. It is not always fully faithful, but it is if $X$ is locally path-connected. In this case it is not always an equivalence of categories, but it is if $X$ is, as they say, "semi-locally simply connected".<|endoftext|> TITLE: Exceptional isomorphism with Spin(6,2)? QUESTION [5 upvotes]: There are all sorts of curios in low-dimensional Lie groups and Lie algebras, many of them due to the presence of the quaternions. There is, I have recently learned, an isomorphism $SO(6,2) \simeq SO(4,\mathbb{H})$ (this isn't listed on Wikipedia, for instance). I'm curious to know is there is any corresponding exceptional isomorphism for $Spin(6,2)$ analogous to the well-known isomorphism $Spin(5,1)\simeq SL(2,\mathbb{H})$? Clearly one cannot make a simple appeal to the Lie algebra isomorphism, but need to pull out some Clifford algebra or spin representation built from quaternions. EDIT: bonus imaginary internet points if this exceptional isomorphism is compatible in some way the one for $Spin(5,1)$. REPLY [7 votes]: To put the problem to a rest, I add my comment as an answer which is, in a nutshell, $Spin(6,2)\cong Spin(4,\mathbb H)$. The existence of this isomorphism follows from the isomorphism of the Satake diagrams and simple connectedness. The tricky thing is the definition of a spin group over the quaternions which is explained in much greater generality in section 13 of "The book of involutions" by Knus at al. The idea is roughly as follows: Given a quadratic space $(V,q)$ over a field $k$ (of characteristic $\ne2$, for simplicity) one constructs the Clifford algebra $Cl(V,q)$. The latter has a $\mathbb Z_2$-grading $Cl=Cl_0\oplus Cl_1$. Then the spin group consists loosely speaking of all $c\in Cl_0$ preserving $V\subseteq Cl_1$. One can do this construction in two steps. First, one assigns to $(V,q)$ the pair $(A,\sigma)$ where $A$ is the algebra $\mathrm{End}(V)$ and $\sigma\in\mathrm{End}(A)$ is taking the adjoint with respect to $q$. This way, one looses information, though: any scalar multiple $cq$ of $q$ gives rise to the same pair $(A,q)$. But the even part $Cl_0$ depends only on $(A,q)$ and that's where the spin group lives in! This observed, one can construct an algebra $Cl_0$ from any pair $(A,\sigma)$ where $A$ is a central simple algebra over $k$ and $\sigma$ is an orthogonal (anti-)involution. This means that $\sigma|_k=\mathrm{id}_k$ and $\dim A_+>\dim A_-$ where $A_\pm$ are the $\pm1$-eigenspaces of $\sigma$. Since $\mathrm{End}(V)=V\otimes V^*=V^{\otimes2}$, the general philosophy is that any construction involving only even tensor powers of $V$ can be done for any pair $(A,\sigma)$, as well. This is the case for $Cl_0$. But also $D=V\otimes Cl_1$ can be defined for any $(A,\sigma)$. Since it possible to formulate the condition "$c\in Cl_0$ preserves $V\subseteq Cl_1$" purely in terms of $D$ (and its $Cl_0$-bimodule structure) alone one arrives at a definition of the group $Spin(A,q)$. In the case at hand, one takes $k=\mathbb R$, $A=M(4,\mathbb H)$, and $\sigma(X)=a\overline X^ta^{-1}$ with $0\ne a\in\mathbb H$ with $\overline a=-a$. Usually, one works with $a=i$. Observe that conjugation $c\mapsto \overline c$ is a symplectic involution of $\mathbb H$. So, the condition $\overline a=-a$ makes $\sigma$ orthogonal. EDIT: Let $L_i$ and $\tilde L_i$ be the Levi part of the $i$-th maximal parabolic $P_i$ of $Spin(6,2)$ and $Spin(4,\mathbb H)$, respectively. Then looking at Satake diagrams the groups $L_1,P_1,\tilde L_4,\tilde P_4$ are defined over $\mathbb R$ and the exceptional isomorphism maps $L_1$ isomorphically onto $\tilde L_4$. So also the commutator subgroups $(L_1,L_1)=Spin(5,1)$ and $(\tilde L_4,\tilde L_4)=SL(2,\mathbb H)$ are mapped to each other. EDIT2: One thing which puzzled me is the way $L_1$ and $\tilde L_4$ projected into $SO(6,2)$ and $SO(4,\mathbb H)$, respectively. In the first case, the image is $SO(5,1)\subseteq SO(6,2)$ in the second it is the $SL(2,\mathbb)\subseteq GL(2,\mathbb H)\subseteq SO(4,\mathbb H)$. The second group is simply connected, the first one is not! The only explanation for this phenomenon is that actually $$ SO(6,2)\not\cong SO(4,\mathbb H). $$ So what is the correct isomorphism? The center of $Spin(6,2)$ is the Klein 4-group. Let $c_+,c_-,c_+c_-$ be its trivial elements such that $Spin(6,2)/c_+c_-=SO(6,2)$. Then $Spin(6,2)/c_\pm=Spin^\pm(6,2)$ are the two "halfspin groups". The correct isomorphisms therefore are $$ Spin^+(6,2)\cong SO(4,\mathbb H)\cong Spin^-(6,2). $$<|endoftext|> TITLE: Matrix concentration bound QUESTION [5 upvotes]: Suppose we have $N$ constant matrices $A_i \in R^{m\times m}, 1\leq i \leq N$. Consider $N$ random rotation-matrices $R_i \in SO(m), 1\leq i \leq N$. Is it possible to obtain a concentration bound on $$ \left\|\sum\limits_{i=1}^{N}R_i A_i\right\| $$ based on $$ \left\|\sum\limits_{i=1}^{N}A_i^{T}A_i\right\| . $$ Note that we do not assume the spectral norm of each individual $A_i$ is bounded by a smaller constant. REPLY [3 votes]: Such a bound follows from Corollary 6.1 here: http://arxiv.org/pdf/1408.3470.pdf (a version of the matrix bounded differences inequality), assuming your matrices $R_i$ are independent. Corollary 13.2 in the same paper gives bounds for the case that that are dependent, in terms of the Dobrushin interdependence matrix. Edit: I was thinking of applying that corollary to $H=\sum_{i=1}^N X_i$, with $X_i = \begin{pmatrix} 0 & R_iA_i \\ A_i^*R_i^* & 0 \end{pmatrix}$ (note that $H$ has the same spectral norm as $M=\sum_{i=1}^N R_iA_i$). However, we can't bound $X_i^2 = \begin{pmatrix} R_iA_iA_i^*R_i^* \\ 0 & A_i^*A_i \end{pmatrix}$ deterministically in terms of the $A_i$. One can show boundedness in expectation: $$\left\|\sum_{i=1}^N\mathbb{E} X_i^2\right\| \le \left\|\sum_{i=1}^N A_i^*A_i\right\|=: \sigma^2 $$ which allows us to apply the matrix Bernstein inequality (see Theorem 1.4 here: http://arxiv.org/pdf/1004.4389.pdf), but this will still require a uniform bound $\max_{i\le N} \|A_i\|\le R$. It gives $$ \mathbb{P}\left( \lambda_{\max}\left(H\right) \ge t\right) \le m \exp\left( \frac{-t^2}{2\sigma^2 + \frac23 Rt}\right).$$ This still offers an improvement over what you get from the assumption $\max_{i\le N} \|A_i\|\le R$ alone when $\sigma^2$ is of lower order than $NR^2$. In general it seems you will need some kind of uniform control on the individual summands to have exponential tail bounds (just as in the scalar case).<|endoftext|> TITLE: Symmetric power series over $\mathbb{F}_2$ QUESTION [5 upvotes]: Consider the symmetric power series $$f = \prod_{i \in I}\left(1+x_i+x_i^2+x_i^4+x_i^8 + x_i^{16} +\ldots \right)$$ in variables $(x_i)_{i \in I}$ over $\mathbb F_2$. Fix some degree $r$, smaller than the number of variables, and denote the degree $r$ part of $f$ by $f^{(r)}$. From looking at small-dimensional examples, I got the impression that the following might be true: If $r$ is odd, then $f^{(r)}$ is divisible by $\sigma_1 = \sum_{i \in I} x_i$. Could somebody provide a proof or counterexample of that statement? It would be interesting to me knowing the answer, either way. REPLY [5 votes]: Let $I=\{i_0\} \cup J$ and $g(x)=1+x+x^2+x^4+\dots$. We show that adding the relation $\sum_{i\in I}x_i=0$, i.e. $x_{i_0}=\sum_{i\ne i_0}x_i$ leads to an even degree power series: $$ f=g(x_{i_0})\prod_{j\in J} g(x_j) =\left(1+\sum_{i\in J} (g(x_i)-1)\right)\prod_{j\in J} g(x_j)\\ = \prod_{j\in J} g(x_j) + \sum_{i\in J}\left(g(x_i)-1\right)g(x_i)\prod_{j\in J\setminus \{i\}}g(x_j)\\ = \prod_{j\in J} g(x_j) + \sum_{i\in J}x_i\prod_{j\in J\setminus \{i\}}g(x_j)= \prod_{j\in J} g(x_j) + \sum_{i\in J}x_i\frac{\partial}{\partial x_i}\prod_{j\in J}g(x_j)$$ Now this is an even degree power series: check that each monomial with an odd number of $x_j^1$ cancels.<|endoftext|> TITLE: Conjecture of relation between residues of Feynman integrals and mixed Tate motives QUESTION [7 upvotes]: In many articles (for example in articles given by M.Marcoli) there is statement that there is the following conjecture Residues of Feynman integrals in scalar field theories are always periods of mixed Tate motives I have three questions related to this conjecture 1) What is the recent knowledge of this conjecture ? Is it proven or partially proven ? (Articles which I read have at least five/six years). 2) One of the supporting evidence were computations collected by Broadhurst and Kreimer. Where can I found more actual data (if such exist) ? 3) Is it a generalization of this conjecture on bosonic/fermionic field theories ? REPLY [15 votes]: 1) Counterexamples were found in the paper Brown, Francis; Schnetz, Oliver: "A $K3$ in $\phi^4$". Duke Math. J. 161 (2012), no. 10, 1817–1862. It is now the general feeling that most $\phi^4$-Feynman integrals are not mixed Tate. 2) In 2008, Schnetz has compiled a list of Feynman integrals in Quantum periods: A census of $\phi^4$-transcendentals. These were obtained using numerical computations. Meanwhile, for all results rigorous proofs have been found. 3) No clue.<|endoftext|> TITLE: Hasse principle for $H^2$ of a maximal torus of a split simply connected group? QUESTION [7 upvotes]: Let $k$ be a number field, and let $G$ be a split simply connected algebraic group over $k$. Let $\Omega_k$ denote the set of places of $k$. Let $T$ be a maximal torus of $G$ (defined over $k$). Consider the second Tate-Shafarevich group $$ Ш^2(k,T)=\ker\left[H^2(k,T)\to\prod_{v\in\Omega_k} H^2(k_v,T)\right].$$ Question. Is it true that $Ш^2(k,T)=1$ for any maximal torus $T$ of any split simply connected $k$-group $G$? Note that the answer is YES for $G=SL_n$. Indeed, then $T=\ker[N: S\to {\mathbb G}_m ]$, where $S$ is a maximal torus in $GL_n$. We have a cohomology exact sequence $$ 1\to H^2(k,T)\to H^2(k,S)\to H^2(k,{\mathbb G}_m)$$ and similar exact sequences for any place $v$ of $k$. We see that $Ш^2(k,T)$ embeds into $Ш^2(k,S)$. Since $S$ is a product of tori of the form $R_{L_i/k}\, {\mathbb G}_{m,L_i}$ where each $L_i$ is a finite extension of $k$, we have $$ Ш^2(k,S)=\prod_i Ш^2(k,R_{L_i/k}\,{\mathbb G}_{m,L_i})=\prod_i Ш^2(L_i,{\mathbb G}_m)=1, $$ hence $Ш^2(k,T)=1$. REPLY [2 votes]: This is a partial answer inspired by comments of Jason Starr. I show that the answer is YES for $G=\mathrm{Sp}_{2n}$ (and also for the classical non-simply-connected groups $\mathrm{SO}_{2n+1}$ and $\mathrm{SO}_{2n}$). Let $T\subset G$ be a maximal torus. The Galois group ${\Gamma}=\mathrm{Gal}(\bar k/k)$ acts on the character group $X(T)$ via $\mathrm{Aut}\, R(G_{\bar k},T_{\bar k})$, and hence, via the Weyl group $W$. We know that $$ W=(\pm 1)^n\rtimes S_n.$$ We have $G\subset \mathrm{GL}(V)$, where $V=k^{2n}$. Our torus has $2n$ different eigenspaces in $V$ with characters $\chi_1,-\chi_1,\chi_2,-\chi_2,\dots$, which are naturally divided into pairs. The Galois group ${\Gamma}$ acts on these characters and on the pairs. Let ${\Gamma}_1$ denote the stabilizer of the pair $(\chi_1,-\chi_1)$ in ${\Gamma}$, and let ${\Gamma}'_1$ denote the stabilizer of $\chi_1$ in ${\Gamma}$. Let $K_1\subset{\bar k}$ and $K'_1\subset {\bar k}$ denote the subfields corresponding to ${\Gamma}_1$ and ${\Gamma}'_1$, respectively, then either $K'_1=K_1$ or $K'_1$ is a quadratic extension of $K_1$. Let $X=X(T_{\bar k})$ denote the character group of $T$, and let $X_1$ denote the subgroup of $X$ generated by the characters $\sigma\chi_1$ for $\sigma\in\Gamma$. Then $X_1$ is a $\Gamma$-invariant direct factor of $X$, and we obtain a direct factor $T_1$ of $T$, where $$ T_1=R_{K_1/k}\, S_1,$$ and $S_1$ is a certain one-dimensional torus over $K_1$. Namely, if ${\Gamma}'_1={\Gamma}_1$, then $S_1$ is a split one-dimensional $K_1$-torus; otherwise it is the nonsplit one-dimensional $K_1$-torus that splits over the quadratic extension $K'_1/K_1$. We see that $T$ is a product of the Weil restrictions $T_i=R_{K_i/k}\, S_i$ of one-dimensional tori $S_i$ corresponding to the orbits of the Galois group in the set of pairs of characters of $T$. Since $Ш^2$ is trivial for a one-dimensional torus, it is trivial for $T$.<|endoftext|> TITLE: A derived category of formal sheaves QUESTION [5 upvotes]: Grothendieck (EGA I 0.7 & 1.10) defined a category of (topologically Noetherian) "formal rings" and a corresponding global category of formal schemes. Roughly, a formal ring is a topological commutative ring $R$ whose topology is $I$-adic for $I\subset R$ an open ideal. Its formal Spec is $Spf(R) : = Spec(R_{red})$ for $R_{red}$ the "maximal topologically reduced quotient", equivalently the reduced quotient of $R/I$. In the special case $I=0$ (discrete topology), we get the category of ordinary schemes as a full subcategory of formal schemes. Given a formal ring $R$, we have a category $Mod(R)$ of suitably nice topological modules; these have a theory of completed tensor product and pullback. Again, this globalizes to a category of formal quasicoherent sheaves on a formal scheme. Now in order to have pushforwards for formal sheaves, we need to allow new topological modules over ordinary rings. The fundamental example is the topological module $k[[t]]$ over $k[t]$ (the pushforward of the constant sheaf on the formal neighborhood of $0$). Since the completed stalk of $k[[t]]$ is zero at all points $0\neq x\in \mathbb{A}^1,$ it makes sense to think of this as a "formal sheaf on $\mathbb{A}^1$ supported at $0$". I would like to understand a category of topological coherent sheaves on an ordinary scheme which includes such "formal" objects, and especially a derived category of such objects (I'm particularly interested in a category of "constructible" formal sheaves, which I would define as repeated extensions of pushforwards of coherent sheaves on formal completions of closed subschemes). For some reason, I can't seem to find anyone who has written about this, except for in the very general context of ind-schemes à la Gaitsgory-Rozenblyum. Do people here know of a good source? REPLY [8 votes]: For a start, you have an approach to ind-coherent sheaves and its derived category in Duality and Flat Base Change on Formal Schemes Contemp. Math. 244 (1999), pp. 3-90. On line version, with some corrections incorporated. The theory gets interesting because this derived category contains two subcategories, one of quasi-coherent torsion sheaves and another one of complete complexes, where Grothendieck duality takes a specially nice form. These categories turn out to be equivalent, but the equivalence does not respect the inclusion functors. It might be interesting to point out that the full subcategory of the derived category of complexes with coherent homology is contained in the subcategory of complete complexes. All of this is discussed in the cited paper.<|endoftext|> TITLE: Solving cubic equations in characteristic 2 QUESTION [9 upvotes]: Consider the cubic polynomial $$ f = x^3+px+q, $$ where $p,q$ are elements of a fixed algebraic closure $\overline{\mathbb{F}}_2$ of $\mathbb{F}_2$. Is there an elegant criterion for deciding whether $f$ has $0$, $1$ or $3$ roots in the field $\mathbb{F}_2(p,q)$? For example, I would consider as "elegant" any criterion stipulating that certain polynomial expressions in $p,q$ be "of a certain form," e.g., are themselves values of certain polynomials (with coefficients in $\mathbb{F}_2$). REPLY [2 votes]: Let $K$, the field generated by the coefficients of $f$, have $t=2^n$ elements. If $p=q=0$, $f$ has a triple root. If $q$ is not $0$, then $f$ is separable. Assume $q$ not $0$. If $p=0$ and n odd, then $f$ has exactly one root in $K$. If $p=0$ and $n$ even then $f$ has $0$ or $3$ roots in $K$. Assume now that $f$ has at least 1 root in $K$. Let $tr$ be the trace of $K$ over the prime field. Put $B=1+\frac{p^3}{q^2}$. Then $f$ has $3$ roots iff $tr(B)=0$. The question if $f$ has no roots (i.e. $f$ irreducible) can be done by Berlekamp's algorithm.<|endoftext|> TITLE: Relationship between parabolic subgroups and parabolic subalgebras over non-algebraically-closed fields QUESTION [5 upvotes]: Let $\mathbb{K}$ be an arbitrary field, and $G$ an algebraic group (or group variety?) over this field. A Borel subgroup of $G$ is a connected solvable subgroup variety $B$ of $G$ such that $G/B$ is complete. A parabolic subgroup of $G$ is a subgroup variety $P$ such that $G/P$ is complete. From section 29.4 of Tauvel and Yu, I know that if $\mathbb{K}$ is algebraically closed and has characteristic zero, then a Borel subalgebra of a finite-dimensional Lie $\mathbb{K}$-algebra $\mathfrak{g}$ (not necessarily semisimple) is defined to be a maximal solvable subalgebra $\mathfrak{b} \leq \mathfrak{g}$. A parabolic subalgebra $\mathfrak{p} \leq \mathfrak{g}$ is any subalgebra containing a Borel subalgebra. Tauvel and Yu prove that under these hypotheses on the field ($\mathbb{K}$ is algebraically closed and has characteristic zero), if $G$ is a connected algebraic group with Lie algebra $\mathfrak{g}$, then parabolic subalgebras of $\mathfrak{g}$ are precisely the Lie algebras of parabolic subgroups of $G$. Now if we generalize so that $\mathbb{K}$ is not algebraically closed (or more generally, but less importantly, when the characteristic is nonzero), I would like to know to what extent these definitions and correspondences carry over? Moreover, if $\overline{\mathbb{K}}$ is the algebraic closure of $\mathbb{K}$, what properties are preserved and reflected by "extending" the scalars? For my particular purposes, I am most interested in the case $\mathbb{K}=\mathbb{R}$. However, it seems that everywhere I look the complex case is emphasized. I like to think very categorically, and in a "top-down" sort of way, so I am hoping to get a good overview of where I am going and what to expect before I slog through the complex theory. If it helps at all, I trying to get a handle on parabolic subgroups and subalgebras so that I can reach a more comprehensive understanding of parabolic Cartan geometries and (eventually) BGG resolutions. Unfortunately for me, my knowledge of algebraic geometry and algebraic groups is very limited, so I apologize if the answer to my question is very obvious. REPLY [4 votes]: I think the best book for you is Parabolic Geometries I : Background and General Theory by Andreas Čap and Jan Slovák. The first half covers the basics of Cartan geometries and classification of real Lie groups. The parabolic Cartan geometries are the content of its second half. As far as I know, there is no book for BGG resolutions as of yet. My feeling is that we will have to wait a few years for new results "to stabilize". E.g. for the most advanced stuff in this direction you can checkout arXiv:1510.03331 and its sequel. Some answers to your questions can be found in the third chapter of the aforementioned book. The main point here is that from the point of view of parabolic geometries you need to care only about subgroups of $G$ that preserve certain filtrations.<|endoftext|> TITLE: Structure of foliations of codimension 2 on three dimensional torus QUESTION [6 upvotes]: Is it possible to have a one-dimensional foliation on three dimensional torus such that the foliation has a trefoil knot as its leaf? Moreover, does a one dimensional foliation on three dimensional torus have both compact and non-compact leaves? Thank you very much! REPLY [11 votes]: [Apologies: the answer I wrote below is for $S^2 \times S^1$, not the 3-torus. Corrections added. The case of a 3-torus, or any 3-manifold, is included at the end.] First, there are different kinds of "trefoil knots" in $M = S^1 \times S^2$. Let's say we are interested in the "split trefoil knot", which lies in a 3-ball in $M$. This does in fact occur as a leaf of a foliation and it is easy to see. To build it, first recall the standard foliation of $S^3$ by circles where the one leaf is the trefoil knot $T$. This can be obtained by taking the orbits of the circle action $(z,w) \mapsto (t^2 z, t^3 w)$, $|t|=1$, on the 3-sphere in $\mathbb C^2$. Next we observe that the trefoil knot $T$ and a "disjoint" (split) unknot $U$ in $S^3$ can arise as the leaves of a foliation of $S^3$. (Split means there is an $S^2$ separating $U$ from $T$). To see this, consider the unknot $U'$ in (2) that arises as the orbit of $(z,w) = (1,0)$. This would almost do, except that $U'$ links $T$, i.e. $U'$ and $T$ are not split. To fix this, we use the fact (mentioned by Zare) that the space of constant slope foliations of a 2-torus is connected (it is just a circle). First adjust the foliation in (2) so it contains a solid torus $D^2 \times S^1$ foliated by $p\times S^1$'s with $U' = 0 \times S^1$. Here $D^2 = D^2(1)$ is the unit disk. By rotating the foliations on the tori $N_r = S^1(r) \times S^1$ as $r$ varies from 0 to 1, we can arrange that for some $r>0$ the torus $N_r$ is foliated by leaves of the form $U = S^1(r)\times p$, $p$ in $S^1$. Each of these leaves is an unknot split from $T$, so we have shown 3. To complete the picture, we now observe that $S^2 \times S^1$ can be obtained from $S^3$ by Dehn surgery along $U$, i.e. by cutting out a solid torus (disjoint from a ball containing $T$) with core curve $U$ and gluing it in again differently. Using the rotating foliation idea again, it is easy to extend the foliation from (3) to a foliation of $M$ with $T$ as a leaf. By construction, $T$ is a split trefoil in $M$. Case of the 3-torus: It is also true that for any closed 3-manifold $M$ and any link $L$ in $M$, there exists a nowhere zero vector field on $M$ such that $L$ is invariant under the flow. One way to see this is to use Zare's construction, which depends on two facts: (a) $M$ can be obtained by Dehn surgery on a link $L'$ in $S^3$ (Lickorish) and (b) any link $L''$ in $S^3$ can be presented as a closed braid (Alexander). This construction will be a little less explicit than the one above, since it uses (a) and (b). The main point is to prove the statement when $M=S^3$. For this one can start with any flow (e.g. the Hopf flow) such that there exists a closed trajectory. Thickening this flow line, we obtain an unknotted torus $S^1\times D^2$ with the product foliation. The foliation is obtained by suspending the identity map $D^2 \rightarrow D^2$. Now let us present $L$ as the closure of a braid $B$ with $n$ strands. Then $B$ can be obtained by suspending a diffeomorphism of $D^2$ to itself, fixing the boundary, and permuting $n$ points. Plug this new foliation into the original $D^2 \times S^1$. We then have a flow on $S^3$ such that $L$ is periodic. For the general case, let $W = L' \cup L''$ such that $M$ is obtained by Dehn surgery on $L'$, and $L''$ becomes the desired link $L$ in $M$ after surgery. Then apply the construction above to $(S^3,W)$, and rotate the foliation near $L'$ as in step (5) above so it is compatible with Dehn surgery.<|endoftext|> TITLE: Profunctorial Grothendieck Construction? QUESTION [9 upvotes]: I've been taking the ideas expressed in "Functors are Type Refinement Systems" seriously lately and it's lead me to a form of the Grothendieck construction I've never seen before. The idea in that paper is to think of a programming language $T$ as a category of terms/substitutions (with objects as types/contexts) and then think of a functor into $T$, say $p : D \to T$ as a type system where the objects of $D$ are refinements of types in $T$ and the arrows are typing derivations. Then the functor can be thought of as "type erasure". The general ideas there are very fibrational, for instance the fiber over a type $A$ in $T$ is a subtyping category over $A$. However, from this perspective type systems that are fibrations or opfibrations are very rare in programming languages, because they correspond to having principal types in a sense. So I started thinking about why the Grothendieck construction fails for a general functor $p : D \to T$. First, we map the an object $A \in T$ to the fiber category $p^{-1}(A)$. For $f : A \to B$, if $p$ is a fibration we get a functor from $p^{-1}(B)$ to $p^{-1}(A)$ by taking the pullback and if its an opfibration we can go the other way by taking the pushforward. However, even assuming neither of these things we get that $f$ gives a profunctor from $p^{-1}(B)$ to $p^{-1}(A)$, where $f(a,b) = \{g \in D(a,b) | pg = f\}$. So this works in general but the problem is that this doesn't generally extend to a functor from $T \to \text{Prof}^{\text{op}}$, because while identity is always preserved, composition may not be. The fact that composition is preserved, unrolling the coend, is equivalent to the statement that for every $\alpha : a \to c$ over $g \circ f$ where $f : A \to B, g : B \to C$ there exists (modulo some equivalence) $b, p(b) = B$ and $\alpha_f:a \to b, \alpha_g : b \to c$ over $f,g$ respectively, with $\alpha = \alpha_g \circ \alpha_f$. I've never seen this property before, but it seems to me to be somewhat natural for programming language type systems since they are basically always presented syntactically in a "bottom-up" way. I'll call a functor that satisfies this property a "bottom-up" functor. So if you have a bottom up functor you get a functor into $\text{Prof}^{\text{op}}$, and you can get back where you started by pulling back along a functor from what I'll call $(\text{Prof}^{op})_*$, whose objects are pairs of a category and an object in it $c \in C$ and arrows $c \in C \to d \in D$ are profunctors $R$ from $C$ to $D$ and an element of $R(d,c)$. Equivalently the objects are representable profunctors from $C \to 1$ and arrows from $c : C \to 1$ to $d : D \to 1$ are profunctors $R : D \to C$ and a natural transformation $\alpha : d \to c \circ R$. Has anyone ever seen this construction before or the corresponding restriction on functors that it requires? And if I didn't mess it up are there any obvious nice consequences of it that I could use to prove things about functors that satisfy the property? REPLY [10 votes]: First of all, the phrase "Grothendieck construction" generally refers to the inverse construction, starting with a functor $T^{\mathrm{op}}\to \mathrm{Cat}$ and constructing a fibration $D\to T$. Now, if you have an arbitrary functor $D\to T$, what you get is actually a normal lax functor $T^{\mathrm{op}} \to \mathrm{Prof}$, and conversely from any normal lax functor you can construct a functor $D\to T$. A proof can be found in Street's paper Powerful functors; he attributes the result to Benabou (unpublished). At the end, Street has a characterization of those functors $D\to T$ that correspond to pseudofunctors into $\mathrm{Prof}$; they have the factorization property you mentioned, plus a uniqueness for that factorization up to zigzags. These functors are also exactly the exponentiable morphisms in $\mathrm{Cat}$, which was proven by Conduche and Giraud; thus they are sometimes called Conduche functors.<|endoftext|> TITLE: What are examples of D-modules that I should have in mind while learning the theory? QUESTION [22 upvotes]: I've been reading about D-modules this summer in preparation for a learning seminar on intersection cohomology. Unfortunately, many of the ideas are not sticking while I learn about the theory. What are examples of D-modules, which demonstrate monodromy D-modules on nontrivial spaces, such as projective curves or surfaces D-modules with support on a singular space (so I can apply Kashiwara's theorem to see what D-modules can say about singularities) D-modules on families of varieties over a fixed base Also, what are some classical systems of differential equations I should be looking at while doing computations? I'm not very familiar with the classical theory. REPLY [3 votes]: I will edit this question as I learn more, but here's one useful example: Consider the $\mathcal{D}_{\mathbb{A}^1}$-module $$ \frac{\mathcal{D}_{\mathbb{A}^1}}{\mathcal{D}_{\mathbb{A}^1}(t\partial_t - \beta)} $$ is a local system on $\mathbb{C}^*$ with monodromy $\text{exp}(2\pi i \beta)$. Apparently this is a useful example for showing the definition of nearby cycles in mixed hodge modules is the right one.<|endoftext|> TITLE: Combinatorial proof of fact about Eulerian numbers? QUESTION [8 upvotes]: Let $A(m,n)$ denote the Eulerian numbers. I'm looking for a simple combinatorial proof of the following fact. Fact. If $p$ is prime and $0\le k < p-1$, then $A(p-1,k) \equiv 1 \pmod{p}$. The closest thing I'm aware of is an argument of S. Tanimoto, An operation on permutations and its application to Eulerian numbers, European Journal of Combinatorics 22 (2001), 569–576 that can be adapted to give a rather complicated proof, but I'm hoping for something simpler and more direct. REPLY [7 votes]: We identify the permutations of $1,2,\dots,p-1$ and the cyclic permutations $c=(c_0,\dots,c_{p-1})$ of $0,\dots,p-1$: if $c_k=0$, $c$ corresponds to $\pi(c):=(c_{k+1},c_{k+2},\dots,c_{k-1})$. There are $p-1$ cyclic permutations which are arithmetic progressions $(0,a,2a,\dots,(p-1)a)$, and other cyclic permutations are partitioned onto orbits of size $p$ of the action of $\mathbb{Z}_p$: $(c_0+a,c_1+a,\dots,c_{p-1}+a)$. The key fact is that in each group $C$ the corresponding permutations $\pi(c)$, $c\in C$, have the same number of ascents (example: for a cyclic permutation $01324$ the orbit contains also $12430$, $23041$, $34102$, $40213$, and permutations of $1234$ which are $1324$, $1243$, $4123$, $2341$, $2134$, all have 2 ascents.) The result follows after we also establish that the arithmetic progression $(a,2a,\dots,(p-1)a)$ (multiplication is taken modulo $p$) has exactly $p-1-a$ ascents. REPLY [3 votes]: You can use the closed-form expression from Wikipedia: $$A(n,m)=\sum_{k=0}^{m+1} (-1)^k\dbinom{n+1}{k}(m+1-k)^n$$ At $k=0$ the binomial is $1$, and Fermat Little Theorem tells us the last expression is $1\pmod{p}$. All the other binomials are divisible by $n+1=p$.<|endoftext|> TITLE: Did Gaston Julia ever get to see a computer-generated image of his eponymous set? QUESTION [44 upvotes]: I learned from Wikipedia that Gaston Julia died in 1978. Is it known if he ever got to see a computer-generated image of the set named after him? REPLY [10 votes]: I have a recollection of having seen a computer picture due to Cherry, an Autralian mathematician, who died in 1966. You would think I would have noted exactly where and when, but I didn't. Does anyone know what paper of his it might be?<|endoftext|> TITLE: Is a "smooth" finite-dimensional algebra separable modulo its radical? QUESTION [12 upvotes]: Let $k$ be a field, and let us write the "unadorned" tensor $\otimes$ in place of $\otimes_k$. For a unital finite-dimensional $k$-algebra $A$, let $A^e = A \otimes A^{op}$ denote the enveloping algebra, so that $k$-central $(A,A)$-bimodules are the same as left $A^e$-modules. Recall that a finite-dimensional $k$-algebra $A$ is separable if it satisfies the following equivalent conditions: $A$ is projective as a left $A^e$-module; $A \otimes E$ is semisimple for every field extension $E/k$; $A^e$ is semisimple; $A \cong \prod_{i=1}^n \mathbb{M}_n(D_i)$ for division algebras $D_i$ whose centers are separable field extensions of $k$. The first item above can be restated as "$A$ has projective dimension 0 as a left $A^e$-module." Thus the following property can be seen as a higher-dimensional generalization of separability. The terminology below is borrowed from "homologically smooth" (dg-)algebras. Definition: Say that a finite-dimensional $k$-algebra $A$ is smooth (of dimension $d$) if $A$ has finite projective dimension (equal to $d$) as a left $A^e$-module. This condition is known to be equivalent to $A^e$ having finite global dimension, and to imply that $A$ itself has finite global dimension. Let $J(A)$ denote the Jacobson radical of $A$. I am curious about the relationship between smoothness of $A$ and of $A/J(A)$. Question: If a finite-dimensional $k$-algebra $A$ is smooth, is the semisimple algebra $S = A/J(A)$ necessarily separable? In case the answer is negative, I would greatly appreciate any further conditions that could be imposed on a smooth algebra $A$ as above to ensure that $S$ is separable. What little I do know is that if a finite-dimensional algebra $A$ is smooth and semisimple, then $A = S$ is separable. (Indeed, both $A$ and $A^{op}$ are finite-dimensional semisimple, hence Frobenius, which implies that $A^e$ is Frobenius. It follows that every left $A^e$-module has projective dimension equal to $0$ or $\infty$. Since $A$ has finite $A^e$-projective dimension, it must be $A^e$-projective, so $A$ is separable.) In case the answer is negative, I would prefer a much better sufficient condition for $S$ to be separable, which allows for examples $A$ that have positive global dimension! REPLY [10 votes]: Let $K$ be an algebraic closure of $k$. The following lemma must surely be well-known, but I haven't found an explicit reference, so I'll include a proof at the end of this post. Lemma. If $S$ is a finite dimensional semisimple $k$-algebra, then $S_K=S\otimes_kK$ is a finite product of matrix rings over commutative local $K$-algebras. Let $A$ be as in the question and let $A_K=A\otimes_kK$. Applying $-\otimes_kK$ to a projective $A^e$-module resolution of $A$ gives a projective $A_K^e$-module resolution of $A_K$, so $A_K$ is also smooth and in particular has finite global dimension. To prove that $S=A/J(A)$ is separable, it suffices to prove that $S_K=S\otimes_kK$ is semisimple. By the lemma, $\operatorname{Ext}^1_{S_K}(U,V)=0$ for non-isomorphic simple $S_K$-modules $U$ and $V$. If there is a non-split extension of $U$ by itself for some simple $S_K$-module $U$, then $\operatorname{Ext}^1_{A_K}(U,U)\neq0$. But since $A_K$ is a finite dimensional algebra of finite global dimension over an algebraically closed field, the "no loops conjecture" implies that this can't be the case. [Kiyoshi Igusa, MR 1086558 Notes on the no loops conjecture, J. Pure Appl. Algebra 69 (1990), no. 2, 161--176.] Since there are no non-split extensions of simple $S_K$-modules, $S_K$ is semisimple. Proof of the lemma. Let $K_s$ be the separable closure of $k$ in $K$. Extending scalars to a separable field extension preserves semisimplicity, so $S_s=S\otimes_kK_s$ is semisimple, and therefore a finite product of matrix rings over division algebras finite dimensional over $K_s$. Since $K/K_s$ is purely inseparable, and hence $K/L$ is purely inseparable for any $L$ with $K_s\subseteq L\subseteq K$, all of these division algebras have separably closed centres. Since the Brauer group of a separably closed field is trivial, $S_s$ is a finite product of matrix rings over finite field extensions of $K_s$. The tensor product of any field extension with a purely inseparable field extension is a commutative local algebra [see for example, Theorem 8.46(3) in Jacobson's Basic Algebra II]. So $S_K=S_s\otimes_{K_s}K$ is a finite product of matrix rings over commutative local $K$-algebras.<|endoftext|> TITLE: Applications of Topological Complexity of configuration space QUESTION [7 upvotes]: I'm starting to work on topological complexity of configuration spaces. Articles say that this field has applications in robotic and control theory. One of the important articles belongs to Michael Farber My questions are : 1) How topological complexity can help to robotic and control theory? 2) Does there exist any applied problems in robotic or control or other subjects that have been solved by notion of Topological complexity? Thanks REPLY [3 votes]: (EDITED) It's a bit late to answer this. However, I have spent around a year working on this topic as an undergraduate and it was my dissertation's topic so I hope I can make my own tiny contribution. Topological Complexity can help you to understand the Gimbal Lock, which is strongly related to the incident happened in some of the Apollo Moon missions: The explanation from the point of view of topological complexity goes as follows: The configuration space of the spacecraft is (topologically): $$X \colon = \mathbb{R}^3 \times SO(3)$$ and $TC(X)=4$ since $X$ is homotopy equivalent to $SO(3)$ and $TC(SO(3))=4$. The key to prove the last fact is a result proved by Michael Farber (thanks to Mark Grant for pointing out a previous mistake here) which says that if $X$ is a topological group, then the L-S category of $X$ and it's topological complexity are equal. (If you want I can send you the full proofs of all this facts -in Spanish-) You can also use Topological Complexity to deal with robot arms in factories and so on....but I don't want to write a book as an answer. Hope that this answers, at least partially, some of your questions. Don't hesitate to ask for details or more information if you wish! --------------------------EDIT------------------------- The way I relate gimbal lock to TC is as follows. First of all, gimbal lock is the loss of control of an airplane or spacecraft. And when restricted to drones, we see it as an overturning. As you know better than me, $TC$ measures the stability of motion planning algorithms, and if $TC(X)$ is grater than one, then the motion planning algorithms over $X$ are unstable. Moreover, the configuration space of the drone is homotopy equivalent to $SO(3)$. And the motion planner needs to check periodically the inclination of the drone (which is alterated due to exterior circumstances -a storm or wind in case of a dron flying close to the land-) and considering the inclination that it should have, to correct it by changing the path in $P(SO(3))=SO(3)^{I}$. The problem about $TC(SO(3))=4$ is that there is no continuous global section of the path fibration, and as a consequence, given a small change in the inlination of the drone, the algorithm can produce a completely wrong path. I have made some pictures to illustrate this. In the first picture we see the drone, (it is supposed to have suffered a small perturbation despite of I have pictured it very big in order to produce a better drawing). It should correct its inclination by rotating clockwise direction. However, due to the instabilities caused by $TC(SO(3))=4$ it rotates the other way around and it overturns, which means we have lost it since it can not recover from that position. I hope that this could help a bit. Feel free to criticize the ''arguments'' or ''explanations'' written above since they are just the way an undergraduate thought about gimbal lock and are far from rigorous or even serious.<|endoftext|> TITLE: Dualizing sheaf after a finite group action QUESTION [9 upvotes]: Let $X\subset \mathbb P^n$ be a surface (possibly singular) and $\omega_X$ be its dualizing sheaf. Let $G$ be a finite group acting on $X$ (possibly with fixed points). We know how to calculate the dualizing sheaf for surfaces in projective space (ref https://mathoverflow.net/q/125724). I want to know what will happen to dualizing sheaf on $X/G$. In particular I want to know how to calculate the dualizing sheaf on $X/G$. Also if $\omega_X$ is ample then under what conditions the dualizing sheaf of $X/G$ is ample. Kindly provide with down-to-earth references and examples. REPLY [9 votes]: If $f:X\to Y$ is a finite morphism, and assuming that both $X$ and $Y$ admit dualizing sheaves, then by duality (see [Hartshorne, Ex.III.6.10]) you have the first map of the following: $$ \eta: f_*\omega_X\to \mathscr Hom _Y(f_*\mathscr O_X,\omega_Y)\to \omega_Y, $$ where the second map is induced by the natural map $$ \varrho: \mathscr O_Y \to f_*\mathscr O_X. $$ Now, if $f$ is the quotient map by a finite group $G$, then $\varrho$ is just the embedding of the set of $G$ invariant sections of $f_*\mathscr O_X$. Similarly, in that case $\eta$ is $G$-invariant and I am guessing that it is the largest $G$-equivariant quotient of $f_*\omega_X$ on which the induced $G$ action is trivial. I suppose this should tell you what the quotient is, but the problem is that this map is not necessarily surjective, the cokernel is contained in $$ \mathscr Ext^1_Y\left( f_*\mathscr O_X\big/\mathscr O_Y, \omega_Y\right). $$ Of course if you can prove that this is trivial, then you're in business. It does happen frequently that $f_*\mathscr O_X$ is free and $\mathscr O_Y$ is a direct summand in which case the above $\mathscr Ext^1$ is zero. But in general this will depend on $G$, the singularities of $X$ and the action.<|endoftext|> TITLE: Is it possible to construct K(G, 1) out of a subgroup and its quotient? QUESTION [7 upvotes]: Suppose we have a short exact sequence $1 \to K \to G \to Q \to 1$ and let $X_K$ and $X_Q$ be $K(K, 1)$ and $K(Q, 1)$, respectively. Is it possible to construct (in a "natural" way) a model of $K(G, 1)$ out of $X_K$ and $X_Q$? For example, if $G$ is a direct product of $K$ and $Q$, then it acts on $\tilde{X}_K \times \tilde{X}_Q$ in a decent way. Is it possible to generalize it somehow? I know that something can be done for cohomology (there is this Lyndon–Hochschild–Serre spectral sequence). But I need a topological space that is a model of $K(G, 1)$ (or rather, I am interested in the universal cover of that, and the action by deck transformations). REPLY [2 votes]: If the total spaces $EQ$ and $EK$ are given as $Q$ and $K$ CW-complexes respectively, you can also use a construction of C.T.C. Wall (see C. T. C. Wall, Resolutions for extensions of groups, Proc. Cambridge Phil. Soc. 57 (1961), 251-255, I also have a short summary in this paper here, in Section 3: http://arxiv.org/pdf/1006.0129v2.pdf) in the following way: The spaces $EQ$ and $EK$ will give you free resolutions over $\mathbb{Z}Q$ and $\mathbb{Z}K$ (which are contractible over $\mathbb{Z}$). Wall then explains how to construct a free resolution for $\mathbb{Z}G$ out of the two resolutions. This is a twisting of the idea that if $G=K\times Q$ you would just pick the product of the two resolutions. In fact, the number of $n$-cells in the new resolution will still be $\sum_{i+j=n} a_ib_{j}$ where $a_i$ is the number of $i$-cells in $EK$ and $b_j$ the number of $j$-cells in $EQ$. In the general case, where the sequence does not split, one needs to be more careful when choosing the differentials. From this free resolution you can construct (I believe) an $EG$ space and then also $BG$.<|endoftext|> TITLE: Ordinary abelian varieties over a finite field QUESTION [6 upvotes]: Let $q$ be a power of a prime $p$. Deligne's paper "Variétés abéliennes ordinaires sur un corps fini" seems to describe an equivalence of categories between ordinary abelian varieties over a finite field $\mathbf F_q$, complex abelian varieties equipped with an endomorphism $\pi$ which is a $q$-Weil number. Is that true? The paper actually states the equivalence of categories between (1) and a category of free Z-modules. It sketches the relation with complex abelian varieties, but does it remain an equivalence of categories? This question has been discussed a bit here: Canonical lifts from $\mathbb F_q$ and CM-theory. The answer seems to be "yes". I still ask this question because of the following, which looks like a contradiction to me: Fix a $q$-Weil number $\pi$, let $K = \mathbf Q(\pi)$, and $\mathcal O_K$ the maximal order in $K$. Then, $\mathrm{Cl}(K)$ acts freely and transitively on the set of isomorphism classes of complex abelian varieties with endomorphism ring $\mathcal O_K$. On the other hand, this action of $\mathrm{Cl}(K)$ on the abelian varieties over $\mathbf F_q$ with endomorphism ring $\mathcal O_K$ is still free but not necessarily transitive (Waterhouse, Abelian varieties over finite fields, Theorem 5.3: the number of orbits is $2^s$ where $s$ is the number of primes factors of $p$ in $K_0$ staying prime in $K$, and $K_0$ is the real subfield of $K$). Did I misinterpret Deligne's paper? Or is this $s$ actually always 0? REPLY [4 votes]: Because the abelian variety is ordinary, $\pi$ has the property that it's $p$-adic valuation at every place is either $0$ or $1$ (this follows easily from condition (IV) on the first page of Deligne's paper.) That means no prime $p'$of $\mathbb Q(\pi)$ lying over $p$ is equal to its complex conjugate, because if the $p'$-adic valuation of $\pi$ is $1$ then the $\overline{p}'$-adic valuation of $\pi$ is $0$, and vice versa. On the other hand, if a prime of $K_0$ remained a prime of $K$ then it would equal its complex conjugate. Hence $s=0$.<|endoftext|> TITLE: Can $C^1$ mappings with derivative of low rank be approximated by smooth maps? QUESTION [12 upvotes]: Asked once on SE-mathematics. Let $U$ be an open subset in $\mathbb{R}^n$, $m\in\mathbb{N}$, $1\leq m TITLE: Gauss' theorem for null boundaries QUESTION [7 upvotes]: Prenote: I have asked this question first on math stackexhange, but a user suggested that mathoverflow might be a better place for this question. Upon thinking about it I have agreed with him and copy-pasted the question here. Note: I have solved this problem on my own, mostly while actually typing it in here, as I was stuck with this problem previously. This is however quite important for my research, so I nontheless would like to verify the correctness of my solution, hence I went forth with posting this "question". If this goes against site principles then please vote close on the question. Introductions: Let $(M,g)$ be a Riemannian manifold with boundary (or a Lorentzian/pseudo-Riemannian manifold whose boundary is timelike or spacelike everywhere). Gauss' theorem then states that $$ \int_M\text{div}(X)\mu_g=\int_{\partial M}\langle X,n\rangle\mu_h, $$ where $\mu_g$ is the volume form associated with $g$, $\mu_h$ is the volume form associated with the induced metric $h$, and $n$ is the unit normal vector field along the boundary. This theorem is also available in an intermediate step between Stokes' theorem and the form I have given above. Since $\mathcal{L}_X\mu_g=\text{div}(X)\mu_g$, but $\mathcal{L}_X\mu_g=\mathrm{d}i_X\mu_g+i_X\mathrm{d}\mu_g=\mathrm{d}i_X\mu_g$, we have $$ \int_M\text{div}(X)\mu_g=\int_M\mathrm{d}i_X\mu_g=\int_{\partial M}\phi^*(i_X\mu_g), $$ where $\phi$ is the inclusion $\phi:\partial M\rightarrow M$. The problem: I wish to use Gauss' theorem on a Lorentzian manifold, whose boundary is a null surface. Then the normal vector is also tangent, the induced metric is degenerate, and the induced volume is zero. Durr. The "intermediate" form of the Gauss' theorem still applies though, so I seek to use it to define a version of Gauss' theorem which, instead of using the normal vector $n$, it will use an arbitrary transverse vector field $N$, which is surely not tangent to $\partial M$. Proposed solution: If $e_1,...,e_{n-1}$ is a positively oriented frame on $\partial M$, then $X$ is locally decomposible as $X=\langle X,N\rangle N+Y$, where $Y=Y^ie_i$, furthermore, I assume if $N$ is globally defined then this decomposition also applies globally in the general form $X=\langle X,N\rangle N+Y$, where $Y$ is tangent to $\partial M$. (Is this assumption correct?). Then plugging $X$ into $\mu_g$ gives $$ i_X\mu_g(e_1,...,e_{n-1})=\mu_g(\langle X,N\rangle N+Y,e_1,...,e_{n-1})=\langle X,N\rangle\mu_g(N,e_1,...,e_{n-1})= \\=\langle X,N\rangle i_N\mu_g(e_1,...,e_{n-1}) $$ where the term involving $Y$ is annihilated, because if $Y$ is tangent to $\partial M$, then the arguments of $\mu_g$ are linearly dependent. So from this we have $$ \int_M\text{div}(X)\mu_g=\int_{\partial M}\langle X,N\rangle \phi^*(i_N\mu_g), $$ so apparantly $i_N\mu_g$ is such a volume form for $\partial M$, that Gauss' theorem with $N$ instead of $n$ applies. I also want to obtain a local coordinate formula for this, so if $(U,x^\mu)$ is a local chart for $M$, and $(V,y^i)$ is a local chart for $\partial M$, then in the relevant chart domains we have $$ i_N\mu_g=i_N(\sqrt{-g}\mathrm{d}x^1\wedge...\wedge\mathrm{d}x^n)=i_N(n!^{-1}\sqrt{-g}\epsilon_{\mu_1...\mu_n}\mathrm{d}x^{\mu_1}\wedge...\wedge\mathrm{d}x^{\mu_n})=\\=i_N(\sqrt{-g}\epsilon_{\mu_1...\mu_n}\mathrm{d}x^{\mu_1}\otimes...\otimes\mathrm{d}x^{\mu_n})=\sqrt{-g}N^{\mu_1}\epsilon_{\mu_1...\mu_n}\mathrm{d}x^{\mu_2}\otimes...\otimes\mathrm{d}x^{\mu_n}=\\=\frac{1}{(n-1)!}\sqrt{-g}N^\mu\epsilon_{\mu\mu_2...\mu_n}\mathrm{d}x^{\mu_2}\wedge...\wedge\mathrm{d}x^{\mu_n}. $$ Now pulling back gives $$ \phi^*(i_N\mu_g)=\frac{1}{(n-1)!}\sqrt{-g}\epsilon_{\mu\mu_2...\mu_n}N^\mu\frac{\partial x^{\mu_2}}{\partial y^{i_1}}...\frac{\partial x^{\mu_n}}{\partial y^{i_{n-1}}}\mathrm{d}y^{i_1}\wedge...\wedge\mathrm{d}y^{i_{n-1}} .$$ Now, the matrix $e^\mu_{(i)}=\partial x^\mu/\partial y^i$ maybe identified (with a bit of abuse of notation regarding tangent maps) as the $\mu$th component of the $i$-th coordinate basis vector field on the boundary $\partial M$, with the components being taken with respect to the coordinate system $\{x^\mu\}$. Then, by the definition of determinants we have $\epsilon_{\mu\mu_2...\mu_n}N^\mu e^{\mu_2}_{i_1}...e^{\mu_n}_{i_{n-1}}=\det(N,e_{(i_1)},...,e_{(i_{n-1})})$, so $$\phi^*(i_N\mu_g)=\frac{1}{(N-1)!}\sqrt{-g}\det(N,e_{(i_1)},...,e_{(i_{n-1})})\mathrm{d}y^{i_1}\wedge...\wedge\mathrm{d}y^{i_{n-1}}=\\=\sqrt{-g}\det(N,e_1,...,e_{n-1})d^{n-1}y, $$ so the modified Gauss' theorem in coordinates is (assuming $M$ can be covered by a single chart) $$ \int_M \text{div}(X)\sqrt{-g}d^nx=\int_{\partial M}X^\mu N_\mu\sqrt{-g}\det(N,e_1,...,e_{n-1})d^{n-1}y $$. Questions: Is this derivation correct? I can work with the form I have gotten, but I don't like the determinant factor in it. Is there any way I can express that in a more pleasant form? In case the derivation is incorrect, how can I get a formula I can work with in coordinate notation that does what I want? REPLY [4 votes]: Gauss' Theorem has nothing to do with the (pseudo-)metric. Is just a consequence of Stokes' theorem. Stokes's theorem says that, for any $n-1$ form $\omega$, $$ \int_M d\omega = \int_{\partial M} \omega. $$ Now fix any smooth measure $\mu$ (i.e. given by a smooth non-vanishing top dimensional form, or a density if $M$ is not orientable). It might be the Riemannian measure of a Riemannian structure, but it's not relevant. Let $X$ be a vector field. If you apply the above formula to the $n-1$ form $\omega:=\iota_X \mu$, you get, using one of the many definitions of divergence (associated with the measure $\mu$) $$\int_M \mathrm{div}(X) \mu = \int_{\partial M} \iota_X \mu, \qquad (\star),$$ where $\iota_X$ is the contraction. This holds whatever is $X$, on any smooth manifold with boundary. Now it all boils down to how you want to define a "reference" measure on $\partial M$. As you propose, you can pick a transverse vector $N$ to $\partial M$, and define a reference measure on $\partial M$ as $\eta:=\iota_T \mu$. Then $$ \int_M \mathrm{div}(X) \mu = \int_{\partial M} f\, \iota_T \mu, $$ where $$f := \frac{\mu(X,Y_1,\ldots,Y_{n-1})}{\mu(T,Y_1,\ldots,Y_{n-1})}. $$ for any arbitrary local frame $Y_1,\ldots,Y_{n-1}$ tangent to $\partial M$ (oriented, otherwise take densities and absolute values). In particular, you can easily compute $f$ by looking at the "$T$" component of "$X$": $$X = f\, T \mod \mathrm{span}\{Y_1,\ldots,Y_{n-1}\}.$$ There is no need to use orthonormality, or a (pseudo)-metric. However, if you are on a Riemannian manifold, $\mu = \mu_g$ is the Riemannian measure, and $N$ is the normal vector to $\partial M$, then $f= g(N,X)$, and $\iota_N \mu$ is Riemannian measure of the induced Riemannian metric on $\partial M$, recovering the classical statement.<|endoftext|> TITLE: For positive definite $A,B$ why does $AB+BA$ tend to be positive definite? QUESTION [9 upvotes]: Let $A$ and $B$ be two positive definite $n \times n$ matrices. It is, of course, not true that $AB+BA$ is necessarily positive definite. Consider, though, the results of the following numerical experiment. I generated $A$ by letting its eigenvalues be random in $[0,1]$, and selecting its eigenvectors by generating a random matrix of standard Gaussians and applying Gram-Schmidt to it. The matrix $B$ is generated in the same way. I did this 1000 times and checked what proportion of times the matrix $AB+BA$ has at least one negative eigenvalue [1]. Here are the results for different dimensions $n$: $n=2, ~~~~94.8 \%$ $n=3, ~~~~89.4 \%$. $n=4, ~~~~78 \%$. $n=5, ~~~~72.7 \%$. $n=10, ~~~40.3 \%$. $n=15, ~~~20.1 \%$. $n=20, ~~~11.4 \%$. $n=50, ~~~0.3\%$. $n=100, ~~0 \%$. This suggests that, as a function of $n$, examples with $AB+BA$ not psd tend to get rarer and rarer. Is it possible to give a proof of this? It may be more natural to consider a different random model of randomly generated psd matrices; I only generated them in the way I described above because it seemed easiest. [1] Actually, I checked if there is an eigenvalue less then $-1 \cdot 10^{-5}$ in MATLAB to account for rounding error. REPLY [3 votes]: It is in theory possible to use free probability to describe the limit eigenvalue distribution (as the dimension tends to $+\infty$) of the anticommutator $AB+BA$ when $A$ and $B$ are independent random matrices which are unitarily invariant (which is the case for the description used in the OP). The same is true actually for any self-adjoint polynomial in $A$, $B$. The drawback of this method it that the limit distribution will be obtained in an indirect way. If one does this for the uniform measure on $[0,1]$ as asked in the OP, it may be hard to decide from this approch whether the limit distribution is supported on $\mathbf{R}^+$ or not. However, such computations are possible in simple cases. Propositon 6.11 from Fevrier, Maxime; Mastnak, Mitja; Nica, Alexandru; Szpojankowski, Kamil, Using Boolean cumulants to study multiplication and anti-commutators of free random variables says the following: if $E$ and $F$ are subspaces of dimension $n/2$ which are chosen independently according to the $O(n)$-invariant measure on the Grassmann manifold, and $A$ and $B$ are the corresponding orthogonal projections, then as $n \to \infty$ the eigenvalue distribution of $AB+BA$ is described by an explicit density which is supported on $[-1/4,2]$ (see also Figure 2 in the same paper). In particular, the probability that $AB+BA$ is positive semidefinite tends to $0$ as $n \to \infty$. For this toy model, this gives a rigorous proof of the fact that the anticommutator of positive matrices is typically not positive. (This is probably overkill, ane can presumably analyze everything in terms of principal angles between the subspaces $E$ and $F$.)<|endoftext|> TITLE: Finite well-completeness and the small object argument? QUESTION [5 upvotes]: I'm reading a few papers on reflective factorization systems and I've just noticed they're all mentioning a procedure which seems very similar to the small object argument. First of all, some background. My first encounter with the small object argument was Garner's paper about which I've asked several questions and learned a lot from. Lately I've been revisiting the basic theory of strong factorization systems in the context of categorical Galois theory and its connection to semi-left-exact reflections. I do not have a good understanding of the transfinite subtleties of small object arguments, but as far as I understand, what they do is freely construct cofibrantly generated factorization systems (weak or strong). This question is about an apparent similarity between the goal of the small object argument, and an important procedure proving the finite well-completeness implies that reflective prefactorization systems are actually factorization systems. Let $F\dashv G$ be an adjunction between $\mathsf C\overset{F}\rightleftarrows \mathsf B$. The prefactorization system on $\mathsf C$ fibrantly generated by the image of $G$ is $(\mathcal L,\mathcal R)=(\operatorname{\!^\perp Im}G, (\operatorname{\!^\perp Im}G)^\perp)$. Generally speaking this is not a strong factorization system. Let $f$ be an arrow. Seemingly the only way to get any interesting factorization of it is to factor the naturality square below through the pullback of $B\rightarrow GFB\leftarrow GFA$. This yields a factorization $f=v\circ w$, where $v$ is the pullback of $GFf$ along $\eta_B$ and $w$ is the unique arrow induced by the universal property of the pullback $P$. By definition $GFf\in \mathcal R$, and since $\mathcal R$ is always universal, so is $v$. Hence, in a sense, we already have "half" the $(\mathcal L,\mathcal R)$-factorization of $f$. $$\require{AMScd} \begin{CD} A @>{\eta_A}>> GFA\\ @V{f}VV @VV{GFf}V\\ B @>>{\eta_B}> GFB \end{CD}$$ $w$ though is not generally in $\mathcal L$, which is always the class of arrows inverted by $F$. However, since $\mathcal R$ is closed under composition, finding an $(\mathcal L,\mathcal R)$-factorization of $w$ itself would suffice. The tentative strategy now is to take the intersection $m$ of all (strong) subobjects of the pullback object $P$ that are also contained in $\mathcal R$. If this creature exists, its universal property implies it must factor $v$ as $v=mg$ for some $g$, while limit closedness implies it's also in $\mathcal R$. Some careful arguments then show $g\in \mathcal L$. The condition on a finitely complete category asking for intersections of strong subobjects to exist is called finite well-completeness. Adámek and Rosicky's book mentions locally $\lambda$-presentable categories as those in which every objects is a directed colimit of $\lambda$-presentable objects of a certain set, which seems at the very least analogous to finite well-completeness. Since locally presentable categories are basically "ones which admit a small object argument", I'd like to understand precisely the relation of the picture I sketched above with small object arguments in general. Also, what can be done in the situation described if instead of an adjunction one just starts with some class of arrows $\Sigma$? REPLY [3 votes]: This isn't a full answer because it isn't completely precise, but I would say that the relationship is between "predicative" and "impredicative" constructions of universal objects. Suppose $P$ is a poset and $f:P\to P$ is monotone and inflationary, i.e. $x\le y \Rightarrow f(x)\le f(y)$ and $x\le f(x)$. And say we have $a\in P$ and we want to construct the least fixed point of $f$ above $a$. (Since $f$ is inflationary, we have $f(x)=x$ iff $f(x)\le x$.) One way to do it is to take the infimum of all fixed points of $f$ above $a$, if it exists. This is clearly still above $a$, and below all fixed points of $f$ above $a$, while it is itself a fixed point of $f$ since $f(\bigwedge x) \le f(x) \le x$ for each such $x$, hence $f(\bigwedge x) \le \bigwedge x$. This in an "impredicative" construction, analogous to the well-completeness argument. Another way to do it is to consider the sequence $a \le f(a) \le f(f(a))\le f(f(f(a))) \le \cdots$ and take its supremum. This is also clearly above $a$, and below all fixed points above $a$ since if $a\le x$ and $f(x)=x$ then $f(a) \le f(x) = x$ too (and so on). And if $f$ preserves this supremum, then it is a fixed point of $f$, since $f(\bigvee_n f^n(a)) = \bigvee_n f(f^n(a)) = \bigvee_n f^{n+1}(a) = \bigvee_n f^n(a)$. (And if $f$ doesn't preserve this supremum, then we can continue into transfinite ordinals in hopes of finding some supremum that $f$ does preserve.) This is a "predicative" construction, analogous to the small object argument. Universal properties in posets often have two constructions like this. However, when generalizing to categories, predicative constructions tend to fare better, because they involve (co)limits whose size is at least potentially smaller than that of the ambient category, while impredicative constructions tend to involve (co)limits of roughly the same size as the ambient category. Such "large limits" do sometimes exist --- well-completeness is one example, another is total categories --- but they are the exception rather than the rule; indeed a classical result of Freyd says that if a category has all limits of the same size of itself, then it must be a poset. Locally presentable categories are those that are "determined by a small amount of data" in a certain precise sense. This ensures that the (co)limits involved in predicative constructions are (usually) in fact small, so that the small object argument works. And it also ensures that certain large limits exist, since they can also be "determined by a small amount of data"; for instance, every locally presentable category is total. So that gives at least an intuitive reason why local presentability seems to make both of these sorts of arguments work in some cases.<|endoftext|> TITLE: A modern interpretation of Quillen's computation of the K theory of finite fields QUESTION [30 upvotes]: In his beautiful paper On the cohomology and K theory of the general linear group over a finite field, Quillen constructs (if I understand correctly) an isomorphism on connected components of K-theory $K(\mathbb{F}_q)\mid_{\ge 1} \cong \text{Fib}(1-\Psi^q:ku\to ku).$ His construction, while very cool, seems to be based on a miracle: that the Brauer lift produces complex representations from mod-$p$ ones. It also uses the +-construction for defining K-theory (though perhaps it's not that hard to update his construction to work with Segal formalism)? I'm curious whether there is a more modern perspective on this equivalence of spectra, maybe using cyclic structure on topological Hochschild homology. REPLY [11 votes]: Firstly, there is a proof using the motivic spectral sequence (the Atiyah-Hirzerbruch style spectral sequence from motivic cohomology to algebraic $K$-theory). This is written in the master's thesis of Gabe Angelini-Knoll. Gabe and Andrew Salch are also working to answer this question and a paper is apparently due. From Andrew's website: "My student Gabe Angelini-Knoll and I have been working on the problem of computing the Waldhausen algebraic K-groups of the algebraic K-theory spectra of certain finite fields. This is an example of "iterated K-theory" and Rognes' redshift conjecture is not known in these cases. Thus far, Gabe and I have (with the aid of a new "THH-May" spectral sequence for computing topological Hochschild homology) computed the homotopy groups of THH(K(F_q)) smashed with the p-primary Smith-Toda complex V(1), for p > 3 and for many (but not all) prime powers q. Gabe is working on the computations of the homotopy groups of the C_p fixed points of this spectrum (this will probably be Gabe's thesis), with the goal of using trace methods to recover the K-groups of K(F_q). We expect to post and submit our first two papers on this topic before the end of summer 2016. "<|endoftext|> TITLE: Squarefree Fibonacci Numbers QUESTION [8 upvotes]: Let $F_n$, $n\geq 0$, be the sequence of Fibonacci numbers, where $F_0=F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\geq 1$. A number is squarefree if it is is not divisible by the square of a prime number. Question: Are there infinitely many squarefree Fibonacci numbers? REPLY [2 votes]: For prime $p$, let $M(p)$ be the least positive $n$ such that $p^2 \mid F_n$. Then $p^2 \mid F_n$ iff $M(p) \mid n$. Thus at most $N/M(p)$ of the first $N$ Fibonacci numbers are divisible by $p^2$. If we could prove that $\sum_p 1/M(p) < 1$, then at least a positive fraction of all Fibonacci numbers will be squarefree. Well, it seems to be true numerically; I don't know if it's provable.<|endoftext|> TITLE: Algebra in a category QUESTION [8 upvotes]: I am try to understand the concept: an algebra in a category. Let $\mathcal{C}$ be a category and $A$ an object in $\mathcal{C}$. $A$ is an algebra in $\mathcal{C}$ means the multiplication $m: A \otimes A \to A$ is a morphism in $\mathcal{C}$? Let $H$ be a bialgebra and $V$ a Yetter-Drinfeld module over $H$. Let $YD_{H}^H$ be the monoidal category of all Yetter-Drinfeld modules over $H$. Since $YD_H^H$ is a tensor category, the tensor algebra $T(V)$ is a Yetter-Drinfeld module over $H$. Therefore $T(V)$ is an object in $YD_H^H$. When we want to show that $T(V)$ is an algebra in $YD_H^H$, we need to show that the multiplication $m: T(V)\to T(V) \otimes T(V)$ is a homomorphsim of $H$-modules? That is, $h.(ab)=(h_{(1)}.a)(h_{(2)}.b)$ for $h \in H, a,b \in T(V)$. Is this correct? Thank you very much. REPLY [4 votes]: The answer to the specific situation you describe is as follows: Yes, $T(V)$ is an algebra in Yetter-Drinfeld modules over $H$ (using the concept that Jacob White mentions in his answer, where in addition we also have a $1$, which is a map from the tensor unit to the algebra satisfying the unital property expressed as a commutative diagram). A reference for this fact for $T(V)$ is, for example, [AS, 2.1]. You need to check two things: The multiplication is a map of algebras (you state this condition), The multiplication is a map of coalgebras (this is missing). In fact, more is true. As $YD_{H}^H$ is a braided monoidal category, we can speak of bialgebra and Hopf algebra objects in it. $T(V)$ is a (graded, braided) Hopf algebra object in this category. This concept is for example defined in [AS, 1.7]. [AS]: Andruskiewitsch, Nicolás; Schneider, Hans-Jürgen. Pointed Hopf algebras. New directions in Hopf algebras, 1--68, Math. Sci. Res. Inst. Publ., 43, Cambridge Univ. Press, Cambridge, 2002<|endoftext|> TITLE: mixing convex and concave for convexity QUESTION [19 upvotes]: Let $n\in\mathbb{N}$ and $0 TITLE: Do copowers commute with k-linear functors? QUESTION [7 upvotes]: Fix a field $k$ and suppose $\mathcal{C}$ and $\mathcal{D}$ are $k$-linear additive categories and are enriched over the category $\mathcal{V}$ of finite-dimensional $k$-vector spaces. So we have copower functors $$ \mathcal{C} \times \mathcal{V} \to \mathcal{C} \quad \text{and} \quad \mathcal{D} \times \mathcal{V} \to \mathcal{D}. $$ Now suppose $F \colon \mathcal{C} \to \mathcal{D}$ is a $k$-linear functor. Does it follow that the diagram $\require{AMScd}$ \begin{CD} \mathcal{C} \times \mathcal{V} @>>> \mathcal{C} \\ @V F \times \text{id}_{\mathcal{V}} VV & @VV F V \\ \mathcal{D} \times \mathcal{V} @>>> \mathcal{D} \end{CD} commutes (where the horizontal arrows are the copower functors)? If not, are there some natural assumptions on $F$ that make this diagram commute? I have a particular example in mind, where $\mathcal{C}$ and $\mathcal{D}$ are categories of modules over finite-dimensional $k$-algebras and $F$ is a functor given by tensoring with a bimodule. So I'd be happy with assumptions that are natural in that setting. REPLY [5 votes]: Yes, this is true (up to natural isomorphism). The simplest way to see this is just to write down what the copowers are explicitly. Let us take a skeleton of $\mathcal{V}$ consisting of all vector spaces of the form $k^n$, and write $\otimes$ for copowers. Then on objects, $A\otimes k^n$ is just a direct sum of $n$ copies of $A$. And on morphisms, given $f:A\to B$ and $T:k^n\to k^m$, the induced map $f\otimes T:A\otimes k^n\to B\otimes k^m$ can be described as follows. The domain is $A^n$ and the codomain is $B^m$, so we can describe $f\otimes T$ as an $n\times m$ matrix of morphisms $A\to B$. The $ij$ entry of this matrix is just $T_{ij}f$, where $T_{ij}\in k$ is the $ij$ matrix entry of $T$. All of this structure depends only on the $k$-linear structure of $\mathcal{C}$, so it is preserved by any $k$-linear functor. Explicitly, morphisms $A^n\to B^m$ are in bijection with matrices of morphisms $A\to B$ by composing with the inclusion and projection maps $A\to A^n$ and $B^m\to B$, and this correspondence is preserved by any additive functor. ($k$-linearity is also needed so that you know the scalar products $T_{ij}f$ are preserved.)<|endoftext|> TITLE: Haar measure on the Grassmannian space QUESTION [7 upvotes]: The grassmannian space $G(n,m)$ may be identified with the quotient space $O(n)/(O(m)\times O(n-m)$. As such, it is endowed with a natural invariant probability measure which I call "Haar measure on $G(n,m)$". Let $\gamma_{n,m}$ be this measure. Let us endow $G(n,m)$ with the usual metric $d(V,W)=\|P_V-P_W\|$, where $P_V$ is the orthogonal projection onto $V$. Question. Is it true that there is some constant $C$ such that $\frac{1}{C} r^{m(n-m)} \leq \gamma_{n,m}(B(V,r)) \leq C\times r^{m(n-m)}$ for any $V \in G(n,m)$ and any $r>0$ small enough? REPLY [2 votes]: Yes, this is true. The answer by Ben Webster gives the idea, but for completion let me give you a list of easy facts that combined together answer the question. The inequalities you seek are correct for any Riemannian manifold with bounds on the curvature (upper/lower bounds for corresponding sides of the inequalities), once you take the Riemannian volume and Riemannian metric. In particular, this is the case for a compact Riemannian manifolds. Our Grassmanian could be seen as a Riemannian manifold, by embedding it in $\text{End}(\mathbb{R}^n)$ as you do, and restrict the standard inner product to the tangent spaces. The obtained Riemannian structure will be $G$-invariant (for $G=\text{O}(n)$ of course). For our manifold, up to normalization, the Riemannian volume coincides with the Haar measure, as both are $G$-invariant, by "uniqueness". For a Riemannain manifold $M$ and a compact submanifold $N$ the two metrics on $N$, the restriction to $N$ of the Riemannian metric of $M$ and the Riemannian metric of $N$ associated with the induced Riemannian structure, are biLipischits (this is an easy excercise). All norms on a Euclidean space are equivalent. Of course, we apply 5 for whatever norm you had in mind in the formulation of the question and the standard inner product on $\text{End}(\mathbb{R}^n)$ and we apply 4 with $M=\text{End}(\mathbb{R}^n)$ and $N$ being the image of the Grassmannian.<|endoftext|> TITLE: On Ray-Singer's proof of the metric invariance of analytical torsion QUESTION [5 upvotes]: The Ray-Singer paper "R-torsion and the Laplacian on Riemannian manifolds" claimed that one may prove the metric invariance of analytical torsion by forming a homotopy between metric $\rho_{0},\rho_{1}$ using $\rho_{\mu}=\mu\rho_{0}+(1-\mu)\rho_{1}$. Then one differentiate $$ \frac{\partial }{\partial u}f(u,s)=\frac{1}{2}\sum^{N}_{q=0}(-1)^{q}q\int^{\infty}_{0}t^{s-1}Tr(e^{t\Delta_{q}(u)}\dot{\Delta_{q}})dt $$ here the $\frac{1}{\Gamma(s)}$ is omitted as it is entire, and if we know the differential of original function with $u$ is $0$, then with respect to $s$ should be $0$ as well. Now the differential of the fundamental solution of the initial value problem ("heat kernel") is given by $$ \frac{d}{d\mu}Tr e^{t\Delta_{q}(\mu)}=-tTr((\delta \alpha d-d\alpha\delta-\alpha \delta d+\alpha d\delta))e^{t\Delta_{q}(\mu)} $$ hence in the first equation we actually have $$ \dot{\Delta}_{q}=\alpha \delta d-\delta \alpha d+d\alpha \delta-d\delta\alpha,\alpha=*^{-1}\dot{*}=*^{-1}(\frac{\partial}{\partial \mu}*) $$ The proof then proceed by substituting the later formula into the above identity, using the fact that the trace of trace-class operator and bounded operator commutes to reduce the sum to a telescoping sum, which then reduces to the following: $$ \frac{1}{2}\sum^{N}_{q=0}(-1)^{q}\int^{\infty}_{0}t^{s}\frac{d}{dt}Tr(e^{t\Delta_{q}(\mu)}\alpha)dt $$ using integration by parts, we gain an additional $s$ on the front, and the desired result $\frac{\partial}{\partial \mu}f(\mu,s)=0$ now follows by regularity of the heat kernel for $\operatorname{Re}(s)$ large enough. I noticed a seemingly discrepancy in the above proof,namely the setting of the theorem is when $W$ is a closed manifold, and the formula (Proposition 6.1 of the paper) is derived in the context of a manifold with two boundary components $M_1,M_2$, on which the Laplacian satisfies certain boundary conditions. In the case of the closed manifold, the formula does not easily reduce from the manifold with boundary case (see the remark in page 195), and the same strategy seems no longer works (possibly) due to the failure of using integration by parts in the second to last step. It seems the proof works mainly because in the detailed computation in page 196 we actually do not have any integral over the boundary components, hence the same strategy would work trivially for a closed manifold. My question is: Is there a simpler proof for this in the case of a closed manifold? I am sure Ray and Singer must have tried and thought about it, but somehow still decided to use the strategy presented in paper to finish the proof. After all the proof presented in page 195-199 is really involved. So if there is a "clear cut" solution, it most likely would not be easy. I checked the literature but I am not sure if someone else has done any work related to it. It really isn't clear to me how to analyze the regularity of the differential of the Laplacian $\frac{\partial}{\partial \mu}\Delta_{q}(\mu)$ when $\operatorname{Re}{s}\rightarrow \infty$. I am not sure if the strategy in page 154 still works in this context. REPLY [5 votes]: Ray and Singer state their independence result in Thm 2.1 for closed manifolds and give a full proof. This uses Proposition 6.1, which is formulated for manifolds with boundary, and the result slightly differs from the one used in chapter 2. However, the only difference is that $\alpha d\delta$ occurs instead of $d\delta\alpha$. Their remark says that one can rotate inside the trace to get back to the formula of chapter 2 only in the case that $W$ is closed. There is a different formula for the analytic torsion in Bismut-Lott, Thm 3.29: $$-\int_0^\infty\left(\operatorname{tr}\left((1-2t\Delta)e^{-t\Delta}\right)-\frac{\chi'(M;F)}2-\left(\frac{\dim M\,\operatorname{rk} F\,\chi(M)}4-\frac{\chi'(M;F)}2\right)\right)\frac{dt}t\;.$$ In fact, this is the degree $0$ component of the analytic torsion form (Def 3.22). Here, $\Delta=(d+\delta)^2$, $\chi(M)$ is the Euler number, and $\chi'(M;F)=\sum (-1)^qq\,b_q(M;F)$ is a "secondary" Euler number for $M$ and the flat vector bundle $F$. The main difference to Ray-Singer's construction is that the integral above converges without further regularisation. Moreover, the formula for the exterior differential of the torsion form implies a variational formula, see Thm 3.24. It implies that the Ray-Singer torsion is independent of the metric as long as $F$ is acyclic and $\dim M$ is odd.<|endoftext|> TITLE: Is Proposition 2.6 in J. Silverman's book Arithmetic of Elliptic Curves correct? QUESTION [11 upvotes]: In J. Silverman's book "Arithmetic of Elliptic Curves" Chapter 2 Proposition 2.6 (a) it is considered a non constant morphism $\Phi:C_{1}→C_{2}$ between two smooth curves defined over a perfect field $K$. I am having trouble understanding why there is no inertia degree appearing in the degree formula $\deg(\Phi)=\sum_{P\in \Phi^{-1}(Q)}e_{\Phi}(P)$ for any $Q\in C_{2}$. I don't see why in general the inertia degree should be 1 and not appear in the well known degree formula as it does in standard Number Theory. This is what I mean: as the curves are smooth the regular functions rings are Dedekind domains and we have the unique factorization $M_{\Phi(P)}=\prod_{P′\in \Phi^{-1}(\Phi(P))}M_{P'}^{e_{\Phi}(P′)}$ being $M_{\Phi(P)}$ and $M_{P}$ the maximal ideals associated to $\Phi(P)$ and $P$ respectively (as $\Phi$ is non constant $\Phi^{*}$ is injective and embeds $K[C_{2}]$ in $K[C_{1}]$). In the degree formula the inertia degree should appear by standard Number Theory, which is not the case. REPLY [15 votes]: The reason is that there are two ways of thinking about "points". Let $A$ be a ring. Then, define: A scheme-theoretic/topological point of Spec $A$ is a prime ideal of $A$. A geometric/functorial point of Spec $A$ is an equivalence class of morphisms Spec $K\rightarrow$ Spec $A$, where $K$ is a field, and two morphisms: $$p_1 : \text{Spec }K_1\rightarrow \text{Spec }A,\qquad p_2 : \text{Spec }K_2\rightarrow\text{Spec }A$$ are equivalent if there exists a field $K$ containing both $K_1,K_2$ as subfields, and a morphism Spec $K\rightarrow$ Spec $A$ making the obvious diagrams commute. Note that if $A$ is a $k$-algebra, then geometric points of Spec $A$ are just morphisms Spec $\overline{k}\rightarrow\text{Spec }A$. Further, if $A$ is a finite type $k$-algebra (for example $k[x,y]/(f)$) then a geometric point of $\text{Spec }k[x,y]/(f)$ is nothing but a pair $(a,b)\in\overline{k}^2$ satisfying the equation $f$, which is precisely how Silverman defines a "point". Here's an example that will explain everything: Let $k = \mathbb{Q}$, and let $A = \mathbb{Q}[x]$, $B = \mathbb{Q}[y]$, and suppose $X = \text{Spec }A$ and $Y = \text{Spec }B$. Consider the map $f : Y\rightarrow X$ given by $x\mapsto y^2$. This map is degree 2, and sends $y = a$ in $Y$ to $x = a^2$ in $X$. First, lets consider the scheme-theoretic/topological point $q\in X$ corresponding to $x = -1$. Thus, $q$ is the prime ideal $(x+1)\subset\mathbb{Q}[x]$. It's not hard to check that the only prime of $B$ lying over $q$ is $p := (x^2+1)$, whose residue field is $\mathbb{Q}(i)$, so $p$ has "inertia degree" 2 over $q$. Now, lets consider the geometric point $Q : $Spec $\overline{\mathbb{Q}}\rightarrow$ Spec $A$ given by sending $x\mapsto -1$. It's not hard to see that sending $y\mapsto i$ and sending $y\mapsto -i$ define two distinct morphisms $P_1,P_2 : \overline{\mathbb{Q}}\rightarrow$ Spec $B$ (ie, two distinct geometric points) which lie above $Q$. Thus, while there is only one topological point above $x = -1$, there are two geometric points, each of which makes an appearance as a summand of the equation in Prop 2.6. This makes up for the fact that the inertia degree doesn't appear. In fact, one may define the inertia degree as the number of geometric points lying over $x = -1$ who have the same image $P\in Y$. Ie, in the situation of our morphism $f : Y\rightarrow X$, the two equivalent ways of writing the formula would be: $deg(f) = 2\cdot e_f(p) = 2\cdot 1$ (2 is the inertia degree), or $deg(f) = e_f(P_1) + e_f(P_2) = 1 + 1$<|endoftext|> TITLE: Can we reconstruct the region in the xy plane by length measurements? QUESTION [6 upvotes]: Consider a closed smooth bounded curve enclosing a region $S$ in the XY-plane $\mathbb{R} ^2$. We define the function $f(x)$, where $x$ is a point on the $x$ axis, as the length of the intersection of the line paralell to the $y$-axis which goes through $x$, with the set $S$, so it can have multiple components and we know the total length. Suppose we know the same function for some coordiante systems rotated relative to the xy system. Can we reconstruct $S$ using only this data? How many coordinate systems do we need? REPLY [3 votes]: A key search term is geometric tomography. The problem posed is a version of Hammer’s X-ray reconstruction problem: Gardner, Richard J., and Markus Kiderlen. "A solution to Hammer's X-ray reconstruction problem." Advances in Mathematics 214.1 (2007): 323-343. PDF download                     Two shapes with the same parallel X-rays.<|endoftext|> TITLE: How can I verify that a given solution of the Quantum Yang-Baxter equation is associated to a given Lie algebra? QUESTION [6 upvotes]: Take, for instance, the $R$ matrix, \begin{equation} R(u)=\begin{pmatrix}u+1 & 0 & 0 & 0\\0 & u & 1 & 0\\0 & 1 & u & 0\\0 & 0 & 0 & u+1\end{pmatrix}, \end{equation} wich satisfies the Yang-Baxter equation \begin{equation} R^{12}(u-v)R^{13}(u)R^{23}(v)=R^{23}(v)R^{13}(u)R^{12}(u-v). \end{equation} This $R$ matrix is associated to the $sl(2)$ Lie algebra, but how can I prove that? Actually I have a very complicated $R$ matrix: I'm looking for a general guide to find the Lie algebra associated to a given $R$ matrix. In other question posted here, (Solutions of the Quantum Yang-Baxter Equation) I found the comment: "The general theory (due to Jimbo) is that each irreducible finite dimensional representation of the quantised enveloping algebra of a Kac-Moody algebra (not of finite type) gives a trigonometric R-matrix." I think I'm looking for the converse of this: given a $R$-matrix satisfying the Yang-Baxter equation, what is the Lie algebra associated to it? Thank you in advance. REPLY [2 votes]: To the best of my knowledge this is a very hard problem and the answer to this question is, unfortunately, open. A famous example that illustrates this in the context of quantum integrability comes from the one-dimensional Hubbard model in condensed-matter physics. Its (quite complicated) $R$-matrix was known since the late eighties, yet the corresponding quantum group was only found in the last decade, "by accident" in a very different context. Namely: the $R$-matrix of the Hubbard model came out of the study of the quantum group associated to a Lie (super)algebra, and the choice of the latter was motivated by string theory (more precisely: the AdS/CFT correspondence). For a bit more about this, and further references, see for example (the introduction of) arXiv:1509.06205.<|endoftext|> TITLE: Constant related to continued fraction of quadratic irrationals QUESTION [5 upvotes]: Let $d$ be a positive, non-square integer, and define $c_d$ to be the smallest positive number with the following property: for all pairs of co-prime integers $(p,q)$ with $q > 0$, the inequality $$\displaystyle \left \lvert \frac{p}{q} - \sqrt{d} \right \rvert > \frac{c_d}{q^2}$$ holds. The existence of such a number follows from Lagrange's theorem that the continued fraction expansion of a quadratic irrational is eventually periodic. It follows from Hurwitz's theorem that $c_d < \dfrac{1}{\sqrt{5}}$ for all $d$. Does anyone have a reference for the size of $c_d$? One can assume that the fundamental solution $(u_0, v_0)$ to the equation $x^2 - dy^2 = \pm 4$ is known. REPLY [6 votes]: This feels like the kind of thing that's easier to figure out than find in the literature. Seems that the optimal $c_d$ comes from the minimal positive solution of $p^2 - d \, q^2 = +1$, and is thus less than $\frac1{2\sqrt d}$ but very close to $\frac1{2\sqrt d}$ if the fundamental solution is at all large. REPLY [5 votes]: Edit: I answered a different question than the one asked. The question asks for the least $c_d = q^2|\sqrt{d}-p/q|$. I evaluated the liminf. If $d = a^2+b$ with $1\le b \le 2a$ then the simple continued fraction for $\sqrt{d}$ is preperiodic, and the period ends with $2a$, which is the largest coefficient, with $a$ being the second largest, and the rest of the period is a palindrome. I believe these classical results are in Hardy and Wright, An Introduction to the Theory of Numbers. We can represent $\sqrt{d}$ as a terminating simple continued fraction if we allow the last coefficient to be irrational, containing the rest of the simple continued fraction, instead of an integer. $\alpha = [a_0;a_1,...,a_n,...] = [a_0;a_1,...,a_n,[a_{n+1};a_{n+2},...]]$. Let $a_{n+1}' = [a_{n+1};a_{n+2},...]$. If $[a_0;a_1,...,a_n] = p_n/q_n$ and then $\alpha = \frac{a_{n+1}'p_n + p_{n-1}}{a_{n+1}'q_n + q_{n-1}}$ so $$\bigg|\alpha - \frac{p_n}{q_n}\bigg| = \frac{1}{q_n(a_{n+1}'q_n + q_{n-1})} = \frac{1}{q_n^2} \frac{1}{a_{n+1}' + \frac{q_{n-1}}{q_n}}$$ So, we want to determine the liminf of $\frac{1}{a_{n+1}' + q_{n-1}/q_n}$ For $\alpha = \sqrt{a^2+b}$, or equivalently, the limsup of the reciprocal $a_{n+1}' + q_{n-1}/q_n$. The only values greater than $2a$ occur just before the period ends, when the next coefficient is $2a$, where $a_{n+1}' = a+\sqrt{a^2+b}$. The continued fraction of $q_{n-1}/q_n$ is $[0;a_n,a_{n-1},...,a_1]$ so the limit is $\sqrt{a^2+b}-a$, which means $$\frac{1}{a_{n+1}' + q_{n-1}/q_n} \to \frac{1}{2\sqrt{a^2+b}} = \frac{1}{2\sqrt{d}} = c_d.$$ If we want the infimum instead of the liminf, we get slightly lower values. Then $$c_d = \frac{1}{\frac{p}{q} + \sqrt{d}}$$ where (p,q) is the smallest nontrivial solution to $p^2-dq^2=1$. This is a convergent to the simple continued fraction just before the end of the first period if the period is even, and just before the end of the second period if the period is odd. For example, $\sqrt{3} = [1; \overline{1,2}]$ has an even period length, so $[1;1]=2 \gt \sqrt{3}$. Then $c_3=1^2|2-\sqrt{3}| = \frac{1}{2+\sqrt{3}} \lt \frac{1}{2\sqrt{3}}$. $\sqrt{2} = [1;\overline{2}]$, so $p/q = [1;2] = 3/2 \gt \sqrt{2}$, and $c_2=2^2(3/2-\sqrt{2}) = \frac{1}{3/2 + \sqrt{2}} \lt \frac{1}{2\sqrt{2}}$. The computation above still applies. The only candidates are convergents just before the period ends. If $p_n/q_n$ is the convergent just before the period ends, then $q_n^2|p_n/q_n - \sqrt{d}| = \frac{1}{a_{n+1}' + q_{n-1}/q_n}$ and $q_{n-1}/q_n$ has a continued fraction that consists of $0$ followed by repetitions of the period of the simple continued fraction for $\sqrt{d}$, with the last $2a$ missing. This is a convergent to the simple continued fraction of the fractional part $\sqrt{d}-\lfloor \sqrt{d} \rfloor$. When $n$ is odd, $q_{n-1}/q_n > \sqrt{d}-\lfloor \sqrt{d} \rfloor$, and the odd convergents decrease toward $\sqrt{d} - \lfloor \sqrt{d} \rfloor$ so the maximum occurs at the first occurence of an odd index just before a period ends.<|endoftext|> TITLE: Graph minors, and Kronecker product QUESTION [6 upvotes]: Let $X$ and $Y$ be graphs and consider the Kronecker product: $Z = X \otimes Y$. Is it true that if $X$ excludes an $M$-minor, $Z$ excludes an $M \otimes Y$ minor? I am particularly interested in the case where $Y$ is just an edge, and $Z$ is just the bipartite double cover of $X$. REPLY [3 votes]: The diamond cubic is a subgraph of the Kronecker product of three infinite paths, and $K\times K\times K$ patches of the diamond cubic are subgraphs of the Kronecker product of three length-$K$ paths. But there are no forbidden minors for diamond cubics (one way to see this is that they have treewidth $\Omega(K^2)$ whereas a graph with $K^3$ vertices in a family with a forbidden minor would necessarily have treewidth $O(K^{3/2})$). So the existence of a forbidden minor is not preserved by arbitrary Kronecker products. As for bipartite covers: the bipartite double cover of an infinite tiling of the plane by equilateral triangles (or a large patch of it) includes subdivisions of any finite cubic graph, obtained by drawing the graph with crossings in the plane, rounding the drawing to the triangle tiling, using the double cover to handle crossings without creating intersections, and adding little triangle loops where necessary to change between the two levels of the graph. But every graph is a minor of a cubic graph. So, again, bipartite double covers do not preserve the existence of forbidden minors.<|endoftext|> TITLE: Small fppf/syntomic/smooth sites? QUESTION [5 upvotes]: I was reading the section about sites in the Stacks Project. Small sites are studied only in few cases, but I don't get why. As some users say, it's because of this. If $f: X\to Y$ is a morphism of schemes, it's known it's not generally true that the small fppf inverse image functor $f^{-1}$ is exact, as opposed to $f_{small, etale}^{-1}$, $f_{small, Zariski}^{-1}$, $f_{small, crystalline}^{-1}$, and all the big inverse images. Any explicit counterexample for $f^{-1}_{small, fppf}$, $f^{-1}_{small, smooth}$, $f^{-1}_{small, syntomic}$, etc. ? Related - Leo Alonso's comment to Etale site is useful - examples of using the small fppf site? Also related - Why is lack of functoriality of the Lisse-Etale topology specific to the Lisse-Etale topology? It'd be nice to collect some examples here. EDIT. Merlin's example says it all. $f^{-1}_{small, *}$ is not exact for $*\in\{smooth, syntomic,fppf, fpqc, Lisse-\acute{E}t, Flat-fppf\}$, so this is why Grothendieck switches to big sites, as Leo Alonso says in his comment (see first link). REPLY [5 votes]: Good point. Meanwhile I'll say a few things, which I plan to make into an answer at some point, though I suggest you wait for a bunch of examples to come from users all over. To fix ideas, we consider the category of all smooth $S$-schemes with arbotrary $S$-morphisms between them, denoted $\mathcal{C}$, and endow it with the smooth topology. The resulting site will be the small smooth site for us $S_{\text{smooth}}$ (details omitted on checking this is indeed a site). Given a morphism of schemes $f : S'\to S$, asking whether $f^{-1}$ is exact is really a question about $\mathcal{C}$ and not the topology $\mathcal{C}$ is endowed with. For $f^{-1}$ to be exact, the base-change functor has to commute with all finite limits, hence finite limits must, first, exist in $\mathcal{C}$. This is equivalent to asking that all finite products and equalizers exist in $\mathcal{C}$. Here we find a first obstacle: if you equalize two $S$-morphisms $a,b: U\to V$ in the category of $S$-schemes, for objects $U\to S$ and $V\to S$ in $\mathcal{C}$, the equalizer $E$ of $a,b$ in schemes over $S$ won't be $S$-smooth in general, ie. $\mathcal{C}$ doesn't contain equalizers, so we're screwed (strictly speaking, $\mathcal{C}$ may still contain equalizers, and they may not agree with equalizers in the category of schemes. You should check this is not the case). For an explicit example, you may take a ring $R$, $S = \text{Spec}(R)$, $U = V = \mathbb{A}^1_S = \text{Spec}(R[t])$, and the maps: $a : U\to V$ given by $t\mapsto 0$, and $b: U\to V$ given by $t\mapsto t^2$. Their equalizer in the category of $S$-schemes is $E := \text{Spec}(R[t]/(t^2))\to S$, which is not $S$-smooth.<|endoftext|> TITLE: Can these two irreducible $GL_n \mathbb Z$-representations be isomorphic? QUESTION [6 upvotes]: Fix $n\in \mathbb N$ and a partition $\lambda$ with at most $n-1$ parts (of length at most $n-1$). Let $V$ be the irreducible $GL_n \mathbb R$-representation with highest weight $\lambda$ and $D$ the determinant representation. Is it possible that $$ Res_{GL_n \mathbb Z} V \cong Res_{GL_n \mathbb Z} (V \otimes D)?$$ REPLY [6 votes]: First of all, if $V$ is not irreducible, this can happen. Every matrix in $GL_n(\mathbb{Z})$ has determinant $\pm 1$ so $D^{\otimes 2}|_{GL_n(\mathbb{Z})}$ is trivial. Therefore, $(1 \oplus D) \cong (1 \oplus D) \otimes D$ on $GL_n(\mathbb{Z})$. However, with the stated hypothesis that $V$ is the irrep of $GL_n$ with highest weight $\lambda$, this is true. Note that the Schur polynomial $s_{\lambda}(x_1, x_2, \ldots, x_n)$ is homogenous of degree $|\lambda|$. Therefore, it cannot be divisible by the inhomogoneous polynomial $1+x_1 x_2 \cdots x_n$. Write $s_{\lambda}$ as a polynomial in the elementary symmetric polynomials $e_1$, $e_2$, ..., $e_n$, we have just shown that it is not divisible by the polynomial $1+e_n$. Therefore, we can find integers $(f_1, f_2, \ldots, f_n)$ with $f_n=-1$ so that $s_{\lambda}$ does not vanish when the $f_j$ are plugged in for the $e_j$. Let $g \in GL_n(\mathbb{Z})$ be a matrix with characteristic polynomial $x^n - f_1 x^{n-1} + f_2 x^{n-2} - \cdots + (-1)^n f_n$ (for example the companion matrix), and let $\alpha_1$, ..., $\alpha_n$ be its roots. Then $\mathrm{Tr} \ \rho_{V}(g) = s_{\lambda}(\alpha_1,\ldots,\alpha_n) \neq 0$. We have $\mathrm{Tr} \ \rho_{V \otimes D}(g) = (\det g) \mathrm{Tr} \ \rho_{V}(g) = -\mathrm{Tr} \ \rho_{V}(g)$. So $V$ and $V \otimes D$ have different characters at $g$ and are not isomorphic.<|endoftext|> TITLE: Fermat's opponents QUESTION [14 upvotes]: It is well known among historians of Fermat that, while his technique of adequality prepared the ground for the general framework later developed by Leibniz and Newton, Fermat himself gave very little in the way of explanation of his technique exploiting a symbol $E$ that appears to us to behave like an infinitesimal. There is an interesting issue concerning Fermat's opponents and whether he may have felt a need to be watchful in his actions correspondingly, and specifically with regard to how he presented his method of adequality that could potentially run the risk of being found heretical because of its connection to the infinitely small, indivisibles and the decomposition of the continuum into indivisibles, and atomism. The latter were thought to be at tension with established canon (specifically Trent 13.2 and the problems atomism constitutes for transubstantiation and eucharist similar to Grassi's criticism of Galileo, as discussed by both Redondi and Festa) and were repeatedly banned; for details see the books by Redondi and Alexander. (1) Fermat was a judge on the Parlement de Toulouse. He must have aggravated some people at the Parlement de Toulouse by seeking out help in Paris to get the appointment on the Chambre d'Edit in Castres. The king overruled the decision of the Parlement and named Fermat instead. (2) The president of the Parlement seems to have been an opponent of Fermat. There is a document that was once secret that contains a character report on Fermat describing him as "interessé" and implying that he was unreliable. The damaging secret report was written by a crony of the President's. (3) Descartes attacked Fermat's technique of adequality. This must have been well known at the time because Descartes was famous. It would be interesting to determine whether Fermat's enemies ever used this against him. (4) According to Barner's interpretation of a capital case at the Parlement de Toulouse involving a priest named Delpoy, the latter was executed apparently over helping a Calvinist girl hide after she escaped from a re-education school. The execution apparently went against Fermat's recommendations. Fermat was reporting on this case but was not the one who ruled to execute the priest. (5) According to Barner, Fermat was shocked by the Delpoy thing and couldn't concentrate on his math for at least a month, according to Kenelm Digby. There was a correspondence between Wallis and Fermat at the time through Digby that came to a standstill following the Delpoy affair. (6) Barner also documents a sharp drop in Fermat's productivity at the Parlement for a month following the execution of Delpoy. (7) As far as Weil's comment with regard to Mahoney's claims concerning Digby's letter to Wallis: Weil wrote: "an inquisitive historian might do worse than try to find out whether the above story was not a figment of [DIGBY'S] lively imagination." While Mahoney did not put in such an effort, Barner did, and found archival evidence that Fermat was in distress over the Delpoy affair. I am seeking such information about Fermat's possible opponents and their possible actions against him in the context of sensitive events such as the indictment and execution of the priest Delpoy in order to gauge whether Fermat may have also been hesitant to be too frank about the foundations of his method of adequality because of possible theological risks involved and the possible use of this by his opponents. Do we have evidence of such actions by his opponents beyond the president's attempt to block Fermat's appointment at Castres, Descartes' attack against adequality, and disagreements over Delpoy? Note 1. See this related thread on Redondi, Galileo, and atomism. Note 2. See a related 2018 publication in Foundations of Science. REPLY [11 votes]: The point (5) is at most doubtful. It was made by Mahoney in his book on Fermat, but Mahoney's interpretation of what Digby writes to Wallis does not correspond to what Digby actually writes, as quoted by Weil in his review cited in comments: I have had nothing from him [Fermat] but excuses . . . . It is true it came to him upon the nick of his removing his seat of Judicature from Castres to Tholose; where he is supreme Judge in the soveraign Court of Parliament. And since that, he hath been taken up with some Capital causes of great importance; in which in the end he hath given a famous and much applauded sentence for the burning of a Priest that had abused his function; which is but newly finished; and execution done accordingly. But this which might be an excuse to many other, is none to Mons. Fermat, who is incredibly quick and smart in any thing he taketh in hand. So Fermat did not answer with his usual celerity to mathematical questions, but there is nothing but Mahoney's imagination that indicates that it was because he was shocked by a judicial decision. The point (3) might be valid, but remember that Descartes had much more serious problems with (ecclesiastic and otherwise) authorities than Fermat ever had (if indeed he had any), so it seems unlikely that his "ennemies" would use Descartes against him. (1), (2) and (4) are unrelated to the mathematical work of Fermat. Given the evidence provided, I don't see any reason to think hat Fermat was afraid of being found heretical for some part of his mathemamatical work. I understand you are asking for more evidence concerning this supposition, which I have not, pointing in one direction or te other, but that supposition seems to me rather arbitrary at this point.<|endoftext|> TITLE: Reference book for Galois Representations QUESTION [8 upvotes]: I am an undergrad. I have taken courses in algebraic number theory and have a basic idea about $p$-adic numbers. I have also read a little bit of infinite Galois theory. But I have no idea about modular forms and automorphic forms and Galois cohomology. I want to study about Galois Representations. All the books I have seen have titles like [some topic] and Galois Representations, where some topic=modular forms or automorphic forms. Is there any textbook or lecture notes which does not require automorphic forms or modular forms as a prerequisite (suited especially for beginners)? REPLY [5 votes]: You can see the book of J.P.Serre "Abelian l-adic representations and elliptic curves" a very nice presentation of Galois representations and properties !<|endoftext|> TITLE: Conformal blocks in genus zero QUESTION [8 upvotes]: In section 10.4 of "Vertex Algebras and Algebraic Curves", Ben-Zvi & Frenkel (second edition), the authors claim that for any vertex algebra V, the space of one-pointed conformal blocks with insertions in $V$ $C(\mathbb{P}^1,0,V)$ is one-dimensional. They justify this by saying that the Lie algebra $U_{\mathbb{P}^1/0}(\mathcal{V}_0)$ is generated by the Fourier coefficients $A_{(n)}$ for $n<0.$ ($U_{\mathbb{P}^1/0}(\mathcal{V}_0)$ is such that $\mathcal{V}_0/U_{\mathbb{P}^1/0}(\mathcal{V}_0)\mathcal{V}_0$ is the space of coinvariants.) I don't see why this is true (and the later calculations in the section don't seem to justify this.) Also the statement as they wrote it seems false, as I believe $A_{(-1)}$ for $A$ the vacuum is the identity map? REPLY [7 votes]: Well, the claim is bogus, so you can't expect the proof to hold much water. On the other hand, it may be instructive to try filling in details to see why it fails. First of all, we can't define conformal blocks for arbitrary vertex algebras - we need Möbius structure for genus zero, and quasiconformal structure for higher genera. Since we are working on $\mathbb{P}^1$ the bundle $\mathcal{V}$ splits as a sum of line bundles $\bigoplus_{n \in \mathbb{Z}} V_n \otimes \Omega^{-n}$, so the Lie algebra is given by global sections of $$(\bigoplus_n V_n \otimes \Omega^{1-n})/(T \otimes 1 + 1 \otimes d)(\bigoplus_n V_n \otimes \Omega^{-n}).$$ Choosing a global coordinate $t$, we find that in order for a polydifferential of the form $A \otimes t^k(dt)^{1-n}$ to be regular at infinity, it is necessary and sufficient that $2n-2-k \geq 0$. That is, we only get an action of the coefficient $A_{(k)}$ for $A$ of degree $n$ when $k \leq 2n-2$, so in particular, the vacuum cannot act as identity. On the other hand, any positive degree vector $A$ can act on the vacuum by $A_{-1}$, so the vacuum axiom says that $A$ lies in the image. We conclude that if $V$ is non-negatively graded and the vacuum spans the degree 0 subspace (i.e., $V$ is of "CFT type"), then we get a 1-dimensional space of conformal blocks. In general, we can identify the 2-point conformal block space with the space of invariant bilinear forms on $V$, and by a 1994 theorem of Li, this is dual to $V_0/L_1V_1$. Using the "propagation of vacuum" property (Theorem 10.3.1), this is also isomorphic to the 1-point conformal block space.<|endoftext|> TITLE: (ir)reducibility of projective representation when restricted to abelian subgroup QUESTION [6 upvotes]: I am looking for a finite group $G$ and an irreducible projective representation $\rho: G \to PGL(\mathbb C^n)$ such that for any abelian subgroup $A\subset G$, the restricted representation $\rho|_A$ is reducible. (EDIT: Moreover $\rho$ should be a `genuine' projective representation, in the sense that it represents a non-trivial cohomology class. In other words I am not allowing $\rho$ to be equivalent to a linear representation.) More prosaically, I'm looking for a projective representation that is truly dependent on the non-abelian structure of the original group. Of course it would be nice if the example were not a horribly complicated group; the 'simpler' the better. Alternatively I would of course greatly appreciate if someone could get me on the way to derive such an example myself. REPLY [7 votes]: The group ${\rm SL}(2,7)$ has a faithful complex irreducible representation of degree $4$. Viewing this as a projective representation of $G = {\rm PSL}(2,7)$ gives an example where every Abelian subgroup of $G$ acts reducibly, but $G$ acts irreducibly. This is because the only non-cyclic Abelian subgroups of $G$ are Klein $4$-groups, and the largest degree of a complex irreducible projective representation of a Klein $4$-group is $2$.<|endoftext|> TITLE: Turan-like bound for $k$-partite graphs QUESTION [8 upvotes]: This is a question I've implicitly asked on AoPS five years ago, and has not been answered. I apologize for its possible simplicity, as I am not a graph theorist. Let $n$ and $k$ be two positive integers. Let $G$ be a graph with vertex set $V$. Assume that $V$ is partitioned into $k$ disjoint subsets $V_1, V_2, \ldots, V_k$, each of which has exactly $n$ elements (so that $\left|V\right| = nk$). Assume that for each $i \in \left\{1, 2, \ldots, k\right\}$, the set $V_i$ is an independent set of $G$ (that is, no two vertices of $V_i$ are connected by an edge). Assume also that each vertex of $G$ has degree $\geq \left(k-2\right)n+1$. Prove or disprove that $G$ has a $k$-clique (that is, $k$ distinct vertices pairwise connected by edges). This is a triviality when $k = 1$ or $k = 2$, and a known contest problem in the case when $k = 3$ (see http://artofproblemsolving.com/community/c6h40982 , and http://artofproblemsolving.com/community/c6h49347 for an extension which is not completely true as stated). In my AoPS thread linked above, I have asked whether the $k = 4$ case is true, but of course the real question is the general case. REPLY [3 votes]: It is not true in general that $G$ under your conditions contains a $k$-clique. For example, if $k = 4$ consider the following construction by Gouping Jin (see the proof of Theorem 3.1 in [1]): Let $V_i = A_i \cup B_i$ for $i \in \{1,2,3,4\}$, where $\left|A_i\right| = \lfloor n/3 \rfloor$ and $\left|B_i\right| = n - \lfloor n/3 \rfloor$. Join vertices so that the following pairs of sets induce complete bipartite subgraphs: $(A_1,V_2 \cup B_3 \cup B_4)$, $(A_2,V_1\cup B_3 \cup B_4)$, $(A_3,V_4\cup B_1 \cup B_2)$, $(A_4,V_3 \cup B_1 \cup B_2)$ and $(B_1 \cup B_2, B_3 \cup B_4)$ (i.e. join every vertex of $A_1$ with every vertex of $V_2 \cup B_3 \cup B_4$, and so on). The resulting graph, $H_n$, is clearly $4$-partite and it is not difficult to see that it has minimum degree $\delta(H_n) = \lfloor (2 + 1/3)n \rfloor$. Furthermore, $H_n$ contains no $4$-clique. Now, since $\lfloor (2 + 1/3)n \rfloor \geq (4-2)n + 1$ for $n \geq 3$ this gives us examples of graphs that satisfy your conditions but do not contain a $4$-clique. In fact, Jin shows in [1] that the best lower bound (to guarantee a $k$-clique) on the degrees of a $k$-partite graph with each part of size $n$ is $\lfloor (2 + 1/3)n \rfloor + 1$ for $k = 4$, $\lfloor (3 + 1/3)n \rfloor + 1$ for $k = 5$ and $\leq \left(k - 1 - \frac{1}{1 + 1/2 + 1/3 + \dots + \frac{1}{k-2}} \right)n + 1$ for all $k \geq 4$. [1] G. Jin, "Complete Subgraphs of $r$-partite Graphs", Combinatorics, Probability and Computing 1, (1992), 241-250.<|endoftext|> TITLE: A strong annulus theorem for 3-manifolds QUESTION [8 upvotes]: A strong version of the loop theorem implies that if an essential closed curve on the boundary of a 3-manifold $M$ is nullhomotopic, then realizing its image in $\partial M$ as a 4-valent graph, you can draw a cycle with no edge repeats that is homotopic on $\partial M$ to the boundary of an essential embedded disk in $M$. (Compare with Bing's "The geometric topology of 3-manifolds", pg 205.) Q: Is there a similarly strong version of the annulus theorem? For instance, suppose you have a (say, hyperbolizable) 3-manifold M with incompressible boundary, and you have two closed curves on the boundary of M that bound an essential singular annulus in M. (E.g., they're homotopic in M but not within its boundary.) Homotoping them to be in general position, can you find two cycles with no edge repeats on the 4-valent graph that is the union of their images, such that the two corresponding curves are homotopic to the two boundary components of an essential embedded annulus in M? The closest thing I've found is Thm 2 in Cannon-Feustel's "Essential embeddings of annuli...", but that requires the two original curves to be disjoint. REPLY [6 votes]: I think this might follow from JSJ theory. Assume that $M$ is irreducible with incompressible boundary. Then any essential annulus is homotopic into an $I$-bundle region or a Seifert-fibered region of the JSJ decomposition. In the Seifert case, the region meets the boundary in annuli, in which case the boundaries of the immersed annulus will be a multiple of the core of the annuli, and I think one can cut and paste to get a simple closed essential core of each annulus, which then cobound an embedded essential annulus. Otherwise, if the annulis lies in a product region, then your immersed annulus goes between two curves on the surface. So this amounts to asking: for a given homotopy class of curve on a surface, can one always cut and paste a subset of the curve to give an embedded curve in the same isotopy class? This seems plausible, e.g., if the curve is not filling (one should be able to cut and paste to get a boundary component of the subsurface that it fills). In any case, it seems that at least this gives a reformulation of the problem (although I'm not sure what happens for twisted $I$-bundles). REPLY [5 votes]: I think that Theorem 15.1.5 in Scott's unpublished book on three-manifolds (email him for a copy) is very close to what you want.<|endoftext|> TITLE: Is forming the Albanese variety a monad? QUESTION [25 upvotes]: I'm trying to understand the idea of an Albanese variety. It reminds me of something simpler: Given a set $X$ with a chosen point $x \in X$, we can form the free abelian group on the pointed set $(X,x)$, which is just the free abelian group on $X$ modulo a relation saying $x = 1$. If we call this group $A(X,x)$, it has a nice universal property: any map of pointed sets $f$ from $X$ into an abelian group $A$ factors uniquely as the obvious inclusion $$i_X \colon X \to A(X,x)$$ followed by some homomorphism $$\overline{f} \colon A(X,x) \to A.$$ So: $$ f = \overline{f} \circ i_X $$ The process of taking the free abelian group on a pointed set defines a functor $$ A \colon \mathrm{Set}_* \to \mathrm{AbGp} $$ which has a right adjoint $$ U\colon \mathrm{AbGp} \to \mathrm{Set}_* $$ sending any abelian group $A$ to its underlying pointed set $(A,1)$. The composite $$ U A \colon \mathrm{Set}_* \to \mathrm{Set}_* $$ is thus a monad, and if I'm not mistaken, the algebras of this monad are just abelian groups. The idea of an Albanese variety seems to be similar, but working with projective algebraic varieties instead of sets. Namely: Given any such variety $X$ with a chosen point $x$ there is an abelian variety called the Albanese variety $A(X,x)$, apparently defined by the following universal property: there is a map of varieties $$i_X \colon X \to A(X,x)$$ such that any map of pointed varieties $f$ from $X$ into an abelian variety $A$ factors uniquely as $i_X$ followed by some map of abelian varieties $$\overline{f} \colon A(X,x) \to A.$$ So: $$ f = \overline{f} \circ i_X $$ (The Wikipedia article on Albanese varieties doesn't clearly require that $\overline{f}$ be a map of abelian varieties, but Ravi Vakil's lecture notes do, so I'm going with that.) So, naturally, I'm wondering: does taking the Albanese variety define a functor from pointed varieties to abelian varieties, which has a right adjoint, which together define a monad on the category of pointed varieties whose algebras are the abelian varieties? The so-called Albanese map $i_X \colon X \to A(X,x) $ would then be the unit of this monad, while the multiplication of the monad would be the map of abelian varieties $\overline{1} \colon A(A(X,x),x) \to A(X,x) $ obtained from the identity map of varieties $1 \colon A(X,x) \to A(X,x)$. REPLY [12 votes]: I'll go ahead and turn my comment into an answer. It does form a monad, but (probably) not a very interesting one. Namely, first note that any pair of adjoint functors $L:\mathcal{C}\leftrightarrows \mathcal{D}:R$ is associated to a monad $RL$ on $\mathcal{C}$. Multiplication is given by the map $RLRL\overset{LR\to 1}{\to} RL$ and unit is $1\to RL$. Here the natural transformations $1\to RL$ and $LR\to 1$ (with $1$ denoting the identity functor) are the unit and counit of the adjunction, adjoint to the identity maps $L\to L, \text{ } R\to R$, respectively. Further, given an object $A$ of $\mathcal{D}$, we get an algebra $R(A)$ over the monad $RL$ with action morphism $RLR(A)\overset{LR\to 1}{\to} R(A)$ induced from the counit. This gives us a functor $\alpha:\mathcal{D}\to Alg(RL)$ to algebras over the monad $RL$. In "most real-life situations" where the functor $R$ is "forgetful" and $L$ is "free", the functor $\alpha$ to algebras is in fact an equivalence of categories. This can be checked formally, using something called the Bar-Beck monoidicity theorem (in a sense this is similar to the condition for an abelian category $\mathcal{A}$ to be the catgeory of modules over an algebra: in fact, modules over an algebra are a special case of algebras over a monad). The monad you are considering is associated to the adjunction $A:Sp_*\leftrightarrows Ab:F$, where $F:Ab\to Sp_*$ is the forgetful functor from abelian varieties to pointed spaces and $A:Sp_*\to Ab$ is the Albanese functor. The monad is then $FA:Sp_*\to Sp_*$. The reason this monad is not very interesting is that the forgetful functor $F$ is fully faithful (this result in itself is very interesting, and encapsulates the rigidity of abelian varieties in algebraic geometry). This implies that the composition $FA:Sp_*\to Sp_*$ is an idempotent functor. In particular, here we can apply the theory of idempotent monads, see e.g. https://ncatlab.org/nlab/show/idempotent+monad. Part 6 of definition 2.1 in the above notes then gives us for free that the adjunction $A:Sp_*\leftrightarrows Ab:F$ is monadic, i.e. the category $Ab$ of Abelian varieties is indeed equivalent to the category of algebras over the monad $FA$.<|endoftext|> TITLE: Hyper-symmetric polynomials (reference request) QUESTION [9 upvotes]: Let $M_n$ be the linear space of $n\times n$ matrices. The product of symmetric groups $S_n\times S_n$ acts naturally on $M_n$, and thus induces an action on the coordinate algebra $k[M_n]$. Is there any research on the invariant algebra $k[M_n]^{S_n\times S_n}$ in the literature? Does it have a name so that I may google it? I am also interested in the invariant algebra for the diagonal action of $S_n$. Thank you. REPLY [4 votes]: Here is a way to obtain some information on the invariant algebra. Let $G$ denote the permutation group isomorphic to $S_n\times S_n$ acting on an $n\times n$ array by permuting rows and columns. If $w\in G$, then let $c_i(w)$ denote the number of cycles of length $i$ in $w$. Let $f(r)$ denote the dimension of the degree $r$ part of $k[M_n]^{S_n\times S_n}$. By Molien's theorem (e.g., https://en.wikipedia.org/wiki/Molien_series), we have $$ \sum_{r\geq 0}f(r)x^r = \frac{1}{n!^2} \sum_{w\in G}\frac{1}{\prod_{i\geq 1}(1-x^i)^{c_i(w)}}. $$<|endoftext|> TITLE: A rare property of Hausdorff spaces QUESTION [25 upvotes]: Is there a Hausdorff topological space $X$ such that for any continuous map $f: X\longrightarrow \mathbb{R}$ and any $x\in \mathbb{R}$, the set $f^{-1}(x)$ is either empty or infinite? REPLY [3 votes]: If $(X,\mathcal{S})$ is a regular space, then the $P$-space coreflection of $(X,\mathcal{S})$ is the space generated by the basis $\{\bigcap_{n\in\omega}U_{n}\mid\forall n\,U_{n}\in\mathcal{S}\}$. A completely regular space $X$ satisfies the property that $f^{-1}(\{x\})$ is empty or infinite for any real $x$ if and only if the $P$-space coreflection of $X$ has no isolated points. There exists infinite regular spaces $X$ where every continuous function $f:X\rightarrow\mathbb{R}$ is constant and in such spaces $f^{-1}(x)$ is always empty or infinite.<|endoftext|> TITLE: "Universal embedding structures" in a general setting? QUESTION [7 upvotes]: There are a number of famous results to the effect that "countable structures" of a certain type have a universal, "homogeneous" structure of the same type into which they all embed with various nice properties. For example, Urysohn's metric space, Rado's graph and Higman's finitely-presented group all follow this pattern. Cameron in "The Rado graph and the Urysohn space" notes some relationships between those two objects, while Aksoy, Glassman, Kosheleva and Kreinovich in "From Urysohn's Universal Metric Space To A Universal Space-Time" show that Urysohn's construction can be converted into a version for a suitable definition of "spacetime". Given the analogies between these constructions, the following questions seem natural: Does a general category-theoretic version exist? Assuming it does, what are the conditions on the "type of structure" giving the category in order for the result to apply? REPLY [7 votes]: The following is essentially a copy-paste from a stackexchange answer I gave here, so I'll make it community wiki. Trevor Irwin's thesis gives a categorical treatment of Fraïssé limits, a construction which can be used to construct the Rado graph, the Higman group, and the rational Urysohn space. It takes a bit of extra work to get the real Urysohn space from the general theory. This also yields algebraic closures of fields, for instance. In model theory, it gives Hrushovski constructions, and with a slight generalization, also saturated models in model theory. The construction is also similar to Quillen's small object argument, which is not surprising once you see that it's about constructing injective objects. Let $\mathcal{C}$ be a category and let $\mathcal{F} \subseteq \mathcal{C}$ be a full subcategory. (The following is most interesting when the morphisms of $\mathcal{F}$ are all monos, so $\mathcal{C}$ and $\mathcal{F}$ will often be non-full subcategories of "embeddings" in some category of primary interest.) We say that $X \in \mathcal{C}$ is $\mathcal{F}$-universal if every $F \in \mathcal{F}$ admits a morphism $F \to X$. We say that $X \in \mathcal{C}$ is $\mathcal{F}$-injective if for every $F' \leftarrow F \to X$ with $F,F' \in \mathcal{F}$ there exists $F' \to X$ making a commutative triangle. We say that $X \in \mathcal{C}$ is $\mathcal{F}$-homogeneous if for every $X \leftarrow F \to X$ with $F \in \mathcal{F}$, there is an automorphism $X \cong X$ making a commutative triangle. Now suppose that $\mathcal{C}$ has colimits of ($\mathbb{N}$-indexed) chains $C_0 \to C_1\to \dots$ (equivalently, $\mathcal{C}$ has countable filtered colimits). $\mathcal{F}$ has the property that if $f: F \to \varinjlim C_i$ is a morphism in $\mathcal{C}$ from an object $F \in \mathcal{F}$ to a colimit $\varinjlim C_i$ of a chain $C_0 \to C_1 \to \dots$ in $\mathcal{C}$, then $f$ factors through some $C_i$ (this follows if every $F \in \mathcal{F}$ is finitely presentable in $\mathcal{C}$). Proposition. If (1) and (2) hold and $X \in \mathcal{C}$ is a colimit of a chain of objects of $\mathcal{F}$, then the following are equivalent and characterize $X$ up to isomorphism: (i.) $X$ is $\omega\mathcal{F}$-universal and $\mathcal{F}$-homogeneous (ii.) $X$ is $\mathcal{F}$-universal and $\mathcal{F}$-homogeneous (iii.) $X$ is $\mathcal{F}$-universal and $\mathcal{F}$-injective. Here $\omega \mathcal{F}$ denotes the closure of $\mathcal{F}$ under colimits of chains. (Irwin assumes an additional "monotonicity" condition, but as far as I can see this is not necessary.) Proof. (i.) -> (ii.) is trivial. (ii.) -> (iii.) requires just a moment's thought. (iii.) -> (ii.) uses a back-and-forth argument, inductively building up both directions of an isomorphism, working up the chain defining $X$. (iii.) -> [the first clause of (i.)] is easy. The fact that (iii.) characterizes $X$ up to isomorphism uses another back-and-forth argument. Let's call an $X$ satisfying the equivalent condtions of the proposition $\mathcal{F}$-saturated. We can get existence of such an $X$ with some mild additional conditions on $\mathcal{F}$: $\mathcal{F}$ has the joint embedding property: any pair of objects $F_1, F_2 \in \mathcal{F}$, admit a cocone, i.e. there is a diagram $F_1 \to F_3 \leftarrow F_2$ in $\mathcal{F}$. $\mathcal{F}$ has the amalgamation property: any span $F_1 \leftarrow F_0 \to F_2$ in $\mathcal{F}$ admits a cocone, i.e. it can be completed to a commutative square like $\require{AMScd} \begin{CD} F_0 @>>> F_1\\ @VVV @VVV \\ F_2 @>>> F_3 \end{CD} $ Theorem. If (1,2,3,4) hold and $\mathcal{F}$ is essentially countable then there exists an $\mathcal{F}$-saturated object $X \in \mathcal{C}$ which is a colimit of a chain of objects of $\mathcal{F}$. Conversely, if there exists an $X \in \mathcal{C}$ which is $\mathcal{F}$-injective and $\mathcal{F}$-universal, and which is a colimit of a chain of objects of $\mathcal{C}$ and (2) holds, then (3,4) hold. Proof. We construct a chain $X_0 \to X_1 \to \dots$ whose colimit is will be $X$. The idea is to solve the lifting problems of $\mathcal{F}$-universality and $\mathcal{F}$-injectivity by using (3) and (4) to extend the chain with solutions to these lifting problems. Then since any lifting problem must factor through a finite stage of the chain, it can be solved at a finite stage and thus be solved in the colimit. Some care must be taken in doing this, see Irwin's thesis or another reference on the Fraïssé construction. The "conversely" statement takes just a moment's reflection. When we take $\mathcal{C}$ to be the category of graphs and embeddings and $\mathcal{F}$ to be the subcategory of finite graphs, we construct the Rado graph this way, and similarly other Fraïssé limits. When we take $\mathcal{C}$ to be the category of rational metric spaces and isometric embeddings and $\mathcal{F}$ to be the subcategory of finite metric spaces, we construct the rational Urysohn space this way. When we take $\mathcal{C}$ to be groups and injective homomorphisms and $\mathcal{F}$ to be the finitely presented groups, we get the Higman group, and so forth. Everything generalizes easily to cardinalities $\kappa$ larger than countable -- in (1) $\mathcal{C}$ will have to be closed under colimits of chains of length $\leq \kappa$, while in (2), $\mathcal{F}$ will have to have this property with respect to chains of length $\kappa$ and moreover be closed under colimits of chains of length less than $\kappa$. (3) and (4) require no modification. In the existence theorem we require that $\mathcal{F}$ have size essentially $\leq \kappa$. By taking $\mathcal{C}$ to be the category of models of a first-order theory and $\mathcal{F}$ to be the models of size $< \kappa$, we recover the construction of saturated models in the usual sense in model theory, with the usual set-theoretical provisos.<|endoftext|> TITLE: Massey Products on a specific space QUESTION [5 upvotes]: Let $a,b$ be the canonical generators of $\pi_1(S^1\vee S^1)$ corresponding to the edges with some choice of orientation. Are there nonzero-Massey products in the cohomology with $\mathbb{F}_2$-coefficients of the space $X=(S^1\vee S^1) \cup_{S^1}D^2$ where the attaching map sends the boundary of $D^2$ to the word $a^2ba^2b^{-1}$? The cohomology ring $H^*(X;\mathbb{F}_2)$ is isomorphic to the $\mathbb{F}_2$-algebra with two generators $a,b$ in degree $1$ and one generator $c$ in degree $2$ and all products $a^2=b^2=ab=0$. (All other products are zero for degree reasons). So there could be a nontrivial triple Massey product, maybe $\langle a,a,b\rangle$ or some other combination. Maybe I should give reasons why $a^2,ab,b^2$ are zero, hoping that those reasons help in computing the Massey Products. $a^2=Sq^1(a)$ is zero because it is the Bockstein. The differentials in both $C^*(-;\mathbb{F}_2)$ and $C^*(-;\mathbb{Z}/4)$ are zero, so the map $H^*(X;\mathbb{Z}/4)\rightarrow H^*(X;\mathbb{F}_2)$ are surjective and hence the Bockstein is zero. To show that $b^2$ and $ab$ are zero one can use group homomorphism from the fundamental group of $X$ to $\langle a,b\mid a^2\rangle\cong \langle a \mid a^2\rangle *\langle b\rangle $. On group cohomology it induces a ring homomorphism $H^*(BG;\mathbb{F}_2)\cong \mathbb{F}_2[a]\oplus \mathbb{F}_2[b]/b^2\rightarrow H^*(X;\mathbb{F}_2)$. REPLY [6 votes]: Here a general way how to compute Massey Products on nice spaces. The first step would be to realize the space as the geometric realization of a simplicial set. I used the simplicial set given by the following picture. It has one 0-simplex $v$, its 1-simplices are $a,b,c,e,f$ and its 2-simplices are $A,B,C,D$ and their boundaries are indicated in the picture. Note that the arrowtips indicate the orientations on the edges. Note that each 2-simplex really is totally ordered, so we really have a simplicial set here. Also note that there are identifications in this picture, for example the edge a appears four times and so on. Now let us have a look at the dual of the simplicial chain complex. Let the dual bases be denoted by $v^*,a^*,...$. Let us read off the boundaries of those basis vectors and note that $\mathbb{F}_2$-coefficients lets us ignore signs. We have: $$\begin{eqnarray*} d(v^*)&=& 0,\\ d(a^*)&=& A^*+B^*+C^*+D^*,\\ d(b^*)&=& A^*+D^*\\ d(c^*)&=& A^*+B^*\\ d(e^*)&=& B^*+C^*\\ d(f^*)&=& C^*+D^*\\ \end{eqnarray*}$$ To find the DGA-structure on that cochain complex, we still have to write down the multiplication. In degree $0$ everything is a multiple of one ($v^*$) and everything above degree $2$ is zero. So it suffices to compute the products of all the basis vectors in degree one. To determine this, we need to recall the definition of the cup product. To compute say $a^*\cdot b^*$,we have to look at the 2-simplices and add up the duals of those which start with $a$ and end with $b$. So we get $$A^* = c^*a^*, \quad B^* = a^*e^*, \quad C^* = f^*a^*, \quad D^* =a^*b^*,$$ and all other products of those base vectors are zero (especially also the ones above after switching the order). Let us first compute the cohomology ring. Since the cohomology in concentrated in degrees $\le 2$ and $H^0=\mathbb{F}_2$, we just have to compute products of classes from $H^1$. The classes $[a^*+c^*+f^*]$ and $[b^*+c^*+e^*+f^*]$ form a basis of $H^1$ (Those are alled $a$ and $b$ in the question). $$\begin{eqnarray*} (a^*+c^*+f^*)^2&=& A^*+C^* = d(c^*+e^*),\\ (b^*+c^*+e^*+f^*)^2 &=& 0,\\ (a^*+c^*+f^*)(b^*+c^*+e^*+f^*)&=& B^*+D^* = d(e^*+f^*),\\ (b^*+c^*+e^*+f^*)(a^*+c^*+f^*)&=& A^*+C^* = d(c^*+e^*).\\ \end{eqnarray*}$$ So we see that the product in cohomology of all those classes is zero (They are boundary). Especially the Massey Product $\langle a,a,b \rangle$ really lives in $H^2$ and not in a quotient of it (We quotient out the zero module). Following the definition of Massey product, we seee that it is represented by the cycle $$ (c^*+e^*)(b^*+c^*+e^*+f^*)+(a^*+c^*+f^*)(e^*+f^*)= B^*$$ and $0\neq [B^*] \in H^2$. One small final remark: That cochain complex also is an algebra over the sequence operad from McClure-Smith and thus we could also compute the Steenrod squares on $H^*$ in a similar fashion.<|endoftext|> TITLE: Union of almost Hamming balls QUESTION [6 upvotes]: This question is connected with my previous question: Union of Hamming balls Let $V \subseteq \{0,1\}^n$, $\log|V| = k < 0.9n$. Harper's theorem states that the set $V_r:= \bigcup_{x \in V} V_r(x)$ has rather large cardinality ($V_r(x)$ is a Hamming full-ball of radius $r$ and center $x$, $|V_r|$ is the smallest when $V$ is a Hamming full-ball). Now consider a set $V_r^a$ with the following property: $|V_r^a \bigcap V_r(x)| \ge a \cdot|V_r(x)|$ for every $x \in V$. For explicitness we may assume that $a = \frac{1}{100}$. What is a lower bound for $|V_r^a|$? Is it true that $\log|V_r^a| - \log|V| = \Omega(n)$ for $k, r = \Omega(n)$ and constant $a$? UPD. Why this question is interesting? Let $x \in \{0,1\}^n$ be a string of Kolmogorov complexity $k$, where $k < n$. Change $r$ random symbols of $x$ and get a string $x'$. It seems that with high probability the complexity of $x'$ is greater than the complexity of $x$, but how to prove it? In fact this question is equivalent to the question above. REPLY [4 votes]: Unfortunately the life is not that good. Take the Hamming ball $B_R$ of radius $R=\beta n$ (where $\beta$ is small but positive) centered at $(0,0,\dots,0)$ and take the union of the Hamming balls of radius $r\ll n$ centered at the points from that ball. Not surprisingly you'll get the Hamming ball of radius $R+r$ and that is the worst case scenario for the full balls. Now notice that if you take a point $x\in B_R$ with at least $R-r\approx\beta n$ ones in it and try to flip $r$ random entries, in the typical case you'll flip $\beta r$ ones and only $(1-\beta)r$ zeroes, so at least one half or so of the Hamming ball of radius $r$ centered at $x$ lies within the Hamming ball centered at the origin of radius $R+(1-2\beta)r$, which has the volume about $\left(\frac{1-\beta}\beta\right)^{2\beta r}$ times smaller than the ball of radius $R+r$ and that factor is exponential in $n$ if $r$ is a small multiple of $n$. Edit To answer the modified question, let us restate it in the following way. Assume that a set $F$ has measure ($2^{-n}$ times the number of points) $\mu(F)\le e^{-sn}$. Consider $r=\frac{1-t}{1+t}n$ with $t\in(0,1)$. Then the set $G$ of points $x$ such that $\mu(V_r(x)\cap F)\ge a\mu(V_r(x))$ has measure $\mu(G)\le e^{-c(s,t)n}\mu(F)$. To prove it, consider the convolution of the characteristic function $f$ of the set $F$ with the kernel $K_t(x)=\prod_{j=1}^n (1+tx_j)$ (I assume that the cube is $\{-1,1\}^n$ and the convolution is multiplicative and associated to the natural group structure on the cube given by the coordinate-wise multiplication). This kernel has total mass $1$ out of which a noticeable part ($1/(n+1)$ for sure but much better bounds are possible) lies on the boundary of $V_r(\pmb 1)$ (that is why we chose such a strange parameterization for $r/n$). Since that boundary also is where $\min K_t$ is attained in $V_r(\pmb 1)$, we conclude that $g=f*K_t\ge \frac an$ on $G$. On the other hand, this convolution corresponds to the multiplier $t^{|S|}$ in the Fourier-Walsh representation. So, if $f=\sum_{k=0}^n f_k$ is the Fourier-Walsh orthogonal decomposition, we have $g=\sum_k t^k f_k$. Now observe that $\|f_k\|_\infty\le {n\choose k}\mu(F)$, which is below $e^{-sn/2}\ll \frac an$ for $k\le \gamma(s,t)n$. Thus, the large values of $g$ on $G$ are due to the tail. However, the $L^2$-norm of that tail is exponentially small compared to the $L^2$-norm of $f$ and the desired result follows. Of course, you can try to choose a better kernel to get sharper bounds though in that case the computation of the corresponding multiplier will be more difficult. I do not know if this simple approach can give you an asymptotically sharp bound however.<|endoftext|> TITLE: Can $\mathsf{Set}$ be seen as a (non-trivial) 2-category? QUESTION [7 upvotes]: I know that $\mathsf{Rel}$ can be seen as a 2-category with inclusion of relations as 2-morphisms, when passing to $\mathsf{Set}$, relations become functions and inclusions are constrained to be "equalities" (see Is $\mathsf{Set}$ a 2-category?), is there any other way where we can see $\mathsf{Set}$ as a non-trivial 2-category? If not, how can we prove that it cannot be other than a trivial 2-category? REPLY [2 votes]: This should be a comment, but it's too long so there it is. You could always define a new concept of arrows that probe deeper inside the inner structures of $Sets$. $Sets$ should actually be called "$Card$" (as category of cardinals). Knowing that elements of sets are sets themselves, its not difficult to come up with some concepts of $2$-functions $f: A \rightarrow B$ seen as a pair $(g, (g_a))$, where $g: A \rightarrow B$ is an usual function of set, and $(g_a)$ is a family of functions such that $g_a:a \rightarrow g(a)$ for all $a$ in $A$. These $2$-functions would play the role of $1$-arrows in your $2$-category $2$-$Sets$. Now, you have to find a natural definition for $2$-arrows (i.e., maps of $2$-functions). It surely can be done (I didn't try), and you could even be able to make $Sets$ into an $\omega$-category, that would actually encodes all the membership tree (thus being the "true" category of set theory). You have to take care of empty set I think, but I guess it can be done. edit: Let $A$ and $B$ be two sets, let us write a 2-map between $A$ and $B$ as $(f, f_a)$ with $f: A \rightarrow B$ and for each $a$ in $A$, a map $f_a: a \rightarrow f(a)$. You can define a composition here as $(g, g_b) \circ (f, f_a) = (g \circ f, (g\circ f)_a = g_{f(a)} \circ f_a)$. This composition is associative, and identity is obviously given by $(Id_A, ({Id_A})_a = Id_a)$, turning the collection of sets where $1$-arrows are $2$-maps into a $1$-category. We now have to define $2$-arrows and show that the composition is functorial. Let us define a $2$-arrow $\eta: (f,f_a) \Rightarrow (g,g_a)$ as a collection of maps $\eta_a: f(a) \rightarrow g(a)$ for each $a$ in $A$, such that $\eta_a \circ f_a = g_a$. This is highly inspired from the concept of natural transformation as you might see. I'm going to check if this turn the above category as a $2$-cat (it should). The vertical composition is simply given by $(\xi \circ_1 \eta)_a = \xi_a \circ \eta_a$, making the collection of arrows between $2$-maps between two fixed sets a $1$-cat as it should be. It is left to define an horizontal composition and verify the functoriality. Note that the initial element is given by the empty set and that if $(f,f_a)$ is a $2$-map between $A$ and $B$ with empty set in the image of $f$, then the empty set is necessarily an element of $A$ with $f(\emptyset) = \emptyset$.<|endoftext|> TITLE: A different avatar of the complexity of a graph QUESTION [12 upvotes]: Let $G$ be a connected, finite graph. (For me a graph is undirected, and it possibly has multiple edges, although the latter is not really crucial for this question). The complexity $c(G)$ (also known as the tree-number of $G$) is defined to be the number of spanning trees of $G$. There is another number $k(G)$ that one can easily (and naturally?) associate to $G$. In all examples that I have studied, $k(G)$ equals $c(G)$, but I am not able to prove this in general. The natural number $k(G)$ is defined to be the cardinalty of a certain set of functions $$\mathcal{F}(G) \subset \left\{ f \colon V(G) \to \mathbb{N}: \sum_{v \in V(G)} f(v) = b_1(G)\right\}\subset \mathbb{N}^{V(G)}$$ where $V(G)$ is the vertex set of $G$ and $b_1(G)$ is the first Betti number of $G$. Here we construct all elements of $\mathcal{F}(G)$. Start by considering the constant zero function on all vertices of $G$. Then fix a spanning tree $\Gamma$ of $G$. For all edges $e$ that are missing from $\Gamma$ in $G$, add a $1$ to the function at exactly one of the two endpoints of $e$, and do this in all possible ways to obtain a set of functions $\mathcal{F}_{\Gamma}(G)$. Then $\mathcal{F}(G)$ is the union of all $\mathcal{F}_{\Gamma}(G)$, for $\Gamma$ that ranges over all spanning trees of $G$. Here are two simple examples where it is immediate to check that $k(G)=c(G)$. 1) Take for $G$ the graph with two vertices $v_1,v_2$ connected by $k$ edges. The functions set $\mathcal{F}(G)$ constructed in the above paragraph consists of the assignments $$\{(k-1, 0), (k-2, 1), ... (0, k-1)\},$$ so $c(G)=k(G)=k$. 2) Take for $G$ the graph that is a planar $k$-gon (with $k$ vertices and $k$ edges). Here $\mathcal{F}(G)$ consists of functions that are constantly zero except for one vertex of $G$ where they equal $1$. Again we have that $c(G)=k(G)=k$. Question: is it true that $k(G)=c(G)$ for all $G$? Even if you do not know the answer, maybe you could point me to the relevant literature. Since I am not myself a graph theory expert nor a combinatorialist, this may very well be well-known or trivial in which case I do apologize. REPLY [10 votes]: You're looking at the "integral break divisors" of the graph $G$. These were introduced by Mikhalkin and Zharkov. For a very clear presentation check out the paper "Canonical representatives for divisor classes on tropical curves and the Matrix-Tree Theorem" by An, Baker, Kuperberg, and Shokrieh, available online at https://arxiv.org/abs/1304.4259. EDIT: I should say, the answer to your question is yes, the number of integral break divisors is always equal to the number of spanning trees of $G$, because these serve as canonical representatives of $\mathrm{Pic}^g(G)$.<|endoftext|> TITLE: Deligne-Scholl's motives for modular forms: Hecke operators vs. transposed Hecke operators QUESTION [8 upvotes]: EDIT: I moved my original question down to the bottom. The question here at the top is related, at least I suppose that the same phenomenon is behind both of them. In the article "Valeurs de fonctions L et périods d'intégrales" (p. 333/334), Deligne sketches how to define the motive $M(f)$ attached to a newform $f=\sum a_nq^n$. The construction involves taking the eigenspace (inside a certain cohomology) where the Hecke operators act with the eigenvalues given by the Fourier coefficients of $f$, i.e. the common kernel of all $T_p-a_p$. He then uses the differential form $\omega_f:=f(z)\,2\pi i\mathrm dz$ to calculate periods (I'm assuming that the weight is $2$ here, for simplicity). Deligne claims that the transposes ${}^tT_n$ of the Hecke operators act on $\omega_f$ by the eigenvalues $a_n$, so that $\omega_f$ represents a class in the de Rham realization of the motive $M(f^*)$ attached to the dual cusp form. Why is this true? Let ${}^N_kW$ be the motive defined by Scholl having $M(f)$ as a submotive. If we view $\omega_f$ as a class in $\operatorname{fil}^1{}^N_kW_{\mathrm{dR}}\otimes\mathbb C\cong S_k(X(N),\mathbb C)$, then the complex comparison isomorphism to ${}^N_kW_{\mathrm B}\otimes\mathbb C$ (which is the Eichler-Shimura isomorphism) sends $\omega_f$ to the usual cocycle given by integrating $\omega_f$ along geodesics on the upper half plane, which corresponds to $f$ (and not $f^*$!). Since the Eichler-Shimura isomorphism is Hecke equivariant, $\omega_f$ should lie in the eigenspace for $f$ and not $f^*$, i.e. it should be just $f$ under the identification $\operatorname{fil}^1{}^N_kW_{\mathrm{dR}}\otimes\mathbb C\cong S_k(X(N),\mathbb C)$. EDIT: This was my original question. I have a question concerning a detail in Tony Scholl's construction of motives attached to modular forms. To my understanding, the construction of Deligne's Galois representation attached to modular forms works as follows. Let $Y_1(N)$ be the modular curve over $\mathbb Q$ classifying elliptic curves with a point of exact order $N$, and let $f\colon E_1(N)\rightarrow Y_1(N)$ be the universal elliptic curve over it. Then one looks at the étale cohomology group $$ {}_k^NW_\ell:=H^1_{\mathrm p}(Y_1(N),\operatorname{Sym}^{k-2}R^1f_*\mathbb Q_\ell).$$ From the Hecke correspondences on $Y_1(N)$ one gets a Hecke action on this $\mathbb Q_\ell$-vector space, and using the (Eichler-Shimura) congruence relation and the "corrected scalar product", or "twisted Poincaré duality" (involving the Atkin-Lehner involution) one can show that $$ \det(1-F_pX,{}_k^NW_\ell)=1-T_p+p^{k-1}\langle p \rangle X^2 $$ where $F_p$ is a geometric Frobenius at $p$. Hence to get Deligne's Galois representation attached to a fixed newform $f$ of weight $k$ and level $N$ with Fourier coefficients in a number field $K$, one just tensors ${}_k^NW_\ell$ over the Hecke algebra $T$ with the morphism $T\rightarrow K_\lambda$ induced by $f$, where $\lambda$ is a prime of $K$ lying above $\ell$ (or equivalently, one projects to the eigenspace of ${}_k^nW_\ell\otimes_{\mathbb Q_\ell}K_\lambda$ where the Hecke operators act by the eigenvalues given by the Fourier coefficients of $f$). Now in the article "Motives for modular forms" Scholl defines a motive over $\mathbb Q$ called ${}_k^NW$ whose $\ell$-adic étale realization is precisely ${}^N_kW_\ell$. He then shows that the Hecke correspondences induce endomorphisms of this motive and that on the Betti realization these endomorphisms are the same as the ones defined by Deligne (Prop. 4.1.1). Hence also on the $\ell$-adic realization these are the same as Deligne's. But then Scholl projects to the eigenspace where the transposes ${}^tT_p$ of the Hecke operators act by the eigenvalues given by $f$ to get his motive attached to $f$. He claims (4.2.2) that on the $\ell$-adic realization we get precisely Deligne's Galois representation attached to $f$. To me this seems to contradict the above, it should rather give the representation attached to the "dual" modular form $f^*$. Also Scholl writes that on the space of cusp forms (which is the intermediate step $\operatorname{fil}^1{}^N_kW_{\mathrm{dR}}=\operatorname{fil}^{k-1}{}^N_kW_{\mathrm{dR}}$ in the Hodge filtration on the de Rham realization of ${}^N_kW$) the endomorphisms of ${}^N_kW$ given by the Hecke correspondences induce the tranposes of the usual Hecke operators. But I do not see why this statement is true. If on the Betti realization they are the same as Deligne's, then by the Eichler-Shimura isomorphism one should see that the are the usual ones on the space of cusp forms... (?) REPLY [3 votes]: You can define Deligne's representation $V_f$ associated to $f$ either a subspace of the \'etale cohomology group $W_{\ell}$ or as a quotient. If you choose to define it as a subspace, then $V_f$ is indeed the subspace on which the usual Hecke operators $T_p$ for $p\nmid N \ell$ acts as multiplication by $a_p(f)$, as you were expecting. If you prefer to see $V_f$ as a quotient, note that there is a Galois-equivariant perfect pairing $$ H^1(Y,Sym^{k-2}(R^1f_*\mathbb{Q}_{\ell})) \times H^1(Y,Sym^{k-2}(R^1f_*\mathbb{Q}_{\ell})^\vee) \rightarrow \mathbb{Q}_{\ell}(1) $$ where $(1)$ denotes the Tate twist by the cyclotomic character. This means that in order to obtain the same $V_f$ as above, it must arise as a quotient of the dual of $H^1(Y,Sym^{k-2}(R^1f_*\mathbb{Q}_{\ell}))$, which the above pairing shows that may be identified with $H^1(Y,Sym^{k-2}(R^1f_*\mathbb{Q}_{\ell})^\vee)(1)$. Under this isomorphism, the Hecke operator $T_p$ on the left-hand side is identified with the adjoint Hecke operator $T_p^*$ on the right-hand side. This explains why the definition of $V_f$ you encounter in Deligne coincides with the one we described above as a subspace.<|endoftext|> TITLE: Alignment of random points QUESTION [13 upvotes]: Whenever I draw randomly about ten points, I see that there will be always 3 points that are "almost" collinear. This observation leads me to considering the following questions: Question 1: Suppose $n$ points are generated uniformly randomly in a square, let $\phi_n$ be the largest angle that is formed by three of them. How quickly does $E(\phi_n)$ tend to $\pi$? Question 2: For $n>2$ and $\epsilon > 0$, estimate $P(\phi_n > \pi-\epsilon)$? REPLY [4 votes]: The intuition in the comment by Ori Gurel-Gurevich appears to be correct. Indeed, let us show that $\pi-\phi_n$ is on the order of $1/n^3$ in probability. Let $T$ denote the set of all triangles with vertices at some of the $n$ points, with $N:=|T|=\binom n3$, the cardinality of $T$. For each $t\in T$, let $X_t$ denote the largest angle in the triangle $t$, so that $\phi_n=\max_t X_t$. It should be not hard to show that (i) the random variable (r.v.) $X_t$ has a density $f$ left-continuous at point $\pi$, with $f(\pi-)=:c\in(0,\infty)$ and (ii) for any distinct $t$ and $s$ in $T$, the random pair $(X_t,X_s)$ has a joint density bounded by some $C<\infty$. Now we can use the key result by Galambos \url{https://www.jstor.org/stable/2239989?seq=1#page_scan_tab_contents}, based on a combinatorial graph sieve theorem by Renyi, with $N$ in place of the symbol $n$ in that paper, and with $E$ defined as the set of pairs $(t,s)\in T^2$ such that the triangles $t$ and $s$ have at least one common vertex. Then clearly $N_E:=|E|=O(n^3\cdot n^2)=o(N^2)$. For any fixed positive real $a$, let $c_N=c_N(a)$ be the root of the equation $$NP(X_t>c_N)=a.$$ Since $P(X_t>x_N)\sim c(\pi-x_N)$ if $x_N\uparrow\pi$, we have \begin{equation} c_N=\pi-\frac{a}{cN}\,(1+o(1)); \end{equation} all the limit relations here are for $n\to\infty$. Then it is easy to see all the conditions in Galambos's theorem hold, with $r(\dots)=0$ in his condition (iii) and $d_k=1$ in his condition (iv). It follows that $P(\phi_na)\to e^{-a}, \end{equation} for each real $a>0$. That is, the r.v. $\frac c{3!}\,n^3(\pi-\phi_n)$ converges to an exponential r.v. in distribution. Additional efforts are needed to show that the corresponding convergence of the means holds, so that $\frac c{3!}\,n^3\mathbb{E}(\pi-\phi_n)\to1$, whence $\mathbb{E}(\pi-\phi_n)\sim\frac{3!}c\,n^{-3}$.<|endoftext|> TITLE: cohomology of the loop space of a Lie group: divided power structure QUESTION [10 upvotes]: Let $G$ be a complex simple simply connected algebraic group and $Gr_G$ be the corresponding affine Grassmannian (it is well known that $Gr_G$ is homotopy equivalent to the base loop space $\Omega K$ of the maximal compact subgroup $K\subset G$). Let $H=H^*(Gr,k)=H^*(\Omega K,k)$ be the cohomology ring where $k$ is a field of characteristic $p>0$. The computation of $H$ by Yun and Zhu in their paper "Integral homology of loop groups..." shows that the commutative $k$-algebra $H$ has a divided power structure, at least when $p$ is not too small. Can this be seen from first principles? In particular, is it related to Cartan's divided power structure on homology of a commutative simplicial $k$-algebra? REPLY [4 votes]: Let me make the following remarks regarding divided power structures and cohomology of loop spaces. Let us fix a field $\mathbb{F}$. 1) Le me recall the Eilenberg-Moore spectral sequence. Let us suppose that we have a fibration $$F\rightarrow E\rightarrow B$$ such that $B$ is $1$-connected, $H^k(B;\mathbb{F})$ and $H^k(E;\mathbb{F})$ are finite dimensional $\mathbb{F}$-vector spaces for any k. Then we have a convergent spectral sequence $$E_2^{p,q}=Tor^{H^*(B;\mathbb{F})}_{p,q}(H^*(E;\mathbb{F}),\mathbb{F})\implies H^*(F;\mathbb{F}).$$ In the case of the path fibration $$\Omega X\rightarrow PX\stackrel{ev_1}{\rightarrow} X$$ we get a convergent spectral sequence $$E_2^{p,q}=Tor^{H^*(X;\mathbb{F})}_{p,q}(\mathbb{F},\mathbb{F})\implies H^*(\Omega X;\mathbb{F}).$$ 2) Behind the Eilenberg-Moore spectral sequence you have the Eilenberg-Moore quasi-isomorphism $$C^*(F;\mathbb{F})\simeq Tor^{C^*(B;\mathbb{F})}(C^*(E;\mathbb{F}),\mathbb{F})$$ in order to take into account of the multiplicative structure given by the cup product we have to be very careful and use the $E_{\infty}$-algebra structure on the singular cochains $C^*(-;\mathbb{F})$ and play with $E_{\infty}$-torsion products. This is very well explained in M. Mandell's paper "E_{\infty}-algebras and $p$-adic homotopy theory" (Topology, 2001, have a look at section $5$). 3) When one can replace the dg-algebra $C^*(X;\mathbb{F})$ by a commutative dg algebra i.e. when it is quasi-isomorphic to a cdga $A^*(X)$ the Torsion product $$Tor^{A^*(X)}(\mathbb{F};\mathbb{F})$$ is a cdga and by results of Cartan (Cartan's seminar 1954-1955, exposé $7$) when $char(\mathbb{F})>0$ its cohomology has a divided power algebra structure. The relationship between this divided power structure and the one coming from the homotopy ring of a simplicial commutative algebra is explained in B. Richter's paper "Divided power structures and chain complexes." Alpine perspectives on algebraic topology, 237–254, Contemp. Math., 504, Amer. Math. Soc., Providence, RI, 2009. The relationship is explained thanks to the Dold-kan correspondence and the classical Bar construction. 4) Last point, when can we replace $C^*(X;\mathbb{F})$ by a cdga? I know at least three cases: when $X$ is formal over the field $\mathbb{F}$, when the field is of characteristic zero, in the Anick's range if $X$ is $r$-connected and $dim(X)\leq r.char(\mathbb{F})$. Conclusion: for any $1$-connected finite CW-complex $X$ the cohomology algebra of $\Omega X$ with coefficients in a field $\mathbb{F}$ has a divided power structure provided $char(\mathbb{F})>>0$. This divided power structure is inherited from the one coming from the classical Bar construction and is of simplicial nature.<|endoftext|> TITLE: Hodge to de Rham spectral sequence for stacks QUESTION [13 upvotes]: For some work I'm doing, I need a version of the Hodge to de Rham spectral sequence for stacks. I am not at all an expert on stacks, so please excuse me if I make minor technical mistakes in stating it. I only need to deal with quotient stacks. Let $X$ be a smooth quasiprojective variety over $\mathbb{C}$ and let $G$ be a finite group acting on $X$. I am trying to understand the cohomology of the quotient stack $X/G$. Sheaves on $X/G$ should be the same thing as $G$-equivariant sheaves on $X$ and sheaf cohomology should be the derived functors of the $G$-equivariant global sections functor. What I think should be true is that there should be a Hodge to de Rham spectral sequence converging to $H^{\ast}(X/G;\mathbb{C})$ with $$E_1^{pq} = H^p(X/G;\Omega^q).$$ Here $\Omega^q$ is the sheaf of $G$-equivariant holomorphic $q$-forms on $X$. This should be able to be proved following the usual proof of the Hodge to de Rham spectral sequence: one first proves that the constant sheaf $\mathbb{C}$ on $X/G$ is quasi-isomorphic to the de Rham complex $\Omega^{\ast}$, and the one looks at the hypercohomology spectral sequence. The only place where I can see any issue is proving the exactness of the de Rham complex, which would require an equivariant version of the Poincare lemma. However, this should be able to be derived from the usual Poincare lemma by averaging. Question: Am I correct that this spectral sequence exists, and if so can anyone give me a reference for it? I can find plenty of papers that investigate situations where a version of the Hodge to de Rham spectral sequence for stacks degenerates, but all of them are working in much more generality than I am and it is hard for me to verify that the spectral sequence they deal with is the same one I (tried) to state above. I don't need the spectral sequence to degenerate, though it presumably does in the situation I am working in. REPLY [15 votes]: Short answer: yes. As I recall, Teleman constructs such a spectral sequence for fairly general stacks. You can look at his paper The quantization conjecture revisited Annals 2000. But this may fall into the category of "more generality than you need". The case you want can probably done by hand. The point is if $G$ is a finite group acting on a complex quasi projective variety $X$, then the quotient variety $X/G$ exists (see Sarah's comment below for an explanation). This is not the same as the quotient stack $[X/G]$ for a nonfree action, but there is relationship. The cohomology of $[X/G]$ with coefficients in a sheaf $\mathcal{F}$ would be equivariant cohomology $H^*_G(X,\mathcal{F})$. This can be taken as the derived functor of $G$-invariants composed with global sections as you said. This yields a spectral sequence $$H^a(G, H^b(X,\mathcal{F}))\Rightarrow H^{a+b}([X/G],\mathcal{F}) $$ Since $G$ is finite, this will collapse to an isomorphism when $\mathcal{F}$ is a sheaf of $\mathbb{C}$-vector spaces by Maschke's theorem. This remark applies to $\mathcal{F}=\mathbb{C}$ or $\Omega_X^q$. In this case, we can further identify $$H^i(X,\Omega_X^q)^G= H^i(X/G,\tilde\Omega_{X/G}^q)$$ where the sheaf on the right is defined in Steenbrink, Mixed Hodge structure on vanishing cohomology. The upshot is that you may as well work on $X/G$, where the details about how to construct Hodge to de Rham can be found in Steenbrink's paper. Added explanation of second isomorphism. By definition $\tilde \Omega^q(U) = \Omega^q(\pi^{-1}U)^G$, where $\pi:X\to X/G$ is the projection. This implies that, we have an isomorphism of Cech complexes $$C(\{U_i\}, \tilde \Omega^q)=C(\{\pi^{-1}U_i\}, \Omega)^G$$ and therefore on cohomology (you need to use exactness of $(-)^G$ here as well).<|endoftext|> TITLE: Reference for ring structure on Thom spectra QUESTION [8 upvotes]: I have been asked to provide at reference for the fact that if $X\to B(BG)$ (where $G$ is the stable auto-homotopy-uquivalences of spheres - or $BG=Pic(S)$ in some peoples terminology) then the Thom spectrum associated to the based loop map $$ \Omega X \to BG=Pic(S) $$ is a ring-spectrum. I was looking around for a good reference, but couldn't find one that was very clean cut (the paper is about something in symplectic geometry - so I would like the reference to be precise). I also don't know where this originally appeared - so I would prefer the original reference if it is clear or a reference that points back to the original. REPLY [2 votes]: Along the great references cited above, I think Mahowald, Mark; Ray, Nigel A note on the Thom isomorphism. (English) Zbl 0469.55007 Proc. Am. Math. Soc. 82, 307-308 (1981). is also a good reference. Theorem 1 of this paper, proved as Corollary 3, is what you are after. I must add that Theorem 1 of this paper was originally proved in "Mahowald. Ring spectra that are Thom complexes" that is cited above. I also like to add that Thom spectrum of a map from a loop space into $BG$ is not necessarily a ring spectrum! The statement is that the Thom spectrum of any $H$-map, in particular loop map, is a ring spectrum.<|endoftext|> TITLE: An "explicit" description of cocomma-categories ? QUESTION [5 upvotes]: Let the cocomma-square $\require{AMScd}$ \begin{CD} A @>T>> Y\\ @V S V V \Uparrow_\beta @VV i_1 V\\ X @>>i_0> (S\star T) \end{CD} where : $Ob\,(S\star T) :=Ob\,X\sqcup Ob\,Y$ and $\hom_{(S\star T)}(r,r') $ is given by : $\hom_{(S\star T)}(r,r'):=\hom_{X}(x,x')$ if $r=x, r'=x' \in X$, $\hom_{(S\star T)}(r,r'):=\hom_{Y}(y,y')$ if $r=y, r'=y' \in Y$, $\emptyset$ if $r\in Y$ and $r'\in X$, for all $a\in A$, a formal arrow : $Sa \overset{\underline{a}}{\longrightarrow} Ta $ such that : $id_{Ta}\circ \underline{a}=\underline{a}\circ id_{Sa}=\underline{a}$ all this arrows subject to the relations : if $r=x\in X$ and $r'=y\in Y$, two parallel arrows of $\hom_{(S\star T)}(x,y)$ : $$x\overset{f}{\longrightarrow}Sa\overset{\underline{a}}{\longrightarrow}Ta\overset{g}{\longrightarrow}y$$ $$x\overset{f'}{\longrightarrow}Sa'\overset{\underline{a}'}{\longrightarrow}Ta'\overset{g'}{\longrightarrow}y$$ are equal if exists a "zig-zag" in $A$ : $$a\overset{q_{1}}{\longleftarrow}a_{1}\longrightarrow...\overset{q_{n}}{\longrightarrow}a' $$ and arrows $f_{i}:x\longrightarrow Sa_{i}$ in $X$ and $g_{i}:Ta_{i}\longrightarrow y$ in $Y$ making the diagram commutes : \begin{CD} x @>f>> Sa @>\underline{a}>> Ta @>g>> y \\ @| @A Sq_1 AA @AA Tq_1 A @| \\ x @>f_1>> Sa_1 @>>\underline{a_1}> Ta_1 @>g_1>> y \\ @| @V Sq_2 VV @VV Tq_2 V @| \\ \cdots @>>> \cdots @>>> \cdots @>>> \cdots \\ @| @V Sq_n VV @VV Tq_n V @| \\ x @>>f'> Sa' @>>\underline{a'}> Ta' @>>g'> y\\ \end{CD} In this description : $\beta_a:=\underline{a}$. Can someone confirm me if it is right or not ? REPLY [6 votes]: Your construction is basically correct. The only slight mistake (probably you had the correct thing in mind and this was just a mistake in phrasing) is that the formal arrows $\underline{a}$ are not sufficient to give the hom-sets $\hom_{S\star T}(r,r')$ when $r\in X$ and $r'\in Y$; you also need the "formal composites" of these arrows with arrows in $X$ and $Y$ on either side, which then get quotiented by the equivalence relation you describe. Here is an abstract argument for why this construction is correct. Every comma object is a discrete two-sided fibration. Dually, every cocomma object is a codiscrete two-sided cofibration. In Cat, every discrete two-sided fibration is the comma of its cocomma, and every codiscrete two-sided cofibration is the cocomma of its comma. This sets up an equivalence between discrete two-sided fibrations and codiscrete two-sided cofibrations, both of which represent profunctors from $Y$ to $X$. Moreover, the comma of the cocomma is a reflection from arbitrary spans into discrete two-sided fibrations, and dually the cocomma of the comma is a coreflection from arbitrary cospans into codiscrete two-sided cofibrations. Thus, to compute an arbitrary cocomma, it suffices to compute the reflection of a span into discrete two-sided fibrations, regard it as a profunctor, and then take the collage of that profunctor. Every span $X \leftarrow A \to Y$ generates a free two-sided fibration $X \leftarrow X^{\mathbf{2}}\times_X A \times_Y Y^{\mathbf{2}} \to Y$, where $X^{\mathbf{2}}$ is the category of arrows in $X$. The objects of this free two-sided fibration are therefore triples $(a,f,g)$ where $a\in A$, $f:x\to S a$, and $g:T a \to y$, which we can regard as the above "formal composites". Now it suffices to take the discrete reflection of this over $X$ and $Y$, which amounts to imposing your zig-zag equivalence relation.<|endoftext|> TITLE: A question about ordinal definable sets and large cardinal axioms QUESTION [5 upvotes]: It is well known that the Large Cardinal Axiom which asserts the existence of a measurable cardinal number is inconsistent with V=L. Do any of the well-known Large Cardinal Axioms contradict V=OD? REPLY [9 votes]: Almost all of the usual large cardinal axioms can be preserved by the reverse-Easton support class forcing to make every set of ordinals coded into the GCH pattern, or the $\Diamond^*$ pattern, and these axioms imply V=HOD. For example, this is true of the supercompact cardinals by the usual Laver argument, and analogues work with many other large cardinals, including much larger cardinals. Therefore, any smaller cardinal notion that is a consequence of supercompact cardinal or these other large cardinals will also be relatively consistent with V=HOD. Meanwhile, there are some large cardinal notions that imply V=HOD outright. For example: Theorem. If there are a proper class of Laver-indestructible supercompact cardinals, then V=HOD. Proof. If $\kappa$ is supercompact and Laver indestructible, then for any bounded subset $x\subset\delta<\kappa$, there is $<\kappa$-directed closed forcing $\mathbb{Q}$ that codes $x$ into the GCH pattern above $\kappa$. If $\kappa$ is still supercompact after this forcing, then it would be $\Sigma_2$-reflecting in the extension, and the existence of a coding place for $x$ would reflect below $\kappa$ into $V_\kappa$, which was not changed by the forcing. Thus, $x$ is already coded in $V$, and hence every bounded subset of $\kappa$ is coded into the GCH pattern below $\kappa$. Since we have a proper class of such $\kappa$, it follows that every set is coded, and thus V=HOD. QED It also works with indestructible strong cardinals and many others; really all you need is that the $\Sigma_2$-correctness is indestructible.<|endoftext|> TITLE: Abelian variety with prescribed endomorphism ring QUESTION [11 upvotes]: Consider the cyclotomic field $L={{\mathbb{Q}}}(\zeta_8)={{\mathbb{Q}}}(\sqrt{2},i)$, where $\zeta_8$ is a primitive 8-th root of unity. Let $\Lambda={{\mathbb{Z}}}[\zeta_8]$ denote the ring of integers of $L$. Question 1. Does there exist an abelian variety $A$ over ${{\mathbb{C}}}$ of dimension two with ${\mathrm{End\,}} A\simeq \Lambda$ ? I expect the answer "no". Note that I want the endomorphism ring to be exactly $\Lambda$, not just contain $\Lambda$. In particular, my abelian variety should be simple. Question 2. Does there exist an abelian variety $A$ over ${{\mathbb{C}}}$ of dimension four with ${\mathrm{End\,}} A\simeq \Lambda$ ? I expect the answer "yes". Question 3. Is it possible to find an abelian variety $A$ over ${{\mathbb{C}}}$ of dimension 2 or 4 with ${\mathrm{End\,}} A\simeq \Lambda$ and such that $A$ is definable (without endomorphisms) over a small number field, say over ${{\mathbb{Q}}}(i)$ or even over ${{\mathbb{Q}}}$? REPLY [9 votes]: The answer to your Question 3 is YES with the ground field $\mathbb{Q}$. Here is a sketch of the proof. For each positive integer $q$ and a "parameter" $t$ (in char 0) consider the smooth projective model $C_{q,t}$ of an affine curve $y^q=x^3-x-t$. Let $P_{8,t}$ be the Prym variety of the double cover $$C_{8,t}\to C_{4,t}, (x,y)\mapsto (x,y^2).$$ Then $P_{8,t}$ is an abelian fourfold provided with an embedding $$\mathbb{Z}[\zeta_8]\hookrightarrow End(P_{8,t}).$$ One may deduce from Theorem 1.5 of arXiv:math/0601072 [math.AG] that if $t$ is a transcendental number then $P_{8,t}$ does not contain positive-dimensional abelian subvarieties of CM type. It follows from Th. 1.1 of a paper by Jiangwei Xue and Chia-Fu Yu, arXiv:1304.6202 [math.NT] that for such $t$ the endomorphism ring $End(P_{8,t})$ coincides with $\mathbb{Z}[\zeta_8]$. Now, by Masser's specialization theorem one may choose a rational number $c$ such that $End(P_{8,c})$ equals $End(P_{8,t})$ and therefore coincides with $\mathbb{Z}[\zeta_8]$.<|endoftext|> TITLE: Mapping Class Groups and torus (JSJ) decomposition of closed 3-manifolds QUESTION [9 upvotes]: I am wondering if some intuitive relation exists between Mapping Class Group (MCG) of a 3-manifold (assume "simple" enough manifolds: closed,compact,irreducible, orientable, non-hyperbolic) and its JSJ decomposition? More generally, I am thinking along some generalization of Dehn-Nielsen-Baer theorem to 3-manifolds: in 2D there exists a minimal set of loops such that Dehn Twists along those loops generates the MCG. In 3D (again, "simple" enough manifolds) does there exist a minimal set of incompressible surfaces whose Dehn twisting generates the MCG? Thank you REPLY [5 votes]: The natural definition of a higher Dehn twist at a surface $F$ is to identify a neighborhood of $F$ homeomorphic to $F\times\left[0,1\right]$, take a loop $\phi_t$ in $Homeo(F)$ based at $id$, and consider the map $f\colon M\to M$ which is $\phi_t$ on $F\times\left\{t\right\}$ and the identity outside the neighborhood. For a surface of genus $\ge 2$, the components of $Homeo(F)$ are contractible, so the Dehn twist at such a surface is homotopic to the identity. Similarly, Dehn twists at 2-spheres have order 2 because of $\pi_1SO(3)=\left\{\pm1\right\}$. Thus only Dehn twists at tori can be interesting. For a closed Haken manifold, Johannson proved that the group generated by Dehn twists at tori has finite index in the mapping class group. (In particular the mapping class group of atoroidal Haken $3$-manifolds is finite.) For a survey you can look at Waldhausen's paper "Recent results on sufficiently large 3-manifolds", the result is the very last corollary on page 37. There are some results for non-Haken manifolds, e.g., Gabai-Kazez have considered atoroidal manifolds with genuine laminations. But the general case seems to be still open(?).<|endoftext|> TITLE: Is this result on tensor products of faithful modules known? QUESTION [5 upvotes]: On the first page of this preprint, I give a proof (a simplification of an argument suggested by Ken Goodearl) that if k is a field, A and B are k-algebras (not necessarily commutative), M is a faithful left A-module, and N is a faithful left B-module, then the left A ⊗k B-module M ⊗k N is again faithful. It is hard to believe this is not classical. Does anyone know a reference? REPLY [7 votes]: See Lemma 1.1 of http://www.math.wisc.edu/~passman/balgebra.pdf by Passman. I think he published this years later in Communications in Algebra.<|endoftext|> TITLE: $S^3$ as cyclic branched cover of itself QUESTION [14 upvotes]: In Chapter One of his notes (March 2002) Thurston says: If $K$ is the trivial knot the cyclic branched covers are $S^3$. It seems intuitively obvious (but it is not known) that this is the only way $S^3$ can be obtained as a cyclic branched covering of itself over a knot. This sentence sounds a little bit enigmatic to me. Is he saying that (we expect that) "if $K \subset S^3$ is such that all its cyclic branched covers are $S^3$ then $K$ is the unknot", or that "if for some $k\geq 2$ the $k$-fold cyclic branched cover over $K$ is $S^3$ then $K$ is the unknot"? I would like to know which is the state of the art this conjecture. In particular if $S^3$ can appear as a branched cover over a non-trivial knot I would like to see some (families of) examples. REPLY [22 votes]: The statement that for arbitrary K in $S^3$, if for some $n \ge 2$, the n-fold cyclic branched cover is $S^3$ (or in some versions, a homotopy 3-sphere) then K is the unknot, was known as the Smith conjecture. It was proved around 1979, combining work of many authors: Thurston, Meeks-Yau, Gordon-Litherland, Bass, Shalen, and perhaps others. Its proof was viewed as an early triumph of Thurston's hyperbolization theorem and more generally the power of geometric methods in 3-manifold topology. The best reference is the 1984 book edited by Morgan and Bass, "The Smith Conjecture". It has lots of background. There are strong generalizations, due to Thurston; google "Thurston's orbifold theorem'.<|endoftext|> TITLE: Does every Shimura variety contain a generic point defined over a number field? QUESTION [10 upvotes]: This question is related to my previous question, to which I got a partial answer. Consider the cyclotomic field $L={{\mathbb{Q}}}(\zeta_8)={{\mathbb{Q}}}(\sqrt{2},i)$, where $\zeta_8$ is a primitive 8-th root of unity. Let $\Lambda={{\mathbb{Z}}}[\zeta_8]$ denote the ring of integers of $L$. Question 1. Does there exist an abelian variety $A$ over $\overline{\mathbb{Q}}$ of dimension 4 (or of any other dimension) with ${\mathrm{End\,}} A\simeq \Lambda$ ? I expect the answer "yes", but have no idea how to prove this. More generally, let $(G,X)$ be a Shimura datum. For any $x\in X$, let $h_x\colon \mathbb{S}\to G_\mathbb{R}$ be the corresponding homomorphism. We define the Mumford-Tate group $\mathrm{MT}(x)$ to be the smallest $\mathbb{Q}$-subgroup of $G$ containing $\mathrm{im\,} h_x$ over $\mathbb{R}$, and we define $\mathrm{MT}(X)$ to be the smallest $\mathbb{Q}$-subgroup of $G$ containing $\mathrm{im\,} h_x$ over $\mathbb{R}$ for all $x\in X$. We say that $x\in X$ is generic if $\mathrm{MT}(x)=\mathrm{MT}(X)$. Let $K\subset G(\mathbb{A}^f)$ be an open compact subgroup. Set $\Gamma=G(\mathbb{Q})\cap K$ (the intersection is taken in $G(\mathbb{A}_f)\,$). We say that $\Gamma$ is a congruence subgroup of $G(\mathbb{Q})$. Set $\mathrm{Sh}_\Gamma=X/\Gamma$, it is a connected Shimura variety. We say that $x\Gamma\in X/\Gamma$ is generic if $x\in X$ is generic. On the other hand, $\mathrm{Sh}_\Gamma$ has a (mysterious) canonical model over $\overline{\mathbb{Q}}$, so the notion of a $\overline{\mathbb{Q}}$-point of $\mathrm{Sh}_\Gamma$ is defined. Question 2. Does $\mathrm{Sh}_\Gamma$ always have a generic $\overline{\mathbb{Q}}$-point? The positive answer to Question 2, even for Shimura varieties of abelian type, would imply the positive answer to Question 1. Indeed, Shimura constructed an abelian scheme over a Shimura variety with prescribed automorphism ring of a generic fiber. If we find a generic $\overline{\mathbb{Q}}$-point of the base, then the fiber over this point will be a desired abelian variety over $\overline{\mathbb{Q}}$ with prescribed automorphism ring. REPLY [10 votes]: The Answer to Question 1 is YES. It follows from Serre's variant of Hilbert's irreducibility theorem (for infinite Galois extensions) combined with the Tate conjecture on homomorphisms of abelian varieties in char 0 (proven by Faltings); see Prop. 1.3 and Cor. 1.5 of 1995 Compositio paper by Rutger Noot (vol 97, Oort Festschrift) for details. The main idea is as follows. If $A$ is an abelian variety over a finitely generated field $K$ of char 0 and $\ell$ is a prime then there is an abelian variety $A_0$ over a number field $K_0$ that is a ``specialization/reduction" of $A$ and such that a canonical (up to a "conjugation") isomorphism of Tate modules $T_{\ell}(A)$ and $T_{\ell}(A_0)$ induces an isomorphism between the corresponding $\ell$-adic images of the absolute Galois groups of $K$ and $K_0$ in the automorphism groups of $T_{\ell}(A)$ and $T_{\ell}(A_0)$ respectively. Since $End(A)$ embeds into $End(A_0)$, one obtains that the endomorphism rings of $A$ and $A_0$ are isomorphic.<|endoftext|> TITLE: How verminous are mice? QUESTION [10 upvotes]: Recall that a mouse is a structure of the form $(J_\alpha[U],\in,U)$ with $U$ being an amenable ultrafilter with some iterability properties. One of the interesting facts about mice is that given two mice, $m$ and $n$, there is an ordinal $\lambda$, such that if we iterate both mice $\lambda$ times, one is going to be an initial segment of the other. This induces a pre-well ordering on the mice of the universe. In a recent course, we saw that there is no maximal element to this order. If you have a mouse, then you can define a slightly larger mouse. This means that if we look at $L[0^\#]$, which is a model with one nontrivial mouse, then there are infinitely nontrivial mice there, which are strictly larger than $0^\#$ itself. Assume that in $V$ the mice pre-well ordering has order type of $\rm Ord$, and let $M$ be an inner model such that the mice in $M$ are only those below $\alpha$ in said order, and $m$ is the least mouse not in $M$. What can we say about the ordering of the mice of $M[m]$? How many new non-equivalent mice have been added? In particular, what is the mice ordering in $L[0^\#]$? Clearly there order type is at least $\alpha+\omega$. But it's probably much more. How much more? REPLY [8 votes]: Looking at your example of mouse, it seems you are going to use, and are asking about, the ‘old-fashioned’ fine-structural mouse, as used originally by Dodd and Jensen in their papers, and by Dodd in his book. It is crucial to say this since the ‘modern’ definition involved a reorganisation of these hierarchies following a suggestion of S. Baldwin in the 1980’s. The ‘modern’ organisation is the current approach to mice in all the various versions of Jensen and Mitchell-Steel. (It essentially ensures that a strong form of GCH applies level-by-level, which the Dodd-Jensen approach did not. This property is called (strong) acceptability.) The upshot is that Dodd-Jensen have mice but a special subclass of these are the ’critical mice’ (and $O^\sharp$ is the first such, $O^{\sharp\sharp}$ is the second such, &c.). A critical mouse engenders an inner model by iterating it’s measure out through the ordinals. A non-critical DJ-mouse does not. (This is a major difference with modern mice: iterating a modern mouse by its topmost active extender repeatedly will always leave behind an inner model.) So mouse from now on means a Dodd-Jensen mouse, and $K$ will be the Diodd-Jensen core model built below the first inner model with a measurable cardinal. One may construct (non-critical) mice from critical mice, so that, as you say in $L[O^\sharp]$ then there are many mice definable from $O^\sharp$. You mention infinitely many but in fact there are a proper class many of such. So the length of the mouse order in $L[O^\sharp]$ is $On$ itself. In $L[O^{\sharp\sharp}]$ it is $On + On$ and so on. More manageably: in any model or V, we may consider the mice in HC, or in $H_\kappa$ for some regular $\kappa$. These can all be iterated to a common hierarchy generated by the c.u.b. filter on $\kappa$, $F=F_\kappa$ say. This common iterate is of the form $J_\theta^F$ and is the direct limit then of all the mice in $H_\kappa$. The ‘strength’ of the mouse ordering can then be gauged by properties of $\theta$, (if it admissible, if it is . . .) but always $cf(\theta) = \kappa$ in any universe. (As an example, in $L[O^\sharp]$, thinking of the mice in $H_{\omega_1}$ they iterate to $J^F_{\omega_1 + \omega_1}$ where $F$ is the c.u.b. filter on $\omega_1$. In $L[O^{\sharp\sharp}]$, $\theta$ would be $\omega_1 \cdot 3$.) Thus, to answer the question specifically, the mouse ordering in $L[O^\sharp]$ of the mice in $H_\kappa$ has length $\kappa$. In $L[O^{\sharp\sharp}]$ it is $\kappa\cdot 2$ &c. In the least model closed under $\sharp$’s it will have length $\kappa\cdot\kappa$. Once the model becomes even a little bit thicker there is no handy description of the length of the ordering in terms of some ordinal arithmetic. (Some of this is summarized in Sect. 3 of my “On unfoldable cardinals, $\omega$-closed cardinals and the beginnings of the inner model hierarchy”, Archive for Math. Logic, vol. 43, No. 4, (2004), pp.443-458.) The common iterate $(J^F_\theta, \in F)$ for $F=F_\kappa, \theta = \theta(\kappa)$ Dodd and Jensen called the “$Q$-structure at $\kappa$” - thus $Q_\kappa$ (but again this has a totally different usage in modern mouse theory). Once the $Q$ structure is admissible interesting things can happen. As a sample (see again cited paper): Theorem (V=K) If there exists a $\kappa$ with $Q_\kappa$ admissible, then $\forall D \subseteq \omega_1( D$ is universally Baire $\leftrightarrow \exists r\subseteq \omega (D \in L[r])$. Theorem (V=K) If $Q_{\omega_1}$ is admissible, then $\delta^1_2=\theta_{\omega_1}.$ These results are not lost by converting to the modern mice concepts. There are reasonably natural analogues. And to emphasise, the Dodd-Jensen $K$ has exactly the same sets as the 'modern' $K$ below a measurable cardinal. Most simply by forgetting about the non-critical mice in the Dodd-Jensen mouse ordering you will get an ordering of the same order type as the modern one of 'active' mice (or equivalently the order type of the inner models of $K$ under comparison).<|endoftext|> TITLE: Is a filtered colimit of rational spaces again rational? QUESTION [30 upvotes]: Let me first explain the statement of the question and then give some indication why the answer might be 'yes'. By a space I mean, say, a simplicial set and by rational I mean rational in the sense of Bousfield, i.e. local with respect to the homology theory $H\mathbb{Q}$. If we denote the category of spaces by $\mathcal{S}$ then the category of rational spaces $\mathcal{S}_\mathbb{Q} \subseteq \mathcal{S}$ is a full subcategory and the question is whether it is closed under filtered colimits. This would in particular imply that the $\infty$-category of rational spaces is compactly generated. Note that it is known that this $\infty$-category is generated by $\kappa$-compact objects for $\kappa$ the successor cardinal of $\omega$ (by results of Bousfield). But whether it is compactly generated is unclear to me and this is really what I want to know, even if the inclusion $\mathcal{S}_\mathbb{Q} \subseteq \mathcal{S}$ is not necessarily $\omega$-continuous. Now let me give some indication why the answer could be 'yes': first a simply connected (or more generally nilpotent) space is rational if and only if its homotopy groups are uniquely divisible. This property is obviously closed under filtered colimits. Thus a filtered colimit of simply connected, rational spaces is again rational. Bousfield characterizes rationality for arbitrary spaces as a property of homotopy groups, but this property is also homotopy theoretic in nature so that I am unable to decide if it is preserved by filtered colimits. The second indication is that one can study the stable analogue of my question: is a filtered colimit of rational spectra again rational. The answer is yes. This is in fact equivalent to the fact that rationalization is a smashing localization. This observation also shows that the class of $\mathbb{S}/p$-local spectra for a prime $p$ cannot be closed under filtered colimits since $p$-completion is not smashing. A consequence is that $\mathbb{F}_p$-local spaces (a.k.a. $\mathbb{S}/p = M(\mathbb{Z}/p)$-local spaces) are not closed under filtered colimits. REPLY [10 votes]: The answer to this, as a result of some discussions with Thomas, turns out to be no. Consider the abelian groups $A_n = \Bbb Q[x]/(x^n)$. Multiplication by $x$ embeds $A_n$ into $A_{n+1}$, and the direct limit of the sequence $$ A_0 \to A_1 \to \dots $$ is the group $A_{\infty} \cong \Bbb Q[x^{\pm 1}] / \Bbb Q[x]$. The group $\Bbb Q$ acts compatibly on the $A_n$ and $A_\infty$ if we define $$ q \ast m = (1+x)^q m $$ where $(1 + x)^q = \sum \binom{q}{n} x^n$ is defined using the binomial theorem. The action of $\Bbb Q$ on each $A_n$ is nilpotent: there is a finite filtration whose layers are acted on trivially. In fact, $\Bbb Q$ acts trivially on the quotient groups $A_n/A_{n-1} \cong \Bbb Q$ because $x$ acts trivially on this group. By contrast, the action of $\Bbb Q$ on $A_\infty$ is perfect: for any element $m \in A_\infty$, we find $$ m = x (x^{-1} m) = 1 \ast (x^{-1} m) - (x^{-1} m) $$ and so the set of elements of the form $q \ast y - y$ generate all of $A_\infty$. Therefore, $A_\infty / \Bbb Q = 0$. There is an associated sequence of Eilenberg--Mac Lane spaces $K(A_n,2)$ and $K(A_\infty,2)$ with compatible actions of $\Bbb Q$, for example by using a functorial construction of E-M spaces. We can form the Borel constructions $X_\alpha = K(A_\alpha,2) \times_{\Bbb Q} E\Bbb Q$. These spaces have the property that $$ \pi_n X_\alpha = \begin{cases} \Bbb Q &\text{if }*=1,\\ A_\alpha &\text{if }*=2,\\ 0 & \text{otherwise}. \end{cases} $$ In particular, $X_\infty$ is the homotopy colimit of the spaces $X_n$. Each $X_n$ is a nilpotent space: it has abelian (hence nilpotent) fundamental group, the fundamental group acts nilpotently on the higher homotopy groups, and the homotopy groups are rational vector spaces. This means that $X_n$ has a Postnikov-type construction formed by iteratively taking fibers of maps to $K(\Bbb Q,m)$, and so $X_n$ is local with respect to rational homology. By contrast, $X_\infty$ is not local with respect to rational homology. The fact that $\pi_1(X_\infty)$ acts perfectly on $\pi_2(X_\infty)$ allows us to use an analogue of Quillen's plus-construction to find a map $X_\infty \to Y$, by attaching $3$-dimensional and $4$-dimensional cells, which is a rational homology isomorphism but which induces the zero map $\pi_2(X_\infty) \to \pi_2(Y)$. In particular, the map $X_\infty \to (X_\infty)_{\Bbb Q}$ to its rationalization factors through $Y$, and so it must send all of $\pi_2$ to zero.<|endoftext|> TITLE: Can any group be embedded in a simple group? QUESTION [25 upvotes]: Any finite group $G$ can be embedded into $A_{|G|+2}$ via Cayley's theorem ($G\hookrightarrow S_{|G|}\hookrightarrow A_{|G|+2}$). If $G$ is not assumed to be finite, is it still always possible to embedd it into a simple group? REPLY [45 votes]: Yes. Assume $G$ is infinite. Cayley still embeds $G$ into the group $S_G$ of permutations of $G$. This group is no longer simple: there is a normal subgroup, call it $N_G$, consisting of all permutations that fix the complement of a subset of $G$ of cardinality smaller than that of $G$. But by a theorem of Baer, Schreier and Ulam, every normal subgroup of $S_G$, other than $S_G$ itself, is contained in $N_G$. Hence $Q_G := S_G / N_G$ is simple. Moreover the composite map $G \to S_G \to Q_G$ is still an embedding because no non-identity element of $G$ has any fixed points. We have thus embedded $G$ into the simple group $Q_G$.<|endoftext|> TITLE: Why does doubling move from product-is-square to each-is-square? QUESTION [11 upvotes]: Let $E_d$ be the group of rational points on the elliptic curve $$y^2=(x-d)x(x+d)$$ It is well-known – and easy to verify by elementary algebra – that $(x,y)\in E_d$ belongs to $2E_d$ precisely when all three of $x-d$, $x$, $x+d$ are squares. This seems too neat to be a mere coincidence. Is there a natural generalisation of this fact? Or a more conceptual way to understand why it ought to be true? REPLY [14 votes]: A geometric view to complement Joe Silverman's cohomological answer: Suppose more generally that $E$ is the "product-is-square" elliptic curve $$ y^2 = (x-e_1)(x-e_2)(x-e_3) $$ for some pairwise distinct $e_i$, and let $C$ be the "each-is-square" curve $$ \begin{cases} y_1^2 = x-e_1, \cr y_2^2 = x-e_2, \cr y_3^2 = x-e_3, \end{cases} $$ Then $E$ is a double cover of the $x$-line, and $C$ is a $\{\pm1\}^3$ cover with $E$ as an intermediate cover with Galois group isomorphic with $({\bf Z} / 2{\bf Z})^2$; explicitly the $4:1$ map $C \to E$ is $$(x,y_1,y_2,y_3) \mapsto (x, y_1 y_2 y_3).$$ By Riemann-Hurwitz, $E$, like $C$, has genus $1$, so this $({\bf Z} / 2{\bf Z})^2$ cover $C \to E$ is unramified. Hence if we choose for the origin of $C$ one of the preimages of the origin of $E$, the cover must be the doubling map. REPLY [13 votes]: It looks as if you're working over $\mathbb Q$. And you are using an elliptic curve in which all of the 2-torsion is rational, so $E[2]$ is isomorphic to (say) $\boldsymbol\mu_2^2$ as a Galois module, where $\boldsymbol\mu_2=\{\pm1\}$ is the group of square roots of 1. Okay, now consider the injection (this comes from basic Kummer theory of elliptic curves) \begin{align*} E(\mathbb Q)/2E(\mathbb Q) &\hookrightarrow H^1(G_{\overline{\mathbb Q}/\mathbb Q},E[2]) \\ &\cong H^1(G_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu_2^2) \\ &\cong H^1(G_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu_2)^2 \\ &\cong \mathbb Q^*/(\mathbb Q^*)^2\times \mathbb Q^*/(\mathbb Q^*)^2. \end{align*} (The final isomorphism is standard Kummer theory.) So this at least tells you that whether $P$ is in $2E(\mathbb Q)$ should depend on whether certain values are squares. If you trace through all of the maps, which requires choosing a basis for $E[2]$, you'll find (for one of the choices) that it is given by $$ P \longmapsto \bigl(x(P),x(P)-d\bigr), $$ so $P$ is in $2E(\mathbb Q)$ if and only if both $x(P)$ and $x(P)-d$ are squares. Of course, from the equation of $E$, any two of $x(P)$, $x(P)-d$ and $x(P)+d$ being a square forces the third one to also be square. Finally, if $x(P)=0$ or $x(P)=d$, one finds during the analysis that one needs to use a different formula. For example, since $x$ and $(x-d)(x+d)$ differ multiplicatively by a square, one has $$ 0 \longmapsto (-d^2,-d) = (-1,-d) \in \mathbb Q^*/(\mathbb Q^*)^2\times \mathbb Q^*/(\mathbb Q^*)^2. $$ To answer your second question, if $E[m]\subset E(K)$, then one similarly gets an injection $$ E(K)/mE(K) \hookrightarrow K^*/(K^*)^m\times K^*/(K^*)^m, $$ so (more or less) there are rational functions $f$ and $g$ in $K(E)$ such that for $P\in E(K)$ we have $$ P \in mE(K) \;\Longleftrightarrow\; \text{$f(P)$ and $g(P)$ are $m$'th powers in $K^*$.} $$<|endoftext|> TITLE: Results on the largest prime factor of $2^n+1$ QUESTION [7 upvotes]: A work of Cameron Stewart (the paper has appeared in Acta Mathematica), proving a conjecture of Erdos, Stewart shows that the largest prime factor of $2^n-1$ is at least $n \times \exp\Big( \frac{\log n}{104 \log \log n}\Big)$ , if $n$ is large enough. Zsigmondy's theorem is on the same topic. I would like to know about the largest prime factor of $2^n+1$. I have searched, but most of the time result of the largest prime factor of $2^n-1$ appears. My questions is- Is there any result regarding the largest prime factor of $2^n+1$ ? if you know anything related to the problem, please inform. REPLY [8 votes]: One has $2^n+1= \frac{2^{2n}-1}{2^n-1} = \prod_{d \mid 2n,~d\nmid n} \Phi_d(2,1)$. In particular, $\Phi_{2n}(2,1)$ divides $2^n+1$, and so the main theorem of Stewart's paper shows immediately that $P(2^n+1) > 2n \exp(\log(2n)/104\log\log(2n))$ for large $n$.<|endoftext|> TITLE: Impact of applying LEM to non-definite statements on definite statements QUESTION [8 upvotes]: Solomon Feferman (1928 – 2016) hold that statements of arithmetic are definite, while "higher-order" notions (such as the set of all subsets of $\mathbb N$) are vague, and questions about them might not be objectively answered. He proposed a formal system of mathematics (or, if you will, of set theory) that allowed to apply the law of excluded middle to arithmetical statements, but not to statements which included vague concepts. That is, in the definite realm we may reason in classical logic, while in the less definite realm one only can use intuitionistic logic (let me call this "semi-intuitionistic"). See the thread Is platonism regarding arithmetic consistent with the multiverse view in set theory? for a discussion on that view. In his paper Is the continuum hypothesis a definite mathematical problem?, Feferman has a section called "3. A proposed logical framework for what’s definite (and what’s not)". There he discusses formal systems that are "semi-intuitionistic" in the sense that these systems somehow distinguish between "definite" and "vague" statements and depending on which of these two types a statement has, one might apply the law of excluded middle to it or not. In this present question I don't want to fix a particular "semi-intuitionistic" formal systems, I just note that it is possible to formalize this idea. Question. Is it possible that there are definite, arithmetical statements that are provable in classical logic + ZFC, but not in such a semi-intuitionistic version of set theory? My motivation of this question are my philosophical thoughts on ZFC and my objection that there might be statements provable in ZFC, but for which this proof isn't satisfying in the sense that it applies classical logic to non-definite statements. In my opinion, these proofs should be treated with caution, since I don't believe in the definiteness of general set theoretic statements. One may critise that my question is vague, for those of you who do, let me put my questions in other words: Are there any results known which demonstrate for a semi-intuitionstic formal system $K$ that there is a definite statement $s$ not provable in this systems, but which is provable in ZFC + classical logic? REPLY [9 votes]: There are no such statements $s$, at least not for a suitably formalized version of your question, using intuitionistic set theory with Peano arithmetic. Suppose we have an arithmetical statement $\phi$, and $ZFC \vdash \phi$. By Shoenfield absoluteness, $ZF \vdash \phi$. Then by results of Friedman, $IZF \vdash \phi^{-}$, the double-negation translation of $\phi$. So $IZF + PA \vdash \phi$. See Michael Beeson, Foundations of Constructive Mathematics (1985), chapter VIII.3. That chapter shows that this phenomenon is fairly robust to the choice of formalization.<|endoftext|> TITLE: maximal tensor product of simple $C^*$algebras is non-simple QUESTION [6 upvotes]: Let $A$ and $B$ simple $C^*$-algebras. One can prove that the minimal tensor product $A\otimes _{min}B$ is simple. This is wrong for the maximal tensor product $A\otimes_{max}B$ . 1.Do you know an explicit example? One idea is to consider (non-nuclear) $C^*$-algebras $A$ and $B$, which are simple and then consider the canonical map $i:A\otimes_{max} B\to A\otimes_{min}B$. One can take $A$ and $B$ such that the map has non-trivial kernel to obtain $\ker(i)\subseteq A\otimes_{max} B$ as a non-trivial closed ideal. Also the example should satisfy: $\ker(i)\neq A\otimes_{max} B$. Therefore if you take this idea, for example, $A=K(H)=B$ for a separable Hilbert space $H$ doesn't work because these $C^*$-algebras are nuclear, simple and the minimal tensor product is simple. Then I tried to take (non-nuclear) group $C^*$-algebras. But I'm not sure for which groups $\Gamma$ the reduced group $C^*$-algebra $C_r^*(\Gamma)$ is non-simple, my knowledge is still very limited. For example, there is a result in a paper that if $\Gamma$ is a nonabelian free group, the $C_r^*(\Gamma)$ is simple (R.T. Powers, "Simplicity of the $C^*$-algebra associated with the free group on two generators."). Therefore, my 2. question is: Do you know a (non-abelian) group $\Gamma$, such that $C_r^*(\Gamma)$ is not amenable and non-simple? REPLY [6 votes]: An explicit example to question 1 is given by $A=B=C^*_r(G)$ where $G$ is the free group on two generators. Takesaki produced the first systematic study of nuclearity (then called Property (T)) in his 1964 paper "On the cross-norm of the direct product of C*-algebras." In Theorem 6 (actually look at the discussion preceding Theorem 6) he shows that $A\otimes_{max} B\neq A\otimes_{min}B.$ And, as you mentioned, Powers showed that $A$ is simple.<|endoftext|> TITLE: Baum Connes conjecture and Atiyah-Singer index theorem QUESTION [10 upvotes]: Baum Connes conjecture is considered as a far generalisation of the Atiyah Singer index theorem (in K-theoretical formulation). I would like to understand how the latter follows from this conjecture. My guess is that one can consider trivial group $G$ but I don;t know how to proceed. REPLY [9 votes]: I find it overwhelmingly difficult to do this question proper justice. So instead I give a highly condensed answer and refer to Connes NCG, Section II.10 for more details (as well as all the literature that has appeared in the meantime concerning the groupoid approach to index theory). In my view a good way to understand how BC is a vast generalization of Atiyah-Singer index theory, is to study the conjecture in its more general form, as stated for Lie groupoids (i.e. groupoids with a smooth structure). Connes gives a very slick proof of AS using a deformation groupoid (called the tangent groupoid). A lot of the following needs "unpacking" and further study in order to be useful. I have certainly glossed over a number of subtleties. Let me just try to give a short recapitulation of Connes idea: Let $M$ be a compact manifold without boundary. Preparations: The first insight is that the Fredholm index of a (pseudo-) differential operator can be rewritten in terms of the tangent groupoid: $G_M^t = M \times M \times (0,1] \cup TM \times \{0\}$. The intuition of this object is to glue the groupoid our operator lives on, namely the pair groupoid $M \times M$, to the tangent bundle $TM$ where its principal symbol lives on (viewed as a bundle of additive groups $(T_x M, +)$, hence a groupoid). The units of this groupoid are $M \times [0,1]$. There are various interpretations on how to obtain a natural smooth structure on $G_M^t$ which I won't go into. One way of looking at it is that $TM$ can be viewed as the normal bundle to the embedding of the diagonal $\Delta_M$ inside $M \times M$ and a smooth structure can be transported using the tubular neighborhood theorem (via the exponential mapping). Consider $C^{\ast}$-algebra of the groupoid and particularly the evaluation at $t = 0$ as $e_0 \colon C^{\ast}(G_M^t) \to C^{\ast}(TM)$ and at $t = 1$ as $e_1 \colon C^{\ast}(G_M^t) \to C^{\ast}(M \times M)$. Then it is not hard to show that in $K$-theory $(e_0)_{\ast} \colon K(C^{\ast}(G_M^t) \to K(C^{\ast}(TM))$ yields an isomorphism (the kernel of $e_0$ is the contractible $C^{\ast}$-algebra $C^{\ast}(M \times M) \otimes C(0,1]$). Note that $C^{\ast}(TM) \cong C_0(T^{\ast} M)$ via the Fourier transform and $C^{\ast}(M \times M) \cong \mathcal{K}$ are the compact operators on $L^2(M)$. Now set: $ind_a := (e_1)_{\ast} \colon (e_0)_{\ast}^{-1} \colon K^0(T^{\ast} M) \to K(\mathcal{K}) \cong \mathbb{Z}$. That $ind_a$ is in fact equal to the Fredholm index follows quickly by a standard device of $C^{\ast}$-algebra $K$-theory (six-term exact sequence). The second insight is a generalized Thom isomorphism of groupoids which transfers the calculation of the index to a groupoid which is Morita equivalent to a space. Here one needs the embedding of $M$ into $\mathbb{R}^N$. Connection to Baum-Connes The Baum-Connes conjecture for a Lie groupoid $G$ asserts that the assembly mapping $\mu \colon K_0(G) \to K_0(C^{\ast}(G))$ is an isomorphism. More precisely, what I want to consider here is the geometric assembly mapping for Lie groupoids. This means the map $\mu_{G}^{geo} \colon K_0^{geo}(G) \to K_0(C^{\ast}(G))$. The precise definition of $K_0^{geo}(G)$ as well as $\mu_{G}^{geo}$ can be found in NCG. I focus here mostly on how they specialise for the tangent groupoid defined above. The connection to BC becomes clear by examining closely the proof of Atiyah-Singer using the tangent groupoid. Steps Embed $M$ into $\mathbb{R}^N$ and assume that $N$ is even. 1) In general $\mu_G^{geo}$ is constructed using a universal $G$-space (for proper $G$ actions) as well as a specially constructed shriek map (using a deformation groupoid associated to the anchor of the action of $G$ on its universal space). Consider $G = G_M^t$, the tangent groupoid from above. Then $\mu_{G_M^t}$ is simply given by the shriek map from a particular action of $G_M^t$ on $Z = M \times [0,1] \times \mathbb{R}^N$. In other words - after some digging through the definitions in NCG - we obtain the well-defined homomorphism $\mu_{G_M^t} \colon K(C^{\ast}(Z \rtimes G_M^t)) \to K(C^{\ast}(G_M^t))$. The definition of the space $Z \rtimes G_M^t$ needs some further unpacking: It is actually the orbit space of the action of the groupoid $G_M^t$ on $Z$ which is obtained with the help of a homomorphism $h \colon (G_M^t, \cdot) \to (\mathbb{R}^N, +)$ defined with the help of the embedding $j$ of $M$ into $\mathbb{R}^N$: $h(x,y,t) = \frac{j(x) - j(y)}{t}$ if $t > 0$ and $h(x,v) = dj(v)$, $v \in T_x M$ if $t= 0$. 2) Key observation: The map $\mu_{G_M^t}$ is the inverse Connes-Thom isomorphism, i.e. the (inverse!) of the equivariant Thom isomorphism: $\mathcal{CT} \colon K(C^{\ast}(G_M^t)) \to K(C^{\ast}(Z \rtimes G_M^t))$. For the latter isomorphism one observes that $C^{\ast}(Z \rtimes G_M^t) \cong C^{\ast}(G_M^t) \rtimes \mathbb{R}^N$. Here $\mathbb{R}^N$ acts on $C^{\ast}(G_M^t)$ via $\alpha_v(f)(\gamma) = e^{i (v \cdot h(\gamma))} f(\gamma), f \in C_c(G_M^t)$. Using that $N$ is even we apply Bott periodicity to arrive at the isomorphism $\mathcal{CT}$. 3) The Connes-Thom isomorphism is nothing but the inverse geometric assembly map. I have omitted the proof of this (it involves yet another deformation). This is the key tool to prove the Atiyah-Singer index theorem: Connes calculates the orbit space $Z \rtimes G_M^t$ (after defining $h$ with the help of the fixed embedding of $M$ into $\mathbb{R}^N$). It turns out that $Z \rtimes G_M^t$ is then another deformation of $\mathbb{R}^N$ into the normal bundle $\mathcal{N}$ to the inclusion of $M$ into $\mathbb{R}^N$, i.e. $BG_M^t = (]0,1] \times \mathbb{R}^N) \cup \mathcal{N}$ as a set. We have the evaluations: $(e_0^h)_{\ast} \colon K(C^{\ast}(Z \rtimes G_M^t) \to K(\mathcal{N})$ and $(e_1^h)_{\ast} \colon K(C^{\ast}(Z \rtimes G_M^t)) \to K(\mathbb{R}^N)$. Together with the Thom isomorphism $K^0(TM) \to K^0(\mathcal{N})$ and the Bott isomorphism (contracting $\mathbb{R}^N$ to a point) $K(\mathbb{R}^N) \cong \mathbb{Z}$ we recover the topological index $ind_t$ of Atiyah-Singer using $Z \rtimes G_M^t$. Recall at this point the definition of $ind_a$. The Connes-Thom isomorphism $CT$ (which specializes to the Thom isomorphism for $t = 0$) - or the inverse of $\mu_{G_M^t}$ - provides the last missing piece to obtain equality of $ind_t$ with $ind_a$ proving the Atiyah-Singer index theorem. General case We can now ask the question: What does the Baum-Connes conjecture tell us about index theory of more general spaces? As observed above, if we consider the tangent groupoid for a compact manifold without boundary, the geometric Baum-Connes assembly map yields basically the Atiyah-Singer theorem. In general the story gets really interesting: We can study manifolds with boundary or more generally foliated manifolds (i.e. manifolds with a family of immersed submanifolds or leaves). There are associated groupoids for such more singular manifolds to which a lot of the machinery of index theory can be adapted. Connes defined potentially useful analogs of the analytical and topological index for the case of foliations. The general analytical index takes values in $K(C^{\ast}(G))$ where $G$ is the holonomy groupoid of the foliation. The Baum-Connes conjecture is in this case indistinguishable from the index problem which states that we ought to find a topological interpretation of this general analytical index (in terms of the $K$-homology of $G$).<|endoftext|> TITLE: order of reality QUESTION [6 upvotes]: Suppose I have a monic polynomial $p(x) \in \mathbb{Z}[x],$ of degree $d,$ and $\alpha$ a nonzero root of $p(x).$ The question is: assuming that $\alpha$ has some real power, what is the maximum order of $\alpha$ in $\mathbb{C}^\times/\mathbb{R}^\times?,$ in terms of $d?$ Obviously, order $d$ is possible and achieved by $p(x) = x^d-1.$ EDIT since there appears to be some confusion, here is a restatement: suppose $\alpha$ is an algebraic integer of degree $d,$ with $\arg \alpha = \frac{2 \pi k}{l},$ with $k, l \in \mathbb{Z}$ in lowest terms. Can we bound $l$ in terms of $d?$ REPLY [8 votes]: The trivial bound is slightly above $d^2$. Note that $\beta=\alpha/\bar\alpha$ is a root of unity of pretty much the same order (give or take a factor of $2$) whose minimal polynomial has degree at most $d^2$. However, with the roots of unity, we know that the order $n$ corresponds to the degree $\varphi(n)$ (cyclotomic polynomials and such) and $\varphi(n)$ cannot be much less than $n$ on the rough power scale (but can be as many times smaller than $n$ as you wish, whence the discrepancy between $2$ and $3$ in Christian's example). I have no idea if the square is actually needed though...<|endoftext|> TITLE: Lelong number of Ricci flat metric QUESTION [5 upvotes]: Let $M$ be a compact Kahler Calabi-Yau variety which admit Ricci flat metric $\tilde\omega$, $Ric(\tilde \omega)=0$, then the Lelong number $\tilde \omega$ is zero? In general if $\omega$ satisfies in the Kahler-Einstein equation $$Ric(\omega)=\lambda\omega$$ then is the Lelong number of $\omega$ vanishes? REPLY [8 votes]: Motto: Canonical metric has vanishing Lelong number or (some times called Lelong-Demailly number) Let me start with the definition of Lelong number. Let $W\subset \mathbb C^n$ be a domain, and $\Theta$ a positive current of degree $(q,q)$ on $W$. For a point $p\in W$ one defines $$\mathfrak v(\Theta,p,r)=\frac{1}{r^{2(n-q)}}\int_{|z-p| TITLE: What tools can I use to compute the cohomology of the fibers of a Lefschetz Pencil? QUESTION [6 upvotes]: I'm learning about Lefschetz pencils and vanishing cycles and have looked at a few sources: http://www.math.purdue.edu/~dvb/preprints/sheaves.pdf http://www3.nd.edu/~lnicolae/Morse2nd.pdf Voisin's Complex Algebraic Geometry and Hodge Theory II I want to be able to start computing explicit examples of vanishing cycles and monodromy, but I need to first learn how to compute the singular homology of complex varieties. What are some tools I can use to compute cohomology of some smooth projective varieties over $\mathbb{C}$? For example, how can I study the cohomology of some Lefschetz pencil for the projective scheme $$ \begin{matrix} \textbf{Proj}\left( \frac{\mathbb{C}[x,y,z,w]}{(x^8 + y^8 + z^8 + w^8 +x^2y^2z^2w^2)} \right) \end{matrix} $$ REPLY [2 votes]: Your example is a smooth hypersurface, which is much easier to understand than a general smooth projective variety. You ask for a computation of the cohomology of a Lefschetz pencil. This could mean either: The cohomology groups of the total space of the pencil. A description of the Gauss-Manin local system on $\mathbb P^1-\Delta$ (e.g. in terms of monodromy action of the generators for $\pi_1$). The Hodge numbers of a hypersurface $X\subset \mathbb P^n$ can be computed using Lefschetz hyperplane away from the middle row, and then Hirzebruch's generating function for the primitive middle Hodge numbers (see http://www.math.purdue.edu/~dvb/preprints/book-chap17.pdf for details). In your example, the Hodge diamond has middle row $(35,232,35)$. The total space of a Lefschetz pencil is the blow up of $X$ at the base locus of the pencil (8 points in your example), so it has $b_2 = 310$. The local system description is more difficult. Let $C_p$ be a smooth octic curve over a general point $p\in \mathbb P^1$. The local system has rank $h^1(C_p)=2g(C_p)= 42$. There are $392$ nodal curves in the pencil. Since $\pi_1(\mathbb P^1 - 392,p)$ is free on 391 generators, you need to find the monodromy for each generator. By the Picard-Lefschetz formula, the monodromy around a given nodal fiber is given by $$\tau(x) = x- \langle x,e \rangle e$$ where $e\in H^1(C_p,\mathbb Z)$ is the vanishing cycle, represented by an embedded $S^1$ which gets contracted to the node.<|endoftext|> TITLE: Poisson cohomology QUESTION [7 upvotes]: Let $\Pi$ be a Poisson structure on a manifold $M$. Then we can define a differential $d$ on the complex $\Lambda^{\bullet}M$ $$ C^{\infty}M \to TM \to...\Lambda^kTM \to... $$ in the following way: $$ d = [\Pi,\cdot], $$ where $[\cdot,\cdot]$ is the Schouten-Nijenhuis bracket. The cohomology of this complex is called Poisson cohomology. I have several question about this construction: Suppose $M=\mathbb R^{2d+s}$ with a Poisson structure $\sum\limits_{i=1}^{d}\frac{\partial}{\partial q_i}\wedge \frac{\partial}{\partial p_i}$. Is it true that Poisson cohomology of $M$ are zero? Is this construction functorial? I mean, let $f\colon M\to N$ be a poisson map, then do we have a map of complexes $f_*\colon \Lambda^{\bullet}M \to \Lambda^{\bullet}N$? Let $F\colon M\times I \to N$ be a homotopy such that for each $t$ $F_t$ is a Poisson map. Is it true that $F_0$ and $F_1$ induce the same map from Poisson cohomology of $M$ to Poisson cohomology of $N$? REPLY [9 votes]: First of all, one can think of Poisson cohomology (or, rather, Poisson cochains) as functions on the $QP$-manifold $T^*[1]M$ with the differential as you describe. In particular, it has a natural odd Poisson ($\mathbb{P}_2$) structure given by the Schouten bracket. You can split $T^*[1](\mathbb{R}^{2d+s}) = T^*[1](\mathbb{R}^{2d})\times T^*[1](\mathbb{R}^s)$ and so the same decomposition applies to Poisson cohomology. Since $\mathbb{R}^{2d}$ is symplectic, its Poisson cohomology coincides with the de Rham cohomology which is quasi-isomorphic to $\mathbb{R}$ in degree 0. Therefore, Poisson cohomology of $\mathbb{R}^{2d+s}$ coincides with the Poisson cohomology of $\mathbb{R}^s$ with the zero Poisson structure, i.e. functions on $T^*[1](\mathbb{R}^s)$. Given a map of manifolds $M\rightarrow N$, there is no natural map on their cotangent bundles. Instead, there is a Lagrangian correspondence $$T^* M \leftarrow T^*N\times_N M \rightarrow T^* N.$$ The same construction works with a shift and so instead of a map one has a dg bimodule between Poisson cohomologies of $M$ and $N$. The underlying graded bimodule is $C^\infty(T^*[1] N\times_N M)$. The differential is constructed as follows. Recall that for any coisotropic submanifold $C\hookrightarrow X$ the graded commutative algebra $\bigwedge^\bullet N_{C/X}$ carries a natural differential (a version of the differential you wrote). The map $M\rightarrow N$ is Poisson iff its graph $M\rightarrow M \times N$ is coisotropic and taking the exterior algebra on the normal bundle of this coisotropic you recover $C^\infty(T^*[1] N\times_N M)$ and that's the required differential. Given such a homotopy, you get a homotopy of (dg) bimodules from point 2, which seems to be the best you can hope for.<|endoftext|> TITLE: Why do homotopy theorists care whether or not $BP$ is $E_\infty$? QUESTION [20 upvotes]: I have often heard that it is not known whether or not the Brown-Peterson spectrum $BP$ is an $E_\infty$-ring spectrum. Though I see that this is a somewhat natural question to ask, I have often wondered what would the practical consequences of an answer to this question would be. In particular, I am wondering what would a positive answer to this question mean for homotopy theory. Are there computations we would be able to do that we couldn't do before? What would be the take away if $BP$ did not admit an $E_\infty$-ring structure? I would also appreciate any references on this problem. Thanks! REPLY [21 votes]: I suppose I should try to answer since the question of whether or not $BP$ is an $E_{\infty}$ ring spectrum was Problem 1 of "Problems in infinite loop space theory'', http://www.math.uchicago.edu/~may/PAPERS/16.pdf, written in 1974, two years after $E_{\infty}$ ring spectra were first defined. At the time, we were hoping for some good way of starting from the relevant group law and manufacturing $BP$ from it. The example "in nature'' that led to the definition of $E_{\infty}$ ring spectra was $MU$, so it was impossible not to be intrigued by the question. Some things proven since with $MU$ might be streamlined if we could start with $BP$, such as the construction of Morava $K$-theory. However, Basterra and Mandell proved that $BP$ is an $E_4$-spectrum, compatibly with $MU$, which is enough for such applications. I'm still intrigued by the question, although I admit to having no applications in mind.<|endoftext|> TITLE: Do singularities of plane curves deform independently? QUESTION [5 upvotes]: If $C\subset\mathbb{P}^2$ is an integral curve of degree $d$, do its singularities deform independently as we vary $C$ over degree $d$ curves? If not, what about in the case $C$ is a nodal curve or $C$ is a general nodal curve? Edit: In other words, is the composition $H^0(C,N_{C/\mathbb{P}^2})={\rm Hom}(N_{C/\mathbb{P}^2}^\vee,\mathscr{O}_C)\rightarrow {\rm Ext}^1(\Omega_C,\mathscr{O}_C)\rightarrow Ext^1(\Omega_C,\mathscr{O}_C)$ surjective? (If I did the reduction correctly, which is not a given, I think this is equivalent to the vector space $V=k[X,Y,Z]_d$ of degree $d$ homogenous polynomials surjecting onto the sheaf associated to the graded module $k[X,Y,Z]/(F,\partial_XF,\partial_YF,\partial_ZF)$, where $C=V(F)$.) REPLY [5 votes]: A good survey on this problem is the paper by Greuel, Lossen and Shustin Equisingular families of projective curves, see http://arxiv.org/pdf/math/0612310.pdf. In particular, at page 5 one can find the reference [Sev68] to the work of Severi showing that singular points of a nodal curve, irreducible or not, can be smoothed, or preserved, independently. This answers the last part of your question. Regarding the possibility of independently smoothing singularities worse than nodes, as explained in J. Starr's comment the answer is in general no, even if this can be hard to check in explicit examples. In fact, at p. 30 of the aforementioned paper the authors state the following more general question for cuspidal curves: Are there $k$ and $d$ such that a curve of degree $d$ with $k$ cusps does exist, but with $k′ < k$ cusps does not ? A candidate are a series of irreducible cuspidal curves constructed by Hirano (reference [Hir92]). They have degree $d= 2 \cdot 3^t$, with $t \geq 1$, and contain precisely $k=9(9t −1)/8$ cusps.<|endoftext|> TITLE: Infinitely many solutions to a particular embedding problem in Galois theory QUESTION [17 upvotes]: Given a Galois extension of number fields $L/K$ and an exact sequence of groups $$1\to \ker \varphi\to G \overset{\varphi}{\to} \text{Gal}(L/K)\to 1$$ where $G$ is a finite group, $\ker \varphi$ is solvable and $(|\ker \varphi|, |G/\ker\varphi|)=1$, then there exists a Galois extension $N/K$ that contains $L$ and such that $G=\text{Gal}(N/K)$ and $\varphi(\sigma)=\sigma|_L$ (see V.V. Ishkhanov, The Embedding Problem in Galois Theory). My question: is it true that in this case the embedding problem has infinitely many solutions? i.e. there exist infinitely many such $N/K$ ? I'm particularly interested in the case $\ker \varphi$ is a $2-$group. And what happens when $\ker \varphi$ is abelian? REPLY [4 votes]: The answer is positive, and the abelian case can be handled even over 'Hilbertian fields'. By the Schur-Zassenhaus Theorem the exact sequence splits, so we get a so called split embedding problem. If the kernel is abelian, the embedding problem is regularly solvable. That is, it is (properly) solvable over the function field $K(T)$. Since $K$ is Hilbertian (i.e. satisfies Hilbert's irreducibility theorem), the solution over $K(T)$ can be specialized to a solution over $K$ in infinitely many distinct ways. In fact, the infinitely many specializations can be taken linearly disjoint over the fixed field of $\mathrm{Ker}(\varphi)$ (i.e. $L$). For more details see the book 'Field Arithmetic' by Fried and Jarden. More generally, consider the embedding problem where $G$ is replaced by its $n$-th fibered power over $\varphi$. The solvability of this embedding problem (which follows from the result stated by the OP) implies that the original embedding problem has $n$ distinct solutions for each positive integer $n$, as required. The $n$-th fibered power $G^n_{\varphi}$ of $G$ over $\varphi$ is the subgroup of all $(g_1, \dots,g_n) \in G^n$ such that $\varphi(g_i) = \varphi(g_1)$ for all $1 \leq i \leq n$. It fits into the exact sequence $$1\to \ker(\varphi|_G)^n \to G^n_{\varphi} \overset{\varphi}{\to} \text{Gal}(L/K)\to 1.$$ Even more generally, it was conjectured that the answer to this question is positive also in the case that the kernel is not solvable. Specifically, Debes and Deschamps conjectured that every finite split embedding problem is solvable over a number field (so in particular, the inverse Galois problem has an affirmative answer).<|endoftext|> TITLE: Günter Tamme's course "Arakelov theory and Grothendieck-Riemann-Roch" QUESTION [21 upvotes]: On chapter III.4 ("Metrized $\mathcal{o}$-modules") of this book on algebraic number theory, Neukirch credits his treatment of the theory of finitely generated $\mathcal{o}$-modules to the course "Arakelov theory and Grothendieck-Riemann-Roch" taught by Günter Tamme. He continues: There, however, proofs were not given directly, as we will do here, but usually deduced as special cases from the general abstract theory. In the references, Neukirch only lists Tamme's book on Étale Cohomology and his article on the proceedings "Beilinson's Conjectures on Special Values of L-functions". Was any of this ever put on writing? Alternatively, is there a good reference where this things are "deduced as special cases from the general abstract theory"? REPLY [3 votes]: Here is a link of Gunter Tamme's course. I take a brief look and it seems centering around proving Gronthendieck-Riemann-Roch using K-theory machinery. I did not see Arakelov theory anywhere. The course notes at here are incomplete (the page 106 is missing, for example). Personally I would guess the "general abstract theory" can be found in Fulton's book and SGA6. Neukirch should be credited for all the concrete proofs he presented in his book. I think the recent expository article by Gerald Montplet in the Springer "Progress in Mathematics" book and his papers are great examples of ""deducing special cases from the general abstract theory". This is usually very difficult in Arakelov theory or even in index theory.<|endoftext|> TITLE: $\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$ is a convex function on $[0,+\infty)$? QUESTION [26 upvotes]: Let $f(x)=\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$, where $u\in\mathbb{Z}^+$ and $\binom{x}{l}=\frac{x(x-1)\dots(x-l+1)}{l!}$ for all $l\in\mathbb{Z}^+$. Then can we prove $f(x)$ is a convex function on $[0,+\infty)$? Updates: 1) It was pointed out by @user44191 that, observing $\binom{x}{i}=\binom{x-1}{i}+\binom{x-1}{i-1}$, the question is equivalent to $\binom{x-1}{1}+\binom{x-1}{2}+\dots+\binom{x-1}{2u}$ is convex on $[0,+\infty)$. 2) Pointed out by @FedorPetrov @GeraldEdgar @H.H.Rugh: For $x<0$ each summand $\binom{x}{2i}$ is obviously convex, thus the question is equivalent to $f(x)$ is convex on $\mathbb{R}$. 3) Pointed out by @WłodzimierzHolsztyń‌​ski: It has $(\Delta^2 f_u)(x) = 1+ f_{u-1}(x-2)$, where $(\Delta f_u)(x)=f_u(x)-f_u(x-1)$. Then we can conclude that $f(x)$ is discrete convex. REPLY [6 votes]: Here's an alternative proof based on probabilistic arguments (showing different aspects). Let $$f_n(x):=\sum_{j=0}^n { x \choose j}=[t^n]\,\frac{(1+t)^x}{1-t}\;\;,$$ and let $^\prime$ denote derivative with resp. to $x$. We have to show that for even $n=2k$ the second derivative $f_{2k}^{\prime\prime}(x)=[t^{2k}] \frac{(1+t)^x}{1-t}\,(\log(1+t))^2$ is nonnegative. The basic observation used below is that $g(t):=\frac{\log(1+t)}{t}$ is the Laplace transform (LT) of a nonnegative random variable possessing all moments. The relation $g(t)=\int_0^1 \frac{1}{1+st}\,ds$ shows that $g$ is the LT of $U\cdot X_1$, where $U$ is uniform on $[0,1]$, $X_1$ is $\Gamma(1,1)=\exp(1)$ distributed (i.e. has LT $\frac{1}{1+t}$), and the factors are independent. Case 1: $x<0$. In this case $(1+t)^x $ is the LT of $\Gamma(1, -x)$ . Thus $\ell(t):=(1+t)^x g(t)^2$ is the Laplace transform of a nonnegative rv and (for $k\geq 1$) $f_{2k}^{\prime\prime}(x)=[t^{2k-2}]\, \frac{1}{1-t} \ell(t)$ is an even degree MacLaurin sum of $\ell$, evaluated at $1$, and therefore exceeds $\ell(1)>0$. Case 2: $x\geq 0$. Since $f_n(x+1)=f_n(x)+f_{n-1}(x)$ it will suffice to show that $f_n^{\prime\prime}(x)\geq 0$ for $x\in [0,1)$ and all $n$. For $x=0$ all derivatives $f_n^{\prime\prime}(0)$ are nonnegative, since $a_r:=[t^r] (g(t))^2=2\,(-1)^r \frac{H_{r+1}}{r+2}$ , and $a_0=1>0$, $a_{2k}+a_{2k+1}\geq 0$. Let $00.12$, all even partial sums are nonnegative. Finally, note that this proof also shows that $f_n$ is convex on $[0,\infty)$ for odd $n$.<|endoftext|> TITLE: Etale and Algebraic K-theory with rational coefficients QUESTION [7 upvotes]: We know that the Quillen-Lichtenbaum conjecture, now proved by Rost, Voevodsky, and Weibel, says that for smooth finite type $k$-schemes $X$, etale and algebraic $K$-theory with finite coefficient $\mathbb{Z}/n$, $n$ invertible in $k$, agree in sufficiently high degree. Could someone please describe what is known and what is conjectured about the relation between $K_*^{et}(X)_{\mathbb{Q}}$ and $K_*(X)_{\mathbb{Q}}$? REPLY [10 votes]: They are always (for any scheme) equivalent using transfer maps as in Proposition 2.14 in Thomason's paper. For smooth qcqs schemes, more is true - the etale and Nisnevich rational motivic stable homotopy category are equivalent (this is essentially because the Nisnevich and etale cohomology coincide rationally) so the equivalence is true for general motivic spectra (like algebraic $K$-theory; in the non-smooth case you get a statement about Weibel's homotopy algebraic $K$-theory).<|endoftext|> TITLE: Original reference for Riemann's inequality QUESTION [14 upvotes]: Let $D$ be a divisor on a compact Riemann surface of genus $g$. The inequality $$ l(D)\geq {\textrm {deg}}(D)-g+1 $$ is called Riemann's inequality. $ \phantom{aaaaaaaa}$In which of Riemann's papers did this inequality first appeared? REPLY [13 votes]: B. Riemann, Theorie der Abel'schen Functionen, Journal für die reine und angewandte Mathematik 54, 101–155 (1857). Here is a description of this contribution, by Jeremy Gray: In this 1857 paper Riemann established the existence of complex functions on a surface with no boundary. He supposed the surface was p-fold connected which means that it is rendered simply connected by p cuts when it forms a p-sided polygon. He showed that there are p linearly independent everywhere holomorphic functions defined inside the polygon by considering what would happen if the real parts of their periods all vanished (using the Dirichlet principle). Later he showed that the differentials of these functions are everywhere defined holomorphic integrands. Then he specified d points at which the function may have simple poles, again imposing the condition that the functions jump by a constant along the cuts. Now he argued that to create functions with only simple poles and constant jumps one should take a sum of p linearly independent functions with no poles plus functions of the form $1/z$ at one of the specified points and add a constant term. The resulting expression depends linearly on $p + d+1$ constants. The jumps therefore depend linearly on $p + d+1$ constants and there are $p$ of them to be made to vanish (if the function is single valued as required). So there will be non-constant meromorphic functions when $p + d+1-2p\geq 2$, i.e. $d>p$. This result, today called the Riemann inequality, says there is a linear space of complex functions of dimension $h^0>d+1-p$ and this contains non constant functions as soon as $d+1-p>1$ or $d>p$. REPLY [10 votes]: Bernhard Riemann "Theorie der Abel'schen Functionen" (1857) You can find a digital copy in here, pages 115 to 155.<|endoftext|> TITLE: Canonical bundle of the Lagrangian Grassmannian QUESTION [5 upvotes]: I work through the paper On branched coverings of some homogeneous space of Kim and Manivel and I came across the definition of the canonical bundle of the Lagrangian Grassmannian $\mathbb{LG}_n$, the set of $n$-dimensional lagrangian subspaces of a $2n$-dimensional symplectic vector space. In the paper it is called 'a fact' that the canonical bundle $K_{\mathbb{LG}_n} = \mathcal{O}_{\mathbb{LG}_n}(-n-1)$, but I'm not able to verify it. Can someone help me in this? Thanks in advance. REPLY [10 votes]: Here is a group theoretical solution which makes it is easy to compute the canonical bundle of any generalized flag variety. The Lagrangian Grassmannian is the homogeneous space $G/P$ where $G=Sp(2n)$ and $P$ is the maximal parabolic subgroup corresponding to the set of simple roots $\{\alpha_1,\ldots,\alpha_{n-1}\}$. Then $Pic(G/P)$ equals $\Xi(P)$, the character group of $P$. The latter is cyclic and its generator $\eta:=\sum_{i=1}^n\epsilon_i$ corresponds to $\mathcal O(-1)$. The canonical bundle corresponds to the determinant character of $(\mathfrak g/\mathfrak p)^*\cong\mathfrak p_u$ and therefore to the sum of roots in $\mathfrak p_u$: $$ \sum_{i TITLE: History of name "Fundamental Theorem on Symmetric Polynomials" QUESTION [6 upvotes]: Do you know the history (esp. first use) of the name "Fundamental Theorem on Symmetric Polynomials" (or "FT of SP", or "FT on S Functions", etc.) for the statement that any polynomial (resp. rational function) symmetric in n variables is uniquely expressible as a polynomial (resp. rational function) in the elementary symmetric polynomials in those variables? I got a comment from a reviewer (for an article about this theorem) asking for a reference for the name. Every 20th century source I consulted (back to van der Waerden) just presents the theorem with this name. The 18th and 19th century sources I've looked at (Waring, Galois, Gauss) do not give a name (nor do they uniformly articulate it as a theorem; Gauss does; Waring kind of does). I assume the theorem was christened with the name sometime between 1850 and 1950. Do you know where, when and by whom? REPLY [5 votes]: A good source is Muir's "The Theory of Determinants in the Historical Order of Development". The whole book has been digitized and put in the public domain (available through University of Michigan Historical Math Collection).<|endoftext|> TITLE: Real orbits of Complex Flag Varieties QUESTION [6 upvotes]: Let $G$ be a semi-simple algebraic group over $\mathbb{C}$ with Borel subgroup $B$ and consider the flag variety $G/B$. If $G_0 \subset G$ is some real form, then $G_0$ acts on $G/B$ and decomposes the space into a disjoint union of orbits. In the case of $SL_2$ one gets 2 open orbits while in $SL_3$ my calculations seem to show a single open orbit. Can one characterize the number of open orbits in general? For what groups is this number larger than 1? REPLY [8 votes]: Even though there is already a good answer to the question I would like to add that Aomoto's paper just marks the beginning of an extensive body of research on open $G_0$-orbits on $G/B$ or, more generally, on $G/P$ for any parabolic $P\subseteq G$. These open orbits are often called flag domains. After Aomoto's paper one should mention a paper by Wolf (BAMS, 1969) in which general parabolics where considered. In particular, he derives a parameterization of flag domains in terms of double cosets inside the Weyl group of $G$. There is also a monograph by Fels-Huckleberry-Wolf on this topic and a survey by Akhiezer (Real group orbits on flag manifolds). Finally, there is the Matsuki correspondence, setting up a bijection between the set of $G_0$- and the set of $K$-orbits in $G/P$. Here $K$ is the complexification of a maximal compact subgroup $K_0\subseteq G_0$. Hereby, open $G_0$-orbits correspond to compact (i.e. closed) $K$-orbits. In general, $K$-orbits on $G/B$ have been studied by Springer (some of them with Richardson) in a series of papers.<|endoftext|> TITLE: In which sense the GNS-construction is a functor? QUESTION [17 upvotes]: I asked this at mathstackexchange a week ago, without success. I think the Gelfand–Naimark–Segal construction must be a functor in some sense, but I can't find an explicit statement anywhere. Can anybody enlighten me? For example, is the following hypothesis true? Let $\varphi:A\to B$ be an involutive homomorphism of C*-algebras, and let $f$ be a state on $B$. Consider the corresponding state $f\circ\varphi$ on $A$ and the GNS-constructions $\pi_{f\circ\varphi}:A\to {\mathcal B}(H_{f\circ\varphi})$ and $\pi_f:B\to {\mathcal B}(H_f)$. Let $\widetilde{A}$ and $\widetilde{B}$ be the von Neumann algebras in ${\mathcal B}(H_{f\circ\varphi})$ and ${\mathcal B}(H_f)$ generated by $\pi_{f\circ\varphi}(A)$ and $\pi_f(B)$ respectively, and let us consider $\pi_{f\circ\varphi}$ and $\pi_f$ as homomorphisms with ranges in $\widetilde{A}$ and $\widetilde{B}$. Is there a homomorphism $\widetilde{\varphi}:\widetilde{A}\to\widetilde{B}$ such that $$ \pi_f\circ\varphi=\widetilde{\varphi}\circ\pi_{f\circ\varphi} \,\,\,\,\,\text{?} $$ I believe, it is not essential here that $A$ and $B$ are C*-algebras, they can just be topological algebras with involution is some sense, and the states can be defined as continuous positive functionals $f:A\to{\Bbb C}$ such that the map $x\mapsto f(x^*\cdot x)$ is also continuous. I need this for my current work, if anybody could help, I would appreciate this very much. REPLY [10 votes]: Although this response is a bit late, perhaps this perspective may help nonetheless. It only addresses the question about functoriality of the GNS construction. The GNS construction is not quite a functor, though as Nik Weaver's response indicates, it has a certain functoriality associated with it. The GNS construction, is, however, (almost) a natural transformation that satisfies a universal property, which is indicated in Westerbaan's response. Let me make this more precise. Background: There are two relevant functors on the category of $C^*$-algebras. One is the functor that associates the set of all states to every $C^*$-algebra and the function that pulls back states to states to every $C^*$-algebra $*$-homomorphism. This is the ``states'' functor $\mathcal{S}:\mathbf{C}^*\text{-}\mathbf{Alg}^{\mathrm{op}}\to\mathbf{Set}.$ There is another functor that sends a $C^*$-algebra to its category of representations and a $*$-homomorphism to a functor between categories of representations $\mathbf{Rep}:\mathbf{C}^*\text{-}\mathbf{Alg}^{\mathrm{op}}\to\mathbf{Cat}.$ A representation is not enough to produce a state, so we add the additional data in the category of representations to include a normalized vector. Morphisms in this category are then taken to be intertwining isometries preserving unit vectors. Thus, we write $\mathbf{Rep}^{\bullet}:\mathbf{C}^*\text{-}\mathbf{Alg}^{\mathrm{op}}\to\mathbf{Cat}$ for this new functor. The codomains of $\mathcal{S}$ and $\mathbf{Rep}^{\bullet}$ are not the same so it does not make sense to compare them. By viewing every set as a discrete category, we can take the codomain of $\mathcal{S}$ to be $\mathbf{Cat}$ as well. To distinguish $\mathcal{S}$ from this functor, let's call the latter $\mathbf{States}.$ Statement: The GNS construction furnishes a pseudo-natural transformation $\mathbf{GNS}^{\bullet}:\mathbf{States}\Rightarrow\mathbf{Rep}^{\bullet}.$ It is not a strict natural transformation. Technically, it is a 1-morphism in the 2-category of functors from $\mathbf{C}^*\text{-}\mathbf{Alg}^{\mathrm{op}}$ to $\mathbf{Cat}.$ The objects here are functors, 1-morphisms are pseudo-natural transformations (natural transformations that satisfy naturality up to a higher coherence), and 2-morphisms are modifications. Universal property: The GNS construction is actually part of an adjunction in this 2-category mentioned in the previous paragraph. It is left adjoint to the natural transformation (which is an honest natural transformation) that takes a representation with a unit vector and simply pulls back the canonical state associated to this unit vector to the underlying $C^*$-algebra. [There is supposed to be an image here---sorry, not enough reputation points---please see the introduction in the paper below] More details can be found in https://arxiv.org/abs/1609.08975 "From observables and states to Hilbert space and back: a 2-categorical adjunction," which was coincidentally posted not long after your question, though I was unaware of this until now.<|endoftext|> TITLE: The automorphism of a group with a given fixed point set QUESTION [8 upvotes]: It is a very known fact that for every group $G$ and an automorphism $\sigma$, the set of fixed points of $\sigma$ is a subgroup of $G$. My question is about the converse. For a finite group $G$ and a subgroup $H$ of $G$, under which conditions, there exists an automorphism $\sigma$ such that the set of fixed points of $\sigma$ is equal to $H$? REPLY [4 votes]: A necessary condition is that $H $ be closed under any function $f $ that is equivariant under automorphisms $\sigma $ of $G $: $f(\sigma (x))=\sigma (f (x)) $ (and similarly for functions of several variables). For instance, taking inverses, products, and $n $th roots. For the latter see also http://groupprops.subwiki.org/w/index.php?title=Fixed-point_subgroup_of_a_subgroup_of_the_automorphism_group#Weaker_properties<|endoftext|> TITLE: Free monad sequence versus colimit over injections of ordered sets QUESTION [7 upvotes]: Kelly describes a constructive procedure for building the algebraically free monad on a pointed endofunctor. Garner gives a concise summary, which I partially review here for convenience. Let $V$ be a cocomplete category (feel free to make other strong assumptions about $V$ if I have omitted them) and $(S,\eta)$ a pointed endofunctor of $V$. Define $X_0=\mathrm{id}_V$, $X_1=S$, and $\sigma_0:SX_0\to X_1$ as the identity. Then assuming $\sigma_0,\ldots,\sigma_i$ and $X_0,\ldots,X_{i+1}$ are defined, define $X_{i+2}$ and $\sigma_{i+1}:SX_{i+1}\to X_{i+2}$ as the coequalizer of the following diagram: $\require{AMScd}$ \begin{CD} SX_i @>\sigma_{i}>>X_{i+1}\\ @VS\eta X_iVV @V\eta X_{i+1}VV\\ SSX_i@>S\sigma_i>>SX_{i+1} \end{CD} Then putting $\{X_i\}$ together via the maps $X_{i}\xrightarrow{\eta X_i} SX_{i}\xrightarrow{\sigma_i} X_{i+1}$, the resulting diagram $$ X_0\to X_1\to \cdots $$ is the (finite part of the) free monad sequence for $(S,\eta)$. If we make what seems to be a fairly strong convergence assumption, then the colimit $X_\omega$ of the sequence is the (algebraically) free monad on $(S,\eta)$. On the other hand, another thing one could thing to do is to consider $(S,\eta)$ as generating a semicosimplicial functor $S_*$ where $S_n= S^n$ and the coface maps are given by $S^i\eta S^j$. Then one could consider the colimit of the diagram made up by coface maps up to a certain point: $$ S_0\to S_1\rightrightarrows S_2\to\cdots \to S_n $$ Call this colimit $Y_n$. Inductively it looks to me like $X_i\cong Y_i$ and so $X_\omega\cong Y_\omega$. Here are my questions. Is there something subtly (or not so subtly) wrong with this observation? The presentation of $Y_n$ and $Y_\omega$ seems "more symmetric." Assuming the answer to question 1. is "no," is there a good reason to prefer the presentation of $X_n$ and $X_\omega$? Is this because things will go badly as soon as one needs to extend to infinite ordinals in the free monad sequence? If the answer to question 1. is "no," is there a reference that uses the $Y_n$ presentation to build this sequence where I can see in detail whether there are secret assumptions I am suppressing? Richard Garner, MR 2506256 Understanding the small object argument, Appl. Categ. Structures 17 (2009), no. 3, 247--285. G. M. Kelly, MR 589937 A unified treatment of transfinite constructions for free algebras, free monoids, colimits, associated sheaves, and so on, Bull. Austral. Math. Soc. 22 (1980), no. 1, 1--83. REPLY [4 votes]: Let's talk about free monoid constructions in a monoidal category $(\mathcal{V}, \otimes, I)$. Free monad constructions are a special case, when we take $\mathcal{V}$ to be a monoidal category of endofunctors. There are at least two flavors: the free monoid on an object $X$, and the free monoid on a pointed object $I \to X$. It's pretty easy to construct the free monoid functor $T: \mathcal{V} \to \mathcal{V}$ in typical monoidal categories -- just take $T: X \mapsto \sum_{n=0}^\infty X^{\otimes n}$, or in the pointed case, $T: I \to X \mapsto \varinjlim_{n\to \infty} X^{\otimes n}$ using the map $I \to X$ to define semicosimplicial connecting maps. The issue is with defining multiplication. We want to say that $TX \otimes TX = \varinjlim_m X^{\otimes m} \otimes \varinjlim_n X^{\otimes n} = \varinjlim_{m,n}X^{\otimes m+n}$, and then define multiplication $\mu_X: T^2 X \to TX$ by the universal property of the colimit, setting $X^{m+n} \to T^2 X \overset{\mu_X}{\to} TX$ equal to $X^{m+n} \to TX$, where the unlabeled maps are the canonical colimit inclusions. If you think about this, you'll see it's the usual formula for multiplication in a free monoid. The diagrams all commute, so we can define this map $\mu_X$ with no problem. The problem is that the domain of $\mu_X$ is $\varinjlim_{m,n} X^{\otimes m+n}$, and in order to identify this with $T^2 X$, we had to commute a colimit past $\otimes$ -- we needed the functor $TX \otimes(-): \mathcal{V} \to \mathcal{V}$ to commute with the colimit defining $TX$. This is fine if $\otimes$ is (right) closed, in which case $TX \otimes(-)$ commutes with all colimits, so these simple constructions work in the familiar cases of free monoids in categories like $\mathsf{Set},\mathsf{Ab}$, etc. But right closedness is simply not true in a category of endofunctors $\mathcal{V} = [\mathcal{C},\mathcal{C}]$. The condition becomes slightly different in this setting -- if you start with $F$ and define $M(C) = \varinjlim_n F^n C$, you want to argue that $M^2(C) = \varinjlim_m F^m (\varinjlim_n F^n(C)) = \varinjlim_{m,n} F^{m+n}(C)$, for which it suffices for $F$ to commute with the colimits defining $M$. Unfortunately, this is not often the case, although I believe the semisimplex category is at least sifted, so if $F$ commutes with sifted colimits you might be in luck. Otherwise I don't know a shortcut to reading Kelly. I think the trick to reading his paper on a first pass must be to keep in mind that you can take the factorization system he uses to be $(M,E)$ = (all maps, isos), and then "preserving $E$-tightness of a cocone" just means preserving a colimit. What I currently understand is that Kelly gives free constructions endofunctor -> pointed endofunctor -> well-pointed endofunctor -> monad. I'm currently most mystified by the middle one.<|endoftext|> TITLE: Bound on the joint distribution of three real random variables with given two dimensional marginals QUESTION [6 upvotes]: Let $X,Y,Z$ be real r.v. with $(X,Y)$, $(Y,Z)$ and $(Z,X)$ centered unit normal. How large can $\mathbb E (XYZ)$ be? REPLY [2 votes]: For a discretized version of this problem, $E(XYZ)$ can be as high as $5/(4\sqrt{3})$, or ~ 0.72. Suppose that $X$, $Y$, $Z$, are all constrained to have values in $\{\pm \sqrt{3}, \pm 1 / \sqrt{3}\}$. Let \begin{align} P(X=x, Y=y, Z=z) = & 1/64 \text{ in the }\text{ 4 cases with } xyz = \sqrt{27}\\ & \ \ 0 \ \ \ \text{ in the 12 cases with } xyz = \sqrt{3}\\ & 3/64 \text{ in the 12 cases with } xyz = 1/\sqrt{3}\\ & 6/64 \text{ in the }\text{ 4 cases with } xyz = 1/\sqrt{27}\\ & \ \ 0 \ \ \ \text{ in the 32 cases with } xyz < 0\\ \end{align} which reduces the calculation of $E(XYZ)$ to multiplying the numbers on each row and adding up the results. The marginal distributions are \begin{align} P(X=x, Y=y) = & 1/64 \text{ in the 4 cases with } xy = \pm 3\\ & 3/64 \text{ in the 8 cases with } xy = \pm 1\\ & 9/64 \text{ in the 4 cases with } xy = \pm 1/3\\ \end{align} \begin{align} P(X=x) = & 1/8 \text{ in the 2 cases with } x = \pm \sqrt{3}\\ & 3/8 \text{ in the 2 cases with } x = \pm 1/\sqrt{3}\\ \end{align} So each of $X,Y,Z$ is a variable with mean 0 and variance 1, and they are pairwise independent. I expect that there are similar continuous distributions with similar values for $E(XYZ)$.<|endoftext|> TITLE: Properties of complexified octonions QUESTION [6 upvotes]: I have following questions about the complexified octonion algebra $\mathbb C \otimes_{\mathbb R} \mathbb O$. Zero divisors are of shape $p+i\otimes q$ (shortly $p+iq$) where $p$, $q$ are perpendicular octonions of the same length. We denote octonion conjugation by an overbar, so $\overline{z \otimes x} = z \otimes \bar x$. An element $x$ of the algebra is invertible when $x\bar{x}$ is a non-zero complex number. Each element of the algebra is either a zero divisor or invertible. If we call the norm of $x$ the value $x\bar{x}$, then zero divisors are elements of norm equal to zero. 1) Is it true that, for $u$ a zero divisor, the span of $\{xu:\text{$x$ invertible}\}$ includes all zero divisors? 2) Is it true that, for each zero divisor $u$, there are zero divisors $a$, $b$ such that $u=a\bar{b}$? 3) For a given zero divisor $a$, what is the dimension of $\{b:ab=0\}$? Probably the set is just $\{\bar {a}x: \text{$x$ invertible}\}$. 4) What is the dimension of $\{(a,b):ab=0\}$? I discovered that elements $b=\bar{p}(qs)+i\otimes s$, for $s \in \mathbb O$, satisfy $ab=0$ for $a=p+iq$, $|p|=|q|=1$, $p\bar {q}$ imaginary. 5) What is the dimension of $\{(a,b,c):ab=0, bc=0, ca=0\}$ ? The answers are needed for my analysis of reflections in $E_6$ Lie group. The definition of reflection is contained in my other question. REPLY [5 votes]: Here are some answers, but let me introduce some notation: What you are calling the norm of $x\in\mathbb{C}{\otimes}\mathbb{O}$, I will denote by $N(x) = x\bar x = \bar x x\in \mathbb{C}$. The complexified octonions satisfy $N(xy) = N(x)N(y)$, and the norm is a nondegenerate quadratic form on $\mathbb{C}{\otimes}\mathbb{O}$, considered as a complex vector space. 1) No. Given a zero divisor $u$, consider the quadratic form $Q_u(x) = N(xu)$ on $\mathbb{C}\otimes\mathbb{O}$. This quadratic form is identically zero, so this implies that the subspace $(\mathbb{C}{\otimes}\mathbb{O})u$ is a totally $N$-null linear subspace of $\mathbb{C}{\otimes}\mathbb{O}$, an $8$-dimensional vector space on which $N$ is nondegenerate. Thus, this subspace has dimension at most $4$ and it contains all of the elements $xu$ where $x$ is invertible. Thus, these latter elements can't span $\mathbb{C}{\otimes}\mathbb{O}$ all zero divisors because the cone $N(u) = 0$ of zero divisors does not lie in any proper linear subspace of $\mathbb{C}{\otimes}\mathbb{O}$. A similar argument applies to the elements $ux$ where $x$ is invertible. 2) Yes. To understand this, you need to realize that there are two kinds of zero divisors. The first kind (generic) are the zero divisors $u$ such that $u+\bar u = \lambda\not=0$, i.e., $u$ is not 'purely imaginary'. For such a $u$, we have $$ 0 = u\bar u = u(\lambda-u) = \lambda u - u^2, $$ so setting $a = u$ and $b = \bar u/\lambda$, we have $u = a\bar b$. The second kind has $\bar u = - u\not=0$, so we can write $u = p + iq$ where $p$ and $q$ are purely imaginary elements of $\mathbb{O}$ that have the same norm, say, $p\bar p = q\bar q = r>0$, and are orthogonal. Then $1,p,q,pq$ span an associative quaternion subalgebra of $\mathbb{O}$. We then compute, using that $p^2 = q^2 = -r$ and $pq=-qp$ that $$ (1+ipq)(p+iq) = (1+r)(p+iq). $$ Dividing by $1+r>0$, we have $p+iq$ written as a product of zero divisors. 3) For any (nonzero) zero-divisor $a$, the subspace consisting of those $b$ that satisfy $ab = 0$ is a (complex) vector space of dimension $4$. Hence the subspace consisting of (left) multiples of $a$ has dimension $8-4 = 4$ as well. The same dimension counts hold true for right multiplications. In particular, by dimension count, the subspace consisting of elements of the form $\bar a x$ is equal to the subspace consisting of elements $b$ that satisfy $ab=0$. 4) As a complex analytic variety, it has dimension 11. The reason is that, in the pair $(a,b)$, as a fixed $a\not=0$ ranges over the $N$-null cone (which has complex dimension $7$), $b$ is only required to lie in the $4$-dimensional subspace that is the kernel of left multiplication by $a$. When $a=0$, you get an $8$-plane, since there is then no restriction on $b$, but this doesn't affect the count. 5) (Corrected after the OP's comment.) The complex dimension of this singular variety is $14$. The reason is that the set of pairs $(a,c)$ that satisfy $ca=0$ has complex dimension $11$, and, for each such pair with neither $a$ nor $c$ equal to zero, the equations $ab=bc=0$ are linear equations for $b$ that, it turns out, have a (complex) $3$-dimensional space of solutions. Thus, the variety has dimension $11+3 = 14$. (The special cases when one of $a$, $b$, or $c$ vanishes only have dimension $11$, so they can't increase the count.)<|endoftext|> TITLE: Completion and algebraic closure QUESTION [8 upvotes]: Following this question: Given a valued field $K$, denote with $\bar{K}$ its algebraic closure and with $\hat{K}$ the completion. Then both $\hat{\bar{K}}$ and $\hat{\bar{\hat{K}}}$ are complete and algebraically closed. If $K=\mathbb{Q}$, these constructions always give $\mathbb{C}$ or $\mathbb{C}_p$ for a prime $p$, i.e. both constructions give the same fields. Is this true for any field $K$? Or is there a field $K$ with a valuation $v$ such that for no extensions of $v$ to $\bar{K}$ we have $\hat{\bar{K}}=\hat{\bar{\hat{K}}}$? REPLY [16 votes]: First you have to observe that since all extensions of the valuation to $\bar{K}$ are conjugate, $\hat{\bar{K}}$ is well-defined up to (non-unique) isomorphism. Now, since $\hat{\bar{K}}$ is complete and $K$ is dense in $\hat{K}$, the inclusion $K\subset \hat{\bar{K}}$ extends continuously to $K\subset \hat{K}\subset \hat{\bar{K}}$ (in fact you can identify $\hat{K}$ with the closure of $K$ in $\hat{\bar{K}}$). Since $\hat{\bar{K}}$ is an algebraically closed extension of $\hat{K}$, it contains a unique copy of $\bar{\hat{K}}$, namely the algebraic closure of $\hat{K}$ in it. Moreover this copy obviously contains $\bar{K}$ which is dense in $\hat{\bar{K}}$, hence it is also dense. Since $\hat{\bar{K}}$ is complete, it is therefore isomorphic to the completion of $\bar{\hat{K}}$.<|endoftext|> TITLE: Is a CohFT completely determined by its high genus values? QUESTION [6 upvotes]: Suppose I have a CohFT, which is basically a vector space $V$ and a collection of maps $V^{\otimes n} \rightarrow H^*(\overline{M_{g,n}})$ satisfying certain properties under pull-backs by the gluing maps. Under which assumptions is it completely determined by its high genus values? High means with respect to the cohomology degree. I.e. for each $k$ I know the map $V^{\otimes n} \rightarrow H^k(\overline{M_{g,n}})$ for $g$ large enough for all $n$. And I want to reconstruct it for all $g$. I saw some argument in a paper by Teleman which begins by saying that we can assume the genus is large, and then use the Madsen-Weiss theorem. So I wonder what's behind this trick and how to justify it properly. REPLY [4 votes]: Teleman's trick works as follows. Suppose I want to deduce $\Omega_{g,n}$ from $\Omega_{g+1,n}$. Note that $\overline{\mathcal{M}}_{g+1,n}$ contains a copy of $\overline{\mathcal{M}}_{g,n}$. It is constructed as follows. Take any fixed elliptic curve $E$ with two marked points. Now consider the locus of curves obtained by gluing $E$ to a stable genus $g$ curve at its $n$th marked point. The factorization property of CohFTs tells you that $\Omega_{g,n}$ is equal to the restriction of $\Omega_{g+1,n}$ to this locus, contracted with the value of $\Omega_{1,2}$ on $E$, which is simply the cohomological degree 0 part of $\Omega_{1,2}$. So the subtlety is this: is the bilinear form $\Omega_{1,2}(E) \in V^* \otimes V^*$ invertible (nondegenerate)? If yes, you can deduce $\Omega_{g,n}$ from $\Omega_{g+1,n}$. If not, in general you cannot. By some linear algebra one checks that the invertibility of $\Omega_{1,2}(E)$ is equivalent to $\Omega$ being semisimple, in other words, the algebra structure on $V$ given by $\Omega_{0,3}$ should be semisimple.<|endoftext|> TITLE: What is special in dimension $2$ (When characterizing isometries using the cofactor matrix)? QUESTION [6 upvotes]: Let $A$ be a real $n \times n$ matrix. Denote by $\operatorname{cof} A$ The cofactor matrix of $A$. By definition, $A (\operatorname{cof} A)^T=\det A \cdot I$. Thus, it is immediate that $A \in \operatorname{SO}_n$ if and only if $$ (**) \operatorname{cof} A =A,\det A =1$$ However, if $n \neq 2$ the condition on the determinant is superfluous: $ \operatorname{cof} A =A \Rightarrow AA^T=\det A \cdot I \Rightarrow (\det A)^2=(\det A)^n \Rightarrow \det A \in \{1,-1\}$. However, Since $\det A \cdot I=AA^T$ is positive semidefinite $\det A \ge 0$ so $\det A = 1$ and $A \in \operatorname{SO}_n$. For the case $n=2$, an easy calculation shows $\operatorname{cof} A=A$ if and only if $A$ is a scaled rotation. Question: While the above derivations are easy to do algebraically, I would like to find a more geometric explanation of these results. I think this amounts to obtaining a better geometric interpretation for the cofactor matrix. (I know it measure in some sense the volume of $n-1$ dimensional parallelepiped, see here). In particular, Is there any geometric intution behind the condition $A \in\operatorname{SO}_n \iff (**) \operatorname{cof} A =A,\det A =1$? Is there any explanation for why dimension $2$ is special? Note: The condition $(**)$ for characterizing matrices in $\operatorname{SO}_n$ is not a mere game. In some contexts this is the only way to show some transformations are indeed isometries. (For instance in proofs of Reshetnyak’s rigidity theorem). REPLY [2 votes]: This is also not a formal answer, but gives another (similar) viewpoint on the difference between dimension 2 and higher dimensions: When $n=2$, the mapping $A \mapsto \text{cof}\, A$ is a linear, and hence the solutions to $\text{cof}\,A = A$ must form a subspace. In particular, it will be invariant to dilations. In higher dimensions $A \mapsto \text{cof}\, A$ is not linear, and in particular $\text{cof}\,\lambda A = \lambda^{n-1} \text{cof}\,A$.<|endoftext|> TITLE: Uniqueness Principle for function types QUESTION [5 upvotes]: I am currently trying to understand the first chapter of the HoTT book and for 1.2 Functions Types of the book, Since it is by definition ``the function that applies $f$ to its argument'' we consider it to be definitional equal to $f$: $$ f\equiv (\lambda x.f(x)) $$ This equality is the uniqueness principle for functions types, because it shows that $f$ is uniquely determined by its values. I can understand its definitional equality as it is by construction. But I do not understand why we need to establish or state this uniqueness principle. I am tempted to think of this uniqueness principle as obvious, but I think there is some underlying reason for this. I am guessing it has to do with the big picture of how each type is introduced systematically by the formation, introduction, elimination, computation rules but I can't seem to relate to them. Thank you very much. REPLY [11 votes]: It is not true "by construction". Remember that at this point in the theory a "function" is an abstract undefined thing, not something "defined by its action on inputs" as it is in set theory. The only thing we can do with a "function" is apply it to an argument; the only way we have to make "functions" is to $\lambda$-abstract take an expression involving a variable. So if I have a "function" $f$, apply it to a variable $x$ to get an expression $f(x)$, then abstract that to get $\lambda x.f(x)$, I have a new "function". There's nothing in this abstract picture (before we assert the uniqueness principle) that says that this new "function" must be the same as the "function" $f$ that I started with.<|endoftext|> TITLE: Differentiability of Distance to a Closed Convex Set QUESTION [7 upvotes]: Let $( \mathbb{R}^d, \| \mathbf{x}\|_2 )$ be a Euclidean Space. For any closed convex set $A\subseteq \mathbb{R}^d$, we define \begin{align} d(\mathbf{x}, A) = \inf \{ \| \mathbf{x} - \mathbf{y} \|_2, ~~\mathbf{y}\in A\}. \end{align} I was wondering how to prove that $d(\mathbf{x}, A)$ is differentiable and that the gradient is Lipschitz? REPLY [5 votes]: Borwein+Lewis Convex Analysis, Section 3.3 Exercise 12(d) shows that for convex $A$ and $x\notin A$, $$\nabla f(x) = f(x)^{-1}(x-P_A(x)),$$ where $f(x)=d(x,A)$ is your function and $P_A(x)$ is the projection of $x$ onto $A$ (i.e., the unique $x^*\in A$ achieving the $\inf$). This shows, in particular, that for $x\notin A$, $||\nabla f(x)||_2=1$. The Lipschitz property follows from Borwein+Lewis Section 2.1 Exercise 8(c.iii), which shows that projection is a contraction. The full book reference: Convex Analysis and Nonlinear Optimization, Borwein, Jonathan, Lewis, Adrian S. http://www.springer.com/us/book/9780387295701 Edit: I forgot to state the additional hypothesis that $A$ is closed.<|endoftext|> TITLE: Reference request: shift invariant measures are (locally exactly) approximable by periodic ones QUESTION [5 upvotes]: Let $A$ be a finite alphabet, let $S = A^{\mathbb{Z}}$ be the set of bi-infinite sequences of characters from $A$, where $A$ is given the discrete topology and $S$ is given the corresponding product topology, and let $\sigma$ denote the right shift operator. In order to solve some other problem, I have recently shown the following: Suppose $\mu$ is a shift invariant probability measure on $S$ (that is, $\mu \circ \sigma = \mu$). Then for every $n \geq 1$, there exists a positive integer $L_n$ and a shift invariant probability measure $\mu_{n}$ such that $\mu_n$ is supported on sequences of period $L_n$ and $\mu_n$'s induced distribution on words of length $n$ is the same as $\mu$'s induced distribution on words of length $n$. What I'm wondering is whether this is a standard result, or at least already known, and if so, where I can find a reference. Many thanks! Note: Each $\mu_n$ has exactly the same distribution on words of length $n$ as $\mu$. There are counterexamples to this claim when $A$ is infinite, for example, when $A = S^1$. REPLY [3 votes]: I don't know wheter this counts as standard, but... There is a paper by Krystyna Ziemian: Rotation sets for subshifts of finite type. Fund. Math. 146 (1995), no. 2, pp. 189--201 containing some general results which after some work yield the conclusion. The reduction of your question to Ziemian's framework is fairly standard, but notationally heavy. Assume that the alphabet has $\alpha$ letters. Let $\mu$ be any shift invariant measure. Fix a positive integer $n$. Consider a directed graph $G$ with edges labelled by all words of length $n$ and directed edges $u\to v$ between each vertices $u=u_1\ldots u_n$ and $v=v_1\ldots v_n$ such that $u_2\ldots u_n=v_1\ldots v_{n-1}$. Denote the set of edges of $G$ by $E$ and the set of vertices of $G$ by $V$. Note that the edges of $G$ are in the one-to-one correspondence with the words of length $n+1$. There are $N:=\alpha^{n+1}$ edges. For $e\in E$ let $\chi_e\colon E\to\{0,1\}$ be the characteristic function of the edge $e\in E$. Let $\varphi\colon E\to\{0,1\}^N$ be the concatenation of all $\chi_e$'s, that is, for a fixed edge $e_0\in E$ we have $\varphi(e_0)=(\chi_e(e_0))_{e\in E}\in\{0,1\}^N$. Now given an infinite sequence of letters $x$ it makes sense to define $\varphi(x)=\varphi(x_1\ldots x_{n+1})$ since the first $n+1$ symbols of $x$ determine uniquely an edge of $G$. Then $$ \lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^{k-1} \varphi(\sigma^i(x)) $$ is the vector whose coordinate given by $e\in E$ describes the limiting frequency of occurrences of the word of length $n+1$ corresponding to $e$ in $x$ (provided the limit exists). In the full shift for every shift invariant measure $\mu$ there is a sequence $x$ such that the above limit exists and equals $(\mu(e))_{e\in E}$, where $\mu(e)$ is the measure of the cylinder set determined by $e$ for each $e\in E$. This result can be found in Sigmund's paper cited below. In Ziemian's terminology this means that the vector $(\mu(e))_{e\in E}\in [0,1]^N$ belongs to the pointwise rotation set of $\varphi$. Here we follow Ziemian and call a loop in $G$ elementary if it is not a concatenation of two shorter loops. Clearly, each loop that is not elementary can be written as a concatenation of two loops, at least one of which is elementary. For each elementary loop $\tau$ of length $L$ in $G$ we can define a periodic sequence $y$ and for that $y$ define its rotation vector $\rho_\tau$ given by $$ \lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^{k-1} \varphi(\sigma^i(y)). $$ This simply gives us a vector which is non-zero at $L$ coordinates each equal $1/L$ and corresponds to the periodic shift-invariant measure determined by $y$. Now by Theorem 3.4 of the Ziemian's article cited above the vector $(\mu(e))_{e\in E}\in [0,1]^N$ belongs to the convex hull of $\rho_1,\ldots,\rho_C$ where $C$ is the number of the elementary loops in the graph $G$, and let $\rho_1,\ldots,\rho_C$ are their rotation vectors. This is the desired combination of periodic measures which has the same distribution on finite words of length $n+1$ as $\mu$ as required. If we only ask about density in the weak star topology of the periodic measures, which means that for each $\epsilon>0$, each shift-invariant measure $\mu$ and any integer $n$ there exists a single periodic orbit with the distribution of finite words of length $n$ which is $\epsilon$-close to the distribution given by $\mu$, then this is a fairly standard result, the oldest reference I know is the paper of Ville: J. Ville Étude critique de la notion de collectif, vol. 218, Theses françaises de l’entre-deux-guerres, Paris, 1939. https://eudml.org/doc/192893 Parthasaraty extended it to not necessarily finite alphabets: K. R. Parthasarathy On the category of ergodic measures. Illinois J. Math. 5 (1961), 648--656. https://projecteuclid.org/euclid.ijm/1255631586 Further generalizations are due to Sigmund: K. Sigmund Generic properties of invariant measures for Axiom ${\rm A}$ diffeomorphisms. Invent. Math. 11 1970 99--109. Some more references can be found here: http://arxiv.org/abs/1404.0456<|endoftext|> TITLE: Does there exist something like an $H_3$ and $H_4$ (icosahedral) Lie algebra or algebraic group? QUESTION [18 upvotes]: The (finite-dimensional) complex simple Lie algebras have been classified by Killing and Cartan a long time ago in the $\mathsf{A}_n,\mathsf{B}_n,\mathsf{C}_n,\mathsf{D}_n$ families and $\mathsf{G}_2,\mathsf{F}_4,\mathsf{E}_6,\mathsf{E}_7,\mathsf{E}_8$ exceptional cases. So the non-crystallographic $\mathsf{H}_3$ and $\mathsf{H}_4$ Coxeter groups (i.e., icosahedral symmetry) do not appear as Weyl groups of such algebras. On the other hand, there is also no such thing as a field with one element, yet it is interesting to construct something which might play that role. Perhaps more relevantly here, there is no such thing as a tiling of the plane with regular pentagons, but there is something of the sort. And the concept of finite-dimensional simple Lie algebra can be generalized in various ways, e.g., Kac-Moody algebras. So I've often wanted to ask: is there something which is remotely like the $\mathsf{H}_3$ and $\mathsf{H}_4$ Lie algebras or algebraic groups? They should probably be $33$- and $124$-dimensional respectively, and certainly the golden ratio $\frac{1+\sqrt{5}}{2}$ should play an important role in their definition (posing as an integer). REPLY [6 votes]: I will address an "easier question", namely analogues of algebraic groups corresponding to finite dihedral groups $I_2(n)$ of non-crystallographic type. At least in this case, we have candidate groups, but their theory was never developed much. First of all, Tits had realized in 1950s that semisimple algebraic groups (more precisely, sets of $F$-points of such groups for fields $F$) can be realized as automorphism groups of certain simplicial complexes, called (thick) spherical buildings modeled on Coxeter complexes of finite Coxeter groups $W$. In 1977 Tits proved existence of such buildings for all finite dihedral groups $I_2(n)$. It is known (it might be already in Tits' 1977 work, I forgot) that buildings for $I_2(n)$ constructed by Tits (provided that some care is taken during the construction) have large groups of automorphisms $G$, namely, they are flag-transitive. At the very least, this follows from the work of Tent. In particular, these groups $G$ have structure of BN pairs, etc. However, algebraic nature (in the sense of connection to algebraic geometry and Lie algebras) of groups $G$ remains a mystery. Some circumstantial evidence that such connection exists appears in my paper with Arkady Berenstein (where you can also find references to the relevant work of Tits and of Tent): A. Berenstein, M. Kapovich, Stability inequalities and universal Schubert calculus of rank 2, Transformation Groups, Vol. 16 (2011) p. 955-1007. Now, the bad news is that Tits also proved that there are no thick spherical buildings for the Coxeter groups $H_3$ and $H_4$. It is quite possible that one needs to relax spherical building axioms but at this stage it is very much unclear how in order to make a meaningful connection to the group theory.<|endoftext|> TITLE: Rule for Finding Largest Possible Perimeter QUESTION [10 upvotes]: Some months ago I was given a high-school level maths question that made me wonder if there was a definitive principle for finding the shape of largest perimeter given a set of dots. The question was worded something like the following; Given a 5 by 5 array of points, and using only straight lines, form a polygon with the highest possible perimeter. I understood I would have to use as many and as space-efficient diagonals as possible to maximize my possible perimeter, but couldn't find a rule or certain answer as to the upper bound. So A) What would be the highest possible perimeter shape (I believe I found somewhere around 40, but others found higher) and B) What constant rule can be followed to find the largest possible perimeter for a definite area. I was considering writing a program to brute-force all possible shapes, but wasn't sure if that was even feasible in a reasonable finite time. EDIT: Polygon must touch all points in the array and can not self-intersect. Sorry for not being clear REPLY [4 votes]: This gives a perimeter of $11 + 3\sqrt{2} + 7\sqrt{5} + 2\sqrt{10} + 2\sqrt{13} \simeq 44.43$.<|endoftext|> TITLE: can another topology be given to $\mathbb R$ so it has the same continuous maps $\mathbb R\rightarrow \mathbb R$? QUESTION [33 upvotes]: We say two topologies $\tau$ and $\rho$ on $X$ are similar if the set of continuous functions $f:(X,\tau) \rightarrow (X,\tau)$ is the same as the set of continuous functions $f:(X,\rho)\rightarrow (X,\rho)$. Does there exist a topology $\tau$ that is similar to the euclidean topology on $\mathbb R$? This was asked here but all we could prove is that $\tau$ must be a refinement of the euclidean topology. Regards. REPLY [15 votes]: This is a special case of much more general results surveyed in the book MR0393330 Magill, K. D., Jr. A survey of semigroups of continuous selfmaps. Semigroup Forum 11 (1975/76), no. 3, 189–282. For example, Theorem 2.3 of this book says that only the abstract semigroup structure (with respect to composition) of the set of continuous maps is sufficient to recover the topology.<|endoftext|> TITLE: Harmonic functions (eigenfunctions of the Laplace-Beltrami operator) of SO(2n)/U(n) QUESTION [11 upvotes]: Have the eigenfunctions of the Laplace-Beltrami operator on $SO(2n)/U(n)$ been worked out explicitly? If not, how does one approach finding them? (I'm thinking of this as in analogy with the spherical harmonics $Y(\theta,\phi)$ being the eigenfunctions of the Laplace-Beltrami operator on $SU(2)/U(1) = \mathbb S^2$.) Context: Brif and Mann (1998) gave a nearly explicit construction for phase space representations of quantum systems with Lie algebraic dynamical symmetries. However, their construction is not all that useful in practice because it lacks some ingredients from representation theory. This is evident in their examples which start with the statements "Group elements can be parameterized in the following way..." and "The harmonic functions on [some coset space] are..." What follows the ellipses are well-known facts and the connection to the general theory is lost on me. I would love to follow their construction for the algebra I'm interested in: $\mathfrak{so}(2n)$. Following their construction leads to the coset space $SO(2n)/U(n)$. Now this is where I am stuck since in the examples they give the parameterization of the group elements and harmonic functions come from elsewhere. REPLY [5 votes]: The pair $(SO(2n), U(n))$ is a symmetric pair. You can have a look at Kraemer's paper (Comp. Math. 1979), where in the table the author lists explicitly the complex representations of $SO(2n)$ on which the fixed point set of $U(n)$ is nontrivial. There you'll find (combinations of) fundamental weights of $so(2n)$. These are the representations that appear in the Peter-Weyl theorem about $L^2(SO(2n)/U(n))$.<|endoftext|> TITLE: Does anyone understand this comment about the continuum hypothesis? QUESTION [8 upvotes]: At 31:37 in his lecture titled What is a manifold? posted on Youtube, Mikhail Gromov states that if we do not allow generic functions to exist then the continuum hypothesis is "obviously" true, and that if we do allow generic functions to exist, then the continuum hypothesis is "obviously" false. What concept is he referring to? It sounds extremely important. Note: this is a follow-up to a question I posted on Math.SE: https://math.stackexchange.com/questions/1887350/what-is-a-generic-genetic-geometric-map-in-the-study-of-manifolds Also I'm not sure how to tag this question, so please feel free to fix the tags as appropriate. EDIT: When Gromov mentions generic functions for the first time in the lecture, around 29:30 if I remember correctly, it seems like it is with regards to generic points (which are generic by Sard's Theorem I believe) for which the Implicit Function theorem can be used to generate a manifold from the equation $f(x)=0$; the functions $f$ for which this holds are what he refers to as "generic". I don't know if this helps anyone at all; I don't really understand Gromov's terminology. REPLY [15 votes]: You might read "if we do not allow generic functions to exist then the continuum hypothesis is obviously true" as a reference to the fact that Borel sets satisfy CH, or possibly to the fact that CH holds in Godel's constructible universe. For the "obviously false" part, I suspect a nod to Paul Cohen's comment: A point of view which the author feels may eventually come to be accepted is that CH is obviously false. The main reason one accepts the axiom of infinity is probably that we feel it absurd to think that the process of adding only one set at a time can exhaust the entire universe. Similarly with the higher axioms of infinity. Now $\aleph_1$ is the set of all countable ordinals and this is merely a special and the simplest way of generating a higher cardinal. The set $c$ is, in contrast, generated by a totally new and more powerful principle, namely the power set axiom. It is unreasonable to expect that any description of a cardinal which attempts to build up that cardinal from ideas deriving from the replacement axiom can ever reach $c$. Thus $c$ is greater than $\aleph_1$, $\aleph_\omega$, $\aleph_\alpha$ where $\alpha = \aleph_\omega$, etc. This point of view regards $c$ as an incredibly rich set given to us by one bold new axiom, which can never be approached by any piecemeal process of construction. I feel disingenuous not adding that my personal view is that $\aleph_1$ and the real line are both proper classes, so that CH is not a meaningful question.<|endoftext|> TITLE: Examples of non-Kähler compact complex manifolds which satisfy the Dolbeault isomorphism QUESTION [6 upvotes]: Let $X$ be a complex manifold and let $$f_X^n : \bigoplus_{p + q = n} H^{p, q}(X) \to H^n_{\mathrm{DR}}(X, \, \mathbb{C})$$ be the natural map from the Dolbeault cohomology to the de Rham cohomology with complex coefficients. Question. Is there an example of a compact complex manifold $X$ which is not Kähler such that $f_X^n$ is an isomorphism for all $n$? REPLY [4 votes]: I am slightly confused by Francesco's comment and answer, because the Frolicher spectral sequence does degenerate for compact complex surfaces. So why isn't a non-Kahler surface an example? Perhaps there is some subtlety in defining the groups $H^{p,q}(X)$ that has been swept under the rug. In any case, because the Frolicher spectral sequence degenerates at the $E_1$-page, Hodge decomposition (without Hodge symmetry!!) should hold, even for non-Kahler surfaces. An example would be the quotient of $\mathbb{C}^2\backslash 0$ by multiplication by $2$. This compact complex manifold is diffeomorphic to $S^1\times S^3$ so is non-Kahler (the first Betti number is odd). But it is the case that $$H^k(X,\mathbb{C})=\bigoplus_{p+q=k} H^{p,q}(X).$$ Also, there is a subtlety to phrasing your question, as in general, there is no natural map from Dolbeault to de Rham cohomology, only a spectral sequence from one to the other.<|endoftext|> TITLE: Are Fourier transforms of L^p stable under diffeomorphisms? QUESTION [15 upvotes]: Let $\xi$ be a compactly supported distribution on $\mathbb R^n$ and assume that its Fourier transform is in $L^p$. Let $\phi:\mathbb R^n\to\mathbb R^n$ be a diffeomorphism. Does the Fourier transform of $\phi^*(\xi)$ always lie in $L^p$? REPLY [8 votes]: This is false in all dimensions, even if $\phi$ is real-analytic. In dimensions $n\ge 2$ Piero D'Ancona already explained why: the Fourier transform of surface measure on a spherical cap in $\mathbb{R}^n$ is bounded by $O(1)|\xi|^{-(n-1)/2}$ (what matters here is curvature), while surface measure on a piece of hyperplane does not decay at all in the direction orthogonal to the hyperplane, so it is in no $L^p$ with finite $p$ (see edit below for explanation). In the line, let $\mu$ be the uniform (Cantor-Lebesgue) measure on the ternary Cantor set. It is well known that $\widehat{\mu}(\xi)$ does not tend to $0$ as $\xi\to\infty$ (since $\widehat{\mu}(3^k\xi)=\widehat{\mu}(\xi)$), so it is in no $L^p$ space with $p<\infty$. However, Kaufman proved that there is $\delta>0$ such that for any $C^2$ map $\phi$ with $\phi''>0$ everywhere, the push down measure $\phi\mu$ satisfies that $$ |\widehat{\phi\mu}(\xi)| \le O(1)|\xi|^{-\delta}, $$ so that $\widehat{\phi\mu}$ is in $L^p$ for sufficiently large $p=p(\delta)$. I believe it is even possible to construct examples of measures $\mu$ and diffeomorphisms $\phi$ such that $\widehat{\mu}\notin L^p$ for any finite $p$ and $\widehat{\phi\mu}\in L^{2+\varepsilon}$ where $\varepsilon>0$ is arbitrarily small. Edit: The Fourier transform of compactly supported probability measures is uniformly continuous (in fact Lipschitz). Then, if $\widehat{\mu}$ does not tend to $0$ at infinity, there is $c>0$ such that $|\widehat{\mu}(\xi)|>c$ has infinite measure, so $\widehat{\mu}$ can be in no $L^p$ space, $p<\infty$.<|endoftext|> TITLE: When is the adjacency algebra of a graph an association scheme? QUESTION [5 upvotes]: The adjacency algebra of a graph is the algebra consisting of all polynomials in the adjacency matrix of the graph. An association scheme is a commutative matrix algebra containing the identity and all ones matrices and which is closed under entrywise product and transposition. An association scheme is symmetric if it contains only symmetric matrices. It is not difficult to see that since an association scheme is closed under entrywise product, it must have an orthogonal (with respect to the Hilbert-Schmidt inner product) basis of 01 matrices. These matrices can be viewed as the adjacency matrices of some (di)graphs and these are often called the classes of the scheme. From here it is easy to see that the adjacency algebra of a graph is a (necessarily symmetric) association scheme if and only if it is closed under entrywise product and contains the all ones matrix (the latter of which is equivalent to the graph being connected and regular). I am interested mostly in sufficient conditions on a graph for its adjacency algebra to be a symmetric association scheme. This is a little open ended, so maybe I should ask something more specific like the following: If $G$ is a connected graph that is a single class in an association scheme, then is it true that the adjacency algebra of $G$ is an association scheme? Two obvious necessary conditions on $G$ are that it must be connected and be the union of classes in a symmetric association scheme, but those are not sufficient. On the other hand it is sufficient for $G$ to be distance regular, but I was hoping for something a bit more general than that. REPLY [4 votes]: I am not sure is this is a useful answer, but it is correct: if the graph is connected and regular, its adjacency algebra is an association scheme if and only if it is closed under Schur product. Note that that it's enough to check that the Schur product of any basis elements is a linear combination of the basis elements. (And there is a basis consisting of powers of the adjacency matrix)><|endoftext|> TITLE: Why is the current math community not contributing to machine learning much? QUESTION [17 upvotes]: This question was inspired from What advantage humans have over computers in mathematics? and the answer of Brendan McKay, part of which is quoted in the below: The day will come when not only will computers be doing good mathematics, but they will be doing mathematics beyond the ability of (non-enhanced) humans to understand. Denying it is understandable, but ultimately as short-sighted as it was to deny computers could ever win at Go. The question and the answer got 42 and 35 votes, respectively, so at least a part of the community acknowledges the importance of AI in the future of mathematics or even the possibility that it will eventually replace mathematicians as they will develop mathematics much faster than humans will do. Recently, machine learning algorithms are getting better at lower-level tasks such as computer vision, motion control and speech recognition, thanks to deep learning, reinforcement learning and massive amount of data available online, which boosted the power of supervised learning. However, without more sophisticated unsupervised learning algorithm (such as adversarial networks as discussed here), higher-level tasks such as reasoning are still difficult, and this is where the research is actively done right now. Most parts of the mainstream ML don't require sophisticated mathematics, and traditionally such areas haven't attracted many mathematicians. However, since ML will eventually achieve human-level AI, which will develop every areas of math like a far-reaching theory, ML should be an interest for mathematicians. (There is nothing that prevents mathematicians from working on any problem whose solution may require non-purely-mathematical process. For example, the proof of four color theorem used a computer, and experimental mathematics has been gaining popularity.) I'm not proposing mathematicians to construct a purely mathematical model of AI such as Universal AI. Rather than such theoretical pursuit or simulating human brain, leading AI researchers argue that the future of AI will be on the line of the current mainstream ML. If mathematicians of this generation will not contribute to ML research, and if ML researchers will achieve human-level AI, that means ML researchers of this generation contributed to the development of math more than we did. This would be a nightmare for us. My question is the following: Why is the math community not spending enough manpower for developing ML, even though the earlier the completion of human-level AI will be, the faster the long-run development of math will be? REPLY [4 votes]: Actually there are a lot of mathematicians working on Machine Learning technics but it is unlikely that you can access their work. In the U.S., the NSA "is actively seeking mathematicians to join a vibrant community of mathematicians, statisticians, computer scientists, and other intelligence professionals to work on some of our hardest signals intelligence and information security problems." They are many mathematicians employed by the NSA, you can apply there if you are interested. Here is an article describing some of the consequences of their researches in the field of machine learning. I guess this answers the firt part of your question: there are a lot of manpower dedicated to ML from a mathematical viewpoint (but not all mathematicians want to be part of it.) You seem to believe that the completion of a human level A.I. is in reach. There is no evidence that this is the case, despite the billions grants national Science agencies are pouring over scientists. We have been there already. Here is a famous quote from Herbert Simon, Nobel Prize-winning AI researcher: "There are now in the world machines that think". This quote is from the sixties and probably got him to harvest a few millions in grants for his researches. This should not prevent you to think on your own, and take these kind of assertions with a grain of salt. Finally, there are many efforts going on now when it comes to automated solvers. ML is one of them but mathematicians have since long learned not to put all their eggs in one basket.<|endoftext|> TITLE: Topological spaces with too many open sets QUESTION [7 upvotes]: Is there a Tychonoff space $X$ without isolated points with the following property: For any $a\in X$ and any function $f : X\longrightarrow \mathbb{R}$, if $f$ is continuous on $X\backslash \{a\}$ then we can redefine $f$ at $x=a$ such that (the new) $f$ is continuous on $X$. REPLY [4 votes]: Consistently, $\omega^*$ has this property. In the paper E. van Douwen, K. Kunen, and J. van Mill, "There can be $C^*$-embedded dense proper subspaces of $\beta\omega - \omega$," Proc. Amer. Math. Soc. 105 (1989), pp. 462-470, available here. it was shown to be consistent that, for every $p \in \omega^*$, every bounded continuous real-valued function on $\omega^* - \{p\}$ can be extended to $p$. This almost gives a consistent answer to the OP's question, the only problem being that the property in the question does not mention boundedness. However, in the context of $\omega^*$ it turns out that adding in boundedness does not hurt anything: Observation: If $p \in \omega^*$, then every real-valued continuous function on $\omega^* - \{p\}$ is bounded. Proof: Suppose $f$ is an unbounded continuous real-valued function on $\omega^* - \{p\}$. Pick a sequence $\langle x_n : n < \omega \rangle$ of points in $\omega^* - \{p\}$ such that $f(x_n)$ converges to infinity. It is well known that $\omega^*$ is compact and contains no nontrivial convergent sequences. Thus there must be a point $q \in \omega^*$ with $q \neq p$ at which the $x_n$ cluster. Since $f(q)$ is some (finite) real number but $f(x_n)$ goes to infinity, this contradicts the continuity of $f$. $\qquad$ QED On the other hand, Fine and Gillman proved that $\omega^*$ consistently fails to have the OP's property (CH implies that it does not). I do not know whether your question has a positive answer in ZFC (but I suspect that it does). Here is a link to a related paper that you might find interesting (it's where I learned about the two results I mention above).<|endoftext|> TITLE: The Halting Problem and Church's Thesis QUESTION [8 upvotes]: In the opening chapters of Hartley Rogers, Jr.'s book Theory of Recursive Functions and Effective Computability, the proofs of the unsolvability of the halting problem and related unsolvability results invoke Church's Thesis. Can Church's Thesis be avoided? Edit added on September 7 while putting a bounty on the question: While in no way whatsoever doubting the diagonal argument per se, I am nevertheless in doubt as to how this is meant to work in the undecidability set up as witnessed also by my probes. One $\textit{evasion-possibility}$ that I, given my befuddlements, am not able to exclude a priori is that there $\textit{is}$ a recursive function that tells us whether a given Turing machine halts, but that a diagonalisation of the Turing machine would - pace Church and Turing - provide an effective non-recursive method beyond that recursive function. I am rather certain that the evasion-possibility does not obtain, but I want to understand precisely why it does not. Somehow, I would have thought that the answer to this were simple. (The last sentence was not meant to contradict or undermine comments that pointed towards a tediousness of carrying these matters through.) REPLY [4 votes]: To address your "evasion-possibility" directly: Let's look again at the overall structure of the proof of the unsolvability of the halting problem. If there were a Turing machine $M$ that could solve the halting problem, then we could construct another Turing machine $M'$ that uses $M$ as a subroutine in a certain way, etc., contradiction. We can explicitly write down the additional instructions that are needed to convert $M$ into $M'$, since this is a fixed piece of code that is independent of $M$, and so in particular there is no reason (other than laziness) to invoke the Church–Turing thesis to establish the existence of this piece of code. So if $M$ is a Turing machine then $M'$ must also be a Turing machine. That's all that we need to prove that there is no Turing machine to solve the halting problem. This proof doesn't exclude the possibility that there is an effective but non-recursive procedure to solve the halting problem; for that conclusion, we would need to invoke the Church–Turing thesis. But the proof does exclude the possibility that there is a Turing machine to solve the halting problem for Turing machines, and that is all that we would expect to be able to prove mathematically anyway.<|endoftext|> TITLE: Singularity theorems for semiclassical gravity QUESTION [5 upvotes]: The semi-classical Einstein equations (without a cosmological constant) are $G^{\mu \nu} = 8\pi \langle T^{\mu \nu} \rangle$. I am told that there are serious objections as to why these equations cannot be a completely correct description of nature. (If they are correct, then there is no need for quantum gravity.) From what I have heard, the objections are 1) Superluminal propagation is possible in some cases. 2) If one takes $\frac{1}{\sqrt{2}} (\vert M \ at \ A \rangle + \vert M \ at \ B \rangle)$, where $M$ is large and $A$ and $B$ are widely separated, one arrives at counter-intuitive conclusions. My question is : Do the Hawking-Penrose singularity theorems still hold in semi-classical gravity ? If so, I shall be happy if pointed out to a rigorous mathematical reference. I have some physics questions too but perhaps another forum is a better venue for these : 1) Is objection 1) true? Are there are any other objections ? 2) So what if superluminal propagation is possible in extreme situations ? What evidence do we have against it anyway (in such extreme cases that is) ? As for objection 2), does the wave function of the universe allow such pathologies ? REPLY [5 votes]: The semi-classical Einstein equations (without a cosmological constant) are $G^{\mu \nu} = 8\pi \langle T^{\mu \nu} \rangle$. I am told that there are serious objections as to why these equations cannot be a completely correct description of nature. (If they are correct, then there is no need for quantum gravity.) From what I have heard, the objections are [...] There are many, many issues with semiclassical gravity. Examples of problems in addition to the two you mention: questions about stability of flat spacetime, problems with the Born rule and Copenhagen interpretation due to nonlinearity. The issue is not whether semiclassical gravity is fundamental but whether it's even a reliable approximation, and, if so, which formulation of semiclassical gravity to use and how to put bounds on the errors. To my mind, there is an overriding physical reason why semiclassical gravity clearly can't be a fundamentally correct theory, which is that it couples a classical field to a quantized field. Physicists realized almost a century ago that this just generically doesn't work. Bohr and his followers wanted to keep the electromagnetic field classical while quantizing the atom. This resulted in the Bohr-Kramers-Slater theory, which lived for only a few years before being disproved experimentally by Bothe. The basic issue is that if you want energy and momentum to be conserved in a quantum context, you need long-range correlations, and those correlations don't exist in a classical field. For example, in the photoelectric effect, you would like the absorbed photon to give its energy to exactly one electron. If it gives its energy to both electron A and electron B, then you've doubled the amount of energy that's present. What prevents this from happening is that the electromagnetic field is quantized, and therefore it can have correlations between what it does at A and what it does at B. Without these quantum correlations, energy and momentum can at best be conserved on a statistical basis. Nonconservation of energy-momentum has never been observed, and if it did happen (locally), it would render the Einstein field equations inconsistent. Do the Hawking-Penrose singularity theorems still hold in semi-classical gravity ? Willie Wong has answered this, but I would like to provide some perspective in addition. Semiclassical gravity, if it works at all as an approximation scheme, is an approximation scheme that applies to weak gravitational fields. Therefore if it told us something about singularities, we wouldn't believe it. Generically, theories of quantum gravity tend to provide mechanisms that prevent the formation of singularities once you get down to the Planck scale. So what if superluminal propagation is possible in extreme situations ? What evidence do we have against it anyway (in such extreme cases that is) I think Willie Wong's answer clarifies why this is not really the right way to frame the issue. However, there is a generic objection to superluminal propagation, which is that if it happens, it typically means we don't have existence and uniqueness of solutions to Cauchy problems, so that physics loses its predictive value.<|endoftext|> TITLE: Is the J homomorphism compatible with the EHP sequence? QUESTION [21 upvotes]: The EHP sequence consists of maps $S^n\stackrel{E}{\to} \Omega S^{n+1} \stackrel{H}{\to} \Omega S^{2n+1}$. This is a homotopy fibration sequence if $n$ is odd. For even $n$ it is a homopy fibration sequence after localizing at $2$. Applying $\Omega^n$, one obtains the sequence $\Omega^n S^n{\to} \Omega^{n+1} S^{n+1} {\to} \Omega^{n+1} S^{2n+1}$ By J homomorphism I just mean the map $O(n)\to \Omega^n S^n$ given by one-point compactification. It is mentioned on page 30 of this book that at least if one localizes at $2$, then the following diagram commutes (up to homotopy?) (EDIT: as John Klein points out, the right square only commutes after being looped once, and this is what is claimed in the linked book of Ravenel). Here the right vertical map is the suspension. $$ \begin{array}{cccccc} O(n) &\to & O(n+1)& \to & S^n \\ \downarrow & & \downarrow & & \downarrow \\ \Omega^n S^n &{\to}& \Omega^{n+1} S^{n+1} &{\to} &\Omega^{n+1} S^{2n+1} \end{array} $$ Question 1: Is there a reference where it is proved that the right square commutes? Is it necessary to localize at $2$ for it? This diagram implies that the induced homomorhism on relative homotopy groups $\pi_k(O(n+1),O(n))\to\pi_k( \Omega^{n+1} S^{n+1}, \Omega^{n} S^{n})$ is an isomorphism roughly for $k < 2n$, after localizing at $2$. In fact, there is a good reason to believe that this is an isomorphism without any localization. If true, this must be well-known. Question 2: What is a good reference to the asserion that this map of relative homotopy groups is an isomorphism in the metastable range $k\le 2n-\epsilon$? REPLY [5 votes]: In my comment to my first answer, I noted that I didn't address Greg's question (1). This answer aims to address that question. I am indebted to Bill Richter for explaining the following argument to me. We are trying to understand whether or not the square $$ \require{AMScd} \begin{CD} O(n+1) @>>> S^n \\ @VVV @VVV \\ F_{n+1} @>> H > \Omega^{n+1} S^{2n+1} \end{CD} $$ commutes. I mentioned in my comment that James proved it commutes after a loop. I will sketch a proof of a related result which is both more and less general than the one of James. Let $\lambda = \Sigma H$ denote the Boardman and Steer Hopf invariant. This is a natural transformation of based function spaces $$ F_\ast(A,B) \to F_\ast(\Sigma^2 A,\Sigma B\wedge \Sigma B) $$ given by suspending the James-Hopf invariant. Let's consider the related problem of whether or not the diagram $$ \begin{CD} G_{n+1} @> \pi >> S^n \\ @VVV @VV E V \\ F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2} \end{CD} $$ commutes. If it does then Greg's question (1) will commute if we replace $H$ by $\lambda$, since the map $O(n+1) \to S^n$ factors through $G_{n+1}$. Claim: The last diagram above homotopy commutes after taking a single loop. Reformulate the claim as follows: let $B = \Sigma A$ be a suspension and let $B\to G_{n+1}$ be any map. Then the claim is equivalent to the assertion that the diagram $$ \begin{CD} B @> \pi >> S^n \\ @VVV @VV E V \\ F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2} \end{CD} $$ commutes (by taking $B = \Sigma \Omega G_{n+1}$). To verify the reformulation of the claim, we will use the Boardman and Steer Cartan formula. Let $X = S^n$, then the action of $G_{n+1}$ on $X$ induces a map $$ F: B \times X \to X\, . $$ The Hopf construction of $F$ is a map $h_F: \Sigma B\wedge X \to \Sigma X$ whose adjoint is the left vertical (inclusion) map of the square. Thus we need to compute $\lambda(h_F): \Sigma B \wedge \Sigma X \to \Sigma X \wedge \Sigma X$. Take the suspension of this map $\Sigma F: \Sigma (B \times X) \to \Sigma X$. Then we have a factorization up to homotopy $$ \begin{CD} \Sigma (B\times X) @>\Sigma F >> \Sigma X \\ @V p_1 + p_2 + p_{12} VV @| \\ \Sigma B \vee \Sigma X \vee \Sigma B\wedge X @>>(f,g,h_F) > \Sigma X \end{CD} $$ where $p_1$ is the projection onto $\Sigma B$, $p_2$ the projection onto $\Sigma X$ and $p_{12}$ is the quotient map onto $\Sigma B \wedge X$. The map $f$ is the suspension of the restriction of $F$ to $B \times \ast$, the map $g$ is the suspension of the restriction of $F$ to $\ast \times X$. Since $\Sigma F$ is a suspension, its Hopf invariant $\lambda(\Sigma F)$ is trivial. So, if we take the Hopf invariant of the composite $$ \Sigma F = (f,g,h_F) \circ (p_1 + p_2 + p_{12}) $$ and use the composition formula and the Cartan formula (in Boardman and Steer), we obtain, after some rewriting, the formula $$ \lambda(h_F) = (f \wedge g) \circ \Sigma p_{12} \qquad (*) $$ Implicit in this calculation, which I've omitted, is the fact that the reduced diagonal maps of both $B$ and $X$ are trivial since these spaces are suspensions. I refer the reader to Boardman and Steer's paper for the details. Returning to our original notation, note that the map $f$ is the suspension of composition $$ \begin{CD} B @>>> G_{n+1} @>\pi >> S^n \, , \end{CD} $$ and the map $g$ is the identity map of $S^{n+1}$. The map $\lambda(h_F)$ is adjoint to the composite $$ \begin{CD} B\to G_{n+1} \to F_{n+1} @>\lambda >> \Omega^{n+2} S^{2n+2}\, \end{CD} $$ which is one of the composites of the reformulated claim. The map $(f \wedge g) \circ \Sigma p_{12}$ is the composite $$ \begin{CD} \Sigma^2 (B \times S^n) @>>> \Sigma^2 (G_{n+1} \times S^n) @>q >> \Sigma^2 G_{n+1} \wedge S^n @>\Sigma \pi \wedge 1 >> \Sigma S^n \wedge \Sigma S^n \end{CD} $$ where $q$ is the quotient map. By definition, the last composition is adjoint to the composition $$ \begin{CD} B @>>> G_{n+1} @> \pi >> S^n @> E >> \Omega^{n+2} S^{2n+2} \end{CD} $$ which is the other composite of the reformulated claim. Thus the claim follows from equation $(\ast)$. Comments: (1). The above argument shows that the diagram of Greg's question (1) commutes after looping. The equation $(\ast)$ isn't generally valid when $B$ (or $X$) isn't a suspension. For general $B$, there is a correction term in the formula involving the reduced diagonal $B\to B \wedge B$. This gives evidence that Greg's diagram will fail to commute before taking loops. A potential counterexample to the diagram commuting is to take $n = 2$ and $B =\Bbb RP^3 = SO(3) \subset O(3)$, since the diagonal $\Bbb RP^3 \to \Bbb RP^3 \wedge \Bbb RP^3$ is stably essential (it's detected by the the cup product). In fact, Bill Richter (unpublished) shows that if $n > 2$ then Greg's diagram fails to homotopy commute (Richter also explicitly identifies the deviation from the diagram commuting). (2). The reader might ask why we've worked with the Boardman-Steer Hopf invariant rather than the one of James. The answer is that James' invariant doesn't satisfy a Cartan formula––––there are "partial" Cartan formulae, which were known to Barratt, but this is lost knowledge... (3). The Boardman and Steer paper I alluded to is: Boardman, J. M.; Steer, B. On Hopf invariants. Comment. Math. Helv. 42 1967 180–221. See also: Boardman, J. M.; Steer, B. Axioms for Hopf invariants. Bull. Amer. Math. Soc. 72 1966 992–994. (4) The result of James about the diagram looping after one suspension isn't stated by James. In effect, James proves the (reformulated) claim in the special case when $B = S^p$ is a sphere (so Greg's diagram will commute on homotopy groups). The result is given by Corollary 15.9 of the paper: James, I. M. On the suspension triad. Ann. of Math. (2) 63 (1956), 191–247.<|endoftext|> TITLE: A topological group which is also a (not necessarily smooth) manifold is orientable QUESTION [8 upvotes]: I am trying to show that a topological group which is also a (not necessarily smooth) manifold is automatically orientable. I know of a proof involving transition functions for smooth manifolds, in which case the object in question is a Lie group. I am using Hatcher's definition of orientability: An $n$-manifold $M$ is orientable if it admits a local orientation $\eta_x$ at each $x\in M$ where $\eta_x$ is a generator of $H_n(M\mid x)\cong \mathbb{Z}$, with the following compatibility property: For each $x\in M$, there is an open ball $x\in B_x\cong \mathbb{R}^n$ so that for every $y\in B_x$, the local orientation $\eta_y$ is the isomorphic image (induced by inclusion of pairs) of the same generator $\eta_{B_x}$ of $H_n(M\mid B_x)$. I have a clear candidate for such a local orientation, but I am having trouble showing the compatibility: Let $e$ be the identity of the topological group $M$. Choose any generator $\eta_e$ of $H_n(M\mid e)$, and for any $g\in M$, let $\eta_g = L^g_*(\eta_e)\in H_n(M\mid g)$ where $L^g:M\to M$ is left multiplication by $g$ ($L^g$ is a homeomorphism, so it certainly induces an isomorphism on homology). To start showing the compatibility condition, given $x\in M$, let $B_x$ be any open neighborhood of $x$ homeomorphic to $\mathbb{R}^n$. We are required to show that the following diagram commutes: $\require{AMScd}$ $$\begin{CD} H_n(M\mid B_x) @>id>\cong> H_n(M\mid B_x) \\ @VV{\cong}V @V{\cong}VV \\ H_n(M\mid x) @>{\cong}>L^{(y^{}x^{-1})}_*> H_n(M\mid y) \end{CD}$$ where the vertical maps are induced by inclusion. Here is where I am stuck. The corresponding diagram on the level of topological spaces certainly does not commute. Any ideas, thoughts, hints, or full solutions are welcome! REPLY [11 votes]: In fact beside your question there is a beautiful theorem in homotopy theory. This theorem due to T. Bauer, N. Kitchloo, D. Notbohm and E. K. Pedersen guarantees that any loop space $X=\Omega B$ where $B$ is a $CW$-complex and such that $H_*(X)=\oplus H_i(X;\mathbb{Z})$ is a finitely generated abelian group is homotopy equivalent to a compact, smooth, parallelizable manifold ("Finite loop spaces are manifolds" Acta Math. 2004). This theorem applies to your case where $X=G\simeq \Omega BG$. The very first step in the proof of this result which is very closely related to your question is to prove that $X$, thus $G$ in your case, is a Poincaré duality space. The proof is completely algebraic and uses the fact that for any field $\mathbb{F}$ the cohomology algebra is a finitely generated, connected graded Hopf algebra. Then you use Borel's classification that tells you that $H^*(G;\mathbb{F})$ is a tensor product of exterior algebras and truncated polynomials algebras and get a family of top classes $[G]_p\in H_{n_p}(G;\mathbb{Z}/p\mathbb{Z})$ for an integer $n_p$. Then you show that all these $n_p$ are equal to a fixed integer $n$ and that $[G]_p$ is the reduction $mod(p)$ of an integral fundamental class whose cap product induces a Poincaré duality isomorphism. All details and references are given in the introduction of Bauer, Kitchloo, Notbohm, Pedersen's paper. Edit: It is also certainly worth recording Gleason, Montgomery-Zippin theorem, namely: "A topological group G underlies a (unique) Lie group structure if and only if the underlying space of G is a topological manifold."<|endoftext|> TITLE: Intuition behind Madsen-Tillmann spectra QUESTION [10 upvotes]: Dan Freed discussed in his notes beneath (20.32) the intuition behind the definition of Madsen-Tillmann spectra as "virtual" Thom spectrum, which I reproduce below I don't think I quite get his idea, like what it means by "virtual" and why this could be more interesting or useful than the real one (I mean the usual Thom prespectrum). Could somebody give an elaborate discussion on his intuition? Of course if you have a different opinion or other intuition please do not hesitate to share with us. PS: the answer given by Gregory below explains the motivating for introducing virtual Thom spectra. It would be even better that somebody can elaborate more on the motivation of using negative-dimensional vector bundles to define Madsen-Tillmann spectra and the specific usage of negative dimension in it. Actually in Freed's notes there is one comment right after this which says that $MT$ does not only stands for "Madsen-Tillmann" but also for "tangential variant" of the Thom spectrum i.e. the Madsen-Tillmann spectra are tangential and unstable. I wonder whether this "tangential" and "unstable" are also tied to the negative dimensionality. REPLY [15 votes]: As I understand, the question is about virtual Thom spectra, rather then specifically the Madsen-Tillmann spectra. For what it is worth, here is how I like to think about this. To understand virtual bundles and their Thom spectra, consider first bundles over the one point space. Such a "bundle" is the same thing as a vector space ${\mathbb R}^n$. The Thom space of this bundle is the sphere $S^n$ - the one point compactification of ${\mathbb R}^n$. In the world of spaces, there is no meaning to the notion of a negative-dimensional sphere. However, if you pass to the world of spectra, negative-dimensional spheres are well-defined. $S^{-n}$ can be defined as the Spanier-Whitehead dual of $S^n$, which is the same as the function spectrum $F(S^n, S^0)$. So we can say that $S^{-n}$ is the Thom spectrum of the "virtual vector space" $-{\mathbb R}^n$. Integers can be defined as formal differences of positive integers. Virtual vector spaces (or more generally virtual bundles) are formal differences of vector spaces (bundles). (So if you want to ask why virtual vector bundles are "useful" you can begin by asking why not necessarily positive integers are useful). In the same vein, the ($\infty$-)category of spectra is obtained from the ($\infty$-)category of spaces by inverting the spheres. The Thom spectrum of a virtual vector space is a formal difference of spheres. Note that the equivalence $S^m \wedge S^n\simeq S^{m+n}$ holds in the category of spectra, for all integers $m, n$. Moving to bundles over a general space $X$. To specify a vector bundle $\eta$ over $X$, say of dimension $n$, is the same as to specify a space $\tilde X$ with a free action of $O(n)$, such that that $X$ is identified with the quotient space of this action and the quotient map $\tilde X\to X$ is a fiber bundle. $\tilde X$ can be constructed as the pullback of the diagram $X \to BO(n) \leftarrow EO(n)$. Then the Thom space of $\eta$ can be defined as the quotient space $\tilde X_+ \wedge_{O(n)} S^n$. The Thom spectrum of $\eta$ is the orbits spectrum $\Sigma^\infty \tilde X_+ \wedge_{O(n)} S^n$. This expression remains meaningful if we replace $S^n$ with any spectrum with an action of $O(n)$, such as $S^{-n}$. So the Thom spectrum of the virtual bundle $-\eta$ can be defined as the spectrum $\Sigma^\infty \tilde X_+ \wedge_{O(n)} S^{-n}$. With this definition, there is an equivalence of spectra $Th(\xi \pm \eta)\simeq Th(\xi)\wedge Th(\pm\eta)$ for any two vector bundles $\xi, \eta$, where $\oplus$ stands for external direct sum. To connect this with Freed's explanation, note that the Thom space of the trivial bundle of dimension $n$ is equivalent to the smash product $X_+\wedge S^n$ and the Thom spectrum of negative the trivial bundle is the smash product of spectra $\Sigma^\infty X_+\wedge S^{-n}$. Now suppose $\eta$ is a vector bundle over $X$ that embeds as a subbundle of a trivial bundle of dimension $n$. Let $\xi$ be the orthogonal complement of $\eta$. Then you have an equivalence of virtual bundles $-\eta=\xi - n$, and an equivalence of Thom spectra $Th(-\eta)\simeq Th(\xi) \wedge S^{-n}$. I suppose the advantage of this presentation is that it only involves inverting the trivial bundle.<|endoftext|> TITLE: What is the status of the Hilbert 6th problem? QUESTION [31 upvotes]: As you know, the Hilbert sixth problem was to axiomatize physics. According to the Wikipedia article, there is some partial succes in this field. For example, Classical mechanics, I believe, can be treated now as an axiomatized discipline since it is properly formalized in the modern theories of Lagrangian and Hamiltonian mechanics (and as a corollary, one can consider it as an extension by definition of the theory of real numbers, which in its turn is an extension by definition of the axiomatic set theory -- and eventually all these theories can be treated as first order theories). I am not a specialist, but the Theory of relativity seems to be formalized as well in terms of differential geometry (is that correct?). On the other hand, as we understood not long ago, the Quantum mechanics is not axiomatized. I often discuss this with my colleagues and students, so I believe this is an important question, it deserves specification and clarification, the Wikipedia article is too short, it must be detailed. I believe there are people here who can explain clearly which parts of physics are axiomatized now, and which are not. For the educational purposes (and I am sure, this will be interesting for specialists as well) it would be good to have a list of these disciplines. In the discussions at the web, for example, in the discussion at MathStackExchange, I don't see this list. If it is possible, I ask specialists to share their knowledge. There is no necessity to list all the disciplines in one answer, each answer can be devoted to one discipline, I only ask people to give the necessary references and to provide some elementary explanations (of course, detailed answers are better) so that non-specialists (and students) could understand. Thank you. EDIT. From Ben Crowell's answer I deduce that there is some discrepancy in understanding of what is meant by the 6th Hilbert problem: some people interpret it as a suggestion to construct a first order theory of a given physical discipline (this means, to build a "completely new theory", formally independent from the other mathematics, with its own language of logic -- like in the modern set theories, ZF, NBG, MK, which are the standard examples of the first order theories), while others interpret this wider, as a possibility to give any system of axioms, not necessarily first order (this can be understood as building an extension by definition of another, already constructed first order theory -- and the standard examples are Hilbert's axioms of Euclidean geometry, Kolmogorov's axioms of probability theory, etc.). As to me I don't see serious reasons to restrict ourselves on first order theories: the "axioms as definitions" are quite satisfactory, moreover, I would say they are preferable, since they do not deflet attention from the essential part of the problem (to the details with the new language of logic, etc.). Actually, if we look at the problem from this "practical" point of view, I do not see the difference between "axiomatization" and "formalization", that is why I referred to the books by V.I.Arnold and A.L.Besse from the very beginning. References: V.I.Arnol'd, Mathematical methods of classical mechanics A.L.Besse, Einstein Manifolds REPLY [2 votes]: I did some work that was quite successful at axiomatis-ing (pardon grammatical abuse!) the principal Hamiltonian operators for all finite quantum theories. My co-worker was able to extend the methodology to continuous operator groups. The basic conclusion was to find a quantum brachistochrone equation- essentially the Heisenberg equation for the Hamiltonian + constraint- and show that all physical matrix groups were given as solutions. So to that extent, I would say that the Hilbert 6th problem has been addressed for quantum mechanics, as all the observed Hamiltonian matrices on SU(2), SU(3),..., SU(N) for N finite come out as natural solutions via this method, and coupled with the infinite dimensional case, all angles are covered and closed out. The essential link was to state that the path of flow on the complex projective manifold was defined by a Lagrangian that implied a least time principle under energetic constraints. Quantum mechanics can seem quite esoteric, and believe me when I say that my professor was not a massive fan of the first step being the derivation of the Hamiltonian operator from first principles. Most quantum physics books just assume that it has one form or another. Personally I find it better to derive than to assume. If there is any interest in exploring this topic, I'm happy to upload some references to some papers and continue the discussion.<|endoftext|> TITLE: Determinant of a specific $4 \times 4$ symmetric matrix QUESTION [8 upvotes]: In a recent research work, I have come across the following nice identity, where the entries $a,b,x$ belong to an arbitrary commutative unital ring: $$\begin{vmatrix} 2 & a & b & ab-x \\ a & 2 & x & b \\ b & x & 2 & a \\ ab-x & b & a & 2 \end{vmatrix}=(x^2-abx+a^2+b^2-4)^2.$$ Note that if the ring has characteristic $2$ then the formula is an obvious application of the Pfaffian. The only way I have been able to check this identity is through a tedious computation (of course, running any formal computing software will do). My question: Is there any elegant way to prove it? REPLY [5 votes]: In short: determinant of every symmetric matrix is a square! Consider arbitrary three matrices $a, b, c \in Sl_2.$ One can wonder, what relations do the traces of their products satisfy. The answer is given by the famous Jimbo-Fricke cubic: \begin{equation} \begin{split} &tr(ab)\ tr(bc) \ tr(ac)+ tr(ab)^2+tr(bc)^2+tr(ac)^2\\ &+tr(a)^2+tr(b)^2+tr(c)^2+tr(abc)^2\\ &-(tr(a)tr(b)+tr(c)tr(abc))tr(ab)\\ &-(tr(b)tr(c)+tr(a)tr(abc))tr(bc)\\ &-(tr(a)tr(c)+tr(b)tr(abc))tr(ac)\\ &+tr(a)tr(b)tr(c)tr(abc)-4=0.\\ \end{split} \end{equation} Every determinant of a symmetric matrix can be written in the following form for some matrices $a, b, c$: $$ G=\begin{vmatrix} 2 & -tr(a) & -tr(b) & -tr(bc)\\ -tr(a) & 2 & -tr(ab)& -tr(abc)\\ -tr(b) & -tr(ab) & 2 & -tr(c)\\ -tr(bc) & -tr(abc) & -tr(c)& 2\\ \end{vmatrix}. $$ The relation above is equivalent to the following: $$(2tr(ac)+tr(ab)tr(bc)-tr(a)tr(c)-tr(b)tr(abc))^2=G.$$ Usually a symmetric determinant is not a square, because $tr(ac)$ is not a polynomial in the entries of $G.$ The case of the matrix in the question corresponds to $c=a^{-1},$ because $tr(c)=tr(c^{-1}),$ $tr(aba^{-1})=tr(a)$ and $tr(ab)+tr(ba^{-1})=tr(a)+tr(b).$ The square root of $G$ is algebraic, because $tr(ac)=2.$ I have seen this presentation of the Jimbo-Fricke cubic only in one place: https://arxiv.org/pdf/1308.4092.pdf, formula (3.9), and I will be really grateful for any references.<|endoftext|> TITLE: Confusion with formally unramified = immersion and formally smooth = submersion QUESTION [5 upvotes]: From this MO question I learned to tentatively think of formally unramified arrows as immersions and of formally smooth arrows as submersions. I'm trying to semi-formally handwave myself into conviction about this but I'm getting the opposite identifications. The setting is a commutative square as below posing a unique lifting problem. $$\require{AMScd} \begin{CD} P @>>> U\\ @VVV @VV{f}V\\ N @>>> X \end{CD}$$ Suppose we start in the category of topological spaces and let $P\rightarrow N$ be the inclusion of a point into an open neighborhood, and let all the other arrows also be inclusions into opens of a space $X$. Then our diagram really lives in the lattice of opens of $X$ containing the point $P$. Since we want to deal with infinitesimal neighborhoods, move to the pro-completion of this lattice and replace $N$ by some formal intersection which isn't really there to get a good notion. Now our square lives in the pro-completion with all arrows still "inclusions of opens". The existence of a diagonal filler signifies that whenever an open $U$ of $X$ contains a point $P$, it also contains every infinitesimal thickening of it. (This fits perfectly with the intuition that open sets contain neighboring points.) Uniqueness follows from the fact all arrows are "on the nose inclusions", or, alternatively, from the fact we're living in the pro-completion of a lattice. So open inclusions are tentatively right-orthogonal to the class of such infinitesimal thickenings of points. The same goes for open embeddings. Now since $N$ is infinitesimal, this lifting problem should only care about local proeprties of $f$. So let $f$ be a topological immersion, i.e a map which is locally-on-the-domain an embedding. I think being an immersion gives existence by precomposing with elements of a good open cover, but by the analogy suggested above, immersions correspond to unramified arrows whose property gives at most one filler. What am I doing wrong here? It seems my mistake must be in saying that open embeddings gives existence, but I don't see why they should give uniqueness (without necessarily existence) instead. REPLY [2 votes]: I can't precisely isolate the wrong step, but it might help to consider a concrete motivational example, where $X$ is the $xy$-plane, $N$ is a very small interval in the $x$-axis, and $P$ is the origin. If $f: U \to X$ is the immersion given by inclusion of an interval in the $x$-axis, then you have a unique lift. If $f: U \to X$ is the immersion given by inclusion of an interval in the $y$-axis, then there is no lift, and passing to an open cover won't help.<|endoftext|> TITLE: Is this a q-count of Alternating Sign Matrices? QUESTION [11 upvotes]: The number of Alternating Sign Matrices of size $n$ is well known to be $\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!}$. Is it known whether the naive q-analog expression $$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{[n+k]_q!}$$ is a polynomial in $q$ with positive coefficients? Does it come from a known statistic on ASM's? REPLY [10 votes]: It's true these polynomials are not unimodal for n=2 and up, since they all start with coefficient sequence 1 0 1 ... (The reason for this is clear from the definition of descending plane partitions.) For n = 4, 5, 6, they are also non-unimodal in the middle. But the sequence of nonzero coefficients seems to be unimodal for n=7 and above (I've calculated up to n=35). I had observed the non-unimodality of n=4,5,6 several years ago, but had thought it would continue to fail, so it's interesting to me that it seems to be (nearly-)unimodal at n=7 and beyond. When the ASM is a permutation matrix, the right statistic is a weighted inversion count: weight each inversion pair by the larger number in the pair and then add up all the contributions. (See St000616 on FindStat.) The generating function for this statistic on permutations is an interesting analogue of the q-factorial. (See https://arxiv.org/pdf/1002.3391v2.pdf Corollary 6.) I also think looking at the Gog or Magog triangles to try and find the right statistic is a good idea, but it's not one of the usual statistics, as far as I can tell. I've thought about this question quite a lot and would be very interested if anyone is able to make progress on finding this statistic on ASM or TSSCPP.<|endoftext|> TITLE: Error in integral in Gradshteyn Ryzhik QUESTION [5 upvotes]: In formula 4.224 12 in the 8. edition of the Table of Integrals, Series, and Products is an error: In the case $a^2=1$ both equations should hold, but they are not equal. I think the second equation (i.e. $\int_0^\pi \ln(1+a\cos x)dx = 2 \pi \ln \frac{|a|}{2}$ ($a^2 \geq 1$) is wrong and it should be $$\int_0^\pi \ln((1+a\cos x)^2)dx = 2 \pi \ln \frac{|a|}{2}$$ This would also solve the problem, that the argument of the logarithm gets negative if $a^2 \geq 1$. Can someone confirm this and/or does someone know how to prove this? REPLY [5 votes]: this is entry 172 in the 2005 Errata, which somehow didn't make it properly into the 8th edition.<|endoftext|> TITLE: Can a zero entropy automorphism of the torus have a Li-Yorke pair? QUESTION [5 upvotes]: If a continuous map of a compact metric space has positive topological entropy, then it has a Li-Yorke pair. In general, the converse is not true, and a zero entropy continuous map can have a Li-Yorke pair. Can a zero entropy automorphism of the torus have a Li-Yorke pair? More generally, can a zero entropy endomorphism of a compact Lie group have a Li-Yorke pair? REPLY [3 votes]: I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$. For toral automorphisms, I think the answer to your question is no. The idea is this: if $T$ is ergodic, then it has positive entropy (every ergodic linear toral automorphism does, see for instance here). If $T$ is not ergodic, you can find a factor of (some power of) $T$ on a torus of lower dimension which is ergodic, which means again that $T$ has positive entropy. Indeed, being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. Moreover you can find an integer eigenvector $v$ with eigenvalue $1$ (because the coefficients of the matrix corresponding to $T$ are integer), so passing to the quotient by $\mathbb{R}v$ you should get a torus of dimension $n-1$. Since $T$ had a Li-Yorke pair, so does the automorphism induced by $T$ on the quotient, so you have reduced the dimension. If the new map is still not ergodic, we may repeat this reduction. But this process has to stop before reaching dimension $1$ since you cannot have a Li-Yorke pair in dimension $1$. Note: I edited this answer since the previous one had a big mistake (the claim was false)<|endoftext|> TITLE: Is the space of countable closed covers of the Cantor set analytic? QUESTION [6 upvotes]: For an uncountable compact metric space $X$ denote by $K(X)$ be the hyperspace of non-empty compact subsets of $X$, endowed with the Vietoris topology (which is generated by the Hausdorff metric). In the countable power $K(X)^\omega$ consider the subspace $$cov(X)=\{(K_n)_{n\in\omega}\in K(X)^\omega:\bigcup_{n\in\omega}K_n=X\}$$ of closed countable covers of $X$. It can be shown that $cov(X)$ is a coanalytic subspace of $K(X)^\omega$. Problem. Is the space $cov(X)$ analytic (equivalently, Borel in $K(X)^\omega$)? I expect that the answer is negative and moreover, $cov(X)$ is $\Pi^1_1$-complete. But how to prove this? Maybe it has been published somewhere? REPLY [7 votes]: For $X=2^\omega$ (and similarly for every uncountable $\sigma$-compact Polish space) this set is indeed $\Pi^1_1$-complete: Take a $G_\delta$ set $B \subset 2^\omega \times 2^\omega$ so that $proj_1(B)$ is $\Sigma^1_1$-complete. Let $S=2^\omega \setminus proj_1(B)$, then of course $S$ is $\Pi^1_1$-complete and the set $A=2^\omega \times 2^\omega \setminus B$ is $K_\sigma$. --Added later following the suggestion of Taras Banakh-- It is not hard to show that if $K \subset 2^\omega \times 2^\omega$ the map $x \mapsto K_x$ from $2^\omega$ to $\mathcal{K}(2^\omega)$ is Borel (where $K_x$ denotes $\{y:(x,y) \in K\}$). Let us denote this map for a compact $K$ by $\overline{K}$. Let $A=\bigcup_{n \in \omega} K_n$ with $K_n$ compact. Thus, we get Borel functions $\overline{K}_n:2^\omega \to \mathcal{K}(2^\omega)$ for $n \in \omega$ such that for every $x \in 2^\omega$ we have $A_x=\bigcup_{n \in \omega} (K_n)_x=\bigcup_{n \in \omega} \overline{K}_n(x)$. Alternatively, one can use a more general theorem of Saint Raymond (Theorem 35.46 in Kechris' book) to derive the same conclusion. Letting $K=(\overline{K}_0,\overline{K}_1,\dots)$ we get a Borel map $K:2^\omega \to (\mathcal{K}(2^\omega))^\omega$. Then $K(x) \in cov(2^\omega)$ if and only if $A_x =2^\omega$ which is equivalent to $x \in S$. Thus, $S=K^{-1}(cov(2^\omega))$, so by the completeness of $S$ the set $cov(2^\omega)$ is also $\Pi^1_1$-complete.<|endoftext|> TITLE: Corson-Lindenstrauss : Weakly compact sets as intersection of finite unions of cells QUESTION [5 upvotes]: A theorem of Corson and Lindenstrauss in: Corson, H. H. and Lindenstrauss, J. “On weakly compact subsets of Banach spaces”. In: Proceedings of the American Mathematical Society 17.2 (1966), pp. 407–412. http://www.ams.org/journals/proc/1966-017-02/S0002-9939-1966-0199669-9/S0002-9939-1966-0199669-9.pdf states that: Theorem: Let $(X, \lVert . \rVert)$ be a separable reflexive Banach space. A subset K of X is weakly compact if and only if it is the intersection of finite unions of cells. in which a cell---also known as a ball---with center $x_0 \in X$ and radius $r \geq 0$ is defined as: $$c(x_0,r) := \{ x \in X \mid \lVert x - x_0 \rVert \leq r\}.$$ In other words, a subset $K$ of $X$ is weakly compact if and only if it can be written as $K = \bigcap_{i \in I} U_i$, in which $I$ is an index set, and each $U_i$ is of the form $U_i = c^{(i)}_1 \cup c^{(i)}_2 \cup \cdots \cup c^{(i)}_{n_i}$, for some $n_i \in \mathbb{N}$, and some cells $c^{(i)}_1, c^{(i)}_2, \ldots, c^{(i)}_{n_i}$. Note that the 'if' direction is kind of expected, and the stronger claim is the 'only if' part. Now take the separable Hilbert space $X := \ell^2$ under its usual norm, and let $E := \{ e_n \mid n \in \mathbb{N} \}$ be the set of elements $e_n \in X$, defined as: $$\forall k \in \mathbb{N} : e_n(k) := \left\{ \begin{array}{ll} 1, & \text{if }n = k,\\ 0, & \text{otherwise}.\end{array}\right.$$ We define the set $C$ as follows: \begin{equation} C := \{ e_n \mid n \in \mathbb{N}\} \cup \{ 0 \}. \end{equation} The set $C$ is sequentially weakly compact, hence by the Eberlein–Šmulian theorem, it is weakly compact (i.e., under the usual `finite open subcover' definition). Hence, by the Corson-Lindenstrauss thoerem, it can be represented as the intersection of finite unions of cells. Question: Can anyone present an example of representing $C$ as the intersection of finite unions of cells? REPLY [4 votes]: Let $x_n=-\frac{\sum_{k=1}^n e_k}{n}$. Then $\|e_i-x_n\|^2=\frac{n+3}{n}$ if $i \leq n$ and $\|e_i-x_n\|^2=\frac{n+1}{n}$ if $i>n$. So your set is the intersection of the sets $B(x_n,\sqrt{\frac{n+1}{n}}) \cup \bigcup_{i=1}^n B(e_i,\frac{1}{n})$. Nice question, btw.<|endoftext|> TITLE: Relatively concise English expositions of the proofs of the various Weil conjectures QUESTION [25 upvotes]: Where can I find relatively concise (i.e. not excessively wordy and waxing poetic about history and intuitions and such, doesn't spend an eternity carefully developing various parts of the theory of étale cohomology, etc.) English expositions of the proofs of the various Weil conjectures? The four Weil conjectures, according to Wikipedia, are as follows. (Rationality) $\zeta(X, s)$ is a rational function of $T = q^{-s}$. More precisely, $\zeta(X, s)$ can be written as a finite alternating product$$\prod_{i = 0}^{2n} P_i(q^{-s})^{(-1)^{i + 1}} = {{P_1(T) \ldots P_{2n - 1}(T)}\over{P_0(T) \ldots P_{2n}(T)}},$$where each $P_i(T)$ is an integral polynomial. Furthermore, $P_0(T) = 1 - T$, $P_{2n}(T) = 1 - q^nT$, and for $1 \le i \le 2n - 1$, $P_i(T)$ factors over $\mathbb{C}$ as $\prod_j (1 - \alpha_{ij}T)$ for some numbers $\alpha_{ij}$. (Functional equation and Poincaré duality) The zeta function satisfies$$\zeta(X, n - s) = \pm q^{{{nE}\over2} - Es}\zeta(X, s)$$or equivalently$$\zeta(X, q^{-n} T^{-1}) = \pm q^{{{nE}\over2}}T^E \zeta(X, T)$$where $E$ is the Euler characteristic of $X$. In particular, for each $i$, the numbers $\alpha_{2n - i, 1}$, $\alpha_{2n - i, 2}$, $\ldots$ equal the numbers $q^n/\alpha_{i, 1}$, $q^n/\alpha_{i, 2}$, $\ldots$ in some order. (Riemann hypothesis) $|\alpha_{i, j}| = q^{i/2}$ for all $1 \le i \le 2n - 1$ and all $j$. This implies that all zeros of $P_k(T)$ lie on the "critical line" of complex numbers $s$ with real part $k/2$. (Betti numbers) If $X$ is a (good) "reduction mod $p$" of a non-singular projective variety $Y$ defined over a number field embedded in the field of complex numbers, then the degree of $P_i$ is the $i$th Betti number of the space of complex points of $Y$. Thanks in advance! REPLY [5 votes]: Sorry for a bit more of self-promotion but I want to share it somewhere (when I was searching for such an article, it was not available). Maybe I will try to publish this as an expository paper (not sure if that is appropriate, should I?). Anyway, I have covered proofs of the Weil Conjectures by Grothendieck and Deligne, explained all of the arguments that might not be obvious right away and also added an overview on Etale cohomology and tried to provide motivation for all of the steps of the proof - https://drive.google.com/file/d/1KQ6k6vtAowCc9WbuqAlEf7aZIQFEv54Z/view?usp=sharing<|endoftext|> TITLE: When is an almost simple group a split extension of its socle? QUESTION [7 upvotes]: Here an almost simple group is a finite group whose socle (product of all minimal normal subgroups) is a nonabelian simple group. As an extension of its socle, an almost simple group could be split or non-split. For example, there are four groups with socle $L=PSL(2,9)$ other than $L$, i.e. $S_6$, $PGL(2,9)$, $M_{10}$ and $P\Gamma L(2,9)$; the former two are split extensions of $L$ while the latter two are not. Now the question is how to determine all the almost simple groups which is a non-split extension of its socle? REPLY [4 votes]: See the following paper: A. Lucchini, F. Menegazzo, M. Morigi. On the existence of a complement for a finite simple group in its automorphism group. Special issue in honor of Reinhold Baer (1902–1979). Illinois J. Math. 47 (2003), no. 1-2, 395–418. MR2031330<|endoftext|> TITLE: Is there a group scheme G such that there is a proper class of non-isomorphic fpqc G-bundles over some fixed base? QUESTION [5 upvotes]: Famously, the fpqc topology is so big that no one set of fpqc covers of an affine scheme is cofinal in all fpqc covers. The Stacks Project gives a construction using field extensions with arbitrarily large infinite transcendence basis, which are all fpcq maps. This means that it is difficult to see whether the collection of isomorphism classes of fpqc-locally trivial principal $G$-bundles, for a fixed group scheme $G$, on a given scheme form a set, rather than a proper class. If $G$ is nice (e.g. finite type/smooth/insert your favourite condition here), then standard reductions show that any fpqc bundle is in fact locally trivial in a far smaller topology, such as the étale topology. I think (from reading this thesis) that even if we are working with a group scheme over a field $k$ such that $G\to Spec(k)$ is fppf, then we should probably only get set-many isomorphism classes, as the stack of $G$-bundles is then algebraic. So any group scheme that would give a proper class of non-isomorphic bundles would have to be rather nasty. (Also, I believe we need to be in positive characteristic, to get maximum nastiness...) My vague plan for constructing an example would be to consider fpqc $G$-bundles on $Spec(k)$ ($G$ a group scheme over $k$) via cocycles over the covers that the Stacks project describes at the above link. If a purely transcendental extension $k\to k\{I\}$ (where $I$ is a set of arbitrary cardinality) made $Spec(k\{I\}) \to Spec(k)$ a torsor (so the extension is 'Galois' for a suitable interpretation of that term), then one could consider homomorphisms $\alpha\colon Gal(k\{I\}) \to G$ and the associated $G$-bundles. It seems possible that one could figure out when two such bundles are isomorphic, possibly using a cardinality bound coming from data attached to $G$. Or not, I really don't know. Can we make a call in either direction? Does the above sketch give proper class-many non-isomorphic bundles, or can we prove that for every fpqc group scheme $G$ and base scheme $X$ there is only a set of isomorphism classes of fpqc principal $G$-bundles on $X$? REPLY [9 votes]: In more down-to-earth terms, you are asking whether there is always a set of $G$-torsors for the fpqc topology over a scheme $S$ such any every $G$-torsor for the fpqc topology over $S$ is isomorphic to one of those in the set. The answer is "yes" when $G$ is flat over $S$ (as is automatic over a field). The question asks about something being a "proper class", which I am told in the comment by Q. Yuan below means "class that is not a set", so the answer to the question in the title is then "no" under that flatness hypothesis on $G$. I don't see a reason for interest in torsors for for the fpqc topology on $S$ relative to a group scheme that isn't flat over the base, so I will regard the $S$-flatness of $G$ as a basic hypothesis one should impose. The real content is in the following simple lemma: Lemma: Let $A$ be a ring, $B$ an $A$-algebra, and $A'$ a faithfully flat $A$-algebra. For an infinite set $I$, if the $A'$-algebra $B' = A' \otimes_A B$ admits a set of at most $|I|$ generators $($equivalently, $B'$ is a quotient of a polynomial algebra $A'[X_i]_{i \in I}$ over $A'$$)$ then so does $B$ over $A$. This lemma and its proof are motivated by the more well-known fact (and its proof!) that if $B'$ is finitely generated (resp. finitely presented) over $A'$ then the same holds for $B$ over $A$ (EGA IV$_2$, 2.7.1.1), but the proof is easier for this controlled "infinitely generated" property. Proof: Pick a set of generators of $B'$ over $A'$ indexed by a subset $J \subset I$. Write each as a finite sum of elementary tensors, and consider the set $S$ of elements of $B$ that arise in those tensors. Since $I$ is infinite, it is clear that $|S| \le |I|$. Hence, it suffices to show that $B$ coincides with its $A$-subalgebra $B_0$ generated by the elements in $S$. By flatness of $A'$ over $A$, the natural map $A' \otimes_A B_0 \rightarrow A' \otimes_A B$ is injective. But by design of $B_0$ in terms of $S$, this latter injection is also surjective. Hence, by faithful flatness the inclusion $B_0 \hookrightarrow B$ is an equality too. QED Lemma Now we consider $G$-torsors for the fpqc topology on $S$, where $G$ is $S$-flat (hence faithfully flat, due to the identity section, but whatever). It suffices to show that for each open affine $U = {\rm{Spec}}(A) \subset S$ (or really just the members of whatever fixed affine open cover you wish) there is a set's worth of isomorphism classes of $G$-torsors for the fpqc topology over $U$. That is, we can assume $S$ is affine, say with coordinate ring $A$. By choosing finitely many affine opens in $G$ which cover the quasi-compact identity section, we can shrink $S$ a bit more so that there exists an open affine $V \subset G$ containing the identity section (so $V \rightarrow S$ is faithfully flat between affines). Let $I$ be an infinite cardinal so that each affine open subscheme of $G$ (e.g., $V$) admits a set of at most $|I|$ generators for its coordinate ring as an $A$-algebra (such an $I$ clearly exists); for what follows it would be enough to pick $I$ that "works" for the members of a single affine open covering of $G$. For any given faithfully flat $A$-algebra $R$, there is a set's worth of isomorphism classes of $G$-torsors for the fpqc topology over $S$ which admit an $R$-point (since by fpqc descent these are controlled by the descent data encoded in ${\rm{H}}^1(R/A, G)$; note that not all such descent may be effective when $G$ isn't relatively affine). Hence, it suffices to find a set's worth of such $R$ so that every $G$-torsor for the fpqc topology on $S$ splits over one of those $A$-algebras $R$ (i.e. admits an $R$-point over $A$). We will show that the set (!) of isomorphism classes of faithfully flat $A$-algebras $R$ that admit a set of $A$-algebra generators of size at most $|I|$ does the job. Let $E$ be a $G$-torsor for the fpqc topology over $S$. Thus, there exists a faithfully flat $A$-algebra $A'$ then $E(A')$ is non-empty, so $E_{A'} \simeq G_{A'}$ as $A'$-schemes. In particular, $V_{A'}$ is an affine open subscheme of $E_{A'}$ admitting at most $|I|$ generators as an $A'$-algebra. Its image in $E$ is quasi-compact, so is contained in a quasi-compact open $\Omega \subset E$. The map $\Omega \rightarrow S = {\rm{Spec}}(A)$ is faithfully flat because after fpqc base change to $A'$ it becomes $\Omega_{A'} \rightarrow {\rm{Spec}}(A')$ that is faithfully flat (as $\Omega_{A'}$ is identified with an open subscheme of $G_{A'}$ containing $V_{A'}$). By quasi-compactness, $\Omega$ is covered by a finite set of affine open subschemes ${\rm{Spec}}(B_j)$ with $1 \le j \le n$. Thus, $B = \prod_{j=1}^n B_j$ is a flat $A$-algebra that is faithfully flat since ${\rm{Spec}}(B) = \coprod {\rm{Spec}}(B_j) \rightarrow \Omega$ is surjective. Hence, $E(B)$ is non-empty with $B$ faithfully flat over $A$. We claim that $B$ admits a set of at most $|I|$ generators as an $A$-algebra (which will do the job). It suffices to do the same for each $B_j$ separately. Now finally we do something that isn't formal nonsense, namely we invoke the Lemma: it suffices to show that $B_j \otimes_A A'$ admits a set of at most $|I|$ generators as an $A'$-algebra. But the spectrum of this $A'$-algebra is an affine open subscheme of $E_{A'} = G_{A'}$, so it suffices to show that every affine open $W$ in $G_{A'}$ admits a set of at most $|I|$ generators for its coordinate ring as an $A'$-algebra. By design, $G$ is covered by affine opens each of whose coordinate rings admits a set of at most $|I|$ generators as an $A$-algebra. Thus, the same holds for $G_{A'}$ relative to $A'$ for the base-change covering, and consequently for any basic affine open of one of those. Recall the fact that for any two affine opens in a scheme, their overlap is covered by affine opens which are simulatenously affine open in each. Hence, the affine $W$ is covered by basic affine opens (relative to $W$!) whose coordinate rings each admit a set of at most $|I|$ generators as an $A'$-algebra. By quasi-compactness, $W$ is covered by finitely many of those. Since $I$ is infinite, the numerators in such fractional expressions give a set of size at most $|I|$ that generate the coordinate ring of $W$ as an $A'$-algebra.<|endoftext|> TITLE: Homeomorphism between del Pezzo surfaces QUESTION [5 upvotes]: Let $X$ and $Y$ be smooth del Pezzo surfaces of the same degree $K_X^2=K_Y^2$. Are the sets $X(\mathbb{C})$ and $Y(\mathbb{C})$ homeomorphic, or at least homotopy equivalent? REPLY [14 votes]: Yes, with precisely one exception. If $K^2 \neq 8$, then the del Pezzo surface is the blow-up of the plane at $9-K^2$ points, so it is homeomorphic to the connected sum of $\mathbb{CP}^2$ with $9-K^2$ copies of $\overline{\mathbb{CP}^2}$. If $K^2=8$, then we have either the quadric $\mathbb{P}^1 \times \mathbb{P}^1$, which is clearly homeomorphic to $S^2 \times S^2$, or the blow-up of $\mathbb{P}^2$ at one point, which is homeomorphic to $\mathbb{CP}^2 \# \, \overline{\mathbb{CP}^2}$. These surfaces are not homeomorphic, since in the former case the class of $K$ is $2$-divisible in (co)homology whether in the latter case is not, and the divisibility of the canonical class is known to be a topological invariant.<|endoftext|> TITLE: Is cohomology of groups all about $H^{i}: -2\leq i\leq 2$? QUESTION [7 upvotes]: I am reading Class field theory - Bonn Lectures by Neukirch. Given a $G$ module $A$ he defines Cohomology groups $H^i(G,A) : i\in \mathbb{Z}$ by considering some complete resolution of $G$ modules and applying hom functor. He then calculated $H^{-1}(G,A),H^0(G,A),H^1(G,A),H^2(G,A)$ It says $H^2(G,A)$ is related to group extensions. Given an abelian group $A$ written multiplicatively and an arbitrary subgroup $G$, we want to find all group extensions $\hat{G}$ of $A$ such that $A$ is normal subgroup pf $\hat{G}$ and $\hat{G}/A\cong G$. He says this is related to $H^2(G,A)$. It says if the group $G$ acts trivially on $A$ then $H^1(G,A)=\rm{Hom}(G,A)$, in particular for $A=\mathbb{Q}/\mathbb{Z}$ we have character group $H^1(G,A)=\chi(G)$. It says $H^0(G,A)=A^G/N_GA$ the norm residue group of $G$ module $A$ with notation $A^G=\{a\in A : \sigma a=a~\forall \sigma\in G\}$ and $N_GA=\{\sum_{g\in G}ga:a\in A\}$. It says this is useful in class field theory. It says $H^{-1}(G,A)$ has some other concrete form. It says $H^{-2}(G,\mathbb{Z})=G^{ab}$ abelianization of $G$. My question is are there any higher cohomology groups that are of interest at least in some special cases? REPLY [6 votes]: Here's a number field example, analogous to Dalawat's local field example. Let $K$ be a number field. One can prove that $H^3(G_{\overline K/K},\overline K^*)=0$. (Although interestingly, there may be finite Galois extensions $L/K$ with $H^3(G_{L/K},L^*)\ne0$.) Anyway, Tate's cohomological construction of the pairing on the Tate-Shafarevich group $III(E/K)$ of an elliptic curve $E/K$ proceeds by first constructing a 2-cochain and then using the fact that $H^3(G_{\overline K/K},\overline K^*)=0$ to find a "correction term" that turns it into a 2-cocycle, thereby giving the desired element of $\text{Br}(K)$. So the whole construction depends on the calculation of an $H^3$.<|endoftext|> TITLE: Relationship between Tate-Shafarevich group and the BSD conjecture QUESTION [6 upvotes]: The finiteness of the Tate-Shafarevich group is known to be equivalent to BSD for elliptic curves over function fields over $\mathbb{F}_{q},$ this result is due to Kato and Trihan if I am not mistaken. Question: Is any result in that direction known to hold for number fields? REPLY [11 votes]: Firstly, the functional field result your state is due to Tate in his Bourbaki talk. In fact he proves that the finiteness of the $p$-primary part of Sha is enough for $p$ different from the characteristic. For elliptic curves over a number field, the finiteness of Sha (over larger fields) gives us the parity on the BSD conjecture, by work of Dokchitsers. The analogue of Tate's method carried over to the number field case is what Iwasawa theory for elliptic curves attempts to do. The corresponding result is considerably weaker. Here an example of what one can prove: Let $E/\mathbb{Q}$. If we can find a prime $p$ of good ordinary reduction (just to make the statement cleaner) such that The $p$-primary of Sha of $E/\mathbb{Q}$ is finite, The canonical $p$-adic height is non-degenerate on $E(\mathbb{Q})$, and The orders of vanishing of the $p$-adic $L$-function and the complex $L$-function agree. Then the order of vanishing of the complex $L$-function is equal to the rank of $E(\mathbb{Q})$. As for the precise formula of the leading term, one only gets the formula for the $p$-adic $L$-function, subject to the first two conditions.<|endoftext|> TITLE: What is the most efficient way to open glass curtains QUESTION [7 upvotes]: This question originated from a real life problem. It has to do with a certain system of sliding glass panels used to close a terrace. Describing exactly how it works would take a long time, so here is a youtube video showing the concept: https://www.youtube.com/watch?v=QMvF3TcPUDA (Disclaimer: I am in no way associated with the company marketing the product advertised in this video) As the video shows, the person has to slide each glass panel to the right hand side of the balcony before being able to rotate it, thus getting it out of the way. Observe that the rotation can only take place when the panel is at the right hand side due to a single slot in the top railing. In order to state my question it is best to imagine a set of glass panels covering a single side of a very long terrace, rather than the one in the video, where two perpendicular sides are involved. In the video, the person opens the glass panels one by one, having to walk further and further in order to bring the next panel towards the rotating position. This is obviously not the most efficient way of doing it (too much walking) since she could easily bring more than one panel each time. However, since the panels are not attached to each other, should she want to bring more than one panel at once, she would have to walk a little further to the leftmost panel about to be slid and then push them all together. She could even walk all the way to the absolute leftmost panel and slide them all together but, after doing so, she would be required to come back to the right hand side to rotate the single panel by the slot in the railing. First question: what is the most efficient method for opening all glass panels with the minimum amount of walking? As far as I know, this is an open problem. I have asked it to a few people in combinatorics but so far I do not have an answer. What do I know about it? I wrote a computer program to find the solution by brute force, which is capable of dealing with up to 16 panels. Once you see the solution, a conjecture immediately pops up in your mind. You do it recursively! Split all panels in two halves, open the rightmost half (using recursion), walk all the way to the left, bring the remaining doors to the right and open them (using recursion again). Second question: is this indeed the best method? REPLY [4 votes]: Suppose there are $n$ $1$s in a row at the start. Let $f(n)$ be the cost of the most efficient way to clear the second position, $n-1$ $1$s followed by a $0$. $f(1) = 0, f(2)=2, f(4) = 8.$ Say that a move $a_i$ (counting from the right) is a record if for all $j \lt i, a_j < a_i$. Claim: In an efficient strategy, just before every record the position is a string of $1$s followed by a string of $0$s (empty on the first move). Proof: If there is a $0$ followed by a $1$ to the right of the leftmost window you are moving, then you could reach the same configuration faster by not creating the leftmost $0$. So, in any efficient strategy, after any record, you clear all of the $1$s you move before the next record. The last record's value is $n$. Consider the second-to-last record, of value $m$, in an efficient sequence. $f(n) = f(m) + n + f(n-m)$. You clear the bottom $m$ places, then shift the remaining $n-m$ $1$s, then clear those $n-m$ $1$s. So, $f(n) = \min_{m\lt n} f(m) + f(n-m)+n$. Claim: $f(n) \ge n \log_2 n$ with equality when $n$ is a power of $2$. Proof: This is trivially true for $n=1$. If each $f(k) \ge k \log_2 k$ for $k \lt n$ then $f(n) \ge n + \min_{0 \lt x\lt n}(x \log_2 x + (n-x) \log_2(n-x)).$ The minimum value is when $x=n/2,$ so $f(n) \ge n+ 2 (n/2) \log_2 (n/2) = n \log_2 n.$ So, by strong induction, $f(n) \ge n \log_2 n$. This is achieved when $n$ is a power of $2$ by the recursive construction so that the penultimate record is $n/2$. Claim: If $n=2^k+a$ with $0 \le a \le 2^k$ then $f(n) = k(2^k) + (k+2)a$. You can choose any penultimate record $m$ so that $2^{k-1} \le m,n-m \le 2^k$. This gives a recursive description of the set of all optimal strategies. Proof: Strong induction. The formula holds if you set the penultimate record $m = \lfloor n/2 \rfloor$. Note that $f(n+1)-f(n) = 2+\lfloor \log_2 n \rfloor$ so decreasing $m \le n/2$ by $1$ and increasing $n-m$ by $1$ only changes $n+f(m)+f(n-m)$ if $m-1$ and $n-m$ are between different powers of $2$.<|endoftext|> TITLE: Are multiplicity-free representations weight multiplicity free? QUESTION [7 upvotes]: A rational representation $(G,V)$ of a complex reductive linear algebraic group is called multiplicity-free if the decomposition of $\mathbb C[V]$ into irreducible $G$-modules contains each irreducible $G$-module with multiplicity at most one. Part of the importance of such representations lies in the fact that they serve to explain several phenomena in invariant theory and harmonic analysis (e.g. Peter-Weyl theorem). There are several different characterizations of multiplicity free representations, among them: any Borel subgroup has an open (dense) orbit; the principal orbits of a compact real form are coisotropic submanifolds with respect to an invariant (real) symplectic structure. If $(G,V)$ is multiplicity free then $G$ has finitely many orbits in $V$. In particular, it is visible in the sense that $V$ has finitely many nilpotent orbits, and hence the multiplicity of any nonzero weight is at most one, see Kac, V. G. Some remarks on nilpotent orbits. J. Algebra 64 (1980), no. 1, 190–213. It is apparent from the classification of (indecomposable) multiplicity free representations that they are weight multiplicity free, namely, the multiplicity of any weight is at most one. Is there a direct proof of this fact? REPLY [10 votes]: It is a theorem of Brion and, independently, of Vinberg that varieties with an open $B$-orbit (a.k.a. spherical varieties) have in fact only finitely many orbits. A shorter argument is due to Matsuki (see his ICM talk) and independently (using the same idea) by me (On the set of orbits $\ldots$). Thus multiplicity free spaces are visible. Kac implies that all non-zero weights have multiplicity one. On the other hand, the map from the zero weight space to the quotient $V/\!\!/ G$ is always finite (see e.g. Luna "Adherence d'orbites"). The latter is a point. So the zero-weight space of a multiplicity free space is in fact zero. Edit 1: I realize that my reference to Adherance is over-kill in this situation but Dadok-Kac is so, as well, since it uses Hilbert-Mumford. There is an easier classical argument, probably due to either Rosenlicht or Borel. Claim: Let $X$ be an affine $G$-variety and $x\in X^T$. Then $Gx$ is closed in $X$. Proof: The key observation is due to Rosenlicht, saying that any orbit of a unipotent group on an affine variety is closed. Let $B=TU\subseteq G$ be a Borel. Then $Ux$ is closed. But then also $Bx=TUx=UTx=Ux$ is closed. The morphism $\phi:G\times^BBx\to X$ is proper since it factors through the closed immersion $G\times^BBx\hookrightarrow G/B\times X:[g,bx]\mapsto (gB,gbx)$ and the proper projection $G/B\times X\to X$. Thus $Gx={\rm image}(\phi)$ is closed. Edit 2: I have erased a previous alternative proof since I couldn't recall all details. I do remember now. So it is possible to prove weight multiplicity freeness from the Local Structure Theorem (Brion-Luna-Vust) using the line of reasoning in my survey Some remarks on multiplicity free spaces.<|endoftext|> TITLE: Are there "principal" bundles $S^1 \to S^3 \to S^2$ other then Hopf's? (They would be necessarily not locally trivial) QUESTION [24 upvotes]: It is well known that the only principal locally trivial fiber bundle $S^1 \to S^3 \to S^2$ is Hopf map $h$ (see, for example, [1]). What if we drop the local triviality but mantain a "principality" hypothesis meaning that the fibers are given by an $S^1$ action? Are there well known families of such bundles $S^1 \to S^3 \to S^2$ whose homotopy class is not that of $h$? In particular, are the homotopy class of the nontrivial multiples $k[h]$ of Hopf's map in $\pi_3(S^2)$, $k \not\in \{ \pm 1,0 \}$, given by such bundles? Such bundles would be necessarily not locally trivial, but hopefully they can be fibrations. REPLY [48 votes]: The 3-sphere has infinitely many Seifert fibrations with generic fiber a torus knot (including the unknot). For a $(p,q)$ torus knot, the Hopf invariant will be $pq$ (up to sign). To see this, note that the fibration has two exceptional fibers which form the Hopf link (these are the red line and yellow circle in the image). The generic fibers wrap $p$ times around one of these, and $q$ times around the other, so have linking number $pq$. In fact, then, one can represent each homotopy class in $\pi_3(S^2)$ by a Seifert fibration with a single exceptional fiber (so all fibers are unknots, generically a $(p,1)$ curve on the Clifford torus). These are (non-locally trivial) ``principal bundles" in the sense that there is an action of $S^1$ whose orbits are the fibers of the fibration. If $S^3\subset \mathbb{C}^2$ as the unit sphere, then the action is $(z_1,z_2)\to (z^pz_1,z^qz_2)$, for $z\in S^1= \{z\in \mathbb{C}, |z|=1\}$.<|endoftext|> TITLE: Solovay-random pairs of reals QUESTION [8 upvotes]: A real $x\in2^\omega$ is random (or Solovay-random) over a model $M$ iff it's obtained via forcing by closed sets of reals, of positive measure, coded in $M$. Similarly, a pair $(x,y)$ of reals is random over a model $M$ iff it's obtained via forcing by closed sets in $2^\omega\times2^\omega$, of positive product measure, still coded in $M$. Thus such a pair is not a product-forcing pair. (To establish this, one has to define a closed set $X\subseteq2^\omega\times2^\omega$, of positive product measure, which does not include any positive-measure rectangle, not an easy exercise!) Rather $(x,y)$ is a random pair over a model $M$ iff $x$ is random over $M$ and $y$ is random over $M[x]$ iff vice versa. Question. In spite of not being product-forcing-generic, prove that if $(x,y)$ is a random pair over a model $M$ then $M[x]\cap M[y]=M$. A proof of $M[x]\cap M[y]\cap2^\omega\subseteq M$ can be manufactured by reduction, via continuous reading of names, to the claim that if $X\subseteq2^\omega\times2^\omega$, is a closed set of positive product measure and $f,g:2^\omega\to2^\omega$ are continuous maps such that $f(x)=g(y)$ for almost all $(x,y)\in X$, then $X=X'\cup\bigcup_nX_n$, where $X'$ is null and each $X_n$ is Borel non-null on which both $f$ and $g$ are constants. What about the general case? REPLY [2 votes]: This is a response to Kanoveri's comment. If $\{\kappa_n : n < \omega\}$ is strictly increasing with limit $\kappa$, then since $A \upharpoonright \kappa_n \in M$ (by minimality of $\kappa$), using ccc, we can find a countable $X \in M$ such that $M[x] \cap M[y]$ contains a new subset of $X$ which translates to a new real.<|endoftext|> TITLE: Convex polygonal tiling of the square such that every line intersects at most k polygons QUESTION [12 upvotes]: Consider a tiling of a square by convex polygons, such that every line through the square intersects at most $k$ polygons. Let $n$ be maximum number of polygons such a tiling can have. What is the dependence of $n$ on $k$? Tiling the square with a grid of $k/2$ by $k/2$ smaller squares gives a simple construction for $\Omega (k^2)$ polygons. But is $n \in O(k^2)$? If each polygon has the same area $A$, the answer is yes. The sum of the horizontal lengths of the polygons cannot exceed $k$, otherwise some vertical line will intersect more than $k$ polygons by pigeonhole. The same follows for the vertical lengths of the polygons. Let $L_i$ be the sum of the horizontal and vertical lengths of polygon $i$, then $L_i \geq 2\sqrt{A}$. We have $n = \frac{1}{A}$ and $\Sigma_i L_i \leq 2k$ so $$2n\sqrt{A} \leq 2k$$ $$\sqrt{n} \leq k$$ $$n \leq k^2$$ This proof only makes use of vertical and horizontal lines intersecting at most $k$ polygons. If a general upper bound exists, the proof must make use of diagonal lines intersecting at most $k$ polygons as well: an arbitrary number of arbitrarily small squares could be placed along a diagonal line, with two large polygons filling the space above and below the line, and in this tiling vertical and horizontal lines would only intersect a small constant number of polygons. Any suggestions or references would be greatly appreciated. REPLY [8 votes]: The correct upper bound seems to be of order $k^k$ in the sense that we can create $\ge k^k$ polygons so that each line intersects at most $Ck$ of them and the number is bounded from above by $k^{Ck}$. I'll prove the second part of this claim here. Definition. Let $A,B$ be two closed convex polygons with disjoint interiors. We say that a finite sequence of closed convex polygons with disjoint interiors also disjoint from the interiors of $A$ and $B$ is a chain connecting $A$ to $B$ if the first polygon has a common point with $A$, each other polygon has a common point with the previous one, and the last polygon has a common point with $B$. Lemma. Suppose that $\mathcal F$ is a family of chains of length at most $L$ connecting $A$ to $B$. Suppose that no line intersects more than $k$ different polygons forming these chains. Then $|\mathcal F|\le (kL)^L$. Viewing the original tiling as a sequence of chains connecting the left side to the right one of length at most $k$ (say, the chains of polygons crossing a fixed horizontal line), we get the upper bound $k^{2k+1}$ in the original problem. Proof. Induction on $L$. If $L=0$, there is nothing to prove (there is only one empty chain). Case $L=1$ is also trivial because each chain consists of exactly one polygon that has to cross any line that separates the interiors of $A$ and $B$. To make an induction step, let us first estimate the number of chains containing a fixed polygon $C$. Notice that for each such chain we can find two numbers $a,b\ge 0$ with $a+b=L-1$ such that our chain (of length $\le L$) consists of a chain of length $\le a$ connecting $A$ to $C$, then $C$ itself, and then a chain of length $\le b$ connecting $C$ to $B$. By the induction assumption, the total number of chains containing $C$ is then at most $$ \sum_{a+b=L-1}(ka)^a(kb)^b=k^{L-1}\sum_{a+b=L-1}a^ab^b\le k^{L-1}L^L $$ Now notice that each chain crosses any line separating the interiors of $A$ and $B$, so at least one polygon in it crosses that line. However, each particular polygon can be found in only $k^{L-1}L^L$ chains, so, unless the line crosses more than $k$ different polygons, we may have only $k\times k^{L-1}L^L=(kL)^L$ chains. Of course, this upper bound can be improved a bit, but, as I said, not by too much. Edit. I guess I simplified my original construction enough to make it postable, so here goes. The starting point is a very long horizontal rectangle $R$ of height $1$ split by $k$ extremely narrow diagonal (angle of 45 degrees with the horizontal axis) paralellograms placed into it at regular (also very long) intervals in the middle half (so we have $2k+1$ pieces in the original picture). At every step, we shrink the picture we already have enormously in the vertical direction, scale it so that it gets width $2$ (something between $\sqrt 2$ and $2\sqrt 2$), rotate it by 45 degrees, place $k$ copies of it the way the narrow parallelograms were placed in the starting picture with some smalll shifts so that no 3 "narrow parallelograms" in different copies are crossing one line (which is possible because those look like points now), and cut out the pieces hanging outside $R$ (see the picture with $k=3$). We do this $k$ times. The main property of this picture is that if we have a line that is almost vertical (say, making the angle of 10 degrees or less with the vertical axis), then it intersects at most 2 big pieces in the top generation and at most one picture in the next generation. Moreover, any line that violates the last property should be nearly horizontal, which means that with respect to the picture of the previous generation it is nearly vertical after we undo the enormous compression in the vertical direction, but being nearly vertical there, it stays nearly vertical for all following generations as well after the corresponding decompressions. Suppose we have any line now. Let's see how it goes across the pictures from the top generation down. For a while it intersects just $\le 2$ pieces in the current generation and at most one picture in the next one. This sort of intersections is harmless because we get 2 pieces per generation only. At some generation $g$ this nice property may get violated. Then the line must be nearly hzontal in the decompressed picture for $g$, so it can cross $\le k+1$ big pieces from generation $g$, $\le 2k$ big pieces from generation $g+1$ and at most $2$ pictures from generation $g+2$. Moreover, it stays nearly vertical for all generations starting with $g+1$, so we return to the rate of 2 pieces per generation (well, it is $4$ now because we crossed $2$ pictures in generation $g+2$, but it is still harmless). So the total number of crossed pieces is still under $Ck$ while we clearly have at least $k^k$ pieces total.<|endoftext|> TITLE: Easy Applications of Model Theory QUESTION [7 upvotes]: I've also posted this question on MathSE. I'm posting it here in hopes of a more comprehensive answer. The question is inspired by the following: Model theoretic applications to algebra and number theory(Iwasawa Theory) MathSE: Applications of model theory to analysis The applications mentioned are usually pretty complicated (except for Ax-Grothendieck, but it seems to be a rare occurrence). My question is; (also omitting these examples) are there easy "applications of model theory". For example, can we use Morley's categorecity theorem to determine a class of structures of interest "outside of model theory" has only unique models in higher cardinalities, where the actual proof of the fact is fairly complicated. I know that this works for $ACF_0$ for example, but it seems this can be established with the same amount of effort (i.e. proving Morley's theorem would be the same level of difficulty) Also are there example of nice categorizations: For example (this may be false but something in the spirit of) "omega-stable countable groups have a nice classification as products of some class of groups". To sum up: I'm looking for "simple" applications that follow from basic results in, say Marker's text on Model Theory for example. REPLY [5 votes]: Well Steinitz's theorem on the existence of an algebraic closure for any field can be proved through an easy application of the compactness theorem for first order logic. Of course the "usual proof" is not so complicated but it requires a bit of attention, whereas with compactness it's obvious. I hope this is the kind of example you were looking for.<|endoftext|> TITLE: Group cohomology and condensed matter QUESTION [69 upvotes]: I am mystified by formulas that I find in the condensed matter literature (see Symmetry protected topological orders and the group cohomology of their symmetry group arXiv:1106.4772v6 (pdf) by Chen, Gu, Liu, and Wen). These formulas have been used in some very interesting work in condensed matter and I would like to know how to understand them. I begin with the simplest case. Let $G$ be a finite group. One is given an element of $H^2(G,U(1))$ that is represented by an explicit $U(1)$-valued group cocycle $\nu(a,b,c)$. This is a homogeneous cocycle, $\nu(ga,gb,gc)=\nu(a,b,c)$ and obeys the standard cocycle condition $\nu(a,b,c)\nu^{-1}(a,b,d)\nu(a,c,d)\nu^{-1}(b,c,d)=1$ for $a,b,c,d\in G$. Let $X=G\times G$ be the Cartesian product of two copies of $G$. We consider $G$ acting on $X=G\times G$ by left multiplication on each factor. The cocycle $\nu$ is then used to define a twisted action of $G$ on the complex-valued functions on $X$. For $g\in G$ and $\Phi: X\to \mathbb{C}$, the definition (eqn. 27 of the paper) is $$\hat g(\Phi)=g^*(\Phi) \Lambda(a,b;g)$$ where $g^*(\Phi)$ is the pullback of $\Phi$ by $g$ and (with $a,b\in G$ defining a point in $X=G\times G$, and $g_*$ an arbitrary element of $G$) $$\Lambda(a,b;g)=\frac{\nu(a,g^{-1}g_*,g_*)}{\nu(b,g^{-1}g_*,g_*)}.$$ It is shown in appendix F of the paper that this does given an action of $G$ on the functions on $X=G\times G$. The authors also describe a version in one dimension more. In this case, $\nu(a,b,c,d)$ is a homogeneous cocycle representing an element of $H^3(G,U(1))$ and satisfying the usual cocycle relation and one takes $X=G\times G\times G\times G$ to be the Cartesian product of four copies of $G$. A twisted action of $G$ on the functions on $X$ is now defined by $$\hat g(\Phi)=g^*(\Phi) \Lambda(a,b,c,d;g)$$ with $$\Lambda(a,b,c,d;g)=\frac{\nu(a,b,g^{-1}g_*,g_*)\nu(b,c,g^{-1}g_*,g_*)}{\nu(d,c,g^{-1}g_*,g_*)\nu(a,d,g^{-1}g_*,g_*)}.$$ It is shown in appendix G that this does indeed give a twisted action of $G$ on the functions on $X$. I presume there is supposed to be an analog of this in any dimension though I cannot see this stated explicitly. Can anyone shed light on these formulas? REPLY [2 votes]: Maybe formula (27) is some kind of re-indexed loop transgression map in the case $n=2$. In general, loop transgression is a map $ \tau : H^n(BG) \rightarrow H^{n-1}(L(BG))$ where $L(X) = Maps(S^1, X)$ is the free loop space of a topological space $X$. Here and elsewhere, I'll take the coefficient group to be $U(1)$. There should be a continuous map $ \psi : B(Y_{G \times G}) \rightarrow L(BG) $. Here, $Y_{G \times G}$ is the groupoid (the "action groupoid") corresponding to the diagonal action of $G$ on $G \times G$ from the left, and $B(Y_{G \times G})$ is its classifying space. The map $\psi$ should be induced by the equivariant map $G \times G \rightarrow G$ given by $(a,b) \mapsto ab^{-1}$. Now, suppose you start with a 2-cocycle $\nu \in H^2(BG)$, and transgress it to $\tau(\nu) \in H^1(L(BG))$ and then pull it back to get $\psi^*(\tau(\nu)) \in H^1(B(Y_{G \times G}))$. Maybe one can interpret their $\Lambda (a,b; g)$ function as an element in $H^1(B(Y_{G \times G}))$, and that in fact $ \Lambda = \psi^*(\tau(\nu)). $ However, I haven't been able to get that to work. If true, it may suggest a similar story for higher $n$.<|endoftext|> TITLE: Conditional convergence of $\sum_{n\geq 1} \frac{\sin(p(n))}{n}$? QUESTION [11 upvotes]: The series $\sum_{n\geq 1} \frac{\sin n}{n}$ is easily seen to be conditionally convergent, e.g. by Abel summation. But how about $\sum_{n\geq 1} \frac{\sin(n^2)}{n}$? (for which Abel summation fails) I came across the problem of showing that $\sum_{n\geq 1} \frac{|\sin (n^2)|}{n}$ is divergent in convergence of $\sum_{n\geq 1}|\sin(n^2)|/{n}$. A natural question is if it converges conditionally without the absolute signs. More generally, if given a polynomial $p(n)$ where the dominant coefficient is rationally independent from $\pi$ then by Weyl, see e.g. Tao-Weyl equidistribution $p(n)$ is equidistributed (mod 1) but is the equidistribution sufficiently rapid for the series $\sum_{n\geq 1} \sin(p(n))/n$ to be conditionally convergent? REPLY [11 votes]: This is something I learned from fedja (artofproblemsolving.com): We will show that if $x$ is not Liouvillian then the sum $$ \sum_{n \geq 1} \frac{\sin (2 \pi x n^{2})}{n} $$ converges. An argument goes like this: rewrite $$ \sum_{k =1}^{n} \frac{\sin (2 \pi x k^{2})}{k} = \frac{1}{n}\sum_{k=1}^{n}\sin (2 \pi x k^{2}) + \sum_{k=1}^{n} \frac{S_{k}}{k(k+1)}=I+II $$ where $S_{k} = \sum_{j=1}^{k}\sin (2 \pi x k^{2})$. The first term is OK (converges to integral). For the second term it is enough to show that $|S_{k}| = O(k^{\delta})$ for some $0<\delta<1$. We use the following trick: $$ S_{n}^{2} \leq \left| \sum_{k=0}^{n} e^{2 \pi i x k^{2}}\right|^{2} = \sum_{k,m} e^{2\pi i x(k-m)(k+m)} = (*) $$ Next change variables $k-m=h$ and $m=m$ then $$ (*) =\sum_{-n \leq h \leq n}e^{2 \pi i x h^{2}} \sum_{1\leq m \leq n, 1 \leq m+h \leq n}e^{2 \pi i \cdot 2hxm} \leq \sum_{-n \leq h \leq n}\left| \sum_{1\leq m \leq n, 1 \leq m+h \leq n}e^{2 \pi i \cdot 2hxm}\right| $$ Clearly (Dirichlet kernel) $$ \left| \sum_{1\leq m \leq n, 1 \leq m+h \leq n}e^{2 \pi im \cdot 2hx}\right| \leq \frac{C}{\|2hx\|} $$ where $\|2hx\| = |2hx \mod 1|$ is the distance to the nearest integer. Since $x$ is not Liouvillian then there exists $0n^{a}$ for some $a>0$. This means that when $h \in [-n, n]$ then $2hx$ is close to the fraction of the type $h p /q$ up to an error $|h/qn|\leq 1/n^{a}$ therefore when $h$ runs over $[-n, n]$ then $2hx \mod 1$ almost runs over the fractions $\ell/q$, $\ell TITLE: Positive Semidefinite Polynomial with Bounded Degree in each Variable QUESTION [7 upvotes]: Given a polynomial $p \in \mathbb{R}[x_1, \ldots, x_n]$, suppose that for each monomial in $p$, $c_\alpha x_1^{\alpha_1} \ldots x_n^{\alpha_n}$ we have $\alpha_i = 0, 1, 2$. I have been calling these polynomials multiquadratic (is there another established name for them?). I am interested in developing an algorithm to decide if given a multiquadratic polynomial $p$ it is positive semidefinite, that is if for all inputs in $\mathbf{x} \in \mathbb{R}^n$ we have $p(\mathbf{x}) \geq 0$. In general I have been told this is a difficult problem unless one can reduce to sum of squares which is not always possible. In the case of multiquadratic polynomials are all positive semidefinite polynomials sum of squares? Edit: I believe the polynomial $p(x,y,z) = x^2y^2 + x^2 z^2 + y^2 z^2 - 4xyz + 1$ is PSD, multiquadratic, but not SOS Is there still a decent algorithm for determining if a given multiquadratic polynomial is PSD? Is this a NP-hard problem? REPLY [2 votes]: The conjectured polynomial $p(x,y,z)$ indeed works. It is PSD by comparing the arithmetic and geometric means of $x^2y^2,\ y^2z^2,\ z^2x^2$ and 1. To show that it is not SOS, assume that $$ p(x,y,z)=\sum_i f_i(x,y,z)^2. $$ Clearly, $\deg f_i\leq 2$, otherwise $\deg p$ would be greater than 4. Notice that $$ p(x,0,0)=1=\sum_i f_i(x,0,0)^2; $$ this means that all $f_i(x,0,0)$ are constants, i.e., no $f_i$ contains a non-constant monomial depending on $x$ only. The same holds for other variables. Therefore, each monomial in $f_i$ is either constant or quadratic. But then the term $-4xyz$ cannot appear in $p$ --- a contradiction.<|endoftext|> TITLE: (Co)completness of the 2-category of ``2-Rings" QUESTION [5 upvotes]: Let $2Ring$ denote the 2-category of cocomplete categories with monoidal structures that preserve colimits in each argument. The morphisms are cocontinuous and strong monoidal functors (which are categorified ring homomorphisms). This notion is described in detail in Martin Brandenburg's thesis Tensor categorical foundations of algebraic geometry, arXiv:1410.1716 (pdf) and many authors use a slightly more/less structured notion as a categorification of the category of rings. Is $2Ring$ is (co)complete? I have in mind strict 2-(co)limits but would be interested in other flavors of 2-categorical (co)limits as well. Additionally, I would also like to consider the same question but replacing $2Ring$ with $Comm2Ring$. $Comm2Ring$ is the same as $2Ring$ except with symmetric monoidal structures. What if we impose that $2Ring$ is the 2-category of locally presentable categories with monoidal structures which preserve colimits in each argument? (This is the approach taken in Chrivasitu and Johnson-Freyd's The fundamental pro-groupoid of an affine 2-scheme, arXiv:1105.3104v4 (pdf).) REPLY [6 votes]: I would argue that the "correct" version of (co)limits is the "pseudo" or "bi" or "weak" or "strong" or "homotopy" version. (As far as I can tell, all of these words mean the same thing.) The bicategories of cocomplete or of locally presentable categories are closed under limits: indeed, you can show by hand that the limit just as categories of cocomplete categories along cocontinuous functors is itself cocomplete, and Bird's thesis verifies that if all constituents are locally presentable, then so is the limit. Lurie has shown the same statement in the $\infty$-categorical world. 2-rings, in either Brandenburg's sense or ours, are therefore the symmetric monoidal objects in a complete bicategory. Just as in the 1-categorical case, I believe it is straightforward (if tedious) to show that the limit of underlying categories carries a canonical symmetric monoidal structure, making it the limit of 2-rings. So this handles the "limit" version of your question. The colimit version I believe also works, although you cannot just take the colimit of underlying categories. I haven't thought as much about the bicategory of all cocomplete categories, but the bicategory of locally presentable categories is known, again by Bird's thesis, to contain all colimits. (They are computed by computing instead the limit along right adjoints to the cocontinuous functors you have in mind.) Then you can present a colimit of 2-rings (in our sense) much as you would present a colimit of commutative algebras: any time you see a coproduct, for example, write down a free commutative algebra; any time you see a quotient, do it symmetric monoidally; etc. for the other types of colimits in the bicategorical world. I haven't done the exercise fully myself, but again at least in the locally presentable case I'm sure you can do it (in terms of sketches if necessary). I don't recall off the top of my head whether colimits of arbitrary cocomplete categories exist. I could imagine that you would run into size problems, but I don't know. I should say, there is also a fair amount of work on the bicategorical version of monads, and (co)limits, in all senses (including lax and oplax), of their algebras. I don't recall the references, but my memory is that there are no surprises from the 1-categorical case.<|endoftext|> TITLE: Broken families QUESTION [5 upvotes]: Assume $\sf GCH$. Let $\kappa$ be a regular cardinal, we say that $\{A_\alpha\mid\alpha<\kappa^+\}\subseteq\mathcal P(\kappa)$ is an almost disjoint family, if whenever $\alpha\neq\beta$, $A_\alpha\cap A_\beta$ is bounded in $\kappa$. A family is MAD if it is a maximal almost disjoint family. Given an almost disjoint family, we say that it is a broken family if whenever $A\subseteq\kappa^+$ is bounded, there is a refinement $B_\alpha\subseteq A_\alpha$ (and $A_\alpha\setminus B_\alpha$ bounded) for $\alpha\in A$, such that $\{B_\alpha\mid\alpha\in A\}$ are a pairwise disjoint family with the property that $A_\xi\subseteq^*\bigcup_{\alpha\in A}B_\alpha$ if and only if $\xi\in A$. We can produce broken families from towers, as the pointwise difference. But those are not maximal. Is it provable, or at least consistent (with $\sf GCH$), that there are almost disjoint families which are not broken? How about MAD families? Can we prove that every MAD family is not broken, or at least consistently obtain broken MAD families? (If the general case is a bit too hard, I'd be interested in the case for $\kappa=\omega$ as a particular case.) REPLY [3 votes]: To answer your Question 1, under GCH there is always an almost disjoint family which is not broken. To construct such a family, begin with a partition $\{A_\alpha \mid \alpha < \kappa\}$ of $\kappa$ into pairwise disjoint, unbounded subsets. Enumerate each $A_\alpha$ in increasing order as $\{\eta^\alpha_\beta \mid \beta < \kappa\}$. By GCH, fix a sequence of functions $\vec{f} = \langle f_\gamma \mid \gamma < \kappa^+ \rangle$ that is increasing and cofinal in ${^\kappa}\kappa$ modulo the bounded ideal. Now, for $\gamma \in [\kappa, \kappa^+)$, let $A_\gamma = \{\eta^\alpha_{f_\gamma(\alpha)} \mid \alpha < \kappa\}$. Since $\vec{f}$ is increasing modulo the bounded ideal, the family $\{A_\alpha \mid \alpha < \kappa^+\}$ is almost disjoint. Suppose for sake of contradiction that it is broken. Then there is a function $g:\kappa \rightarrow \kappa$ such that, setting $B_\alpha = \{\eta^\alpha_\beta \mid g(\alpha) \leq \beta < \kappa\}$, we have that $\{B_\alpha \mid \alpha < \kappa\}$ witnesses brokenness for the set $A = \kappa$. But then, if $\gamma \in [\kappa, \kappa^+)$ is such that $g <^* f_\gamma$, we have $A_\gamma \subseteq^* \bigcup_{\alpha < \kappa} B_\alpha$. This seems like sort of a cheap example, though, since, as far as I can tell, the family might be made broken simply by removing the first $\kappa$ elements. Perhaps a more involved version of this construction could produce a more robust failure of brokenness.<|endoftext|> TITLE: Classification of sections of free loop fibration over the two-sphere QUESTION [7 upvotes]: For any space $X$ there is a fibration $$ \Omega X\to LX\stackrel{ev}{\to} X $$ where $LX=Map(S^1,X)$ is the free loop space, $\Omega X = Map_*(S^1,X)$ is the based loop space, and $ev:LX\to X$ is the evaluation map sending $\omega$ to $\omega(1)$. There is a canonical section $\sigma_0:X\to LX$ sending $x\in X$ to the constant loop at $x$. I'm interested in the classification of sections of this fibration up to fiberwise homotopy, which seems to be a difficult problem in general. In particular, the standard obstruction theory methods (such as Corollary VI.6.16 on p.302 of Whitehead's "Elements of homotopy theory") don't seem to apply. So I'm asking about what I believe to be the simplest example where this fibration does not split, namely $X=S^2$. Have the sections of $ev: LS^2\to S^2$ been classified up to fiberwise homotopy? The only reference I could find on this problem was Samson Saneblidze, MR 1255935 On the homotopy classification of sections in the free loop fibration, J. Pure Appl. Algebra 91 (1994), no. 1-3, 317--327. from which I can probably deduce an answer for the rationalization $S^2_\mathbb{Q}$. In case the answer turns out to be "no", let me ask an easier question. There is an action $S^1\times S^2\to S^2$ given by rotation around an axis. The adjoint of this action gives a section $\sigma: S^2\to LS^2$. Is the section $\sigma$ fiberwise homotopic to the canonical section $\sigma_0$? If not, how do I see this? Added later: I think I see a way to distinguish between $\sigma$ and $\sigma_0$. Denote the adjoints of these sections by $\sigma',\sigma_0':S^1\times S^2\to S^2$. Let $S^1\subseteq S^2$ be a diameter which passes through the two fixed points of the rotation. Then $\sigma'$ and $\sigma_0'$ when restricted to $S^1\times S^1\subset S^1\times S^2$ have different degrees (1 and 0), hence they are not homotopic. I wonder if this can be formalized somehow. REPLY [6 votes]: The space of sections of $\Lambda S^2\to S^2$ is a subspace of the space of all maps $$S^2\to \Lambda S^2=Map(S^1,S^2).$$ When viewed as a subspace of the space of all maps $$S^1\to Map(S^2,S^2),$$ it becomes $\Omega_1 Map(S^2,S^2)$, the based loopspace of the space of self-maps of $S^2$, with the identity map as basepoint. Thus the problem is to compute $\pi_1(Map(S^2,S^2),1)$. I claim that it has order two. The space of all degree one maps $S^2\to S^2$ is sometimes denoted by $SG(3)$; it is the "structure group" for oriented spherical fibrations of rank $3$. It is part of a fibration sequence $$ SF(2)\to SG(3)\to S^2 $$ with $SF(2)$ the space of degree one based maps $S^2\to S^2$, a component of $\Omega^2S^2$. Mapping into this is another fibration sequence $$ SO(2)\to SO(3)\to S^2. $$ Thus $SG(3)/SO(3)$ is homotopy equivalent to $SF(2)/SO(2)$. The latter is $1$-connected and has finite $\pi_2$. This follows as soon as one knows that $\pi_1SO(2)\to \pi_1SF(2)=\pi_3S^2$ is an isomorphism, which I think is not hard to see. Thus $\pi_1SG(3)$ is the cokernel of the map $\pi_2(SG(3)/SO(3))\to \pi_1SO(3)=\mathbb Z/2$. This map is zero because it factors through a map $\pi_2(SG(3)/SO(3))\to \pi_1SO(2)$ from a finite group to $\mathbb Z$.<|endoftext|> TITLE: What's the stabilization of the $\infty$-category of $\infty$-categories? QUESTION [16 upvotes]: $\require{AMScd}$One nice thing about $\infty$-categories is that spaces are themselves $\infty$-categories. What's the analogue for spectra? Presumably this would be the stabilization of the $\infty$-category of $(\infty,1)$-categories, fitting into a cube: $$ \begin{CD} ILS_* @>>> \text{Spaces}_* @>>> \text{$(\infty,1)$-Cats}_* @<<< \text{S.M. $(\infty,1)$-Cats}_*\\ @V\simeq VV @V\Sigma^\infty VV @VV\Sigma^\infty V @VV\simeq V\\ \text{ConnSpectra} @>>> \text{Spectra} @>>> ? @<<< Conn ? \end{CD} $$ where the far left/right entries of the top row are respectively symmetric monoidal $(\infty,0/1)$-categories, and the $*$s denote that we have chosen a basepoint. Is this just the $\infty$-category of spectrally enriched categories? edit if this idea doesn't work out, I'm more interested in "What's the analogue for spectra?" than the second question. REPLY [21 votes]: In a project in progress with Matan Prasma and Joost Nuiten concerning the abstract cotangent complex formalism we compute the stabilization of the $\infty$-category $\infty\mathrm{Cat}_{/C}$ of $\infty$-categories over a fixed $\infty$-category $C$, and show that it is equivalent to the $\infty$-category of functors $\mathrm{Tw}(C) \to \mathrm{Spectra}$ from the twisted arrow category of $C$ to spectra. In particular, the stabilization of $\infty\mathrm{Cat}$ is equivalent to the $\infty$-category of spectra and the stabilization of $\infty\mathrm{Cat}_{/\Delta^1}$ is equivalent to the $\infty$-category of (co)spans of spectra, as suggested by Charles Rezk in the remarks above. Another way of phrasing it is to say that the data of a spectrum object in $\infty\mathrm{Cat}_{/C}$ is equivalent to the data of decorating, for each $x,y \in C$, the mapping space $Map_C(x,y)$ with a parametrized spectrum over it, in a way that is suitably functorial in $(x,y) \in C^{op} \times C$. In this language one can phrase and prove the result in the more general setting of enriched categories. We expect to upload this preprint to the arXiv very soon. Edit: The paper is now on the arXiv and can be found here.<|endoftext|> TITLE: If $\mathcal{A} \equiv \mathcal{B}$ and $\mathcal{A} \not \cong \mathcal{B}$, is it possible that $\mathcal{A}$ and $\mathcal{B}$ are bi-embeddable? QUESTION [10 upvotes]: Are there structures $\mathcal{A}$ and $\mathcal{B}$ such that $\mathcal{A}$ and $\mathcal{B}$ are elementary equivalent, not isomorphic but bi-embeddable? REPLY [14 votes]: Let $\mathcal{A}=(\mathbb{R},<)$ and $\mathcal{B}=(\mathbb{R}\setminus\{0\},<)$. They are elementarily equivalent being both dense linear orders without end-points, clearly each one can be embedded in the other, but $\mathcal{A}$ is a complete order while $\mathcal{B}$ is not. REPLY [4 votes]: Here's another example, with a computability-theoretic motivation and a set-theoretic proof: for a countable linear order $\alpha$, let $H(\alpha)$ be the linear order $\alpha+\alpha\cdot\eta$, where $\eta$ is the order type of the rationals. So, $H(\alpha)$ looks like a copy of $\alpha$ followed by a dense family of copies of $\alpha$. The "$H$" stands for "Harrison," who investigated the case where $\alpha=\omega_1^{CK}$ is the first non-computable ordinal. That linear order has a number of interesting computability-theoretic properties. Now, since $\eta$ embeds into $H(\alpha)$ for all $\alpha$, we have that $H(\alpha)$ and $H(\beta)$ are bi-embeddable for any countable linear orders $\alpha$ and $\beta$ (this is just Cantor's theorem that any countable linear order embeds into the rationals). On the other hand, $H(\alpha)\cong H(\beta)\iff \alpha\cong\beta$ is easily checked if $\alpha$ and $\beta$ are ordinals (that is, linear orders with no infinite descending chain). So we just need to find two non-isomorphic ordinals whose $H$-versions are elementarily equivalent. Note that a counting argument does not immediately do the job: there are continuum-many nonisomorphic linear orders, but there are also continuum-many different complete first-order theories in the language $\{<\}$. In fact, if we restrict attention to ordinals, there may be fewer countable ordinals (up to isomorphism, there are $\aleph_1$) than there are complete theories (there are always $2^{\aleph_0}$-many of those, and assuming that the Continuum Hypothesis fails that's more than $\aleph_1$). So we need to be a bit clever. There are many ways to do this, but here's one that I like. It's not hard to show that for each sentence $\varphi$ in the language of linear order, the set $$S_\varphi=\{\alpha: \alpha\mbox{ is a countable ordinal and }H(\alpha)\models\varphi\}$$ is $\Pi^1_1$ (or rather, the set of codes for elements of $S_\varphi$ is a $\Pi^1_1$ set of reals). Now, it's a nice fact that any such set either contains or is disjoint from a club ( = closed unbounded set of countable ordinals). So for each of the countably-many sentences $\varphi_i$ in the language of linear order, we may find a club $C_i$ of countable ordinals such that for all $\alpha,\beta\in C_i$, $H(\alpha)$ and $H(\beta)$ agree on whether $\varphi_i$ is true. Now, the intersection of countably many clubs is a club, so we get a club $C$ of ordinals such that for every sentence $\varphi$ in the language of linear order, and every $\alpha,\beta\in C$, either $H(\alpha)$ and $H(\beta)$ both satisfy $\varphi$ or they both satisfy $\neg\varphi$; that is, $H(\alpha)\equiv H(\beta)$ for all $\alpha,\beta\in C$. Now just use the fact that a club has more than one element. EDIT: Note that this in fact provides an uncountable family of countable structures which are pairwise elementary equivalent and nonisomorphic, any one of which can be embedded into any other. With a bit more work, we can turn this into such a family of size continuum, which is of course the best possible, by consider the notion of a club of countable linear orders and performing basically the same analysis. NOTE 1: We could have made do with a much simpler construction - let $I(\alpha)=\alpha+\eta$. This would be enough. However, the $H(\alpha)$s are a natural class of structures, so that seemed worth the extra complexity. NOTE 2: large cardinal axioms imply that more complicated sets of countable ordinals contain or are disjoint from a club; that is, that the club filter is "closer to being an ultrafilter." So in cases where we have a construction more complicated than $\alpha\mapsto H(\alpha)$, we can still get away with the same argument . . . if we have the right set-theoretic hypotheses.<|endoftext|> TITLE: Partial $\mathsf{T}$-algebras, where $\mathsf{T}$ is a Lawvere theory QUESTION [5 upvotes]: Question. Has this proposal (or anything like it) for a category of partial $\mathsf{T}$-algebras been considered anywhere, and if so, where, and if not, does it have any obvious flaws or deficits? Let $\mathsf{T}$ denote a Lawvere theory with generating object $\Box$. I was thinking that a reasonable definition of "partial $\mathsf{T}$-algebra" might go something like this. Firstly, let $\mathbf{PSet}$ denote the category of pointed sets. View $\mathbf{PSet}$ as a symmetric monoidal category with $\otimes$ as smash product, and note that there's reasonable-looking choices for projection maps like $X \otimes Y \rightarrow X$ and diagonals like $X \rightarrow X\otimes X$. Secondly, define that a morphism of pointed sets $f : (X,\bot_X) \rightarrow (X,\bot_Y)$ is tight iff $f^{-1}(\bot_Y) = \{\bot_X\}$. See here for relevant nonsense. Thirdly, make $\mathbf{PSet}$ into a (thin) $2$-category by defining that for each pair of morphisms $$f,g : (X,\bot_X) \rightarrow (Y,\bot_Y),$$ we have $f \leq g$ iff $$\forall x \in X(f(x) \neq \bot_Y \rightarrow f(x)=g(x)).$$ Fourthly, define that a partial $\mathsf{T}$-algebra is an oplax functor $F : \mathsf{T} \rightarrow \mathbf{PSet}$ with data $F(\Box^n) \cong (F\Box)^{\otimes n}$, and which turns each canonical projection $\pi_i : \Box^n \rightarrow \Box$ into the corresponding canonical projection $\pi_i:(F\Box)^{\otimes n} \rightarrow F\Box$ and the same for the diagonals. And fifthly, define that a morphism $X \rightarrow Y$ of partial $\mathsf{T}$-algebras is just a lax natural transformation $f:X \rightarrow Y$ whose component at $\Box$ is tight. By "lax", mean that $$f_\Box \circ X(j) \leq Y(j) \circ f_{{\Box}^{\otimes n}}.$$ It would be nice, for example, if the partial $\mathsf{T}$-algebras for $\mathsf{T}$ equal to the Lawvere theory of commutative monoids were precisely the partial commutative monoids. This is probably too much to hope for, but at least its a guiding example. REPLY [2 votes]: The closest reference to what you are asking for that I know is Erik Palmgren and Steve Vickers: Partial Horn logic and cartesian categories. Annals of Pure and Applied Logic. Volume 145, Issue 3, March 2007, Pages 314-353. (DOI, PDF) To get the universal partial algebra going one has to address partiallity in a serious way. For instance, it matters how complicated the domains of definition may be, or to put it another way, what is the language for expressing definedness conditions? The above paper addresses this, which is why it the title talks about "Horn logic" but is really (also) about universal partial algebra. Another place to look is essentially algebraic theories.<|endoftext|> TITLE: Is it possible to construct an action of an $E_\infty$ operad on $BU$ that respects filtration by $BU(n)$? QUESTION [23 upvotes]: It is well known that $BU$ is an infinite loop space, and as such it has an action of an $E_\infty$ operad. An explicit construction of such an action is given, for example, in an answer to this MO question . My question is whether it is possible to construct an action that respects the filtration of $BU$ by $BU(n)$. Let me spell out the question more explicitly: is it possible to construct a filtered space $X_1 \subset X_2 \subset \cdots X$ such that for each $n$ $X_n\simeq BU(n)$, and there is an action of an $E_\infty$ operad $P$ on $X$ such that operad structure maps like $$P(i)\times X^i\to X$$ restrict to maps $$P(i)\times X_{n_1}\times\cdots \times X_{n_i}\to X_{n_1+\cdots+ n_i}$$ REPLY [17 votes]: Unfortunately there is no such filtration. At first, this looks very similar (but not as strong as) asking for a map of $E_\infty$ spaces $\coprod BU(n) \to BU$ which would become a splitting map $ku \to bu$ of spectra. We know that doesn't happen, because there's a nontrivial $k$-invariant. We can look more carefully at how this $k$-invariant works and it leads us to a method to give an actual contradiction. The $k$-invariant is detected by the fact that the generator $1_{ku}$ of $\pi_0 ku$ is annihilated by the Hopf element $\eta$, and the Toda bracket $\langle 2, \eta, 1_{ku}\rangle$ contains the Bott element $\beta$ and does not contain zero. This is spelled out in the following way on the level of $E_\infty$ spaces. Suppose $X$ is an $E_\infty$ space with multiplication $\smile$ and associated spectrum $KX$, and let $\alpha \in \pi_0(X)$ have image $[\alpha] \in \pi_0(KX)$. Up to translating path components, the element $\eta[\alpha] \in \pi_1(KX)$ lifts (up to changing path components) to the element $\alpha \smile_1 \alpha \in \pi_1(X, \alpha \smile \alpha)$. We can find a canonical nullhomotopy of the path composite of $\alpha \smile_1 \alpha$ with itself, expressing the identity $2 \eta [\alpha] = 0$; if we also have a nullhomotopy of $\alpha \smile_1 \alpha$, then we can use this to construct a representative for the bracket. If you carry this out for the basepoint of $BU(1)$ using the standard map $E\Sigma_2 \times_{\Sigma_2} BU(1)^2 \to BU(2)$, you find that you get the generator of $\pi_2 BU(2) = \Bbb Z$. However, in $bu$ this would mean that there was a bracket $\langle 2,\eta, 0\rangle$ which did not contain zero; that would be bad. Of course, it turns out to be a lot easier to appeal to some machinery. Kochman calculated the Dyer-Lashof operations on $H_* BU = \Bbb F_2[x_1, x_2, \dots]$. His calculations show, for example, that $Q^4 x_1 = x_1^3 + x_1 x_2 + x_3$. As a result, the map $H_6 (E\Sigma_2 \times_{\Sigma_2} BU(1)^2) \to H_6(BU)$ is surjective, but the map $H_6(BU(2)) \to BU$ is not (it is missing $x_1^3$). This means that our operad structure map $P(2) \times X_1 \times X_1 \to X$ could never land in a subobject $X_2$ as you want. (This assumes that I've read and understood correctly in calculating the operations.) (I actually find Priddy's method for calculating the Dyer-Lashof operations here a little easier than trying to understand Kochman's algorithm: Priddy calculates the Dyer-Lashof operations on $H_*(\coprod BU(n)) = \Bbb F_2[a_0,a_1,\dots]$ and then you can deduce the operations on the generators $x_i = a_i a_0^{-1}$ of $H_* BU$ by the Cartan formula. The inverse screws up the property of preserving homogeneous degree and it's what's messing us up here.)<|endoftext|> TITLE: Any reference for the series expansion of $\Bigr[-\log(1-t)\Bigr]^x$? QUESTION [6 upvotes]: Any reference that we can find the following $$\Bigr[-\log(1-t)\Bigr]^x = t^x + x t^x \sum_{k=0}^\infty \psi_k(x+k)\,t^{k+1}; \quad \mbox{for all} \, x\in \mathbb R, \, |t|<1$$ where $\psi_k(.)$ are the Stirling Polynomials (Stirling Polynomials). Thank in advance REPLY [4 votes]: It should be remarked that the expansion is an immediate consequence of the Lagrange Inversion Formula, and of the definition of the Stirling polynomials via their generating function (as given e.g. in the Wolfram MathWord link you gave). Let $f=\sum_{k=1}^\infty f_kt^k $ and $g=\sum_{k=0}^\infty g_kt^k $ be formal power series inverse of each other, with $f_1=g_1=1$; let $k\in\mathbb{N}$, $x\in\mathbb{C}$ and $y\in\mathbb{C}$ with $k+x+y=0$. Denoting as usual $[t^k]F$ the coefficient of $t^k$ in $F$, the Lagrange Inversion Formula may be written in the form: $$\frac{1}{x}[t^k]\left( \frac{f}{t}\right)^x+\frac{1}{y}[t^k]\left( \frac{g}{t}\right)^{y}=0$$ or in the equivalent, less symmetric form $$[t^k]\left( \frac{f}{t}\right)^x=\frac{x}{x+k}[t^k]\left( \frac{t}{g}\right)^{k+x}.$$ If we take $f(t):= -\log(1-t) $ then its inverse is $g(t)=1-e^{-t}$, so $$[t^k]\left( \frac{-\log(1-t)}{t}\right)^x=\frac{x}{x+k}[t^k]\left( \frac{t}{1-e^{-t}}\right)^{k+x}.$$ By the definition of the Stirling polynomials (with $x+k-1$ in place of $x$) $$[t^k]\left( {t \over {1-e^{-t}}} \right) ^{x+k}= { S_k(x+k-1) \over k!} $$ so we get $$\big( -\log(1-t) \big)^x=xt^x \sum_{k=0}^\infty \frac{S_k(x+k-1)}{x+k}{t^k \over k!} $$ which is the expansion you want. Incidentally, the expansion $$\log(-\log(1-t))=\log t-\sum_{k=1}^\infty {S_k(k-1)\over kk!}t^k,$$ can be obtained differentiating w.rto $x$ at $x=0$.<|endoftext|> TITLE: Reconstructing an oriented matroid from its deletion and contraction QUESTION [5 upvotes]: Suppose that $\mathcal{M}_1$ and $\mathcal M_2$ are two oriented matroids on the same ground set $E$. Under what conditions on $\mathcal{M}_1$ and $\mathcal{M}_2$ is there an oriented matroid $\mathcal{M}$ on the ground set $E\cup \{e\}$ such that $\mathcal{M}-e=\mathcal{M}_1$ and $\mathcal{M}/e=\mathcal{M}_2$? I feel like I've seen the answer to this in the book, and that the necessary and sufficient condition was $\mathcal{T}(\mathcal{M}_2)\subset \mathcal{T}(\mathcal{M}_1),$ where $\mathcal{T}(\mathcal{M})$ is the set of topes of $\mathcal{M}$. However, I've been trying to find it in the book again without any success for a very long time. REPLY [3 votes]: Found the answer in a paper by Ziegler & Richter-Gebert, see Theorem 4.1. Basically, they show that if two oriented matroids $\mathcal{M}_1$, $\mathcal{M}_2$ of ranks $r$ and $r-1$ respectively satisfy $\mathcal{L}(\mathcal{M}_2)\subset \mathcal{L}(\mathcal{M}_1)$ then there exists a (essentially unique) oriented matroid $\mathcal{M}$ with $\mathcal{M}\setminus e=\mathcal{M}_1$ and $\mathcal{M}/e=\mathcal{M}_2$. So the necessary and sufficient condition is the inclusion of covectors.<|endoftext|> TITLE: Littlewood–Richardson rule and the Harish-Chandra-Itzykson-Zuber integral QUESTION [13 upvotes]: The Littlewood–Richardson rule states that the product of two Schur polynomials can be written as a finite weighted sum of Schur polynomials. More precisely $$ s_\lambda s_\mu = \sum_\nu c_{\lambda,\mu}^\nu s_\nu $$ where $c_{\lambda,\mu}^\nu$ is equal to the number of Littlewood–Richardson tableaux of skew shape $\nu /\lambda$ and of weight $\mu$. Linked with some explicit computations in random matrix theory, I am hopping that a similar product rule may hold for the functions $\varphi_a(s)$ defined by $$ \varphi_a(x)=\frac{\det[e^{a_ix_j}]}{\prod_{i TITLE: Decidability of (restricted) periodicity of Wang tilings QUESTION [5 upvotes]: Consider a Wang tiling (given a subset of $C^4$ for a finite set $C$ of colours, e.g.). It's well-known to be undecidable whether there exists a tiling, and also whether there exists a periodic tiling. It's also well-known that there exists a periodic tiling iff there exists a doubly periodic tiling. On the other hand, for any given n, then it's decidable whether there exists a periodic tiling with period $(n,0)$: construct a directed graph with vertex set $C^n$ and an edge from $v$ to $w$ if there exists a tiling of the $n\times1$ rectangle with identical left and right labels and $v,w$ at the bottom and top respectively. Then there exists such a periodic tiling iff this graph has a loop. Here is a question I can't answer: for any given $n$, is it decidable whether there exists a periodic tiling with period $(n,m)$ for some $m$? REPLY [2 votes]: For $m$ fixed, you can consider all proper tilings of a $1 \times m$-rows. The number of such rows is crudely bounded above by $C^{2m}$. Then, we pretend each such row is an individual Wang tile, and have thus reduced it to the $n \times 1$ case.<|endoftext|> TITLE: Can the trivial module be stably free for a monoid ring? QUESTION [5 upvotes]: Let $M$ be a non-trivial monoid and $\mathbb ZM$ its monoid ring. All modules are left modules in what follows. Suppose that $M$ contains a zero element (or absorbing element) $z$. That is $mz=z=zm$ for all $m\in M$. For example the zero element of a ring is absorbing for multiplication. It is well known that the trivial $\mathbb ZM$-module $\mathbb Z$ is projective in this case. Indeed, $z$ is a central idempotent and $\mathbb ZMz\cong \mathbb Z$. Therefore, $\mathbb ZM=\mathbb ZMz\oplus \mathbb ZM(1-z)$ and hence $\mathbb Z$ is projective. Moreover, the above direct sum is a ring direct product decomposition. Question 1. Can the trivial module $\mathbb Z$ be stably free? Recall that a finitely generated projective module $P$ is stably free if $P\oplus F\cong F'$ for some finitely generated free modules $F,F'$. Note that in our setting if $\mathbb Z\oplus F\cong F'$ and $r$ is the rank of $F$ and $r'$ is the rank of $F'$, then $$\mathbb Z^{r'}\cong zF'\cong \mathbb Z\oplus zF\cong \mathbb Z^{r+1}$$ and so $r'=r+1$ if this happens. Also, $R=\mathbb ZM(1-z)$ is a ring, known as the contracted monoid ring of $M$. It is obtained by identifying the zero of $M$ with the zero of $\mathbb ZM$ and it has the property that $R$-modules are precisely those $M$-modules annihilated by the zero of $M$. Then note that $$R^{r+1}\cong (1-z)F'\cong (1-z)F\cong R^r$$ and hence if $\mathbb Z$ is stably free, then $R$ does not have the Invariant Basis Number Property, which says that finite rank free modules "know" their rank. Question 1'. Can a contracted monoid ring fail to have the Invariant Basis Number Property? If either $M$ is finite or $M\setminus \{z\}$ is a submonoid, then the contracted monoid ring has the Invariant Basis Number Property. It has been suggested to me by both George Bergman and Jason Bell (independent private communications) that the polycyclic inverse monoid (or Cuntz monoid) $$P_2=\langle x,x^*,y,y^*,z\mid xx^*=1,yy^*=1, xy^*=z=yx^*, z=0\rangle,$$ where $z=0$ is shorthand for $z$ is the absorbing element, may be relevant here for Question 1'. The point is that if $R$ is the contracted monoid ring of $P_2$, then $R^2$ is isomorphic to a direct summand in $R$. One notes that $x^*x,y^*y$ are orthogonal idempotents and so $e=x^*x+y^*y$ is an idempotent and $(a,b)\mapsto ax+by$ is a module isomorphism $R^2\to Re$ with inverse $c\mapsto (cx^*,cy^*)$. Thus $Re$ is stably isomorphic to $R$. Since $e\neq 1$, this doesn't contradict the invariant basis number property, but it hints that the $K$-theory of this ring may not be so nice. Note that $R/R(1-e)R=R/R(1-(x^*x+y^*y))R$ is a Leavitt algebra which fails to have the invariant basis number property because the free modules of rank $1$ and $2$ are isomorphic. The extension of scalars of this algebra over any field is simple and this ring is known to contain interesting groups inside of its group of units like the simple group Thompson's group $V$. REPLY [4 votes]: I've had a little more time to think about your questions, and I believe the following is true: Contracted monoid rings have IBN. This might appear implicitly in Cohn's paper, but I couldn't find it on a causual reading. We will need the following lemma, which appears (with proof) as Exercise 1.5 in "Exercises in Modules and Rings": Suppose a ring $R$ admits an additive group homomorphism $T$ into an abelian group $(A,+)$ such that $$ (\ast)\quad T(cd)=T(dc)\ \text{ for all }\ c,d\in R. $$ If $T(1)$ has infinite addivite order in $A$, then $R$ must have IBN. Now, we will proceed to prove the main claim, and assume that $R$ is a contracted monoid ring. Since we will be concerned with constructing an additive group homomorphism from $R$, we are concerned with the additive group structure of $R$, which is given as a free $\mathbb{Z}$-module with basis $M':=M\setminus\{z\}$. Define an equivalence relation on $M$ by first saying $a,b\in M$ are "basically related" if there exist $x,y\in M$ with $a=xy$ and $b=yx$, and then taking the transitive closure of this relation. Define $T:R\to \mathbb{Z}$ by first defining it on $M'$ and then extending $\mathbb{Z}$-linearly. We must have $T$ send $1_M$, and anything related to it (by a finite sequence of basically related elements), to $1_{\mathbb{Z}}$. We send everything else in $M'$ to $0$. This is a well-defined additive group homomorphism. All that remains is to check that it satisfies $(\ast)$. Let $c,d\in R$. We can then write $$ c=\sum_{m\in M'}a_m\cdot m,\qquad d=\sum_{n\in M'}b_n\cdot n $$ for some unique $a_m,b_n\in \mathbb{Z}$ (for each $m,n\in M'$), all but finitely many of which are zero. Since $R$ is a contracted monoid ring, we have either $mn\in M'$ or $mn=z$, and similarly for $nm$. Thus $$ T(cd)=\sum_{m,n\in M'}a_mb_nT(mn),\qquad T(dc)=\sum_{m,n\in M'}a_mb_nT(nm), $$ where we define $T(z)=0$. Thus, these sums will be equal unless $z\in M$ is in the equivalence class of $1_M\in M$. Suppose, by way of contradiction, that this were the case. Then we would have $$ 1_M=x_1 y_1, y_1x_1=x_2y_2,\cdots, y_n x_n=z. $$ Hence $$ 1_M = 1_M^{n+1} = (x_1y_1)^{n+1} = x_1 (y_1x_1)^{n}y_1 = x_1(x_2y_2)^{n}y_1 = x_1 x_2(y_2x_2)^{n-1}y_2y_1 = \cdots $$ $$ = x_1 x_2 \cdots x_n (y_n x_n) y_n y_{n-1} \cdots y_1 = z $$ contradicting the fact that $M$ is a nontrivial monoid.<|endoftext|> TITLE: Traces of powers of integral marices QUESTION [10 upvotes]: I have encountered the following linear algebra/number theory question in my work (low-dimensional topology), so I thought I should ask the experts. Let $A \in SL(n,\mathbb{Z})$ be a matrix , $n \geq 3$. One can easily show that there is some $j \in \mathbb{N}$ such that $Tr(A^j) \geq 3$ (proof given at the end). My question is that how long does it take for this to happen? Question 1 Given $n \geq 3$, is there a number $k \in \mathbb{N}$ (depending only on n) such that for all $A \in SL(n,\mathbb{Z})$ there is some $1 \leq i \leq k$ such that $Tr(A^i) \geq 3$? Question 2: If the answer to the above question is yes, then how $k$ depends on $n$? Proof of the claimed fact: Assume the contrary that $Tr(A^i) \leq 2$ for all natural numbers $i$. Let $\lambda_1 , ... , \lambda_n$ be the eigenvalues of $A$, therefore $T_i := Tr(A^i) = \lambda_1 ^i + ... + \lambda_n ^i$. Since $T_i$ are bounded above and $\lambda_1 ... \lambda_n =1$ then we should have $|\lambda_i| = 1$ for each $1 \leq i \leq n$. This implies that all $\lambda_1 , ..., \lambda_n$ are roots of unity (because all of their Galois conjugates are on the unit circle) and therefore for some $j \in \mathbb{N}$ we have $\lambda_1 ^j = ... = \lambda_n ^j = 1$. This implies $T_j \geq n \geq 3$. Contradiction. PS: Replace the assumption $A \in SL(n,\mathbb{Z})$ by $A \in Sp(2n,\mathbb{Z})$, if it makes things easier. REPLY [4 votes]: Yes: for every $d\ge 3$ and all $A\in\mathrm{GL}_d(\mathbf{Z})$, there exists $n\in\{1,\dots,12^d\}$ such that the trace of $A^n$ is $\ge 3$. (This is probably far from sharp, and I don't know if we can do better than exponential.) First part: I claim that if $t=(t_1,\dots,t_d)$ is a $d$-tuple in the unit circle, then there exists $n\in\{1,\dots,12^d\}$ such that all $t_1^n\dots t_d^n$ have argument in $[-\pi/6,\pi/6]$. Indeed otherwise, denote $P_\theta$ as the set of $d$-tuples in the unit circle of argument in $\mathopen]-\theta,\theta\mathclose[$. Then for all $m\neq n$ with $m,n\in\{-12^d/2,\dots,12^d/2\}$, we have $t^{m-n}$ not in $P_{\pi/6}$. Define $V_n=t^nP_{\pi/12}$. Then the $V_i$, for $i\in\{-12^d/2,\dots,12^d/2\}$, are pairwise disjoint. If we endow the torus with the normalized Haar measure, the measure of $P_{\pi\theta}$, for $0\le\theta\le 1$ is equal to $\theta^d$. Hence the measure of $V_i$ is $(1/12)^d$. So $(12^d+1)/12^d\le 1$, that is, $1+12^{-d}\le 1$, contradiction. I conclude with the following claim. Consider a matrix in $A=\mathrm{GL}_d(\mathbf{R})$ with $d\ge 3$, $|\det(A)|=1$, and all eigenvalues having argument in $[-\pi/6,\pi/6]$. Then the trace of $A$ is $>2$. In particular, if $A$ is integral then its trace is $\ge 3$. Indeed, let $\lambda_1,\dots,\lambda_d$ be the eigenvalues, of modulus $r_1\ge r_2\ge \dots \ge r_d$ (we can suppose that conjugate eigenvalues are written consecutively). If $r_3\ge 1$, then the trace is $\ge (\sqrt{3}/2)(r_1+r_2+r_3)\ge 3\sqrt{3}/2> 2$. Now suppose $r_3<1$. Note that $\prod r_i=1$, so $r_1r_2=\prod_{i\ge 3}r_i^{-1}>1$, so $r_2>1/r_1$. If $\lambda_1$ and $\lambda_2$ are real, then the trace is $>r_1+r_2>r_1+r_1^{-1}\ge 2$. If $\lambda_1$ is real and $\lambda_2$ is nonreal, then $r_2=r_3$ and hence the trace is $\ge r_1+2(\sqrt{3}/2)r_2>r_1+r_2>r_1+r_1^{-1}\ge 2$. If $\lambda_1$ is non-real and $\lambda_3$ is real, then $r_1=r_2$ and the trace is $>\sqrt{3}r_1+r_2\ge r_1+r_1^{-1}\ge 2$. Finally if both $\lambda_1$ and $\lambda_3$ are non-real, then $k\ge 4$, $r_1=r_2$, $r_3=r_4$ the trace is $\ge \sqrt{3}(r_1+r_3)$; we have $r_1^2r_3^2\ge 1$ implying $r_3\ge r_1^{-1}$ and hence the trace is $\ge \sqrt{3}(r_1+r_1^{-1})\ge 2\sqrt{3}>3$. REPLY [4 votes]: Thue's lemma allows to get an exponential bound for $k(n)$, though it seems to be not sharp. More specifically, idea is the following. We partition the unit circle onto three equal arcs $A,B,C$ and for each $k$ encode a sequence $(\lambda_1^k,\dots,\lambda_n^k)$ by a sequence of $n$ letters A,B,C which correspond to arcs containing arguments of $\lambda_i^k$. By pigeonhole principle two sequences concide for some positive integers $k,m$ not exceeding $3^n+1$. It means that for $h:=|k-m|$ real parts of all numbers $\lambda_i^h$ are positive and at least half of their absolute values, thus $\sum \lambda_i^h\geqslant \sum |\lambda_i|^h/2\geqslant n/2$.<|endoftext|> TITLE: Connections with the same geodesics QUESTION [6 upvotes]: Suppose there are two connections over the tangent bundle of a smooth manifold with the same geodesics. What is the relation between the curvature tensors, the Ricci curvature and the scalar curvature of these two connections? Note that the question gives no address to the metric, such as Riemannian or the Levi-Civita. REPLY [6 votes]: You are looking for the theory of projective connections. See the paper of Molzon and Mortensen, The Schwarzian derivative for maps between manifolds with complex projective connections, Trans. Amer. Math. Soc., Volume 348, Number 8, August 1996 which is surely the best survey. It treats the complex case, but the details are the same as for the real case. You might also look at the paper of Kobayashi and Nagano, On projective connections.<|endoftext|> TITLE: a question on symbolic power QUESTION [5 upvotes]: Let $R$ be a Noetherian ring and $P$ a prime ideal. Then the $n$-th symbolic power of $P$ is $$P^{(n)} = P^n R_P \cap R = \{ f \mid sf \in P^n \text{ for some } s \in R - P\}$$ (cf. wiki). We have $P^{(n)}$ is just the $P$-primary component of $P^n$. Now, we consider $R = \mathbb{C}[X_1, \ldots, X_d]$ and $V = V(P)$ the variety defined by $P$. For each $a \in V$ we set $\mathfrak{m}_a$ the maximal ideal corresponding with $a$. We have $P = \cap_{a \in V} \mathfrak{m}_a$, since $R$ is a Jacobson ring. $P^{(n)}$ consists of functions with zeros of order $n$ along $V$. Question 1. With above setting. Are the following equivalent: $f \in P^{(n)}$; $f \in \mathfrak{m}_a^n$ for all $a \in V$; $f, \frac{\partial^{\alpha} f}{\partial X^{\alpha}}$ vanish on $V$ for all $| \alpha | \le n-1$. It is clear that $(2) \Leftrightarrow (3)$ and we can prove that $(1) \Rightarrow (3)$. Question 2. Suppose Question has a positive answer. Can we have $(1) \Leftrightarrow (2)$ in more general rings (as Jacobson rings)? REPLY [4 votes]: Question 1 is answered affirmatively in Eisenbud's "Commutative Algebra with a View Toward Algebraic Geometry" (p. 106, attributed to Zariski, Nagata). The text refers to "A Nullstellensatz with Nilpotents" [1979] by Eisenbud and Hochster for generalizations.<|endoftext|> TITLE: An H-group that's not a loop space? QUESTION [10 upvotes]: Recall that an H-group is a space $X$ (in the sense of homotopy theory, so say CW complex) that is a group object in the homotopy category. I.e., there's a multiplication map $X \times X \rightarrow X$ which is associative up to homotopy and with an inverse map again up to homotopy. The standard example of an H-group is a loop space $\Omega X$. What is a simple example of an H-group that is not a loop space? If I understand the language right, I'm asking for a group-like $A_3$ algebra that's not $A_\infty$. If I were asking for $A_2$ but not $A_3$ then I know $S^7$ is a good example. My motivation is illustrating some of the subtleties in defining $\infty$-groups for a HoTT seminar. REPLY [10 votes]: The first example appears in the paper "Homotopy associativity and finite CW complexes" Topology vol.9 (1970) 121-128 by Alexander Zabrodsky. With present knowledge it is possible to construct many examples. Here are some examples. Consider Sp(n) with n 3 or more, or G_2. Construct X from a pullback diagram of p-local spaces, where X localized at 2 or 3 is the Lie group and localized at the remaining primes is the product of local sphere having the same type as the Lie group. Then X is homotopy associative because each of its localizations is homotopy associative. The space X is not a mod p group for p between 5 and 2n if Sp(n) is used and not a mod 5 group if G_2 is used in the construction. The latter claim is based of a theorem proved by Clarence Wilkerson, "K-theory Operations in mod-p Loop Spaces" Math.Z. 132, 29-44, (1973)<|endoftext|> TITLE: Homotopically trivial vs isotopically trivial diffeomorphisms QUESTION [7 upvotes]: Let $M$ be a manifold. Let's say $M$ is smooth, connected, oriented. We can also assume that $M$ is closed if that makes things easier. Let $\mathit{Diff}(M)$ denote the group of diffeomorphisms of $M$ and $\mathit{Diff}_0(M)$ denote its identity component, consisting of the isotopically trivial diffeomorphisms of $M$. Let us also denote by $\mathit{Diff}_1(M)$ the subgroup of homotopically trivial diffeomorphisms of $M$. I know that $\mathit{Diff}_0(M) \subsetneq \mathit{Diff}_1(M)$ in general and that they are the same for surfaces and some hyperbolic $3$-manifolds, but that's about all I know. Can we say more? In particular, I would like to know if $\mathit{Diff}_0(M)$ is always a subgroup of $\mathit{Diff}_1(M)$ such that the quotient is discrete? (maybe with assuming some extra conditions on $M$?) Thank you for your insights. REPLY [14 votes]: The quotient group $Diff_1(M)/Diff_0(M)$ is a discrete group since $ Diff_0(M)$ is a path component of $Diff(M)$, hence also a connected component since $Diff(M)$ is locally path-cconnected, and $Diff_1(M)$ is a union of components of $Diff(M)$, making the quotient discrete. [It is easy to see that $Diff_0(M)$ is a normal subgroup of $Diff(M)$, hence also of $Diff_1(M)$, so the quotient $Diff_1(M)/Diff_0(M)$ is indeed a group.] For closed 3-manifolds it is usually true that $Diff_0(M)=Diff_1(M)$, but there are some exceptions such as the connected sum of two suitably chosen lens spaces, if I'm remembering correctly. Dimension 4 is far more subtle and much less is known. I have a dim recollection that examples are known where $Diff_0(M)\neq Diff_1(M)$, but someone like Danny Ruberman would know for sure. In dimension 5 there are many simply-connected closed manifolds for which $Diff_0(M)= Diff_1(M)$ such as the sphere $S^5$, but this often fails in the nonsimply-connected case. For example when $M$ is the 5-dimensional torus the quotient group $Diff_1(M)/Diff_0(M)$ is the direct product of an infinite (but countable) number of cyclic groups of order 2. However for the connected sum of the 5-torus and $S^2\times S^3$ the quotient $Diff_1(M)/Diff_0(M)$ is trivial. These 5-dimensional phenomena persist in higher dimensions as well, and there are additional reasons why $Diff_0(M)$ can differ from $Diff_1(M)$ due to the existence of exotic spheres. For example $Diff_1(S^6)/Diff_0(S^6)$ is isomorphic to the group of exotic 7-spheres, a cyclic group of order 28. More generally $Diff_1(S^n)/Diff_0(S^n)$ is isomorphic to the group of exotic $(n+1)$-spheres whenever $n\geq 5$, and it seems that this group is almost always nontrivial. The known exceptions where no exotic $n$-spheres exist are $n\leq 3$ and $n=5$, $6$, $12$, $56$, and $61$. This list is complete up to $n=61$ except that the case $n=4$ is still unknown. The case $n=56$ was only added to the list in the past year due to recent calculations of Dan Isaksen and Zhouli Xu. There is a nice discussion of this question in an article by Milnor in the 2011 Notices of the A.M.S. called "Differential topology 46 years later", though this predates the discovery that 56 is also an exceptional dimension.<|endoftext|> TITLE: Why are two "random" vectors in $\mathbb R^n$ approximately orthogonal for large $n$? QUESTION [20 upvotes]: I saw that two random independent vectors are approximately orthogonal in high dimensional space. How can I prove this? And is there an intuitive explanation? Thank you. REPLY [4 votes]: One way to come at this is to try to stretch your intuition even farther, toward the Johnson-Lindenstrauss lemma, which says that while we can only fit $n$ orthogonal vectors into $\mathbb{R}^n$, we can easily (randomly) fit exponentially many almost-orthogonal vectors. The precise statement of the lemma includes the result you mention as a special case, so it's not really fair to say it provides the intuition. Rather I mention it because if you can fit it into your toolbox for thinking about high dimensions it will make other things more intuitive.<|endoftext|> TITLE: Is it possible to classify finite dimensional vector bundles in terms of Fredholm operators? QUESTION [13 upvotes]: Let ${\mathcal K}$ be the space of Fredholm operators on a Hilbert space. It is well known that ${\mathcal K}$ represents $K$-theory. Let ${\mathcal K}_0$ be the path component of ${\mathcal K}$ of operators of index zero. Then ${\mathcal K}_0$ is a model for the space that algebraic topologists usually call $BU$ - the classifying space of the infinite unitary group. My question is: is it possible to realise the filtration of $BU$ by the subspaces $BU(n)$ in terms of Fredholm operators? An obvious idea is to consider the subspace of ${\mathcal K}_0$ consisting of operators whose kernel/cokernel has rank at most $n$. Is this space weakly homotopy equivalent to $BU(n)$? Hopefully this question is interesting enough on its own (I am prepared to find out that it is stupid). My motivation is to understand better Jesse McKeown's answer to my previous question. This is my attempt to understand the statement that the space of subspaces of a Hilbert space of corank at most $n$ is a model for $BU(n)$. If there are other ways to make it precise, I would be very interested in learning about it. EDIT: I think that Tyler Lawson's negative answer to the previous question makes it very likely that the answer to this question is negative as well. REPLY [3 votes]: The problem with my remark from a couple of years ago is that the cokernel can also "move around". Let us therefore fix this. Let $W\subset \mathbb{H}$ be a finite, say $l$, dimensional subspace of a separable infinite dimensional Hilbert space. I assume it is known that the space of $l$ dimensional subspaces of the Hilbert space is a model for $BU(l)$. Let $\mathcal{O}^W=\{A\in \mathcal{K}_0\,|\, A(\mathbb{H})+W=\mathbb H\}$ be the space of Fredholm operators whose image is transverse to $W$. The space $\mathcal{O}^W$ is an open subspace of the space of Fredholm operators. There is a map $\mathcal{O}^W\rightarrow BU(l)$, that maps an operator $A$ to the subspace $A^{-1}(W)$. Note that the operator $A$ sends $A^{-1}(W)^\perp$ isomorphically to $W^\perp$. There are $GL(A^{-1}(W)^\perp,W^\perp)$ many such choices to do this. This suggest that $$ GL(\mathbb H)\rightarrow \mathcal O^W\rightarrow BU(l) $$ is a fiber bundle. But by Kuiper's theorem $GL(\mathbb H)$ is contractible, so we have the required (weak) homotopy equivalence. Now how does this fit into a grander story? You can take a sequence of finite dimensional $W_i$ that include in each other and $\mathbb{H}=\overline{\bigcup W_i}$. For example one can take $W_i$ to be the span of the first $i$ vectors in a basis of $\mathbb H$. Let us assume this. Then $\mathcal O^{W_i}\subset \mathcal O^{W_{j}}$ if $i\leq j$. This is the analogue of $BU(i)\subset BU(j)$. And $\bigcup _i \mathcal O^{W_i}=\mathcal K_0$. I should credit Dan Freed's lecture notes https://web.ma.utexas.edu/users/dafr/M392C-2015/index.html.<|endoftext|> TITLE: Relation of SFT and Gromov-Witten theory QUESTION [9 upvotes]: There's something that's been confusing me about symplectic field theory, and I'm not sure if I can formulate it exactly, but I'll try asking an imprecise question. Symplectic field theory, as described in the seminal paper (https://arxiv.org/abs/math/0010059) of Eliashberg-Givental-Hofer, appears superficially very similar to Gromov-Witten theory, in that the mathematical definitions of SFT correlators involve integration over moduli spaces of pseudoholomorphic curves. And in fact, SFT provides new ways of computing correlators in Gromov-Witten theory. If you told me that you had some sort of "field-theoretic" construction similar to Gromov-Witten theory which counted holomorphic curves, I might guess that this thing had the structure of a 2d field theory, as Gromov-Witten theory does. (Gromov-Witten theory has the structure of a 2D "cohomological field theory," as described in http://www.ihes.fr/~maxim/TEXTS/WithManinCohFT.pdf. As one ought to expect, this is a 2-dimensional sort of structure structure, having to do with the moduli space of stable curves.) But SFT is not a 2-dimensional field theory; in fact, SFT is exactly what it sounds like: it is a functor from a certain category of $2n$-dimensional symplectic cobordisms. In other words, SFT is a $(2n)$-dimensional symplectic field theory, where we don't seem to have specified anywhere a choice of $n.$ This strikes me as really weird! My imprecise question is twofold: first, where did the "2-dimensional" aspect of the theory go? And second, how can it be that the constructions in SFT produce a $2n$-dimensional "field theory" for all $n$? From my experience with TFT and QFT, I expect that interesting theories live in a single specified dimension and have something to do with that definition. What makes SFT so different? I have one guess, which is that in Gromov-Witten theory & related constructions, we specify a space $X$, and then to a surface $\Sigma$ the theory assigns a count of holomorphic maps $\Sigma\to X,$ and this has the locality properties on $\Sigma$ which are necessary to construct a topological field theory (or some related structure); but in SFT, somehow we are using a sort of locality property on the symplectic manifold $X$ which has something to do with contact hypersurfaces in $X$. If this is what's going on, then a slightly more precise version of my question is: what structure of a symplectic manifold allows us to do this? And why is SFT the only place where I've seen some version of this construction? REPLY [4 votes]: What follows is a guess. I don't know enough about Symplectic Field Theory to be sure. In perturbative string theory, one describes physics in the target spacetime $X$ by summing over maps from Riemann surfaces into $X$. These sums have an interpretation in terms of a 2d CFT living on the Riemann surfaces, but for physicists, the 2d constructions are auxiliary. They're primarily interested in physics in the target space (and indeed, there are constructions of string theory, such as Matrix theory and AdS/CFT, which make no mention of worldsheets). The physics on the worldsheet is local on the worldsheet, as you point out. The physics on the target spacetime $X$ isn't local in general -- that doesn't seem to be compatible with what we know of gravity -- but it does reduce to local quantum field theory on $X$ in some limits and special situations. The most obvious example is the low-energy limit, where the target space string theory can be approximated by a gauge theory on $X$ or some compactification of $X$. N=4 super Yang-Mills theory describes life in the 10d string theory spacetime as perceived by a 3-dimensional D3-brane embedded in the spacetime. (The 6 scalar fields on the 4d worldvolume parametrize the transverse fluctuations of the brane inside the 10d spacetime.) Perhaps more interesting to mathematicians: Witten's realization of Chern-Simons theory on $M$ as the string theory whose worldsheet CFT is the open-string topological A-model with target space $X = T^*M$ (and appropriate boundary conditions!). This is physically degenerate, but mathematically rather nice: The target space physics is not just local, but actually topological! You find this sort of construction everywhere in the string theory literature. (Since real string theory calculations are rather hard, most string theory 'consistency checks' involve finding two quantities that can be computed in QFT limits and checking that they match.) I suspect that symplectic field theory is an example of the third sort: A topologically twisted A-model CFT with appropriate boundary conditions gives rise to a target space string theory which is actually a TQFT, namely symplectic field theory. I'd be more certain of this if I knew exactly which boundary conditions were involved. The topological twist in the worldsheet CFT also explains the lack of dependence on the target dimension. For untwisted theories, there's a strong constraint on target space dimension. The 'conformal anomalies' of the 2d sigma model and 2d gravity have to sum to zero. In untwisted theories, the anomaly of the 2d sigma model is 3/2 $d$, with $d$ the dimension of the spacetime and the anomaly of 2d gravity is $-15$, leading to $d=10$. In topologically twisted theories, both anomalies vanish, so there's no constraint implied on target space dimension.<|endoftext|> TITLE: Reduced resultants and Bezout's identity QUESTION [16 upvotes]: Introduction: Given two univariate and coprime integer polynomials $f(x), g(x)$, we can always write \begin{equation} u(x)f(x)+v(x)g(x)=c \end{equation} for a unique pair of integer polynomials $u(x), v(x)$, and a unique positive integer $c$, such that $\deg u<\deg g$ and $\deg v<\deg f$. (This is a consequence of cancellation in the denominators in Bezout's identity, applied to $f$ and $g$ as elements of $\mathbb{Q}[x]$ and $c$ is the $lcm$ of the denominators of the unique pair of rational polynomials determined by Bezout's identity). However, the positive integer $c$ need not be the minimal integer satisfying such a relation—there may be integer polynomials $p(x)$ and $ q(x)$ of arbitrary degrees such that $$ p(x)f(x)+q(x)g(x)=d $$ with $d TITLE: What is meant by the inverse function theorem in algebraic geometry? QUESTION [19 upvotes]: I heard several times the inverse function theorem fails in algebraic geometry. Now I realize I'm pretty confused by this. This question has two parts. The first part asks for the correct formulation of synthetic inverse function theorems. The second asks for clarifications and intuitions about Penon's paper. Detailed references would be much appreciated - I only know about Kock's two texts, Lavedhomme's book, and Kostecki's notes on SDG and these do not discuss these things, and I don't know the algebraic geometry literature well enough to find this stuff... Apologies if this is all too elementary! Part I In Synthetic Geometry of Manifolds, Kock writes the inverse function theorem takes us "from infinitesimal invertibility to local invertibility", which sounds morally right. For "local invertibility" I can only think of one interpretation - $f$ étale at a point $x$ implies an open $U\ni x$ such that $f|_U$ is an isomorphism. For "infinitesimal invertibility" I can think of two options: (2a) the usual unique lifting property against infinitesimal neighborhoods; (2b) the differential $\operatorname{d}_x\!f$ being an isomorphism at all points. I'll go with the second option trying to parallel classic differential geometry. Following Kock's words, am I right to call the following condition "an inverse function theorem"? Condition IFT1. (2b)$\implies$(1). Or is this the wrong idea completely? Also, is the Henselian property all about the implication (2a)$\implies$(1)? I ask because of the following excerpt from section 2.3 of Néron Models, which, at least for schemes over fields, seems similar. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $k$. Let $S$ be the affine (local) scheme of $R$, and let $s$ be the closed point of $S$. From a geometric point of view, Henselian and strictly Henselian rings can be introduced via schemes which satisfy certain aspects of the inverse function theorem. Definition 1. The local scheme $S$ is called Henselian if each étale map $X\to S$ is a local isomorphism at all points of $X$ over $s$ with trivial residue field extension $k(x)=k(s)$. If, in addition, the residue field $k(s)$ is separably closed, $S$ is called strictly Henselian. Also, for a morphism $f:X\to S$ for schemes, $f$ is smooth at $x$ iff it's étale-locally a projection, i.e there's an open neighborhood $U\ni x$ such that $f|_U$ factors through an étale morphism followed by a canonical projection from $\mathbb A_S^n$. Replacing smooth with étale looks like an implicit function theorem and I think this is precisely what is meant by the Henselian property over a field. Is this correct, or is some other notion of "local isomorphism" is meant in the excerpt above? Part II This paper by Penon formulates a synthetic inverse function theorem and moreover claims that a morphism of equidimensional varieties is étale in several equivalent senses iff it satisfies the usual synthetic definition - the square below is a pullback. $$\require{AMScd} \begin{CD} TM @>{df}>> TN\\ @VVV @VVV\\ M @>>{f}> N \end{CD}$$ I don't read French and I can't make much sense of the pullback square he describes, but I know it only involves infinitesimal objects. Hence, it seems Penon's paper couldn't possibly address "local invertibility" in the sense of open neighborhoods. If so, what's the point of the paper? Note his notion are also used in this paper by Marta Bunge. Added. Here's Penon's Thesis. Again, from what I can make out, there's no mention of neighborhoods in the local context. REPLY [9 votes]: Penon's thesis "De l'infinitésimal au local" is often cited in this context. A rough overview of some things he does there (which I got from skimming through): He defines the infinitesimal nbhd of $x_0\in X$ to be all $x\in X$ such that $\neg \neg (x=x_0)$ on p.51. (I'm assuming there's a typo when he defines $\neg \neg \{x\}$, it should probably read $x'\sim x$, not $x' \neq x$.) He says $f:X\to Y$ is infinitesimally invertible at $x_0\in X$ if the restriction of $f$ to the infinitesimal nbhds of $x_0$ and $f(x_0)$ is bijective (p. 52). On p. 59 he shows that infinitesimal invertibility is equivalent to bijectivity of the tangent map under certain assumptions. He defines intrinsic (local) nbhds of $x_0$ on p. 84 to be subset $V\subset X$ s.t. $\forall x\in X \colon (\neg(x = x_0)\vee x \in V)$. Penon translates this into everyday language as "$x$ is discernible from $x_0$ or it is in V". He calls $f:X\to Y$ locally invertible at $x_0\in X$ if there exists an intrinsic nbhd. $V \subset X$ of $x_0$ s.t. $f(V)$ is an intrinsic nbhd. of $f(x_0)$ and $f|_V$ is injective. Here $f|_V$ is the restriction of $f$ to $V$. This is on p. 92. On p. 125 he proves that infinitesimal invertibility implies local invertibility for spaces that are opposites of finitely presented smooth rings. This seems to be inverse function theorem. Hope this helps for the first part of the question. I haven't read the other paper by Penon. You might also be interested in an upcoming book in english by Bunge "Synthetic Differential Topology", although a quick search only gave me a place where the inverse function theorem is used, no statement of it.<|endoftext|> TITLE: Is a Lebesgue measurable subgroup of $\mathbb{R}$ a Borel measurable set? QUESTION [12 upvotes]: Assume that $H$ is a Lebesgue measurable additive subgroup of $\mathbb{R}$. Is $H$ necessarily a Borel subset of $\mathbb{R}$? REPLY [15 votes]: No. Let $\langle A \rangle$ be a proper Borel subgroup of ${\mathbb R}$ generated by an algebraically independent Borel set $A$ with the cardinality of the continuum (such a set was constructed in J. v. Neumann, MR 1512442 Ein System algebraisch unabhängiger zahlen, Math. Ann. 99 (1928), no. 1, 134--141. , see also the discussion in Barthélemy Le Gac, MR 687640 Some properties of Borel subgroups of real numbers, Proc. Amer. Math. Soc. 87 (1983), no. 4, 677--680. ). By the Steinhaus lemma, $\langle A \rangle$ has measure zero, thus $\langle A' \rangle$ is a Lebesgue measurable subgroup for every $A' \subset A$. But this gives $2^{\mathfrak c}$ such subgroups, more than the cardinality of the Borel $\sigma$-algebra, so at least one of these subgroups must be non-Borel.<|endoftext|> TITLE: Permanent of a complete graph with negative cliques QUESTION [6 upvotes]: Let $K_n$ be the complete graph on $n$ vertices. Inside $K_n$ there are $k$ negative(edge weight is equal to -1) complete subgraphs $K_{n_1}, K_{n_2},...,K_{n_k}$, which are vertex disjoint. Let $A(G)$ denotes the adjacency matrix of graph $G$. What will be the permanent of $A(K_n)$?. With suitable relabeling of vertices we can write $A(K_n)=\begin{bmatrix} -A(K_{n_1}) & \\ & -A(K_{n_2})& & J\\ & & \ddots\\ J ^T& & & A(K_m) \end{bmatrix},$ where, $m=n-\sum_{i=1}^{k}n_i$ and $J$ denote space filled with all 1. REPLY [4 votes]: Using John Machacek's notation, which differs from the OP's, application of Ryser's formula gives $$\mathrm{perm}\; A_{m,n} = \sum_{s=0}^m \sum_{t=0}^n (-1)^{s+t} \binom ms\binom nt (t-s+1)^s (t-s)^{m-s} (t+s-1)^t (t+s)^{n-t}.$$ It might readily simplify to John's formula; I didn't try. I don't see how to reduce it to a 1-dimensional sum. The case with $k$ blocks instead of 2 blocks can be written as a similar $k$-fold sum by the same method.<|endoftext|> TITLE: If a Hopf Algebra has a nontrivial, finite-dimensional right ideal, then it is finite dimensional QUESTION [5 upvotes]: Let $H$ be a Hopf Algebra (over a field $K$), with comultiplication $\Delta$, counit $\varepsilon$, and antipode $S$. A $K$-subspace V is said to be: A right ideal if $VH \subseteq V$ A right coideal if $\Delta (V) \subseteq V \otimes H$ (some call this a left coideal...) Suppose that there is a nonzero right ideal $J$ in $H$, such that $\dim_K (J)$ is finite. Then $\dim_K (H)$ is finite. The proof breaks in two parts: If $V$ is a nonzero $K$-subspace of a Hopf algebra $H$ that is both a right ideal and a right coideal, then $V=H$. If $J\subseteq H$ is a finite-dimensional, nonzero right ideal of $H$, then $N:= H^* J = \{ f\cdot x = \sum x_{(1)}f(x_{(2)}) : f\in H^* , x\in J\}$ is finite dimensional, nonzero, and it is both a right ideal and a right coideal (thus $H=N$ is finite dimenisonal) Now, the first part is easy (you prove that there is $x\in V$ such that $\varepsilon (x) = 1$, then $1_H = \varepsilon(x)1_H = \sum x_{(1)}S(x_{(2)}) \in V$). My problem is the second part. I can prove that $N$ is a finite dimensional subspace: since $J$ is finite dimensional, we can assume it is contained in a finite dimensional subcoalgebra $D$ of $H$; then $N\subseteq D $. Does anyone have suggestions on how to prove that $N$ is a right ideal and a right coideal? REPLY [2 votes]: This is a part of Lemma 5.3.1 in Hopf Algebra: an introduction. Dascalescu, Sorin, Constantin Nastasescu, and Serban Raianu. Hopf algebra: An introduction. CRC Press, 2000.<|endoftext|> TITLE: Enumerative characterisation of boolean lattices QUESTION [5 upvotes]: The boolean lattice of rank $n$ (noted $B_n$) is the subset lattice of $\{1,2, \dots , n \}$. See the Hasse diagram of $B_3$ below: The Hasse diagram of $B_n$ is of length $n$, with $2^n$ vertices and $n2^{n-1}$ edges. Question: Is the converse true? In other words: Is a lattice with Hasse diagram of length $n$, with $2^n$ vertices and $n2^{n-1}$ edges, always boolean? If it is not true, it would be useful (to me) to know up to which rank $n$ it is true. REPLY [4 votes]: Define a ground set $X$ of size $2^{n-1}$. Now choose $2^{n-1}-(n-1)$ subsets of $X$, each of size at least 2, such that the sum of their sizes is $(n-2)2^{n-1}+2$ (so the average size is slightly more than $n-2$), and such that no two sets overlap in more than one element. Now take the poset consisting of the empty set, the singletons, the chosen subsets, and the full set $X$, ordered by inclusion, and then a chain upwards from $X$ so that the overall poset has height $n$. Because no two sets overlap in more than one element, the join of any two singletons in well-defined, and it follows immediately that the join is always well-defined. Since the poset is also bounded, it is a lattice. By construction, it has $2^n$ elements and $n2^{n-1}$ edges. We now go back and check for what $n$ it will be possible to choose a set system such as we have described. This isn't hard. For odd $n$, we can consider our ground set to be points in the plane $\mathbb F^2_{2^{(n-1)/2}}$. There are more than $2^{n-1}$ affine lines in this plane, no two intersect in more than one point, and each contains at least $2^{(n-1)/2}$ points of our ground set. (This produces sets much bigger than we need, but if the sum of the sizes of our sets is too big, we can throw away elements from the sets until they are small enough. The intersection property will obviously be preserved.) Since $2^{(n-1)/2}$ is larger than $n-1$ if $n$ is at least 7, we are certainly done for all odd $n$ at least 7. A one-off construction for $n=5$ seems easy enough. Even $n$ should also be straightforward. I don't know whether or not my construction works for $n=4$, but it should have no trouble starting at $n=6$. Edited to add: In rank 0, there is one vertex (the empty set). In rank 1, there are $2^{n-1}$. In rank 2, $2^{n-1}-(n-1)$. In ranks 3 to $n$, there is one each. The total number of vertices is therefore $2^n$.<|endoftext|> TITLE: Ideal generated by two univariate, coprime, integer polynomials QUESTION [5 upvotes]: Let $f(x)$, $g(x)$ be two univariate, coprime, integer polynomials and let $I=\big(f(x),g(x)\big)$ the ideal of $\mathbb{Z}[x]$ generated by $f, g$. Let $I \cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which can be expressed as linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. $I \cap \mathbb{Z}$ is clearly an ideal in $\mathbb{Z}$. The following argument shows that, given two coprime integer polynomials $f,g$, the ideal $I \cap \mathbb{Z}$ of $\mathbb{Z}$ has always non zero elements: If $f,g$ are considered as elements of $\mathbb{Q}[x]$, Bezout's identity tells us that there exists a pair of unique rational polynomials $U(x)$, $V(x)$ with $\deg U<\deg g$, $\deg V<\deg f$ such that $$ U(x)f(x)+V(x)g(x)=1 $$ Thus, clearing denominators in the above identity, we get that there exists a pair of unique integer polynomials $u(x)$, $v(x)$ with $\deg u<\deg g$, $\deg v<\deg f$ such that $$ u(x)f(x)+v(x)g(x)=c $$ where $c$ is the $lcm$ of the denominators of $U,V$ and: $u=cU$, $v=cV$. Thus: $0 \neq c\in I \cap \mathbb{Z}$. Question $1$: Can we determine a sufficient and necessary condition such that: $\mathbb{Z}\subset I \cap \mathbb{Z}$? Question $2$: Is there a general method for determining the least positive generator of $I \cap \mathbb{Z}$ ? References: Reduced resultants and Bezout's identity The resultant and the ideal generated by two polynomials in $\mathbb{Z}[x]$ Reduced resultant of monic polynomials P.S.: The second question above, is actually part of the question posted in the first of the references above. However, it is posted here rather as a problem of commutative algebra than a problem of number theory. REPLY [5 votes]: For question 2, compute a Gröbner basis over $\mathbb{Z}$ for the ideal generated by $f(x)$ and $g(x)$, which gives the required generator. You can do this easily in SageMathCloud (available free to use). The algorithm behind this computation is exactly what you would be doing by hand to find the generator. This also answers question 1, if you allow an algorithm to provide a necessary and sufficient condition.<|endoftext|> TITLE: Which reals are "hyperarithmetic modulo ordinals"? QUESTION [8 upvotes]: The context for this question is the theory ZFC + a measurable cardinal, although answers not in this context would also be interesting to me. In a project I'm working on, the following class of reals has emerged, and I'd like to understand it better: Say that a real $r$ is hyperarithmetic modulo ordinals (and write $r\in\Delta_1^{1, ord}$) if $r$ can be defined in a $\Delta^1_1$ way relative to some ordinal parameters. Formally, $r$ is $\Delta_1^{1, ord}$ if there is a tuple $\overline{\alpha}$ of ordinals and a pair of formulas $\varphi(x, \overline{y}), \psi(x, \overline{y})\in\Sigma^1_1$ in only the displayed parameters such that, whenever $\overline{c}$ is a tuple of reals coding copies of $\overline{\alpha}$, we have $$r=\{n: \varphi(n, \overline{c})\}=\{n: \neg\psi(n, \overline{c})\}.$$ Now, this class of reals is much bigger than $\Delta^1_1$. For example, Kleene's $\mathcal{O}$ is $\Delta_1^{1, ord}$: $\Phi_e$ is well-founded iff it embeds into $\omega_1^{CK}$. EDIT: as soon as I posted this, I realized that this can be pushed further: unless I'm missing something, every constructible real is $\Delta_1^{1, ord}$. So the right question now is: Is $\Delta_1^{1, ord}=L\cap\mathbb{R}$? Currently I suspect the answer is "yes" - note that $\Delta_1^{1, ord}$ is forcing absolute! - but I don't see how to prove it. A more conservative question is: Is there a $\Delta^1_3$ real which is not $\Delta^{1, ord}_1$? In general, any information about this class (and its obvious variations - e.g. $\Sigma^{1, ord}_n$) would be valuable to me. I suspect this is all very well-known in the descriptive set-theory community, so I've added the "reference-request" tag. REPLY [9 votes]: In fact $\Delta^{1, ord}_1 = \mathbb{L} \cap \mathbb{R}$. For suppose $(\phi(x, \overline{y})$, $\psi(x, \overline{y})$, $\overline{\alpha})$ is as you describe, defining $r \subset \omega$. We show $r \in \mathbb{L}$. For convenience we suppose $\overline{\alpha} = \alpha$ is a single ordinal; this is no loss, by coding. Now if $\alpha$ is countable in $\mathbb{L}$ we are done by Shoenfield. If $\alpha$ is not countable in $\mathbb{L}$ we proceed as follows. Let $\mathbb{P}$ be the usual forcing notion for adding a surjection $f: \omega \to \alpha$. Let $G$ be $\mathbb{P}$-generic over $\mathbb{V}$; then $G$ is also $\mathbb{P}$-generic over $\mathbb{L}$ so we can consider $\mathbb{L}[G] \subset \mathbb{V}[G]$. Now, by Shoenfield, we have that in $\mathbb{V}[G]$, whenever $c$ codes $\alpha$, then $\{n: \phi(n, c)\} = r$. By Shoenfield again, this remains true in $\mathbb{L}[G]$. Henceforth we work in $\mathbb{L}$. Since $G$ could have been chosen to contain any given $p \in \mathbb{P}$, we have shown that for each $n$, $\mathbb{P}$ forces that $\phi(n, \dot{c})$ holds iff $n \in r$ (where $\dot{c}$ is a generic code for $\alpha$). Hence $r$ is definable in $\mathbb{L}$.<|endoftext|> TITLE: Violating the Lebesgue density theorem QUESTION [8 upvotes]: Can anyone exhibit a finite-dimensional metric space (preferably, $R^d$) equipped with a measure that does not satisfy the conclusions of the Lebesgue Density Theorem? Such examples exist in infinite-dimensional spaces (e.g., Hilbert space with the Gaussian measure) but what about a finite-dimensional one? REPLY [13 votes]: It is a theorem of Besicovitch that measures on $\mathbb R^d$ do satisfy the density theorem. Fremlin, Measure Theory, Chap. 47 added Besicovitch, around 1930, extended his density properties of sets to those of finite Hausdorff measure. source next: D. G. Larman, "A new theory of dimension", Proc. London. Math. Soc. 17 (1967) 178-192 Def: a metric space is finite-dimensional in the sense of Larman iff there is a constant $K$ such that every ball of radius $2R$ can be covered by at most $K$ balls of radius $R$. Larman proves that such finite-dimensional spaces have a Vitali-type property, which of course implies the density theorem for all measures. (Lebesgue's and Besicovitch's proofs used Vitali coverings.) My student Manav Das investigated metric spaces with various Vitali-type properties. For example: Nonlinear Anal. 46 (2001) 457-463 Real Analysis Exchange 27 (2001/02) 7-15<|endoftext|> TITLE: Higher Chow groups: a basic case QUESTION [6 upvotes]: Let $Y$ be a smooth irreducible variety over a field $k$. I know that $\mathcal{O}_Y^\times$, the group of invertible functions on $Y$, embeds in the higher Chow group $\mathrm{CH}^1(Y,1)$ [or the first graded piece of $K_1(Y)$]. Is this always an isomorphism? I know it is when $Y=\mathrm{Spec}(k)$. If it is always an isomorphism, can you provide a reference? If it is not, can you provide a counterexample? REPLY [2 votes]: See the precise statement in Corollary 4.2 and Theorem 19.1 in Mazza-Voevodsky-Weibel. Lecture notes on Motivic Cohomology.<|endoftext|> TITLE: Is an ordinary scheme in Borger's Absolute Geometry the same as a "scheme over F1" with a map to Spec(Z)? QUESTION [11 upvotes]: $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\F}{\mathbb{F}_1}$ $\newcommand{\spec}{\operatorname{Spec}}$ If I understand correctly, in Borger's paper about the field with one element, the category of "affine schemes over $\spec\F$" is just the opposite of the category of $\Lambda$-rings. The base change from $\spec\F$ to $\spec\Z$ is forgetting the $\Lambda$-structure. i.e. we can view a $\Lambda$-structure on a commutative ring as a descent data from $\spec \Z$ to $\spec\F$ (there is ageneralization to the non-affine case, but let's keep it simple for now). The base change has adjoints from both sides. The one that comes from the composition $$\operatorname{Spec}R \to \operatorname{Spec}\mathbb{Z} \to \operatorname{Spec}\mathbb{F}_1$$ Takes a ring $R$ to the $\Lambda$-ring of (big) Witt vectors over $R$ denoted $W(R)$. Now, it seems to me that a desirable property of this setup would be that an (affine) scheme over $\spec\Z$ would be "the same as" an (affine) scheme over $\spec\F$ with a morphism to $\spec\Z$ over $\spec\F$. So for a $\Lambda$-ring $R$, a map of $\Lambda$-rings $W(\Z)\to R$ should be induced by a ring $R_0$ by applying $W$ to the structure morphism $\Z \to R_0$. Moreover, there should be a natural bijection between the two types of data (up to isomorphisms in the obvious way). It seems that the elements of $R_0$ should be something like elements of $R$ of rank $\le 1$ (i.e. those for which $\lambda^n$ vanish for n $> 1$) and the sum should be something like the "rank one approximation of the sum in $R$). Question: Is this indeed true? Perhaps I am wrong in assuming that this is desirable, so an explanation of that would be great too. By the way, I am not an expert on $\Lambda$-rings and I ask this purely out of curiosity. REPLY [15 votes]: I guess I should be the one to answer this! Unfortunately, the answer is no. There is a natural adjunction between the two types of data: given a $\Lambda$-ring $R$, set $R_0$ to be $R\otimes_{W(\Z)}\Z$, where the map $W(\Z)\to\Z$ is the projection on the first component, i.e. the co-unit of the adjunction between $W$ and the forgetful functor from $\Lambda$-rings to rings. (This is as in Will Sawin's comment.) For instance, if $R$ is already of the form $W(S)$, then $R_0$ would be $W(S)\otimes_{W(\Z)}\Z$, which does indeed map to $S$ but not isomorphically, in general. It is if $S$ is etale over $\Z$, but it fails for $S=\Z[x]$, if I remember. (Also, note that $R_0$ is naturally a quotient ring of $R$, not a subring, as you proposed in your question. This is as it should be, since $S$ is a quotient ring of $W(S)$, not a subring.) But it's also true that there are some $R$'s that are not of the form $W(S)$ for any $S$. To see this it's nice to warm up with a toy example. Instead of considering $\Lambda$-structures on rings, let's consider $G$-actions, where $G$ is a monoid. Then the analogue of the Witt vector functor is given by $S\mapsto S^G$, where $G$ acts on $S^G$ by translation in the exponent. (In fact, this is more than an analogy. If $G$ is the monoid $\Z_{\geq 1}$ of integers $\geq 1$ under multiplication, then a $G$-action is also called a $\Psi$-ring structure, and the analogue of the Witt vector functor $S\mapsto S^G$ is the ghost-component functor.) Then a $G$-equivariant $S^G$-algebra $R$ is canonically of the form $R=\prod_{g\in G} R_g$ and an element $h\in G$ sends $R_g$ to $R_{hg}$. If $G$ is a group, then these maps are isomorphisms and hence all the $R_g$ can be recovered from $R_1$. More precisely, $R_1$ is exactly $R\otimes_{\Z^G}\Z$, and the map $R\to (R\otimes_{\Z^G}\Z)^G$ is an isomorphism. But if $G$ is not a group, then the maps $R_g\to R_{hg}$ are not in general isomorphisms. For instance, you can take $G=\mathbb{N}$ (under addition) and $R_0$ a nonzero ring but $R_n$ the zero ring for all $n\geq 0$. In fact, as alluded to above, a $\Lambda$-structure on $R$ is just an action of the monoid $\Z_{\geq 1}$ such that certain congruence conditions are satisfied. (More precisely, this is true if $R$ is torsion-free.) So if $R$ is a $\mathbb{Q}$-algebra, then all congruence conditions are vacuously true, and so a $\Lambda$-structure is equivalent to an action of the monoid $\Z_{\geq 1}$. So then we're really in monoid land, and we can make a counterexample as above, by setting $R_1=\mathbb{Q}$ and $R_n=\{0\}$ for all $n>1$, and $R=\prod_{n\geq 1} R_n$. Then the map $R\to (R\otimes_{\Z^G}\Z)^G$ is identified with the diagonal map $\mathbb{Q}\to \mathbb{Q}\times\mathbb{Q}\times\cdots$. Another way of putting this is that the answer to your question is already no if we work with the ghost-component functor, instead of the Witt vector functor, because the Adams operators / Frobenius maps are not necessarily isomorphisms. But you might then ask your question again in the world of 'perfect $\Lambda$-rings' and 'perfect Witt vectors'. These are where you require the Adams / Frobenius maps to be isomorphisms. I don't see the answer immediately, but it shouldn't be too hard to work out.<|endoftext|> TITLE: Reference request to proof that H$^2(\Gamma, \mathbb{Q}/\mathbb{Z}) = 0$ QUESTION [6 upvotes]: Does anyone maybe have a reference to the proof of the following result by Tate? Let $\Gamma$ be the absolute Galois group of the rationals. Then the second cohomology group (for trivial $\Gamma$-action) H$^2(\Gamma, \mathbb{Q}/\mathbb{Z})$ is trivial. Unfortunately I couldn't find it online or in the library. Any help would be greatly appreciated! Kind regards! REPLY [6 votes]: By the Galois cohomology long exact sequence, this is isomorphic to $\operatorname{H}^3(\Gamma,\mathbb{Z})$, and the vanishing of this is Chapter I, Corollary 4.17 in Milne's Arithmetic Duality Theorems.<|endoftext|> TITLE: Condition for quasi-splitting of special orthogonal group QUESTION [7 upvotes]: Let $\text{SO}_{p,q}$ be the special orthogonal group of type $(p,q)$. In D. Prasad's lecture notes on algebraic groups the statement $$\text{"The group $\text{SO}(p, q)$ is quasi-split if and only if $|p − q| ≤ 2$."}$$ is given without proof or further explanation. Does anyone have a reference to the proof of this result? Any help would be greatly appreciated. N.B.: I have already (unsuccesfully) asked a similar question on math.stackexchange REPLY [2 votes]: Since the OP asked for a reference, I would like to provide some. First of all, a group is quasi-split iff its Satake diagram does not contain black dots. The list of Statake diagrams can be found in many places but most of them are difficult to understand out of context. Of the more useful kind are the tables in Onishchik, A. L.; Vinberg, Ėrnest Borisovich (1994), Lie groups and Lie algebras III: structure of Lie groups and Lie algebras For the theory, I found Section 29 of Daniel Bumps book on Lie groups quite useful (see in particular p. 294ff)<|endoftext|> TITLE: The "right" topological spaces QUESTION [22 upvotes]: The following quote is found in the (~1969) book of Saunders MacLane, "Categories for the working mathematician" "All told, this suggests that in Top we have been studying the wrong mathematical objects.The right ones are the spaces in CGHaus." CGHaus is the category of compactly generated Hausdorff spaces. It is advocated that it is a better category than Top because it is cartesian closed. Almost 50 years later, what is the general feeling on that point? Is CGHaus the "right" topology to study algebraic topology (which is what MacLane is interested in)? Is there a better choice? Or no consensus on the question? Is CGHaus used in other fields of mathematics? REPLY [12 votes]: I'm a little late to the party. I agree with Dennis Nardin's comment that really any convenient category of spaces should be good for homotopy theory. By this, I mean any complete and cocomplete cartesian closed full subcategory of $\mathsf{Top}$ which includes the CW complexes and whose limits and colimits are not too far from those in $\mathsf{Top}$; in particular, homotopy groups of CW complexes should not change from those in $\mathsf{Top}$. $k$-spaces and its variants are rather large categories as these convenient categories go.[1] But sometimes you actually want to work with a smaller category of topological spaces -- usually asking for there to be a universal way to turn any topological space into one in your category without changing the weak homotopy type. For example, sequential spaces or delta-generated spaces, or variants with separation conditions, are good options for point-set level categories. These smaller categories have the advantage, unlike $k$-spaces and variants, of being locally presentable, which gives even better categorical control, and in particular allows all the techniques of combinatorial model categories to be brought to bear. An additional advantage of delta-generated spaces is that connected components are the same as path components, and there is a connected component functor, which is left-adjoint to the discrete space functor. There is a general recipe which allows one to cook up convenient categories of topological spaces. Start with a subcategory $\mathcal{C} \subseteq \mathsf{Top}$, and consider the category $\mathsf{Top}_\mathcal{C}$ which is the closure of $\mathcal{C}$ under colimits in $\mathsf{Top}$ [2]. If $\mathcal{C}$ consists of locally compact Hausdorff spaces [3] and is closed under finite products in $\mathsf{Top}$ [4], then $\mathsf{Top}_\mathcal{C}$ will be cartesian closed, and coreflective in $\mathsf{Top}$, so its colimits are computed as in $\mathsf{Top}$, and limits are computed by taking the ordinary limit and then making the topology finer to land back in $\mathsf{Top}_\mathcal{C}$. So as long as $\mathsf{Top}_\mathcal{C}$ includes the unit interval, it will also contain all CW complexes and have the right homotopy groups thereof, and hence be a convenient category of topological spaces. If in addition $\mathcal{C}$ is small, then $\mathsf{Top}_\mathcal{C}$ will have the added benefit of being locally presentable. If desired, additional separation conditions can also be added without much fuss. Delta-generated spaces are $\mathsf{Top}_\mathcal{C}$ when $\mathcal{C}$ is the category of topological simplices [5]. Sequential spaces are $\mathsf{Top}_\mathcal{C}$ when $\mathcal{C}$ consists of all metric spaces [6]. And $k$-spaces are $\mathsf{Top}_\mathcal{C}$ when $\mathcal{C}$ is all compact Hausdorff spaces. [1] To be fair, it's maybe an unnatural restriction to require the category to be a subcategory of $\mathsf{Top}$ - e.g. a supercategory like quasitopological spaces might do as well. But I'll let others discuss these options. [2] Another way to describe $\mathsf{Top}_\mathcal{C}$, familiar in the case of $k$-spaces, is the following. If $X$ is a space, say that $U\subseteq X$ is $\mathcal{C}$-open iff for every $C \in \mathcal{C}$ and map $f: C \to X$, the set $f^{-1}(U)$ is open. Then $X \in \mathsf{Top}_\mathcal{C}$ iff every $\mathcal{C}$-open subset of $X$ is open. An equivalent description (again familiar in the case of $k$-spaces) says that $X \in \mathsf{Top}_\mathcal{C}$ iff for every $Y \in \mathsf{Top}$ and every function $f: X \to Y$, if the induced map $\mathsf{Top}(C,X) \to \mathsf{Top}(C,Y)$ is continuous for every $C \in \mathcal{C}$, then $f$ is continuous. Finally, $\mathsf{Top}_\mathcal{C}$ is also equivalently described as the closure of $\mathcal C$ in $\mathsf{Top}$ under coproducts and quotients. [3] Or more generally, exponentiable spaces. [4] This condition can be weakened to say that finite products in $\mathsf{Top}$ of objects of $\mathcal{C}$ are at least in $\mathsf{Top}_\mathcal{C}$. [5] Or equivalently, you could take $\mathcal{C}$ to be the category of CW complexes, or the category of manifolds, or you even just the one-object category consisting of the unit interval, or the real line; in all these cases $\mathsf{Top}_\mathcal{C}$ will still be exactly the delta-generated spaces. [6] Or equivalently, you could take $\mathcal{C}$ to be the category of first-countable spaces, or the category of second-countable spaces, or just the one-object category consisting of the one-point compactification of the natural numbers.<|endoftext|> TITLE: Enumerative characterisation of boolean lattices II QUESTION [7 upvotes]: This is a sequel of this post. The boolean lattice $B_n$ is graded with rank numbers $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$, and $n2^{n-1}$ edges. Question: Is a graded lattice with the above rank numbers and $n2^{n-1}$ edges, equivalent to $B_n$? If it is not true, it would be useful (to me) to know up to which $n$ it is true. REPLY [3 votes]: True for $n \le 4$. For $n \le 3$ it is straightforward. For $n=4$, inspect all 16-element vertically indecomposable graded lattices (listed here). The list has 1900132 lattices; of them, 103411 have rank sequence 1,4,6,4,1, and 249 have exactly 32 edges. But only 1 lattice has both, that's the $B_4$.<|endoftext|> TITLE: Is hyperbolicity a Zariski open condition? QUESTION [7 upvotes]: Let $f:X\to T$ be a smooth family of complex projective varieties, and assume that the special fiber $X_{t_0}$ is hyperbolic for $t_0\in T$. Is $X_t$ also hyperbolic for all $t\in U$ in a Zariski open neighborhood $U\subseteq T$ of $t_0$? Here we say that $X$ is (Brody) hyperbolic if there is no non-constant holomorphic map $f:\mathbb C\to X$. REPLY [5 votes]: Kobayashi hyperbolicity (or Brody hyperbolicity, the two notions coincide for compact complex spaces), is an open condition with respect to the analytic topology. Thus, for instance, once you find an example of hyperbolic compact complex space, you also get that small deformations are hyperbolic, too. This fact is a direct consequence of the celebrated Brody's lemma. A proof can be found, as already said, in the comprehensive book by Kobayashi or in the Santa Cruz lecture notes by Demailly. There is another notion of hyperbolicity, which as been introduced by Demailly in his Santa Cruz lecture notes, whose name is algebraic hyperbolicity. This notion is weaker than Kobayashi's one. Roughly speaking, a compact complex space $X$ is algebraically hyperbolic if you can control from below minus the Euler characteristic of (the normalization of) every compact complex curve in $X$ by a (universal constant times) its degree (with respect to some polarization, or more generally, to some hermitian metric). This notion also satisfies an openness condition. Namely, let $\mathcal X\to S$ be an algebraic family of projective algebraic manifolds (i.e. given by a projective morphism). Then the set of $s\in S$ such that the fiber $X_s$ is algebraically hyperbolic is open with respect to the “countable Zariski topology” of $S$ (by definition, this is the topology for which closed sets are countable unions of algebraic sets). Quoting Demailly himself, it would be interesting to know whether algebraic hyperbolicity is open with respect to the euclidean topology; still more interesting would be to know whether Kobayashi hyperbolicity is open for the countable Zariski topology (of course, both properties would follow immediately if one knew that algebraic and Kobayashi hyperbolicity coincide, but this seem otherwise highly non trivial to establish). The latter openness property has raised an important amount of work around the following more particular question: is a (very) generic hypersurface $X\subset\mathbb P^{n+1}$ of degree $d$ large enough (say $d\ge 2n + 1$) Kobayashi hyperbolic? Again, “very generic” is to be taken here in the sense of the countable Zariski topology. Very recently, this question seems to have been answered in the affirmative in this paper by D. Brotbek. Perhaps, it worths noticing that Brotbek obtains the hyperbolicity for the generic hypersurface, and not only very generic. This might be a peculiar property of families of projective hypersurfaces, but it is not completely clear to me how to fit it in the general picture.<|endoftext|> TITLE: Amenable inverse limits of torsionfree amenable groups QUESTION [8 upvotes]: Let $$ \cdots \to \Gamma_n \to \Gamma_{n-1} \to \cdots \to \Gamma_0$$ be an inverse system countable groups and let's assume (for this post) that all homomorphisms in such an inverse system are surjective. We say that $\Gamma$ is an inverse limit if there exists compatible surjective homomorphisms $\varphi_n \colon \Gamma \to \Gamma_n$, such that $\varphi_n(g)\neq e$ for every $g \neq e$ eventually. Question: Is there an inverse system of torsionfree amenable, such that all inverse limits are non-amenable? What about the inverse system of free nilpotent or free solvable groups? I conjecture that the answer to the question is yes, but it seems hard to find an example. REPLY [8 votes]: Suppose that $\Gamma_n$ is the free nilpotent group on $k$ generators of nilpotency class $n$. Any finite $k$-generated $p$-group ($p$ a prime) is nilpotent, hence the free $k$-generated pro-$p$ group is surjected by the (universal) inverse limit of $\Gamma_n$. Hence any inverse limit of $\Gamma_n$ maps to a dense subgroup of the free pro-$p$ group on $k$ generators. But the free pro-$p$ $k$-generator group surjects a non-solvable linear group for large enough $k$ (say $ker\{ GL_m(\mathbb{Z}_p)\to GL_m(\mathbb{Z}/p)\}$, $m\geq 2$). This group is linear and not virtually solvable, hence any of its dense subgroups has a free subgroup by the Tits alternative. So any inverse limit of the free $k$-generated nilpotent groups will have a free subgroup, hence is not amenable. Addendum: In fact, we may take $k=2$ above, since any finite-index subgroup of $SL_n(\mathbb{Z})$, $n\geq 3$, contains a finite-index rank 2 subgroup.<|endoftext|> TITLE: Does flipping Young diagrams has anything to do with Fourier? QUESTION [5 upvotes]: Here's a chaser to this question. Recall the proof that the number of partition of an integer $n$ into at most $k$ addends is the same as the number of partition of an integer $n$ into integers no larger than $k$: you flip the corresponding Young's diagrams (drawn in French notation) over the main quadrant diagonal. Is this flipping operation related to a Fourier transform of something out there? You can also add a couple of binary operations on Young diagrams: "multiplication" and "convolution". "Multiplication" draw the 1st diagram above the 2nd one and drops the columns to close the gaps: $(a*b)_i=a_i+b_i$, where $a_i,b_i$ are $i^{th}$ addends in the partitions. Similarly, "convolution" draw the 1st diagram on the right of the 2nd one and "pulls" rows to the left until the gaps are closed. Clearly, the above "flipping operation" interchanges multiplication and convolution. Is that property enough to model the "flipping operation" as a Fourier transform on something? REPLY [7 votes]: A sort of answer: There are two ways to realise the Springer correspondence in the language of perverse sheaves: one by restriction (see [BM]), and one by Fourier transform (see [HK]). [BM] Borho, Walter; MacPherson, Robert Partial resolutions of nilpotent varieties. Astérisque 101-102, 23-74 (1983). [HK] Hotta, R.; Kashiwara, M. The invariant holonomic system on a semisimple Lie algebra. Invent. Math. 75, No. 2, 327-358 (1984). It turns out that these two correspondences are related by tensoring with the sign representation which in type A is "flipping the young diagram". Thus it is not too much of an exaggeration to say that under the Springer correspondence for $S_n$, flipping the Young diagram is the Fourier transform. PS: I haven't thought if this explains your "multiplication" and "composition" operations...<|endoftext|> TITLE: Homotopy function complex for quasi-categories QUESTION [6 upvotes]: The model structure on sSet for quasi-categories (the Joyal model structure) is not enriched over sSet with the Quillen model structure, so ordinary internal Hom of simplicial sets is not a correct homotopy function complex of quasi-categories. We can give a correct definition using either cosimplicial or simplicial resolutions, which can be defined by the small object argument as usual. But we can give a more explicit description of such resolutions. For example, we can define a cosimplicial frame on the terminal simplicial set by taking nerves of groupoids $\{ 0 \simeq \ldots \simeq n \}$. Then a cosimplicial frame on a simplicial set can be defined as the cartesian product with this cosimplicial object. This construction can be found, for example, in [1]. The problem with this construction is that it produces too large function complex. For example, a 1-simplex of $\mathrm{hMap}(\Delta^0,X)$ is a 1-simplex of $X$ together with an inverse and higher simplices that guarantee that they are inverses. It seems that there should be a definition of a homotopy function complexes such that 1-simplices in the example above are just 1-simplices of $X$ which are equivalences. The obvious way to define such homotopy function complex is to take the largest Kan complex contained in the internal Hom. It seems that we also can define a simplicial resolution which produces the same homotopy function complex. Can we construct homotopy function complexes in such a way? Were such constructions described in the literature? [1] Daniel Dugger and David I. Spivak, MR 2764043 Mapping spaces in quasi-categories, Algebr. Geom. Topol. 11 (2011), no. 1, 263--325. REPLY [8 votes]: Yes, you can compute the mapping spaces in ∞-categories by taking the biggest Kan subcomplex of the internal hom. The trick is not to use the Joyal model structure, but instead the model structure on marked simplicial sets defined in Higher Topos Theory, proposition 3.1.3.7 (in the case $S=\Delta^0$). By proposition 3.1.4.1 the fibrant objects are exactly the quasicategories with the equivalences marked, by corollary 3.1.4.4 it is a simplicial model category and its mapping spaces are defined exactly as you describe and by theorem 3.1.5.1 this is Quillen equivalent to the Joyal model structure in a way that does the obvious thing on fibrant objects. REPLY [6 votes]: You certainly can take the maximal Kan complex of the quasicategory of maps between quasicategories (note that the mapping quasicategory already has all of the information you want -- you're just forgetting the non-invertible 1-morphisms in order to get at the underlying space). The standard way of replacing $s\mathcal{S}et_{\mathrm{Joyal}}$ with a Quillen equivalent model category enriched over $s\mathcal{S}et_\mathrm{Quillen}$ (i.e. a simplicial model category) is to use "marked simplicial sets," described e.g. in Riehl's Categorical Homotopy Theory. A marked simplicial set is a simplicial set along with a distinguished collection of 1-simplices (always including the degenerate simplices) which are regarded as "equivalences". We turn quasicategories into marked simplicial sets by marking the 1-simplices which were already equivalences in the quasicategory. The reason why marked simplicial sets form a simplicial model category essentially follows from the fact that marked simplicial sets are a subcategory of the category of presheaves on a certain indexing category $\Delta^+$, which looks just like $\Delta$ except we factor $s_0:[1]\to[0]$ through an intermediary object $e$. 1-simplices in the image of $X_e\to X_1$ are understood to be the "equivalences". Note that in this case, our cosimplicial frame is the standard one in simplicial sets, with each 1-simplex marked. As you're probably aware, substituting the Joyal model structure for a Quillen equivalent simplicial model category gives us access to a number of technical tools and constructions; for example, we can use weighted limits to calculate homotopy limits as Riehl describes in her book. I've left a lot of details hazy. If you have further questions, I'll try to clear them up, but you may just want to consult the book I've mentioned, which is available here: http://www.math.jhu.edu/~eriehl/cathtpy.pdf<|endoftext|> TITLE: Dyadic Hilbert symbols and higher unit groups QUESTION [10 upvotes]: Let $F$ be a local dyadic number field, $\mathfrak{p}$ its maximal ideal, $(*,*)_F$ its quadratic Hilbert symbol and $e$ its ramification index (i.e. $\mathfrak{p}^e$ is exact divisor of $2$). Fix an even $s\le 2e$. What is then the smallest $t\ge 0$ such that $(U_s,U_t)_F=1$ (i.e. such that for all $a\in U_s$ and $b\in U_t$, we have $(a,b)_F=1$). Here $U_s$ are the $s$-units, i.e. $U_s=1+\mathfrak{p}^s$ for $s\ge 1$, and $U_0$ are the units of $K$. By the Local Square Theorem the elements of $U_{2e+1}$ are squares, hence such a $t$ exists. By the very last exercise in Serre's "Local Fields" we know that $t\le 2e-s$ (in fact, I do not have a proof for this and would be grateful for a complete reference). Is always $t=2e-s$ (which is true for $\mathbb Q$), or are there fields and even $s<2e$ with $t<2e-s$? I know that there are explicit formulas for dyadic Hilbert symbols (Vostokov/Letsko, Henniart, ...) which possibly would enable me to work out an answer to my question. However, to become comfortable with these formulas seems to be not so obvious and I would be happy for a reference or any hint to a more conceptual proof avoiding such formulas. REPLY [5 votes]: More generally, allow $p$ to be any prime and let $K$ be a finite extension of $\mathbf{Q}_p$ containing a primitive $p$-th root $\zeta$ of $1$ (which is automatic when $p=2$). The ramification index $e$ of $K$ over $\mathbf{Q}_p$ is divisible by $p-1$ (because $K$ contains $\mathbf{Q}_p(\zeta)$ by hypothesis); define $e_1$ by $e=(p-1)e_1$. A little work as in Chapter 15 of Hasse's Number Theory (or as in Section V of Local discriminants) allows you to determine the structure of the filtered $\mathbf{F}_p$-space $K^\times\!/K^{\times p}$. The filtration on this quotient comes from the filtration $$ \cdots U_2\subset U_1\subset\mathfrak{o}_K^\times\subset K^\times $$ on $K^\times$, where $\mathfrak{o}_K$ is the ring of integers of $K$, with unique maximal ideal $\mathfrak{p}_K$, and, for every $i>0$, $U_i=1+\mathfrak{p}_K^i$ is the kernel of $\mathfrak{o}_K^\times\to(\mathfrak{o}/\mathfrak{p}_K^i)^\times$. Denote the image of $U_i$ in $\overline{K^\times}=K^\times\!/K^{\times p}$ by $\bar U_i$. Then the image of $\mathfrak{o}_K^\times$ is $\bar U_1$, we have $\bar U_{pe_1+1}=\{1\}$, and the filtration on $\overline{K^\times}$ looks like $$ \{1\} \subset_1\bar U_{pe_1} \subset_f\bar U_{pe_1-1} \cdots \subset_f\bar U_{pi+1} =\bar U_{pi} \subset_f\cdots \subset_f\bar U_1 \subset_1\overline{K^\times}. $$ Here, $i$ is any integer in the interval $[1,e_1[$ (which is empty when $e_1=1$), an inclusion $E\subset_rE'$ means that $E$ is a codimension-$r$ subspace of $E'$, and $f$ is the residual degree of $K$ over $\mathbf{Q}_p$. We have the hilbertian pairing $\overline{K^\times}\times\overline{K^\times}\to{}_pK^\times$, where ${}_pK^\times$ is of course the group of $p$-th roots of $1$ in $K$. The orthogonal complement of the subspace $\bar U_i$ for the hilbertian pairing is precisely $\bar U_{pe_1-i+1}$, for every $i\in[0,pe_1+1]$, provided we adopt the convention $\bar U_0=\overline{K^\times}$. It is amusing to try to figure out the analogue of all this when $K$ is a finite extension of $\mathbf{F}_p((\pi))$, where $\pi$ is transcendental. Addendum 1 Okay, here is a brief sketch of the proof. First, before the hilbertian pairing, there is the kummerian pairing : $$ \overline{K^\times}\times\mathrm{Gal}(M|K)\to{}_pK^\times, $$ where $M$ is the maximal abelian extension of $K$ of exponent $p$. The group $G=\mathrm{Gal}(M|K)$ comes with a natural filtration : the ramification filtration in the upper numbering. One may ask : how is the filtration on $\overline{K^\times}$ related to the filtration on $G$ ? Answer : The two filtrations are orthogonal to each other in an appropriate sense. See for example Section IX of Local discriminants. Secondly, we have the reciprocity isomorphism $\rho:\overline{K^\times}\to G$ (with a normalisation which doesn't affect anything here), and the hilbertian pairing is obtained from the kummerian pairing via this isomorphism. Moreover, the filtration on $G$ is the image of the filtration on $\overline{K^\times}$ by $\rho$. Putting these two things together gives you the result. Addendum 2 What happens when the local field $K$ has characteristic $p$ ? Kummer theory has to be replaced by Artin-Schreier theory, so we have to first understand the filtration on $\overline{K^+}=K^+/\wp(K^+)$, where $K^+$ is the additive group of $K$ and $\wp(x)=x^p-x$. Denoting the image of $\mathfrak{p}_K^i$ by $\overline{\mathfrak{p}^i}$, it turns out that $\overline{\mathfrak{p}}=\{0\}$, and the analogous picture is $$ \{\bar0\}\subset_1 \overline{\mathfrak{p}^0}\subset_f \overline{\mathfrak{p}^{-1}} \cdots\subset_f \overline{\mathfrak{p}^{pj+1}} = \overline{\mathfrak{p}^{pj}} \subset_f \overline{\mathfrak{p}^{pj-1}} \cdots\subset K^+\!/\wp(K^+). $$ See for example Further remarks. Let $M$ be the maximal abelian extension of $K$ [edit of exponent $p$] and $G=\mathrm{Gal}(M|K)$. We have the analogous pairing $$ \overline{K^+}\times G\to\mathbf{F}_p, $$ and we still have the ramfication filtration (in the upper numbering) on $G$. It turns out that the two filtrations are orthogonal to each other under this pairing. As before, putting $\overline{K^\times}=K^\times\!/K^{\times p}$, we have the reciprocity isomorphism $\rho:\overline{K^\times}\to G$, and it carries the filtration on $\overline{K^\times}$ (which is no longer finite) onto the filtration on $G$. Putting these two facts together gives the analogous result in characteristic $p$. I leave for you the pleasure of working out the details. Addendum 3 (2016/09/06) Still not convinced ? Some more details can be found in my Note arXiv:1609.01160.<|endoftext|> TITLE: Canonical English edition of Dellacherie and Meyer's "Probabilities and Potential" QUESTION [9 upvotes]: Probabilities and Potential by Dellacherie and Meyer is a "bible" of probabilistic potential theory, Markov processes, and many related topics. I want my library to acquire it, but I am a bit confused by the several volumes and editions. What specific books constitute the "canonical" English edition of this work? Does all of the original French material currently exist in English translation? REPLY [8 votes]: The original French edition of Probabilities and Potential contains 24 chapters spread over five volumes, as outlined here. The English translation of chapters 1 through 13 is spread over three volumes in the series "North-Holland Mathematics Studies". Chapters 14 through 24 on Markov processes have not yet been translated into English, as far as I am aware. Since North-Holland was absorbed by Elsevier, that publisher now distributes the English translations. You can browse them on Amazon, volume A (chapters 1-4), volume B (chapters 5-8), and volume C (chapters 9-13). I would refer to the "North-Holland Mathematics Studies" as the "canonical" edition, in that there are no other English translations (and I presume Elsevier would not allow them).<|endoftext|> TITLE: Interesting implications on the theory of motives if the Hodge conjecture holds QUESTION [6 upvotes]: For example, Under the Hodge conjecture the Motivic galois group coincides with Mumford-Tate group. The Hodge conjecture implies the Lefschetz and Kunneth standard conjectures, as well as conjecture D (for singular cohomology) over fields of characteristic 0. REPLY [6 votes]: If we assume the Hodge or the Tate conjecture, then the functor H∗MR is fully faithful on the category of Grothendieck motives (with homological or, under these assumptions equivalently, numerical equivalence).Hence it gives a linear algebra description of the conjectural abelian category of pure motives. Under the Hodge conjecture, the period conjecture can be reformulated in terms of Mumford-Tate group. The Hodge conjecture implies that the functor from the category of Grothendieck motives to the category of André motives and the functor from the category of André motives to the category of pure absolute hodge motives are equivalences of semi-simple Abelian categories. Periods and Nori motives, Huber and Stach<|endoftext|> TITLE: Is there a strictly increasing sequence such that it is o(2^n) and any term cannot equal the sum of any unrepeated predecessors? QUESTION [14 upvotes]: Does there exist a strictly increasing sequence $\{a_n\}_{n\in N}$ of natural numbers such that the following two requirements hold: 1, For all $n\in N$, there is NO subset $M$ of $\{0,\cdots ,n-1\}$ such that $a_n=\Sigma\{a_m\ |\ m\in M\}$. 2, $\{a_n\}_{n\in N}$ is $o(2^n)$, i.e. $\lim_{n\to\infty}\frac{a_n}{2^n}=0$. REPLY [18 votes]: Such sequences are called sum free sequences. In the paper "On a question about sum-free sequences", Deshouillers, Erdős and Melfi construct a sequence where $a_n$ is $o(n^{3+\epsilon})$. Luczak and Schoen, later improved this to a construction of a sum free sequence with $a_n=o(n^{2+\epsilon})$, and show that the exponent $2$ is minimal. I should mention that the construction, referenced above is related to finding suitable perturbations starting from the set of cubes of integers. Of course for the purposes of the particular question in the OP, one can be more explicit at the cost of not being optimal. For example we can define a sequence $\{a_n\}$ as $$a_1=2 \, ,\quad a_{2n}=a_{2n-1}+1\, , \quad a_{2n+1}=1+\sum_{i=1}^{2n}a_i$$ which is obviously a sum free sequence and grows as $\sim (\sqrt{3})^n$.<|endoftext|> TITLE: How many colors do we need to avoid bichromatic triangles? QUESTION [10 upvotes]: Ramsey theory studies whether a monochromatic subgraph (more generally, structure) appears when we color the edges of a complete graph with some colors. I wonder if the following type of question has been studied before, where monochromatic is replaced by bichromatic (which now I use in the sense of having at most two colors). How many colors do we need to color the edges of a given graph to avoid a certain bichromatic subgraph? We could define the multicolor biRamsey number of $G$, denoted by $biR_k(G)$, as the least integer $n$ such that in any $k$-coloring of $K_n$ there is a bichromatic copy of $G$. Probably the simplest example is $biR_3(K_3)=5$, as a coloring of a complete graph has no bichromatic triangle if and only if every monochromatic component is a matching, and such a $3$-coloring exists for $K_n$ if and only if $n\le 4.$ Obviously, we have $biR_k(G)\le R_k(G)$, but I don't see any general lower bound for $biR_k$. Has this parameter ever been studied? (Possibly under a less brilliant name.) Motivation: I would need something like in any $n$-coloring of a graph on $n$ vertices and $\Omega(n^2)$ edges there is a trichromatic $K_{3,3}$, or something similar, where I have other conditions, like every color class is a matching. Update: As you can see from the answers of Ilya and Jan, the answer to the motivation is negative, but I can produce other similar conditions, so I am still interested whether the problem has been studied. REPLY [8 votes]: Problems of this variety have been studied, beginning with a paper of Erdős and Gyárfás, 'A variant of the classical Ramsey problem'. In that paper, they define a function $f(n,p,q)$ to be the smallest number of colours $k$ needed to produce a $k$-colouring of the edges of $K_n$ such that every $K_p$ contains at least $q$ distinct colours. The study of this function and its variants (such as replacing $K_p$ with a general graph $H$) leads to a great number of interesting questions. For example, even the behaviour of $f(n,4,3)$ is poorly understood.<|endoftext|> TITLE: Zariski closure is semisimple QUESTION [6 upvotes]: Suppose that $\Gamma$ is a finitely generated semi-group of $SL(n,\mathbb{Z})$ which acts strongly irreducible on $\mathbb{R}^n$, i.e. there is no finite union of proper nonzero linear sub-spaces of $\mathbb{R}^n$ which is invariant under $\Gamma$. It is easy to see that the zariski closure of $\Gamma$ is a reductive group. Now how to prove that this group is actually semi-simple? (Actually this is lemma 8.5. in this paper by Benoist-Quint, but I can not understand their proof for finiteness of center!) REPLY [7 votes]: $\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\Qbar{\overline{\mathbb{Q}}}$Let $G$ be the Zariski closure of $\Gamma$. We may (and do) replace $G$ by the connected component of the identity $G_0$ and replace $\Gamma$ by $\Gamma \cap G_0$, in order to assume that $G$ is connected. Note that this is the place where we needed strong irreducibility: An irreducible representation may not stay irreducible when restricted to a finite index subgroup, but a strongly irreducible representation stays strongly irreducible. (For example, the group of matrices of the form $\left( \begin{smallmatrix} z & 0 \\ 0 & z^{-1} \end{smallmatrix} \right)$, $\left( \begin{smallmatrix} 0 & z \\ z^{-1} & 0 \end{smallmatrix} \right)$ acts irreducibly, but becomes reducible when restricted to the index two group of diagonal matrices. But neither action is strongly irreducible.) Therefore, we now assume $G$ is connected. Let $Z(G)$ be the center of $G$ and let $G^{ab}$ be the abelianization. By structure theory of reductive groups, the composite $Z(G) \to G \to G^{ab}$ is an isogeny. (Not true if $G$ is disconnected! See above example.) Let $H$ be the (finite) kernel of this isogeny, so we have $1 \to H \to Z(G) \to G^{ab} \to 1$. Our goal is to show that $Z(G)$, and equivalently $G^{ab}$, is finite. Suppose for contradiction that $G^{ab}$ is infinite. Since $\Gamma=G(\ZZ)$ is Zariski dense in $G$, we know that $G^{ab}(\ZZ)$ is Zariski dense in $G^{ab}$. Since we are assuming $G^{ab}$ is infinite, we can find $g \in G^{ab}(\ZZ)$ of infinite order. Replacing $g$ by $g^{|H|}$, we may (and do) assume that $g$ lifts to $Z(G)$. So, we now have $g \in Z(G)(\ZZ) \subset G(\ZZ) \subset SL_n(\ZZ)$ of infinite order. Let $P(t)$ be the minimal polynomial of $g$, and let $P(t)$ factor over $\mathbb{R}[t]$ as $\prod P_i(t)^{a_i}$. Since $g$ is central in $\Gamma$, we see that $\Gamma$ preserves $\mathrm{Ker}(P_i)$. So, if $P(t)$ is not irreducible (as a polynomial in $\mathbb{R}[t]$) then $\mathbb{R}^n$ is not irreducible (as a $\Gamma$ representation.) Thus, $P(t)$ is either linear, or a quadratic with imaginary roots, and its constant term is $\pm 1$ since $g \in \mathrm{SL}_n(\ZZ)$. So $P(t)$ is $t \pm 1$, $t^2+1$ or $t^2 \pm t + 1$. But the roots of any of these are torsion, so $g$ is torsion, a contradiction. The OP asks why I can lift from $G^{ab}$ to $Z(G)$. Let $\phi : T_1 \to T_2$ be an isogeny of algebraic tori with kernel $H$, a group scheme of length $N$. (In case this is a source of confusion, note that the kernel of $z \mapsto z^3$, from $\mathbb{G}_m \to \mathbb{G}_m$ is length 3, not 1, even if our ground field does not contain cube roots of $1$.) Let $g \in T_2(\mathbb{Z})$. The claim is that $g^N$ lifts to $T_1(\mathbb{Z})$. Since $H$ is a commutative group scheme of order $N$, it is annihilated by $N$. Let $K \subset T_1$ be the kernel of multiplication by $N$. So $H \subseteq K$ as group schemes, and we have $T_1 \overset{\phi}{\longrightarrow} T_2 \overset{\psi}{\longrightarrow} T_1$ where $\psi \circ \phi$ is multiplication by $N$. It must also then be true that $\phi \circ \psi$ is multiplication by $N$. So $\psi(g)$ is a lft of $g^N$.<|endoftext|> TITLE: Fundamental group of $\mathrm{Sym}^2 (C_g)$ minus the diagonal QUESTION [7 upvotes]: Let $C_g$ be a compact Riemann surface of genus $g$, and let $X:=\mathrm{Sym}^2(C_g)$ be its double symmetric product and $\Delta \subset X$ the diagonal. Question. What is the topological fundamental group $\pi_1(X - \Delta)?$ I'm primarily interested to the case $g=2$, but I would like to know the general answer. I'm aware that this is probably standard material for the experts, so any reference will be greatly appreciated. REPLY [2 votes]: $\newcommand{\Sym}{\operatorname{Sym}}$The answer is above, in the comment by Stefan Behrens -- I am just adding this to make some robot happy. The $k$-fold symmetric product $\Sym^k$ of a space $C$ (minus the big diagonal $\Delta$) is the "configuration space" of distinct unlabelled $k$-tuples of points in $C$. After fixing a basepoint $(c_i)$, a loop in $\Sym^k(C)$ is a one-parameter family of $k$-tuples - so it is a union of $k$ non-intersecting arcs in $C$ that start and end at the $c_i$. Accordingly we call $\pi_1(\Sym^k(C) - \Delta)$ the $k$-strand braid group in $C$. Note that there is a homomorphism from $\pi_1(\Sym^k(C) - \Delta)$ to the symmetric group on the indices $\{1, 2, \ldots, k\}$. If $C$ is a surface, then sometimes the phrase surface braid group is used. If $C$ is the complex plane, then we obtain the standard braid group.<|endoftext|> TITLE: What is the mirror of symplectic field theory? QUESTION [17 upvotes]: Mirror symmetry is, very roughly, a relation between symplectic geometry on one side and complex/algebraic geometry on the other side. For example, counts of pseudoholomorphic spheres in a closed symplectic manifold (Gromov-Witten invariants) are related to variations of the complex structure on a "mirror" complex manifold. A natural generalization of the study of pseudoholomorphic curves in closed symplectic manifolds is given by the study of pseudoholomorphic curves in symplectic manifold with boundaries. From the point of view of symplectic geometry, it is natural to assume that these boundaries are contact manifolds and natural boundaries for pseudoholomorphic curves are provided by Reeb orbits in contact manifolds. Rather than to only consider closed symplectic manifolds and to associate numbers to them, it is therefore more natural to consider the category whose objects are contact manifolds and whose morphisms are symplectic cobordisms, and to associate vector spaces to contact manifolds ("contact homology") and to recover numbers associated to closed symplectic manifolds via cutting along contact submanifolds and gluing formulas, in accordance with the general philosophy of field theory. This program is called symplectic field theory (SFT) and was initiated in https://arxiv.org/abs/math/0010059 . My question is: what is the mirror of symplectic field theory? For example, what is the mirror of a contact manifold? what is the mirror of the operation of cutting a symplectic manifold along a contact submanifold? what is the mirror of contact homology? These questions are maybe too naive and certainly too general (even in the "classical case", I don't know what mirror symmetry is (or even shoud be) in general). Any insights and/or concrete examples would be appreciated. REPLY [9 votes]: A partial answer is as follows. In order to make use of the theory of holomorphic curves (with boundaries or asymptotics, whatever), we should restrict ourselves to symplectic manifolds with convex boundary, or more restrictively, Liouville domains or Weinstein domains. These domains then admit completions into Liouville/Weinstein manifolds by attaching cylindrical ends. (Strictly speaking, the notion of a Liouville manifold should mean something more general, and even symplectic manifolds which are of infinite topological type.) In the case of Weinstein domains, the relation between SFT and wrapped Fukaya categories is well-known. Explicitly, let $(M,\theta,\phi)$ be a Weinstein domain, with $\theta$ a contact form on $\partial M$ and $\phi:M\rightarrow\mathbb{R}$ the Morse function which determines the structure of $M$. Then $M$ can be realized as a handle body decomposition $M=M_0\cup_{i=1}^k H_i,$ where $M_0=N^{2n-2}\times D^2$ (with its corners rounded off) is a subcritical Weinstein domain (with $N$ a $2n-2$-dimensional manifold with boundary), and $H_i$ are critical handles (in this case, $n$-handles). For every $H_i$, there is a Legendrian attaching sphere $\Lambda_i$, so all together we get a Legendrian link $\Lambda=\cup_{i=1}^k\Lambda_i$. For $\Lambda$, there is a well defined $\mathbb{Z}$-graded DG-algebra over a semisimple ring $\mathbf{k}=\oplus_{i=1}^k\mathbb{K}$, namely the Chekanov-Eliashberg DG-algebra $\mathit{LHA}^\ast(\Lambda)$. One important result that can be extracted from the paper Effect of Legendrian surgery due to Bourgeois-Ekholm-Eliashberg is that $\mathit{LHA}^\ast(\Lambda)$ is quasi-isomorphic, as an $A_\infty$ algebra over $\mathbf{k}$, to the wrapped Fukaya category $\mathscr{W}(M)$ of the Weinstein domain $M$. This quasi-isomorphism can be interpreted in another way, in the sense of Abouzaid's geometric generation criterion. Namely the Lagrangian cocores $L_1,\cdot\cdot\cdot,L_k\subset M$ associated to the handles $H_1,\cdot\cdot\cdot,H_k$ together generate the wrapped Fukaya category $\mathscr{W}(M)$. Now this fact has a number of applications. For example, in the case when $M$ carries an exact symplectic Lefschetz fibration $\pi:M\rightarrow D^2$ (in fact, by a theorem of Giroux-Pardon, this is always the case for Weinstein manifolds), it's natural to take $M_0$ to be $F\times D^2$, where $F$ is a smooth fiber of $\pi$, then the above result can be applied to verify that $\mathscr{W}(M)$ is the localization of $\mathscr{F}(\pi)$ along a natural transformation $T:\mu\rightarrow\mathrm{id}$. Here $\mathscr{F}(\pi)$ is the directed $A_\infty$ category associated to the Lefschetz fibration $\pi$, $\mu$ is the autoequivalence on $\mathscr{F}(\pi)$ induced by the global monodromy of $\pi$. In the context of homological mirror symmetry, $\pi:M\rightarrow D^2$ is usually mirror to a complete algebraic variety $X$ over a specified field $\mathbb{K}$ together with a (usually singular) divisor $D\subset X$. Denote by $Y$ the non-compact algebraic variety $X\setminus D$, which should be the putative mirror of $M$. The localization result above then reduces the verification of the triangulated equivalence $D^\pi\mathscr{W}(M)\cong D^b\mathit{Coh}(Y)$ to two (usually much easier to verify) triangulated equivalences $D^b\mathscr{F}(\pi)\cong D^b\mathit{Coh}(X)$ and $D^\pi\mathscr{F}(F)\cong\mathit{Perf}(D)$, where $\mathit{Perf}(D)\subset D^b\mathit{Coh}(D)$ is the triangulated category of perfect complexes. As $D$ is usually singular, it's usually not equivalent to $D^b\mathit{Coh}(D)$. Conclusion. What I want to say is that most versions of mirror symmetry makes sense only in the homological setting (with the Gross-Siebert program as an exception), namely you should deal with invariants which come from Lagrangian Floer or SFT rather than working on the geometric level directly. Once you group these invariants together, those come from Floer theory and SFT are actually compatible with each other, and in many cases, they are just different ways of understanding the same thing, and combining these two viewpoints together do give you something interesting. On the other hand, if you think directly on the geometric level (say asking what is the mirror of certain surgeries in the symplectic or contact category) is kind of wild or unprofessional, and I don't think this is the correct way of doing things. What you should really look at is the effect of these surgeries on invariants, and try to figure out what kind of operations on the mirror have similar effect, then you would probably get some kind of surgery or operation which would conjecturally fit into the mirror correspondence. There are considerations of this flavor due to Katzarkov, Kerr, etc., they considered what happens to a Landau-Ginzburg model as its mirror goes through a birational cobordism.<|endoftext|> TITLE: Chern classes via degeneracy loci QUESTION [7 upvotes]: According to book of Eisenbud-Harris Page 332 and the following summary http://pbelmans.ncag.info/blog/2014/10/09/what-are-chern-classes/ one can describe Chern classes in terms of degeneracy loci. Quote from the link: "Recall that we were trying to measure in which sense a vector bundle is non-trivial, and for ease of statement we will assume that our vector bundle is globally generated. If $r$ is the rank, then for any $k=1,\ldots,r$ we can consider global sections $s_1,\ldots,s_{r−k+1}$. If we evaluate these in a point we get $r−k+1$ vectors of length $r$, hence we can ask whether these are linearly (in)dependent. The degeneracy locus of our set of global sections is then exactly the set of points in which the evaluations become linearly dependent, hence they degenerate. Of course one has to choose these sections sufficiently generally in order to make the codimension of the degeneracy locus correct, but one can prove that if the codimension is correct and the vector bundle is globally generated, then the degeneracy locus is independent of the choice! This can then be one way of defining the Chern classes of a vector bundle." Question: In the case of smooth complex vector bundles, there are plenty of generic sections, so I was wondering if the argument has an analogue in the smooth case (i.e. if it is written with details somewhere). REPLY [3 votes]: The construction you want via degeneracy loci definitely goes through in the general setting of a complex vector bundles over a smooth compact oriented manifold. It is often referred to as Gauss-Bonnet type formulae. You can find a nice account about that in "Principle of Algebraic Geometry" by Griffiths and Harris, Chapter 3, Section 3.<|endoftext|> TITLE: Conjugacy classes of involutions in compact simple Lie group QUESTION [6 upvotes]: Is there a known set $S=\{x \in G: x^2=1, x\ne1\}$ of elements of a simple compact Lie group $G$ ? By simple compact Lie group I consider $SO_n$, $SU_n$, $Sp_n$, $G_2$, $F_4$, $E_6$, $E_7$, $E_8$. (Feel free to extend this definition in case you find it useful). What are the conjugacy classes in this set and their dimensions ? By "conjugacy class" I mean any set $\{gxg^{-1}\}_{g\in G}$ for $x$ in above set $S$. For example in $SO_8$ the dimensions are 0, 12, 16, 12. In $O_8$ we should add 7, 15, 15, 7. These are dimensions of grassmanians $G_{k,8-k}$. For spin groups it would be oriented grassmanians. For given grassmanian element $K$ we have the map in $O_8$ which reverse $K$ and fix perpendicular space. I am especially interested in the answer for exceptional Lie groups $F_4, E_6, E_7, E_8$. Elements which square to one I would call "octonion reflections" there. The conjecture is that the conjugacy classes of involutions correspond to Riemanian symmetric spaces. I would add later more information on how to define such elements in spin groups. By another occasion I am also interested in set $\{x^2=-1\}$. Element $-1$ in the group can be defined as order 2 element in the center. This question is only valid for groups having $-1$. This should be another question, I suppose. For mentioned group $SO_8$ this set is 12-dimensional, equal to complex structures on $\mathbb R^8$. By looking at Clifford algebra $C_6$ we can prove that it is equal to oriented grassmanian $G_{2,6}^+$. EDIT: I am mainly interested in answer for groups $SO_n$, $SU_n$, $Sp_n$ and exceptional ones listed above. The "conjecture" is that for three families there are subsets of dimensions $n, 2n, 4n$ corresponding to projective spaces $\mathbb RP^n, \mathbb CP^n, \mathbb HP^n$. For exceptional ones the sets correspond to exceptional symmetric spaces embedded in the group (all of them or some of them ?). EDIT 2: By looking at this wikipedia list of all Riemannian symmetric spaces we can check whether each such space can be embedded in proper Lie group and what are the properties of this set in the group. For example let's count dimension of quaternion structures on $\mathbb C^4$. Fix the point $v_0$ in $\mathbb CP^3$. Quaternion structure is the element $j$ in $U_4$ which square to $-1$ and map each complex plane to perpendicular one. We have 4 dimensions to select $w$ in $\mathbb CP^2$ perpendicular to $v_0$. Additional $5^{th}$ dimension is allowed to rotate image of $v_0$ in $w$. In space $\mathbb C^2$ perpendicular to $$ map $j$ has also one more dimension of freedom, but we should subtract one dimension, because the circle $e^{i\alpha}j$ gives the same quaternion structure (is it clear ?). This gives $AII$ symmetric space $SU_4/Sp_2$. Here is link to my previous question in this subject. Regards, Marek REPLY [7 votes]: For the case of a connected semisimple compact Lie group $G$, see this preprint, Section 3, where, following ideas of Kac and Vinberg, we describe set of conjugacy classes of $n$-th roots of a given central element $z$ of $G$. You should take $n=2$ and $z=1$ (or $z=-1$ in your sense). EDIT: I can prove your conjecture that the conjugacy class of an involution $x$ in a connected, semisimple, compact Lie group $G$ is a Riemannian symmetric space of $G$. It suffices to assume that $x^2\in Z(G)$ but $x\notin Z(G)$, where $Z(G)$ denotes the center of $G$. The conjugacy class $X$ of $x$ in $G$ is clearly $G/H$, where $H$ is the centralizer of $x$ in $G$. Consider the automorphism $i_x$ of $G$ defined by $$ i_x(g)=xgx^{-1}.$$ Our conditions on $x$ guarantee that that $(i_x)^2=1$, but $i_x\neq 1$. Clearly $$ H=\{g\in G\ |\ i_x(g)=g\}. $$ Thus $H$ is the subgroup of fixed points of an involutive automorphism of $G$, which means that $X=G/H$ is a Riemannian symmetric space of $G$. In this way we obtain all Riemannian symmetric spaces of $G$ of inner type, i.e., the homogeneous spaces $X=G/H$, where $H$ is .the group of fixed points of some inner involutive automorphism of $G.$ See Helgason's book "Differential geometry, Lie groups, and symmetric spaces" for a list of inner and outer involutive automorphisms and the corresponding symmetric spaces. Note that if G is one of the exceptional simple groups $E_7,\, E_8,\, F_4,\, G_2$, then there are no outer involutive automorphisms of $G$ (because the Dynkin diagram of $G$ has no involutive automorphisms), and therefore, in this way we obtain all Riemannian symmetric spaces of $G$. The same refers to ${\rm Sp}_n$ and ${\rm SO}_{2n+1}$.<|endoftext|> TITLE: Distinguishing homology $S^1 \times S^2$'s which bound homotopy $S^1$'s QUESTION [6 upvotes]: Due to Mazur, Akbulut and Kirby and many others, there are many examples of integer homology 3-spheres which bound contractible 4-manifolds given by attaching a single 2-handle to $S^1 \times D^3$ which algebraically intersects the one-handle once but maybe geometrically many times. Many of these can be distinguished from the 3-sphere by things like the Casson invariant and other invariants specifically developed for integral homology $S^3$'s for instance in https://arxiv.org/pdf/1508.01491.pdf I would like to know how to distinguish the analogous constructions (which don't reduce to the above case) for homology $S^1 \times S^2$'s bounding homotopy $S^1 \times D^3$'s (homotopy equivalent not homeomorphic). Specifically are there any examples of knots in $S^1\times S^2 \# S^1 \times S^2$ which algebraically intersect both one handles once but not geometrically, that surger to (hopefully many or even infinitely for the application I have in mind) non-trivial examples of homology $S^1 \times S^2$'s? And if so, how does one distinguish these? Note by construction these all bound homotopy $S^1 \times D^3$'s. It's very easy to draw such candidates knots, but they end up having a decent amount of crossings, and even then I don't know what tools I would use to distinguish their surgeries from $S^1 \times S^2$ or each other. REPLY [7 votes]: A standard construction would be to take a 3-manifold $Y$ given by 0-surgery on a knot $K$. If $K$ is the boundary of a slice disk $D \subset B^4$, then the complement, say $W$ of a neighborhood of $D$ is a homology $S^1 \times B^3$ with boundary $Y$. But you are asking for more, since $W$ should be a homotopy $S^1 \times B^3$; this is equivalent to its fundamental group being ${\bf Z}$. This would require that the Alexander polynomial of $K$ be equal to $1$. There are plenty of examples of Alexander polynomial 1 knots bounding a disk with $\pi_1(B^4-D) = \bf{Z}$. Three nice ones are given in the paper Doubly slice knots with low crossing number (New York J. Math. 21 (2015) 1007–1026) of Livingston and Meier. See section 4.2. If you are willing to accept a topological (rather than smooth) $W$, then Freedman has shown any homology $S^1 \times S^2$ with Alexander polynomial equal to 1 bounds such a $W$. Lots of $S^1 \times S^2$s with Alexander polynomial equal to 1 do not bound a smooth homotopy $S^1 \times B^3$.<|endoftext|> TITLE: On an asymptotic formula of Keating and Snaith involving the Riemann zeta function QUESTION [21 upvotes]: Keating and Snaith have a famous conjecture on the asymptotics of the integral $\int_0^T |\zeta(\frac 12+it)|^{2k}\, dt$, where $\zeta$ denotes the Riemann zeta function. See page 510 of the book review by Brian Conrey of H. Iwaniec, Lectures on the Riemann zeta function in Bull. Amer. Math. Soc. 53 (2016), 507--512 (http://www.ams.org/journals/bull/2016-53-03/S0273-0979-2016-01525-4/S0273-0979-2016-01525-4.pdf). This conjecture involves a certain number $g_k$. This number is equal to the number of standard Young tableaux whose shape is a $k\times k$ rectangle. Equivalently, this is the degree of the irreducible representation of the symmetric group $S_{k^2}$ corresponding to this shape. Is there any (conjectural) explanation for this connection between $\zeta(s)$ and the representation theory of $S_{k^2}$? Does the degree of other irreps of $S_n$ (for suitable $n$) also have connections with the asymptotics of $\zeta(s)$? REPLY [16 votes]: There is a connection! (Though see the edit below.) Keating and Snaith make their conjecture by modeling the distribution of $\zeta(s)$ by the distribution of the characteristic polynomial of a random unitary matrix, distributed according to Haar measure. This is connected to symmetric function theory by Schur-Weyl duality. To my knowledge the first time this connection was used to compute moments (of a characteristic polynomial) was in a beautiful paper of Bump and Gamburd. See in particular Prop 4 (and perhaps especially the quick proof) and Corollary 1. As to other connections, slightly different averages of the zeta function, such as $\frac{\zeta(\alpha+s)\zeta(\beta+\overline{s})}{\zeta(\gamma+s)\zeta(\delta+\overline{s})}$ are also be conjectured to be related to analogous averages in random matrix theory, the computation of which can be reduced to symmetric function theory in a similar way. This too is done in later sections of the Bump and Gamburd paper. There is also a very nice paper of Dehaye that you might find interesting expressing lower order terms in conjectures for the moments of $\zeta(s)$ in terms of counts of taubleaux. Edit: it occurs to me that I may be leaving something unsaid. In particular you ask specifically about $S_{k^2}$. The paper of Bump and Gamburd shows in a pleasant and transparent enough way that $$ \int_{U(n)} |\det(1-g)|^{2k}\, dg = s_{\langle n^k \rangle}(1^{2k}). $$ To see what this should say about the zeta function we need to know that $s_{\langle n^k \rangle}(1^{2k}) \sim g_k n^{k^2}/(k^2)!$, where $g_k$ let's say is defined as the count of tableaux you give. While it is plausible that once symmetric function theory is involved more symmetric function theory will come into play, and while there are various ways to prove this, I don't know of any proof that makes this asymptotic relationship entirely transparent. I'd be happy to learn of one. It may also be that you're wondering whether there is a way to get from number theory to the symmetric function theory without passing through the middleman of random matrix theory. I think this is a very interesting question but I don't have anything conclusive to say about it. One remark that can be made is that the conjectures of Conrey, Farmer, Keating, Rubinstein, and Snaith that Dehaye uses in his paper came out of a recipe that keeps track of various symmetries in products of the approximate functional equation for the zeta function (identity (36) in Dehaye's paper).<|endoftext|> TITLE: class number of prime degree field with prime conductor QUESTION [8 upvotes]: Let $K$ be an finite abelian extension of $\mathbf{Q}$ conductor $p$, where $p$ is an odd prime. That is, $K \subset \mathbf{Q}(\mu_ p)$, the $p$-th cyclotomic field. Let $h_K$ be the class number of $K$. If $[K:\mathbf{Q}]=2$, we know that $K=\mathbf{Q}(\sqrt{p})$ or $\mathbf{Q}(\sqrt{-p})$. Gauss's genus theory tells us that $h_K\equiv 1 \pmod 2$. If $[K:\mathbf{Q}]=3$, a paper of G.Gras https://eudml.org/doc/151629 (Page 94 line 1-2) says that $h_K \equiv 1 \pmod 3$. Unfortunately, I did not find the proof of this result. I am very appreciated if someone gives references or ideas. So my question is that if $[K:\mathbf{Q}]=n$ is a prime number, then do we have that $h_K \equiv 1 \pmod n$? REPLY [10 votes]: Maybe I am making a mistake here, but let me try: Let $H$ be the Hilbert class field of $K$. Then $H\cap \mathbb{Q}(\zeta_p)=K$ as otherwise one prime in there should be totally ramified and unramified at the same time. This shows the triviality of the maximal quotient of the class group $C$ on which the Galois group $G=\operatorname{Gal}(K/\mathbb{Q})\cong \mathbb{Z}/n\mathbb{Z}$ acts trivially. Then the action of $G$ has no non-trivial fixed points on $C$, which means each orbit of $G$ on $C$, apart from the neutral element, is of size $n$, because $n$ is prime. Hence $\vert C\vert \equiv 1 \pmod{n}$.<|endoftext|> TITLE: Finding a short path using $(0.99n)!$ permutations QUESTION [7 upvotes]: Suppose I have $n$ points $x_1,\dots,x_n$ that are all independent uniform samples in the unit square, and I'd like to find a short path (in terms of Euclidean length) that touches all of them (a traveling salesman path, in other words). Obviously, one (inefficient) way to do this would be to write down every permutation of $\{1,\dots,n\}$, record the length of the path that visits the $x_i$'s in that order, and select the permutation that gave the shortest path. Now suppose that, before the $x_i$'s are sampled, I am allowed to write down not all $n!$ permutations, but only a subset of permutations of size (say) $(0.99n)!$. Is there a "clever" choice of permutations that is likely to contain a good (i.e. short) path (in the sense of Euclidean length) through the $x_i$'s, in the limit as $n$ becomes large? REPLY [9 votes]: That is highly unlikely. Take any such system $S$ of permutations. Take any configuration of points obtained by sampling and note that we can also get it with the points re-enumerated in any way. Now let us look at how many reasonably short path enumerations are there at all. For each short path, there is a sequence of integers $a_1,\dots,a_{n-1}$ with sum $\le Cn$ such that the $k$-th step jump is at most $a_k n^{-1/2}$. There are only exponentially (in $n$) many such sequences. Now, for a typical configuration, the number of points in any disk of radius $Rn^{-1/2}$ is at most $CR^2\log n$ (the exponential tail estimate for the Poisson distribution), so at each step we have just $Ca_k^2\log n$ choices at most, giving at most $e^{Cn}(\log n)^n$ permutations total that can work in principle. Call that set $K$. We must have $S\pi\cap K\ne\varnothing$ for most permutations $\pi$, whence $|S||K|\ge n!/2$. So, we can have only $(C\log n)^n$ gain (actually even less; I guess the exponential gain is the right one), which is quite short of $n^{0.01n}$ you want.<|endoftext|> TITLE: How slow are direct solutions of NP-complete problems on computers? QUESTION [7 upvotes]: Sometimes I see that people call a problem NP-hard and because of that refuse to create computer algorithms that directly solve it. I think I've never read actual benchmark results for such problems. Are there any? Are there any NP-complete problems with not very bad performance? REPLY [6 votes]: For Boolean Satisfiability (SAT), there is Sat Competition: http://www.satcompetition.org/ State of the art SAT solvers often perform very efficiently, far faster than exponential. At the above link there should be results to see how large instances were solved. Sucess story: Using SAT solver, important partial results about a conjecture of Erdos were found. The formula was quite big, for details see here: http://cgi.csc.liv.ac.uk/~konev/SAT14/ Sometimes heuristic algorithms perform very fast on graph problems. For SAT, if you generate random SAT instance, very likely state of the art SAT solver will solve it very fast. (There was very similar question about this here).<|endoftext|> TITLE: A criterion for real-rooted polynomials with nonnegative coefficients QUESTION [12 upvotes]: Let $P \in \mathbb{R}[X]$, with $\deg P = n$. Is it true that $P$ has only real roots $\quad \Longleftrightarrow \quad P\cdot P'' + (\frac{1}{n}-1)P'^2 \leq 0$ ? The direct implication can be shown by using Cauchy-Schwarz inequality, but I still lack the reverse (if true). I apologize to those who have already answered, but a new condition arise: what if $P$ has only nonnegative coefficients? REPLY [19 votes]: Also simply for $P(x)=x^4-1$ we have $$P\cdot P''+\left(\frac{1}{4} -1 \right)\left(P'\right)^2=-12x^2\leq 0$$ and $P$ has imaginary roots. The stronger version is also false. Take $P(x)=x^5-x$ which satisfies $$P\cdot P''+\left(\frac{1}{5} -1\right)\left(P'\right)^2=-12x^4-\frac45<0$$ yet it has complex roots.<|endoftext|> TITLE: Number of Lyndon words of given weight QUESTION [5 upvotes]: Consider the alphabet consisting of two letters $a$ and $b$, and put the lexicographic order in which $a TITLE: When does isometric projection respect multiplication? QUESTION [5 upvotes]: Every $A \in \text{GL}_n(\mathbb{R})$ has a unique orthogonal polar factor $O_A=A(\sqrt{A^TA})^{-1}$, ( $A=O_AP_A$, $O \in \operatorname{O}_n, P \in \operatorname{Psym}_n$see Polar decomposition). I am interseted in characterizing the set $W:=\{(A,B)\in \text{GL}_n^+ \times \text{GL}_n^+|\,O_{AB}=O_AO_B \}$. I have made some progress, but am still lacking a unifying necessary and sufficient criterion. Even the special case where $A=B$ does not seem trivial. Any advancement would be welcome. Edit: Another natural question is: How rare is it for a pair $(A,B)$ to be in $W$? (we can take the uniform distribution on $\text{GL}_n^+ \times \text{GL}_n^+$ and ask what is the probability of a random pair to be in $W$). Motivation: This question has a geometric nature, since $O_A$ is the closest matrix in $\text{SO}_n$ to $A$ (w.r.t to the Frobenius norm). Thus, it reflects the "rotational part" of $A$, after the scaling of the different directions is removed; This is very clear when looking at the SVD of $A$, $$A=U\Sigma V^T \Rightarrow O_A=UV^T$$ So, we can think of $O_A$ as the best "isometric approximation" (or projection) of $A$. If we know the two approximations for $A,B$ it is natural to compose them and to ask if we obtained in this way the best approximation to $AB$. (This comes up in some applications of geometry and analysis). What I have so far: (0) The relation is not symmetric, i.e $(A,B) \in W \nRightarrow (B,A) \in W$. (1) A sufficient condition for $(A,B) \in W$ is that $P_A$ commutes with $O_B,P_B$. (This is far from being a necessary condition, see below). (2) $\forall A \in \text{GL}_n^+ \,: \, (A,A^{-1}) \in W$. (3) If $B \in \operatorname{Psym}_n$, then $$(A,B) \in W \iff [P_A,B]=[P_A,P_B] =0$$ (see here for details ) In particular, for $A,B \in \operatorname{Psym}_n$: $$(A,B) \in W \iff [A,B]=0$$ (4) None of these implications hold in the general (proofs at the end): (a) $(A,B) \in W \nRightarrow [A,B]=0.$ (b) $[A,B]=0 \nRightarrow (A,B) \in W.$ (c) $(A,B) \in W \nRightarrow [P_A,P_B]=0.$ (d) $[P_A,P_B]=0 \nRightarrow (A,B) \in W.$ (5) If $A$ is a normal matrix, then $(A,A) \in W$. (This is easy to see since $A$ is normal $\iff$ $[O_A,P_A]=0$, via the uniqueness of the positive square root). Is it true that $(A,A) \in W$ implies $A$ is normal ? (I suspect the answer is negative) Proofs (for the statements in (4)): (a) Take $B \in \operatorname{Psym}_n,O \in \operatorname{SO}_n$ such that $[B,O] \neq 0$, and define $A=OB$. Then $(A,B) \in W$ but $\, AB=OB^2 \neq BOB=BA$. (b) Assume by contradiction $[A,B]=0 \Rightarrow (A,B) \in W$. Then $[A,B]=0 \Rightarrow [O_A,O_B]=0$; Indeed, $$[A,B]=0 \Rightarrow [B,A] =0 \Rightarrow (B,A) \in W,$$ so $$O_BO_A=O_{BA}=O_{AB}=O_AO_B \Rightarrow [O_A,O_B]=0.$$ However, this does not always hold: Let $A \in \text{GL}_2^+,B \in \text{GL}_2^-$ be commuting matrices. Note that $[O_A,O_B] \neq 0$ since for any two elements $X \in \operatorname{O}_2\setminus \operatorname{SO}_2,Y \in \operatorname{SO}_2\setminus\{\pm Id\}$, $[X,Y] \neq 0$. However, I want counter-examples where $A,B$ are both in $\text{GL}_n^+$. This is impossible to find for $n=2$ since $\operatorname{SO}_2$ is abelian. However, we can take $4 \times 4$ matrices, by "doubling" $A,B$ in two diagonal blocks. (For a concrete example see here) (c) See counter-example here. (d) See counter-example here. REPLY [2 votes]: If a decomposition writes $GL_n$ as the product of two subgroups (for example the Iwasawa decomposition), then the group multiplication in $GL_n$ can be expressed via a "automorphically knitted pair of actions" also called a Zappa-Szep product, see the paper: Peter W. Michor: Knit products of graded Lie algebras and groups. Rendiconti Circolo Matematico di Palermo, Serie II, Suppl. 22 (1989), 171--175, arXiv:math.GR/9204220. (pdf) Maybe you can use the formulas and concepts there. As pointed out by the OP, his case is different, since the second factor is not a group. Consider the mappings $A\mapsto O_A$ and $A\mapsto P_A$. They are not rational (because of the square root), but real analytic, and they are submersions. One can try for the codimension of $W$ which is probably quite small.<|endoftext|> TITLE: On the local automorphic components of classical Siegel modular forms QUESTION [6 upvotes]: I am looking for a dictionary that relates the level of a classical genus 2 Siegel modular form and the local components of the corresponding automorphic representation of $Gsp_4(\mathbb{A}_{\mathbb{Q}})$ (Relative to the Sally-Tadic classification of the non-supercuspidals). The notes of J. Booher compile the relevant information for classical elliptic modular forms (http://stanford.edu/~jbooher/expos/adelic_mod_forms.pdf) and I would love to have a similar breakdown for Siegel if such a treatise exists. For instance: If I start with a Siegel modular form $F$ of level $N$, and $p \vert N$, what representations (of type I-XI in S-T) should I expect to see and when? REPLY [3 votes]: Look at the tables of Ralf Schmidt and Brooks Roberts: Tables for Representations of GSp(4) Section 2 also tells you which representations you should find in Saito-Kurokawa versus non-Saito-Kurokawa spaces. Section 5 tells you about what representations can occur for prime level for various congruence subgroups. They also do higher level for paramodular groups (see also their book). I saw a similar table in a talk by Ralf 2 days ago for higher level Siegel congruence subgroups--I don't think it's public yet, but I expect it will appear in a future paper.<|endoftext|> TITLE: Any open Langlands Conjectures for GL_1? QUESTION [23 upvotes]: Are there any general conjectures/properties (in the Langlands Program) for automorphic representations of $GL_n$ which are still open for $n=1$? REPLY [40 votes]: The question is subject to interpretation, because the Langlands program was never a clearly delimited set of conjectures to begin with, and moreover many things were added during the following decades, and the answer depends if you want to consider these innovations as part of the Langlands program or not. I believe nevertheless that essentially everything in the original Langlands program (of the 70's) is known for $GL_1$, and actually was known long before Langlands began working on his program. Every instance of Langlands functoriality involving only $GL_1$ results easily from Class field theory. As for reciprocity, if the existence of the Langlands group is still unknown, only its abelianization matters for $GL_1$, and we know, again by CFT, what this abelianization is. If we add to the Langlands program the correspondance of algebraic automorphic forms with Galois representations, as for example formulated in Clozel's 1988 paper "Motifs et Formes Automorphes" in Clozel-Milne, then again the case of $GL_1$ was well-known long before, and due to André Weil (that's his theory of algebraic Grössencharacters). But if you are willing to consider the theory of $p$-adic automorphic forms and their families as part of the Langlands program, then there is one thing that is still unknown: the dimension of the eigenvariety of automorphic forms for $GL_1$ over a number field. Determining this dimension is equivalent to proving Leopoldt's conjecture.<|endoftext|> TITLE: Who proved that two homotopic embeddings of one surface in another are isotopic? QUESTION [14 upvotes]: If $\Sigma_1$ and $\Sigma_2$ are two compact topological surfaces with boundary and $\phi, \psi : \Sigma_1 \hookrightarrow \Sigma_2$ are two orientation-preserving embeddings that are homotopic, then they are also isotopic. This is a consequence of a result of Maxime Fortier Bourque on conformal embeddings, but it feels like it ought to have been proved earlier. Is there an earlier reference? If $\Sigma_1$ is an annulus, then this is really a question about simple curves, and it appears in a paper by Hass, Freedman, and Scott. If $\Sigma_1$ and $\Sigma_2$ are closed surfaces (so that $\phi$ and $\psi$ are homeomorphisms), this is a consequence of the Dehn-Nielsen-Baer theorem. Maxime Fortier Bourque, MR 3358269 The holomorphic couch theorem, ISBN: 978-1321-75413-1. Michael Freedman, Joel Hass, and Peter Scott, MR 671777 Closed geodesics on surfaces, Bull. London Math. Soc. 14 (1982), no. 5, 385--391. REPLY [15 votes]: First, for simple closed curves, this was known long before Freedman-Hass-Scott. For closed surfaces, it was first proved by Baer in Baer, R., Kurventypen auf Flächen. J. reine angew. Math., 156 (1927), 231–246. and Baer, R., Isotopie von Kurven auf orientierbaren, geschlossenen Flächen und ihr Zusammenhang mit der topologischen Deformation der Flächen. J. reine angew. Math., 159 (1928), 101–111. For general surfaces (not necessarily compact, and also possibly with boundary), it was proved by Epstein in Epstein, D. B. A. Curves on 2-manifolds and isotopies. Acta Math. 115 1966 83–107. Epstein's paper is also the first one to prove a general homotopy implies isotopy theorem for essentially all surfaces (though as you indicated this was proved earlier by Baer for closed surfaces following work of Dehn and Nielsen; the references are the ones above). There are a couple of silly counterexamples to this that you have to rule out (e.g. the homeomorphism of an annulus that flips the boundary components is homotopic to the identity but not isotopic to the identity). Now, on to your question. I believe that the embedding question can be easily reduced to the case of homeomorphisms of surfaces (and thus should be mostly attributed to Epstein, though I am not aware of a reference that points this out). The idea is as follows: You first reduce to the case where the images of your embeddings $\phi,\psi\colon \Sigma_1 \rightarrow \Sigma_2$ are contained in the interior of $\Sigma_2$. For the next step, let's regard $\Sigma_1$ as a subsurface of $\Sigma_2$ via $\phi$. Thus $\phi$ is the restriction of the identity map $\Sigma_2 \rightarrow \Sigma_2$ to $\Sigma_1$. It is then easy to see that you can extend $\psi$ to a homeomorphism $\Psi\colon \Sigma_2 \rightarrow \Sigma_2$ that is homotopic to the identity (which you can certify by ensuring that $\Psi$ acts as the identity on the fundamental group). There are a few cases to check here, but it isn't so hard. You then quote homotopy implies isotopy for $\Psi$ and deduce that $\Psi$ is isotopic to the identity. The desired isotopy between $\psi$ and $\phi$ is the restriction of this isotopy to $\Sigma_1$.<|endoftext|> TITLE: Detailed proof that $\circ_i$ operads are the same as operads QUESTION [9 upvotes]: The original definition of operads involves maps $$ \gamma\colon C(n) \times \prod_{i=1}^n C(k_i) \to C(\sum_ik_i) $$ There is an alternative definition in terms of maps $$ \circ_i \colon C(n) \times C(m) \to C(n+m-1). $$ It is not hard to outline an argument that the two definitions are equivalent (subject to some assumptions about $C(0)$ and $C(1)$; various choices of details are possible). Have the details been spelled out carefully somewhere? I have seen this called an "observation", but that seems a little blasé to me. REPLY [7 votes]: There is a detailed discussion of the "partial" composition operations of an operad in Volume 1, Chapter 2 of Fresse's monograph: Benoit Fresse, Homotopy of Operads & Grothendieck-Teichmüller Groups. (book project web page) (author pdf for volume 1) The equivalence of the two definitions of operad is stated as Theorem 2.1.10, and the proof is worked out in Appendix A (see A.2.10).<|endoftext|> TITLE: Combinatorics of integral over simplices QUESTION [6 upvotes]: Let $\Delta_N := \{0 \leq \tau_1 \leq \dots \leq \tau_N \leq 1\}$ be the $N$-simplex. Let $a_j: [0, 1] \rightarrow \mathcal{A}$, $j=1, \dots, N$ be continuous functions with values in some (non-commutative) finite-dimensional Banach algebra (e.g. some matrix algebra). Consider the $\mathcal{A}$-valued integral $$ \mathrm{Int} = \sum_{\sigma \in \mathcal{S}_N} \mathrm{sgn}(\sigma) \int_{\Delta_N} a_{\sigma_N}(\tau_N) \cdots a_{\sigma_1}(\tau_1)\, \mathrm{d} \tau_1 \cdots \mathrm{d} \tau_N, $$ where $\mathcal{S}_N$ denotes the group of permutations of the numbers $\{1, \dots, N\}$. Suppose that the index set $\{1, \dots, N\}$ splits as the disjoint union of non-empty sets $I_1, \dots, I_k$ such that we elements from different sets $I_u$ anti-commute pairwise, i.e. for $i \in I_u$, $j \in I_v$, one has $$ a_i(t)a_j(s) = -a_j(s)a_i(t), ~~~~~~~~ \forall s, t \in [0, 1].$$ Clearly, if $k=N$, i.e. each set has only one element so that all the $a_j$ anti-commute, then $$\mathrm{Int} = \prod_{j=1}^N \int_{\Delta_1}a_j(\tau) \mathrm{d}\tau.$$ In the general case, we guess that one has $$ \mathrm{Int} = \prod_{j=1}^k \sum_{\sigma \in \mathcal{S}_{I_j}} \mathrm{sgn}(\sigma) \int_{\Delta_{|I_j|}} \prod_{i\in I_j} a_{\sigma_i}(\tau_i) \mathrm{d}\tau_i,$$ where $\mathcal{S}_{I_j}$ denotes the group of permutations of the set $I_j$ and the latter product is run through from the largest to smallest number $i \in I_j$. We were unable to prove this in full generality so far, because we didn't find a way to efficiently handle the combinatorics. Does anybody have hints how to tackle this problem or probably people already looked at this so that there is a reference? REPLY [2 votes]: For a $\sigma\in S_N$, let $\Delta_N^{\sigma}=\{(\tau_1, \ldots, \tau_N)\mid 0\le \tau_{\sigma_1}\le\cdots\le\tau_{\sigma_N}\le 1\}$. Clearly, $\Delta_N^{\sigma}$ is a space homeomorphic to $\Delta_N$, and there is a decomposition $$I^N=\bigcup_{\sigma\in S_N} \Delta_N^{\sigma}.$$ The points of $I^N$ that belong to more than one of the $\Delta_N^{\sigma}$s form a subset of measure zero (it is the set of $N$-tuples that are not all distinct). In what follows, when I say something is a function on $I^N$, it is only required to be defined unambiguously outside this set of measure zero. With this caveat in mind, let $\tilde\sigma\colon I^N \to S_N$ be the "function" that assigns to an $n$-tuple $(t_1, \ldots, t_N)$ the permutation $\sigma$ for which $t_{\sigma_1}\le\cdots \le t_{\sigma_N}$. Then your Int can be rewritten as an integral over $I^N$, as follows $${\mbox{Int}}= \int_{I^N} {\mbox {sgn}}(\tilde\sigma) \cdot a_{{\tilde \sigma}_N}(\tau_{{\tilde \sigma}_N})\cdots a_{{\tilde \sigma}_1}(\tau_{{\tilde \sigma}_1})\, d\tau_1\cdots d\tau_N.$$ Let us denote the integrand function by $f_N(\tau_1, \ldots, \tau_N)$. Now let us suppose that $I_1=\{1, \ldots, i_1\}$, $I_2=\{i_1+1,\ldots i_1+i_2\}$ and so on up to $I_k$. There is a canonical identification $I^N\cong I^{I_1}\times \cdots \times I^{I_k}$. Moreover, your hypothesis about anti-commuting elements is equivalent to saying that the function $f_N$ splits as a product $$f_N(\tau_1, \ldots, \tau_N)= f_{I_1}(\tau_1, \ldots, \tau_{i_1})\cdot f_{I_2}(\tau_{i_1+ 1},\cdots, \tau_{i_1+ i_2})\cdots f_{I_k}(\cdots).$$ It follows that $$\int_{I^N}f_N = \int_{I^{I_1}}f_{I_1}\cdots \int_{I^{I_k}}f_{I_k}.$$ I think this is equivalent to the formula you are asking.<|endoftext|> TITLE: Best non-lattice sphere packings QUESTION [9 upvotes]: Consider a dense sphere packing in $\mathbb{R}^n$, i.e. an arrangement of mutually disjoint solid open spheres, all of the same radius. In dimensions $2, 3, 8,$ and $24$, it is known that lattice packings (packings where center of the spheres form a discrete subgroup of $\mathbb{R}^n$) are optimal. It is widely believed that, in high enough dimensions, the best packings will be non-lattice packings. Is anything known about the best non-lattice packings? Do the methods of Viazovska et al. and Hales give values for the density of the best non-lattice packing? Is anything known (or conjectured) about the the ratio of the density of the best non-lattice packing to the density of the best lattice packing? REPLY [5 votes]: I recently posted a preprint together with Alexei Andreanov in which we enumerate all the locally optimal 2-periodic sphere packings (also known as double-lattice packings) in dimensions up to $d=5$. This enumeration includes lattices, since they can be represented as 2-periodic arrangements by using a sublattice of index 2 as the unit cell. For the optimal packing density (among 2-periodic packings), we found that in $d=3$, there is one non-lattice that achieves it (the hexagonal close packing), in $d=4$ there are no non-lattices that achieve it, and for $d=5$ there are two. The highest locally optimal non-lattice double-lattice density in $d=4$ is the intriguing $\delta = 1/(2 \sqrt{9+4\sqrt{5}}) \approx 0.118$ (compared to $\delta = 1/8 = 0.125$ for the $D_4$ lattice).<|endoftext|> TITLE: A question about Julia set for quadratic family QUESTION [7 upvotes]: Let $P_{c}(z)=z^2+c$. It seems from the software that the map between the parameter $c$ and the Julia set $J(P_c)$ is an injective map. Is there some reference about it? Any comments and reference will be appreciated. REPLY [4 votes]: For $z^2+c$ it is really easy: the answer is yes. In general, if Julia sets of polynomials coincide, these polynomials are equal unless their common Julia set has a rotation symmetry, or polynomials are permutable. (Baker, Eremenko, Ann. Acad. Sci. Fenn., 12 (1987) 229-236.) http://www.acadsci.fi/mathematica/Vol12/vol12pp229-236.pdf For $z^2+c$ we never have rotational symmetry and never have permutable polynomials. The result cited above was reproved and generalized by many other people. For rational functions, the question in full generality is still open.<|endoftext|> TITLE: What is the $p$-adic Langlands conjecture for $\mathbf{GL}_1$? QUESTION [21 upvotes]: In the Boston conference on Fermat's Last Theorem (Summer 1995), Barry Mazur said (around 15m into the video) about class field theory that If you are a number-theorist and you want to cheer yourself up, you can say that we really, we completely understand this.. by class field theory.. If you want to pull yourself down a peg, you can say that even here we are not all the way home, we don't know Leopoldt's conjecture, for example. More recently, the ever-helpful Joël Bellaïche wrote here that [I]f you are willing to consider the theory of $p$-adic automorphic forms and their families as part of the Langlands program, then there is one thing that is still unknown: the dimension of the eigenvariety of automorphic forms for $\mathbf{GL}_1$ over a number field. Determining this dimension is equivalent to proving Leopoldt's conjecture. Question. What is the precise formulation of the (global) $p$-adic Langlands conjecture for $\mathbf{GL}_1$, what is the eigenvariety for $\mathbf{GL}_1$, and how are they related to the Leopoldt conjecture ? A short essay or references to the literature for the benefit of humankind will be greatly appreciated. Presumably, the local $p$-adic Langlands conjecture for $\mathbf{GL}_1$ is already a theorem, because it is a theorem for $\mathbf{GL}_2$ over $\mathbf{Q}_p$. What is the statement (for $\mathbf{GL}_1$ over a local field with finite residue field) ? REPLY [17 votes]: I am not sure it makes sense to ask "what is the p-adic local Langlands conjecture for $\mathrm{GL}_1$". Nobody has succeeded in even formulating a reasonable candidate for a p-adic LLC for $\mathrm{GL}_n$, so you can't just plug $n = 1$ in and see what it says. Morally, the conjecture should be something like this: for $K$ a p-adic field, there is a bijection between continuous representations $\mathrm{Gal}(\overline{K} / K) \to \mathrm{GL}_n(E)$, $E$ a p-adic field, and continuous admissible unitary p-adic $E$-Banach-space representations of $\mathrm{GL}_n(K)$, satisfying [some properties]. The problem is: what properties? Saying "there is a bijection" is just claiming that the sets have the same cardinality, which is trivial, and clearly not what is intended here (Serre makes exactly this point in his lecture How To Write Mathematics Badly, which is on YouTube). So one needs to spell out what the properties are, and this is hard to do. In this $\mathrm{GL}_1$ setting you can cheat, because there is already a candidate for this bijection (any continuous unitary character $K^\times \to E^\times$ extends uniquely to the profinite completion, and $\widehat{K^\times}$ is the abelianization of $\mathrm{Gal}(\bar{K} / K)$ by class field theory). But one doesn't know how to uniquely pin down what the question is for $\mathrm{GL}_n$, which is one reason why it's hard to answer! Even for $\mathrm{GL}_2 / \mathbf{Q}_p$, Colmez has constructed a bijection between p-adic Banach reps of $\mathrm{GL}_2(\mathbf{Q}_p)$ and 2-dimensional Galois reps, which has a long list of very nice properties; but as far as I'm aware there's not a theorem which says "Colmez's correspondence is the unique correspondence with properties X, Y and Z". EDIT. Let me add some more about the global side. Here it's even less clear what the question is. For $\mathrm{GL}_n$ over a number field $K$, the statement should probably be that there should be a bijection between: $n$-dimensional $p$-adic representation of $\mathrm{Gal}(\overline{K} / K)$ which are continuous and unramified almost everywhere; and "p-adic automorphic representations of $\mathrm{GL}_n$". But it's not at all clear what the definition of the latter class of objects should be for $n \ge 2$. The eigenvariety for $\mathrm{GL}_1 / K$, $K$ a number field, is rather well understood. I'll just summarize Buzzard's construction briefly. Choose some open compact subgroup $U$ of $\left(\mathbf{A}_K^{(p\infty)}\right)^\times$ (the finite adeles away from $p$). Let $K_\infty^\circ$ be the totally positive elements of $(K \otimes \mathbf{R})^\times$. Then the quotient $$ G(U) = \mathbf{A}_K^\times / \overline{ U \cdot K_\infty^\circ \cdot K^\times } $$ is a pretty friendly abelian group; it has a subgroup of finite index isomorphic to $\mathbf{Z}_p^n$ for some integer $1 \le n \le [K : \mathbf{Q}]$. Moreover, Leopoldt's conjecture is very easily seen to be equivalent to determining $n$ (it says that $n = 1 + r_2$, where $r_2$ is the number of complex places of $K$). Since $G(U)$ is such a friendly group, it's easy to see that $p$-adic characters of $G(U)$ naturally biject with points of a rigid space $\mathcal{E}(U)$ and that's your eigenvariety. Now, the points of $\mathcal{E}(U)$ can be viewed as continuous $p$-adic characters of $\mathbf{A}_K^\times / K^\times$ with prime-to-$p$ ramification bounded by $U$, or (via class field theory) as 1-dimensional Galois representations, again with prime-to-$p$ ramification bounded by $U$. That gives a bijection between these classes of objects, to which it seems natural to give the name "p-adic Langlands for $GL_1$". So there's the relation between the objects in Chandan's question: p-adic Langlands for $GL_1$ is a bijection between two classes of objects, "automorphic" and "Galois", which is an easy consequence of class field theory; both of these classes of objects are naturally parametrised by a $p$-adic rigid space, which we call the eigenvariety; Leopoldt's conjecture is a statement about the dimension of the eigenvariety, which involves the same objects as $p$-adic Langlands but is logically independent of it. As soon as you step away from $\mathrm{GL}_1$, though, eigenvarieties get less central to the $p$-adic Langlands picture (although they're still clearly very important). The issue now is that eigenvarieties are expected to parametrise Galois representations which are fairly "degenerate" locally at $p$: technically they should be trianguline, a notion introduced by Colmez. For $\mathrm{GL}_1$ everything is (vacuously) trianguline, but this is further and further away from being true as $n$ grows; e.g. a Galois representation with finite image is only trianguline if it's a direct sum of characters. Trianguline representations should correspond under p-adic Langlands to representations of $\mathrm{GL}_n$ induced from the Borel subgroup -- so in a sense these are the easiest ones!<|endoftext|> TITLE: Regular Dirichlet form and the associated transition kernel QUESTION [5 upvotes]: I am reading a paper by Fukushima "On a stochastic calculus related to Dirichlet forms and distorted Brownian motions" and support it by a book "Dirichlet forms and symmetric Markov processes" by Fukushima, Oshima and Takeda. Let me also remind that a dense subset $\tilde{C}$ of $C_0(X)$ is said to be a core of a Dirichlet form $\mathcal{E}$ if $\tilde{C}$ is $\mathcal{E}_1$-dense in $D[\mathcal{E}]$. A Dirichlet form possessing a core is called regular. In the paper, there are the following 2 sentences: "Let $\mathcal{E}$ be a Dirichlet form and $T_t$ the associated semigroup of Markovian symmetric operators on $L^2(X;m)$. If $\mathcal{E}$ is regular, then $T_t$ can be realized as \begin{equation} T_t\,f(x) = \int_X P_t(x, dy) f(y) \end{equation} by a transition function $P_t(x, E)$ on $X$ which is $m$-symmetric in the sense that $\int_XP_t\,f(x)\,g(x)\,m(dx) = \int_X P_tg(x)\,f(x)\,m(dx)$." I could not find the above result in the book and I do not understand why we need the regularity assumption for that. Namely, we know that the space $C_0(X)$ of continuous functions with compact support is dense in $L^2(X,m)$, so we by Riesz-Markov-Kakutani there exists a transition function $P_t(x,dy)$ such that the above equation holds for $f\in C_0(X)$, and by density we can extend it to $L^2(X;m)$. Please tell me why this is wrong. Thank you very much. REPLY [2 votes]: Although the regularity imposed in Fukushima's theory is quite natural, it is a bit of overkill if all one wants is a transition semigroup $(P_t)_{t\ge 0}$. There is work culminating in a short paper of S.E. Kuznetsov (1987-Prob. Theory and its Application; http://epubs.siam.org/doi/pdf/10.1137/1131031) that can be applied to a general Dirichlet form, the only extra condition being needed is that the state space is "nice" -- standard Borel is enough. Picking an initial distribution $\mu$ that is equivalent to the symmetry measure $m$, one can use the $L^2$ operators $T_t$ to define the finite dimensional distributions of a stochastic process, and then invoke Kolmogorov's theorem to gain the existence of such a process, call it $X=(X_t)$ on some probability space $(\Omega,\mathcal F,\Bbb P)$ such that $\Bbb P[X_0\in B]=\mu(B)$, $$ \Bbb P[X_s\in A,X_t\in B]=\int_X T_s(1_AT_{t-s}1_B)\,d\mu,\quad 0\le s TITLE: Sum zero problems on the poset of structures of a combinatorial species QUESTION [5 upvotes]: Consider a finite graded poset $\Gamma$ and assign to each maximal element $z\in \Gamma$ a variable $\mu(z)$. I want to solve the system of equations (minimally, I want to compute its rank, ideally, obtain an explicit basis) $$\sum_{w\leqslant z} \mu(z) = 0.$$ That is, for each $w\in \Gamma$, the weight assigned to the maximal elements that lie over it must sum to zero. (I'm excluding the maximal elements themselves, of course.) Note that one can write any linear system in terms of a bipartite graph by assigning the equations into one group of vertices of the graph, the variables to the other, and joining an equation to a variable if such variable appears in the chosen equation. One then chooses $\mu(z)$ accordingly and the equations above are precisely those of the linear system (consider a bipartite graph as a poset in the obvious way). In such a sense, this problem is as broad as it gets. The following extra information may (or may not) help narrowing it down. To be more concrete, the problem arises as follows. Fix a combinatorial species $P:\mathsf{Set}^\times\longrightarrow \mathsf{Set}$ (a functor) with restrictions, that is: For each finite set $I$ and each subset $S$ of $I$, there is an arrow $$(?)_S : P(I)\to P(S)$$ These arrows are compatible in the sense they form a sheaf over finite sets and injections of subsets, and bijections: If $T\subseteq S\subseteq I$ then $((?)_S)_T = (?)_T$ For every $I$, $(-)_I ={\rm id}_I$ For every bijection $\sigma : I \longrightarrow J$ and every subset $S$ of $I$, $$(P(\sigma)(?))_{\sigma(S)} = P(\sigma)((?)_{S})$$ Using this data one can form, for any finite set $I$, a poset $\Gamma(I)$ as follows. The underlying set of $\Gamma(I)$ is the collection of $P$-structures on subsets of $I$: $$\Gamma(I) = \bigcup_{S\subseteq I} P(S).$$ We define a partial order on $\Gamma(I)$ so that $z\leqslant w$ if The support of $z$ is contained in that of $w$: $z\in P(S)$ and $w\in P(T)$ with $S\subseteq T$, $z$ is obtained by restricting $w$ to $S$: $z= (w)_S$ Note that $\Gamma(I)$ is graded by the cardinality of the support of a structure. Thus, for each subset $S$ of $I$ I want to solve the system of equations $$\sum_{z:(z)_S=w} \mu(z) = 0$$ as $z$ ranges through $P(I)$. To give even a more concrete example, I have computed a few values of the dimension $d_n$ ($n=\# I$) of solutions of the above for $L$ the species of linear orders, and $d_n$ coincides with the derrangement sequence $1,0,1,2,9,44,\ldots$. Variants of this poset have been studied here https://arxiv.org/pdf/0902.4011v3.pdf for example. One option is inverting (after suitably defining $\mu(w)$ for every other $w\in \Gamma$) the equations above to obtain recurrences, but I would expect that the calculation of the Möbius function of an arbitrary poset like the above is not easy at all (as the cited paper shows). One can fix a ring $k$ and linealize the category of species with restrictions by postcomposition with the free functor to $k$-modules. This category is in fact the category of (left) comodules over a fixed (linealized) species, the exponential species, and it is monoidal with a particular product (the Cauchy product). The problem above amounts to computing the cotensor product of $P$ with the unit for this monoidal structure. REPLY [2 votes]: A comment on the problem. This doesn't solve the problem but was too long for a comment. Let $\mathrm{FGP}$ denote the category of finite graded posets with morphisms the graded morphisms of posets and $\mathrm{BG}$ be the full subcategory whose objects are "graded" bipartite graphs, i.e. the grading can only take two values, $0$ or $1$. Given a finite graded poset $\Gamma$, one can associate a graded bipartite graph $B(\Gamma)$ with the same underlying set but the poset relation is given by $x <_{B(\Gamma)} y \quad \Longleftrightarrow \quad x <_{\Gamma} y$ and $y$ is a maximal element of $\Gamma$. Since the equations you build are not affected by relations between non-maximal elements, the set of equations given by $\Gamma$ and $B(\Gamma)$ is the same. Furthermore, $B : \mathrm{FGP} \to \mathrm{BG}$ can be turned into a functor in the obvious way. Let $B_* : \mathrm{BG} \to \mathrm{FGP}$ denote the inclusion map. If we add the condition in this definition of a graded poset that the minimal value of the rank is $0$, then $B \circ B_* = \mathrm{id}_{\mathrm{BG}}$, so $B$ is a retraction which preserves your problem. You can turn the pair $(B_*,B)$ into an adjoint pair if you choose a relaxed definition for the morphisms in $\mathrm{FGP}$, namely that a map of posets $f : P \to Q$ is a morphism in $\mathrm{FGP}'$ if it maps maximal elements to maximal elements and non-maximal elements to non-maximal elements. This is the same as being a graded morphism of posets for a map between two bipartite graphs, but you will have the problem that if $X$ is finite graded bipartite and $Y$ is finite graded not bipartite (i.e. the rank function takes at least $3$ values), there are no morphisms from $X$ to $Y$ which sends maximal elements of $X$ to those of $Y$, hence $\mathrm{Hom}_{\mathrm{BG}}(X,B(Y))$ might not be empty but $\mathrm{Hom}_{\mathrm{FGP}}(B_*(X),Y) = \varnothing$. Taking $\mathrm{FGP}'$ instead of $\mathrm{FGP}$, the pair $(B_*',B')$ where $B' : \mathrm{FGP'} \to \mathrm{BG}$ and $B_*' : \mathrm{BG} \to \mathrm{FGP'}$ are defined analogously is adjoint. This is not natural from the point of the grading but it is from the point of view of your problem. I could have just said "you can restrict your attention to bipartite graphs", but since you deal with species, I wanted to show you it can actually be done functorially. What you mentioned in your question is that if $X$ is a finite graded poset and $F(X)$ the associated linear system, then $F(X) = F(X')$ for some graded bipartite graph $X'$. What I'm saying is that $F(X) = F(B(X))$, so I'm a bit more specific (but I'm essentially saying the same thing in a different language). Hope that helps,<|endoftext|> TITLE: How hard is it for a teaching-oriented lecturer to get tenure in mathematics in the United States? QUESTION [5 upvotes]: My questions is different from this post. Suppose a math's PhD is 1) not very successful in his PhD program research and fails to find an ideal postdoc.(But he is still eligible to gain a PhD degree) or, 2) not very successful in his research as a postdoc and fails to find a tenure-track position like an assistant professor. and he chooses to be a lecturer for a couple of years just to get rid of the heavy pressure of publishing papers but he is still ambitious about becoming a mathematician. My question is, is it extremely difficult for a teaching-oriented lecturer to get a tenure-track position in mathematics(assistant professor at least)? BTW, the only example I know about this is Yitang Zhang, whose experience was so special. But in general, I don't know any other examples of a lecturer becoming an assistant/associate professor in U.S. REPLY [5 votes]: There are a lot of 2-year and 4-year colleges in the US where nowadays the mathematics department prefers to hire Ph.D.s, but where no research is required for promotion and tenure. So I may disagree with Nate's "most" assertion. Now, if you still intend to do research, though: Such places probably have greater teaching duties than a research university, leaving you less time to do research.<|endoftext|> TITLE: Upper bound of the largest eigenvalue of a PSD block matrix in terms of blocks QUESTION [5 upvotes]: Let $\mathbf A=\left[\begin{matrix}\mathbf A_{11}&\mathbf A_{12}\\ \mathbf A_{21}&\mathbf A_{22}\end{matrix}\right]$ be a positive semi-definite matrix, $\mathbf A_{ij}\in\mathbb C^{n\times n}$ and $rank(\mathbf A)=n$. Prove or disprove that $$ \lambda_{max}(\mathbf A) + \lambda_{max}(\mathbf A_{11}+\mathbf A_{22}) \leq \lambda_{max}\left(\left[\begin{matrix}tr(\mathbf A_{11})&tr(\mathbf A_{12})\\ tr(\mathbf A_{21})&tr(\mathbf A_{22})\end{matrix}\right]\right) + tr(\mathbf A_{11}+\mathbf A_{22}), $$ where $\lambda_{max}$ denotes the largest eigenvalue and $tr$ denotes the trace. I tried small values of $n$. For $n=1$ we have "$=$" (the same terms in the LHS and RHS). I have a rather long proof for $n=2$ (in my proof I also replaced $\mathbb C$ by $\mathbb R$). It seems that some results on (completely) positive maps are somehow related to the inequality ( indeed, all the maps $$ \mathbf A\rightarrow \mathbf A_{11}+\mathbf A_{22},\qquad \mathbf A\rightarrow \left[\begin{matrix}tr(\mathbf A_{11})&tr(\mathbf A_{12})\\ tr(\mathbf A_{21})&tr(\mathbf A_{22})\end{matrix}\right],\qquad \mathbf A\rightarrow tr(\mathbf A_{11}+\mathbf A_{22}) $$ are positive). By Theorem 2.3.7 (The Russo-Dye Theorem) from the book R. Bhatia, Positive definite matrices, $$ \lambda_{max}\left(\left[\begin{matrix}tr(\mathbf A_{11})&tr(\mathbf A_{12})\\ tr(\mathbf A_{21})&tr(\mathbf A_{22})\end{matrix}\right]\right)\leq n\lambda_{max}(\mathbf A), $$ which is useless since we need inequality of the type $\lambda_{max}(\mathbf A)\leq\dots$ REPLY [3 votes]: Write $A=\sum w_i\otimes w_i$ where $w_i$ are orthogonal and $\|w_1\|=\max_i\|w_i\|$. Also write $w_i=(u_i,v_i)$. Then the inequality in question is just $$ \|u_1\|^2+\|v_1\|^2+\max_{e:\|e\|=1}\sum_i[|(u_i,e)|^2+|(v_i,e)|^2] \\ \le \left\|\begin{pmatrix}\sum_i\|u_i\|^2&\sum_i(u_i,v_i)\\\sum_i(v_i,u_i)&\sum_i\|v_i\|^2\end{pmatrix}\right\|+\sum_i(\|u_i\|^2+\|v_i\|^2) $$ Now, cancel the first term with the corresponding term on the RHS and estimate $ \sum_{i\ge 2}[|(u_i,e)|^2+|(v_i,e)|^2] $ by $ \sum_{i\ge 2}(\|u_i\|^2+\|v_i\|^2) $. Also, note that the norm of the sum of several positive definite matrices is not less than the norm of each of them, so we can remove all $i\ge 2$ in the matrix and reduce the problem to showing that $$ |(u_1,e)|^2+|(v_1,e)|^2\le \left\|\begin{pmatrix}\|u_1\|^2&(u_1,v_1)\\(v_1,u_1)&\|v_1\|^2\end{pmatrix}\right\| $$ However, we can write the matrix on the right as the sum of positive definite matrices of the kind $\begin{pmatrix}|(u_1,e_j)|^2& (u_1,e_j)(e_j,v_1)\\(v_1,e_j)(e_j,u_1)&|(v_1,e_j)|^2\end{pmatrix}$ where $e_j$ is an orthonormal basis with $e_1=e$. Again, we can remove all matrices with $j\ge 2$, after which we get an identity.<|endoftext|> TITLE: Comparing sizes of sets of natural numbers QUESTION [16 upvotes]: It seems natural to consider $\lim_{q \rightarrow 1^-} \sum_{n \in S} q^n - \sum_{n \in T} q^n$, when it exists, as a way of comparing the sizes of two sets $S,T \subseteq {\bf N}$ that have the same density; for instance, $\{0,2,4,...\}$ and $\{1,3,5,...\}$ both have density 1/2, but the first set might be said to contain "half an element more" than the second, based on the fact that $1-q+q^2-q^3+...$ converges to 1/2 as $q$ goes to 1. Has this (non-Archimedean, non-translation-invariant) refinement of the concept of density, or anything similar to it, been developed anywhere? Put $|S|_q = \sum_{n \in S} q^n$. Just as $|\{0,2,4,6,...\}|_q-|\{1,3,5,7,...\}|_q \rightarrow 1/2$ as $q \rightarrow 1^-$, it seems empirically that $$|\{0,1,4,9,16,...\}|_q-|\{0,2,6,12,20,...\}|_q \rightarrow 1/2,$$ $$|\{0,1,5,12,22,...\}|_q-|\{0,2,7,15,26,...\}|_q \rightarrow 1/3,$$ and $$|\{0,1,3,6,10,..,\}|_q - \sqrt{2} |\{0,1,4,9,16,...\}|_q \rightarrow \sqrt{2}/2.$$ Is there a place in the literature where results like this can be found, as part of a general setup for measuring sets that "sees" things at a finer level than mere density? To give one last example, racing "evil" versus "odious" numbers, it is easy to prove that $$|\{0,3,5,6,9,10,12,15,...\}|_q-|\{1,2,4,7,8,11,13,14,...\}|_q \rightarrow 0.$$ [ADDED ON MAY 17, 2017: There is now (at least) one article that uses analytic tools to generalize the notion of cardinality of sets of natural numbers: my preprint "One-Dimensional Packing: Maximality Implies Rationality", available at https://arxiv.org/abs/1704.08785 ] REPLY [4 votes]: Your post expresses the same idea about comparison of infinite sets as in my previous post with the difference being that you use Abel summation while I did use Ramanujan's summation and Zeta regularization. To be more precise, I used Faulhaber's formula for summation because it gives the same results as Ramanujan's summation (I do not know a proper term for this summation method): $$\sum _{x\ge0}^\Re f(x)=-\sum _{n=1}^\infty \frac {f^{(n-1)}(0)}{n!} B_n(0)$$ Now, addressing your concerns, expressed in the comments to the other answer, there is a set of similar summation methods that are mostly compatible with each other. One can see that some of them depend only on the values of the series at the integer points while others involve integrals and derivatives. So, it clearly means that there can be constructed examples where these methods give different results. My strong conviction, even if I do not have a proof, is that all these methods should give the same results for "well-behaving" functions. By "well-behaving" I mean functions that are equal to their Newton series: $$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$ I call such functions "discrete-analytic", "Newton-analytic" or "Newtonian". The criterion can also be written in the following form: $$f(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f(k)}{(x-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(x-k) k!(n-k)!}}$$ Some functions would have this series divergent, but their two-directional expansion would converge: $$f(x)=\lim_{n\to\infty}\frac{\sum _{k=-n}^n \frac{(-1)^k f(k)}{(x-k) (k+n)! (n-k)!}}{\sum _{k=-n}^n \frac{(-1)^k}{(x-k) (k+n)! (n-k)!}}$$ I would include such functions into "Newtonian" as well.<|endoftext|> TITLE: Find structure geometry of $A_1, A_2,...,A_n$ such that $\prod_{i0$. The Thompson problem corresponds to the minimization of the electrostatic potential energy when $s=1$. According to P.D. Dragnev, Log-optimal configurations on the sphere, Contemporary Mathematics Vol. 661 (2016) the Whyte problem is rigorously solved for the following 7 values of $n$ only, namely, $n=1,2,3,4$ : solutions are the ``trivial'' ones $n=5$ : two points at the Poles and three points forming an equilateral triangle on the equator (P.D. Dragnev, D.A. Legg, and D.W. Townsend 2002) $n=6$ : regular octahedron (A.V. Kolushov and V.A. Yudin 1997) $n=12$ :regular icosahedron (N.N. Andreev 1996) For large values of $n$, the numerical determination of optimal points is difficult as the number of local minima grows exponentially with $n$. Designing a fast algorithm to compute in polynomial time “nearly optimal” logarithmic energy points is Problem #7 in Smale's list of problems for this century. Configuration of 1600 points and Voronoi cells, near optimal with respect to the $s$-energy for $s=1$ (Image by R. Womersley from this paper)<|endoftext|> TITLE: Matrix integrals in combinatorics, for dummies QUESTION [5 upvotes]: This is actually about one particular question that I posted a while ago, "Special" meanders. Among several approaches tried is a huge subclass of approaches which can be generated from approaches to "usual" meanders. "Our" meanders seem to be essentially simpler - roughly, they relate to ordinary meanders in the same way as powers of two relate to Catalan numbers. So every time there is some approach in the general case, there is a hope it simplifies in our case. So much for a motivation. Now one of these approaches is something called the method of matrix integrals. This method seems to have impressive applications in combinatorics. I've tried to find out how to apply this method in our case, and failed miserably. What is maddening is that I found several really excellent explositions - e. g. in the book Graphs on Surfaces by Lando and Zvonkin there is a very informative chapter about it. A survey about meanders by Michael La Croix also contains a separate chapter. Finally, there are several papers by Di Francesco and collaborators where this method is systematically used to obtain important estimates and conjectures about general meandric numbers. And still when I start trying to apply this method to our particular problem, I am stuck from the very beginning. Maybe I am really stupid, I don't know - I simply cannot figure out how to start. So the question is: does there exist in the literature some sort of cookbook for the matrix integral method? Something for dummies, some collection of algorithms, explicit steps, with an input in form of some combinatorial data? REPLY [5 votes]: I'm not sure this qualifies as a "cookbook", but it's a tutorial introduction on the application of matrix integrals to combinatorics, with a quite complete overview of applications in a physics context: Formal matrix integrals and combinatorics of maps, Bertrand Eynard (2006).<|endoftext|> TITLE: Can an analytic function defined on a maximal torus be extended analytically to all the Lie group? QUESTION [15 upvotes]: Let $G$ be a compact group and $T$ a maximal torus on $G$. Suppose $f$ is an analytic function defined on $T$. Is there an analytic function $F$ on $G$ whose restriction agrees with $f$ on $T$? REPLY [21 votes]: This seems to be consequence of the paper Cartan, Henri: Variétés analytiques réelles et variétés analytiques complexes. Bull. Soc. Math. France 85 1957 77–99 Cartan shows more generally (see sections 6 and 7), that a real analytic function on a real analytic submanifold of $\mathbb R^n$ can be extended to a real analytic function on all of $\mathbb R^n$.<|endoftext|> TITLE: Are the Eigenvalues of the Frobenius on Crystalline cohomology bounded by degree? QUESTION [5 upvotes]: Excuse me if this question is trivial or trivially false, or not at this sites level. Lets work over an algebraically closed field of characteristic $p>0$, say $k$. The action of the Frobenius on $H^r_{crys}(X/W(k))$ gives us a sequence of positive rational numbers $a_i$ by taking the order of the eigenvalues. Now if $X$ has a good lift to $W(k)$, $Y$, say with torsion free $H^p(Y, \Omega_Y/W)$, I believe we can decompose the Frobenius morphism using Hodge decomposition to obtain an inequality $a_i\le r$. My question is whether this is true in general. This may be trivial, but I am a bit new to Crystalline cohomology so I can't figure it out. REPLY [10 votes]: I assume that you mean $H^r_{crys}(X/W(k))$ because $H^r_{crys}(X/k)$ is just the de Rham cohomology over $k$. Next, over arbitrary base field this definition of $a_i$'s(they are called slopes of Frobenius) does not make sense because Frobenius is only $p$-linear and its eignevalues a priori depend on the choice of a linearization(see Example 8.1.3 here). Over a finite field one can use that some power of Frobenius is linear, so eigenvalues $\beta_i$ of $F^n$ are well-defined and we put $a_i=\mathrm{ord}_p (\beta_i)/n$. Over arbitrary $k$ one should use Dieudonne-Manin classification(Thm 8.1.4 in the linked notes). Now, let us see what restrictions we have on these slopes. I know to proofs of the fact that on $H^r$ slopes are in the interval $[0,r]$ for $r\leq n$ and are in the interval $[r-n,n]$ for $r\geq n$. First one works in the case $X$ is projective. In what follows, let us denote by $H^r$ the cohomology $H^r_{crys}(X/W(k))/torsion$. First, there is a perfect pairing given by Poincare duality $$H^r\otimes H^{2n-r}\to W(k)(n)$$ where $W(k)(n)$ means $1$-dimensonal module with action of $F$ via $p^n$. This means that if $a_1,\dots,a_k$ are slopes of $F$ on $H^r$ then $n-a_1,\dots,n-a_k$ are slopes of $F$ on $H^{2n-r}$. So, we already see that slopes are in the interval $[0,n]$. For a proof of Poincare duality see, for example, Berthelot's book "Cohomologie Cristalline des Schemas de Characteristique p>0". Next, for porjective $X$ we have Hard Lefschetz theorem(it holds only rationally):$$L^{n-r}:H^r[1/p]\cong H^{2n-r}[1/p]$$ where $L$ is multiplication by hyperplane section, the Lefschetz operator. For $x\in H^r[1/p]$ we have $F(L^{n-r} x)=F(L^{n-r})F(x)=p^{n-r}L^{n-r}F(x)$ because Frobenius on projective space multiplies hyperplane section by $p$. This means that slopes on $H^{2n-r}$ are obtained fromf slopes on $H^{r}$ by adding $n-r$, so $\{a_1+n-r,\dots,a_k+n-r\}=\{n-a_1,\dots,n-a_k\}$. In other words, for every $a_i$ there exists $a_j$ such that $a_i+n-r=n-a_j$ so $a_i+a_j=r$ and thus all slopes $a_i$ are less or equal than $r$. The proof of Hard Lefschetz can be found in Katz and Messing "Some Consequences of the Riemann Hypothesis for Varieties over Finite Fields" Another proof which works, more generally, for arbitrary proper $X$ can be obtained using de Rham-Witt complex. A great account for this theory is given in the original paper by Illusie "Complexe de de Rham-Witt et cohomologie cristalline". In short, crystalline cohomology can be computed as hypercohomology of certain complex $W\Omega^{\bullet}$ of sheaves of abelian groups on $X$. It turns out that, modulo torsion, the corresponding spectral sequence degenrates(as in Hodge theory in characteristic zero) giving, in particular, that the subspace of $H^r[1/p]$ with slopes in the interval $[i,i+i[$ is equal to $H^{r-i}(W\Omega^i_X)[1/p]$ which gives that there are no slopes on $H^r$ less than $r-n$ which is exactly what we wanted combined with Poincare duality. REPLY [6 votes]: Here is an alternative way to prove that for $X$ projective of dimension $n$, the slopes on $H^r$ are in $[0,r]$, using instead of the Hard Lefshetz theorem (used in Sasha's answer), the Lefschetz hyperplane theorem, which is more elementary. By Poincaré duality, we know that the slopes are in $[0,n]$, so it is enough to consider $r \leq n$. We proceed by induction on the dimension $n$ of $X$. The result is obvious for $n=0$. For $X$ of dimension $n$, the result is true for $r=n$ by Poincaré duality. Fix a projective embedding of $X$ and let $Y$ be a generic hyperplane section. By Lefschetz hyperplane theorem, the natural maps $H^r(X) \rightarrow H^r(Y)$ are isomorphisms for $r TITLE: Recurrence of Legendre polynomial roots/ quadrature points QUESTION [6 upvotes]: Consider Legendre polynomials $p_n (x)$ on $[-1,1]$. For each $n \in \mathbb{N}$ we denote the zeros of $p_n (x)$ by $\left( x_j ^{(n)} \right) _{j=1} ^n$. We know that these roots are distinct, and we know that for all $n\in \mathbb{N}$ then $x_j ^{(n)} \neq x_i ^{(n+1)}$ for all $i$ and $j$. In fact, we even now that they are interlaced i.e. $x_i ^{(n+1)} < x_i ^{(n)} < x_{i+1} ^{(n+1)}$. On the other hand, we know for example that for all odd $n=2k+1$ we have that $x_{k+1} ^{(2k+1)} = 0$, and so I arrive at my question: Question 1: Are there any rules as to if, when, and how roots recur, i.e. $x_i ^{(n)} = x_j ^{(m)}$ for some relevant indices? Question 2: Apart from the well known Chebyshev polynomials, are there any other orthogonal polynomials with high rate of roots recurrence? REPLY [6 votes]: It is a conjecture of Stieltjes, apparently still open, see T.J. Stieltjes, Letter No. 275 of Oct. 2, 1890, in Correspondance d'Hermite et de Stieltjes, vol 2, Gauthier-Villars, Paris, 1905. that Legendre polynomials of different degrees have no common roots, except $x=0$ when both degrees are odd. Laguerre polynomials $L_{n}^{\alpha}$, $\alpha>-1$, orthogonal with respect to the weight function $x^{\alpha}e^{-x}$ on $(0,\infty)$, may have common roots. For instance, as observed in K. Driver, K. Jordaan, Stieltjes interlacing of zeros of Laguerre polynomials from different sequences. Indag. Math. (N.S.) 21 (2011), 204-211. $L_{2}^{23}(x)$ and $L_{4}^{23}(x)$ have $x=30$ as a common root. It seems also to be an open question how many common zeros are possible in general for $L_{n}^{\alpha}$ and $L_{n+m}^{\alpha}$, $n\geq 0$ and $m>1$. The corresponding question for Hermite polynomials about a common root other than zero seems to be unanswered as well. Finally, a classical theorem of Stieltjes states that if $p_{0},p_{1},p_{2},\ldots$ is any sequence of orthogonal polynomials then the zeros of $p_{k}$ and $p_{n}$, $k TITLE: Definability of defining classes QUESTION [6 upvotes]: Suppose that $M$ is a transitive class, denote by $\mathrm{HOD}(M)$ the class of all those sets which are hereditarily definable from ordinals and parameters in $M$. Some trivial examples include $\mathrm{HOD}(V)=V$, and $\rm HOD(Ord)=HOD$. Note that this model might not satisfy the axiom of choice, for example if $M=V$ and $V$ does not satisfy the axiom of choice. Is it possible for $M$ to be an inner model of $\sf ZFC$ such that $M$ is not definable (with parameters) in $\mathrm{HOD}(M)$? Alternatively, we can consider $L(M)$ as the smallest model of $\sf ZF$ such that $M\cap V_\alpha\in L(M)$ for all $\alpha$. Is $M$ always a class of $L(M)$? REPLY [6 votes]: Yes, this can happen. Let's start in $L$ and first perform Easton forcing to add a Cohen subset to every regular cardinal, giving rise to the forcing extension $L[G]$. Next, over $L[G]$ we force to code every set into the GCH pattern, for example, by iterating the forcing that forces GCH or its negation with a lottery sum at each such cardinal stage. In the resulting forcing extension $V=L[G][H]$, we therefore have $\newcommand\HOD{\text{HOD}}\HOD^V=V$, since every set is coded. Next, over $V$, force to add a $V$-generic class $A$ of the regular cardinals. This forcing adds no sets, and $A$ is not a definable class in $V$, but we may consider it in GBC. Let $M=L[G\upharpoonright A]$ be the inner model arising by taking only the sets of $G$ at coordinates in $A$. So this is a transitive inner model of $V$, but it is not first-order definable in $V$, since $A$ is generic over $V$. But meanwhile, $\text{HOD}(M)$ as computed in $V$ will be the same as $\HOD$, since it is already all of $V$ and so the extra parameters of $M$ didn't add anything. So we have an inner model $M$ that is not definable in $\HOD(M)$, as desired. Let's now modify the example to produce a model $V$ of ZFC with a definable inner model $M$, such that $M$ is not definable in $\HOD(M)$. Start in $L$ as before, and this time, add the class $A$ first, over $L$. In particular, every initial segment of $A$ is in $L$. Now, over the GBC model $L[A]$, force to add Cohen sets at every uncountable regular cardinal, forming $L[A][G]$. Finally, using product forcing rather than an iteration, we may force to code every set in $L[G]$, and also the classes $G$ and $A$, into the GCH pattern, producing an extension $V=L[A][G][H]$ in which $\HOD=L[G]$. (For example, the methods of my paper G. Fuchs, J. D. Hamkins, J. Reitz, Set-theoretic geology show how to do this, while also controlling the mantle and the generic $\HOD$.) Let $M=L[G\upharpoonright A]$. This is a transitive inner model of ZFC. It is definable in $V$, since the class $G\upharpoonright A$ is definable there. Further, since every initial segment of $A$ is in $L$, it follows that $L[G\upharpoonright A]\subset L[G]$, which is the $\HOD$ of $V$, and so $\HOD(M)=\HOD=L[G]$ in $V$. But $A$ is not definable in $L[G]$, since $A$ and $G$ are mutually generic. So $M$ is not definable in $L[G]$. Which is to say, $M$ is a definable transitive inner model of $V$, but it is not definable in $\HOD(M)$, as you desired.<|endoftext|> TITLE: Extending the tangent bundle of a submanifold QUESTION [8 upvotes]: Let $X$ be a complex manifold, and $Y\subset X$ a compact submanifold. Is it true that the tangent bundle $TY$ may be extended (as a holomorphic vector bundle) to some open neighbourhood of $Y$ in $X$? P.S. This looks like a very simple question, but somehow I am stuck. REPLY [3 votes]: Let us discuss first the differntial geometry case. Here, you can use the tubular neighbourhood theorem to extend the vector bundle. Explicitly, you have a neighbourhood $U$ of $Y$ in $X$ which is isomorphic to a neighbourhood of the zero section of the normal bundle. So you have a projection $$ \pi\colon U \to Y $$ and use it to pull-back $TY$ to $U$. In complex geometry, the tubular neighbourhood theorem holds just locally, because you do not have partition of unity. However, under some apporpiate condition, this theorem holds, so you can perform the same construction. This is a classical result of Grauert. See for instance the survey http://w3.impa.br/~hossein/myarticles/imca03.pdf and references therein.<|endoftext|> TITLE: Ehresmann's theorem over the $p$-adics QUESTION [19 upvotes]: I am looking for a version of Ehresmann's theorem for analytic manifolds over the $p$-adic numbers $\mathbb{Q}_p$ or, more generally, local fields. I follow the conventions from Serre's book "Lie algebras and Lie groups" concerning analytic manifolds over local fields. Recall that Ehresmann's theorem states that a proper submersion between smooth manifolds is a locally trivial fibration. Does a version of this hold for analytic manifolds over $\mathbb{Q}_p$? Namely, is a proper submersion between analytic manifolds over $\mathbb{Q}_p$ a locally trivial fibration? REPLY [3 votes]: It seems to me that the following lines essentially show that the answer is yes. Let $f\colon X\to Y$ be a proper submersion of $p$-adic manifolds, we want to show that $f$ is locally trivial on the target: every point $y$ has an open neighborhood $V$ such that $f_V\colon f^{-1}(V)\to V$ is isomorphic to the projection $p_2\colon U\times V\to V$ of a product. By the inverse function theorem, $f$ is locally trivial on the source: for every point $x\in X$, there exists an open neighborhood $W$ of $x$ such that $f|_W\colon W\to f(W)$ is isomorphic to a product. Let $y\in Y$ and let us cover the compact manifold $f^{-1}(y)$ by a finite family $(W_i)$ of open subsets on which $f$ is isomorphic to a product $U_i\times V_i \to V_i$. Let us refine it to a disjoint finite covering, still denoted by the same letter, of the same form. The union $W=\bigcup W_i$ is an open neighborhood of $f^{-1}(y)$. Since $f$ is proper, there exists an open neighborhood $V$ of $y$ such that $W$ contains $f^{-1}(V)$. Replacing $V_i$ by $V$, we are reduced to the case $V_i=V$ for all $i$. It is now visible that $f^{-1}(V)$ is isomorphic to the projection of the product $U\times V\to V$, where $U$ is defined as the disjoint union of the $U_i$.<|endoftext|> TITLE: Series solution of the trinomial equation QUESTION [5 upvotes]: The roots of trinomial equations $x^p+x-q=0$ ($p\in\mathbb{N}$) can be expressed in terms of the hypergeometric functions. I am wondering if at least one real root, for instance given by the following iterative solution $x_n=\frac{q - (1 - p) x_{n-1}^p}{ 1+p x_{n-1}^{(p - 1)}}$, $x_0=q^{1/p}$ can be expressed in a similar form for $p\in\mathbb{R}$, $p\ge2$. REPLY [13 votes]: An analogous formula does hold, although the corresponding functions are not hypergeometric if $p$ is irrational. For given $p\in\mathbb{R}$, $p>1$, consider the power series $$h(z)=\sum_{k=0}^{\infty} \frac{(-1)^k}{pk-k+1}\binom{pk}{k}\, z^k$$ with radius of convergence $R=(p-1)^{p-1}/p^p.$ Then, for $0\le y\le R^{1/(p-1)}$, the function $g(y):=yh(y^{p-1})$ is the inverse function of $f(x):=x+x^p$. $$*$$ [edit] There is also an analogous inversion formula for three or more terms, to invert e.g. $f(x)=x+ax^p+bx^q$ with real exponents $p>1$ and $q>1$. If $H=H_{p,q}$ is the analytic function $$H(u,v)=\sum_{i\ge0,j\ge0}\frac{(-1)^{i+j}}{ (p-1)i+ (q-1)j+1} {pi+qj \choose i,\, j}u^iv^j,$$ then $g(y):=yH(ay^{p-1},by^{q-1})$ is the local inverse of $f$ at $0$ (the multinomial coefficient in the double series is ${pi+qj \choose i,\, j}:=\frac{(pi+qj)(pi+qj-1)\dots(pi+qj-i-j+1)}{i!j!}$ .)<|endoftext|> TITLE: Voevodsky's Triangulated Categories of Motives and their Relationships QUESTION [17 upvotes]: As we know, Voevodsky constructed several candidates for the triangulated category of motives using different constructions and topologies (h, qfh, etale, and Nisnevich). I would like to know what motivated these specific topologies, and the relationship between these different categories (I know the ones mentioned in Voevodsky's papers, but I want a big picture answer). In particular, I want to know the reason(s) one expects these topologies to naturally give rise to the triangulated category that is supposed to be the derived category of the conjectured abelian category of motives. I should also add a related question: why consider the etale topology if Algebraic K-theory does not satisfy etale descent? REPLY [12 votes]: I won't embark on the difficult question of what one wants out of a category of motives, but I can make some comments on what might motivate the various choices of topologies. Nisnevich (aka completely decomposed, aka completely decomposed étale). If one defines cohomology with support of a closed subspace $Z \subseteq X$ in such a way to induce long exact sequences $\dots \to H^n(X) \to H^n(X{-}Z) \to H^{n-1}_Z(X) \to H^{n-1}(X) \to \dots$, then the Nisnevich topology (on the category of smooth schemes) makes sure that for any regular closed immersion of smooth $k$-varieties $Z \to X$, there is an isomorphism $$ H_Z^\bullet(X) \cong H_{\mathbb{A}^{d-c}}^\bullet(\mathbb{A}^d) $$ when $X$ and $Z$ are of pure dimension $d$ and $d-c$ respectively. (Notice that this statement needs smooth schemes to work). One example of a very serious consequence of this is the Gersten exact sequence for homotopy invariant sheaves with transfers [Voevodsky, Coh.th.of.presh.w.transfers, Theorem 4.37]. Others include Localisation [Morel-Voevodsky, $\mathbb{A}^1$-hom.th.of.sch., Theorem 3.2.21], and Purity [Morel-Voevodsky, $\mathbb{A}^1$-hom.th.of.sch., Theorem 3.2.23]. For other benefits of the Nisnevich topology (cohomological dimension $=$ Krull dimension, $K$-theory has descent, direct image is exact) see [Morel-Voevodsky, $\mathbb{A}^1$-hom.th.of.sch., Beginning of Section 3]. Finite (a.k.a.fs, a.k.a. finite surjective). This is the topology on the category $Sch/k$ of separated finite type schemes whose covering families are jointly surjective families of finite morphisms. It is a way to build in transfers, in the sense that the canonical functors $$Shv_{fs}(Cor/k, \mathbb{Z}) \stackrel{\sim}{\to} Shv_{fs}(Sch / k, \mathbb{Z})$$ $$PreShv(Cor/k, \mathbb{Q}) \stackrel{\sim}{\leftarrow} Shv_{fs}(Sch / k, \mathbb{Q})$$ are equivalences of categories. (The situation is more subtle if one wants to use only smooth schemes). One important use of this is to compare "cycle theoretic" theories (like Suslin homology, or Bloch's higher Chow groups) with "sheaf theoretic" theories (like étale cohomology). Cf. [Suslin-Voevodsky, Sin.hom.of.ab.alg.var.] where they prove that algebraic singular homology $H^{sing}_i$ (now called Suslin homology $H^S_i$) can calculate the classical topological singular homology (which can be calculated via sheaf cohomology) for $\mathbb{C}$-varieties $X$. $$ H_i^{S}(X, \mathbb{Z} / n) \cong H_i(X(\mathbb{C}), \mathbb{Z} / n) $$ Cf. also Suslin's article "Higher Chow groups and étale cohomology" where he shows that Bloch's higher Chow groups calculate étale cohomology with compact support for equidimensional quasi-projective varieties over algebraically closed fields $k$ with char.$k \nmid m$, $i \geq d = $dim $X$. $$CH^i(X, n; \mathbb{Z} / m) \cong H_c^{2(d-i) + n}(X, \mathbb{Z}/m(d-i))^\# $$ The qfh-topology. This is the topology generated by the Zariski topology, and the finite topology. $$ \mathrm{qfh} = \langle \textrm{finite surjective, Zariski} \rangle. $$ Remark: It is finer than the étale topology, so we have qfh $= \langle $fs, Zar$ \rangle = \langle $fs, Nis$ \rangle = \langle $fs, étale$ \rangle$. Proper cdh (which should be called the cdp $=$ completely decomposed proper topology). Every scheme admits a Zariski covering by affine schemes. So properties of Zariski sheaves are completely determined by their values on affine schemes. Analogously, (if the base field admits resolution of singularities) every (finite type) $k$-scheme admits a proper cdh covering by smooth $k$-schemes. So just as we have $$ Shv_{Zar}(Aff/k) = Shv_{Zar}(Sch /k) $$ we have (if res.of.sing. holds for $k$) $$ Shv_{cdp}(Sm/k) = Shv_{cdp}(Sch /k). $$ (the sites on the left are equipped with the induced topologies). Cohomologically, (and in general, i.e., without any dependance on res.of.sing) for any proper birational morphism (i.e., blowup) $X' \to X$ which is an isomorphism outside of a closed subscheme $Z \subset X$, setting $Z' = Z \times_X X'$ there is canonical long exact sequence $$ \dots \to H^n_{cdp}(X, F) \to H^n_{cdp}(X', F) \oplus H^n_{cdp}(Z, F) \to H^n_{cdp}(Z', F) \to H^{n + 1}_{cdp}(X, F) \to \dots $$ for any cdp sheaf $F$, and the same is true of any topology finer than the cdp topology (e.g., cdh, h, see below). In fact, the cdp topology is the coarsest topology with this property. In fact, if a strong version of resolution of singularities holds, the cdp topology is the coarsest topology with the above long exact sequence for blowups of smooth schemes in smooth centres. cdh (a.k.a., completely decomposed h-topology). The cdh topology is generated by the proper cdh topology and the Nisnevich topology. $$ \textrm{cdh} = \langle \textrm{Nisnevich}, \textrm{proper cdh} \rangle. $$ So we get the above cohomology with supports property for smooth closed immersions, the blowup long exact sequence, and, assuming res.of.sing. the power to reduce arguments to smooth schemes. Inconsequential remark: The name is a slight misnomer. We have $h = \langle$ étale, proper $\rangle$, and $cdh = \langle$ comp.dec.étale, comp.dec.proper $\rangle$, but while every cdh covering is a completely decomposed $h$-covering, the converse is not true, comp.dec.h $\neq$ cdh. There are completely decomposed flat coverings which are not cdh coverings. Proper. The topology on $Sch/k$ whose coverings are jointly surjective families of proper morphisms. Without assuming resolution of singularities, a theorem of de Jong on alterations implies that every $k$-scheme admits a proper covering by smooth schemes. $$ Shv_{\textrm{proper}}(Sm/k) = Shv_{\textrm{proper}}(Sch /k).$$ The h-topology. The $h$-topology is generated by the proper topology and the Zariski topology. $$h = \langle \textrm{proper}, \textrm{Zariski} \rangle.$$ It is generated by a number of other combinations of the previously mentioned topologies, e.g., $h = \langle$cdh, finite$\rangle = \langle $proper, étale $\rangle$. $L$dh (a.k.a. $L$-decomposed topology) The proper topology has the disadvantage from the cdh topology that many interesting cohomology theories satisfying cdp descent, only satisfy proper descent after passing to rational coefficients. Between cdp and proper lies the $L$dp-topology. If $L$ is a collection of primes, then one defines a morphism $f: Y \to X$ to be $L$-decomposed if for every (not necessarily closed) point $x \in X$, there exists a point $y \in f^{-1}\{x\}$ such that $k(y) / k(x)$ is a finite field extension of degree prime to every element of $L$. If $L = \{\ell\}$ then we just write $\ell$dp. We have $\varnothing$dp $=$ proper, and $\mathbb{P}dp = cdp$ if $\mathbb{P}$ is the set of all primes. It follows from a theorem of Gabber that every for any choice of $\ell \neq $char.$k$, every (finite type) $k$-scheme admits an $\ell$dp-covering by smooth $k$-schemes. Many cohomology theories of interest satisfy $\ell$dp descent after passing to $\mathbb{Z}_{(\ell)}$-coefficients, so we can still obtain information about their $\ell$-torsion using this topology. Just as we have $cdh = \langle $Nis, cdp$\rangle$ and h $= \langle$ Nis, proper $\rangle$, we could define $$ \ell dh = \langle \textrm{Nis}, \ell dp \rangle. $$ However, in practice, it is much more useful to use the definition $\ell$dh $ = \langle$ cdh, fps$\ell' \rangle$ where fps$\ell'$ is the topology whose coverings are $\ell$-decomposed families of finite flat surjective morphisms, see below. This way questions about $\ell$dh descent be be broken up into a cdh descent part and an fps$\ell'$ descent part. Here is the relationship: $$ \begin{array}{cccc} \textrm{Set of primes} & \textrm{Topology} & \textrm{Coefficients} & \textrm{Theorem} \\ \hline \textrm{all primes} & cdh & \mathbb{Z} & \textrm{Hironika's res.of.sing.} \\ \{\ell\} & \ell dh & \mathbb{Z}_{(\ell)} & \textrm{ Gabber's theorem on alterations } \\ \varnothing & h & \mathbb{Q} & \textrm{ de Jong's theorem on alterations } & \end{array} $$ The fps$\ell'$ topology. This is a version of the finite topology. It is generated by morphisms which are finite flat surjective of degree prime to $\ell$, where $\ell$ is a prechosen prime number. The point is that it can be much easier to show that a cohomology theory has "trace" morphisms for finite flat morphisms, than that it has transfers. If one defines an appropriate notion of presheaves with traces (cf. https://arxiv.org/abs/1305.5349) then it is easy to show that every presheaf of $\mathbb{Z}_{(\ell)}$-modules with traces is an fps$\ell'$ sheaf, and one can leverage this to show an equivalence of categories $$ Shv_{cdh}(Cor/k, \mathbb{Z}_{(\ell)}) = Shv_{\ell dh}(Cor/k, \mathbb{Z}_{(\ell)}) $$ Note, the same is true with h and $\mathbb{Q}$ replacing $\ell$dh and $\mathbb{Z}_{(\ell)}$.<|endoftext|> TITLE: Finitely generated soluble groups with uncountably many maximal subgroups QUESTION [12 upvotes]: Is there a finitely generated soluble group with uncountably many maximal subgroups? Any classification of such groups? REPLY [17 votes]: Yes, there is a finitely generated soluble group with uncountably many maximal subgroups. Fix an odd prime $p$. Denote by $F_p$ the field on $p$ elements, $C_2$ the cyclic group on $2$ elements. Let $A$ be the set of functions from $\mathbf{Z}$ to $\{-1,1\}$. Let $M$ be the free $F_p$-module on the generators $(e_i)_{i\in\mathbf{Z}}$. Consider the lamplighter group $L=C_2\wr\mathbf{Z}$, with generators $u,t$ with $u^2=1$ and $u$ commuting with its conjugates. For every $f\in A$, endow $M$ with a structure of an $L$-module, where $t$ acts by the shift $e_i\mapsto e_{i+1}$ and $u$ acts diagonally by $u\cdot e_i=f(i)e_i$. We write it as $M_f$ to emphasize the dependency on $f$. Each $f\in A$ can be viewed as a bi-infinite word in the letters $\pm 1$. Let $A'$ be the set of "universal" sequences, that is, those sequences for which every finite word occurs at least once. Then if $f\in A'$, the module $M_f$ is a simple module. Indeed, consider any nonzero element $w=\sum_{i=k}^l a_ie_i$ with $a_k,a_l$ nonzero. Then by the assumption, we can find some conjugate $u'$ of $u$ mapping $e_i$ to $-e_i$ for all $i = k, \dots, l - 1$ and $e_l$ to $e_l$. Then $u'w + w = 2 a_l e_l$ generates $F_p e_l$. In turn using $t$ we get all the other basis elements. Let $N=F_p[L]$ be the group algebra. Then mapping $1$ to $e_0$ defines a surjective $F_p[L]$-module homomorphism $p_f:N\to M_f$. Let $I_f$ be its kernel. We claim that Elements of $L$ can be written as $g=(t^n,P)$ with $P$ a finite subset of $\mathbf{Z}$. Let $\delta(g)$ be the corresponding basis element of $N$. Then $$p_f(\delta(t^n,P))=p_f((t^n,P)\delta(1,\emptyset))=(t^n,P)e_0$$ $$=(t^n,\emptyset)(\prod_{i\in P}f(i))e_0=(\prod_{i\in P}f(i))e_n.$$ In particular, $p_f(\delta(1,\{i\}))=f(i)e_0$, and $p_f(\delta(1,\{i\})-\delta(1,\emptyset))=(f(i)-1)e_0$. In particular, $\delta(1,\{i\})-\delta(1,\emptyset)\in I_f$ iff $f(i)=1$. This shows that $f\mapsto I_f$ is injective. Since $A'$ is clearly of continuum cardinality, this shows that $(I_f)_{f\in A'}$ is a family of continuum cardinality of maximal submodules of $N$. Therefore, in the group $N\rtimes L$ (which is solvable of length 3, generated by 3 elements), we have a family $(I_f\rtimes L)_{f\in A'}$ of maximal subgroups, of continuum cardinality. Of course we can pull this back to any group having $N\rtimes L$ as a quotient, such as the free solvable group of any length $\ge 3$ on $\ge 3$ generators. Note: 3 is the minimal derived length. Indeed, finitely generated metabelian groups have maximal subgroups of finite index, hence there are at most countably many. This holds more generally in finitely generated nilpotent-by-polycyclic groups.<|endoftext|> TITLE: Incomparable dense linear orderings extending $\langle \mathbb{R},< \rangle$ QUESTION [7 upvotes]: Where $a TITLE: Multiplicativity twisted Hochschild Kostant Rosenberg isomorphism QUESTION [15 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{C}$. I call (following Swan) Hochschild cohomology of $X$ the graded algebra: $$ \mathrm{HH}^{\bullet}(X) := \mathrm{Ext}^{\bullet}_{X \times X}(\Delta_* \mathcal{O}_X, \Delta_* \mathcal{O}_X),$$ where $\Delta : X \rightarrow X \times X$ is the diagonal embedding. By adjunction, we have $\mathrm{HH}^{\bullet}(X) = \mathrm{Ext}^{\bullet}_{X}(\Delta^* \Delta_* \mathcal{O}_X, \mathcal{O}_X)$. Using the Koszul complex locally for $X \subset X \times X$ and $N_{X/X \times X} \simeq \Omega_X$, one proves: $$\mathrm{HH}^{\bullet}(X) \simeq \bigoplus_{p+q = \bullet} H^p(X, \bigwedge^q T_X).$$ This isomorphism of graded vector spaces is often called the Hochschild-Kostant-Rosenberg isomorphism. Note that this is NOT a ring isomorphism in general. One of Kontsevich's (numerous) insights was that twisting it with the square root of the Todd class of $X$ makes it a ring isomorphism. I know that there are many references on the subject (Markarian, Ramadoss, Caladararu, Yekutieli, Calaque-VandenBergh etc...), but I am not able to follow their proofs. Could someone give some hints on how to prove that twisting with the square root of the Todd class makes this isomorphism multiplicative? If possible, I would like to have a proof which is as low-tech as one can be. I really thank you in advance for your help! REPLY [7 votes]: I am far from being expert in this subject, but I will try to present my understading of there this multiplicativity comes from. I wiil refer to authors you mention but only to the parts which I hope you will find readable. First, let me give an interpretation of isomorphism $$\mathrm{HH}^{\bullet}(X) \simeq \bigoplus_{p+q = \bullet} H^p(X, \bigwedge^q T_X).$$ due to Markarian(https://arxiv.org/abs/math/0610553). Shifted tangent bundle $T_X[-1]$ is a Lie algebra object, the bracket given bu Atiyah class. Markarian shows that its universal enveloping algebra is precisely the algebra of Hohschild cochains $$U(T_X[-1])=\mathcal{RHom}^{\bullet}(\Delta_*\mathcal{O}_X,\Delta_*\mathcal{O}_X)$$ He defines universal enveloping algebra as an algebra admitting PBW isomorphism(Def 4) and in this case PBW is precisely the quasi-isomorphism between $Sym(T_X[-1])=\bigoplus_i\Lambda^i T_X[-i]$(symmetric power becomes exterior because of sign rule applied to shift by $(-1)$) and $\mathcal{RHom}^{\bullet}(\Delta_*\mathcal{O}_X,\Delta_*\mathcal{O}_X)$ that you sketched. See Section 7 of https://arxiv.org/abs/math/0602653 for an elaboration of Markarian's arguement. There is also proven a general statement that a(we do not a priori know that it is unique) universal enveloping algebra in the sense of Markarian actually satisfies the usual universal property. Now, let us interpret taking hypercohomology in terms of Lie algebras. Any object $\mathcal{F}\in D(X)$ carries a functorial action of $T_X[-1]$ given by Atiyah class $$T_X[-1]\otimes \mathcal{F}\to \mathcal{F}$$ In particular, сonsidering $\mathcal{O}_X$ as a module over $T_X[-1]$ we get $$R\Gamma(X,\mathcal{F})=RHom_{T_X[-1]}(\mathcal{O}_X,\mathcal{F})$$ The RHS is precisely the Lie algebra cohomology of $T_X[-1]$ with coefficents in $\mathcal{F}$ because $Hom$ space from $\mathcal{O}_X$ is equal to $T_X[-1]$-invariants. This reduces our problem to the following purely algrebaic problem: for a Lie algebra $\mathfrak{g}$ how one should change the PBW isomorphism so that it induces an isomorphism of cohomology rings $$H^{\bullet}(\mathfrak{g},S(\mathfrak{g}))\cong H^{\bullet}(\mathfrak{g},U(\mathfrak{g}))$$ The faсt that twisting by square root of Todd class does the work is proven in M. Pevzner and C. Torossian, "Isomorphisme de Duflo et cohomologie tangentielle", see http://math.univ-lyon1.fr/~calaque/LectureNotes/LectETH.pdf for an exposition.<|endoftext|> TITLE: A simplified proof of a BSD inequality in char $p$? QUESTION [5 upvotes]: One of the main results of the BSD conjecture in characteristic $p>0$ is the fact that the algebraic rank is less or equal than the analytic rank and that there is an equality iff the $\ell$-torsion ($\ell\neq p$) of the Tate-Shafarevich group is finite. The proof that I know of this fact (which goes back to Tate) uses the elliptic surface associated to the elliptic curve, proves an analogue of this result for the surface and goes back to the elliptic curve. Does there exist a proof of this result without using the elliptic surface? REPLY [6 votes]: Let me briefly sketch Tate's proof for context. If $X$ is a surface over $\mathbb{F}_q$ then Frobenius acts on the Neron-Severi group $NS(X)$ by multiplication by $q$. Using the cycle map from the theory of etale cohomology $NS(X)\otimes \mathbb{Q}_{\ell}$ injects into $H^2_{et}(\bar{X},\mathbb{Q}_{\ell})$ and, by the first statement, lands in the $q$-eigenspace of the action of Frobenius and the rank $r$ of $NS(X)$ is at most the multiplicity of this eigenvalue. An obvious consequence of this is the inequality $r \le \dim H^2_{et}(\bar{X},\mathbb{Q}_{\ell}) = b_2(X)$. This latter inequality was proved by Igusa before Tate (Igusa, Jun-ichi, Betti and Picard numbers of abstract algebraic surfaces. Proc. Nat. Acad. Sci. U.S.A. 46 1960 724–726.). This proof uses a fibration of the surface as an essential tool and can perhaps be turned into a proof using elliptic curves. I don't know if you can do the same with the full Tate inequality. Whether one would want to do this is another question.<|endoftext|> TITLE: Is the triadic odometer isomorphic to its square? QUESTION [5 upvotes]: While doing some simulations dealing with the triadic odometer $T$, I observed that the results were the same when I replace $T$ with $T^2$. Therefore I'm wondering whether $T$ is isomorphic to $T^2$. Is it true? REPLY [7 votes]: Yes. More generally, if $T$ is a $k$-odometer, and $r$ is relatively prime to $k$, then $T^r$ is conjugate (that is, measure-theoretically isomorphic) to $T$. Relative primeness is necessary in order that $T^r$ be ergodic. One way to prove this (among several) is to use the characterization of odometers by their spectral properties. Another is to go directly to the adding machine, and notice that $T^r$ represents adding $r$, and then drawing the corresponding diagram (I haven't a clue about how to put a commuting diagram here) to show (topological, with this method) conjugacy. In more detail (for the latter method), let $A$ be the $k$-adics. Viewed as a ring, $A$ has $r$ as a unit; hence multiplication by $r$, $p:= \times r: A\to A$, is a homeomorphism (continuous, invertible, measure-preserving, everything, etc). Then $pT =T^r p$, yielding a conjugacy. And yet another way to see this, is to realize $T$ and $T^r$ as matrix-valued walks, and there is a not-so-difficult argument to show their Poisson boundaries are equal as $Z$-spaces. [This refers to the classification result of Giordano and me, Matrix-valued random walks and variations on property AT, Münster J Math 1 (2008) 15-72. This paper is too much to go through merely to obtain this result; however, it can be used to also deal with the case of other odometers, corresponding to supernatural numbers. $k$-odometers form a very small subclass of these.]<|endoftext|> TITLE: Outer Automorphisms and the Mapping Class Group of Lie Groups QUESTION [7 upvotes]: Let $G$ be a connected Lie Group. We will denote the outer automorphism group of $G$ by $Out(G)$. Consider $\pi_0(Diffeo(G))$, the mapping class group of the underlying manifold of $G$, and the map, $k:Out(G)\rightarrow \pi_0(Diffeo(G))$ This map is well defined as follows: Given $\phi_1, \phi_2\in Aut(G)$ such that $\phi_1=g\phi_2g^{-1}$, we may construct a path $\alpha: [0,1]\rightarrow G$ such that $\alpha(0)=g$ and $\alpha(1)=e$, the identity in $G$. Then $H(w,t)=\alpha(t)\phi_2(w)\alpha(t)^{-1}$ is an isotopy with $H(w,0)=g\phi_2g^{-1}=\phi_1$ and $H(w,1)=\phi_2$. Consider the case where $G=T^2$, the $2$-torus. Then $Out(T^2)\cong GL_2(\mathbb{Z})$ and $\pi_0(Diffeo(T^2))\cong Out(\pi_1(T^2))\cong GL_2(\mathbb{Z})$ via the Dehn-Nielsen-Baer Theorem. In this case, $k$ is an isomorphism. Another case is the $3$-sphere, in which $Out(S^3)\cong 0$ and $\pi_0(Diffeo(S^3))\cong\mathbb{Z}_2$. In this case, $k$ is injective. Is there a case where $k$ is surjective? If not, what is the obstruction? Moreover, are there higher dimensional examples of the bijective case? It seems that for higher dimensions, the smooth structures may make this problem computationally intractable. REPLY [4 votes]: Your question is largely intractable with present technology. Little is known about smooth mapping class groups of high-dimensional manifolds. But in dimensions 3 and below, there is considerable knowledge. . . unfortunately there are not many Lie groups in those dimensions. The mapping class group of $SO_3$ is trivial. I am not certain of who proved it originally, but the uniqueness of the genus 1 Heegaard splitting would be one way to prove it. This approach would be due to Bonahon. So for $SO_3$ you have surjectivity. I believe there are no other Lie Groups where there any kind of upper bound on the elements of $\pi_0 Diff(G)$ known, so to get any results of the kind you are interested in would require some substantial work.<|endoftext|> TITLE: Explicit 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}, U(1)]$ QUESTION [7 upvotes]: I would like to know the explicit expression of 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}, U(1)]$ with $U(1)\equiv \mathbb{R}/\mathbb{Z}$ coefficient, or namely $H^2[Q_8 \times \mathbb{Z}_2, U(1)]$, here we denote the cyclice group $\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_2$, and the $Q_8$ is the order 8 quaternion group. What I had computed (from Kunneth formula, Universal Coefficient Theorem) and what I had known is that: $$H^2[Q_8 \times \mathbb{Z}_2, U(1)]=\mathbb{Z}_2 \times \mathbb{Z}_2.$$ Now I would like to know the two generators of $H^2[Q_8 \times \mathbb{Z}_2, U(1)]=\mathbb{Z}_2 \times \mathbb{Z}_2$ in terms of explicit 2-cocycles. Any Reference and any partial answer are welcome. Additional remarks and guides: What I also computed and knew are:$H^1[Q_8,U(1)]=\mathbb{Z}_2 \times \mathbb{Z}_2$, $H^2[Q_8,U(1)]=0$, $H^3[Q_8,U(1)]=\mathbb{Z}_8$. I also know that the explicit 3-cocycle $\omega_3((A,a),(B,b),(C,c)) \in H^3[Q_8,U(1)]=\mathbb{Z}_8$ as: $$\omega_3((A,a),(B,b),(C,c))$$ $$= \exp\left( \frac{2\pi i p}{8} \{ (-)^{B+C}a \left( (-)^C b + c- [(-)^C b + c +2BC] \right) - 2 ABC \} \right)$$ where $p \in \mathbb{Z}_8$. The rectangular brackets means the modulo $4$ in the range $-1,0,1,2$. Here we denote the elements of $Q_8$ by the 2-tuples $$ (A,a) := X^A R^a \qquad \qquad \mbox{with $A \in 0,1$ and $a \in -1, 0, 1, 2$. } \qquad $$ with $R^{4} = e, \qquad X^2 \; = \; R^2, \qquad XR \; = \; R^{-1} X.$ So for instance the identity is $e=(0,0)$. The multiplication law of $Q_8$ then reads $$ (A,a) \cdot (B,b) = ([A+B], [(-)^B a+b+2AB]), $$ REPLY [2 votes]: For any two finite groups $G$ and $H$ the Kuenneth formula will give you a homomorphism $$H^1(G,U(1))\otimes_{\mathbb{Z}} H^1(H,U(1))\to H^2(G\times H,U(1)).$$ We can describe this map explicitly. Let $\phi:G\to U(1)$ and $\psi:H\to U(1)$ be group homomorphisms. For some $n$ and some $n$-th root of unity $\zeta$ we can write $\phi(g) = \zeta^{f(g)}$ and $\psi(h) = \zeta^{r(h)}$ where $f:G\to \mathbb{Z}/n$ and $r:H\to\mathbb{Z}/n$ are group homomorphisms (in the case of $Q_8\times \mathbb{Z}/2$ $\zeta$ will just be $-1$). The cocycle which corresponds to $(\phi,\psi)$ will then be given by $$\alpha((g_1,h_1),(g_2,h_2)) = \zeta^{f(g_2)r(h_1)}.$$ For the example that interests you, you can write explicitly all group homomorphisms from $Q_8$ and from $\mathbb{Z}/2$ to $U(1)$ and get the isomorphism you are interested in. We have the following ring theoretic interpretation of this: the data of a two cocycle on a group is the same as describing the twisted group algebra $\mathbb{C}^{\alpha} G\times H$. In this group algebra we will have the rules $U_{g_1}U_{g_2} = U_{g_1g_2}$, $U_{h_1}U_{h_2} = U_{h_1h_2}$ and $U_hU_g = \zeta^{f(g)r(h)}U_gU_h$. In principal, what Kuenneth Formula tells us for the two dimensional case is the following thing: in order to describe a two cocycle on $G\times H$ we need to describe three things: Its restriction to $G$. Its restriction to $H$. The way that ot alters the commutativity between $G$ and $H$. In this case we know that the second cohomology groups of $G$ and of $H$ are trivial, so we are left only with the third option.<|endoftext|> TITLE: Geometric Meaning of Different K-theories QUESTION [14 upvotes]: There is a whole collection of algebraically defined K-theories. My understanding is that algebraic K-theory is a presheaf of spectra $K$ on $\textbf{Sch}/S$ such that the homotopy groups of $K(X)$ give the algebraic K-groups of the $S$-scheme $X$. What we can do is give $\textbf{Sch}/S$ a grothendieck topology, and find a fibrant replacement of $K$ in the local stable model structure on presheaves of spectra on $\textbf{Sch}/S$. These give us different K-theories. For example, if we take the etale topology (and take $K/n$) we get etale K-theory (with mod $n$ coefficients). What is this process of finding a fibrant replacement geometrically? What is happening, and what does etale K-theory, say, measure? REPLY [11 votes]: Running off Denis' answer: the geometric content of Thomason's answer is surprising and he says this himself in the introduction to that legendary paper. Let me say some stuff that I've unpacked from his comments: On the one hand, one can recover the intersection theory of algebraic varieties from $K$-theory: the Gersten resolutionin $K$-theory tells us that if $X$ is a regular $k$-scheme $X$, then $H^n_{Zar}(X; \mathcal{K}_n) \simeq CH^n(X)$. However, one consequence of Thomason's theorem is that $K/\ell^{\nu}[\beta^{-1}](X) \simeq K^{Top}/\ell^{\nu}(X)$ whenever $X$ is $\mathbb{C}$-scheme. This tells us that the process of Bott-inversion, which gives us etale descent, somehow reduces the subtleties of algebro-geometric intersection theory to the cruder topological case (someone more knowledgeable about the Hodge conjecture can perhaps comment on this). If you believe that $K$-theory is about vector bundles, then the above theorem tells us (vaguely, I guess) that the "difference" between topological and algebraic vector bundles lie in $\beta$-torsion phenomenon. Since $\beta \in K^2(X, \ell^{\nu})$, the similarity between algebraic and topological vector bundles seems to lie only in $K_0, K_1$, i.e., they are the "same up to codimension one" (to paraphrase Thomason) by the formula $H^n_{Zar}(X; \mathcal{K}_n) \simeq CH^n(X)$ above Just as Denis indicated, after Thomason's theorem, we get a spectral sequence that makes no reference to etale $K$-theory: its $E^2$-page is $H^p_{et}(X; \mathbb{Z}/\ell^{\nu}(q))$ and it converges to $K/\ell^{\nu}_{p-q}(X)[\beta^{-1}]$. Since the positive homotopy groups of etale $K$-theory and Bott-inverted $K$-theory are the same, it tells us that mod $\ell^{\nu}$-connective $K$-theory has a strong relationship with etale cohomology! This seems to be a general philosophy. For example, although $K$-theory is not $\mathbb{A}^1$-invariant on smooth schemes, Weibel proved that mod $\ell^{\nu}$-K-theory is (not saying that you can reproved this theorem using Thomason, however). This seems to indicate that, the fact that torsion etale cohomology is much better behaved than integral etale cohomology, is also shared by $K$-theory. Lastly, I seem to be indicating that the distance between etale $K$-theory and usual $K$-theory is this Bott inversion. However, the recent work of Clausen, Mathew, Noel, Naumann tells us that this isn't really the case - the distance between etale $K$-theory and usual $K$-theory is covered by "telescopic localization" where Bott-inversion is just one case of telescopic localization.<|endoftext|> TITLE: Decoding a Remark of Gödel on Complexity Theory QUESTION [5 upvotes]: In Gödel's Collected Works (Vol 2), there is a discussion of von Neumann which was brought about by a query, made to Gödel, concerning the existence of a Turing machine which is so complex that its total halting behavior is "asymptotically more complicated to describe than the machine itself". To this, Gödel replied: "What von Neumann had in mind is probably best understood through the notion of a Universal Turing machine $U$. There, it might be said that a complete description of its behavior is infinite because, in view of the non-existence of a decision procedure predicting $U$'s behavior, any $\textit{complete}$ description of $U$ could only be given by an enumeration of all instances. Thus, the Universal machine, where the ratio of the two complexities is infinite, might then be considered as a limiting case for all Turing machines." [italics mine]. What I am puzzled about is the precise statement of what Gödel asserts as "the ratio of the two complexities" which tends to infinity. From the context, I gather that this is the ratio between (1) the procedure which supplies the "complete enumeration of $U$" and (2) $U$ itself with respect to its termination. Clearly, he considers this as a type of worst-case analysis for $U$, but unfortunately I cannot gather a sharp statement about the procedure which supplies the complete enumeration so that I could prove that this is indeed a worst-case analysis or simply something else that I have missed. Is there a proof related to this fact? REPLY [3 votes]: I just read that as saying U's description is finite in size, but describing U's behavior across all inputs needs an infinite description. There is no finite algorithm to say which inputs halt.<|endoftext|> TITLE: Estimate number of solutions in the Roth's theorem QUESTION [10 upvotes]: There is a fundamental theorem in Diophantine approximation : For all algebraic irrational $\alpha$ $$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^{2 + \epsilon}}$$ with $\epsilon>0$, has finitely many solutions. can we estimate number of solutions $N_{\alpha}(\epsilon)$? for instance : what is the upper bound of $N_{\sqrt[3]{2}}(1)$? number of solutions for $\sqrt[3]{2}$, with $\epsilon=1$. $$\displaystyle \left \lvert \sqrt[3]{2}- \frac{p}{q} \right \rvert < \frac{1}{q^{3}}$$ REPLY [11 votes]: I believe that there is a completely explicit upper bound for $N_\alpha(\epsilon)$ (more generally counting in relative number fields and using more then one, possibly non-archimedean, absolute value) in the following paper: Robert Gross, A note on Roth's theorem. J. Number Theory 36 (1990), no. 1, 127–132. MR1068678 Of course, it's not going to be pretty.<|endoftext|> TITLE: Defining hom spaces in the derived category as limits of hom spaces in the homotopy category QUESTION [5 upvotes]: Let $C$ be an abelian category and $K(C)$ the homotopy category of complexes in $C$. I've seen the following claimed in several sources (without proof): A. The following isomorphisms hold: $$\lim_{X' \underset{qis}\to X} Hom_{K(C)}(X',Y) \widetilde\to \lim_{X' \underset{qis}\to X,Y \underset{qis}\to Y'} Hom_{K(C)}(X',Y') \widetilde\leftarrow \lim_{Y \underset{qis}\to Y'}Hom_{K(C)}(X,Y') $$ The limits (actually colimits in this case) are taken over all quasi-isomorphisms. Why is this true? And how can one prove this? Can this statement be upgraded to a statement about the internal Hom bifunctor? Something along the lines of: B. The following quasi-isomorphisms hold: $$\lim_{X' \underset{qis}\to X} Hom^{\bullet}(X',Y) \widetilde\to \lim_{X' \underset{qis}\to X,Y \underset{qis}\to Y'} Hom^{\bullet}(X',Y') \widetilde\leftarrow \lim_{Y \underset{qis}\to Y'}Hom^{\bullet}(X,Y')$$ REPLY [6 votes]: This is actually a standard property of the homotopy category of complexes on which construction of the derived category is based, formulated in unusual way. For a quasi-isomorphism $s:X'\to X$ and $f\in Hom(X',Y)$ denote by $fs^{-1}$ the image of $f$ in the first colimit(in a second we will justify the choice of notation). By description of the filtered colimit, two images $fs^{-1}$ and $gt^{-1}$(for $t:X''\to X,g\in Hom(X'',Y)$) coincide iff there exists $X'''$ and qisms $p:X'''\to X',q:X'''\to X''$ and a map $r\in Hom(X''',Y)$ such that $sp=tq,fp=gq$. This is exactly the condition for fractions $fs^{-1}$ and $gt^{-1}$ to be equivalent in the right localization by qisms(see e.g. Weibel, 10.3). Analogously, the third colimit can be identified with the set of equivalence classes of left fractions. Now, Ore conditions guarantee that these two sets of equivalence classes coincide, which is precisely the isomorphism you are looking for. Edit: As Denis-Charles Cisinski explains, the answer to B is negative. Indeed, let's pick a non-zero object $A$ in $C$ and consider, for instance, $X=A[0], Y=A\xrightarrow{1} A$ where $Y$ is a contractible complex concentrated in degrees $-1,0$. The second and third colimits are zero as the category of quasi-isomorphisms $Y\to Y'$ has a final object $Y\to 0$. But the first colimit is not zero: for $X'=X$ there is a non-zero element $Id_A\in Hom(A,A)=Hom^0(X,Y)$ which survives in the colimit because for a quasi-isomorphism $X'\to X$ the map $(X')^0\to A$ has to be non-zero.<|endoftext|> TITLE: What's the (monoidal) image of a monoidal functor? QUESTION [8 upvotes]: For an ordinary functor $F\colon \mathcal{C} \to \mathcal{D}$ of categories, there is a construction $\operatorname{im} F$, the image of $F$, which is again a category, and $F$ factors through that image. Is there anything vaguely like the image of a strong monoidal functor, which should be monoidal again? What certainly doesn't work straightforwardly is just taking the image of the underlying functor and somehow putting a monoidal structure on it. Given $X, Y\colon \mathcal{C}$, how would you define $FX \otimes FY$? If you take the tensor product in $\mathcal{D}$, the result might not be in the image of $F$, but neither can you define it to be $F(X \otimes Y)$, since $F$ might no be injective on objects (so you can't find out a unique object to start with, and there is no canonical way to choose). I'm utterly surprised I've never encountered such a construction. The only thing I've come across is for the case of fusion categories. There, the full subcategory spanned by summands of objects in the image of $F$ can be defined. The only thing I can come up with for plain monoidal categories is the category spanned by the image of the underlying functor and all isomorphic objects. But this doesn't have good properties if I want to extend structures on $\mathcal{C}$ to $\mathcal{D}$. For example, if I have a braiding $c_{X,Y}\colon X \otimes Y \to Y \otimes X$ on $\mathcal{C}$, I can transport it onto objects of the form $FX$, but not onto all objects isomorphic to an $FX$, since I don't know which isomorphism to transport it along. REPLY [4 votes]: Let $F : \mathcal{C} \to \mathcal{D}$ be a strong monoidal functor. Since monoidal categories are non-strict structures in the sense that e.g. associativity of $\otimes$ only holds up to isomorphism, it makes sense to look at those objects of $\mathcal{D}$ which are isomorphic to objects of the form $F(X)$, where $X$ is an object of $\mathcal{C}$. These constitute a full subcategory of $\mathcal{D}$, which contains the unit object and is closed under tensor products, hence inherits a monoidal structure. If (only) $\mathcal{C}$ has a braiding, there is no reason to expect that it induces a braiding on the image, since this image depends on $F$. For this, we should require that $\mathcal{D}$ is also braided and that $F$ is a braided monoidal functor. It follows that the image, as defined above, carries a braiding induced by $\mathcal{D}$, and that $F$ factors as a braided monoidal functor into it.<|endoftext|> TITLE: What is Kept Fixed for Flexible Spheres QUESTION [10 upvotes]: For background to this question much recent exciting related things, see this videotaped lecture by Alexander Gaifullin. Consider a triangulation $K$ of a two-dimensional sphere and consider maps from the vertices of $K$ to $R^3$. A one-parameter family $\phi_t$ of such maps is called a flexing if the distance $d(\phi_t(v),\phi_t(u))$ is constant whenever $v$ and $u$ are adjacent vertices. A flexing is trivial if it keep the distances between every two vertices. We can extend the maps of the vertices linearly to a map from $K$ considered as a geometric simplicial complex to $R^3$. When $\phi (K)$ is an embedding we can consider the solid body $L$ inside it. The following fundamental facts are known: 1) (Bricard) If $K$ is the boundary complex of the Octahedron then there are maps of the vertices of $K$ with non-trivial flexes. (However when you extends linearly these maps to $K$ this is no longer an embedding.) 2) (Gluck) based on Alexandrov and Stainitz) A generic map of $K$ to $R^3$ is rigid. 3) (Connelly) There are embedded flexible spheres 4) (Sabitov) The volume of $L$ along a flexing of a sphere is constant for every flexing. (This extends to the case where we don't have an embedding by regarding the volume as a signed sum of volumes of bounded complements of the image of $K$.) I would like to know what else might be fixed for flexing of 2-spheres into $R^3$. Q1: (Perhaps this was asked by Bob Connelly) Are Dehn-invariants fixed for a flexing. In other words, are two embedded spheres along the flexing scissor-congruent? Q2: Are the eigenvalues of the Laplacian of the (silid 3D-dimensional) body constant along the flexing? Q3: Are lengths of closed geodesics in $L$ constant for a flexing? Explanation (Added later, see Joe's comment): Here we consider $L$ as a 3-dimensional Billiard "table" and by "geodesics" we refer to billiard paths. (Of course, positive answers to Q1 and Q2 would be a far-reaching if not far-fetched extension of Sabitov's theorem, and Q3 is also rather far-fetched.) I don't know the answers to Q1, Q2, Q3 even for the known examples of flexible spheres. I dont know if Q2,Q3 can be extended to the non-embeddable case but I suspect Q3 could. REPLY [10 votes]: I think the answer to Q3 is negative. Take a flexible polyhedron and choose two adjacent faces $F_1$, $F_2$ such that the angle between them changes during the deformation. Now attach to $F_2$ a small tetrahedron so that one of its faces is parallel to $F_1$. If this is done near the edge $F_1 \cap F_2$, then there appears a short closed billiard trajectory (of period $2$). This trajectory is destroyed during the deformation. The answer to Q2 seems to be negative as well (the idea came from a discussion with Daniel Grieser). The Laplacian eigenvalues determine the simple part of the length spectrum (that is those lengths that are realized by a unique billiard trajectory). In the above example there are infinitely many geodesics between the parallel planes, therefore it doesn't work. Instead of adding a tetrahedron one should create a chamber that has a unique short closed billiard trajectory (with one of the collision points in the face $F_1$), and such that after rotating $F_1$ around the edge $F_1 \cap F_2$ the length of this trajectory changes. Since the simple length spectrum changes, the Laplacian eigenvalues change as well.<|endoftext|> TITLE: Is there an explicit formula for the hessian of "Determinant"? QUESTION [29 upvotes]: Let $f: G= \mbox{GL}(n,\mathbb{R}) \to \mathbb{R}$ be the determinant function. Then $\mbox{Hess} (f)$ is a two linear map on $M_{n}(\mathbb{R})\simeq T_{e}(G)$ where $e$ is the neutral element of $G$, the identity matrix. What is an explicit formula for this Hessian? (In terms of matrix terminologies) REPLY [52 votes]: The formula you're looking for can be obtained by differentiating Jacobi's formula $$ \frac{\mathrm{d}}{\mathrm{d}t} \det A(t) = \det A(t) \cdot \operatorname{tr}\left( A^{-1} \frac{\mathrm{d}A}{\mathrm{d}t} \right) $$ with respect to a second parameter, say $s$: \begin{multline} \frac{\partial^2}{\partial s \partial t} \det A(s,t) = \det A(s,t) \cdot \bigg[ \operatorname{tr}\left( A^{-1} \frac{\partial A}{\partial s} \right) \operatorname{tr}\left( A^{-1} \frac{\partial A}{\partial t} \right) \\ + \operatorname{tr}\left( A^{-1} \frac{\partial^2 A}{\partial s \partial t} \right) - \operatorname{tr}\left( A^{-1} \frac{\partial A}{\partial s} A^{-1} \frac{\partial A}{\partial t} \right) \bigg] \end{multline} Now take $s = A_{ij}$ and $t = A_{kl}$, so $\frac{\partial A}{\partial s} = E_{ij}$ is the matrix with a one in its $(i,j)$-entry, and zeros elsewhere. Similarly $\frac{\partial A}{\partial t} = E_{kl}$, and $\frac{\partial^2 A}{\partial s \partial t} = 0$. The desired Hessian is then \begin{align*} \operatorname{Hess}(\det)_A(U,V) &= U_{ij} V_{kl} (\det A)\bigg[ \operatorname{tr}\left( A^{-1} E_{ij} \right) \operatorname{tr}\left( A^{-1} E_{kl} \right) - \operatorname{tr}\left( A^{-1} E_{ij} A^{-1} E_{kl} \right) \bigg] \\ &= U_{ij} V_{kl} (\det A)\bigg[ (A^{-1})_{mn} (E_{ij})_{nm} (A^{-1})_{pq} (E_{kl})_{qp} \\ &\hspace{4cm} - (A^{-1})_{mn} (E_{ij})_{np} (A^{-1})_{pq} (E_{kl})_{qm} \bigg] \\ &= U_{ij} V_{kl} (\det A)\bigg[ (A^{-1})_{mn} \delta_{in} \delta_{jm} (A^{-1})_{pq} \delta_{kq} \delta_{lp} \\ &\hspace{4cm} - (A^{-1})_{mn} \delta_{in} \delta_{jp} (A^{-1})_{pq} \delta_{kq} \delta_{lm} \bigg] \\ &= U_{ij} V_{kl} (\det A)\bigg[ (A^{-1})_{ji} (A^{-1})_{lk} - (A^{-1})_{li} (A^{-1})_{jk} \bigg] \\ &= \det A \bigg[ U_{ij} (A^{-1})_{ji} V_{kl} (A^{-1})_{lk} - U_{ij} (A^{-1})_{jk} V_{kl} (A^{-1})_{li} \bigg] \\ &= \det A \bigg[ \operatorname{tr}(U A^{-1}) \operatorname{tr}(V A^{-1}) - \operatorname{tr}(U A^{-1} V A^{-1}) \bigg] \\ \end{align*} with the Einstein summation convention in full force throughout. By evaluating this formula at $A = e,$ the $n \times n $ identity matrix, we obtain the desired special case $$ \operatorname{Hess}(\det)_e(U,V) = \operatorname{tr}(U) \operatorname{tr}(V) - \operatorname{tr}(U V). $$<|endoftext|> TITLE: Lowenheim numbers for ordinary language quantifiers QUESTION [7 upvotes]: (First posted in StackExchange) I am interested in the Löwenheim numbers associated with quantifiers that are prevalent in ordinary language use, such as "more" and "most". Definition: The Löwenheim number of a logic L, $\ell$(L), is the least cardinal $\mu$ such that any satisfiable sentence in L has a model of cardinality less or equal to $\mu$ if such exists. Otherwise, $\ell$(L)=$\infty$. For a logic L and a quantifier $Q$, let L($Q$) be the logic L with $Q$ added to its language. Let $\mathcal{L}$ denote standard first order logic. So, for instance: Let $Q_\alpha$ be the unary monadic quantifier ``there are at least $\aleph_\alpha$ many''. We have $\ell$($\mathcal{L}(Q_\alpha)$)=$\aleph_\alpha$ for each ordinal $\alpha$. Another example: Let the Härtig quantifier $I$ be the binary monadic quantifier stating equal cardinality of sets: $I$ is a binary monadic quantifier such that $M\models Ix(\varphi x,\psi x)$ iff $|(\varphi x)^M|=|(\psi x)^M|$. $\ell(\mathcal{L}(I))$ is very high, and is independent of ZFC. $\ell(\mathcal{L}(I))$ is a fixed point of the function $\alpha\mapsto\aleph_\alpha$, and further, Magidor and Väänänen showed that it is consistent with ZFC both that $\ell(\mathcal{L}(I))$ is under the first weakly inaccessible cardinal and that it is above the measurable cardinal (Magidor and Väänänen, 2011). Are there known results for "most" and "more"? I could only find the following bounds: Let $Most$ be the binary monadic quantifier such that $M\models Most\ x (\varphi x, \psi x)$ iff $|(\varphi x)^M\backslash (\psi x)^M|<|(\varphi x)^M\cap(\psi x)^M|$. Then I've found the following lower bound: $\ell(\mathcal{L}(Most))\geq\aleph_\omega$. Let $More$ be the binary monadic quantifier such that $M\models More\ x (\varphi x, \psi x)$ iff $|(\varphi x)^M|>|(\psi x)^M|$. Then we have both $\ell(\mathcal{L}(More))\geq\ell(\mathcal{L}(I))$ and $\ell(\mathcal{L}(More))\geq\ell(\mathcal{L}(Most))$ (due to considerations of expressive power). Are there better results? And are there other quantifiers from ordinary language that have interesting results? REPLY [5 votes]: $\def\L#1{\mathcal L(#1)}\let\fii\varphi\let\eq\leftrightarrow\DeclareMathOperator\I{I}\DeclareMathOperator\most{Most}\DeclareMathOperator\more{More}$I don’t know what other quantifiers from ordinary language are there in the wild, but the three mentioned in the question have the same Löwenheim number: Theorem: $\ell(\L\I)=\ell(\L\most)=\ell(\L\more)$. As already noted in the question itself, the inequalities $$\ell(\L\I)\le\ell(\L\more),\qquad\ell(\L\most)\le\ell(\L\more)$$ follow immediately from the fact that the quantifiers $\I x$ and $\most x$ are definable from $\more$ using $$\begin{align*} \I x\,(\fii x,\psi x)&\eq\neg\bigl(\more x\,(\fii x,\psi x)\lor\more x\,(\psi x,\fii x)\bigr),\\ \most x\,(\fii x,\psi x)&\eq\more x\,(\fii x\land\psi x,\fii x\land\neg\psi x). \end{align*}$$ For the $$\ell(\L\more)\le\ell(\L\most)$$ inequality, note that $\more$ is almost definable from $\most$: if $(\fii x)^M$ or $(\psi x)^M$ is infinite, then $$M\models\more x\,(\fii x,\psi x)\eq\most x\,(\fii x\lor\psi x,\neg\psi x).$$ Rather than trying to figure out if we can fix this definition to work for finite sets as well, let me use a different approach: we expand the models in such a way that we can define disjoint union. Let $\alpha$ be an $\L\more$-sentence. Choose a fresh binary function symbol $p(x,y)$ and constants $a,b$. Let $\theta$ be the FO sentence stating that $p$ is a pairing function, and $a\ne b$. Let $\alpha^*$ be the sentence $\alpha$ in which each subformula of the form $$\more x\,(\fii x,\psi x)$$ is replaced with $$\most z\,(\fii'z\lor\psi'z,\fii'z),$$ where $$\begin{align*} \fii'z&\eq\exists x\,(z=p(x,a)\land\fii x),\\ \psi'z&\eq\exists x\,(z=p(x,b)\land\psi x). \end{align*}$$ Clearly, any infinite model in the signature of $\alpha$ can be expanded to a model of $\theta$, and if $M\models\theta$, then $M\models\alpha\eq\alpha^*$. Thus, for any infinite $\kappa$, $\alpha$ has a model of cardinality $\kappa$ if and only if the $\L\most$-sentence $\theta\land\alpha^*$ has a model of cardinality $\kappa$. Since $\alpha$ was an arbitrary $\L\more$-sentence, $\ell(\L\more)\le\ell(\L\most)$ follows. For the remaining $$\ell(\L\more)\le\ell(\L\I)$$ inequality, we can use a similar expansion approach. Let $\alpha$ be an $\L\more$-sentence, and let $<$ be a fresh binary relation symbol. We define $\L\I$-sentences $\alpha^\#$ and $\xi$ as follows: $\alpha^\#$ is $\alpha$ with each subformula $$\more x\,(\fii x,\psi x)$$ replaced with $$\exists v\,(\I x\,(\psi x,x TITLE: Best known bounds on (border) ranks of small matrix multiplication tensors? QUESTION [8 upvotes]: The $(m,n,p)$-matrix multiplication tensor is a representation of the bilinear map $T\colon\mathbb{R}^{m\times n}\times\mathbb{R}^{n\times p}\rightarrow\mathbb{R}^{m\times p}$ given by $T(A,B)=AB$. Any low-rank decomposition of this tensor leads to a faster-than-naive matrix multiplication algorithm (à la the Strassen algorithm). Actually, to get a fast algorithm, it suffices for arbitrarily small perturbations of this tensor to have low rank; the minimum of such ranks is called the border rank. Where can I find a recent table of the best known upper and lower bounds on the rank and border rank of the $(m,n,p)$-matrix multiplication tensor for moderately small values of $m$, $n$ and $p$? Slide 24 here is close to what I want, though I would prefer explicit citations and larger $m$, $n$ and $p$. Here's a related question that focuses on the $(3,3,3)$ case. REPLY [3 votes]: There is a table at http://cristal.univ-lille.fr/~sedoglav/FMM/ that (try to) gather such bounds<|endoftext|> TITLE: Generalization of cycle decomposition to Coxeter groups QUESTION [12 upvotes]: I'm looking for a generalization of cycle decompositions for permutations to elements of Coxeter groups. (For the purposes of this question, any conjugate of a parabolic subgroup is also a parabolic subgroup.) One can think of an $r$-cycle in $S_n$ as a (conjugate of a) Coxeter element for some parabolic subgroup $S_r$. This means one can think of the cycle decomposition as finding a parabolic subgroup - or equivalently a decomposition of the root space as a direct sum of root subspaces - and realizing your permutation as a Coxeter element for that parabolic subgroup. In the Coxeter group of type B, one can realize elements as signed permutations and take a signed cycle decomposition; the cycles with an odd number of signed elements are Coxeter elements in a parabolic subgroup of type B and the cycles with an even number of signed elements are Coxeter elements in a parabolic subgroup of type A. As far as I can tell, this doesn't seem to work so well for type D. Is there some other way to recover this idea, ideally for all Coxeter groups, but I'd be happy with finite reflection groups or even just type D? If it helps, the property of Coxeter elements I'm interested in is that they have maximal reflection length. Uniqueness of decomposition would be nice but might not be necessary. REPLY [5 votes]: Following Nathan's advice let me elaborate a bit on my comment and also provide an answer. As pointed out it is not true in general that any element $w$ in a finite Coxeter group is a Coxeter element in some reflection subgroup (see the counterexample above). Actually I have been thinking recently about this question of generalizing the cycle decomposition to arbitrary finite Coxeter groups and here is how I did it. I am stating the result here and briefly explaining. The details and proofs can be found on a preprint which I just put on my webpage http://www.mathematik.uni-kl.de/fileadmin/AGs/agag/gobet/ex/cycle.pdf (it will appear on arxiv soon but there is some additional stuff which I have not written yet which will be added). Any element $w$ in a finite Coxeter group $(W,S)$ has a generalized cycle decomposition in a suitable sense, but unicity fails in general. It works as follows. Let $(W,S)$ be finite and let $w\in W$. Denote by $T$ the set of reflections of $W$, that is, the set of all conjugates of the elements of $S$, and by $\ell_T$ the length with repect to the generating set $T$ and by $\leq_T$ the absolute order on $W$ induced by $\ell_T$. Write $\mathrm{Red}_T(w)$ for the set of reduced $T$-decompositions of $w$ (i.e., smallest length factorizations of $w$ as products of reflections). I denote by $P(w)$ the parabolic closure of $w$, that is, the smallest parabolic subgroup containing $w$. I am considering the following condition on $w$: (Condition A) There exists $(t_1, \dots, t_k)\in\mathrm{Red}_T(w)$ such that the reflection subgroup $\langle t_1, t_2, \dots, t_k\rangle$ is parabolic. Then I claim the following Theorem [Generalized cycle decomposition] Let $w\in W$ satisfying Condition A. Then there exist $x_1,x_2,\dots, x_m\in W$ such that $x_i x_j= x_j x_i$ for all $i, j=1,\dots, m$, $w=x_1 x_2\cdots x_m$ and $\ell_T(w)=\ell_T(x_1)+\ell_T(x_2)+\cdots+\ell_T(x_m)$, Each $x_i$ admits a reduced $T$-decomposition generating an irreducible parabolic subgroup $P_i$ (in fact, $P(x_i)$) of $W$ and $$P(w)=P_1\times P_2\times\cdots\times P_m=P(x_1)\times P(x_2)\times\cdots\times P(x_m).$$ Moreover, such a decomposition of $w$ is unique up to the order of the factors. In the case of the symmetric group of course the $x_i$'s are just the cycles occuring in the cycle decomposition of $w$. Moreover in type $A$ every element satisfies Condition $A$. Now let me say a few words about elements for which Condition A fails. In fact, elements satisfying Condition A above are the so-called "(parabolic) quasi-Coxeter elements", which we characterized together with Baumeister, Roberts and Wegener as the elements $w$ for which the Hurwitz operation is transitive on $\mathrm{Red}_T(w)$ in this paper http://www.degruyter.com/view/j/jgth.ahead-of-print/jgth-2016-0025/jgth-2016-0025.xml which Christian pointed out. Now if $w$ does not satisfy Condition A, it precisely means that the Hurwitz operation on $\mathrm{Red}_T(w)$ has several orbits. Taking a reduced expression of $w$ in one of the orbits, $w$ is a quasi-Coxeter element in the reflection subgroup generated by the reflections in this reduced expression (which, as any reflection subgroup, is a Coxeter group), hence $w$ has a generalized cycle decomposition in that subgroup. In this situation $w$ will have as many generalized cycle decompositions as the number of Hurwitz orbits on $\mathrm{Red}_T(w)$. So in fact, somewhat surprinsingly, any (parabolic) quasi-Coxeter element (as the element in $D_4$ above which yielded a counterexample to Nathan's claim) still has a generalized cycle decomposition (I mean: including unicity). The obstruction to unicity comes from elements where no $T$-reduced expression generates a parabolic subgroup, such as for instance the longest element in type $B_2$. Finally, note that this heavily requires properties which only hold in finite Coxeter groups (such as the characterization of parabolic subgroups as centralizers of subspaces), hence I have no idea about possible generalizations to arbitary Coxeter groups... It would be very interesting to be able to characterize parabolic subgroups as centralizers of subspaces in a suitable faithful representation of $W$ in the infinite case... EDIT: I realized that Brady and Watt is not needed for unicity - unicity is already clear from the direct product decomposition.<|endoftext|> TITLE: Classes of Numbers that are easy to factorize using Classical Computers? QUESTION [5 upvotes]: The Carmichael Numbers can be factorized in polynomial time. Are weak fermat pseudoprimes (w.r.t. a given base) easy to factorize as well? What are some large/broad classes of numbers that are easy to factorize (i.e. in polynomial time). This part is pretty broad, so i would be thankful if someone can provide some examples of these classes and more importantly a few good up-to date sources, that studies/lists the classes of numbers and their factorization difficulty ? REPLY [3 votes]: Here is a large class of numbers that can be factored in polynomial time: products of a smooth number and large prime. More precisely, a number less than $x$ whose second largest prime factor is less than $\exp(O((\log \log x)^2/\log \log \log x))$ can be factored in polynomial time. There are about $$x(\log\log x)^2/(\log x \log\log\log x)$$ such numbers, so this is quite a large class. You can read more in Granville's survey Smooth numbers: computational number theory and beyond.<|endoftext|> TITLE: A group with more than one Hausdorff minimal nontrivial group topologies QUESTION [5 upvotes]: I have a few examples of a group $G$, equipped with a Hausdorff minimal nontrivial group topology $\cal T$. This means that $\cal T$ is Hausdorff and for any nontrivial (not necessarily Hausdorff) group topology $\cal S$ on $G$ with $\cal S\subseteq T$ we have $\cal S = T$. However, in these examples $\cal T$ is unique. Is there an example of a group $G$ such that there are more than one Hausdorff minimal nontrivial group topology on $G$? REPLY [5 votes]: Yes, many. Here is an example. Take two different locally compact groups with minimal Hausdorff topologies, say $G_1=\text{PSL}_2(\mathbb{R})$ and $G_2=\text{PSL}_2(\mathbb{Q}_p)$ for some prime $p$ (endowed with their usual topologies, which are minimal by Theorem 5.3 of http://arxiv.org/pdf/1408.4217.pdf), and take a group $\Gamma$ that injects densely into both, say $\Gamma=\text{PSL}_2(\mathbb{Q})$ (you could also take $\Gamma$ to be a free group here). Take $T_i$ on $\Gamma$ to be the pull back of the topology of $G_i$. Note that for each $i=1,2$, the group $G_i$ is the completion (wtr the two sided uniform structure) of $(\Gamma,T_i)$, by the fact that $\Gamma$ is dense and $G_i$ is complete (as it is locally compact). In particular, $T_1\neq T_2$. We are left to show that the topologies $T_1$ and $T_2$ are minimal. Assume that for some $i\in\{1,2\}$, $S\lneq T_i$ is a weaker topology. Let $H$ be the completion of $(\Gamma,S)$. The map $\Gamma\to H$ extends canonically to $G_i\to H$ and we get by minimality of $G_i$ that $H$ is trivial. Thus $S$ is trivial. As a reply to a comment below, note that the groups $G_i$ are necessarily simple: otherwise they would have a weaker (non-Hausdorff) topologies.<|endoftext|> TITLE: Is the mapping torus of an automorphism of a free group virtually an amalgamated product? QUESTION [5 upvotes]: Let $F$ be a nonabelian finitely generated free group, let $\tau \in \mathrm{Aut}(F)$ be an element of infinite order, and set $G = F \rtimes \mathbb{Z}$, where the action of $\mathbb{Z}$ on $F$ is given by $\tau$. Must there be a finite index subgroup $U$ of $G$, such that $U \cong A*_CB$ where $C$ is a proper subgroup of both $A$ and $B$, and its index in either $A$ or $B$ is at least $3$? What happens if $\tau$ is inner? REPLY [3 votes]: If you drop "virtually" from your question, then this has been analyzed very thoroughly by Brinkmann. Whether this is directly relevant to your question is, of course, open to debate. Peter Brinkmann, MR 1934698 Splittings of mapping tori of free group automorphisms, Geom. Dedicata 93 (2002), 191--203.<|endoftext|> TITLE: Perfect chain complexes QUESTION [5 upvotes]: In Thomason-Trobaugh in Remark 2.4.4 it is written: "On a general scheme, the perfect complexes are locally finitely presented objects in the "homotopy stack" of derived categories." I was wondering if someone here could explain this a little bit to me (i.e. roughly has the knowledge of chapters II and III of Hartshorne + knows the definition of stack). In particular I would like to know what is meant by "homotopy stack". Thanks! REPLY [14 votes]: In modern language, one would say that $D_{qcoh}(-)$ is a sheaf of $(\infty,1)$-categories on the scheme $X$ (so "homotopy stack" = "sheaf of $(\infty,1)$-categories"). If $X$ is affine, or more generally has an ample family of line bundles, the perfect complexes on $X$ are exactly the finitely presented objects (aka compact objects) in $D_{qcoh}(X)$: this is what Thomason proves. Nowadays, this is sometimes taken as the definition of perfect complex in the affine case, see for instance: Lurie, Higher Algebra, section 7.2.4. Thomason's assertion can be rephrased by saying that the sub-presheaf $D_{perf}(-) \subset D_{qcoh}(-)$ is in fact a sheaf. Equivalently, if $F\in D_{qcoh}(X)$ is such that $F|U_i$ is perfect for every $U_i$ in some open covering of $X$, then $F$ is perfect. This is obvious with Thomason's definition of "perfect". In particular, $F$ is perfect iff $F|U\in D_{qcoh}(U)$ is finitely presented for every affine $U\subset X$. Another point of view on perfect complexes is that they are precisely the dualizable objects in the symmetric monoidal $(\infty,1)$-category $D_{qcoh}(X)$. One way to prove this is to check it for $X$ affine and observe that dualizable objects also form a subsheaf (for categorical reasons: the functor that sends a symmetric monoidal $(\infty,1)$-category to its subcategory of dualizable objects is limit-preserving). A reference for this is Proposition 6.2.6.2 in Lurie's book Spectral Algebraic Geometry.<|endoftext|> TITLE: Joint measurability of metric QUESTION [6 upvotes]: I am trying to understand in which metric spaces the metric is jointly measurable. There exist a metric space $(X,d)$ for which the Borel $\sigma$-algebra, does not coincide with the product Borel $\sigma$-algebra, that is $\mathcal{B}(X)\otimes \mathcal{B}(X) \subsetneq B(X\times X)$. Every construction, I have encountered, of such a (non-separable) metric space one considers a metric space of cardinality strictly greater than the continuum $\mathfrak{c}$. For in such a metric space the diagonal $I=\{(x,x):x\in X\}\in B(X\times X)$, is not an element of $\mathcal{B}(X)\otimes \mathcal{B}(X)$ (c.f. Nedoma's pathology). This also proves that $d$ is not jointly measurable, since $d^{-1}(\{0\})=I\not\in\mathcal{B}(X)\otimes \mathcal{B}(X)$. For metric spaces $(X,d)$ with $\text{card}(X)\leq \aleph_0$, we obviously have that $d$ is jointly measurable, but the area of interest is : What happends when the cardinality is $\aleph_0 < \text{card}(X) \leq \mathfrak{c}$? More specifically, I have been wondering about the following questions: Let $(X,d)$ be a metric space, Is the metric $d$ jointly-measurable, if $\aleph_0 < \text{card}(X) \leq \mathfrak{c}$? If not, is there equivalence between joint measurability of $d$ and separability of $X$? If not, do you know property of $(X,d)$, weaker than separability, that ensures joint-measurability of $d$? REPLY [6 votes]: One can probably not go too far from separability, there must always exist a sub-$\sigma$-algebra that looks like the Borel $\sigma$-algebra of a separable metric space. If $\mathcal{A}$ is a family of subsets of a set $X$ and $A\in\sigma(A)$, then there exists a countable family $\mathcal{C}\subseteq\mathcal{A}$ such that $A\in\sigma(\mathcal{C})$. In particular, if the diagonal $D\subseteq X\times X$ is in the $\sigma$-algebra generated by measurable rectangles, then the countable family of sets $\mathcal{C}$ must have the property that for each $x,y\in X$ with $x\neq y$, there is $C\in\mathcal{C}$ such that $x\in C$ and $y\notin C$ or $x\notin C$ and $y\in C$. Let $\langle C_n\rangle$ be a sequence that lists all elements of $\mathcal{C}$. Then the Marczewski-funtion $g:X\to\mathbb{R}$ given by $$g(x)=\sum_n 1/3^n 1_{C_n}(x)$$ has the property that $\sigma(\mathcal{C})$ consists exactly of inverse images of Borel sets in $\mathbb{R}$. Since $\mathcal{C}$ separates points, it is injective. Then $d:X\times X\to\mathbb{R}$ given by $d(x,y)=|g(x)-g(y)|$ turns $X$ into a separable metric space with Borel $\sigma$-algebra $\sigma(\mathcal{C})$. However, separability is not necessary. Endow $\omega_1$ with the discrete metric, so it becomes a non-separable metric space with Borel $\sigma$-algebra $2^{\omega_1}$. In this case, the metric is still jointly measurable, for one can show that $2^{\omega_1}\otimes2^{\omega_1}=2^{\omega_1\times\omega_1}$. The proof can be found in: V. Rao, On discrete Borel spaces and projective sets Bull. Amer. Math. Soc. Volume 75, Number 3 (1969), 614-617.<|endoftext|> TITLE: Bundle over $\mathbb{C}^{n}\setminus{0}$ QUESTION [5 upvotes]: Let $\pi:\mathbb{C}^{n}\setminus{0}\rightarrow\mathbb{CP}^{n-1}, n\geq 3$ be the projection from affine space without the origin to the projective space. If we pull back the tangent bundle of $\mathbb{CP}^{n-1}$ we would get a nontrivial bundle over $\mathbb{C}^{n}\setminus{0}$. Now my question would be: what is $H^{1}(\mathbb{C}^{n}\setminus{0},\pi^{*}\mathcal{T}_{\mathbb{CP}^{n-1}})$? REPLY [9 votes]: $\newcommand{tot}{\mathbb{C}^n\setminus 0}\newcommand{tan}{\mathcal{T}_{\mathbb{P}^{n-1}}}$ Since morphism $\pi$ is affine, for any quasicoherent sheaf $\mathcal{F}$ on $\mathbb{C}^n\setminus 0$ its higher direct images $R^{>0}\pi_*\mathcal{F}$ vanish, so from Leray spectral sequence we get $H^i(\tot, \mathcal{F})=H^i(\mathbb{P}^{n-1},\pi_*\mathcal{F})$. Applying this to $\pi^*\mathcal{T}_{\mathbb{P}^{n-1}}$ we get $$H^1(\tot,\pi^*\tan)=H^1(\mathbb{P}^{n-1},\pi_*\pi^*\tan)=H^1(\mathbb{P}^{n-1},\tan\otimes\pi_*\mathcal{O}_{\tot})$$ by projection formula. Note that $\pi$ is the projection from the total space of $ \mathcal{O}_{\mathbb{P}^{n-1}}(-1)$ minus zero section.(the following is corrected thanks to Jason Starr). The direct image of the structure sheaf of the total space itself is $Sym(\mathcal{O}(-1))=\bigoplus\limits_{n\geq 0}S^n\mathcal{O}(-1)=\bigoplus\limits_{n\geq 0}\mathcal{O}(-n)$. So, if we throw away the zero section, sections of structure sheaf are allowed to have a pole along the zero section, so $\pi_*\mathcal{O}_{\tot}=\bigoplus\limits_{n\in\mathbb{Z}}\mathcal{O}(-n)$. So, the problem is reduced to computing $H^1(\mathbb{P}^{n-1}, \tan\otimes \mathcal{O}(-k))$. From Euler exact sequence we get a long exact seqeunce $\dots\to H^1(\mathbb{P}^{n-1},\mathcal{O}(1-k)^{\oplus n})\to H^1(\mathbb{P}^{n-1},\tan\otimes \mathcal{O}(-k))\to H^2(\mathbb{P}^{n-1},\mathcal{O}(-k))\to \dots$ Both left and right groups are zero, because line bundles on $\mathbb{P}^{n-1}$ can have nonzero cohoomology only in degrees $0,n-1>2$ so $H^1(\tot,\pi^*\tan)=0$<|endoftext|> TITLE: Does a minimum area disk that is bounded by a cycle $C$ continuously deform in $R^3$ as $C$ moves in $R^3$? QUESTION [6 upvotes]: Let $C_1=(v_1,v_2,\ldots,v_{i-1},v_i)$ and $C_2=(v_1,v_2,\ldots,v_{i-1},v'_i)$ be two cycles that are drawn in $R^3$ in the shape of an unknot (not knotted) with straight line segments as their edges (note that $C_1$ and $C_2$ differ from each other in one vertex: Vertex $v_i$ of $C_1$ is replaced with vertex $v'_i$ in $C_2$). Let $D_1$ and $D_2$ be two disks with minimum area that are bounded by $C_1$ and $C_2$ respectively and let $l$ be the line that passes through $v_i$ and $v'_i$. Also assume that: $C_1$ has a unique minimum area disk, we can choose $v'$ to be arbitrarily close to $v$. Now assume that we start to move $v_i$ on $l$ towards $v'_i$ (we gradually change the position of $v_i$ towards $v'_i$). Under these conditions, can we guarantee that $D_1$ continuously deforms in $R^3$ until $D_1 = D_2$? Any references that contains a theorem that proves the continuous deformation of such a surface? REPLY [8 votes]: This is a standard question. Look at the following image from Morgan's "Geometric measure theory". It should convince you that the answer is no. The curve admits two area minimizing discs and it admits arbitrary small perturbations so that just one of them stay area minimizing.<|endoftext|> TITLE: Commuting pairs $(a,b)$ with $a$ not in the center QUESTION [8 upvotes]: In a group $G,$ the centralizer of $a \in G$ is the subgroup $$C(a)=\{g \mid ga=ag\}.$$ In a non-abelian group with $|G|=n$ and center $Z$ we have for each $a \notin Z$ that $Z \subsetneqq C(a) \subsetneqq G.$ This shows that $|Z| \le \frac{n}{4}$ and that, when $|Z|=\frac{n}{4},$ each $a \notin Z$ commutes with exactly half the elements so $$\sum_{a \notin Z}|C(a)|=\frac{3n^2}8.$$ My question is if the converse is true: If $\sum_{a \notin Z}|C(a)| = \frac{3n^2}8$ does it follow that $|Z|=\frac{n}{4}?$ Can it happen that $\sum_{a \notin Z}|C(a)| \gt \frac{3n^2}8?$ So really that is the converse of $$ |Z|=\frac{n}4 \text{ implies } \sum_{a \notin Z}|C(a)| \ge \frac{3n^2}8. $$ A near miss is the semi-dihedral group SD16 which has size $16$ and center of size $2$. It has a cyclic subgroup of size $8$ whose $6$ non-central members have $|C(a)|=8$ and the other $8$ elements have order $4$ and $|C(a)|=4$ for a total of $$\sum_{a \notin Z}|C(a)|=6\cdot 8+8 \cdot 4=\frac{5n^2}{16}.$$ A group with $|Z| \lt \frac{n}4$ and all other elements having $|C(a)|=\frac{n}{2}$ would provide an example with $\sum_{a \notin Z}|C(a)|=\frac{(n-|Z|)n}{2} \gt \frac{3n^2}8.$ REPLY [3 votes]: Firstly, any finite non-Abelian $2$-group $G$ as in the last remark of the question has centre of index $4$. Suppose otherwise and set $Z= Z(G).$ Suppose that each element of $G \backslash Z$ has centralizer of index $2$. If $G$ has two different Abelian maximal subgroups $M$ and $N$ then $M \cap N = Z$ and has index $4$. If there is a unique Abelian maximal subgroup $M$ of $G$, then for any $x$ outside $M$, we see that $ C_{M}(x)$ is Abelian, is different from $M$ and has index $2$ in $G$, a contradiction. Hence we may suppose that $G$ has no Abelian maximal subgroup. By induction, we may suppose that $[M:Z(M)] = 4$ for each maximal subgroup $M$ of $G$, so every Maximal subgroup of any such $M$ is Abelian. But as above , we then see that for any $x$ outside $M$, the centralizer of $x$ in $G$ is an Abelian maximal subgroup of $G$, contrary to assumption. Hence we do have $[G:Z] = 4, $ as claimed. Note thatvweay suppose that $Z \leq M, $ otherwise $G = ZM,$ and then $ZX$ is a central subgroup of $G$ of index $4,$ where $X = Z(M).$ Here is a sketch proof that a group $G$ satisfying the first condition must be a $2$-group- well, strictly a direct product of a 2-group and an Abelian group of odd order. You want $G$ to have $|Z|+ 3|G|/8$ conjugacy classes, where $Z=Z(G)$. If we choose $G$ of minimal order subect to having this property, then $G$ can not be expressed in the form $A \times H$ for $A,H$ proper with $A$ Abelian, so suppose that this is the case. Suppose now that $G$ is not a $2$-group. Then $G = SN$ where $S$ is a $2$-group and $N$ is a non-trivial normal $2$-complement. If $[S,N]= 1,$ then $G=S \times N,$ contrary to hypothesis. Hence $[S,N]$ has order $3$ or $5$, in which case $S$ is Abelian as $[G,G]$ has order at most $5$. It follows that $N$ is either a $3$-group or a $5$-group. Also $N$ must be Abelian, since we can't have $[S,N] \leq [N,N]$ by properties of coprime automorphisms. Now $G = (S[N,S]) \times C_{N}(S)$ so $C_{N}(S) = 1$ and $Z = Z(G)$ is a $2$-group, and now $N = [N,S]$ has order $3$ or $5$. Suppose that $|N| = 5.$ Now $G$ has $|S|$ linear characters and $\frac{15|S|}{8}+ |Z|- |S|$ non-linear irreducible characters, so that $5|S| \geq |S| + d^{2}(\frac{7|S|}{8}+ |Z|)$, where $d$ is the smallest degree of a non-linear irreducible character of $G$, which is easily seen to force $d=2.$ This in turn forces $[S:Z]= 2$, which easily yields a contradiction. Hence in the case under consideration, we must have $|N| = 3,$ and then again $[S:Z]=2.$ But now, each non-linear irreducible character of $G$ has degree $2$ by a Theorem of Ito, and we have $3|S| = |S| + 4(\frac{9|S|}{8} -\frac{|S|}{2}),$ which is a contradiction.<|endoftext|> TITLE: Two points of view about Borel-moore homology QUESTION [25 upvotes]: They are several ways to define the Borel-Moore homology on a locally compact space $X$. The first one is by analogy with the singular homology but instead of using finite chains, we use locally finite chains. Let us note $H_p^{lf}(X,\mathbb{Z})$ the associated homology groups (coeff in $\mathbb{Z}$ for convenience). Now, here is a way to define it thanks to sheaf theory. Let us note $a_X$ the map $a_X : X \to \{pt\}$. The dualizing complex of $X$ is defined by $$\omega_X = a_X^!\mathbb{Z}$$ where $\mathbb{Z}$ is the constant sheaf and $!$ given by the Poincaré-Verdier duality. Now we define the Borel-Moore homology $$H_p^{BM}(X,\mathbb{Z})=H^{-p}R\Gamma(X,\omega_X)$$ with the formalism of derived functors. We have the following theorem $$H_p^{BM}(X,\mathbb{Z}) \simeq H_p^{lf}(X,\mathbb{Z}).$$ I was quite surprised to see that this "well-known" fact is not really proved in any book. The usual reference is Bredon, but Bredon defines the Borel-Moore homology thanks to cosheaves and then he writes a remark, saying that this is the same that locally finite chains (without arguments). Moreover the link with the dualizing complex is not clear in his presentation. So my question is quite simple : With the two definitions I gave, is there a rather straightforward way to prove the isomorphism $$H_p^{BM}(X,\mathbb{Z}) \simeq H_p^{lf}(X,\mathbb{Z})?$$ Any help will be much appreciated. REPLY [18 votes]: I'll have more time to write and provide a more thorough answer later, but I think the most straightforward proof (which I agree is hard to find) comes via sheaf theory: On the one hand, there is a sheaf of locally finite singular chains whose hypercohomology is your $H^{lf}$. I work out the details in the setting of intersection homology on pseudomanifolds in Section 3 of the following paper, but the ordinary (non-intersection) lf homology on locally compact (paracompact?) spaces is essentially a special case via basically the same arguments: "Singular chain intersection homology for traditional and super-perversities" Transactions of the American Mathematical Society 359 (2007), 1977-2019 On the other hand, you can find a good treatment of the dualizing complex in Borel's book on intersection homology, section V.7. The idea then is to show that these two sheaves are quasi-isomorphic. This is the part I'll have to think about and add latter, though I'll note that it's not so hard for manifolds: in that case, it's not so hard to show by hand that both have to be quasi-isomorphic to the orientation sheaf (shifted appropriately). That's shown in Borel for the dualizing complex, and follows for the sheaf of locally finite singular chains via the local homology computation on the manifold. Sorry I don't have time to write more now, but hopefully that's a start on the ideas. Added later: Oookay, this was harder than I thought. I’m not sure I can give you a really satisfying answer, but I think I can convince you that what you want is contained in Bredon as Corollary V.12.21. This corollary pretty much says that if $X$ is nice enough (though still pretty general), then ${}_SH_p^{\Phi}(X;\mathcal B)\cong H_p^{\Phi}(X;\mathcal B)$, where here $\mathcal B$ is a sheaf and $\Phi$ is a paracompactifying family of supports. For what we want, let’s take $\Phi$ to be all the closed sets, so we need to assume $X$ paracompact. We’re also going to let $\mathcal B$ be the constant sheaf with stalk the ground ring $L$. Let’s call that $\mathcal R$ (since $\mathcal L$ is used for lots of other things in these sections of Bredon). In all of the definitions we’ll talk about, $\mathcal B$ would come in as a tensor product factor, so letting it be constant basically lets us ignore those factors. Now, on the left we have ${}_SH_p^{cld}(X;\mathcal R)$, which is Bredon’s singular homology with (now constant) sheaf coefficients. He gives a pretty good explanation for this back on pages 287-289. The key point connecting this to what I wrote previously is the following. Suppose you start with the (homologically graded) presheaf he calls $\Delta_*^c(X,X-\bar U;R)$, which is just the usual relative singular chain complex. Since our spaces are locally compact, this sheafifies to the same sheaf as $\Delta_*^{lf}(X,X-\bar U;R)$, the sheaf of locally finite singular chains, whose hypercohomology (or hyperhomology with this indexing) computes the locally finite singular homology (more details about this can be found elsewhere in Bredon or in the Transactions paper of mine that I already mentioned). Though actually here you don’t even need the hypercohomology; on the top of page 289 he just takes the homology of the complex of sections (which is the same as the hyperhomology because the sheaf is homotopically fine – shown earlier in the book). The argument on page 288-289 further shows that the resulting homology groups are isomorphic to singular homology as he defined it on page 288, using basically the same sheaf but also employing the direct limit over subdivisions. So, the upshot is that the term on the left is locally finite singular homology. For the term on the right, we need to trace through the definitions starting on page 292. By the boxes at the top, $$H_p^{cld}(X;\mathcal R)= H_p(C_*^{cld}(X;\mathcal R))= H_p(\Gamma(\mathcal C_*(X;\mathcal R))) =H_p(\Gamma(\mathcal D_*(\mathcal J^*(X;L)))\otimes \mathcal R),$$ where $L$ is our ground ring. So if $X$ is locally connected, we can ignore the $\mathcal R$ and we’re just taking the homology of the global sections of $\mathcal D_*(\mathcal J^*(X;L))$. I claim this is just the Verdier dualizing sheaf with homological indexing. To see this, we unwrap more definitions! By the top of page 292, $\mathcal J^*(X,L)$ is just the canonical injective resolution of the constant sheaf (he also calls the constant sheaf $L$ and I have called it $\mathcal R)$. Let’s just call this injective (and hence also c-soft) resolution $\mathcal K^*$. So we need to understand $\mathcal D(\mathcal K^*)$. By the definitions on pages 289-290, $\mathcal D$ should take a second argument $M$, but there’s a note on page 290 saying it’s omitted from the notation if $M=L$. So we’re really after $\mathcal D(\mathcal K^*;L)$. By the second box on page 290, $\mathcal D(\mathcal K^*;L)$ means $\mathcal D(\Gamma_c\mathcal K^*;L)$, where $\Gamma_c\mathcal K^*$ is the cosheaf given by $(\Gamma_c\mathcal K^*)(U)=\Gamma_c(\mathcal K^*|U)$ by the definition part of Proposition V.1.6. But now piecing together the meaning of Definition V.2.1, which is given explicitly on the top of page 291, $\mathcal D(\Gamma_c\mathcal K^*;L)(U)=Hom(\Gamma_c\mathcal K^*(U),L^*)$, where $L^*$ is an injective resolution of $L$ according to the beginning of Section V.2. So, altogether, $\mathcal D(\Gamma_c\mathcal K^*;L)$ is just the sheaf $U\to Hom(\Gamma_c(\mathcal K^*|U),L^*)$. But this is exactly the definition of the dualizing sheaf given by Borel on page 119 of Intersection Cohomology. That this is $f^!$ of the constant sheaf on a point is on Borel page 131. So there you go. Tracing through the Bredon definitions is a nightmare, and not a satisfying conceptual thing at all, but it’s a proof. To say just a couple of words about the proof of the Corollary, the point is that he defines at some point the idea of a ``flabby quasi-n-coresolution of L,’’ where here $L$ is the notation for the constant sheaf again. The point of these is that if you have one, say $\mathcal L_*$, and want to compute $H_*(X;\mathcal B)$ for some sheaf $\mathcal B$, then instead of going through the horrible definitions we just unwound, you have $H_*(X;\mathcal B)$ is just $H_*(\Gamma(\mathcal L_*\otimes \mathcal B))$. That’s Theorem V.12.20, and this is sort of like what happens in Borel V.6.7 (or other places) when you make a sheaf complex c-soft by tensoring it with a flat c-soft resolution of the constant sheaf. But a bit earlier in Theorem V.12.14 it’s shown that the same sheaf used to compute singular homology (the one using the subdivisions, called $\mathfrak S_*$) is such a flabby quasi-n-coresolution of $L$. That actually just follows from the definitions and the properties on the space. So it follows that $H_*(X;\mathcal B)\cong H_*(\Gamma(\mathfrak S_*) \otimes \mathcal B))$, which is the definition of singular homology with coefficients in $\mathcal B$. It’s really a shame this isn’t all brought out more clearly in Bredon’s exposition. Whew.<|endoftext|> TITLE: $G$ cocycle split and trivialized to a coboundary in $J$, given a group homomorphism $J \overset{r}{\rightarrow} G$ QUESTION [10 upvotes]: Consider a generic nontrivial 3-cocycle $\omega_3^G(g_1,g_2,g_3) \in H^3(G,U(1))$ in the cohomology group of $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the 3-cocycle $\omega_3^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega_3^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition. We like to ask how can we trivialize (or split) the 3-cocycle $\omega_3(g_1,g_2,g_3)$ of $G$ into 3-coboundary if we lift $G$ into a larger group $J$, and given that we know the group homomorphism $r$: $$J \overset{r}{\rightarrow} G,$$ so that $$\omega_3^J(j_1,j_2,j_3)=\omega_3^G(r(j_1),r(j_2),r(j_3))=\omega_3^G(g_1,g_2,g_3) \text{ is trivial in } H^3(J,U(1)).$$ Namely $\omega_3^G(r(j_1),r(j_2),r(j_3))$ becomes a 3-coboundary in $H^3(J,U(1))$ for the cohomology group of $J$, but $\omega_3^G(g_1,g_2,g_3)$ originally wass not a 3-coboundary but was a 3-cocycle for the cohomology group of $G$. Namely, we can explicitly write $$ \omega_3^G(g_1,g_2,g_3)=\omega_3^G(r(j_1),r(j_2),r(j_3))= \frac{\beta_2^J(j_2,\; j_3)\beta_2^J(j_1,\; j_2\cdot j_3)}{\beta_2^J(j_1 \cdot j_2,\; j_3) \beta_2^J(j_1,\; j_2)}. $$ Here $\beta_2^J(j_1,\; j_2)$ is a 2-cochain for $j_1, j_2, j_3 \in J$, and that $g_1=r(j_1)$, $g_2=r(j_2)$, $g_3=r(j_3) \in G$. question 1: For example, can we try explicitly that $G=\mathbb{Z}/2 \mathbb{Z}=\mathbb{Z}_2$ the cyclic group, and $J=Q_8$ the quaternion group. There are $\mathbb{Z}_2 \times \mathbb{Z}_2$ types of group morphism $r(j)=g$ of $J \overset{r}{\rightarrow} G$ as $Q_8 \overset{r}{\rightarrow} \mathbb{Z}_2$. If we know the explicit form of 3-cocycle of $G$, $\omega_3^{G}(g_1,g_2,g_3) = \exp \left( \frac{2 \pi i}{2^{2}} \; g_1(g_2 +g_3 -[(g_2 +g_3)\mod 2]) \right)$, how can we determine the analytic explicit form of 2-cochain $\beta_2^J(j_1,\; j_2)$ for the quaternion $J=Q_8$ such that it can trivialize the 3-cocycle of $G$ to 3-coboundary of $J$? The key question for this post is asking the demonstration of this specific example: What is the explicit form of $\beta_2^{Q_8}(j_1,\; j_2)$? - question 2: Comments about the general procedure to trivialize (split) 3-cocycle of $G$ to 3-coboundary of $J$, given that $J \overset{r}{\rightarrow} G$. One can test that NOT all the $J$ with group homomorphism $J \overset{r}{\rightarrow} G$ can trivialize $G$-cocycle. What conditions are necessary to find such $J$? Any comments and partial answers are more than welcome, please! REPLY [5 votes]: Here is an answer for question 2 on which homomorphisms $J\xrightarrow{r} \mathbb{Z}_2$ will trivialize $\omega_3^G$. Your cocycle takes values in the two-element subgroup $\{1,-1\}\subset U(1)$ (aka $\{0,\frac{1}{2}\}\subset \mathbb{R}/\mathbb{Z}$). Since all two-element groups are isomorphic, you could of course identify this subgroup with $\mathbb{F}_2=\{0,1\}$ under addition. But why would you help? Well, $\mathbb{F}_2=\{0,1\}$ also has a multiplication (that's why I'm using a different letter for it), and so you can ask whether your cocycle factors as a cup product of two smaller cocycles. In this case, we'll see that it does! Let's see what happens in this case (this will also illustrate what a cup product is, if you're not familiar with this). To start, we need to rewrite your formula $$\omega_3^{G}(g_1,g_2,g_3) = \exp \left( \frac{2 \pi i}{2^{2}} \; g_1(g_2 +g_3 -[(g_2 +g_3)\mod 2]) \right)\in U(1)$$ to take values in $\mathbb{F}_2=\{0,1\}$. Note that the inputs $g_i$ in your formula live in $\mathbb{Z}_2=\{0,1\}$. What we get (for now) is $$\omega_3(g_1,g_2,g_3)=g_1\cdot \frac{g_2 +g_3 -[(g_2 +g_3)\mod 2]}{2}\in \{0,1\}=\mathbb{F}_2.$$ The property we want to highlight is that this is a product of two simpler functions: $$\omega_3(g_1,g_2,g_3)=\alpha(g_1)\cdot \beta(g_2,g_3)$$ where $$\alpha(g_1)=g_1\quad\text{ and }\quad\beta(g_2,g_3)=\frac{g_2 +g_3 -[(g_2 +g_3)\mod 2]}{2}.$$ What does this tell us? Well mathematically, we would write $r^*(\omega_3^G)\in H^2(J)$ for what you are calling $\omega_3^J$. And if $w_3\in H^3(\mathbb{Z}_2;\mathbb{F}_2)$ is the cohomology class represented by $\omega_3^G$, then $r^*(w_3)\in H^3(J;\mathbb{F}_2)$ is represented by $\omega_3^J$: so your question about trivializing the cocycle is whether $r^*(w_3)\overset{?}{=}0\in H^3(J;\mathbb{F}_2)$. Note that $\alpha$ and $\beta$ are both cocycles, so they represent cohomology classes $a\in H^1(\mathbb{Z}_2;\mathbb{F}_2)$ and $b\in H^2(\mathbb{Z}_2;\mathbb{F}_2)$. The factorization $$\omega_3(g_1,g_2,g_3)=\alpha(g_1)\cdot \beta(g_2,g_3)$$ then tells us that $w_3$ is the cup product of the cohomology classes $a$ and $b$: we have $w_3=a\cdot b$ (see my answer here for details). Cup product is an operation on cohomology that takes $x\in H^i$ and $y\in H^j$ and combines them to give $x\cdot y\in H^{i+j}$. It satisfies $y\cdot x = \pm x\cdot y$ where $\pm=(-1)^{ij}$. The reason this is useful is that cup products are always preserved by $r^*$: there is a universal formula $$r^*(x\cdot y)=r^*(x)\cdot r^*(y)\quad\text{in cohomology}.$$ This means that ONE way you could hope to show that $r^*(w_3)=0$ would be to show that either $r^*(a)=0$ or $r^*(b)=0$. Actually in 1st cohomology $r^*(a)$ will never be 0 unless the map $J\xrightarrow{r}\mathbb{Z}_2$ is trivial (an uninteresting case, as you discuss in comments), but you could hope that $r^*(b)=0$. [Side note: in your particular example in Q1, I checked and $r^*(b)\neq 0$.] But actually there is a much better way to look at this. It comes from the fact that you wrote your cocycle in a way that hides its symmetry. I claim that if you work out the values of your cocycle $$\omega_3(g_1,g_2,g_3)=g_1\cdot \frac{g_2 +g_3 -[(g_2 +g_3)\mod 2]}{2}\in \{0,1\}=\mathbb{F}_2$$ on the 8 different possible inputs, you'll see that it can instead be written as $$\omega_3(g_1,g_2,g_3)=\begin{cases}1&\text{ if $g_1=1$, $g_2=1$, and $g_3=1$}\\0&\text{else}\end{cases}\in \{0,1\}=\mathbb{F}_2.$$ In other words, $$\omega_3(g_1,g_2,g_3)=g_1\cdot g_2\cdot g_3=\alpha(g_1)\cdot \alpha(g_2)\cdot \alpha(g_3).$$ This means that $w_3=a\cdot a \cdot a=a^3$. So not only can your cocycle be written in a simpler form, it actually is the cup product of a single class in 1st cohomology with itself three times. Let us write $R\in H^1(J;\mathbb{F}_2)$ for $R=r^*(a)$. Then we have $r^*(w_3)=r^*(a)^3=R^3$, so the question of whether $r^*(w3)=0$ is simply the question of whether $R\in H^1(J;\mathbb{F}_2)$ satisfies $$R^3=0\in H^3(J;\mathbb{F}_2).$$ Moreover, an element $R\in H^1(J;\mathbb{F}_2)$ is equivalent to a homomorphism $J\xrightarrow{r}\mathbb{Z}_2$ (that's why I used the same letter). So for a given $J$, your question is equivalent to finding all $R\in H^1(J;\mathbb{F}_2)$ satisfying $R^3=0$. Now, that's not going to necessarily be easy! But it is more likely to be something you can look up somehow. (And in some cases, you may be able to show that $R^2=0$, which would be enough.) One last comment: you might be concerned that I was working with $\mathbb{F}_2$ coefficients here, whereas you had general $U(1)$ coefficients. Here is an argument why this is not so bad. Say we're given some class $z\in H^1$. If we take $x=z$ and $y=z$ in the formula $y\cdot x = \pm x\cdot y$ from above, it tells us $z\cdot z = - z\cdot z$. In other words, we always have $z^2+z^2=0\in H^2$. This means that if you have coefficients in $\mathbb{Q}$ or $\mathbb{Z}_3\subset U(1)$ or anywhere you can divide by 2, then $z^2$ would ALWAYS be 0 (and thus $z^3$ would be as well). This isn't quite a proof that you can always stick to $\mathbb{F}_2$ coefficients (for example, you might need to worry about $\mathbb{Z}/4\mathbb{Z}\subset U(1)$ as well) but as a general rule it argues that you can probably ignore factors other than $2$. EDIT: Here is a concrete example of how to apply this to $Q_8$. According to (eq uation 2.4.2) of this survey by Benson, the cohomology ring of $Q_8$ is $$H^*(Q_8;\mathbb{F}_2)\cong \mathbb{F}_2[x,y,z]/(\,x^2+xy+y^2,\,x^2y+xy^2\,)$$ where $x\in H^1$, $y\in H^1$, and $z\in H^4$. Therefore $H^3(Q_8;\mathbb{F}_2)$ is spanned by the four elements $x^3$, $x^2y$, $xy^2$, and $y^3$ subject to the three relations $x^3+x^2y+xy^2=0$, $x^2y+xy^2+y^3=0$, and $x^2y+xy^2=0$. (The first two are $x(x^2+xy+y^2)$ and $y(x^2+xy+y^2)$.) Subtracting the last relation from the first two gives a basis of relations as $x^3=0$, $y^3=0$, and $x^2y+xy^2=0$. You can now read off easily exactly which $R\in H^1(Q_8;\mathbb{F}_2)$ have $R^3=0$. Namely, if $R=x$ we have $R^3 = x^3 = 0$; if $R=y$ we have $R^3 = y^3 = 0$; and if $R=x+y$ we have $R^3 = (x+y)^3 = x^3 +3x^2y+3xy^2+y^3 = x^3 + (x^2y+xy^2) + y^3 = 0$. Therefore we conclude that every homomorphism $Q_8\to \mathbb{F}_2$ splits your cocycle! You can do this for lots of other groups, if you can find where someone has already computed the $\mathbb{F}_2$-cohomology ring. For example, (eq. 2.8.1) of that same survey says that the "semidihedral group $SD_{2^n}$ of order $2^n$", whatever that is, has (for $n\geq 4$) the cohomology ring $$H^*(SD_{2^n};\mathbb{F}_2)\cong \mathbb{F}_2[x,y,z,w]/(\,xy,\,y^3,\,yz,\,z^2+x^2w\,)$$ where $x\in H^1$, $y\in H^1$, $z\in H^3$, and $w\in H^4$. We find that $H^3(SD_{2^n};\mathbb{F}_2)$ is spanned by $x^3,x^2y,xy^2,y^3,z$ but with relations $x^2y=0$, $xy^2=0$, and $y^3=0$. Therefore we conclude that $R=x$ has $R^3=x^3\neq 0$ and $R=x+y$ has $R^3 = (x+y)^3 \equiv x^3 \neq 0$ but $R=y$ has $R^3=y^3=0$. So only one of the three nontrivial homomorphisms $SD_{2^n}\xrightarrow{r}\mathbb{Z}_2$ will split your cocycle. Finally, you only have to look at groups $J$ whose order is a power of 2. This follows from a theorem of Cartan-Eilenberg, although it's a bit tricky to explain how (see Theorem 4.1 of these lectures by Adem). In any case, here is a precise statement you can use: Theorem: Fix a homomorphism $J\xrightarrow{r}\mathbb{Z}_2$. For any subgroup $H TITLE: Is there a database for tracking the dependencies of mathematical theorems? QUESTION [93 upvotes]: Given a proof for a result, one could denote the proof as a node on a graph, and then draw arrows to the node from axioms and previous results that the proof uses, and then draw arrows from the node to results that the result is used to prove. The sources of such a graph would be mathematical axioms and the sinks would be results which do not yet have corollaries. This is related to the facts that finite categories can be drawn as directed graphs and that proofs form a category. Question: Is there an existing ontological/relational database somewhere that attempts to make use of such an idea to formally encode some of present-day mathematics? It can often be difficult to fully trace all of the logical dependencies of a given result. For example, how can one show whether existing proofs of the Implicit Function Theorem use the Law of the Excluded Middle or not? Rather than spend a couple of days going through tens of textbooks and trying to parse their proofs, not for their content but just to see what assumptions they are using, wouldn't it be easier to ask a computer to do that task? Such a system for tracking the logical dependencies of known proofs for existing theorems could be useful not only for beginners ("What important theorems use the completeness property of the real numbers? I'll check the database."), but it could also be useful for constructivists or intuitionists who want to keep track of which existing results make use of the Axiom of Choice or the Law of the Excluded Middle. And maybe it would be possible to use a sufficiently complete database to run tests to see whether the foundations of mathematics are consistent? I don't actually know, but it sounds on a naive level something that would be useful, even if it would take a long time to build. Note: I am not sure what to tag this question or if it is on-topic. REPLY [18 votes]: Not an answer, just a diagram from the Stacks Project, mentioned by Paul Siegel, illustrating dependencies of "the results needed to prove Chow’s Lemma" (the Noetherian case):                 (Image from MathBabe blog.) And here's a link to a larger image.<|endoftext|> TITLE: Relationships between different classes of L-Functions QUESTION [6 upvotes]: There are many types of zeta (L) functions floating around. Lets consider $\zeta_K(s)$ - the Dedekind Zeta Function of a number field $L(\rho,s)$ - The Artin L-function $\rho:G_{\mathbb{Q}}\to GL_n(\mathbb{C})$ $L(X,s)$ - the Hasse-Weil zeta function of a (suitable) variety/scheme $X$. One of the goals of Number Theory is to show these are actually $L(\pi,s)$ where $\pi$ is a suitable automorphic representation (Artin-Conjecture, generalization of Taniyama Shimura) There are instances where these should agree (correct me if I am wrong) such as the variety $x^2-p$ and the (induction of the trivial) galois representation of $Gal(\mathbb{Q}[\sqrt{p}]/Q)$, and the Dedekind zeta function of $\mathbb{Q}[\sqrt{p}]$. My question is: where do these classes of zeta functions overlap and where do they disagree? How can I go from one to another? I think I understand how the Dedekind zeta function sits in each $\zeta_K(s)=L(Ind^K_{\mathbb{Q}}(1_K),s)$, where $1_K$ is the trivial representation (when $K$ is Galois) $\zeta_K(s)=L(X,s)$ where $X=Spec(\mathcal{O}_K)$ REPLY [6 votes]: This question is more or less answered by the Tate conjecture, at least in the smooth projective case. Namely, the Hasse-Weil $L$-function of a smooth projective variety $X$ should be a product of ratios of shifts of Artin $L$-functions if and only if all its $\ell$-adic cohomology groups are generated by algebraic cycles. The argument in what follows is slightly vague as all of this is entirely conjectural anyway. The Hasse-Weil $L$-function of $X$ can be written as a product of ratios of $L$-functions of the $\ell$-adic cohomology groups of $X$. One can isolate Artin representations amoungst all Galois representations as being those with finite image. However, by the Tate conjecture, the Galois representations coming from the cohomology of an algebraic variety have finite image (after an appropriate Tate twist) if and only if they are genererated by algebraic cycles.<|endoftext|> TITLE: Chern class on a symplectic manifold QUESTION [9 upvotes]: Let $(X,\omega)$ be a closed symplectic manifold. Can we always write $c_1(TX) = [ f \omega ]$ for some function $f: X \to \mathbb{R}$? REPLY [8 votes]: Your question is correspond to weak Kahler-Einstein metric(I know for Kahler case). Let me explain in more details The Hermitian metric $H$ along the fibers of the holomorphic vector bundle $E$ over a K\"ahler manifold $(M,\omega)$ is weak Hermitian-Einstein if $$\sqrt[]{-1}\Lambda_\omega F_H = \phi I$$ at every point of $M$, where $\gamma$ is a function on $M$ and $I$ is the identity endomorphism of $E$ and $F_H$ is the Chern curvature form of $(E, h)$. Here $\Lambda_\omega$ is the contraction with $\omega$ which means $$F_H\wedge\omega^{n-1}=(n-1)!\Lambda_\omega F_H\frac{\omega^n}{n!}$$ If $\Phi$ is constant $\lambda$ then we call it Hermitian-Einstein metric. As additional remark of the sign of $\lambda$, If the metric $h$ admit hermitian-Einstein metric $Λ_ωF_h=λI $, and if $λ<0$ there is no holomorphic section, and if $λ=0$ then all holomorphic sections are parallel. Since $$\Delta |s|^2=|\nabla s|^2-<(\Lambda_\omega F_h)s,s>\geq 0$$ where $s$ is a section and hence $|s|$ is bounded. But maximum principle for subharmonic functions, so $|∇s|^2=λ|s|^2$. Hence if $λ<0$ there is no holomorphic section, and if $λ=0$ then all holomorphic sections are parallel. Now take $E=TM$, then we can define weak Kahler-Einstein metric as $Ric(\omega)=\phi \omega$ and hence $c_1(TM)=[Ric(\omega)]=[\phi \omega]$ When $\phi$ is globally constant, then we call it Kahler-Einstein metric and when it is positive we call $M$ is Fano and when it is negative we call $M$ is of general type and it is zero $M$ is called Calabi-Yau variety From Kobayashi book (complex vector bundle ) we have the following theorem for more general setting of your question Theorem: An Hermitian vector bundle $(E, h)$ over a Riemann surface $(M, g)$ satisfies the weak Einstein condition if and only if it is projectively flat. Note that in general one direction always works Theorem: Let $(M, g)$ is a Kahler manifold, Let $(E, h)$ be a projectively flat vector bundle over $M$. Then, for any Hermitian metric $g$ on $M$,$(E, h)$ satisfies the weak Einstein condition. We have also the following theorem which tell us that any weak Hermitian-Einstein metric after conformal change is Hermitian-Einstein metric Theorem: If an Hermitian vector bundle $(E, h)$ over a compact Kahler manifold $(M, g)$ satisfies the weak Einstein condition with factor $φ$, there is a conformal change $h −→ h′ = ah$ such that $(E, h′ )$ satisfies the Einstein condition with a constant factor $c$. Such a conformal change is unique up to a homothety. Now you can take $E=TM$ and this tells us that any weak Kahler-Einstein there is a conformal change $h −→ h′ = ah$ such that $(TM, h′ )$ is Kahler-Einstein metric<|endoftext|> TITLE: Easiest example where pseudo-isotopy fails to be the same as isotopy? QUESTION [14 upvotes]: This question concerns diffeomorphism of manifolds. Let $f: M \to M$ be a self-diffeomorphism. We will say that it is isotopic to the identity if there is a continuous one-parameter family of diffeomorphisms $$ f_t: M \to M $$ paramatrized by $t \in [0,1]$ such that $f_0 = id$ and $f_1 = f$. We will say that $f$ is pseudoisotopic to the identity if there is a diffeomorphism $$ F: M \times I \to M \times I$$ which restricts to the identity map on the top face $ M \times \{ 0\}$ and to $f$ on the bottom face $M \times \{1\}$. isotopy implies psuedoisotopy, but not necessarily the other way around. Cerf famously proved that in high dimensions if M is also simply connected then psuedoisotopy is the same as isotopy. I would like to better understand how this can fail. What is the easiest example where psuedoisotopy differs from isotopy? REPLY [16 votes]: In high dimensions ($\geq 5$) the most basic examples arise on manifolds with nonempty boundary, where one requires that diffeomorphisms restrict to the identity on the boundary. The simplest case is a diffeomorphism of $S^1\times D^{n-1}$ that is pseudoisotopic to the identity but not isotopic to the identity (always fixing the boundary). I described this construction in a paper called "Concordance spaces, higher simple-homotopy theory, and applications" that appeared in AMS Proc. Symp. Pure Math. 32 (1978). The paragraphs describing the construction are copied almost verbatim below, and a scanned version of the paper is available on my webpage. The construction is due originally to Tom Farrell (unpublished) but the fact that it has the properties claimed is an application of an old theorem of mine that had a gap in its proof that was later filled in by Kiyoshi Igusa. To get examples in closed manifolds $M$ one can embed $S^1\times D^{n-1}$ in $M$ so as to represent a nontrivial element of $\pi_1(M)$, then extend the diffeomorphism via the identity map on the complement. The resulting diffeomorphism of $M$ may or may not be isotopic to the identity, but there are some cases when it is definitely not isotopic to the identity, e.g., $M=S^1 \times S^{n-1}$ or $M=T^n$, the $n$-torus. The latter case is described in the paper, while the former case is more elementary using the fact that the space of embeddings $S^1\to S^1\times S^{n-1}$ homotopic to a given embbedding is simply-connected if $n$ is not too small. Now for the construction of the diffeomorphism $f\colon S^1\times D^{n-1} \to S^1\times D^{n-1}$, as described in the paper: To construct $f$ we will perform two embedded surgeries on the interior of the codimension one slice $D^{n-1}_0 = * \times D^{n-1}$, producing a new disk $D^{n-1}_1\subset S^1\times D^{n-1}$ with $\partial D^{n-1}_0=\partial D^{n-1}_1$ and with the complement of $D^{n-1}_1$ still an $n$-ball, so that $D^{n-1}_1 = f(D^{n-1}_0)$ for some homeomorphism (in fact diffeomorphism) $f$ of $S^1\times D^{n-1}$ fixing the boundary. In a neighborhood of $D^{n-1}_0$ label the two sides of $D^{n-1}_0$ as $+$ and $-$. On the $+$ side, attach an embedded $i$-handle $D^i\times D^{n-i}$ to $D^{n-1}_0$ in the trivial way. This effects a surgery on $D^{n-1}_0$ to $\chi(D^{n-1}_0)$, say. We could undo the effect of this surgery by now attaching an embedded $(i+1)$-handle $D^{i+1}\times D^{n-i-1}$ on the $+$ side of $\chi(D^{n-1}_0)$, in the trivial way so that the surgered $\chi(D^{n-1}_0)$ would be an $(n-1)$-disk isotopic to $D^{n-1}_0$. (All of this would occur near $D^{n-1}_0$.) The Farrell construction is to take instead a new embedding of the $(i+1)$-handle, but attached to $\chi(D^{n-1}_0)$ in the same way so that $\chi(D^{n-1}_0)$ is again surgered to a disk, this time the desired $D^{n-1}_1$. The new $(i+1)$-handle is obtained from the old one by replacing the old core $D^{i+1}$ by $D^{i+1}\# S^{i+1}$, the (interior) connected sum with a certain $S^{i+1}\subset (S^1\times D^{n-1})-\chi(D^{n-1}_0)$. This $S^{i+1}$ is constructed as follows. The core $D^i$ of the $i$-handle can be completed to a sphere $S^i$ by adding another $i$-disk on the $-$ side of $D^{n-1}_0$. In a neighborhood of this $S^i$, embed $S^{i+1}$ so as to represent the Hopf map $S^{i+1}\to S^i$. For this we must assume $i\geq 2$ and $n$ large enough to get $S^{i+1}$ actually embedded. Finally, to form the connected sum of $D^{i+1}$, which is on the $+$ side of $\chi(D^{n-1}_0)$, with $S^{i+1}$, which is on the $-$ side, we connect $D^{i+1}$ to $S^{i+1}$ by an arc which circles around the $S^1$ factor of $S^1\times D^{n-1}$. The construction actually gives a concordance from $D^{n-1}_0$ to $D^{n-1}_1$ in $S^1\times D^{n-1} \times I$, namely the trace of the two surgeries. This then extends to a concordance (pseudoisotopy) between $f$ and the identity, using the h-cobordism theorem to guarantee that the complement of the codimension-one concordance is a product (or maybe this can be seen directly).<|endoftext|> TITLE: Reducing largeness notions, uniformly QUESTION [7 upvotes]: This question is a particular take on the following theme. Suppose $A$ and $B$ are two notions of "large subset of $\omega^\omega$;" when is there a uniform method for turning an element of $A$ into an element of $B$? We work in ZF (although results under strengthenings of ZF are also interesting). For $U, V\subseteq\mathcal{P}(\omega^\omega)$, say $U$ spreads onto $V$ if there is some $F:\omega^\omega\rightarrow\omega^\omega$ such that $F(X)\in V$ for every $X\in U$, where $F(X)=\{F(x): x\in X\}$. The original motivation for this was in the context of working with cardinal characteristics of the continuum without choice - about which I have asked some questions before. This also was a natural, and hopefully less trivial, outgrowth of this question. However, especially in light of question 2 below, it is unclear whether there's actually any relation. So now it's just a pure curiosity question. Here's a trivial example. If $A=\{$dominating families$\}$ and $B=\{$escaping families$\}$, then $A$ spreads onto $B$ via the identity map. Here's a less trivial - maybe even interesting! :P - example. Let $C=\{$pointed perfect sets$\}$ (where a pointed perfect set is a nonempty closed subset $K$ of $\omega^\omega$ with no isolated points, such that every element of $K$ computes the tree representation of $K$ - that is, the set of strings $\sigma\in\omega^{<\omega}$ such that $\exists f\in K(\sigma\prec f)$). Then $C$ spreads onto $A$ (where $A$ is the set of dominating families as above). This argument hinges on two observations: There is a definable way to assign to each perfect $K\subseteq\omega^\omega$ a surjection $s_K: K\rightarrow \omega^\omega$. (Look at the tree representations.) Every real computes only countably many tree representations of perfect sets. We combine these facts as follows. Given a real $f\in\omega^\omega$, let $T_i$ ($i\in\omega$) be the perfect sets whose tree representation is computable in $f$ (these come ordered by index of Turing reduction), and let $g_i=s_{T_i}(f)$. Then let $F$ be defined as $$F(f)(n)=1+\sum_{i=0}^ns_{T_i}(f)(n).$$ That is, $F$ takes in $f$, and spits out a real dominating each of the countably many reals which $f$ could correspond to in any of the perfect sets $f$ computes. Now suppose $T$ is any pointed perfect set; I want to argue that $F(T)$ is a dominating family. Since $s_T$ is surjective, it's enough to show that for each $f\in T$, there is some $g\in F(T)$ such that $g$ dominates $s_T(f)$. But since $T$ is perfect, $f$ computes the tree representation of $T$; so we may take $g=F(f)$, and problem solved. However, each of the two results above has a natural follow-up question: Question 1. Does $B$ (escaping) spread onto $A$ (dominating)? Of course the answer is consistently "no," since it is consistent that the dominating number is strictly greater than the bounding number. However, I would like a ZF answer. Less valuable, but still very interesting, would be answers in either ZFC or ZF+AD+etc (which don't decide whether $\mathfrak{b}<\mathfrak{d}$ and which prove a weak CH, respectively - so the question is not trivial in either case). Question 2. Can we drop "pointed" in the second example? That is, does the family of merely perfect sets spread onto $A$? I don't see how to do this even in ZFC; we seem to need some way to assign to each real $f$ a family of "few" (specifically, smaller than the escaping number) reals which it needs to dominate. Via pointedness, we were able to cheat - every real is in only countably many pointed perfect sets. However, without this I don't see how to get the argument to go through. (Note that there are perfect sets whose tree representations aren't computed by any of their elements!) Actually, I'm beginning to think that the right approach for this type of question is to restrict attention to "pointed subsets of $\omega^\omega$" - that is, dominating/escaping/perfect/whatever sets, every element of which computes "the right representation" of the set in question. However, I still don't know quite what I mean by that. Any thoughts on the matter, while not the main focus of this question, would be welcome! REPLY [2 votes]: I hope that I didn't misunderstand the question. If escaping family means a family of reals which is unbounded in $\leq^*$ then an Erdős-Sierpinski-type proof gives that in ZFC+CH we have that $B$ spreads onto $A$ as follows: From an enumeration of the Borel non-dominating sets $\{B'_\alpha:\alpha<\mathfrak{c}\}$ taking unions of the initial segments (and regrouping if necessary) we can get an increasing sequence of non-dominating Borel sets $\{B_\alpha:\alpha<\mathfrak{c}\}$ such that for every $\alpha$ the ring $B_{\alpha+1} \setminus B_\alpha$ has cardinality $\mathfrak{c}$. Do the same with non-escaping Borel sets and obtain the enumeration $\{C'_\alpha:\alpha<\mathfrak{c}\}$ and the increasing sequence $\{C_\alpha:\alpha<\mathfrak{c}\}$ with $|C_{\alpha+1} \setminus C_\alpha|=\mathfrak{c}$. (We can also assume that $|C_0|=|B_0|=\mathfrak{c}$). Fix bijections $(f_\alpha)_{\alpha<\mathfrak{c}}$ between the sets $B_0$ and $C_0$, and $B_{\alpha+1} \setminus B_\alpha$ and $C_{\alpha+1} \setminus C_\alpha$, respectively. Let $F:\omega^\omega \to \omega^\omega$ be the function which consists of these bijections, i. e., $F(x)=f_\alpha(x)$ iff $x \in B_{\alpha+1} \setminus B_\alpha$. We claim that a set $S$ is dominating if and only if $F(S)$ is escaping. So suppose that $S$ is dominating but $F(S)$ is not escaping. Then $F(S)$ is covered by a Borel non-escaping set, which must be enumerated in $\{C'_\alpha:\alpha<\mathfrak{c}\}$ therefore $F(S)$ is covered by $C_\alpha$ for some $\alpha<\mathfrak{c}$. But $F^{-1}(C_\alpha)=B_\alpha$ by definition of $F$, so $S \subset B_\alpha$, contradicting the assumption that $S$ was dominating. The other direction is the same. This proof works in a more general context, namely for $\sigma$-ideals which have cofinality $\mathfrak{c}$.<|endoftext|> TITLE: Chow rings of smooth toric varieties QUESTION [12 upvotes]: In his 1987 article The geometry of toric varieties Danilov gives a combinatorial presentation of Chow rings of complete smooth toric varietes. Given a complete unimodular fan $\Sigma$ we have $$ A^*(X_\Sigma) \cong \mathbb Z[U]/(I+J), $$ where $U$ is the set of rays in $\Sigma$, $I$ the Stanley-Reisner ideal of the associated simplicial complex (on the rays) and $J$ an ideal depending on the coordinates of the minimal ray generators. In the recent preprint Hodge theory for combinatorial geometries by Adiprasito, Huh and Katz, the authors state that this isomorphism still holds when $\Sigma$ isn't complete. They cite the following two articles for this: Bifet, De Concini, Procesi, Cohomology of regular embeddings, 1990 Brion, Piecewise polynomial functions, convex polytopes and enumerative geometry, 1996 However, I'm unable to extract this result from those articles. Maybe someone can point out why this is implied by the results in the cited articles or possibly knows another reference? REPLY [2 votes]: If I understand correctly the result you want is the corollary at the end of section 3.1 of the Brion article you mention which states $$R_{\Sigma} / M R_{\Sigma} \cong A^*(X).$$ Notice in section 1.3 of the Brion article at the bottom of page 28 it states $R_{\Sigma}$ is isomorphic to the Stanley-Reisner algebra when $\Sigma$ is a simplicial fan. Also in section 1.4 on page 29 we see that $M R_{\Sigma}$ denotes the ideal generated by linear forms, which is the $J$ in the question.<|endoftext|> TITLE: Does nonexpanding map between manifolds decrease volume? QUESTION [7 upvotes]: (This question is a special case of a question I asked at SE, which got no answer there) Let $M,N$ be diffeomorphic connected compact Riemannian manifolds, and let $f:M \to N$ be a surjective nonexpanding map (i.e Lipschitz with constant $1$). Assume that $f$ is strictly nonexpanding, i.e there exists $p,q \in M$ such that $$d(f(p),f(q)) < d(p,q)$$ Is it true that $\operatorname{Vol}(N)<\operatorname{Vol}(M)$? Edit: I want $M,N$ to be smooth manifolds with or without boundary. Nik's answer gives a counter example (when $\operatorname{Vol}(N)=\operatorname{Vol}(M)$), but I am not sure it is smooth in the sense mentioned above. (The same problem comes up in this example given at S.E). Note that if we do not assume $M,N$ are diffeomorphic then the answer is negative: A counter-example is $f:[0,2\pi] \to \mathbb{S}^1, f(t)=e^{it}$. Partial result: In the case the manifolds are one-dimensional the answer is positive: Assume otherwise. Then $\operatorname{Vol}(N)=\operatorname{Vol}(M)$. Since every two compact connected one-dimensional Riemannian manifolds with equal volumes are isometric, there exists an isometry $\phi:N \to M$. Thus, $f \circ \phi:N \to N$ is surjective and nonexpanding, hence an isometry. REPLY [2 votes]: I don't think so, if "compact manifold" means "compact manifold with boundary" (as it seems to from your example) and "diffeomorphic" means "with diffeomorphic interiors". My previous example wasn't right, as Willie Wong pointed out, because after gluing the result wasn't Riemannian. But we can fix that problem by a simple modification: let $M$ be the unit disk minus the region $y > \sqrt{|x|}$, and let $f$ be the map that zips up the slit. That works, doesn't it?<|endoftext|> TITLE: Counting the orbits of a set of tabloids under the action of a Young subgroup QUESTION [5 upvotes]: Let $\lambda = (\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_k)$ and $\mu = (\mu_1 \geq \mu_2 \geq \cdots \geq \mu_\ell)$ be partitions of a positive integer $n$. As in Fulton's book on Young tableaux, recall that a tabloid is an equivalence class of numberings (with the distinct numbers $1,2,\ldots,n$) of a Young diagram, two being equivalent if corresponding rows contain the same entries. The symmetric group $S_n$ then acts in a natural way on tabloids. Let $M^\lambda$ denote the complex vector space with basis the tabloids of shape $\lambda$. Since $S_n$ acts on tabloids, $M^\lambda$ is naturally an $S_n$-representation. Now let $S_\mu$ denote the Young subgroup of $S_n$ corresponding to the partition $\mu$. For example if $n=7$ and $\mu=(3,2,2)$ then $S_\mu = S_{\{1,2,3\}} \times S_{\{4,5\}} \times S_{\{6,7\}} \cong S_3 \times S_2 \times S_2$. I am interested in computing the dimension of the $S_\mu$-invariant subspace of $M^\lambda$, i.e. $dim (M^\lambda)^{S_\mu}$. This amounts to counting the number of orbits under $S_\mu$ in the set of tabloids of shape $\lambda$. A simple reformulation in terms of ``balls and boxes'' would be: with $\lambda$ and $\mu$ the fixed partitions as above, I have $n$ balls in $\ell$ different colors, where there are $\mu_i$ many balls of the $i$-th color. I want to count the number of ways to distribute these balls into $k$ many boxes, where the $j$-th box can hold at most $\lambda_j$ many balls. Does anyone know a formula for this number? Alternatively, any concrete information about these numbers would be appreciated. REPLY [4 votes]: The number of ways to place the balls so that the $j$th box holds exactly $\lambda_j$ balls is the number $N_{\lambda\mu}$ of matrices of nonnegative integers with row sum vector $\lambda$ and column sum vector $\mu$. There is no simple formula for these numbers. For connections with symmetric functions, see for instance Prop. 7.5.1, equation (7.31), Cor. 7.12.3, and Exercise 7.77(c) of Enumerative Combinatorics, vol. 2. Note that if the $j$th box holds at most $\lambda_j$ balls then it holds exactly $\lambda_j$ balls since $\sum \mu_i = \sum \lambda_i$.<|endoftext|> TITLE: Proof of an identity involving $\int \exp(-|x-s|)dx$ over an even sphere QUESTION [22 upvotes]: I want to prove the following identity calculating the integral of an exponential over an even dimensional sphere in terms of functions $\chi_i(R)$ and $\tilde\psi_i(s)$ (described below) which are essentially modified spherical Bessel functions. First set up some notation. Let $p>0$ be an integer and $R>0$. Let $S_R^{2p}\subset \mathbb{R}^{2p+1}$ denote the radius $R$ sphere. Let $\mathrm{s}\in \mathbb{R}^{2p+1}$ be a point inside the sphere and let $s=|\mathrm{s}|$ so $0\le s0$ we have the following integral forms for the functions. \begin{align*} \chi_i(R) &= \frac{e^R R^{2i}}{2^{i-1} (i-1)!}\int_{t=0}^\infty e^{-R\cosh t}\sinh^{2i-1}t\, \mathrm d t\\ &= \frac{e^R R}{2^{i-1} (i-1)!}\int_{y=R}^\infty e^{-y}(y^2-R^2)^{i-1}\, \mathrm d y. \end{align*} \begin{align*} \tilde\psi_i(s) &= \frac{(-1)^{i}}{2^{i} (i-1)!}\int_{\theta=0}^\pi e^{s\cos \theta}\sin^{2i-1}\theta\, \mathrm d \theta\\ &= \frac{(-1)^{i}}{2^{i} (i-1)!s^{2i-1}}\int_{x=-s}^s e^{x}(s^2-x^2)^{i-1}\, \mathrm d x. \end{align*} REPLY [4 votes]: Following on from the answer of Sam Dolan, I generalized my conjecture to the following, which I prove as Theorem 4 in my paper The magnitude of odd balls via Hankel determinants of reverse Bessel polynomials. [The statement in the question is case where $j=0$.] Theorem. For $0\le j \le p$, $\mathbf{s}\in \mathbb{R}^{2p+1}$, with $s=|\mathbf{s}|$ and $R>s\ge 0$ \begin{equation*} \int_{\mathbf{x}\in S^{2p}_R} \psi_j(\left | \mathbf{x}-\mathbf{s}\right|)\,\mathrm{d}\mathbf{x} = (-2\pi)^p 2 e^{-R}\sum_{i=0}^{p-j} \binom{p-j}{i}\chi_{i+p}(R) \tilde\psi_{i+j}(s). %\label{eq:GeneralizedKeyIntegral} \end{equation*} Here $\psi_i$ (as opposed to $\tilde\psi_i$) can be defined by $\psi_i(r)=r^{-2i}e^{-r}\chi_i(r)$; it is related to a modified spherical Bessel function of the second kind. The sequence of such functions begins in the following way: \begin{align*} \psi_0 (r) &= e^{-r}\\[0.5em] \psi_1 (r) &= e^{-r} \left( \frac{1}{r} \right)\\ \psi_2 (r) &= e^{-r} \left( \frac{1}{r^{2}} + \frac{1}{r^{3}} \right)\\ \psi_3 (r) &= e^{-r} \left( \frac{1}{r^{3}} + \frac{3}{r^{4}} + \frac{3}{r^{5}} \right)\\ \psi_4 (r) &= e^{-r} \left( \frac{1}{r^{4}} + \frac{6}{r^{5}} + \frac{15}{r^{6}} + \frac{15}{r^{7}} \right) . \end{align*} This theorem is stated in terms of Bessel functions in another question.<|endoftext|> TITLE: Restricting a continuous positive-semidefinite function to a finite subset QUESTION [6 upvotes]: The following is a toy version of something I've been fiddling with and I thought it might be more efficient to post it as a question here. Just to fix definitions: for me, a complex-valued function $f$ on a group $G$ is said to be positive-semi-definite if, for every choice of points $x_1, \dots, x_n\in G$, the $n\times n$ matrix $[f(x_i^{-1}x_j)]_{i,j=1}^n$ is PSD in the usual sense. Note that in this definition I do not assume $G$ is countable, nor do I assume $f$ is continuous. Question 1. Let $f$ be a continuous PSD function on $\mathbb R$, and pick a finite subset $F\subset R$; then let $E=\{-x+y \mid x,y \in F\}$. Does there always exist a PSD function $h$ on $\mathbb R$, with finite or countably infinite support,such that $h\vert_E = f\vert_E$ ? Obviously, if we knew that the ``truncation'' of $f$ to the set $E$ was itself a PSD function on $\mathbb R$, then the answer to Q1 would be positive. But I don't see why this truncated function would always be PSD. Question 2. The analogue of Q1 with $\mathbb R$ replaced by $\mathbb R^d$ for $d\geq 2$. My personal suspicion is that Q2 will have a negative answer but this is based on pessimism rather than any genuine intuition. REPLY [3 votes]: The answer for countably infinite has been given by fedja and Uri Bader in the comments and is yes: put $f$ to $0$ outside of the subgroup generated by $E$, and leave $f$ unchanged on this subgroup. For finite, the answer is no. For an example, take $f=1$ and $E=\{-1,0,1\}$. Since the only PSD function of the form $\begin{pmatrix} 1 & 1 & a \\ 1&1&1\\ b &1 &1\end{pmatrix}$ is the matrix $\begin{pmatrix} 1 & 1 & 1 \\ 1&1&1\\ 1 &1 &1\end{pmatrix}$, we have that any positive-semidefinite function whose restriction to $E$ is $1$ has to be constant equal to $1$ on $\mathbf Z$.<|endoftext|> TITLE: What is the story behind the Chebyshev polynomials? QUESTION [15 upvotes]: Is there anything reliable known about who actually discovered the Chebyshev polynomials and what the motivation and circumstances were? The reason why I am interested in knowing, is that I needed a solution for a variant of those polynomials: instead of all extrema having the same magnitude, I wanted to have them attain predefined values in a fixed order (I have found a solution for that problem, but involves a system of polynomial equations) and I wonder, whether the definition of the Chebyshev polynomials has been "guessed" or developed for a specific problem. Edit: at the request of @Hans, here is formal definition of my problem: given a sequence $(y_1,\ ...\ y_{n-1}), (y_{i+2}-y_{i+1})(y_{i+1}-y_i)<0$ of values, determine a polynomial $p(x)$ of degree $n$ and, $\ n$-$1$ abszissas $\ \xi_1 <,\ ...,\ <\xi_{n-1}$, so that $\ p(\xi_i)=y_i, p'(\xi_i)=0$ It should be noted that the polynomials that I am looking for, have no special properties, except for the predefined values in the extrema. The leading coefficient can be set to $1$ and the constant term to $0$. $$\ $$ Construction of polynomials with predefined sequence of function-values for its local extrema: we can w.l.o.g. assume that the sought polynomial has leading coefficient $1$, a local extremum in the origin and, that all other local extrema are located at positive abszissas. Then polynomial is $$p(x) =\frac{1}{n}\int x\prod_{i=2}^{n-1}(x-\xi_i)$$ and $$p(\xi_i)=y_i$$ would a be system of polynomial equations for determining the $\xi_i$ and thus $p(x)$; the only problem being that, because of the symmetry, in the current formulation there is no control over the ordering of the $y(\xi_i)$. That can however easily be fixed by defining $$\xi_k=\sum_{i=2}^{k}a_i^2$$ and solving the system of polynomial equations $$p(\sum_{i=2}^{k}a_i^2)=y_k$$ REPLY [3 votes]: An earlier, but different, approach to the questions posed and solved by next-to-be Tchebyshev Polynomials was given by Augustin-Louis Cauchy in his Cours d'Analyse (1821). See p. 230 and ff.<|endoftext|> TITLE: What is the first Pontryagin class of the $n$-dimensional representation of $S_n$? QUESTION [14 upvotes]: The symmetric group $S_n$ has an $n$-dimensional defining representation, which splits as $n = (n-1) + 1$. Although this representation exists integrally, I would like to think of this as a real representation $S_n \to O(n)$, or equivalently as an $n$-dimensional real vector bundle on the classifying space $BS_n$. Real vector bundles $V \to X$ have Pontryagin classes. The first Pontryagin class $p_1(V)$ lives in $H^4(X;\mathbb Z)$. When $n \gg 0$, $H^4(BS_n;\mathbb Z) \cong \mathbb Z/2 \times \mathbb Z/2 \times \mathbb Z/12$. $p_1$ is additive for direct sum of real vector bundles, and vanishes on trivial bundles. What is the value of the first Pontryagin class of the defining rank-$n$ bundle on $BS_n$? I am happy with answers just in the stable range, but of course feel free to share the unstable answers as well. (I say stable, but of course I expect $p_1$ depends on $n$, just not $H^4$.) REPLY [11 votes]: If I am not confused, $p_1(V)=-c_2(V\otimes {\mathbb C})$. According to Theorem 7.1 in the book Characteristic Classes and the Cohomology of Finite Groups by Charles Thomas, $c_2$ of the standard representation of $S_n$ has order $12$, for $n$ large enough.<|endoftext|> TITLE: Examples of compact complex manifolds for which the $dd^c$ lemma does not hold QUESTION [8 upvotes]: The well known $dd^{c}$ lemma in complex geometry claimed that Let $X$ be a compact Kähler manifold. Let $p,q\ge 1$. Let $\eta$ be a $(p,q)$ form on $X$ and assume $\eta$ is $d$-exact. Then there exists a $(p-1,q-1)$ form $\beta$ such that $$ \eta=dd^{c}\beta $$ If $p=q$ and $\eta$ is real, then we may take $\beta$ to be real. (Source, Griffths and Harris, page 149, sign of $\beta$ corrected). The traditional proof breaks down for non-Kähler complex manifolds because in this case $\Delta_{\overline{\partial}}=\Delta_{\partial}=\frac{1}{2}\Delta_{d}$ may no longer holds. And then the lemma seems too good to be true in general. I am wondering if we have examples of compact complex manifolds for which we know the $dd^{c}$ lemma does not hold. Here is a counter-example for an open domain. The Hodge decomposition still holds for any elliptic pseudodifferential operator, and we can still take its complex conjugate. But I am kind of lost how to construct a specific counter-example. The topic is classical, so I am not sure if this is common sense among expert circle already. If the statement is true for certain non-Kähler complex manifolds, then I suspect we need to show that $$ (\eta, (\partial\overline{\partial}^{*}+\overline{\partial}^{*}\partial)\eta)\le 0, \forall \eta\in \mathcal{A}^{p,q}(M) $$ in these cases. I am wondering what does this imply for $M$ from an analysis perspective. REPLY [2 votes]: Below are just some more examples illustrating Donu's answer. In the paper A new geometric construction of compact complex manifolds in any dimension Laurent Meersseman constructed a large family of complex non-Kahler manifolds. Each of this manifolds is (a deformation of) a principle torus bundle over toric variety. The same manifolds are known in the toric topology under the name moment-angle-manifolds (see, e.g., Real quadrics in ℂn, complex manifolds and convex polytopes and Complex-analytic structures on moment-angle manifolds) and it is has been demonstrated that many of them have non-trivial triple Massey products.<|endoftext|> TITLE: Generators of a convex cone defined by a differential inequality QUESTION [5 upvotes]: Consider the cone of continuously twice differentiable functions mapping positive reals to itself (i.e., $f\in C^2(\mathbb R_{++})$ and $f\colon \mathbb R_{++}\to\mathbb R_{++}$) that satisfy \begin{align} f(x)&\geq 0\\ f(x) - xf'(x)- x^2f''(x)&\geq 0 \end{align} for all $x\in\mathbb R_{++}$. Let us observe that this is indeed a convex cone in that scaling and adding two functions together would preserve these two inequalities. My question is whether it would be possible/feasible to find the extreme rays (also called generators, I believe) of this convex cone. One can observe that each power function $f(x) = x^p$ for $p\in[-1,1]$ is in this cone, as they are positive valued and satisfy the differential condition. However it is not clear at all that these functions span this convex cone. Should one expect to have an explicit set of extreme rays? REPLY [2 votes]: You are looking at the intersection $C = C_1 \cap C_2$ of two cones, $C_1$ defined by the condition $f(x) \ge 0$ and $C_2$ defined by the condition $Df(x) \ge 0$, where $D=(1+x\partial_x)(1-x\partial_x)$. In a sense, $C_2 = D^{-1} C_1$, where the right-hand-side is interpreted as the pre-image of $C_1$ with respect to $D$. Now, $C_1$ does not have extreme rays (those that do not belong to the sum of any two different rays from the same cone), because any non-negative function with non-trivial support can always be written as a convex linear combination with positive coefficients of two other non-negative functions with smaller supports. If you embed $C^2(\mathbb{R}_{++})$ in distributions, then then the closure of $C_1$ does have extreme rays, which are the Dirac $\delta$ distributions, $\delta(x-y)$. Next, look at $C_2$. Using an explicit right-inverse $D^{-1}_0$ of the operator $D$, we can write $C_2 = \ker D + D^{-1}_0 C_1$, where $\ker D = \mathbb{R}_+ x + \mathbb{R}_+ x^{-1}$. A right inverse can be given, if I got all the calculations right (edit: turns out that I did miss an overall minus sign in the previous version), by the integral operator \begin{multline*} D^{-1}_0[g](x) = -\int_0^x \frac{(x-y)(x+y)}{2xy^2} \chi_0(y) g(y) \, dy \\ - \int_x^\infty \frac{(y-x)(x+y)}{2xy^2} \chi_\infty(y) g(y) \, dy , \end{multline*} where $\chi_0(y)$ and $\chi_\infty(y)$ are any two non-negative continuous functions such that $\chi_0(y) + \chi_\infty(y) = 1$, with $\chi_0(y)$ zero on a neighborhood of $0$ and $\chi_\infty(y)$ zero on a neighborhood of infinity. The way it's written, the integrals are well-defined for any continuous $g(x)$. Note that, because the integral kernel is non-positive, $g(x) \ge 0$ implies that $D^{-1}_0[g](x) \le 0$. Note that, because the integral kernel is non-negative, $g(x) \ge 0$ implies that $D^{-1}_0[g](x) \ge 0$. Finally, with the above observations, we can conclude that $$ C = C_1 \cap (\ker D + D^{-1}_0 C_1) = D^{-1}_0 C_1 + \ker_{\ge} D , $$ where $\ker_{\ge} D = \mathbb{R}_+ x + \mathbb{R}_+ x^{-1}$. Now, $D^{-1}C_1$ does not have any extreme rays either, by the same argument as for $C_1$. On the other hand, $\ker_{\ge} D$ does have extreme rays, those generated by $x$ and $x^{-1}$. Moreover, since $\ker_{\ge} D \cap D^{-1}_0 C_1 = \varnothing$, these two rays remain extreme when we take the sum of these two cones to get $C$. So, it seems that the conclusion is that the cone $C$ has only the two extreme rays generated by $x$ and $x^{-1}$ and no other ones. Clearly, $C$ is very far from being even the closure of the convex linear combinations of its extreme rays.<|endoftext|> TITLE: Define $\mathbb{N}$ in the ring $\mathbb{Z}$ without Lagrange's theorem QUESTION [30 upvotes]: It is well-known that the set of nonnegative integers $\mathbb{N}$ is definable in the ring of integers $\mathbb{Z}$. Indeed, by Lagrange's four squares theorem we have $\mathbb{N} = \{n \in \mathbb{Z} : \varphi(n)\}$, where $\varphi$ is the formula $$\varphi(x) := \exists a\, \exists b\, \exists c\, \exists d \; x = a^2 + b^2 + c^2 + d^2$$ However, Lagrange's theorem is not so trivial, so I wonder: Is there a more elementary and self-contained proof of the definability of $\mathbb{N}$ in the ring of integers? Thank you. REPLY [35 votes]: Raphael Robinson, in a paper entitled Arithmetical Definitions in the Ring of Integers, gives a definition of $\mathbb{N}$ in $\mathbb{Z}$ using only two existential quantifiers. He does not use Lagrange's Theorem, but he does use the fact that the equation $y^2-az^2=1$ has infinitely many solutions $y$ and $z$ whenever $a$ is a positive nonsquare integer. Whether this is "easier" than Lagrange's Theorem is debatable. Robinson proves that $x\in \mathbb{N}$ if and only if $$\exists y\exists z \left(x=y^2\vee (y^2=1+xz^2\wedge y^3\ne y)\right).$$ He also proves that no definition using just one existential quantifier is possible.<|endoftext|> TITLE: Crux of Dwork's proof of rationality of the zeta function? QUESTION [24 upvotes]: As the question title suggests, what is the crux of Dwork's proof of the rationality of the zeta function? What is the intuition behind the proof, what are the key steps that the proof boils down to? REPLY [7 votes]: Sasha has already pointed you to the primary source I used some years ago for my "expository" undergraduate thesis on Dwork's Theorem: Koblitz's p-adic Numbers, p-adic Analysis, and Zeta-Functions, which is available in 166 searchable pages here. I've uploaded a copy of my thesis, which should constitute a relatively easy to digest ~ 50 page write-up of the proof. There is no significant difference from Koblitz's book; I worked out some of the pieces that were left as exercises or examples (e.g., showing that the zeta function for $f = x_1 x_4 - x_2 x_3 - 1$ is rational i.e. $\frac{1-qT}{1-q^3 T}$, which I re-produced in MO 117904) and omitted some of the proofs of cited theorems (e.g., the $p$-adic Weierstrass Preparation Theorem) with pointers to Koblitz's book or Gouvêa's intro to $p$-adic numbers when necessary. As I understood matters almost a decade ago, there were a couple of important linear maps mentioned by Koblitz (included in my section 3.2) $G$ and $T_q$, which were composed to give the trace operator, $\Psi$, already mentioned. I introduce a slight bit of terminology ("admissible" in Definition 3.14) to discuss the trace of a linear operator from a certain infinite dimensional vector space to itself, and one of the key facts drawn upon in working towards the end of the proof is the power series identity: $$ \det(1 - \psi T) = \exp \left(-\sum_{s=1}^{\infty} \text{Tr}(\psi^{s})T^{s}/s\right) $$ for a linear operator $\psi$ going from a vector space to itself. I re-mention that identity here, because I asked about its history in MSE 277124 for which Markus Scheuer gave a great answer. This identity arises in proving that a sufficient-to-consider modification of the zeta function $Z'(T)$ is $p$-adic meromorphic, which is established through a bit of induction, analysis of the above identity, and the inclusion/exclusion principle (the last of which re-arises, as Sasha mentions, in proving Dwork's theorem for affine hypersurfaces can be extended to projective varieties). I do not believe that there is anything novel in my undergrad write-up, but I think it is possible that its presentation as being intended for college juniors/seniors could make the proof, hence the underlying intuition, easier to grok.<|endoftext|> TITLE: Weaker version of Dwork's rationality of zeta function, what is needed to beef up into a complete proof? QUESTION [14 upvotes]: This is a followup to my question here. Here is a note of Michael Larsen where he gives a very simple proof of a slightly weaker result than Dwork's rationality of the zeta function. http://mlarsen.math.indiana.edu/~larsen/papers/zeta.pdf As the question title suggests, what is needed to beef this up into a complete proof a la Dwork of rationality of the zeta function? REPLY [3 votes]: Not quite an answer, but a heuristics from the point of view of Weil philosophy about why the rationality mod $p$ is much easier. Weil reduced the rationality of zeta-function to the existence of a good cohomology theory. The crucial property of a cohomology which satisifies the Weil axioms is that it has zero-characteristics coefficients. This is needed because Lefshetz formula is an identity in the coefficients ring, so to compute the number of $\mathbb{F}_q$-points precisely, we need that coefficents ring does not have any $\mathbb{Z}$-torsion. But if you are interested only in the number of points modulo $p$, a cohomology theory with $\mathbb{F}_p$-coefficients which satisefies the Lefshetz trace foormula would fit(all the formulas for zeta functions should just be reduced modulo $p$). It is very easy to give an example of such cohomology theory. It is $X\mapsto H^i(X,\mathcal{O}_X)$. Holomorphic Lefshetz trace formula applied to Frobenius gives $$\sum\limits_{x\in X(\mathbb{F}_{p^n})}\frac{1}{\det(1-Fr|_{T_{x,X}})}=\sum\limits_i(-1)^i\mathrm{Tr}\, Fr|_{H^i(X,\mathcal{O}_X)}$$ But Frobenius is inseparable, so its differential is zero and the LHS is stil $\# X(\mathbb{F}_{p^n})$ so we get the desired trace formula modulo $p$. So, the rationality mod $p$ should be much easier. You can also trace the analogy between Chevalley-Warning method and holomorphic Lefshetz trace formula in the proof of equivalence of different definitions of supersingular elliptic curve(see i.e. Hartshorne). Historically, the rationality mod $p$ was certainly known already to Serre, since he considered Witt cohomology $H^i(X,W\mathcal{O}_X)$ as a candidate for Weil cohomology.<|endoftext|> TITLE: Formal power series is Taylor expansion of rational function iff Hankel determinants vanish? QUESTION [9 upvotes]: Let $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants $$\det(a_{i + j + M}){_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M\gg0$? REPLY [2 votes]: This is Lemma III, attributed to Kronecker, on page 5 of Salem, Algebraic Numbers and Fourier Analysis. The book has been posted to the web by Plouffe, at http://plouffe.fr/simon/math/Algebraic%20Numbers%20and%20Fourier%20Analysis.pdf REPLY [2 votes]: I am not going to add anything new, just what is my take on it. The proof consists of four really simple steps. Step I. Determinant is zero iff there exist numbers $b_0,\dots,b_N$, not all of them zero, such that $$\sum_{i=0}^N a_{M+i+j}b_i=0$$ for all $0\le j\le N$. (Proof: Basically, the definition of determinant.) Step II. Determinant is zero iff there exists a nonzero polynomial $B(T)=b_0T^N+\dots+b_N$ such that in the series $$B(T)u(T)=\sum_{n=0}^\infty c_n T^n$$ we have $$c_{M+N}=c_{M+N+1}=\dots=c_{M+2N}=0.$$ (Proof: Multiplication of series.) Step III. The series $u(T)$ is rational iff there exists $N>0$ and a nonzero polynomial $B(T)=b_0T^N+\dots+b_N$ such that in the series $$B(T)u(T)=\sum_{n=0}^\infty c_n T^n$$ we have $$c_{M+N}=c_{M+N+1}=\dots=c_{M+2N}=0$$ for $M\gg 0$. (Proof: just the definitions!) Step IV. The series $u(T)$ is rational iff there exists $N>0$ such that the determinant is zero for $M\gg 0$. Now, I cheated in the step IV: even if the determinant is zero, the polynomial $B(T)$ may actually depend on $M$, which is wrong! Actually, this is the hard part. The standard way to fix this is to chose the smallest possible $N$, as it is done in Koblitz.<|endoftext|> TITLE: Infinite limit of ratio of nth degree polynomials QUESTION [11 upvotes]: The Problem I have two recursively defined polynomials (skip to the bottom for background and motivation if you care about that) that represent the numerator and denominator of a factor and I want to find the limit of that factor as n goes to infinity. $$n_0 = d_0 = 1$$ $$n_n = d_{n-1}x - n_{n-1}$$ $$d_n = d_{n-1}(x-1)-n_{n-1}$$ I was able represent this recursive relationship as a matrix and use eigenvalue matrix decomposition to find a closed form for $d_n$ and $n_n$. They are not very pretty: $$d_n = \frac{2^{-n-1}}{\sqrt{x-4}} \left(\left(\sqrt{x-4}+\sqrt{x}\right) \left(x+\sqrt{x-4} \sqrt{x}-2\right)^n+\left(\sqrt{x-4}-\sqrt{x}\right) \left(x-\sqrt{x-4} \sqrt{x}-2\right)^n+\frac{2 \left(\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n-\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n\right)}{\sqrt{x}}\right)$$ $$ n_n = 2^{-n-1} \left(\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n+\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n+\frac{\left(x+\sqrt{x-4} \sqrt{x}-2\right)^n-\left(x-\sqrt{x-4} \sqrt{x}-2\right)^n}{\sqrt{\frac{x-4}{x}}}\right)$$ The first few terms of the ratio ($r_n = n_n/d_n$) are: $$r_1 = \frac{x-1}{x-2}$$ $$r_2 = \frac{x^2-3 x+1}{x^2-4 x+3}$$ $$r_3 = \frac{x^3-5 x^2+6 x-1}{x^3-6 x^2+10 x-4}$$ $$r_4 = \frac{x^4-7 x^3+15 x^2-10 x+1}{x^4-8 x^3+21 x^2-20 x+5}$$ And so on. All of these equations seem to have poles located solely at $0\le x\le 4$, moreover the denominator and numerator seem to never have imaginary roots, but I haven't proven that. $x$ is assumed to be a positive real number, and because there are no poles $x > 4$ the limit seems to be well behaved and equal to 1(?) outside of this range. But within this range, I am extremely curious about what the limit is, if it even exists or is possible to evaluate. Can someone let me know if it's possible to evaluate this limit in this range? And if so can you let me know how to go about it? Background and Motivation An ideal transmission line can be modeled as an inductor and capacitor, the inductor is in series with the load and the capacitor is in parallel. The impedance of an inductor is given by $Z_L = i \omega L$ and the impedance of a capacitor is $Z_C = \frac{1}{i \omega C}$. If we string several transmission lines together and end in an open circuit then we can start evaluating from the end using the impedance addition laws. We first add $Z_L + Z_C$ and then to combine with the second to last capacitor we must $\frac{1}{\frac{1}{Z_L+Z_C}+\frac{1}{Z_C}}$. When we do that we find, interestingly that the ratio between $Z_C$ and this value is our real valued ratio $r_1$. We can then solve for $r_n$ by breaking up the ratio into a numerator and denominator and starting from the (n-1)th iteration, and find the relationship given at the top of the question, with $x = \omega ^2 L C$. Because a series of short transmission lines strung together is the same thing as a long transmission line, I expected the limit of this ratio to be equal to 1. However, when attempting to evaluate this limit it seems that it is not that simple. REPLY [11 votes]: Here is an explicit formula for your ratio $r_n=\frac{n_n}{d_n}$: $$r_n= \frac{\sum_{k=0}^n\binom{n+k}{2k}(-x)^k} {\sum_{k=0}^n\binom{n+k+1}{2k+1}(-x)^k}.$$ Let $P_n(x)$ and $Q_n(x)$ be the numerator and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad \text{and} \qquad Q_{n+2}+(x-2)Q_{n+1}+Q_n=0.\tag1$$ They differ only in the initial condition where $P_0=1, P_1=1-x$ while $Q_0=1, Q_1=2-x$. The importance of such a description is that (1) the original recursive relations are decoupled here; (2) it is more amenable to an asymptotic analysis; (3) it reveals the roots being in $[0,4]$ due to the interlacing property of three-term recurrences. Note. The original numerator and denominator differ by $\pm$ sign from $P_n$ and $Q_n$, but this makes no difference for the ratio $r_n$. In fact (Fedor!), $$r_n=\frac{P_n(x)}{Q_n(x)}=\sqrt{x}\,\frac{U_{2n}(\sqrt{x}/2)}{U_{2n+1}(\sqrt{x}/2)}$$ where $U_n(y)$ are Chebyshev polynomials of the 2nd kind, expressible as $$U_n(y)=\frac{(y+\sqrt{y^2-1})^n-(y-\sqrt{y^2-1})^n}{2\sqrt{y^2-1}}.$$ If $y\geq1$, or equivalently $z=y+\sqrt{y^2-1}\geq1$, then $$\lim_{n\rightarrow\infty}\frac{U_n(y)}{U_{n+1}(y)}= \lim_{n\rightarrow\infty}\frac{z^n-z^{-n}}{z^{n+1}-z^{-n-1}}=\frac1z=y-\sqrt{y^2-1}.$$ If $0 TITLE: Product of a Laver table and a Hadamard matrix has mostly 0 rows QUESTION [14 upvotes]: I recently noticed (while playing around) that the product of a Laver matrix with a Hadamard matrix gives a very sparse matrix. In particular, all but logarithmically few rows are all zero. The nonzero rows all seem to occur near powers of 2, or multiples of large powers of 2. For instance, multiplying the 256x256 Laver table by the 256x256 Hadamard matrix, the only nonzero rows are {1, 2, 4, 8, 9, 16, 32, 33, 64, 65, 96, 97, 128, 129, 160, 161, 192, 193, 224, 225, 256} Is this a known phenomenon? Is there an explanation for it, other than the high periodicity of the Laver table? In particular since all current proofs of the unbounded growth of the period of the first row of a Laver table rely on strong cardinal axioms, I wonder if understanding this phenomenon could help us understand it. REPLY [2 votes]: The sparsity of the product of a Laver table and a Hadamard matrix only follows from the periodicity of the Laver tables since the non-distributive but still periodic fake Laver tables (see this question for a more general construction) still exhibit this phenomenon. I have used computer calculations to test this phenomenon with the fake Laver tables. My computer calculations indicate that the product with a fake Laver table matrix and a Hadamard matrix is just as sparse as the product of a real classical Laver table and a Hadamard matrix.<|endoftext|> TITLE: Selberg Zeta Function and Fenchel-Nielsen Coordinates QUESTION [7 upvotes]: According to Uniformization theorem every compact Riemann surface $\Sigma$ of genus $g\ge2$ is isomorphic to a space that can be obtained by the action of a Fuchsian group on upper half plane $\mathbb{H}$ $$\Sigma\simeq\frac{\mathbb{H}}{\Gamma}$$ On the Teichmüller or moduli space of such Riemann surfaces, one can consider Fenchel-Nielsen coordinates $\{\ell_a,\tau_a\}_{a=1}^{3g-3}$. On the other hand, Selberg zeta function for a compact Riemann surface can be written as: $$Z(s)\equiv\prod_{\{\gamma_p\}}\prod_{n}\left(1-e^{-(n+s)\ell_{\gamma_p}}\right)$$ In which $\{\gamma_p\}$ is the set of primitive elements of Fuchsian group $\Gamma$ and $\ell_{\gamma_p}$ is the length of the corresponding simple closed geodesic with respect to hyperbolic metric on $\Sigma$ induced from Poincare metric on $\mathbb{H}$. There are several questions: What is the number of primitive elements of $\Gamma$? (There are supposedly infinitely many simple closed geodesics on $\Sigma$ so it should be infinite.) What is the number of generators of $\Gamma$? Is there any relation between number of generators of $\Gamma$ and genus of the surface? There are some quantities on the Riemann surface that can be expressed in terms of Selberg zeta function. For example Determinant of Laplacian acting on various tensor fields on the Riemman surface can be written in terms of Selberg zeta function. These quantities are thus dependent to the Teichmüller/moduli parameters. So the natural question is that what is the relation between lengths $\{\ell_{a}\}$, the Fenchel-Nielsen coordinates and $\{\ell_{\gamma_p}\}$ in Selberg zeta function? Regarding the last question, it seems that we can consider $3g-3$ primitive elements $\{\hat{\gamma}_{p_i}\}_{i=1}^{3g-3}$ of $\Gamma$ and identify their lengths with Fenchel-Nielsen length coordinates $\{\ell_a\}_{a=1}^{3g-3}$: $$\ell_{\hat{\gamma}_{p_i}}\equiv\ell_i \qquad i=1,\cdots,3g-3$$ If that is the case: What is the role of other lengths? Are they related to the set $\{\hat{\gamma}_p\}$ by the action of Mapping class group? Following Quillen, The determinant of Dirac operator is defined through the corresponding Laplacian which in turn can be expressed in terms of Selberg zeta function. Regarding the fact that FN coordinates can not give a complex structure to Teichmuller space, is there any notion of "holomorphic factorization" which makes it possible to define the determinant of Dirac operator directly in terms of Selberg zeta function? If that is not the case, How is it possible to obtain dependence of determinant of the Dirac operator on Fenchel-Nielsen coordinates? REPLY [2 votes]: I can answer some of the questions: there are infinitely many primitive elements, the so called Prime Geodesic Theorem tells you how the number of primes with bounded length growth as the bound tends to infinity. The number of generators is $2g$. This is elementary surface theory. There are generators $\gamma_1,\dots,\gamma_{2g}$ with the generating relation: $$ [\gamma_1,\gamma_2]\cdots[\gamma_{2g-1},\gamma_{2g}]=1, $$ where $[a,b]=aba^{-1}b^{-1}$.<|endoftext|> TITLE: Algorithm for finding quiver algebras QUESTION [9 upvotes]: Im looking for an algorithm that does the following in a quick way: Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output: Finds all two-sided ideals in $J^2/J^s \subseteq \mathbb{F}_q \langle x_1,...,x_r\rangle/J^s$ ,where $A=\mathbb{F}_q \langle x_1,...,x_r\rangle $ is the noncommutative polynomial ring in r variables over the finite field with q elements and $J=(x_1,...,x_r)$ is the ideal generated by $x_1, ... , x_r$. How many ideals are there? And how many up to isomorphism? (Two ideals $I_1$ and $I_2$ are isomorphic iff $A/ I_1$ and $A/I_2$ are isomorphic as $K$-algebras) It might be also interesting to replace the noncommutative polynomial ring by a commutative one. Note that the problem can also be formulated as to find all admissible ideals containing $J^2$ of a quiver algebra with one point and $r$ loops over a finite field. As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (lets say $r=2,3$ and $s=2,3,4$ and choosing $q=2,3,5$). It should be programmed with the GAP-packet qpa. edit: I rewrote the question to make it shorter.Easiest special case is a question on stackexchange: https://math.stackexchange.com/questions/2322214/number-of-ideals-with-gap . REPLY [4 votes]: Here are some observations: $A$ is just a tensor algebra $T(V)$ for $V=K^r=span\{x_1,\ldots,x_r\}=J/J^2$. Let's work with that because choosing bases is evil. Every morphism $\phi: A/I \to A/I'$ lifts to a (in fact multiple) morphism(s) $\Phi: A\to A$ which maps $I$ into $I'$ simply because polynomial algebras are free algebras. But it is not clear that $\Phi$ needs to be an isomorphism as well and I think it might not be true in general. But! It would also be nice to have a lift in between so to speak which is an automorphism of $A/J^s$ instead and that works out nicely because everything is local and finite-dimensional. Lemma 1: Every isomorphism $\phi: A/I\to A/I'$ lifts to an automorphism $\widehat{\phi}$ of $A/J^s$ which sends $I$ to $I'$. Proof: You require $I$ and $I'$ to be contained in $J^2$, the radical of $A/I$ is $J/I$, every isomorphism of algebras sends the radical to the radical and induces an automorphism of $A/J^2=K\oplus V$. Therefore $\forall x\in J: \Phi(x)\equiv \alpha(x) \mod J^2$ for some $\alpha\in GL(V)$. In particular $\Phi(J)\subseteq J$, thus $\Phi(J^i)\subseteq J^i$ for all $i$ and we get a well-defined endomorphism $\widehat{\phi}$ of $A/J^s$. Also $im(\widehat{\phi})+J^2/J^s=A/J^s$ so that $\widehat{\phi}$ is surjective by Nakayama's lemma and therefore bijective as a map $A/J^s\to A/J^s$. QED. Alright, we now face the problem of describing and distinguishing the orbits of $\operatorname{Aut}(A/J^s) \curvearrowright \operatorname{Ideals}(A/J^s)$. Since automorphisms respects the radical, we have a natural filtration here which gets respected by all automorphisms. This first distinguising invariant of ideals we should consider is where in the filtration $A \supseteq J \supseteq J^2 \supseteq \ldots$ we first see $I$, i.e. what is the smallest degree of a monomial occuring in a generating set $I$, i.e. the smallest number $d(I)\in\mathbb{N}$ such that $I\not\subseteq J^{d+1}$. A second invariant is the subspace $I+J^{d+1} \leq J^d/J^{d+1}$, more specifically its orbit under the $Aut(A/J^s)$-action on $J^d/J^{d+1}$. Contained in your problem is the classification of all ideals with $d(I)=s-1$, i.e. those $I$ that are sandwiched betweens $J^s$ and $J^{s-1}$. Since $Aut(A/J^s)$ acts on $J^{s-1}/J^s$ like $GL(V)$ on $V^{\otimes(s-1)}$, we find ourselves with Problem A: For all $s\in\mathbb{N}_{\geq 2}$ classify the orbits of $GL(V)$ on the subspaces of $V^{\otimes s}$. Ideally, find something like a normal form algorithm. This alone seems incredibly hard. It smells like there is a lot of non-trivial algebraic geometry involved here. I may, computationally it is easy to program in GAP, because one can write down everything completely explicit. The number of such orbits probably does not depend on $q$ at all, but is probably polynomial in $r$ with degree depending on $s$. Even just the one-dimensional case is non-trivial, even the case of pure tensors. Since $P(V)\times\cdots\times P(V) \to P(V\otimes\cdots\otimes V)$ is injective, the $GL(V)$-orbits on pure tensors in $V^{\otimes s}$ are the same as the $GL(V)$-orbits on $P(V)^s$ which already gives you a very crude, but very disheartening lower bound for the number of isomorphism classes of ideals: Corollary: There are at least $2^{\lfloor\frac{s-1}{2}\rfloor}$ isomorphism classes of one-dimensional ideals of $A/J^s$. Proof: $GL(V)$ has two orbits on $P(V)\times P(V)$ (all pairs of linearly dependent vectors and all pairs of linearly independent vectors respectively). For any $G$-sets $X,Y$, the number of $G$-orbits on $X\times Y$ is at least the number of $G$-orbits on $X$ times the number of $G$-orbits on $Y$ so that $GL(V)$ has $\geq 2^{s/2}$ orbits on $(P(V)\times P(V))^{s/2}$ so that there are at least $2^{\lfloor\frac{s-1}{2} \rfloor}$ many one-dimensional ideals generated by pure tensors in $V^{\otimes(s-1)}$. QED. EDIT 03.07.: Speaking of algebraic geometry: The Grassmannian $Gr_k(V^{\otimes s})$ is a $k(r^s-k)$-dimensional smooth projective variety while $GL(V)$ is $r^2$-dimensional. The orbits can therefore not be more than $r^2$-dimensional. The orbit space itself isn't a scheme itself, I think, but maybe it's a stack or something. Nevertheless my heurestic is that we should expect it to be $\geq (k(r^s-k) - r^2)$-dimensional at least at some points if $s>2$. In particular, there will be something like $q^{\frac{1}{3}r^{3s}+\Theta(r^{2s})}$ isomorphism classes (at least!) of subspaces. And that's even worse than exponentially w.r.t. $s$. It is so much worse that we are now in the vicinity of the number of elements of $A$ itself. Even in your extremely limited set of cases, this works out the be a massive number! If $(q,r,s)=(5,3,4)$, then $q^{\frac{1}{3}r^s}=5^{3^3}\approx 7.4\cdot 10^{18}$. There is no way you can enumerate all of that. Anyway... Assuming for the moment that somehow we find a way to solve Problem A. In our situation we can decide whether the two invariants $d(I)$ and $d(I')$ as well as the subspaces $I+J^{d(I)+1}$ and $I'+J^{d(I')+1}$ are the same. Ideally we solve problem A in such a way that if they are the same, we get an explicit $\alpha\in GL(V)$ which actually maps these two subspaces onto on another so that we can simplify the problem a bit as we move on to more complex invariants. More generally we can use a solution to Problem A to distinguish the graded structure which comes with any admissible ideal $I$: We have a filtration of $I$: $I \supseteq I \supseteq I \supseteq I\cap J^3 \supseteq I \cap J^4 \supseteq ...$ which gives us a graded ideal $gr(I) = \bigoplus_{k=2}^{s-1} I\cap J^k/I\cap J^{k+1}$ in $gr(A/J^s) = \bigoplus_{k=0}^{s-1} J^k/J^{k+1}$. Note that $Aut(A/J^s)$ respects the radical filtration and therefore induces automorphisms of the graded algebra. In fact the graded algebra is just the tensor algebra $T(V)$ (up to degree s) once again, but now $Aut(A/J^s)$ acts as $GL(V)$ on it so that we can use Problem A to use this invariant. This gives a whole bunch of invariants more.<|endoftext|> TITLE: Pushouts of injective monoid homomorphisms QUESTION [6 upvotes]: Given a pushout square in the category of monoids $$\begin{array}{ccc}A & \rightarrow & M \\ \downarrow && \downarrow \\ N & \rightarrow & P\end{array}$$such that $A \to M$ and $A \to N$ are injective, is it possible to deduce that $M \to P$ and $N \to P$ are injective, too? The answer is yes if the $A$-actions on $M$ and $N$ are free. In fact, in that case there is even a normal form for the elements of $P$ (Bourbaki, Algebra, Chapter I, §7.3). This happens in particular if $A$ is a group. What happens in the general case? For example, what about $A=(\mathbb{N},+,0)$? REPLY [5 votes]: No. Mark Sapir and Marcel Jackson even showed it is undecidable if the factors embed in an amalgamated free product of finite monoids. See the intro of Jackson, Marcel. "The embeddability of ring and semigroup amalgams is undecidable." Journal of the Australian Mathematical Society 69.2 (2000): 272-286. for an easy example of finite semigroups for which the factors don't embed and add identities. See also Sapir, Mark V. "Algorithmic problems for amalgams of finite semigroups." Journal of Algebra 229.2 (2000): 514-531..<|endoftext|> TITLE: Perfectoid universal covers QUESTION [39 upvotes]: It is often said, with varying degrees of rigor or enthusiasm, that every rigid space (say over $\mathbb{C}_p$) has a pro-etale cover which is 'topologically trivial' in some sense. For example, this is hinted at (although never said directly) in section 5.7 of Jared Weinstein's Reciprocity laws and Galois representations: recent breakthroughs (pdf). I was wondering if someone could answer the following questions which, while perhaps naive, are of great interest to me. If this is too many questions I can break it up into multiple posts. 1) Let $X$ be the `perfectoid closed unit disk' given by $\text{Spa}(\mathbb{C}_p\langle T^{\frac{1}{p^\infty}}\rangle,\mathcal{O}_{\mathbb{C}_p}\langle T^{\frac{1}{p^\infty}}\rangle)$. Is it true that $X-\{0\}$ is 'simply connected' in the sense that $\pi_1^{\acute{e}\text{t}}(X-\{0\})=0$? One might imagine that this is something like a pro-etale universal cover of the usual punctured closed unit disk over $\mathbb{C}_p$ since it's at least $\sim$ (in the language of Scholze) to an inverse limit of the finite etale covers $x\mapsto x^{p^n}$ of the punctured closed disk over $\mathbb{C}_p$. I suspect not since covers of the disk (like Artin-Schreier covers) exist. So, is there a 'universal cover' in this case? EDIT: As mentioned below, I was being hasty and should have said something like 'the maximal $p$-quotient of $\pi_1^{\acute{e}\text{t}}$ is zero. 2) If $X$ is any rigid space over $K$ (a $p$-adic field, or perhaps $\widehat{\overline{K}}$ for a $p$-adic field) then is there a 'universal cover' of $X$. The precise definition of this is open for me--is there a particularly good notion, and if so, when does it exist. For example, if there is a pro-etale cover $\{U_i\}$ of $X$ and an adic space $\widetilde{X}$ such that $\widetilde{X}\sim \varprojlim U_i$ and $\widetilde{X}$ is 'simply connected' (i.e. that $\pi_1^{\acute{e}{t}}(\widetilde{X})=0$ or, perhaps even better, $\pi_1^{\text{pro}\acute{e}\text{t}}(\widetilde{X})=0$). EDIT: As mentioned below, I was being hasty and should have said something like 'the maximal $p$-quotient of $\pi_1^{\acute{e}\text{t}}$ is zero and similarly for the proetale fundamental group. 3) I have the sense that perfectoid spaces have 'simpler geometry' (for example, I think this can be seen by the almost zeroness of their cohomology with values in $\mathcal{O}_X^+$ and how this relates to their cohomology with values in $\mathbb{F}_p$ by the AS sequence), but I don't know a precise statement of this. Namely, how is the etale topology of a perfectoid space simpler than, say, a general rigid variety? 4) It is sometimes said that one can use perfectoid geometry to try and compute things like the etale cohomology of some rigid variety by computing the Cech cohomology of some pro-etale perfectoid cover. What is the precise statement of this? I think that one can compute the cohomology of $\mathcal{O}_X^+$ almost (in the technical sense) from a pro-etale perfectoid cover. Does one then try to compute cohomology with coefficients in $\mathbb{F}_p$ almost-ly by using the AS sequence again? Any answers to any of these questions would be greatly appreciate--as well as any other insight someone might want to add. Thanks! EDIT: I guess I should add that one probably doesn't expect universal covers in terms of 'pure topological trivialness' (or, rather, pure topological trivialness in-so far as $\pi_1$ or other '1-dimensional topology' is concerned) but rather in the sort of topological trivialness (again in degree $1$) concerned with $p$-torsion or pro-$p$-torsion coefficients. So, as Will Sawin mentions below, you can still likely make covers with prime-to-$p$ degree of the 'universal cover'. In summary, perhaps '$p$-universal cover' is better... REPLY [16 votes]: Lol @"varying degrees of enthusiasm" ;-). And sorry for the late answer... Let me try to answer your questions. First, for any connected analytic adic space $X$, say, with a geometric point $\overline{x}\to X$, one can define $\pi_1^{\mathrm{et}}(X,\overline{x})$ just like in SGA1 for schemes, by looking at the Galois category of finite etale covers of $X$. In particular, passing to an inverse limit of all such finite etale covers equipped with a lift of $\overline{x}$, one can define a (profinite) "universal cover" $\tilde{X}\to X$. If $X$ lives over $\mathbb Z_p$ and is affinoid (probably Stein is enough) then $\tilde{X}$ is perfectoid; see for example Lemma 15.3 here (the funny phrasing there is solely due to the desire to also handle the case that $X$ is not connected). This largely answers question 2). Unfortunately, I don't know how to define a pro-etale fundamental group in the spirit of my paper with Bhatt. There we handle the case of schemes that locally have only a finite number of irreducible components. This is a very mild condition for schemes, but for analytic adic spaces, the condition is much too strong, see Example 7.3.12 of our paper. That example shows that the formalism actually does not work in the same way for analytic adic spaces, and I don't know how to correct it. So I will only use the usual $\pi_1^{\mathrm{et}}$. For question 1), the answer is actually No. Using Artin-Schreier covers, there are lots and lots of finite etale covers beyond the ones one might think about, so in particular the perfectoid closed unit disc has very large $\pi_1^{\mathrm{et}}$ (even (or especially) pro-$p$). What one might reasonably hope is that any finite etale cover of degree $p$ of the punctured perfectoid closed unit disc extends to a finite etale cover of the perfectoid closed unit disc. For this precise question, I'm actually confused: If the finite etale covers comes from some finite stage, it follows from some classical results in rigid geometry that it extends to a finite, possibly ramified cover, over the puncture, and then by Abhyankar's lemma this becomes trivial after passing to the perfectoid cover. However, I believe that at infinite level, one will get new, more nasty covers, that do not come from finite level. About question 3): One key fact is that affinoid perfectoid spaces have etale $p$-cohomological dimension $\leq 1$, i.e. for etale $p$-torsion sheaves, etale cohomology sits in degrees $\leq 1$. This in fact reduces by tilting to the case of characteristic $p$, where it follows from Artin-Schreier theory. Combining this with some interesting examples of perfectoid towers, one can get interesting vanishing results. In fact, these can usually be slightly improved upon by using $\mathcal O_X^+$-cohomology, the primitive comparison theorem, and the (almost) vanishing of $\mathcal O_X^+$-cohomology on affinoid perfectoids. This has been applied for example to Shimura varieties, abelian varieties [Well, the written version of that paper actually doesn't use this method, but our original approach did use it, see the discussion on page 1], and moduli spaces of curves. This ties in with question 4). What one usually does is the following. Say $\ldots\to X_2\to X_1\to X_0$ is some tower of proper rigid-analytic varieties over $\mathbb C_p$ with perfectoid limit $X_\infty$. For each $X_n$, the primitive comparison theorem says that $$H^i(X_n,\mathbb F_p)\otimes \mathcal O_{\mathbb C_p}/p\to H^i(X_n,\mathcal O_{X_n}^+/p)$$ is an almost isomorphism, where both sides are etale cohomology. (The proof of this uses some Artin-Schreier theory, and one could also formulate an Artin-Schreier sequence, but this tends to give weaker results.) Passing to the colimit over $n$ (so the limit on spaces), one sees that also $$H^i(X_\infty,\mathbb F_p)\otimes \mathcal O_{\mathbb C_p}/p\to H^i(X_\infty,\mathcal O_{X_\infty}^+/p)$$ is an isomorphism. Now on perfectoid $X_\infty$, the group on the right behaves like coherent cohomology, in particular it can (almost) be computed on the analytic side, and in fact by a Cech complex. This shows in particular that it (almost) vanishes in degrees larger than $\dim X_\infty$. In particular, $H^i(X_\infty,\mathbb F_p)$ vanishes in degrees larger than $\dim X_\infty$, which gives the vanishing theorems I mentioned.<|endoftext|> TITLE: Vector bundle over an oriented manifold with non-vanishing w_2w_3 QUESTION [7 upvotes]: I am looking for an example of an oriented rank 5 (or lower) real vector bundle $V$ over an oriented manifold such that the cup product $w_2(V) w_3(V)$ of Stiefel-Whitney classes does not vanish. It would be best if the manifold had dimension 7 or lower. REPLY [11 votes]: As far as I know the Wu manifold $X=SU(3)/SO(3)$ is orientable and has mod 2 cohomology ring $H^*(X;\mathbb{Z}_2)=\Lambda(\omega_2(X),\omega_3(X))$. Thus $\omega_2(X)\cdot\omega_3(X)\neq 0$, and in fact generates $H^5(X;\mathbb{Z}_2)$. REPLY [2 votes]: I think you can take the manifold to be $M = \mathbb{RP}^5$ and the vector bundle $V$ to be the direct sum of $3$ copies of the tautological line bundle $\gamma$ on $M$. Let $x = w_1(\gamma)\in H^1(M,\mathbb{Z}/2\mathbb{Z})$. Then by the Whitney product formula you get $w_2(V) = x^2$ and $w_3(V) = x^3$, so $w_2(V)w_3(V) = x^5$. But this is non-zero in the mod 2 cohomology of $\mathbb{RP}^5$, which is, as an algebra, isomorphic to $\mathbb{Z}/2\mathbb{Z}[x]/(x^6)$.<|endoftext|> TITLE: Does Chabauty-Coleman method give an algorithm for finding rational points? QUESTION [7 upvotes]: Let $X$ be a curve of genus $g\geq 2$ over a number field $K$. If $\mathrm{rk} \,\mathrm{Jac}\, X$ is less than $g$ there is a $p$-adic method of bounding $\# X(K)$ due to Chabauty and Coleman (see http://www-math.mit.edu/~poonen/papers/chabauty.pdf). Does this method give an algorithm for computing $X(K)$ (say, their coordinates)? I agree to assume BSD and finitness of Shafarevich-Tate group. Trying to transform their proof to an algorithm, I met the following bottlenecks. First, we need to compute $\log \overline{\mathrm{Jac}(K)}$ (bar means $p$-adic closure) in order to obtain a $1$-form whose integral will vanish on rational points, which is weaker than computing $\mathrm{Jac}(K)$ but I still do not know any non-conjectural algorithms. Next, the proof requires choosing base points in each residue class and I do not know how to find them algorithmically. Maybe, a solution can be obtained by extending the base field to where it is easy to find a base point. Finally, we need an argument for computing zeroes of $p$-adic analytic functions. There is an argument, à la Hensel's lemma, for computing zeroes sign-by-sign, which probably terminates if we search for points over number fields, but I am not sure. REPLY [10 votes]: Conjecturally, yes. Check out Section 4.4 of this paper by Nils Bruin and myself. The point is to combine Chabauty-Coleman with the "Mordell-Weil Sieve". In the following, I will assume for simplicity that the Jacobian of your curve $X$ is simple and that $K = \mathbb Q$. I will also assume that one rational point $P_0$ on $X$ is known, which we use to embed $X$ into $J = \operatorname{Jac} X$. In any case, assuming finiteness of Sha (say), one can in principle determine the rank $r$ of $J(\mathbb Q)$ and find $r$ independent points. Given this information and a prime $p$ of good reduction, one can determine the space of regular 1-forms $\omega$ over ${\mathbb Q}_p$ on $X$ that kill $J(\mathbb Q)$ under the Chabauty-Coleman pairing to any desired $p$-adic precision. In particular, one can compute the image of this space in the 1-forms mod $p$. If $p > 2$ and there is such a 1-form mod $p$ that does not vanish at any ${\mathbb F}_p$-point of $X$ (which heuristically happens for a set of primes $p$ of positive density), then one can show that each residue class mod $p$ can contain at most one rational point. So we now just have to decide, for each residue class, whether there is a rational point or not. If there is a point, then we will eventually find it by searching for it. If there is no point, then (again heuristically) we should be able to prove this using the Mordell-Weil Sieve as explained in the paper. Basically, one tries to use information obtained from reduction modulo a bunch of further primes $q$, together with the knowledge of a finite-index subgroup of $J(\mathbb Q)$, to get a contradiction. So when $X$ and $J$ satisfy finiteness of Sha (or BSD) there exist $p$ of good reduction and a "good" differential on $X$ mod $p$ that does not vanish in $X(\mathbb F_p)$ the "main" conjecture of this paper then the procedure outlined above will terminate and determine $X(\mathbb Q)$. Also, if the procedure terminates, then it will produce the correct result. So for practical purposes, we can just use it and hope that it finishes. I have implemented it for curves of genus 2 (with Jacobian of rank 1) over $\mathbb Q$ in Magma. It works very well in practice, even though there is no proof so far that it will always work.<|endoftext|> TITLE: Reference for $E_{\infty}$-ness of the Chern Character QUESTION [5 upvotes]: I would like a reference/proof for the fact that the Chern character map: $$KU_{\mathbb{Q}} \rightarrow H\mathbb{Q}[u, u^{-1}]$$ is an $E_{\infty}$-ring map. Thank you in advance! REPLY [3 votes]: The answer really depends on one's desired choice of definitions for KU, HQ, and the Chern character itself; some definitions allow one to produce a very short definition of the Chern character as an E_∞-ring map. For example, start with the Chern-Weil morphism (Vect^∇,⊕,⊗)→(Ω[u],+,∧), which gives a morphism of stacks of E_∞-monoids in symmetric monoidal groupoids. Using Segal's functor, we promote this to a stack of E_∞-rings. (Here u denotes the formal variable in degree 2 used in the Chern-Weil map.) Using the universal property of the homotopy group completion, extend this to a morphism (VirtVect^∇,⊕,⊗)→(Ω[u],+,∧), where VirtVect^∇ stands for virtual vector bundles with connection, which are defined as the homotopy group completion of Vect^∇. Now apply the concordification a.k.a. homotopification functor F↦hocolim_k F(Δ^k). (Here F(Δ^k) denotes the E_∞-ring given by evaluating the stack F on the smooth manifold Δ^k=R^k, the extended k-simplex.) The source becomes ku, the connective K-theory, and the target becomes HR[u], the periodic real Eilenberg-MacLane spectrum. We have KU=ku[β^{−1}], so to extend this morphism from ku to KU it suffices to observe that the image of the Bott class β is invertible in the target. The advantage of the above approach with smooth stacks is that one can instantly extract variants of the E_∞-Chern character for differential cohomology, equivariant cohomology, and twisted cohomology (and any combination thereof, such as twisted equivariant differential cohomology) by replacing the last step (concordification) with the appropriate construction (the Bunke-Nikolaus-Völkl construction for differential cohomology or evaluation on orbits for equivariant cohomology) and adjusting the first step (using modules over E_∞-rings) for twisted cohomology.<|endoftext|> TITLE: Is the area of the Mandelbrot provably computable? QUESTION [33 upvotes]: Recall the Mandelbrot set $M$ is the set of points $c$ in the complex plane such that the sequence $z_0 = 0, z_{n+1} = z_n^2 + c$ is bounded. It is well-known that $M$ is a compact set of positive area. Is the area $A$ of the Mandelbrot set provably a computable real number? That is, is there an algorithm (however slow, complex, or memory intensive) that would estimate $A$ to within any desired precision? Can we prove this algorithm will work? We do known a formula for $A$: $$A = \pi (1 - \sum_{n=0}^\infty n b_n^2)$$ Here $b_n$ are "the coefficients of the Laurent series about infinity of the conformal map $\psi$ of the exterior of the unit disk onto the exterior of the Mandelbrot set" (see here). The sequence $b_n$ is computable as it is given by a (messy) recursive formula (formula 7 in here). Is there a way to compute the rate of convergence of $\sum_{n=0}^\infty n b_n^2$? Remarks: By computable I mean in the sense of computability/recursion theory (also computable analysis). However, all I am asking for is an algorithm which, when given a natural number $k$ as input, outputs a rational approximation $A_k$ which is within $2^{-k}$ of $A$. The formula above is not enough. While we can compute partial sums $A_N = \sum_{n=0}^N n b_n^2$ to get an upper bound on $A$, we need a way to compute the rate of convergence of the series to actually compute $A$. One possible way to do this is to bound $n b_n^2 \leq c_n$ where the sequence $c_n$ is computable and $\sum_{n=0}^\infty c_n$ is finite and computable. It is easy to show that $A$ is computable from above (a.k.a. upper semi-computable, right-c.e., computable in the upper reals). One can use the formula above or one can just use the standard way of drawing the Mandelbrot set: keep removing pixels where every point in that pixel escapes the circle $|z| \leq 2$. (For logicians: $M$ is $\Pi^0_1$.) One could try fancier methods of drawing the Mandelbrot set, such as distance estimation where one draws a pixel iff it contains a point in $M$. One problem with this, is that we don't actually know that this method is computable? (It depends on the hyperbolicity conjecture. See here, here, and here.) A bigger problem is that this method still only gives us an upper estimation of $A$, since if we draw a pixel we don't know how much of the area of the pixel contains $M$. As for estimating $A$ from below, one could enumerate all the hyperbolic components of $M$---those are the iconic cardioids and disks found in the Mandelbrot set (including the mini Mandelbrot islands). Then one can add up all their areas. In this way one can get a lower bound. Hertling did show the hyperbolic components are enumerable in a computable way. (For logicians: the union of the hyperbolic components is $\Sigma^0_1$.) However, there are two problems. It is unknown if the hyperbolic components make up the total of the interior (although it is widely conjectured to be true), and it is not known if the boundary of $M$ has zero area. There are many estimates of the area of the Mandelbrot set in the literature. (See here for a(n outdated) summary.) They seem to group into four methods: Using the series formula above to give an upper estimate. (For newest estimate see here.) Adding up known parts of the interior to give a lower estimate. (Statistical) Pixel counting. It seems like this method could both over count and under count a pixel depending on how it is drawn. (While they give precise answers I don't know how mathematically rigorous the error values given by this method are.) (I think this newer boundary counting method is a type of pixel counting as well.) (Statistical) Monte Carlo. While this is a rigorous method for measuring area, one still has to know if the point is in the set. For example, if the boundary of $M$ has positive area, then this method would have problems since with positive probability one would get a point on the boundary and it would be impossible to determine if it was in the set or not. Therefore, I don't know how mathematically rigorous the error values and confidence interval of this Monte Carlo analysis are. As for statistical methods, surprisingly this does provide a possible rigorous route to showing that the area of the Mandelbrot set is computable. If one can give a (provably correct) algorithm which takes in a rational $\epsilon > 0$ a probability $p < 1$ and a source of randomness, and outputs an approximation $A_{\epsilon, p}$ for which $|A_{\epsilon, p} - A| < \epsilon$ with probability $p$, then $A$ is computable. (The main idea is that any number computable with positive probability is computable outright.) However, it seems to me that the only ways to make these statistical methods rigorous is to know that the boundary of $M$ has area zero and that one can determine (in a computably enumerable way) if a point is in the interior of $M$, both of which are open problems. (However, there maybe some important theorem I don't know about which is why I am asking.) REPLY [8 votes]: Your statement that the area of the union of bounded hyperbolic components is lower-computable is very plausible. So if, as conjectured, hyperbolic parameters have full measure, then the area is indeed computable. Without the assumption of this conjecture, I believe that the question is open. While it seems unlikely to be able to make any rigorous statement here, it seems intuitively clear that we cannot give any rigorous estimates without knowing the answer. Indeed, if there was a non-hyperbolic interior component, this would belong to a sequence of little Mandelbrot copies that do not shrink. Our algorithm somehow has to account for this area? For example, I believe it is not even known whether the Mandelbrot set itself is computable (i.e., whether the distance to the set is a computable function). Now, admittedly sometimes there can be miraculous results using some series formulae. However, the corresponding formulae for the area of filled-in Julia sets can be shown to not converge nicely/quickly in any way, see Dezotti, Convergence properties of the Gronwall area formula for quadratic Julia sets, arXiv:1405.1933. So I would be surprised if a lower bound can be deduced from the series expansion.<|endoftext|> TITLE: Objects of a category of homological dimension 1 is a smooth stack? QUESTION [7 upvotes]: I heard a reference to a statement like: Suppose $A$ is an (Abelian?) category of homological dimension one, then the stack of objects of $A$ is smooth. (I am not really sure what the stack of objects refers to exactly ... I guess it means that there is some stack which naturally equivalent to the underlying groupoid of $A$ (forgetting all non-isomorphisms), or something.) Here homological dimension refers to the vanishing of Ext groups. So in this case, $Ext^i(M,N) = 0$ if $i \geq 2$. Examples that were given were: quiver representations and coherent sheaves on a curve. Could someone link me to a reference to this? Edit: I heard it in this talk: Victor Ginzburg, Geometry of Quiver Varieties I at about 9:50. REPLY [2 votes]: It follows from general deformation theory. Infinitesimal automorphisms of an object $M$ are given by $Hom(M,M)$, first order infinitesimal deformations of $M$ are given by $Ext^1(M,M)$ and obstructions to lift some infinitesimal deformation to the next order live in $Ext^2(M,M)$. So if $Ext^2(M,M)=0$, these obstructions vanish and so the moduli space of objects is (formally) smooth at $M$, formal locally modeled on the quotient stack $Ext^1(M,M)/Hom(M,M)$.<|endoftext|> TITLE: Can we algorithmically contract loops in a simply connected space? QUESTION [8 upvotes]: It is well-known that the question whether a given connected simplicial complex (or simplicial set) is simply connected, is algorithmically undecidable as it can model the word problem. Assuming that $X$ is simply connected, is there an algorithmic way how to contract loops? One way how this can be formalized is as follows. $X$ can be a simply connected simplicial set (with either finitely many nondegenerate simplices in each dimension, or at least with effective homology...) and $\tilde{GX}$ the Kan model of the loop space, such as described here. The loop contraction would than be a map $c: \tilde{GX}_0\to \tilde{GX}_1$ such that $d_0 c(x)=x$ and $d_1 c(x)=1$. Is there any hope to have a general algorithm for evaluating $c(x)$? REPLY [8 votes]: I am assuming that you have a complex with finite 2-dimensional skeleton. There is a silly algorithm for contracting loops which is even linear in the combinatorial length of the loop. Start with defining the "standard presentation" for $\pi_1(X)$, namely, construct a maximal subtree $T\subset X^1$. Generators of $\pi_1(X)$ are represented by edges in $X^1 -T$ connecting vertices of $X$. Let $g_i, i=1,..,n$ be the corresponding loops in $X^1$ (based at a vertex $o\in T$). Since $X$ is simply-connected, there exist combinatorial maps $e_i: D^2\to X$ with boundary maps $\partial e_i=g_i$: You find these by listing all combinatorial maps $D^2\to X$ and examining them one-by-one. Now, each based combinatorial loop $g$ in $(X^1,o)$ is a product of the generators $g_i$ and their inverses: $$ g= g^{\pm 1}_{i_1}....g^{\pm 1}_{i_k} $$ This is again completely algorithmic since it amounts to reading off the product of the oriented edges in $X^1-T$ as they appear in $g$. Lastly, you read off the maps $e_i$ (and their pre-composition with suitable reflections in $D^2$) according to the word $$ w= g^{\pm 1}_{i_1}....g^{\pm 1}_{i_k}. $$ This gives you a combinatorial map $e: D^2\to X$ with $\partial e$ given by $w$. I will leave you the part dealing with constructing a homotopy between the loops $g$ and $w$ inside of $X^1$ (this essentially amounts to contracting $T$ to $o$ and then cancelling the backtracking in $g$).<|endoftext|> TITLE: Identification of a Serre Spectral Seq. via Thom Isomorphism with the Atiyah-Hirzebruch Spectral Seq QUESTION [10 upvotes]: Let $\xi_n$ be an orientable $n$-dimensional vector bundle over a pointed space $B_n$. We can consider the relative Serre Spectral Sequence $$ H_p(B_n; h_q(D(\xi_n|\ast),S(\xi_n|\ast))\Rightarrow h_{p+q}(D(\xi_n),S(\xi_n)) $$ which can be rewritten as $$ H_p(B_n; \tilde{h}_q(S^n))\Rightarrow \tilde{h}_{p+q}(M\xi_n) $$ where $M\xi_n$ is the Thom Space of the vector bundle $\xi_n$. Via the suspension isomorphism we have equivalently $$ H_p(B_n; \tilde{h}_{q-n}(S^0))\Rightarrow \tilde{h}_{p+q}(M\xi_n) $$ using the Thom-Isomorphism $H_p(M\xi_n)\cong H_{p-n}(B_n)$ we get $$ H_{p+n}(M\xi_n; \tilde{h}_{q-n}(S^0))\Rightarrow \tilde{h}_{p+q}(M\xi_n)$$ which seems the Atiyah-Hirzebruch Spectral sequence for $M\xi_n$ in the sense that I'm unable to prove that under these transformations the differentials coincide with the one that comes from the AHSS, in other words that what I got is really the AHSS. How can I show that this is indeed the case? Since these transformations are not given by a geometrical map, I'm unable to boil everything down to a simply naturality argument. I'm interested in this fact because I'm reading this paper about the construction of the James Spectral Sequence where this fact is crucial for the identification of the second differential of this spectral sequence (Prop 1) and seems to be trivial. REPLY [6 votes]: NOTE For simplicity of notation I'm going to work only with ordinary cohomology, although it doesn't really matter (any cohomology theory will do). The best way to see if two spectral sequences are the same is to compare their exact couples. If they originate from the same exact couple, they are the same spectral sequences, differential and all. Fix a CW structure on the base $B$. Then the exact couple originating the Serre spectral sequence is $$ \require{AMScd} \begin{CD}\cdots @>>> H^*(p^{-1}B^{m})@>>> H^*(p^{-1}B^{m-1})@>>> \cdots\\ {} @VVV {} @VVV {}\\ {} @.H^*(p^{-1}B^{m+1},p^{-1}B^m) @. H^*(p^{-1}B^m,p^{-1}B^{m-1)}) \end{CD}$$ (sorry I can't do diagonal arrows, but they should go one step up and to the left). To write down the exact couple for the AHSS you need to choose a CW structure on $M\xi$. If we could choose one such that the $m$-skeleton is $M(\xi|_{B^{m-n}})$, then the corresponding exact couple would be $$ \require{AMScd} \begin{CD}\cdots @>>> \tilde H^*(M(\xi_{B^{m-n}}))@>>> \tilde H^*(M(\xi_{B^{m-n-1}}))@>>> \cdots\\ {} @VVV {} @VVV {}\\ {} @.H^*(M(\xi_{B^{m-n+1}}),M(\xi_{B^{m-n}})) @. H^*(M(\xi_{B^{m-n}}),M(\xi_{B^{m-n-1}})) \end{CD}$$ which by the Thom isomorphism is isomorphic to the previous one, thus proving our thesis. So the problem boils down to construct a suitable CW structure on $M\xi$, or more precisely to prove that $M\xi_{B^{m+1}}$ can be constructed by $M\xi_{B^m}$ by attaching $(m+n+1)$-cells. This is very easy, since you can work one cell at a time on $B^{m+1}$, and it is left as an exercise for the reader. For more detail on this see also this question.<|endoftext|> TITLE: Is there a singularity theorem in higher-dimensional Newtonian gravity? QUESTION [14 upvotes]: In classical Newtonian gravity with 3 spatial dimensions, it's hard to get two particles to exactly collide, since at short distance the centrifugal force (~1/$r^3$) beats the gravitational attraction (~$1/r^2$). As a consequence, two particles can collide only if the angular momentum is exactly zero, which is measure zero. In Newtonian gravity with $d$ spatial dimensions, the centrifugal force still goes like ~$1/r^3$ but the gravitational attraction goes like ~$1/r^{d-1}$. This means that for $d>4$, if two particles have negative total energy, they will collide regardless of their angular momentum. For two particles, having negative total energy is a sufficient condition to guarantee a collision after finite time. My question is whether this is still true with more than two particles? Conjecture: in $d>4$-dimensional Newtonian gravity, a collection of any number of point particles with negative total energy will experience a "singularity" (a distance=0 collision of at least two particles) within a finite time. REPLY [3 votes]: Following the lines Willie followed, but allowing for unequal masses $m_i$, set $I(x) = \langle x, x \rangle $ where $\langle v, w \rangle = \Sigma m_i v_i \cdot w_i$ is the so-called mass metric built so that the kinetic energy is $K =\frac{1}{2}\langle v, v \rangle$ with $v = \dot x$. Here $x$ and $v$ lie in N copies of a d-dimensional Euclidean space. Then Newton's equations read $\ddot x = - \nabla V (x)$ where the gradient $\nabla$ is with respect to the mass metric. Suppose now that the potential $V$ is homogeneous of degree $-\alpha$. Using Euler's identity $-\alpha V (x) = \langle x, \nabla V (x) \rangle$ a fun elementary computation yields the virial identity $\ddot I = 4H +(2 \alpha - 4) V$, where $H = K +V$ is the total energy, which is conserved. (This identity is called by mathematical celestial mechanicians the Lagrange-Jacobi identity.) Now for $d$-dimensional ``Newtonian' gravity we have $V < 0$ and $\alpha = d-2$ so that Lagrange-Jacobi reads $\ddot I = 4H +(8-4d)U$, with $U = -V > 0$. Conclusion: for $d > 4$ and $H \le 0$ we have $\ddot I < 0$ and thus all such solutions either begin and end in total collision $I =0$, (or have singularities before these total collision times), as you conjectured. (Asides. If $d< 4$ then the virial identity implies that $\ddot I > 0$ for $H \ge 0$ and consequently bounded solutions have $H < 0$. The case $d =4$ is a wonderful bounding case between these extremes. At the critical dimension $4$ solutions must lie on spheres $I = const.$ and have energy $H =0$ in order to be bounded.)<|endoftext|> TITLE: Surprisingly simple minimum of a rational function on $\mathbb R_+^n$ QUESTION [6 upvotes]: Motivation: The following problem has occurred in a study of energy dissipation in a chain of coupled, damped oscillators. The problem: Let me define specific rational functions $f$, $g$, and $h$ from $\mathbb R_+^n$ into $\mathbb R_+$ ($n \geq 2$) by the following expressions: \begin{align*} f(x) & = 1 + \frac{x_2}{x_1} + \frac{x_2 x_4}{x_1 x_3} + \cdots = 1 + \frac{x_2}{x_1} \left( 1 + \frac{x_4}{x_3} \left( 1 + \frac{x_6}{x_5} \cdots \right) \right) \ , \tag*{(1a)} \\ g(x) & = 1 + \frac{x_3}{x_2} + \frac{x_3 x_5}{x_2 x_4} + \cdots = 1 + \frac{x_3}{x_2} \left( 1 + \frac{x_5}{x_4} \left( 1 + \frac{x_7}{x_6} \cdots \right) \right) \ , \tag*{(1b)} \\ \\ h(x) & = x_1 f(x)\, g(x) \\ & = x_1 + \cdots + x_n + \frac{x_1 x_3}{x_2} + \frac{x_2 x_4}{x_3} + \cdots + \frac{x_{n-2} x_n}{x_{n-1}} \\ & \qquad + \frac{x_1 x_3 x_5}{x_2 x_4} + \frac{x_2 x_4 x_6}{x_3 x_5} + \cdots + \frac{x_{n-4} x_{n-2} x_n}{x_{n-3} x_{n-1}} \\ & \qquad + \frac{x_1 x_3 x_5 x_7}{x_2 x_4 x_6} + \cdots \quad \text{ etc. } \ . \tag*{(1c)} \end{align*} The dots in (1a-c) indicate a continuation of the respective patterns, with as many terms as possible as long as the indices $i$ on all involved $x_i$ remain within $\{1,...,n\}$. I am interested in a meaningful "minimization" of the function $h$. From its definition above it is obvious that $h$ is both strictly positive and a homogeneous function of degree 1. For any $t\in\mathbb R_+$ then $h(t\,x) = t\,h(x)$, which shows that $h$ has arbitrarily small values close to the origin of $\mathbb R^n$, and so it cannot have a global minimum in $\mathbb R_+^n$. An obvious way around this problem is to impose an additional constraint on $x$. Because of the homogeneity of $h$, the natural choice is a constraint of the form \begin{align*} \prod_{i=1}^n x_i = c \ , \tag*{(2)} \end{align*} with some constant $c>0$. Without loss of generality, we can choose any fixed value, for instance $c=1$; function values for any $c>0$ are (by homogeneity of $h$) related to those for $c=1$ by $h(t\,x) = t\,h(x)$, with $t=c^{1/n}$. We know that, on the hypersurface $\mathbb H_c^n = \{ x \in \mathbb R_+^n \! : \prod_{i=1}^n x_i = c \}$ defined by the constraint (2), $h$ must have a minimum since it is continuous (even $C^\infty$) in $\mathbb R_+^n$ and (Proposition 1:) if $x_i \to \infty$ for some $i$, or if $x \in \mathbb H_c^n$ approaches the boundary of $\mathbb R_+^n$, then $h(x) \to \infty$. Proposition 1 is easily proved as follows. Noting that $h(x) > x_1 + \cdots + x_n$, the first part is obvious. For the second part, let us assume that $x_i < \epsilon$ for some $i$. Then due to the constraint (2) there is at least one $j$ such that $x_j > (c/\epsilon)^{\frac1{n-1}}$, whence $h(x) > (c/\epsilon)^{\frac1{n-1}}$. Consequently, if $\epsilon \to 0$ then $h(x) \to \infty$. The question now is, at which point(s) $x_{\rm min} \in \mathbb H_c^n$ is this minimum located, and what is the value $h_{\rm min} = h(x_{\rm min})$? Any minimum of $h$ on $\mathbb H_c^n$ must also be a stationary point of $h$ on $\mathbb H_c^n$ (i.e., the gradient of $h$ in $H_c^n$ must vanish at that point). I believe that for any given $n$ there is only one stationary point of $h$ on $\mathbb H_c^n$, which then also must be its (unique) minimum. Unfortunately, I cannot prove this yet, but I can prove the existence of a remarkably simple stationary point: Proposition 2: Let us choose $c = \frac14$ in the constraint (2). The point $x$ with components \begin{align*} x_1 = x_n = \frac12 , \quad x_2 = \cdots = x_{n-1} = 1 \tag*{(3)} \end{align*} is a stationary point of $h$ on $\mathbb H_{1/4}^n$, with corresponding stationary value $h(x) = \frac{n^2}4$. This extremely simple result suggests that there may be a simple reason why it is true, but so far I have only a (relatively) complicated proof which I won't write down here. Question: Why is the form (3) of the stationary point this simple? Is (3) the only stationary point (and thus the global minimum) of $h$ under the constraint (2) with $c = \frac14$? REPLY [8 votes]: I'm not sure about stationary points, but the global minimum is certainly there. Let's do it for $n=5$. Write $$ f(x)=1+\frac{x_2}{2x_1}+\frac{x_2}{2x_1}+\frac{x_2x_4}{2x_1x_3}+\frac{x_2x_4}{2x_1x_3}\,, \\ g(x)=\frac 12+\frac 12+\frac{x_3}{2x_2}+\frac{x_3}{2x_2}+\frac{x_3x_5}{x_2x_4} $$ Now use Cauchy-Schwarz ($(\sum a)(\sum b)\ge (\sum\sqrt{ab})^2$) honestly coupling the terms as presented and then apply AM-GM.<|endoftext|> TITLE: Models for equivariant genuine commutative ring spectra QUESTION [8 upvotes]: The following question is an attempt at understanding various flavours of equivariant commutative ring spectra; it may not be suitable level for this forum. Let $\mathcal{C}(G)$ be a symmetric monoidal homotopical category such that $Ho(\mathcal{C}(G))$ is the category $SH(G)$ of genuine $G$-equivariant spectra. Here $G$ is a finite group. The examples I have in mind are the category $\mathcal{C}_O(G)$ of $G$-objects in orthogonal spectra, with the G-equvariant stable weak equivalences (Schwede) or the category $\mathcal{C}_\Sigma(G)$ of T-symmetric spectra in the category of $G$-simplicial sets, with $T$ an appropriate simplicial version of the regular representation sphere (Mandell). Going back to general $\mathcal{C}(G)$, we may then form the homotopical category $CMon(\mathcal{C}(G))$ of commutative, unital monoid objects in $\mathcal{C}(G)$ (with weak equivalences the underlying weak equivalences). Furthermore, given an operad $O$ with values in $\mathcal{C}(G)$ we can talk about the homotopical category $O-Alg(\mathcal{C}(G))$. Note that $CMon(\mathcal{C}(G))$ is just $O-Alg(\mathcal{C}(G))$ for $O$ the commutative operad (all spaces just the tensor unit). Any choice $\mathcal{C}(G)$ should admit a symmetric monoidal functor from the category $sSet(G)$ of $G$-simplicial sets. Then, given any operad in $G$-simplicial sets, we can talk about the induced operad in $\mathcal{C}(G)$. In particular we have "the" classical E-infinity operad $E \in Operads(sSet) \subset Operads(sSet(G))$ (consisting of spaces with trivial action). As far as I understand there is also the "genuine G-equivariant E-infinity operad" $E_G \in Operads(sSet(G))$, and in fact many operads between those two (see Blumberg-Hill). So we now have six homotopical categories: $E-Alg(\mathcal{C}_O(G)), E_G-Alg(\mathcal{C}_O(G)), CMon(\mathcal{C}_O(G))$ and $E-Alg(\mathcal{C}_\Sigma(G)), E_G-Alg(\mathcal{C}_\Sigma(G)), CMon(\mathcal{C}_\Sigma(G))$. My question is, which of these are known (or expected) to have equivalent homotopy categories? Remark 1 I'm phrasing this question in terms of homotopical categories because there are many different model structures, and there are many subtle issues regarding these, but as far as I can tell my question is not about this. Remark 2 As far as I understand, if $G$ is the trivial group, all of these categories model commutative ring spectra and have equivalent homotopy categories. Remark 3 If I understand correctly the work of Blumberg-Hill (and many others), then $Ho(E-Alg(\mathcal{C}_O(G))) \ne Ho(E_G-Alg(\mathcal{C}_O(G)))$ because the objects on the right have "norm maps" but on the left not. A related thing to say is that the homotopy groups on the right are tambara functors and on the left they have less structure. I think it is also mentioned in loc. cit. that $Ho(E_G-Alg(\mathcal{C}_O(G))) = Ho(CMon(\mathcal{C}_O(G)))$. Remark 4 There are various articles proving that $Ho(E-Alg(Spt(\mathcal{M},T))) = Ho(CMon(Spt(\mathcal{M},T)))$ for rather general model categories $\mathcal{M}$, see e.g. Pavlov-Scholbach. This seems to suggest to me that $Ho(E-Alg(\mathcal{C}_\Sigma(G))) = Ho(E_G-Alg(\mathcal{C}_\Sigma(G))) = Ho(CMon(\mathcal{C}_\Sigma(G)))$ (but checking the detailed list of requirements for their theorem is non-trivial). Note that this would be in stark contrast to the case of orthogonal spectra! So my best guess is that loc. cit. does not apply in our situation? Some references Schwede on equivariant orthogonal spectra: http://www.math.uni-bonn.de/people/schwede/equivariant.pdf Mandell on equivariant symmetric spectra: http://pages.iu.edu/~mmandell/papers/gssfinal.dvi Blumberg-Hill on equivariant E-infinity operads: https://arxiv.org/pdf/1309.1750v3.pdf Pavlov-Scholbach: http://wwwmath.uni-muenster.de/sfb878/publications/files/phpimKZBl5582.pdf REPLY [3 votes]: For the questions asked here, there is no difference between orthogonal $G$-spectra, symmetric $G$-spectra of either $G$-spaces or $G$-sSets, or EKMM $G$-spectra. For the first two, the nonequivariant arguments in Mandell-May-Schwede-Shipley http://www.math.uchicago.edu/~may/PAPERS/mmssLMSDec30.pdf generalize directly, once one corrects Lemma III.8.4 in Mandell-May http://www.math.uchicago.edu/~may/PAPERS/MMMFinal.pdf, as is done in HHR. That lemma should say that for positive cofibrant $G$-spectra $X$ (in any good category of $G$-spectra), the natural map $E(G,\Sigma_n)\wedge_G X^n \to X^n/\Sigma_n$ is a weak equivalence, where $E(G,\Sigma_n)$ is the universal principal $(G,\Sigma_n)$-bundle. This implies that the homotopy categories of $E_G$-algebras and of commutative monoids are equivalent for any genuine $E_{\infty}$ $G$-operad $E_G$. As David White says, his thesis improves these equivalences to Quillen equivalences of model categories. The "canonical" choice of $E_G$ is in the eyes of the beholder: there are many different natural choices with different applications of each: infinite Steiner or little discs $G$-operads, linear isometries $G$-operads, permutativity (or Barratt-Eccles) $G$-operads, etc. There are analogous equivalences using naive $E_{\infty}$-$G$-operads, which are nonequivariant $E_{\infty}$ operads regarded as $G$-trivial. That is the weakest in the hierarchy of point-set level kinds of genuine $G$-spectra with commutative ring structures given by Blumberg-Hill,whereas commutative monoids is the strongest.<|endoftext|> TITLE: Asymptotic estimate of double summation QUESTION [5 upvotes]: I would like to determine an asymptotic expansion for the following double summation: $$\sum_{a=1}^{N/\sqrt {j}} \sum_{b=a}^{ja} \frac{1}{ab}$$ where $j$ is a real number $\geq 1$ and $N$ tends to $\infty$. In practice, the summation includes all pairs of integers $a,b$ (with $a \leq b \leq ja $) such that the product $ab$ is $\leq N^2 $. For $j=1$, the summation yields the infinite harmonic sum of squares, i.e. $\pi^2/6$. For $j>1$, the asymptotic expansion has the form $\log(N) \log(j) + O(1)$, where the $O(1)$ term converges to a value $k$ that depends on $j$. Interestingly, plotting $k$ vs $j$, the resulting function is discontinuous, with the most evident discontinuities occurring for $j$ integer. Is there any way to express this value explicitly? I am not necessarily searching a closed form, which may probably not exist. Rather, I would be interested to know potential alternative ways (e.g., using functions, series, and so on) to express the value of $k$, different from the trivial definition given by the difference between the double summation and the log term. This question is a general formulation of a problem that is related to this and this other questions previously posted on MSE, for which no conclusive answer was provided despite multiple bounties. REPLY [2 votes]: Maple 2016 finds it for concrete values of $j$. For example, asympt(sum(sum(1/(a*b), b = a .. 3*a, parametric), a = 1 .. N/sqrt(3)), N, 2) $$\ln \left( 3 \right) \ln \left( N \right) +\ln \left( 3 \right) \ln \left( 1/3\,\sqrt {3} \right) +1/18\,{\pi}^{2}+$$ $$\sum _{a=1}^{ \infty }\,{\frac {\Psi \left( a+2/3 \right) +\Psi \left( a+1/3 \right) -2\,\Psi \left( a \right) }{3a}}+\ln \left( 3 \right) \gamma + {\frac {\,\ln \left( 3 \right) \sqrt {3}-1/3\,\sqrt {3}}{2N}}+O \left( {N}^{-2} \right),$$ where $\Psi(x)$ is the digamma function (http://www.maplesoft.com/support/help/Maple/view.aspx?path=Psi&term=Psi) and $\gamma=0.57721...$ is the Euler-Mascheroni constant. PS. Also the Mathematica command Series[Sum[Sum[1/(a*b), {b, a, 3 a}], {a, 1, N/Sqrt[3]}, Assumptions -> N > 0], {N, Infinity, 1}] does that in a somewhat more complicated form.<|endoftext|> TITLE: How to prove the monotonicity of the following function? $f(x) = \frac{\sum\limits_{i=1}^K\ln(1+a_ix)}{\sum\limits_{i=1}^K\ln(1+\frac{1}{2}a_ix)}$ QUESTION [6 upvotes]: How to prove the monotonicity of the following function? $f(x) = \frac{\sum\limits_{i=1}^K\ln(1+a_ix)}{\sum\limits_{i=1}^K\ln(1+\frac{1}{2}a_ix)}$. where $a_i>0$, $\forall i$, $x>0$. I have proved that $\frac{h_i(x)}{g_i(x)}=\frac{\ln(1+a_ix)}{\ln(1+\frac{1}{2}a_ix)}$ is a monotonically decreasing function with $x$. Actually, $f(x)$ is also monotonically decreasing with $x$, which can be proved by simple simulation. But I can't give the rigorous proof. Any help is appreciated. Thanks a lot! REPLY [2 votes]: OK, I put the remark yesterday at night just to prevent people from wasting time on attempts to prove the statement. Now it is time to post the explanation. Note that if $\frac{F(x)}{G(x)}$ is non-increasing and $F,G>0$, then $\frac{F'(x)}{F(x)}\le \frac{G'(x)}{G(x)}$ and, in particular, if $G'(x)-cG(x)\le 0$ for some positive $c$ and $x$, then $F'(x)-cF(x)\le 0$. Now let us look at what functions we have. Repeating the summands, if necessary, and approximating reals by rationals, we can put $F(x)=\int\log (1+2ax)\,d\mu(a)$ and $G(x)=\int\log(1+ax)\,d\mu(a)$ where $\mu$ is any finite sum of Dirac point masses with positive coefficients. So, if our inequality holds, we must have that $\int \left[\frac{a}{1+ax}-c\log(1+ax)\right]\,d\mu(a)\le 0$ implies that $\int \left[\frac{2a}{1+2ax}-c\log(1+2ax)\right]\,d\mu(a)\le 0$ for every fixed $x,c>0$ and every $\mu$ of the above kind. We'll take $x=1$, $c\in(0,1)$. Then, by the general abstract duality nonsense, for any finite set $A$ of points $a>0$, we must be able to produce a non-negative $\lambda$ such that $$ \frac{2a}{1+2a}-c\log(1+2a)\le\lambda\left[\frac{a}{1+a}-c\log(1+a)\right] $$ for all $a\in A$. Now just take $A$ consisting of $2$ points: $a_1\approx 0$ and $a_2\approx+\infty$. Looking at $a_1$, we get essentially $2a_1(1-c)\le \lambda a_1(1-c)$, so $\lambda$ cannot be noticeably less than $2$. Looking at $a_2$, we get essentially $-c\log a_2\le \lambda(-c\log a_2)$, so $\lambda$ cannot be noticeably greater, than $1$. But this leaves us no room!<|endoftext|> TITLE: Does the least prime factor have a mean of some sort? QUESTION [16 upvotes]: Define $l(n)$ to be the least prime factor of $n$ and, say, $l(1)=0$ for simplicity. Obviously we have $2\leq l(n)\leq n$ for $n\geq 2$. There appears to be very little information about the asymptotic behaviour of $l(n)$ available. One may observe that $$\sum_1^{\infty}\frac{l(n)}{n^s}=\zeta(s)\sum_p\frac{1}{p^{s-1}}\prod_{q2$, where $p,q$ are prime and the empty product is unity. It seems a fair bet that the Dirichlet series on the left has a meromorphic continuation to the region $\sigma>1$ but I haven't proved this. It certainly is singular at $s=2$, thus so is the sum on the right. The type of singularity is not at all obvious, so without further investigation little more information is available from this naive approach. Is $l(n)$ known to have some kind of mean value? If so, how is it calculated? REPLY [21 votes]: Note that $l(n) \le \sqrt{n}$ unless $n$ is prime. This makes it easy to show that $\sum_{n \le x} l(n) \sim \sum_{p \le x} l(p) = \sum_{p \le x} p \sim \frac{1}{2} \frac{x^2}{\log x}$, as $x\to\infty$. (The last asymptotic formula comes from the prime number theorem and partial summation.) This was noted by Kalecki: On certain sums extended over primes or prime factors. (Polish. Russian, English summary) Prace Mat. 8 1963/1964 121–129.<|endoftext|> TITLE: Trace of a nonlinear matrix equation QUESTION [5 upvotes]: Let $X_0$ be a trace-one positive definite matrix, i.e. $X_0>0$, $\mathrm{tr}(X_0)=1$. Let $A>0$ and consider the following iteration $$ X_{k+1} = X_k^{1/2}AX_k^{1/2},\quad k\geq 0,\quad (\star) $$ where $X_k^{1/2}$ denotes the (principal) square root of $X_k$. My question: Is it true that the above iteration is trace-preserving starting from all $X_0$'s as above, i.e. $\mathrm{tr}(X_{k+1})=\mathrm{tr}(X_{k})=1$ for all $k\geq 0$, (if and) only if $A=I$? N.B. After a discussion with Nawaf Bou-Rabee, I decided to restore the original formulation of my question and accept his answer. However, I opened a new OP where you can find the edited version, which is still open. REPLY [5 votes]: If $A=I$, then it follows from the iteration rule that $\mathrm{tr}(X_{k})=1$ for all natural number $k \ge 0$. If $\mathrm{tr}(X_{1})=\mathrm{tr}(X_{0})$, then the given iteration rule implies that: $$ \mathrm{tr}((A - I) X) = 0 \quad \forall X : X > 0 ~\&~ \mathrm{tr}(X) = 1 \;. \tag{$\star$} $$ If $X$ was allowed to be an arbitrary matrix, it would immediately follow that $A=I$. Still, as we will see, there is enough slack in the constraints on $X$ to still obtain $A=I$. To see this, let $\{ e_i \}$ be the standard basis on $\mathbb{R}^n$ and for any $i \in \{1, \cdots, n \}$ define $$ X_{ii} = (1-\epsilon (n-1)) e_i e_i^T + \sum_{j \ne i} \epsilon e_j e_j^T $$ where $\epsilon>0$ is a small parameter. Note that the sum of the eigenvalues of the matrix $X_{ii}$ is one, and if $\epsilon$ is sufficiently small, then $X_{ii}>0$. More importantly, letting $X=X_{ii}$ in ($\star$) gives $$ \mathrm{tr}((A - I) X_{ii}) = (1-\epsilon (n-1) ) (A_{ii}-1) + \epsilon \sum_{j \ne i} (A_{jj}-1) =0 $$ which implies that: $$ A_{ii} = 1 - \frac{\epsilon}{1-\epsilon (n-1)} \sum_{j \ne i} (A_{jj}-1) $$ Since $\epsilon$ can be made arbitrarily small, it follows that the diagonal entries of $A$ are all equal to one. One can similarly treat the off-diagonal terms by, e.g., defining \begin{align*} X_{ij} &= \frac{1}{2} (1-\epsilon (n-1) ) \left( e_i e_j^T + e_j e_i^T \right) \\ &+ (1-\epsilon (n-1) ) e_i e_i^T + (1-\epsilon (n-1) ) e_j e_j^T + \sum_{k \ne \{i,j\}} \epsilon e_k e_k^T \end{align*} which has unit trace and is a symmetric strictly diagonally dominant matrix, and hence, positive definite. Then setting $X=X_{ij}$ in ($\star$) isolates the $A_{ij}$ term for $i \ne j$.<|endoftext|> TITLE: Is there any structure theorem for piecewise linear functions? QUESTION [6 upvotes]: I was wondering if such statements are known like "any piecewise linear function from $\mathbb{R}^d \rightarrow \mathbb{R}$ can be written as $\sum_{i=1}^k \alpha_i (\text{ some $2$ piece linear function})$" (for some positive integer $k$ that depends on the given function and some real numbers $\alpha_i$ and maybe the $2$ piece function is different for each $i$). Is there something in the theory of ``tropical varieties/geometry" which relates to such a thing? REPLY [6 votes]: If you weren't assuming the piecewise linear function $f: \mathbb R^d \to \mathbb R$ is continuous, then I think this is true: we can just choose functions $$ f_i = \begin{cases} \text{(some linear piece of $f$)} & \text{if in domain of linear piece} \\ 0 & \text{otherwise}, \end{cases}$$ and choose all coefficients $\alpha_i = 1$, so $f = \sum_i f_i$. If you were assuming that the function $f$ is continuous, and you want to write it as a sum of continuous two-piece linear functions, then this is false when $d\geq 2$. Note that if $f_i$ is a two-piece linear function which is continuous, then the set of points where $f_i$ is discontinuous (i.e. the ''break locus'' of $f_i$) must be a hyperplane in $\mathbb R^d$. Thus if we take a finite sum $\displaystyle\sum_{1\leq i\leq k} \alpha_i f_i$ over such functions, the points of discontinuity will form a hyperplane arrangement in $\mathbb R^d$ with at most $k$ hyperplanes. (If ''cancellation'' occurs in the locus of discontinuity, it will happen along the whole hyperplane.) However, when $d\geq 2$ there are many possible piecewise linear continuous functions $f:\mathbb R^d \to \mathbb R$ whose break locus is not a hyperplane arrangement. The simplest example is probably $$f(x,y) = \max\{0,x,y\} : \mathbb R^2 \to \mathbb R,$$ whose break locus is the tropical line: [image from Frank Sottile's website ] If we only consider the case $d = 1$, then it is true that any piecewise linear continuous function may be expressed as a sum of two-piece such functions. In light of the above counterexample it may be tempting to conjecture that the correct generalization is that a function $\mathbb R^d \to \mathbb R$ may be expressed as a sum of $(d+1)$-piece functions, but this is not true either, e.g. for the plane conic in $\mathbb R^2$ cut out by $$f(x,y) = \max\{ 0, 1+x, 1+y, 1+x+y, 2x, 2y\}.$$ To explain this behavior there is a perfect analogy with polynomials over complex numbers (or any algebraically closed field), via tropical geometry / algebra: The fundamental theorem of algebra states that any polynomial $f \in \mathbb C[x]$ may be factored as a product of linear polynomials, but it is not true that any polynomial $f \in \mathbb C[x_1,\ldots,x_d]$ has a factorization into linear polynomials when $d\geq 2$.<|endoftext|> TITLE: Frobenius norm of the principal submatrix of a uniformly distributed random orthonormal matrix QUESTION [7 upvotes]: Suppose that we have a uniformly distributed $d\times d$ random orthonormal matrix $\mathbf{X}$. Here "uniform" is defined in the sense of Haar measure, i.e., the distribution does not change up to any rotation of basis (e.g., multiplication with any arbitrary orthonormal matrix). Let $\mathbf{Z}$ be the upper-left $k\times k$ principal submatrix of $\mathbf{X}$. What is the distribution of $\| \mathbf{Z} \|_F^2$, e.g., its probability density function and cumulative distribution function? (We may be also interested in their asymptotic versions.) This also relates to the so-called multivariate beta distribution. Let $\mathbf{Y}\in \mathbb{R}^{d\times k}$ be a random matrix with i.i.d. standard Gaussian entires, and $\mathbf{Y} = \mathbf{Q}\mathbf{R}$ be its QR decomposition. Let $\mathbf{Q}'$ be the top $k\times k$ submatrix of $\mathbf{Q}$. Then $\mathbf{Q}'$ follows the multivariate beta distribution. An equivalent question is as follows. What is the (asymptotic) distribution of $\|\mathbf{Q}'\|_F^2$? REPLY [3 votes]: This problem has been studied in the physics literature as the distribution of the thermal conductance of a superconducting quantum dot. Let me explain the relationship: The $d\times d$ orthogonal matrix $X$ corresponds to the scattering matrix $S$, the $k\times k$ upper-left principal submatrix $Z$ corresponds to the reflection matrix $r$, the Frobenius norm squared $||Z||_F^2$ is related to the dimensionless thermal conductance $G$ by $$G=k-{\rm tr}\,rr^{\rm T}=k-||Z||_F^2.$$ (Superscript T is the transpose and tr is trace.) The distribution of $G$, and hence of the desired Frobenius norm, follows from the distribution of the socalled transmission eigenvalues $T_n=1-R_n$, with $R_n$ an eigenvalue of $ZZ^{\rm T}$. The joint probability distribution of the $R_n$'s is known exactly for any $k$ and $d=k+k'$. In the case $k\leq k'$ it is given by $$P(R_1,R_2,\ldots R_k)=C \sqrt{\prod_{n=1}^k\frac{(1-R_n)^{k'-k}}{(1-R_n)R_n}}\;\prod_{i=1}^j\prod_{j=1}^k |R_i-R_j|,$$ with $C$ a normalization constant and all $R_n\in(0,1)$. If $k>k'$ the above result still holds with $k$ and $k'$ interchanged, and then there are an additional set of $k-k'$ of the $R_n$'s pinned at unity. The desired probability distribution of $$||Z||_F^2=\sum_{n=1}^k R_n$$ now follows upon integration. If $d\gg k$ an approximate result follows by taking the matrix elements of $Z$ to be independent Gaussians, of zero mean and variance $1/d$, resulting in a chi-squared distribution with $k^2$ degrees of freedom. If knowledge of the first two moments is sufficient, then there are exact results for any $k,d$, $$E(||Z||_F^2)=k^2/d,\;\;{\rm Var}\,(||Z||_F^2)=\frac{2k^2(d-k)^2}{d^2(d-1)(d+2)}$$ I note that there are similar results when $X$ is complex unitary, rather than real orthogonal, reviewed here.<|endoftext|> TITLE: morphisms representable by algebraic spaces vs morphisms representable by schemes QUESTION [24 upvotes]: So I've been working with moduli stacks in algebraic geometry for a while now, with no formal training in the technicalities of the theory of algebraic stacks (ie, I've read a few articles and I learn what I need, without having spent much time doing exercises or working through examples/counterexamples). One of the difficulties I've experienced is that while discussing morphisms of algebraic stacks, some authors only require certain morphisms to be representable by algebraic spaces, while others require them to be representable (by schemes). Of course I've always found the latter to be more comprehensible (since it saved me from having to learn too much about alg. spaces on the way to working with alg. stacks). On the other hand, all the really comprehensive references on the subject usually fall into the former category. For example, the stacks project's definition of an algebraic stack only requires the diagonal to be representable by alg spaces, while Gomez's article https://arxiv.org/pdf/math/9911199v1.pdf requires that it be representable (by schemes). So far, whenever I read "representable by alg spaces", I sort of "pretend" that it says "representable by schemes" and proceed with that assumption. My questions are: What is the difference between requiring the diagonal of an algebraic stack to be representable by spaces vs representable by schemes? When are the two notions equivalent? What pitfalls are there to avoid? In general, what are some good illustrative examples of morphisms of alg stacks which are only representable by spaces, but not by schemes? When are the two notions equivalent? What pitfalls are there to avoid? REPLY [2 votes]: This may not directly answer your question. However, one reason to introduce the notion of an algebraic space is that sometimes it is easier to show that the diagonal is representable by an algebraic space than by a scheme. Consider the following example. Let $G$ be a smooth algebraic group over a field $k$. Consider the classifying stack BG that parametrizes principal bundles that are locally trivial in the fpqc topology. Since $G$ is smooth over a point, and smooth surjective morphisms \'{e}tale locally admit a section, we note that these bundles are a fortiori trivial in the \'{e}tale topology. Let $T$ be a scheme and suppose we have a morphism $$T \stackrel{(P_1, P_2)}{\longrightarrow} BG \times BG.$$ Here the $P_i$'s are two principal bundles over the scheme $T$. The fiber product $$T \times_{BG \times BG} BG \cong \underline{\operatorname{Isom}}_T(P_1, P_2)$$ where the right hand side is the functor of isomorphisms between $P_1$ and $P_2$. Now choose an \'{e}tale cover $T' \to T$ that trivializes both the $P_i$'s (for instance if $T_i \to T$ trivializes $P_i$ then we can just take $T' := T_1 \times_{T} T_2$). Hence $$\underline{\operatorname{Isom}}_T(P_1, P_2) \times_T T' \cong \underline{\operatorname{Isom}}_{T'}(G_{T'}, G_{T'}) \cong h_{G \times T'}.$$ This shows that the base change of the Isom functor by an \'{e}tale cover is a scheme, from which we conclude that Isom is an algebraic space.<|endoftext|> TITLE: Integrals involving the Lambert function W QUESTION [5 upvotes]: I'm actually struggling on a calculation of an integral involving the Lambert function W. Let $\tilde{w}$>0 a parameter that I will tune to $0^+$ at the end of my calculation. I'm interested in the function $\Psi$ defined by : $\forall z \in \mathbb{R}^+$, $\Psi(z)=\frac{1}{\pi} \int_{0}^{+\infty} db\sqrt{W(\tilde{w}^2e^{b})} \ln(1 + z \tilde{w}^{-2}e^{-b})$. If the Lambert function were simply replaced by a logarithm, this integral would give me a polylogarithm but the Lambert function messes up everything ... I cannot find the asymptotics of the function or an expansion in $\tilde{w}$. I tried to define the change of variable $x=\sqrt{W(\tilde{w}^2e^{b})}$ but then I got this integral : $\Psi(z) = \frac{1}{\pi} \int_{W(\tilde{w}^2)}^{+\infty} dx \frac{1+x}{\sqrt{x}} \ln\left(1 + \frac{z e^{-x}}{x}\right)$ Because of the ratio $\frac{e^{-x}}{x}$ I cannot expand the logarithm into a series and an integration by part is also not so conclusive ... Thanks a lot for your help concerning the asymptotics of the function ! Alex REPLY [3 votes]: This may only be a long comment. Since you are going to tune $\tilde{w} \to 0^+$, why not set it to $0$ right away? Since $W(\tilde{w}^2)$ is continuous and zero and $W(\tilde{w}^2) = \tilde{w}^2 - \tilde{w}^4 + O(\tilde{w}^6)$, this can be done in the second form of the integral you gave. With the change of variables $x=zy$, it becomes $$ \Psi(z) = z^{1/2} \int_0^\infty \frac{\ln(1+e^{-zy}/y)}{\sqrt{y}} \, dy + z^{3/2} \int_0^\infty \sqrt{y} \ln(1+e^{-zy}/y) \, dy . $$ You didn't say what is it that you want to know about $\Psi(z)$ (barring an "exact" formula, which seems unlikely). Depending on what you want, the above integral can be transformed and approximated in various ways. Edit: For small $z\ll 1$, you need two separate (uniform) approximations, for small $y\ll 1/z$ and for large $y\gg 1$: $$\tag{$*$} \ln(1+e^{-zy}/y) \sim \begin{cases} \ln(1+1/y) - \frac{zy}{y+1} + O(zy)^2 & (y \ll 1/z) \\ e^{-zy}/y - e^{-2zy}/(2y^2) + O(y)^2 & (1 \ll y) \end{cases} $$ These approximations have an overlap region $1 \ll y \ll 1/z$. Then, picking a point within the overlap region, like $y=z^{-1/2}$, you can integrate one approximation over $(0,z^{-1/2}]$ and the other over $[z^{-1/2},\infty)$. If you only integrate one of them over the full $(0,\infty)$ interval, then you get divergences either at one or at the other end. The two separate integrals will still have divergent terms coming from the $y=z^{-1/2}$ cut, but they will cancel when added together. The integrals with the logs and square roots can be computed using elementary functions (a computer algebra system should do this easily). The integrals involving exponentials will be expressible in terms of incomplete $\Gamma$ functions, whose asymptotic expansions can be found here. Putting the results together, I get \begin{multline*} \Psi(z) = z^{1/2} (2\pi -2\sqrt{\pi} z^{1/2} + \pi z + \cdots) + z^{3/2} (\sqrt{\pi} z^{-1/2} + \frac{2}{3}\pi + \cdots) \\ = 2\pi z^{1/2} -\sqrt{\pi} z + \frac{5}{3} \pi z^{3/2} + O(z^{3/2}) . \end{multline*} Note that my estimate for the error term is just a guess (it could be $O(z^2)$ instead). It would need to be checked by computing the leading correction coming from the next order expansions in $(*)$. For large $z\gg 1$, the same method doesn't quite work, since the approximations $(*)$ will be valid in the regions $y\ll 1/z$ and $1 \ll y$, which no longer overlap. Perhaps a third approximation that covers the intermediate region and overlaps with the other two could be found. Then the same procedure would work again.<|endoftext|> TITLE: On entire functions with real and simple zeros QUESTION [8 upvotes]: There is a nice necessary and sufficient criterion for a polynomial $p_n$ of degree $n$ with real coefficients to have all zeros real and simple. Namely, it says: $p_n$ has $n$ (distinct) real roots $x_{1},\dots,x_{n}$ iff the Hankel matrix $$S_{n}=\begin{pmatrix} s_{0} & s_{1} & \dots & s_{n-1}\\ s_{1} & s_{2} & \dots & s_{n}\\ \vdots & \vdots & \ddots & \vdots\\ s_{n-1} & s_{n} & \dots & s_{2n-2} \end{pmatrix}, \quad s_{k}:=\sum_{j=1}^{n}x_{j}^{k},$$ is positive definite; see, for example here. My first question asks if there is some "infinite analogue" of the above equivalence where a polynomial is replaced by an entire function. More concretely, let me assume the following. Let $f$ is a given entire function with real Taylor coefficients. Next, let $\{p_{n}\}_{n\geq1}$ be a sequence of polynomials (with increasing degree) approximating $f$. For example, we may assume $p_{n}\to f$, for $n\to\infty$, uniformly on compact subsets of $\mathbb{C}$. (One can take $p_n$ to be the truncated Taylor expansion of $f$, for instance.) Assume further that $x_{1}(n),\dots,x_{n}(n)$ are zeros of $p_n$ and that there exist limits $$\hskip100pt \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}\left(x_{j}(n)\right)^{k}=\alpha_{k},\hskip100pt (*)$$ for all $k\in\mathbb{N}$. The second question is: if the Hankel matrix $A$, where $A_{i,j}:=\alpha_{i+j}$, $i,j\geq0$, is associated with a positive moment functional, i.e., all the principal sections of $A$ are positive definite, can one say something about zeros of $f$? Are they real and simple, for example? The third question: Let me add one last related question. Assume that we have the sequence of polynomials $p_{n}$ as above, so the limits in $(*)$ hold, but we have no limit function $f$. The sequence $\{p_{n}\}$ is not necessarily convergent. Is it true that, if $A$ has all the principal sections positive definite, then the sequence $\{p_{n}\}$ is asymptotically real-rooted, that is, all the limit points of zeros are real? A slight reformulation of the question as well as an addition of some other assumptions (for example, on the order of $f$, the rate of the convergence in the limits $(*)$, etc.) is possible. REPLY [4 votes]: It seems that the magic words are Laguerre-Polya Class. (the link is to an article by Tom Craven and George Csordas, which has many references.)<|endoftext|> TITLE: Difference between rationally connected and $\mathrm{CH}_0\simeq \mathbb Z$ QUESTION [12 upvotes]: For a projective variety $X$ over an algebraically closed field of characteristic $0$, why the condition $\mathrm{CH}_0(X)\simeq \mathbb Z$ does not imply that $X$ is rationally (chain) connected ? REPLY [19 votes]: I hope that my answer will make it clearer. If $\mathrm{CH}_0X=\mathbb{Z}$, not all points are necesserily connected by rational curves. For instance, fake projective planes satisfy conditions of Bloch's conjecture, but they don't contain rational curve at all (they are quotients of the complex ball). However, points should be "connected by rational curves" in some symmetric power of $X$. I mean that for every two points $P,Q$ of $X$ there are an effective cycle $T$ of degree $n$ and a map $f:\mathbb{P}^1 \to X^{(n+1)}$, such that $f(0)=T+P$ and $f(\infty)=T+Q$. (See Mumford "Rational equivalence of 0-cycles on surfaces" and Fulton "Intersection theory"). Let me give a simple example of a non-rationally connected surface with trivial Chow group. I am going to prove Bloch's conjecture for Enriques surfaces (it will be little different from the original proof by Bloch, Kas, Lieberman in "Zero cycles on surfaces with $p_g = 0$"). Let $X$ be an Enriques surface. It is well-known that $X$ admits an elliptic fibration of index 2 and its Jacobian fibration $J$ is a rational surface [See e.g. Beaville's book on algebraic surfaces]. There is a map $\pi:X^{(2)}_{\mathbb{P}^1} \to J$ from the second relative symmetric power to the relative Jacobian with fibers $\mathbb{P}^1$. Thus, by Graber-Harris-Starr theorem $X^{(2)}_{\mathbb{P}^1}$ is rationaly connected. We can map it to $X^{(2)}$ ( non-relative symmetric power) and we can embed $X$ to $X^{(2)}$ as the diagonal. The variety $X^{(2)}_{\mathbb{P}^1}$ is a rationaly connected divisor in $X^{(2)}$ which contains the diagonal. So, points of $X$ are connected by rational curves in the second symmetric power. Precisely, for every $P,Q \in X$ we have $2P \sim 2Q$. By Roitman's theorem (proof in Bloch's lectures on cycles) $\mathrm{CH}_0X$ is torsion free. Hence, $P \sim Q$ and $\mathrm{CH}_0X= \mathbb{Z}$.<|endoftext|> TITLE: Rationality of zeta function and Grothendieck-Lefschetz fixed point formula, cohomology can be computed as the de Rham cohomology QUESTION [13 upvotes]: Trivial example. First, suppose $X$ is finite. Then we have a finite set $S := X(\overline{\mathbb{F}}_q)$ with an action of $\text{Fr}_q$. How can one explain why the rationality of the zeta function should be true in this case (it is still nonobvious from the definition of $Z(X, t)$)? Let $\mathbb{Q}[S]$ be the vector space spanned by $S$; then we get an induced linear action of $\text{Fr}_q$ on $\mathbb{Q}[S]$. We have$$Z(X, t) = \det(1 - t \cdot \text{Fr}_q; \mathbb{Q}[S])^{-1}.\tag*{$(1)$}$$This is seen by showing that the logarithmic derivatives of the two sides are the same. Let $\{\alpha_i\}$ denote the collection of the eigenvalues of $\text{Fr}_q$ acting on $\mathbb{Q}[S]$, counted with their algebraic multiplicities. Then$$\det(1 - t \cdot \text{Fr}_q) = \prod_i (1 - \alpha_i t).$$Applying $t \cdot {{\text{d}\log}\over{\text{d}t}}$ to both sides of $(1)$, we obtain$$\sum_{n = 1}^\infty |X(\mathbb{F}_{q^n})|t^n = \sum_i \sum_{n = 1}^\infty (\alpha_i t)^n = \sum_{n = 1}^\infty \text{Tr}(\text{Fr}_q^n) \cdot t^n.$$However, it is very easy to see that$$\text{Tr}(\text{Fr}_q^n; \mathbb{Q}[S]) = \left|S^{\text{Fr}_q^n}\right|,$$and so we get $(1)$ as desired. A very different, but morally analogous, story. Let$$X = \text{compact }C^\infty\text{ manifold},$$$$\text{Fr} = \text{a self-diffeomorphism of }X,\text{ such that }\forall n \in \mathbb{N}, \text{ }\left|X^{\text{Fr}^n}\right| < \infty.$$Assume that $X$ is orientable, and that for any fixed point $x \in X$ of $\text{Fr}^n$ (for some $n \in \mathbb{N})$, the condition $\det(1 - \text{Fr}^n; T_xX) > 0$ holds. Then $\deg(1 - \text{Fr}^n; T_xX) \neq 0$ for all $x \in X^{\text{Fr}^n}$ is equivalent to all fixed points of $\text{Fr}^n$ being nondegenerate, i.e. the corresponding intersection points of $\Delta_X$ and $\Gamma(\text{Fr}^n)$ are transverse (here, $\Delta$ means diagonal, $\Gamma$ means graph). Now we can define the zeta function $Z((X, \text{Fr}), t)$ in the same way as above:$$Z((X, \text{Fr}), t) = \prod_{i = 0}^{\dim X} \det(1 - t \cdot \text{Fr}^*; H^i(X, \mathbb{Q}))^{(-1)^{i + 1}}.\tag*{$(2)$}$$If $X$ is finite, then $H^0(X, \mathbb{Q}) \cong \mathbb{Q}[X]^*$, and we are essentially reduced to the previous example. Anyways, applying $t \cdot {{\text{d}\log}\over{\text{d}t}}$ to both sides of $(2)$, we can reduce it to the following identity, the well-known Grothendieck-Lefschetz fixed point formula: for all $n \in \mathbb{N}$,$$\left|X^{\text{Fr}^n}\right| = \sum_{i = 0}^{\dim X} (-1)^i \text{Tr}\left(\text{Fr}^n; H^i(X, \mathbb{Q})\right).\tag*{(3)}$$ What I want. I want to see how one can guess the Lefschetz fixed point formula, which entails an informal argument which can be turned into a rigorous proof. The idea to imitate the argument given above in the case where $X$ is finite. I want to use the fact that cohomology can be computed as the de Rham cohomology. Question. Pretend that all the spaces of differential forms $\Omega^i(X)$ are finite dimensional, replace $H^u(X)$ by $\Omega^i(X)$. How do I see $(2)$ by using the same method that I used in the case where $X$ is finite? What are the extra nuances I have to take into consideration/take care of? REPLY [7 votes]: The following is sections 3-4 of van der Put's The cohomology of Monsky and Washnitzer with all $p$-adic issues and analytic subtleties (necessary to make the argument correct, of course!) removed. Let $X$ be a smooth manifold of dimension $n$ and let $F: X \to X$ be a degree $q^n$ finite map. (You can just view $q$ as a real number, the $n$-th root of the degree; of course, I am calling it that because it is the order of the ground field in the Frobenius setting.) We want to compute $$\sum (-1)^j \mathrm{Tr}\ F^{\ast} H^j_c(X) \to H^j_c(X)$$ where $H^j_c$ is compactly supported cohomology. Using Poincare duality, we may instead compute $$q^n \sum (-1)^j \mathrm{Tr}\ (F^{\ast})^{-1} H^j(X) \to H^j(X).$$ In van der Put, this is given as the starting goal without motivation (equation 1.2). (This part is section 3.2.) Let $\psi : \Omega^j(X) \to \Omega^j(X)$ be push down along $F$: The value of $\psi(\omega)$ near a point $x \in X$ is the sum of $\omega$ over the $q$ preimages of $x$ (computed with multiplicity, if there is ramification. We observe: $\psi \circ d = d \circ \psi$ (because $d$ is linear) so $\psi$ passes to cohomology. We have $\psi(F^{\ast} \omega) = q^n \omega$ (pulling back and pushing forward adds up the same thing $\deg F = q^n$ times). So $q^{-n} \psi$ is a left inverse of $F^{\ast}$ on $H^{\ast}(X)$. As $\dim H^{\ast}(X)$ is finite, this just means that $\psi = q^n (F^{\ast})^{-1}$. (At the time van der Put was writing, Monsky-Washnitzer cohomology was not known to be finite dimensional, so he has a trick to get around this -- Theorem 3.2.(iv). Finiteness was proved by Berthelot and Mebkhout.) So we can rewrite the desired sum as $$ \sum (-1)^j \mathrm{Tr}\ \psi: H^j(X) \to H^j(X).$$ For future use, we note that $\psi(F^{\ast}(a) \omega)=a \psi(\omega)$ for $a \in \mathcal{O}$ and $\omega \in \Omega^j$. This is also clear: If $a(x)=\alpha$, then multiplying $\omega$ by $F^{\ast}(\omega)$ multiplies by $\alpha$ at each point in $F^{-1}(x)$, so the sum defining $\psi(F^{\ast} \omega)$ is multiplied by $\alpha$ at $x$. I pause for an example: Let $F : \mathbb{C}^{\ast} \to \mathbb{C}^{\ast}$ be the $q$-th power map, so the fixed points are the $q-1$ many nonzero roots of $z^q=z$. Let $\zeta$ be a primitive $q-1$ root of unity. Then $\psi(g)(z) = \sum_{k=1}^{q-1} g(\zeta^k z)$ for any function $g$ on $\mathbb{C}^{\ast}$. In particular $$\psi(z^a) = \begin{cases} q z^{a/q} & a \equiv 0 \bmod q \\ 0 & a \not \equiv 0 \bmod q \end{cases}.$$ Working a bit harder, $$\psi(z^a dz/z) = \begin{cases} z^{a/q} dz/z & a \equiv 0 \bmod q \\ 0 & a \not \equiv 0 \bmod q \end{cases}.$$ No, I am not missing a factor of $q$. The push forward of $dz/z$ along the $q$-th power map is $dz/z$. Let's write $w$ for the coordinate on the source space and $z$ on the target, so $F^{\ast} z=w^q$ and $F^{\ast} (dz/z)=q dw/w$. Then $$\psi(dw/w) = \psi((1/q) dz/z) = \sum (1/q) d(\zeta^k z)/(\zeta^k z) = q\cdot(1/q) dz/z=dz/z.$$ Morally, the trace of $\psi$ on $\Omega^0$ wants to be $q$, and on $\Omega^1$ wants to be $1$. When we pass to $H^0$ and $H^1$, this is true with no analytic subtlties. (So $q-1$ is the number of fixed points, as desired.) To make the argument on $\Omega^j$, one uses the word "nuclear" a lot. I don't understand this part, but I believe that one of the reasons to use $\psi$ instead of $F^{\ast}$ is that $\psi$ is nuclear. So, we continue with the general argument. We now reduce to the case where there are no fixed points. Let the fixed points be $z_1$, ..., $z_N$ and let $U = X \setminus \{ z_1, \ldots, z_N \}$. In the case of Frobenius, $F$ maps $U$ to $U$. There is a Gysin sequence relating $H^{\ast}(X)$, $H^{\ast}(U)$ and a simple contribution for each fixed point, and we can see that the alternating sums of traces of $F$ on $H^{\ast}(U)$ and $H^{\ast}(X)$ differ by $N$. I think (this part isn't in van der Put) that one can still make this argument even if $F(U)$ isn't $U$. Let $U' = F^{-1}(U)$. We can map $\Omega^j(U) \to \Omega^j(U') \overset{\psi}{\longrightarrow} \Omega^j(U)$ where the first map is restriction and the second makes sense because $U' \to U$ is finite. I think we can make the same argument with this map. This seems to be another advantage of the pushforward $\psi$ over the pull back $F^{\ast}$. I was also glossing over the difference between the characteristic $p$ variety and its $p$-adic lift two paragraphs back. (Because I decided to shift to the case of a manifold, where we don't have this issue.) In the $p$-adic lift, we are removing $F$-invariant balls around each fixed point, not just points. (This is the proof of 4.10) Assume that $F$ has no fixed points. So, as $a$ ranges over $\mathcal{O}(X)$, there is no place where all the functions $F^{\ast}(a) - a$ vanish. So the $F^{\ast}(a)-a$ generate the unit ideal; say $\sum (F^{\ast} a_i -a_i) b_i =1$. So, for any $\omega \in \Omega^j$, we have $$\sum \psi((F^{\ast} a_i) b_i \omega - b_i a_i \omega) = \psi(\omega).$$ (We have used that multiplication is commutative.) Since $\psi(F^{\ast}(a) \eta) = a \psi(\eta)$, we deduce $$\sum a_i \psi(b_i \omega) - \psi(b_i a_i \omega) = \psi(\omega).$$ In other words, the summand is the commutator of the operations $\omega \mapsto a_i \omega$ and $\omega \to \psi(b_i \omega)$. Since commutators always have trace zero, we "deduce" that $\psi: \Omega^j \to \Omega^j$ has trace zero, and hence the alternating sum of traces is zero.<|endoftext|> TITLE: Trace of a nonlinear matrix equation (cont'd) QUESTION [7 upvotes]: Let $X_0$ be a trace-one positive definite matrix, i.e. $X_0>0$, $\mathrm{tr}(X_0)=1$. Let $A>0$ and consider the following iteration $$ X_{k+1} = X_k^{1/2}AX_k^{1/2},\quad k\geq 0,\quad (\star) $$ where $X_k^{1/2}$ denotes the (principal) square root of $X_k$. My question: Is it true that if there exists $X_0$ as above such that $(\star)$ is trace-preserving starting from $X_0$, i.e. $\mathrm{tr}(X_{k+1})=\mathrm{tr}(X_{k})=1$ for all $k\geq 0$, then $A=I$? If $X_0$ and $A$ are scalars this is clearly true. Moreover, it is easy to see that $A$ cannot be such that $A> I$ or $A< I$, but I cannot quite prove that $A$ must be the identity. Thanks for your help. Addendum 1. Note that by applying to $(\star)$ an orthogonal change of basis $T$ which diagonalizes $A$, we can rewrite $(\star)$ as $$ \tilde{X}_{k+1}= \tilde{X}_k^{1/2} D \tilde{X}_k^{1/2} \quad (\star\star) $$ where $\tilde{X}_k:=T^\top X_k T$ and $D:=T^\top AT>0$ is diagonal. Addendum 2. A "simpler" version of this question was answered in the affirmative here. Addendum 3. My attempts so far were based on working with the "simplified" dynamics $(\star\star)$. Namely, consider the partition $D=\left[\begin{smallmatrix}D_1 & 0 \\ 0 & D_2\end{smallmatrix}\right]$ with $D_1>I_{n_1}$ and $D_2 I_{n_1}$, $D_2=I_{n_2}$, it is easy to show that $\mathrm{tr}(\tilde{X}_1)\neq 1$). Now by partitioning $\tilde X_0^{1/2}$ and $\tilde X_1$ accordingly to the block decomposition of $D$, we get $$ \tilde{X}_1=\begin{bmatrix}(\tilde X_1)_{11} & (\tilde X_1)_{12} \\ (\tilde X_1)_{12}^\top & (\tilde X_1)_{22}\end{bmatrix} = \begin{bmatrix}(\tilde X_0^{1/2})_{11}D_1(\tilde X_0^{1/2})_{11} + (\tilde X_0^{1/2})_{12}D_2(\tilde X_0^{1/2})_{12}^\top & (\ast) \\ (\ast)^\top & (\tilde X_0^{1/2})_{12}^\top D_1(\tilde X_0^{1/2})_{12} + (\tilde X_0^{1/2})_{22}D_2(\tilde X_0^{1/2})_{22} \end{bmatrix}, $$ with the constraint $\mathrm{tr}(\tilde X_0)=1$ which now reads as $$ \mathrm{tr}\left((\tilde X_0^{1/2})_{11}(\tilde X_0^{1/2})_{11} + (\tilde X_0^{1/2})_{12}(\tilde X_0^{1/2})_{12}^\top\right)+\mathrm{tr}\left((\tilde X_0^{1/2})_{12}^\top (\tilde X_0^{1/2})_{12} + (\tilde X_0^{1/2})_{22}(\tilde X_0^{1/2})_{22}\right)=1. $$ Now my conjecture is that $\mathrm{tr}(\tilde X_2)\neq 1$ for all $\tilde{X_0}>0$, $\mathrm{tr}(\tilde{X_0})=1$, (indeed, it is easy to find examples for which $\mathrm{tr}(\tilde X_1)= 1$). My idea is to use the same block decomposition for $\tilde X_2$ and then exploit some "trace inequalities" applied to the diagonal blocks. However I didn't manage to conclude anything so far. Addendum 4. A more general version of this question which I suspect hold true (actually, for the $2\times 2$ case, it does hold true) is the following one: Is it true that if there exists $X_0\ge 0$, $\mathrm{tr}(X_0)=1$, such that $(\star)$ is trace-preserving starting from $X_0$, i.e. $\mathrm{tr}(X_{k+1})=\mathrm{tr}(X_{k})=1$ for all $k\geq 0$, then $X_{k+1}=X_k^{1/2}AX_k^{1/2}=X_k$ for all $k\geq 0$? REPLY [5 votes]: Actually, you have completely solved it yourself, just didn't dare to acknowledge it. In my notation, you have $(X\circ X^T)v(A)=(Y\circ Y^T)v(I)$ when $Y^2=XAX$. Similarly, $(Z\circ Z^T)v(I)=(Y\circ Y^T)v(A)$ when $Z^2=YAY$. Taking the trace, we must have $$ 1=\langle(X\circ X^T)v(I),v(I)\rangle=\langle(Y\circ Y^T)v(I),v(I)\rangle=\langle(Z\circ Z^T)v(I),v(I)\rangle\,. $$ However, $$ \langle(Y\circ Y^T)v(I),v(I)\rangle=\langle(X\circ X^T)v(A),v(I)\rangle $$ and $$ \langle(Z\circ Z^T)v(I),v(I)\rangle=\langle(Y\circ Y^T)v(A),v(I)\rangle \\ = \langle(Y\circ Y^T)v(I),v(A)\rangle=\langle(X\circ X^T)v(A),v(A)\rangle $$ so $$ \langle(X\circ X^T)(v(A)-v(I)),(v(A)-v(I))\rangle=0 $$ whence $X\circ X^T$ and, thereby, $X$ must be degenerate unless $v(A)=v(I)$, i.e., $A=I$. This story definitely has a few morals but I'll abstain from spelling them out :-).<|endoftext|> TITLE: Is a space with p-norm a Finsler manifold? QUESTION [5 upvotes]: Suppose $\mathbb{R}^n$ is equipped with the p-norm $\left\Vert x \right\Vert_p$. Let $x\in \mathbb{R}^n$ and let $y$ be in a neighborhood of $x$. The distance between $x$ and $y$ can be defined as $\left\Vert x-y \right\Vert_p$. Is this setting an example of a Finsler manifold? If so, it should be a very simple example of a Finsler manifold. Can you point out some references about the study of the geometry of this type of p-norm spaces, like the curvature and geodesics? (Apparently when $p=2$, it is the trivial case of a Euclidean space, which is a trivial case of Riemannian manifold, which is a special case of Finsler manifold. But what about other $p\neq 2$?) REPLY [4 votes]: It is a Finsler metric for $p>1$. I don't know a reference. But you can easily see that its indicatrix is smooth and strictly convex and symmetric. The triangle inequality says that the geodesics are the usual straight lines. The Finsler flag curvature vanishes by the Jacobi equation. I don't know the complete description of the curvature.<|endoftext|> TITLE: Plurisubharmonic function and complete Kähler metric on certain Kähler manifold QUESTION [6 upvotes]: Given a compact Kähler manifold $M$, let $D$ be an effective divisor on $M$. Is $M\setminus D$ pseudoconvex? That is, can we find a smooth plurisubharmonic function that exhausts $M\setminus D$ ? Can we find a complete Kähler metric on $M\setminus D$ ? If $1$ and $2$ are not true, can we find any obstructions? Or necessary and sufficient conditions? Note: if $D$ is an ample divisor, we can choose an Hermitian metric on $[D]$ which is the associated divisor bundle to $D$, then we can take $-ln|s_{D}|^{2}_{h}$ to be the strongly smooth exhaust plurisubharmonic function on $M\setminus D$. Surely, this is just a Stein (or Affine) manifold. Given such a plurisubharmonic function on $M\setminus D$ we can easily contruct a compelte Kähler metric. REPLY [5 votes]: Question 1: Plurisubharmonic functions extend across codimension 2 subvarieties . Let X be the complex projective plane blown up at one point and D be the exceptional divisor then any plurisubharmonic function on the complement of D in X extends to X and is therefore a constant.<|endoftext|> TITLE: Does there exist a stochastic time derivative? QUESTION [6 upvotes]: The Setup Suppose I have a stochastic process $f(Z_t)$ where $Z_t$ solve the $d$-dimensional SDE $$ dZ_t = \mu(t,Z_t)dt + \sigma(t,Z_t)dW_t $$ and $f$ is a smooth function. My Question Is there a notion of time-derivative "$d_t$" of the process $f(Z_t)$ which satisfies: Some sort of chain rule like $$ d_tf(Z_t) = \partial_t f(Z_t) d_t(Z_t), $$ where $\partial_t$ is the usual derivative wrt $t$. If $Z_t$ is deterministic (ie: $\sigma(t,Z_t)=0$) and $\mu(t,z)$ is $C^1$ in $t$ then $$d_t=\partial_t,$$ ie: $d_t$ reduces to the usual derivative when $f(Z_t)$ is a smooth function of $t$. REPLY [5 votes]: I think the chain rule $d[f(Z_t)]=f'(Z_t)\circ dZ_t$ is valid when the product $\circ$ is defined as in Stratonovich stochastic integral (while the SDE uses Ito's). Note that $g(Z_t)\circ dt$ doesn't differ from the ordinary product $g(Z_t)\ dt$, but $g(W_t)\circ dW_t$ is Ito's $g(W_t)\ dW_t$ plus $\frac12 g'(W_t)\ dt$.<|endoftext|> TITLE: Interactive model of the hyperbolic plane for a general public lecture QUESTION [48 upvotes]: The following is not quite a research level question, but I still find this site appropriate for asking it. I hope I get it right here. I am preparing a talk for a general public and I want to discuss some hyperbolic geometry. I wish I had a good illustration device. I imagine a dynamical version of one of Escher's tessellations of the Poincare model (e.g the one in How might M.C. Escher have designed his patterns?) which changes isometrically when I slide the computer mouse. Question: Could you please make any recommendation regarding any device of a nature similar to the one I describe above? In fact, I will be happy to have whatever interactive model of whatever geometry, not necessarily hyperbolic. Subquestion: if you're kind enough to make a recommendation, could you also advice regarding copyright issues (if applicable)? Sidequestion: Any other recommendation regarding presentation of geometry will be appreciated. Please note that my concern is more about the quality of the presentation than the actual mathematical content... UPDATE: Thank you! I am thrilled to get in less than 24 hours so many excellent answers and comments. Fortunately, Arnaud Chéritat provided EXACTLY what I asked for, and I happily accept his answer. Indeed, I am going to use his tool for my presentation. However, there are other excellent tools here which could be useful elsewhere. It seems to me a good idea to keep collecting those and MOF is an excellent platform for that. I suppose one should post one of these big-list questions which has a broader scope than this one (but not too broad), something like "Visualizing tools for lectures on geometry". I am not sure how this is done, so you can pick up the glove! REPLY [2 votes]: I've used my program MagicTile during a few presentations, and it has many capabilities that make it a nice choice. Supports all regular 2-dimensional tilings across the 3 geometries (spherical, Euclidean, and hyperbolic). Many models: Poincare, Klein, Upper Half Plane, Orthographic, and their analogues in spherical geometry (Stereographic, Gnomonic, etc.), and some less common models... All these models are dynamically pannable. I've found this a nice way to demonstrate classes of conformal Möbius transformations. Spherical/Euclidean geometries also include quotient surface views, including the non-orientable Klein bottle and Boy's surface. It is a very colorful program (as it is centered around analogues of Rubik's cube), which could heighten public interest. It is completely free, in active development since 2009, and the opensource code has the most permissive GitHub license. One downside is that it is a Windows program, though some have it running on Macs as well. This would also be a good place to recommend Henry Segerman and Saul Schleimer's physical models, which they have proved work well for public lectures. For an example presentation, see their YouTube video Illuminating hyperbolic geometry, which has links to purchase the models.<|endoftext|> TITLE: How to construct a constructive proof from a non-constructive proof using prime ideals? QUESTION [19 upvotes]: The sum of two nilpotent elements of a commutative ring is nilpotent. This can be checked by a direct calculation using the binomial theorem. In fact, this calculation shows the stronger statement $x^n=y^m=0 \Rightarrow (x+y)^{n+m-1}=0$. But we can also give a more sophisticated proof: If $x,y$ are nilpotent, they are contained in every prime ideal. Hence, the same is true for $x+y$. Hence, $x+y$ is nilpotent: otherwise, the localization at $x+y$ would be non-zero and therefore have a prime ideal, but this corresponds to a prime ideal in the given ring not containing $x+y$. (In short: The set of nilpotent elements is the intersection of all prime ideals, hence closed under addition.) The general existence of prime ideals is equivalent to the Boolean Prime Ideal Theorem and therefore the proof above is not constructive. The proof shows nothing about the nilpotence exponent of the sum. On the other hand, it is quite elegant and it is really a no-brainer if you are used to commutative algebra. Moreover, it can be made "more constructive" (not really constructive, as Matt F. points out), or at least provable in $\mathsf{ZF}$, as follows: We restrict our attention to the subring generated by $x,y$. This ring is countable. The same is true for the localization at $x+y$. There is a constructive proof that every non-trivial countable commutative ring has a maximal ideal, hence has a prime ideal. And now we may proceed as before. So we have two constructive proofs: (a) the direct calculation using the binomial theorem, (b) the proof using prime ideals. The question is: Assume that we know the proof (b), is there a general method how to produce the proof (a) from it? Perhaps even including the stronger statement about the nilpotence exponent? This is just a toy example for the general question how to get rid of prime ideals in proofs in commutative algebra where we would expect to have, or already know, more direct proofs. I have only picked this toy example because I hope that the general method can be easily explained with it. Another toy example: How to produce the direct proof of $I+J=A \Rightarrow I^n+J^n=A$ for ideals $I,J \subseteq A$ from the proof using prime ideals? A more sophisticated example can be found here, where I have no idea how a direct calculation looks like (perhaps I will ask this in a separate question). I know that Thierry Coquand and Henri Lombardi have worked on related questions, but after some skimming through their work I couldn't find an answer to my question. REPLY [3 votes]: Since this is somewhat hidden in the comments, let me give the following answer: The statement that the sum of two nilpotents is nilpotent is so basic that it seems to be used in the construction and the verification of the Zariski locale/topos/lattice. I don't think that constructive algebra can prove this without circular arguments. However, the statement $I+J=A \Rightarrow I^n+J^m=A$ can be proven by working in the lattice of radical ideals: $$\sqrt{I^n+J^m}=\sqrt{I^n} \vee \sqrt{J^m}=\sqrt{I} \vee \sqrt{J}=\sqrt{I+J}=A.$$ Whenever one uses the open subset $D(I)$ in a proof, one may simply replace it by the radical ideal $\sqrt{I}$<|endoftext|> TITLE: Average decay of Fourier coefficients of continuous measures along the sequence $\lfloor n^{3/2}\big\rfloor$ QUESTION [7 upvotes]: Let $\mu$ be a continuous measure on $[0,1]$ (i.e. each individual point has $0$ measure). As usual, denote by $\hat\mu(n)=\int_0^1e^{2\pi inx}d\mu(x)$ the Fourier transform of $\mu$, and let $\lfloor x\rfloor$ denote the floor of $x\in\mathbb R$. Is it true that $$\lim_{N\to\infty}sup_{M\in\mathbb N}\frac1N\sum_{n=M}^{M+N}\left|\hat\mu\Big(\big\lfloor n^{3/2}\big\rfloor\Big)\right|=0~?$$ Observe that if $\mu$ is absolutely continuous, then the answer is yes by the Riemann-Lebesgue lemma. If instead of $\hat\mu(\lfloor n^{3/2}\rfloor)$ we consider $\hat\mu(n)$, then the answer is again yes (it follows from Wiener's theorem). If we don't take a supremum over all shifts of the interval $\{1,\dots,N\}$, then the result is again well known because the sequences $n\mapsto\lfloor n^{3/2}\rfloor x$ are uniformly distributed for any irrational $x$. The motivation for this question comes from the fact that $$\lim_{N\to\infty}sup_{M\in\mathbb N}\frac1N\sum_{n=M}^{M+N}\left|\hat\mu\big(n^2\big)\right|=0.$$ This follows easily from the fact that the sequences $n\mapsto n^2 x$ are well distributed for every irrational $x$, but this is not the case for the sequences $n\mapsto\lfloor n^{3/2}\rfloor x$. REPLY [4 votes]: Really, you have almost answered it yourself, just left the very final words out. Take $M=4N^4$. Then $\lfloor (M+n)^{3/2}\rfloor=8N^6+3N^2n$ for $n=1,\dots,N$, so $\frac 1N\sum_{n=1}^N e^{-2\pi i \lfloor (M+n)^{3/2}\rfloor x}$ is $1$ when $x=q/N^2$ and nearly $1$ on a small open neighborhood $U_N$ of those points. Now take some very fast increasing sequence of $N$ so that $\cap_N U_N$ contains a Cantor-type set and put the measure on it. The exact formulae do not matter, of course. What mattered a bit was having long linear pieces of arbitrarily large slope, but even that wasn't crucial. As you noticed yourself, the (particular kind of) absence of good distribution was the key.<|endoftext|> TITLE: Finite groups whose prime graphs are complete QUESTION [8 upvotes]: For a finite group $G$, the prime graph of $G$ is an undirected graph such that its vertices are all prime divisors of $\vert G\vert$ and two distinct vertices $p$ and $q$ are adjacent when there is an element in $G$ of order $pq$. Which finite groups does have a complete prime graph? i.e I am looking for finite groups in which for every two distinct prime divisors of $\vert G\vert$ say $p$ and $q$, there exists at least one element of order $pq$. For instance every Dedekind group has this property. REPLY [3 votes]: To understand the "minimal troublemakers" in the case of solvable groups, I would start with a solvable group $G$ such that every proper section of $G$ has a complete prime graph but $G$ does not. Recall that a section of $G$ is a group $X/Y$ where $Y \lhd X$ and $X$ is subgroup of $G$. Such a group $G$ must be a $\{p,q\}$-group for a pair of distinct primes $p$ and $q$ (for otherwise $G$ has a Hall $\{p,q\}$-subgroup $H$ which is proper and $H$ contains an element of order $pq).$ Also, a Sylow $p$-subgroup of $G$ must be a maximal subgroup of $G$, and likewise a Sylow $q$-subgroup of $G$ must be maximal. Now $G$ can't have both a normal Sylow $p$-subgroup and a normal Sylow $q$-subgroup. However, we have either $O_{p}(G) \neq 1$ or $O_{q}(G) \neq 1.$ Label so that $O_{p}(G) \neq 1.$ Then $G/O_{p}(G)$ must be a $q$-group (otherwise it contains an element of order $pq$, and then so does $G$). Hence $O_{p}(G)$ is a normal Sylow $p$-subgroup of $G$, and is hence a maximal subgroup of $G$, so that $[G:O_{p}(G)] = q.$ Hence $G$ is a Frobenius group with kernel $O_{p}(G)$ and cyclic complement of order $q$. We can go a little further since a Sylow $q$-subgroup $Q$ of $G$ is a maximal subgroup. For then $G = QM$ for some minimal normal subgroup of $G$ contained in $O_{p}(G),$ and we deduce that $|G|$ has the form $qp^{e}$ where $e$ is the smallest positive integer such that $q$ divides $p^{e}-1.$<|endoftext|> TITLE: When is a sheaf coherent if its image under a Fourier-Mukai transform is coherent? QUESTION [6 upvotes]: Let X and Y be to varieties and $F\colon D\mathrm{QCoh}(X) \to D\mathrm{QCoh}(Y)$ a continuous functor between the corresponding unbounded derived categories of quasi-coherent sheaves (given by a kernel on X×Y). Assume that $F(D^b\mathrm{Coh}(X)) \subseteq D^b\mathrm{Coh}(Y)$ and that F is conservative. Are there any conditions that ensure that $F(\mathcal{G}) \in D^b\mathrm{Coh}(Y)$ implies $\mathcal{G} \in D^b\mathrm{Coh}(X)$? The corresponding statement for the abelian categories is easy (if $\mathcal G$ is not coherent, then there is an infinite increasing sequence of coherent subsheaves giving rise to such a sequence of subsheaves of $F(\mathcal G)$, which has to stabilize), but I don't seem to be able to transport this proof to the derived setting. REPLY [3 votes]: Your proof works in the derived case as well. That is, assume smoothness so that $D^bCoh$ is identified with the full subcategory of compact objects (in general the argument will apply to the subcategory of perfect complexes). Then every object $\mathcal{G}$ of $DQCoh$ can be written as a filtered homotopy colimit $colim \mathcal{G}_\alpha$ of objects of $D^bCoh$, as there is an equivalence of infinity-categories $DQCoh(X) = Ind(D^bCoh(X))$. If $F(G) = colim F(\mathcal{G}_\alpha)$ is compact, then you get $F(G) = F(G_\beta)$ for some $\beta$, and use conservativity of $F$ to conclude.<|endoftext|> TITLE: Is the map $\mathrm H^4(S_{24}) \to \mathrm H^4(M_{24})$ surjective? QUESTION [19 upvotes]: The group $S_{24}$ of permutations of $24$ things has fourth integral cohomology $\mathrm H^4(S_{24};\mathbb Z) \cong \mathbb Z/2 \oplus \mathbb Z/2 \oplus \mathbb Z/12$. According to Sikiric and Ellis the largest Mathieu group $M_{24}$ has $\mathrm H^4(M_{24};\mathbb Z) \cong \mathbb Z/12$. The Mathieu group is defined in terms of a permutation representation on $24$ things (namely, the coordinate vectors in the extended binary Golay code), and so there is a restriction map $\mathrm H^4(S_{24};\mathbb Z) \to \mathrm H^4(M_{24};\mathbb Z)$. The number $12$ being somewhat magical, I expect that this map is a surjection. Is it? Is the answer to this question known? My impression of the literature is that $M_{24}$ is just beyond where current technology can fully work out its (2-local, or even $\mathbb F_2$) cohomology. Compare this older MO question. REPLY [11 votes]: The answer to my question is No. The generator of the $\mathbb Z/12$ part of $H^4(S_{24})$ is $p_1$ of the permutation representation. That representation restricts to $M_{24}$ to a Spin representation, i.e. one with $w_1=w_2=0$. For any such representation, $p_1$ is automatically even. Thus the map $H^4(S_{24}) \to H^4(M_{24})$ has image within the $\mathbb Z/6 \subset \mathbb Z/12$.<|endoftext|> TITLE: Branch locus of projection of cubic surface QUESTION [6 upvotes]: I think it's well known that if $X\subset\mathbb{P}^3$ is a smooth cubic surface and we take the projection $\pi: X\rightarrow \mathbb{P}^2$ from a point off the surface, then it's branched over a sextic curve with 6 cusps. Why is this true? In particular, I'm not seeing the 6 cusps. Example of a statement of the fact: the Problem of Existence of Algebraic Functions by Zariski (1929), pg 320 REPLY [10 votes]: A slightly simpler way: if you project from $(0,0,0,1)$, after a change of coordinates you can write the equation of your surface as $T^3+PT+Q=0$, where $P$ and $Q$ are forms of degree $2$ and $3$ in $X,Y,Z$. The branch curve is given by $4P^3+27Q^2=0$, and it is fairly easy to see that the 6 points given by $P=Q=0$ are cusps.<|endoftext|> TITLE: The Hessian of invariant functions on a Lie group QUESTION [7 upvotes]: Assume that $G$ is a Lie group with Lie algebra $\mathfrak{g}$. We fix an invariant Riemannian metric on $G$ and fix its corresponding $LC$ connection. Consider the natural right action of $G$ on its Lie algebra $\mathfrak{g} \simeq \{X \in \chi^{\infty}({G}) \mid R_{g}^{*} X=X\}$, the space of smooth vector fields which are invariant under right multiplications. In fact the right action is defined as follows: For $g\in G$ and $X\in \mathfrak{g}$ define $X.g=L_{g}^{*} X$ where $L_{g}$ is the left multiplication by $g$. So there is a natural (component wise) action of $G$ on $\mathfrak{g} \times \mathfrak{g}$ A smooth function $f:G \to \mathbb{R}$ is called $G$-invariant if $f(g^{-1}hg)=f(h)$ for all $g,h \in G$. For example $Det: Gl(n, \mathbb{R}) \to \mathbb{R}$. A $2$- linear map $T: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ is $G$-invariant if $T((v,w).g)=T(v,w)$ Example The $2$-linear map $tr(u)tr(v)-tr(uv)$ defined on $M_{n}(\mathbb{R}) \times M_{n}(\mathbb{R})$ is a $Gl(n, \mathbb{R})$ invariant map, with the natural (conjugate) action of $Gl(n, \mathbb{R})$ on its Lie algebra $M_{n}(\mathbb{R})$ as described above. Recall that for a Riemannian manifold with the corresponding $LC$ connection $\nabla$, the Hessian of a function $f$ defined on the manifold, is a two linear map on the tangent space with the formula $Hess (f).(V, W)=\nabla ^ {\nabla f}_{V}.W$ Question: Let $G$ be a Lie group and $f:G \to \mathbb{R}$ be a $G$-invariant smooth function. Is its Hessian $Hess(f)$ a $G$-invariant $2$-linear map on the Lie algebra $\mathfrak{g}$ of $G$? This question is motivated by the following post. Is there an explicit formula for the hessian of "Determinant"? REPLY [7 votes]: Yes. In what follows, I use standard notation for the derivative; see, e.g., $\S$2.3 of Banach Spaces and Differential Calculus (Chapter 2) of the book referenced below. First Derivative of $f$. As the OP stated, a function $f: G \to \mathbb{R}$ is $G$ invariant means that $$ f(A) = f(g A g^{-1}) \quad \forall A, g \in G \tag{$\star$} $$ If we differentiate this relation once we obtain $$ D f(A) \cdot B_1 = \frac{d}{d\lambda} \left. f( g (A+\lambda B_1) g^{-1} ) \right|_{\lambda=0} = D f( g A g^{-1} ) \cdot g B_1 g^{-1} \tag{$\star \star$} $$ where $B_1 \in T_A G$. In the special case of the determinant on $G = Gl(n,\mathbb{R})$, this is simply saying that: $$ D \det(A) \cdot B_1 = \det(A) \operatorname{trace}(A^{-1} B_1) = \det( g A g^{-1} ) \operatorname{trace}(g A^{-1} g^{-1} g B_1 g^{-1} ) $$ which follows from basic properties of the determinant and trace of a matrix. Second Derivative of $f$. Now differentiate ($\star \star$) to obtain: $$ D^2 f(A) \cdot (B_1, B_2) = D^2 f( g A g^{-1} ) \cdot ( g B_1 g^{-1}, g B_2 g^{-1} ) \tag{$\star \star \star$} $$ where $B_1, B_2 \in T_A G$. In the special case of the determinant on $G = Gl(n,\mathbb{R})$, \begin{align*} D^2 \det(A) \cdot (B_1, B_2) &= \frac{d}{d \lambda} \left. D \det(A+\lambda B_2) \cdot B_1 \right|_{\lambda=0} \\ &= \frac{d}{d \lambda} \left. \det (A + \lambda B_2) \operatorname{trace} ( (A+\lambda B_2)^{-1} B_1 ) \right|_{\lambda=0} \\ &= \det(A) \operatorname{trace} (A^{-1} B_2) \operatorname{trace}(A^{-1} B_1) - \det(A) \operatorname{trace}(A^{-1} B_2 A^{-1} B_1 ) \end{align*} which is clearly $G$-invariant in the sense of relation ($\star \star \star$). Higher Derivatives of $f$. This result for the first and second derivative seems to be true for higher derivatives of a $G$-invariant function. In other words, differentiation preserves $G$-invariance in the sense given above. Reference For background info see, e.g., An Introduction to Lie groups Chapter 5 of Jerrold E. Marsden, Tudor Ratiu, and Ralph Abraham. Manifolds, Tensors, and Applications. Second Edition. Vol. 75. Springer Science & Business Media, 2012.<|endoftext|> TITLE: Harmonic map proof of Riemann mapping theorem QUESTION [6 upvotes]: Is there a way to prove the Riemann mapping theorem using the theory of harmonic maps (in the sense of "Harmonic Mappings of Riemannian Manifolds" by Eells and Sampson)? REPLY [11 votes]: Yes, there is a classical proof of the Riemann mapping theorem using harmonic maps and the Dirichlet problem. Riemann's original assumption of boundary smoothness can be removed using Perron's method and a simple argument due to Osgood. For the detailed proof, see this note by Greene and Kim.<|endoftext|> TITLE: Deligne's letter to Millson QUESTION [20 upvotes]: The deformation theoretic principle that any reasonable deformation problem should be governed by a dg-Lie algebra seems to come from a letter of Deligne to Millson. It is clear how the Maurer-Cartan formalism indeed produces a deformation problem and conversely for most deformation problems the dg-Lie algebra is well known. In the derived setting this principle has even become a theorem following Lurie-Pridham. However, I was wondering if anyone could explain the original intuition Deligne had behind this principle? REPLY [8 votes]: I suspect, Deligne's intuition went along the following lines. Deformation theory describes the tangent cone at a point $x$ of the moduli space $M$ of the problem. The tangent cone is Spec Gr $\mathcal{O}_{M,x}$, where the associated graded Gr is taken with respect to powers of the maximal ideal of the local ring $\mathcal{O}_{M,x}$, the stalk of the structure sheaf $\mathcal{O}_{M}$ at $x$. Usually, this cone is singular. You may resolve this singularity within derived algebraic geometry, for instance, find a free dg-commutative algebra whose cohomology is Gr $\mathcal{O}_{M,x}$. A free dg-commutative algebra is equivalent to an $L_\infty$-algebra. Then you take a quasi-isomorphic dg-Lie algebra, and you are done.<|endoftext|> TITLE: Elliptic-curve related equivalence between fields of different characteristic? QUESTION [9 upvotes]: Conjecture: If I have an elliptic curve with j-invariant 0 of the form $y^2 = x^3 + b$ over some prime-order field $\mathbb{F}_p$ (where $p$ is not 2 or 3), and the group of rational curvepoints has prime order $q$ (which is not 2 or 3), then there is some curve of the form $y^2 = x^3 + b'$ over $\mathbb{F}_q$ with prime order $p$. I'm not sure about the characteristic 2-or-3 case, I guess I could check that exhaustively, but it's less interesting to me and I haven't taken the time to do it yet. For larger primes I've found a few dozen cases with small primes, a couple with large primes, and no counterexamples. I have 3 questions: Is this conjecture true? What does it mean? This seems to give an equivalence relation between ground fields of different characteristic, which seems really weird to me... what am I looking at here? (Sorry this is so vague.) What is the natural way to generalize this to curves whose groups of rational points has composite order, and/or fields with non-prime order? REPLY [12 votes]: If $\# E(\mathbb{F}_p) = q$ and $j=0$, then the endomorphism ring is an order in the field of third roots of unity so $(p+1-q)^2 - 4p = -3u^2$ for some integer $u$. Now note that $(p+1-q)^2 - 4p$ is symmetric in $p$ and $q$. Hence, if there is an elliptic curve at all over $\mathbb{F}_q$ with $p$ points, then it automatically has endomorphism ring by an order in the field of third roots of unity and thus has $j=0$ and is of the required form. So the remaining issue is whether $p$ lands in the Hasse interval for $q$. But, let's say $p < q$, then $q < (\sqrt{p}+1)^2$ so $\sqrt{p} > \sqrt{q}-1$ and $p$ is in the Hasse interval for $q$. The case $q3$ have order $p+1$, hence not prime, this is not an issue but can become an issue when considering non-prime fields. Edit: To get $j=0$ the endomorphism ring has to be the maximal order. I think my proof is incomplete but hopefully is OK.<|endoftext|> TITLE: A Cohen real which always amalgamates with others QUESTION [12 upvotes]: I heard the following fact a while back from Joel Hamkins, who told me at the time that he learned it from Hugh Woodin: Observation: Let $M$ be a countable transitive model of $\mathsf{ZFC}$. Then there are forcing extensions $M[c]$ and $M[d]$, each by adding a Cohen real, which are non-amalgable. By this, I mean that if $N \supseteq M$ is a model of $\mathsf{ZFC}$ has the same ordinals as $M$ then $N$ cannot contain both $M[c]$ and $M[d]$. (We don't actually need that $M$ is transitive, but let's throw it in to make things easier.) This is not hard to prove so I'll include a sketch of the argument. Proof sketch: Since $M$ is countable it's coded by some real $m$. We will construct $c$ and $d$ so that together they allow us to build $m$. Hence, any $N \supseteq M[c] \cup M[d]$ with the same ordinals as $M$ will see that all its ordinals are all countable, and hence cannot be a model of $\mathsf{ZFC}$. Enumerate the countably many dense sets for Cohen forcing over $M$ as $D_0, D_1, \ldots$. We build $c$ and $d$ in $\omega$ many stages. Start with $c_0$ which meets $D_0$. Take $d_0$ to be $0$s up to the length of $c_0$ followed by a $1$ followed by $m(0)$, then extend to meet $D_0$. We can then keep going: $c_{n+1}$ extends $c_n$ by $0$s until it reaches the length of $d_n$ followed by a $1$ then extend to meet $D_{n+1}$. Next, $d_{n+1}$ extends $d_n$ with $0$s up to the length of $c_{n+1}$ followed by a $1$ followed by $m(n+1)$. At the end, $c$ and $d$ are the unions of, respectively, the $c_n$s and the $d_n$s. They are generic because they meet every dense set. If we have both $c$ and $d$ in our model of $\mathsf{ZFC}$ then we can reconstruct $m$ by looking at the blocks of $0$s alternating between the two reals and use that to find $m(0)$, then $m(1)$, and so on. ∎ This construction relies upon being able to choose both $c$ and $d$. My question is whether we can get this sort of non-amalgability if we're only allowed to choose one of them. Question: Let $M \models \mathsf{ZFC}$ be countable (and transitive, if we like) and let $c$ be a Cohel real generic over $M$. Is there a Cohen real $d$ generic over $M$ so that $M[c]$ and $M[d]$ are non-amalgable? Phrased in the negative, this question is: Question: Let $M \models \mathsf{ZFC}$ be countable (and transitive). Is there a Cohen real $c$ generic over $M$ so that if $d$ is any Cohen real generic over $M$ then $M[c]$ and $M[d]$ are amalgable? REPLY [11 votes]: Let $c$ be Cohen over $M$. Construct $d$ as follows. Let $X$ be an infinite set of integers all of whose infinite subsets compute the height of $M$. Inductively construct $d$ such that $d$ lies in every open dense set coded in $M$ and the set of positions at which $c, d$ disagree is an infinite subset of $X$.<|endoftext|> TITLE: Non uniquely ergodic interval exchange transformations QUESTION [6 upvotes]: Consider an interval exchange transformation that is, a bijective piecewise continuous map $[0,1] \rightarrow [0,1] $ whose restriction to its continuity intervals are translations. Assume that it is minimal and non uniquely ergodic. Do the ergodic probability measures necessarily give the same weight to each of its continuity intervals? I would naively expect the answer to this question to be negative, but since I have no example to test it Thank you all! Selim Edit As Uri Bader pointed out in the comments, conjugating by a well chosen IET, any minimal non-uniquely ergodic IET provides a negative answer to the question, but the price to pay is to increase the number of continuity intervals involved. The 'good' question concerns then the IETs whose number of interval is minimal in its conjugacy class. REPLY [8 votes]: The space of invariant probability measures of a minimal interval exchange transformation $T$ on $d$ intervals $I_1, \dots, I_d$ is parametrized by a subsimplex $\mathcal{M}$ of the standard simplex $\Delta=\{(x_1,\dots,x_d)\in\mathbb{R}^d_+: \sum x_i=1\}$ in the sense that: 1) a point $(x_1,\dots, x_d)\in\mathcal{M}$ has the form $(x_1,\dots, x_d) = (\mu(I_1),\dots,\mu(I_d))$ for a $T$-invariant probability measure $\mu$. 2) the ergodic $T$-invariant probability measures correspond to the extremal points of $\mathcal{M}$. This fact is explained in Section 4 of Ferenczi's notes (as pointed out by Uri Bader in the comments) and also Yoccoz's 'Pisa lecture notes' (https://www.college-de-france.fr/media/jean-christophe-yoccoz/UPL15305_PisaLecturesJCY2007.pdf). Very roughly speaking, $\mathcal{M}$ is obtained as a decreasing intersection of the successive images of the standard simplex $\Delta$ under the projective actions of the matrices prescribed by repeated iteration of the Rauzy-Veech algorithm starting with $T$. In the case of a minimal non-uniquely ergodic interval exchange transformation $T$, the distinct ergodic invariant probability measures correspond to distinct points in $\Delta$ and, hence, it is always the case that some (actually, all but possibly one) ergodic probability measure does not give the same weight to all (continuity) intervals.<|endoftext|> TITLE: Are the terms of a linear recurrence integral? QUESTION [11 upvotes]: Given rational numbers $a_1,\ldots, a_k$ and $u_0, \ldots, u_k$, let $(u_n)_{n \geq k}$ be the linear recurrence defined by $$u_n := a_1 u_{n-1} + \cdots + a_k u_{n-k}, \text{ for } n \geq k .$$ Obviously, $u_n \in \mathbb{Q}$ for any $n \geq 0$. My question is: Is there an effective way to decide if it is true that $u_n \in \mathbb{Z}$ for all sufficiently large $n$ ? REPLY [13 votes]: The problem is effectively decidable. To test whether $u_n$ is eventually integral, first use the recurrence relation for $u_n$ to construct relatively prime polynomials $A,B\in \mathbb{Z}[x]$ such that the rational function $A/B$ has power series expansion $\sum_nu_nx^n$. (Here relatively prime will always mean no common factor in the ring $\mathbb{Z}[x]$ besides $\pm1$.) Then the following Lemma provides an effective test on $A$ and $B$ that determines whether or not $u_n$ is eventually integral. Lemma. Suppose that $A,B\in \mathbb{Z}[x]$ are relatively prime, and that $A/B$ has power series expansion $\sum_{n\ge0}u_nx^n$. Let $B=cC$, where $c$ is the gcd of the coefficients of $B$. Then the sequence $u_n$ is eventually integral if and only if $B(0)=\pm c$. $A$ is an element of the ideal of $\mathbb{Z}[x]$ generated by $c$ and $C$. Proof of the Lemma. The if direction: we assume that Conditions 1 and 2 hold, and prove that the sequence $u_k$ is eventually integral. By Condition 2, $$A=cD+CE,$$ for some choice of $D,E\in \mathbb{Z}[x].$ Dividing by $B$, and using $B=cC$, $$\tag{*}\dfrac{A}{B}=\dfrac{D}{C}+\dfrac{E}{c}.$$ But Condition 1 implies that $C$ has the form $\pm(1-xC_1)$, for some $C_1\in \mathbb{Z}[x]$. Therefore $D/C$ has the form $$\pm D(1+(xC_1)+(xC_1)^2+\ldots).$$ It follows that the power series for $D/C$ has all integral coefficients. Since $E/c$ is a polynomial with rational coefficients, it follows from ($*$) that the power series $\sum u_kx^k$ for $A/B$ eventually has integral coefficients. The only if direction: We assume that the sequence $u_k$ is eventually integral, and verify Conditions 1 and 2. Remark. At this point it will be convenient to extend the usual notion of the content of a polynomial to power series $f=\sum_nu_nx^n$ with eventually integral coefficients. Define $$\gamma(f)=\prod_{p \text{ prime}}p^{\min_n(v_p(u_n))},$$ where $v_p(u_n)$ is the exponent to which $p$ appears in the rational number $u_n$. If $P\in \mathbb{Z}[x]$ then the product $Pf$ again has eventually integral coefficients, and it holds that $\gamma(Pf)=\gamma(P)\gamma(f)$. The proof is similar to the case of two polynomials. Proof of Condition 1. Since $A$ and $B$ are relatively prime, there are polynomials $U,V\in \mathbb{Z}[x]$ and an integer $m\ne0$ such that $AU+BV=m$. Let $f=\sum_nu_nx^n$ be the power series expansion of $A/B$. By factoring out $B$, write the equation $AU+BV=m$ in the form $$\tag{**}B(fU+V)=m.$$ Since $c=\gamma(B)$, the multiplicativity of the content function $\gamma$ implies that $$c\gamma(fU+V)=m.$$ But ($**$) implies that $B(0)t=m$, where $t$ is the constant term of $fU+V$. Therefore $B(0)t=c\gamma(fU+V)$, or equivalently $$ \dfrac{B(0)}{c}\cdot \dfrac{t}{\gamma(fU+V)}=1.$$ Since the two factors are both integers, it follows that $B(0)=\pm c$. This proves Condition 1. Proof of Condition 2. As before, let $f$ denote the power series for $A/B$. We note that the power series $cf$ has all integer coefficients. Indeed, taking the content of both sides of the equation $A=Bf$, we have $$\gamma(A)=c\gamma(f)=\gamma(cf).$$ The equation implies that $cf$ has integer content. Therefore all coefficients of $cf$ are integers. It follows that $f$ has the form $D/c+g$, for some $D\in \mathbb{Z}[x]$ and some power series $g$ with integer coefficients. Multiplying the equation $A/B=D/c+g$ by $B$ and using the definition $B=cC$, we conclude that $$A=CD+c\cdot(Cg).$$ But $Cg$ is visibly a polynomial, being a difference of two polynomials. Therefore $A$ is an element of the ideal of $\mathbb{Z}[x]$ generated by $c$ and $C$. This completes the proof of the Lemma. Notes. For the connection between generating functions of recurrence sequences and rational functions, see Chapter 4 of Richard Stanley's book Enumerative Combinatorics, v1. Concerning Condition 2 of the Lemma, an algorithm for ideal membership in the ring $\mathbb{Z}[x]$ is given in Chapter 10 of Ideals Varieties and Algorithms by Cox, Little and OShea. But anyway in this case we need only determine whether $C$ divides $A$ in the ring $\mathbb{Z}/c\mathbb{Z}[x]$. The proof of Condition 1 in the only if part of the Lemma is substantially the same as in the solution to Exercise 2a in Chapter 4 of Stanley's book. Condition 1 in the Lemma by itself is equivalent to the requirement that the $u_n$ have bounded denominators, that is, that there is a nonzero integer $M$ such that for all $n$, $Mu_n\in\mathbb{Z}$.<|endoftext|> TITLE: Brauer group of a product of curves QUESTION [5 upvotes]: By a famous theorem of Tate, we know that the Tate conjecture holds for a product of curves over a finite field. But this implies that the Brauer group of a product of curves (over finite field) is finite. What do we know about this Brauer group apart from the fact that it's finite ? Does there exist some computation done, for example for $\mathbb{P}^1 \times \mathbb{P}^1$ (or any other simple example) ? REPLY [2 votes]: Suppose that $k$ is a finite field with $q$ elements. Let $C_i$, $i = 1, 2$ be two smooth projective geometrically irreducible curves over $k$. Set $X = C_1 \times_{\text{Spec}(k)} C_2$. In principle we know how to compute the Brauer group of $X$ in terms of the action of Frobenii on the \'etale cohomology of $C_1$ and $C_2$. Let's assume there is no nontrivial isogeny from the Jacobian of $C_1$ to the Jacobian of $C_2$, or equivalently that the characteristic polynomials $P_1, P_2$ of the Frobenius endomorphisms of these Jacobians are relatively prime. This means that the Picard group of $X$ is generated by pullbacks of invertible sheaves from $C_1$ and $C_2$. Thus for $n$ prime to $q$ the exact sequence $$ \text{Pic}(X) \to H^2_{\acute{e}t}(C_1 \times C_2, \mu_n) \to Br(X)[n] \to 0 $$ coming from the Kummer sequence and cohomology is (in principle) computable. Working through what is known about \'etale cohomology of curves over finite fields you'll get $$ \left( H^1_{\acute{e}t}(C_{1, \overline{k}}, \mathbf{Z}/n\mathbf{Z}) \otimes H^1_{\acute{e}t}(C_{2, \overline{k}}, \mathbf{Z}/n\mathbf{Z}) \otimes \mu_n \right)^{\text{Gal}(\overline{k}/k)} $$ for the $n$ torsion in the Brauer group of $X$. If there are nonzero isogenies between the Jacobians, then you'll just get a quotient of this. If $n$ is not coprime to $q$, then you'll have to look at papers of Milne to see what happens. Now it is clear that you can certainly get lot's of Brauer classes. In fact, you can show from the above that if $P_1$ and $P_2$ have a factor in common modulo $\ell$, then there'll be an $\ell$-torsion class in $Br(X)$ for $\ell$ a prime not dividing $q$. And given a situation as above and an $\ell$, you can always find a finite extension $k'$ of $k$ such that this thing happens over $k'$. Enjoy!<|endoftext|> TITLE: Positivity of a finite sum involving Stirling numbers QUESTION [15 upvotes]: In my research in theoretical physics, I have arrived at some coefficients $a_{n,m}$ depending on two integers, $n\geq 1$ and $0\leq m\leq n$: $$ a_{n,m}=\sum_{j=0}^{n-1} {2j \choose j+m} \left(\frac{n}{4}\right)^j s(n,j+1) $$ where $s(n,j+1)$ are Stirling numbers of the first kind. Although this expression does not make it manifest (to me), these coefficients are zero when $n+m$ is even. For physical reasons, I am convinced that when $m+n$ is odd, $a_{n,m}>0$, but I haven't been able to prove it. I would like to find a proof that $a_{n,m}\geq 0$. I have found various ways to rewrite these coefficients. For instance, in terms of power series involving modified Bessel functions of the first kind $I_m$: $$ a_{n,m}= \left. \frac{d^{n-1} \,\left( (1-z)^{\frac{n-2}{2}} \, I_m\left(-\frac{n}{2} \log (1-z)\right) \right)}{d z^{n-1}} \right|_{z=0} $$ but I am not able to conclude that $a_{n,m}\geq 0$ from this expression either. Alternatively, when $n$ is a multiple of 4, these coefficients are integers, so I was hoping that there might be a combinatorial argument for this particular case. However, I have been unable to produce it. Any help would be greatly appreciated. REPLY [16 votes]: The numbers $a_{n,m}$ are in fact the Fourier coefficients of the polynomial $$P_n(x)=\prod_{j=1}^{n-1} \Big( \frac{nx}{2} + \frac{n}{2}-j\Big) $$ with respect to the Chebyshev measure $d\sigma:=(1-x^2)^{-1/2}dx$ on $[-1,1]$, and its orthogonal bases of the Chebyshev polynomials of the first kind. Precisely, for $0\le m\le n$ $$a_{n,m}=\frac{1}{\pi} \int_{-1}^1 P_n(x)T_m(x)d\sigma \ .$$ Changing variable, we have a trigonometric version: $$a_{n,m}=\frac{1}{\pi}\int_0^\pi P_n(\cos \theta)\cos (m\theta)d\theta \ .$$ Note that the polynomials $P_n$ and $T_m$ are odd resp. even, according to the parity of $n-1$, respectively $m$, so the integrand $P_n(x)T_m(x)$ has the same parity of $n+m-1$. On the other hand, the Chebyshev measure is symmetric, which explains the vanishing property $a_{n,m}=0$ whenever $n+m$ is even. Moreover, for odd $n+m$ the integrand is positive and concentrated about $\pm 1$; this should hopefully yield to the desired estimate $a_{n,m}>0$. I'll try some computation and in case add details later. To compute the integral we may also use the Chebyshev-Gauss Quadrature formula on $N$ nodes, which is exact on polynomials of degree less than $2N$. Therefore, for $2N\ge n+m$ we have $$a_{n,m}=\frac{1}{N}\sum_{k=1}^{N} P_n\Big(\cos\big( \frac{2k-1}{2N}\pi \big) \Big)\ \cos\big( m\frac{2k-1}{2N}\pi\big) \ .$$<|endoftext|> TITLE: Constructive proof that a kernel consists of nilpotent elements QUESTION [17 upvotes]: I am interested in the following innocent looking statement: Let $A \leftarrow R \rightarrow B$ be two homomorphisms of commutative rings. Assume that their kernels consist of nilpotent elements. Then, the kernel of $R \to A \otimes_R B$ consists of nilpotent elements, too. Geometrically, this means that if $X \to S$ and $Y \to S$ are two quasi-compact morphisms of schemes with dense image, then the image of $X \times_S Y \to S$ is dense, too. We can prove this as follows (using the axiom of choice many times): It suffices to show that the kernel of $R \to A \otimes_R B$ is contained in any minimal prime ideal $P \subseteq R$. Since the kernel of $R \to A$ is contained in $P$, we have $A_P \neq 0$, and likewise $B_P \neq 0$. Choose prime ideals in these rings. Their preimages are prime ideals $I \subseteq A$, $J \subseteq B$ such that $I \cap R = P$ (since $I \cap R \subseteq P$ and $P$ is minimal) and $J \cap R = P$. The kernel of $R \to A \otimes_R B$ is contained in the kernel of $R \to Q(A/I) \otimes_{Q(R/P)} Q(B/J)$, which equals the kernel of $R \to Q(R/P)$, i.e. $P$, since $Q(R/P) \to Q(A/I) \otimes_{Q(R/P)} Q(B/J)$ is injective. In fact, one can prove this statement in $\mathsf{ZF}$, i.e. the axiom of choice is not necessary. The trick is to use filtered colimits (or the explicit construction of the tensor product) to reduce to the case that $R$ is of finite type over $\mathbf{Z}$, hence countable, and to the case that $A,B$ are finite type over $R$, hence countable too. In Kostas Hatzikiriakou, "Minimal Prime Ideals and Arithmetic Comprehension", it is proven (in a fragment of $\mathsf{ZF}$) that every non-trivial countable commutative has a minimal prime ideal. Moreover, $\mathsf{ZF}$ proves that vector spaces are flat: Again, using filtered colimits, it suffices to prove this for finitely generated vector spaces, but these are free and hence flat. This proves in particular that the tensor product of two non-trivial vectors is a non-trivial vector. In particular, $Q(R/P) \to Q(A/I) \otimes_{Q(R/P)} Q(B/J)$ is injective. The proof is still not very satisfying. We have a rather elementary statement about tensor products of algebras, do we really need prime ideals? Is it possible to simplify the proof further? Also notice that the proof above is not constructive, i.e., it uses the law of the excluded middle. Question. Is there a constructive proof of the statement? If yes, how does it look like? Notice that we may assume that $R,A,B$ are reduced. In that case, the question becomes: If $R \to A$ and $R \to B$ are injective, why is $R \to A \otimes_R B$ injective, too? The special case that $R$ is a field was already discussed above, and this part was constructive, I guess. I have recently asked a similar yet broader question. I am not sure how to handle minimal prime ideals here. There is just a very short chapter about minimal prime ideals in the book by Lombardi and Quitté on constructive commutative algebra, and they "only" apply this to give a constructive proof of Traverso-Swan's theorem characterizing seminormal rings. But I would prefer a constructive proof of the statement which is really down-to-earth and can be therefore understood without any prerequisites on constructive mathematics (like the proof for the case that $R$ is a field). Added 1. Using localization at some element in the kernel, it suffices to prove $A \otimes_R B = 0 \Rightarrow R = 0$ when $R \to A$, $R \to B$ are as above. Therefore, a related statement is the following: If $M,N$ are finitely generated $R$-modules, then it is known $\sqrt{\mathrm{Ann}(M \otimes_R N)} = \sqrt{\mathrm{Ann}(M)+\mathrm{Ann}(N)};$ this follows from $\mathrm{supp}(M \otimes_R N) = \mathrm{supp}(M) \cap \mathrm{supp}(N)$ as subsets of $\mathrm{Spec}(R)$. In particular, $\mathrm{Ann}(M) \subseteq \sqrt{0}$ and $\mathrm{Ann}(N) \subseteq \sqrt{0}$ imply $\mathrm{Ann}(M \otimes_R N) \subseteq \sqrt{0}$, thus $M \otimes_R N = 0 \Rightarrow R =0$. Again, I don't know a constructive proof. (But this probably won't help here since $A,B$ may just be assumed to be finitely generated $R$-algebras, not finitely generated $R$-modules). Added 2. Here is a first simplification: We may assume that $R$ is reduced. Let $P$ be a minimal prime ideal of $R$. Then $R_P$ is a reduced zero-dimensional local ring, i.e. a field. The $R_P$-algebras $A_P$ and $B_P$ are non-trivial, hence $R_P \to A_P \otimes_{R_P} B_P$ is injective. It follows that the kernel of $R \to A \otimes_R B$ is contained in the kernel of $R \to R_P$, which is $P$. Now, Lombardi and Quitté write in their book in section XIII.7: "It is a fact that the use of minimal prime ideals in a proof of classical mathematics can in general be made innocuous (i.e. constructive) by using $A_{\mathrm{min}}$". Here, $A_{\mathrm{min}}$ denots a rather peculiar looking commutative ring constructed in Theorem 7.8. Does this mean that we can somehow prove that $R_{\mathrm{min}} \to A_{\mathrm{min}} \otimes_{R_{\mathrm{min}}} B_{\mathrm{min}}$ is injective? This would suffice since $R \to R_{\mathrm{min}}$ is injective when $R$ is reduced. Added 3. Here is another proof, which seems to be more suitable for constructivization. We may assume that $A,B$ are of finite type over $R$ with $A \otimes_R B = 0$ and that $R$ is reduced, the goal is $R=0$. By generic freeness, there is some open dense subset $U \subseteq \mathrm{Spec}(R)$ such that $A_f$ is free over $R_f$ (hence, flat) for every $D(f) \subseteq U$. It follows that $A_f = A_f \otimes_{R_f} R_f$ injects into $A_f \otimes_{R_f} B_f = 0$, i.e. $A_f=0$. Since $R \to A$ is injective, this means $f=0$. This shows $U=\emptyset$, and therefore $R=0$. Theorem 2.45 in "Computational Methods in Commutative Algebra and Algebraic Geometry" by Wolmer Vasconcelos seems to be a constructive proof of Generic Flatness at least for Noetherian domains $R$. So we would have to generalize this to reduced commutative rings $R$, and avoid the usage of the prime spectrum. REPLY [3 votes]: Let us use $k_\min$. The references we will give are in the book Commutative Algebra. Constructive methods (Lombardi-Quitté) (arXiv:1605.04832v1). We have to prove the following. Let $k \to A$ and $k\to B$ be two injective morphisms of commutative reduced rings, if $A\otimes_k B=0$ then $k=0$. 1) This is true when $k$ is a discrete field or, more generally, a zerodimensional reduced ring (von Neuman regular). Indeed $k\to A$ is faithfully flat (VIII-6.2) and $k \to B$ is injective, so $ A \to A \otimes_k B$ is injective, thus $A=0$, thus $k=0$ 2) This is true when $k$ is a pp-ring (IV-6). Indeed, let $S$ be the monoid of regular elements, $k_S$ is von Neuman regular, and $k_S\to A_S$ remains injective: if $x \in k$ and $x.1_{A_S} =_{A_S} 0$ we have an $u \in S$ such that $ u.x.1_A =_A 0$, thus $ux=_k0$, thus $x=0$ in $k_S$. Similarly $k_S \to B_S$ is injective. If $A\otimes_k B=0$ then $A_S\otimes_{k_S}B_S\simeq k_S \otimes_k (A \otimes_k B)=0$ thus $k_S = 0$ by Item 1, thus $k = 0$. 3) General case. We use $k_\min$. Since $k_\min$ is a pp-ring and $k \to k_\min$ injective ($k$ is reduced), it is sufficient to prove that $k_\min \to k_\min \otimes_k A$ is injective (same thing for $k_\min \to k_\min \otimes_k B$). Indeed letting $A'=k_\min \otimes_k A$ and $B'=k_\min \otimes_k B$, we have $A'\otimes_{k_\min} B'\simeq k_\min \otimes_k(A\otimes_k B)$, so $A'\otimes_{k_\min} B'=0$ which implies $k_\min=0$ by Item 2, which implies $k=0$. Viewing the construction of $k_\min$, and using the notation of the book it is sufficient to show that $k_{\{u\}} \to k_{\{u\}} \otimes_k A$ is injective for any $u \in k$. We write $I^*=(0:I)_k=\{x\in k\,;\, xI=0\}$ for an ideal $I$ of $k$ and $u^*$ for $(u)^*$. By definition $k_{\{u\}} = k/u^* \times k/(u^*)^*$, thus $k_{\{u\}} \otimes_k A = A/(u^*A) \times A/((u^*)^*A)$. Let $x \in k_{\{u\}}$, $x = ( y \mod u^*, z \mod (u^*)^* )$, $y,z \in k$. Let $y'=y1_A$ and $z'=z1_A$ in $A$. We assume $y' \mod u^*A = 0$, $z' \mod ((u^*)^*)A = 0$ and we have to show that $y \mod u^* = 0$, $z \mod (u^*)^* = 0$. $y'=\sum y_i$ where $y_i \in u^* A$. We have $y_i=\sum a_{ij} y_{ij}$ with $a_{ij}\in k$, $ua_{ij} =_k0$, $y_{ij}\in A$. Thus $u y_i =_A 0$, thus $uy'=_A0$. So $u y1_A=_A0$, so $uy =_k 0$, so $y\in u^*$, i.e. $y\mod u^*=0$. $z'=\sum z_i$ where $z_i \in (u^*)^* A$. We have $z_i=\sum b_{ij} z_{ij}$ with $b_{ij} \in (u^*)^*$. Let $v\in u^*$. We get $v b_{ij} = 0$, $v z_i = 0$, $ v z' = 0$ and $vz = 0$. So, for any $v \in u^*$ we have $vz = 0$, thus $z \in (u^*)^*$ i.e. $z \mod (u^*)^* =0$ – Henri Lombardi<|endoftext|> TITLE: The number $\pi$ and summation by $SL(2,\mathbb Z)$ QUESTION [114 upvotes]: Let $f(a,b,c,d)=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}-\sqrt{(a+c)^2+(b+d)^2}$. (it is the defect in the triangle inequality) Then, we discovered by heuristic arguments and then verified by computer that $$\sum f(a,b,c,d)^n = 2-\pi/2$$ where the sum runs over all $a,b,c,d\in\mathbb Z$ such that $a\geq 1,b,c\geq 0, ad-bc=1$ and $n=2$. It seems that when $n=1$, we obtain $2$ in the right hand side. We have failed to guess the result for $n>2$. So the question is: can you prove the result for $n=2$? (We can, but in a rather unnatural way. We will write this later and now we want to tease the community, probably somebody can find a beautiful proof.) Added: we have two answers for the above question, so the rest is: Question: Can you guess the result for $n>2$? I tried http://mrob.com/pub/ries/ but nothing interesting was revealed. PS. In case it can help someone, below are these sums of powers ($n=1,2,3,4,5$), calculated by computer: $1.9955289122768913 = 2$ $0.4292036731309361 = 2 - \pi/2$ $0.21349025954227965 = $ ? $0.11983665032283052 = $ ? $0.06933955916793563 = $ ? Added: this and something more can be found in https://arxiv.org/abs/1701.07584 and https://arxiv.org/abs/1711.02089 REPLY [26 votes]: I add our explanation and the origin of the problem. To obtain the formula we just need to verify the following lemma (by a straightforward computation). Magic Lemma. Let $(a,b),(c,d)$ be as in our summation in the main posting. Draw tangents to the unit circle which are orthogonal to the directions $(a,b),(c,d),(a+c,b+d)$. Then the triangle in the intersection has the area $\frac{f(a,b,c,d)^2}{2}$. Now consider the circle given by $x^2+y^2=1$. At every rational point on it we draw the tangent line. Clearly, our disk is the intersection of all the half-planes given by these lines. So we can approximate the area of the disk by the following sequential cutting. We start with the square $[-1,1]\times [-1,1]$. It has four points of tangency with our circle. We will construct a sequence of polygons converging to the circle. Initially we have sides of directions orthogonal to $(1,0),(0,1),(-1,0),(0,1)$. At every step, we take two vectors $v_1,v_2$(in the same quadrant) which give the basis of $\mathbb Z^2$ and add to our polygon a new side, which is given by the tangent line of the direction orthogonal to $v_1+v_2$. For example, the first step will be cutting the square by the line $x+y=\sqrt 2$, if we take the vectors $(1,0),(0,1)$. The area of the eliminated triangle is $\frac{(\sqrt2-1)^2}{2}=\frac{f(1,0,0,1)^2}{2}$. It seems that this procedure can be generalized to higher dimensions — at least a preliminary computation shows that something like the above magic lemma should happen.<|endoftext|> TITLE: Is it always possible to write a scheme as a colimit of affine schemes? QUESTION [18 upvotes]: My question is: Is it possible to write any scheme as a (1-categorical) colimit of a diagram of affines? If no, what are some examples? I ask this question because I have read that one can write any derived scheme as a colimit (in the $\infty$-categorical sense) over a diagram of affine derived schemes. So I am curious what is the analogous statement is in the classical setting. REPLY [23 votes]: Yes, this is just a basic fact in category theory, if interpreted correctly. For $C$ any category, and $F$ any preheaf on $C,$ $F$ is the colimit in presheaves of the diagram $C/F \to C \stackrel{y}{\hookrightarrow} Psh(C),$ which sends a morphism $f:y(C) \to F,$ to $y(C),$ where $y$ is the Yoneda embedding. This follows immediately from the Yoneda lemma. Now, if $F$ is a sheaf for some Grothendiek topology, then since the sheafification functor $a$ is a left adjoint, it preserves all colimits, so we also have that $F$ is the colimit of the diagram $C/F \to C \stackrel{y}{\hookrightarrow} Sh(C).$ Note that this diagram consists entirely of representables (provided the Grothendieck topology is subcanonical, i.e. each representable is a sheaf). Now, take $C$ to the the category of affine schemes, and let the Grothendieck topology be the Zariski topology. The functor of points of any scheme, in particular, is a sheaf. The result now follows. REPLY [10 votes]: Actually, every scheme is the canonical colimit of all affine schemes mapping into it: $$X = \underset{\substack{U \to X\\U \text{ affine}}}{\mathrm{colim}} U$$ In order to avoid size issues, we can in fact restrict ourselves to open affine subschemes $U \hookrightarrow X$ here. (So the diagram scheme doesn't really depend on the choice of an open affine cover of $X$. But the proof, that $X$ is the colimit, clearly requires such a choice.) The statement above is equivalent to the statement that the functor of points $\mathsf{Sch} \to [\mathsf{CRing},\mathsf{Set}]$ is fully faithful. In functorial algebraic geometry, one defines $\mathsf{Sch}$ as a certain full subcategory of $[\mathsf{CRing},\mathsf{Set}]$, so in that setting there would be nothing to prove.<|endoftext|> TITLE: Tell me something about these "component tensor" TQFT's QUESTION [5 upvotes]: I noticed that there is a class of TQFT's that exists for every dimension $n\geq1$. It's probably well-known because it's quite simple, but I'm looking for a standard name or a better way to think about it. Let $n\text{-}\mathrm{Cob}$ denote the symmetric monoidal category of (unoriented) closed $(n-1)$-manifolds $M,N$ and cobordisms $M\sqcup N=\partial B\subseteq B\;$ between them. Let $\mathrm{Cospan}_{\mathrm{Fin}}$ denote the symmetric monoidal category of finite sets $X,Y$ and cospans between them $X\to C\to Y$. In both case the monoidal product is given by disjoint union. Consider the monoidal functor $$\pi_0: n\text{-}\mathrm{Cob}\to\mathrm{Cospan}_{\mathrm{Fin}}.$$ where $\pi_0(M)$ is the set of connected components in $M$. On morphisms it is given by $$\pi_0(M\to B\leftarrow N)=(\pi_0M\to\pi_0B\leftarrow\pi_0N).$$ For every field $k$ and natural number $r$, there's a strong monoidal functor $$\mathrm{tens}_r: \mathrm{Cospan}_{\mathrm{Fin}} \to \mathrm{Vect}_k$$ that sends a finite set $X$ to the $r^X$-dimensional vector space of order-$X$ tensors $T^{r,\ldots,r}$. For example, if $X=\{1,2\}$ then $\mathrm{tens}_r(X)\cong\mathrm{Mat}_{r\times r}\;\;$. The assignment $\mathrm{tens}_r$ is functorial via tensor multiplication, a straightforward generalization of matrix multiplication. The functor $\mathrm{tens}_r$ is strong monoidal because there is a natural isomorphism $$\mathrm{tens}_r(n)\otimes\mathrm{tens}_r(n')\cong\mathrm{tens}_r(n+n').$$ The composite $\mathrm{tens}_r\circ\pi_0\colon n\text{-}\mathrm{Cob}\to\mathrm{Vect}_k\;\;$ is a strong monoidal functor for any $r,n\in\mathbb{N}$. Questions: Is there a standard name for this sort of TQFT? Does it have any universal or other nice properties? Is it degenerate in any particular way? REPLY [5 votes]: The theory you describe is Dijkgraaf-Witten theory with target space a discrete set with $r$ elements. In general, if $X$ is a $\pi$-finite space (i.e., a space with finite homotopy groups, all but finitely many of which are non-trivial), then Dijkgraaf-Witten theory with target space $X$ is the functor $n\text{-}\mathrm{Cob} \to \mathrm{Vect}_k$ which associates to an $(n-1)$-manifold $M$ the vector space of functions $\pi_0\mathrm{Map}(M,X) \to k$, where $\mathrm{Map}(M,X)$ is the mapping space from $M$ to $X$ (here the assumption that $X$ is $\pi$-finite insures this mapping space is $\pi$-finite, and hence has only finitely many connected components). If $W$ is a cobordism from $M$ to $N$ and $\iota_M: M \hookrightarrow W$, $\iota_N: N \hookrightarrow W$ are the inclusions of $M,N$ in the boundary of $W$, then the associated linear map $$ T_W: k^{\pi_0\mathrm{Map}(M,X)} \to k^{\pi_0\mathrm{Map}(N,X)} $$ is defined as follows: for a function $f: \pi_0\mathrm{Map}(M,X) \to k$ define $T_W(f): \pi_0(N,X) \to k$ by the formula $$ T_W(f)(C) = \sum_{D \in \pi_0\mathrm{Map}(W,X) | \iota_N^*D = C}f(\iota_M^*D) $$ for $C \in \pi_0\mathrm{Map}(N,X)$, where $\iota_M^*: \pi_0\mathrm{Map}(W,X) \to \pi_0\mathrm{Map}(M,X)$ and $\iota_N^*: \pi_0\mathrm{Map}(W,X) \to \pi_0\mathrm{Map}(N,X)$ are the maps induced by pre-composition with $\iota_M$ and $\iota_N$ respectively. As explained in section 3 of this paper of Freed, Hopkins, Lurie and Teleman, this topological field theory should actually extend all the way down to $0$-manifolds, yielding a fully extended topological field theory. This can definitely be considered as a nice property. I'm not sure about any universal properties, but this theory is definitely related to a lot of interesting mathematics. As for degeneracy, the only form of degeneracy that comes to my mind is that the value associated to $M$ depends only on the homotopy type of $M$, so in terms of producing manifold invariant it cannot go beyond homotopy type. Note that when $X$ is a discrete set of size $r$ then $\pi_0\mathrm{Map}(M,X) = X^{\pi_0(M)}$ is a set of size $r^{\pi_0(M)}$ and the associated vector space is the space of tensors of order $\pi_0(M)$ and degree $r$, which is the case described in the question. Another case of special interest is when $X = BG$ is the classifying space of a finite group.<|endoftext|> TITLE: subgroups of $p$-groups and elementary $p$-groups QUESTION [6 upvotes]: Let $G$ be a $p$-group and $A$ an elementary $p$-group. Suppose that $\sharp G=\sharp A$, where $\sharp (-)$ denotes the order of $(-)$. Write $C_{d}(G)$ (resp. $C_{d}(A)$) for the cardinality of the set of subgroups of $G$ (resp. $A$) whose order are equal to $d$. My question is that $C_{d}(G) \leq C_{d}(A)$? REPLY [8 votes]: Yes. Moreover, assuming that $d$ is a power of $p$ and $p TITLE: Does the "propositions-as-types" paradigm match mathematical practice? QUESTION [10 upvotes]: Does the "propositions-as-type" paradigm matches mathematical practice? Or is logic in mathematical practice a layer beneath mathematics, and mathematics built on top? For example, should we distinguish between the meta-"or" (which we use when formulating theorems) and the function $\mathrm{Bool}\times\mathrm{Bool}\to\mathrm{Bool}$ on $\mathrm{Bool}:=\{0, 1\}$ that maps $(p, q)$ to $\mathrm{max}(p, q)$? I think that in classical foundations such as FOL+ZFC, logic is a separated layer, but in modern foundations of mathematics such as type theory, the logical connectives which we are use when formulating theorems are themselves mathematical objects. Is that true? REPLY [25 votes]: There are many aspects to the question "does a logical formalism reflect mathematical practice?" I will focus just on a very simple but important detail that every mathematician is familiar with. In mathematical practice we differentiate between $\phi$ or $\psi$, and we know which one, and $\phi$ or $\psi$, but we may not know which one. We also differentiate between there is a given $x$ such that $\theta(x)$, and there is $x$ such that $\theta(x)$, but we may not be given one. Let me call the first kind the concrete disjunction an existential, and the second kind the abstract disjunction and existential. (There is no established terminology.) Thus, "concretely $\exists x \,.\, \theta(x)$" is meant to convey that I have a particular $a$ such that $\theta(a)$, while "abstractly $\exists x \,.\, \theta(x)$" is meant to convey that we know there is an individual satisfying $\theta$, but we may not have a specific one. First-order logic formalizes the abstract version, because the inference rule for existentials forgets the witness $a$: $$\frac{\phi(a)}{\exists x \,.\, \theta(x)},$$ Martin-Löf type theory formalizes the concrete version because the witness $a$ is recorded in the proof term: $$\frac{p : \theta(a)}{(a,p) : \sum_{x : A} \theta(x)}.$$ A formalism which captures both is Martin-Löf type theory with propositional truncation. This is an operation which hides witnesses of statements. If you are familiar with type theory then you can look it up in homotopy type theory. If you are an ordinary mathematician, then it can be described as a quotient: given a type $A$, its propositional truncation $|A|$ is the quotient $A/{\sim}$ by the trivial equivalence relation $\sim$ which relates every $x$ and $y$ in $A$. Thus, if $A$ has an element then $|A|$ has one element. If $A$ is empty, then $|A|$ is empty as well. With propositional truncation we can get all four variants: concrete disjunction is $\phi + \psi$ abstract disjunction is $|\phi + \psi|$ concrete existential is $\sum_{x : A} \theta(x)$ abstract existential is $|\sum_{x : A} \theta(x)|$ Propositional truncation can be defined in homotopy type theory as a higher inductive type. Thus, homotopy type theory reflects mathematical practice (in one respect) better than logic and Martin-Löf's propositions-as-types. You are asking about "layers" of logic and mathematics, so let me address this as well. In mathematical practice we prove statements and perform constructions (an early reference of such activity would be Euclid's Elements). Moreover, deduction and construction steps are intertwined. During a proof we often consturct auxiliary objects, and a complicated construction may require justification of certain steps. (For example, while constructing a solution of a differential equation, we may have to argue that a certain sequence converges.) Traditional foundations layer a term language at the bottom and first-order logic on top. This only captures the case of a logical deduction during which we perform axuliary constructions (expressed as terms of the first-order language) that require no further justifications. However, in practice we need more than that. For example, in the theory of fields it is natural to think of inverse as an operation. Every time we write $x^{-1}$ we need to argue that $x \neq 0$, so inverse is a construction which requires justification. If you look at how the theory of fields is formalized, you will see that it is unsatisfactory: either inverses are treated abstractly as existentials (instead of having the operation $x \mapsto x^{-1}$ we only state $\forall x \,.\, x \neq 0 \Rightarrow \exists y \,.\, x y = 1$), or the justification of $x \neq 0$ is ignored and the operation $x \mapsto x^{-1}$ is made everywhere defined by fiat by stipulating that $0^{-1} = 0$ or some such non-sense. There are real-world consequences of the failure of logic to correctly capture the multi-layered nature of constructions and deductions: mathematicians often formulate constructions inappropriately as abstract existential statements (as there is no other kind in first-order logic). They say Theorem 4.2: There exists $x$ such that $\theta(x)$. Proof. (Construction of $x$ is given here.) QED and later refer to Theorem 4.2 as if it were a construction. It is as if they were ashamed of saying Problem 4.2: Give $x$ such that $\theta(x)$. Construction. (Construction of $x$ is given here.) DEFINED We should all go back to reading Euclid. In contrast, Martin-Löf type theory is a theory of constructions, so there the problem arises when we want to state something without giving a specific construction. Once again, homotopy type theory comes to the rescue with the idea of propositional truncation. And we can precisely explain which bits of a statement are to be read as constructions and which ones as abstract existence. An example may help here: $\prod_{n : \mathbb{N}} \sum_{b : \{0,1\}} \sum_{k : \mathbb{N}} n = 2 k + b$ means "a construction which decomposes a number into its least significant bit and the rest of the number". $\prod_{n : \mathbb{N}} \sum_{b : \{0,1\}} \big|\sum_{k : \mathbb{N}} n = 2 k + b\big|$ means "a construction which calculates the least significant bit of a number". $\prod_{n : \mathbb{N}} \big|\sum_{b : \{0,1\}} \sum_{k : \mathbb{N}} n = 2 k + b\big|$ means "every number is (abstractly) even or odd".<|endoftext|> TITLE: Formal theory of (some) generating functions in $t$ and $t^{-1}$? QUESTION [6 upvotes]: I am interested in using series of the form $\sum_{n=-\infty}^{\infty} a_nt^n$ (where $a_n\in\mathbb C$) as generating functions. In general, multiplication of such series goes against the "formal power series" philosophy since the coefficients of the product are themselves infinite sums. However, the results are sometimes intelligible. Here is an example: Let $G(t) = \frac 1{2-t} = \frac 12 + \frac 14t + \frac 18t^2 + \cdots$ be the usual probability-generating function for a random variable with Geometric($\frac 12$) distribution. If we have two independent Geometric($\frac 12$) random variables, $X$ and $Y$, then we can correctly read the distribution of $X-Y$ from the expansion of $G(t)\cdot G(t^{-1})$: $$G(t)\cdot G(t^{-1}) = \cdots + \frac 1{12}t^{-2} + \frac 16t^{-1} + \frac 13 + \frac 16t + \frac 1{12}t^2 + \cdots$$ Although each coefficient is an infinite sum, those sums are absolutely convergent and so there is no doubt about how to evaluate them. We can go further with this example. Suppose we write \begin{align*} G(t)\cdot G(t^{-1}) &= \frac 1{2-t}\cdot\frac 1{2-\frac 1t} \\ &= \frac 1{5-2t-\frac 2t} \\ &= \frac 15\cdot\frac 1{1-\frac 25(t+\frac 1t)} \\ &= \frac 15\sum_{n=0}^\infty \left[\frac 25\left(t+\frac 1t\right)\right]^n \end{align*} and attempt to extract the coefficients. Then the results appear to be accurate; for example, the constant term is $\frac 15\sum_{n=0}^\infty \binom{2n}n\left(\frac 25\right)^{2n}$, which does equal $\frac 13$. But this is a delicate game: if we multiply by $\frac tt$ at some point in the computation above, and try to expand $\frac t{5t-2t^2-2}$ as an ordinary power series, the results are (obviously) different. Here's my question: Is there a theory that formalizes useful computations with series in $\mathbb C[[t,t^{-1}]]$ while excluding contradictory computations (much like the usual theory of formal power series does for computations in $\mathbb C[[t]]$)? References gratefully accepted. REPLY [5 votes]: There is a natural space in which such operations can be carried out; however, necessarily, its definition is functional-analytic rather than algebraic. Let $H=L^2(S^1)$. Any $f\in H$ has a unique Fourier expansions of the form $$ f(\theta)=\sum_{n\in\Bbb{Z}}a_n t^n, \quad \theta\in\Bbb{R}/2\pi\Bbb{Z}, \quad t=e^{i\theta}\tag1 $$ whose coefficients satisfy $$ \sum_{n\in\Bbb{Z}}|a_n|^2<\infty.\tag2 $$ Conversely, every formal expansion of the form (1) whose coefficients satisfy (2) defines $f\in H$. In fact, $H$ is a commutative Banach algebra and it is a standard fact that addition and multiplication can be carried out at the level of formal expansions (1). Thus for $$ g(\theta)=\sum_{n\in\Bbb{Z}}b_n t^n,\tag3 $$ the expansion of the product $fg$ is given by $$ f(\theta)g(\theta)=\sum_{n\in\Bbb{Z}}\left(\sum_{k\in\Bbb{Z}}a_k b_{n-k}\right)t^n,\tag4 $$ where each coefficient in the expansion has a well-defined complex value due to the absolute convergence of the corresponding series (the latter property follows from the Cauchy-Schwartz inequality). Clearly, the multiplication is associative, due to the corresponding property in $H=L^2(S^1)$ and uniqueness of the Fourier expansion. On the other hand, let us start with a general expansion of the form $(1)$ and impose a fairly mild requirement that the constant term of the product $$ \left(\sum_{n\in\Bbb{Z}}a_n t^n\right)\left(\sum_{n\in\Bbb{Z}}a_n t^{-n}\right),\tag5 $$ where the second factor is obtained by substituting $t^{-1}$ for $t$ in $(1)$, be given by an absolutely convergent series. Using $(4)$ with $b_n=a_{-n}$, one easily sees that this requirement translates precisely to condition $(2)$ on the coefficients. So if we are interested in the products of type $(5)$, as in the example in the original posting, we might as well restrict attention to the elements of $H$ (i.e. impose condition $(2)$ on the coefficients) from the get-go. This indicates that $H$ is indeed a natural realm for usual aritmetic operations with such doubly-infinite expansions. To summarize, we found a natural space $H$ where addition and multiplication of formal expansions of type $(1)$ and $(3)$ are well-defined. The definition of $H$ involves analytic condition $(2)$. The resulting object is a commutative Banach algebra.<|endoftext|> TITLE: Is being an NDR a local property? QUESTION [6 upvotes]: I've asked this on MathSE without success: https://math.stackexchange.com/questions/1929559/is-being-an-ndr-a-local-property A pair of topological spaces $(X,A)$ is an NDR (neighborhood deformation retract) pair if there are continuous maps $u:X\to I$ and $h:X\times I\to X$ satisfying, (1) $u^{-1}(0)=A$, (2) For all $x\in X$, $h(x,0)=x$, (3) For all $a\in A$ and $t\in I$, $h(a,t)=a$, (4) If $u(x)<1$, then $h(x,1)\in A$. By exercise I.E.6 in Spanier, a theorem on p. 43 of May's Concise Course, or Thm. 7.1.10 in Selick's Intro To Homotopy Theory, $(X,A)$ is an NDR pair if and only if $A\to X$ is a cofibration. My question is: is being an NDR pair a local property? That is, if there is an open cover $\{U_i\}$ of $X$ with each pair $(U_i,A\cap U_i)$ an NDR pair, can we conclude that $(X,A)$ is an NDR pair? I think this can be shown to hold if the open cover is finite and $X$ satisfies some separation axioms by using a partition of unity together with the deformation retraction appearing in the NDR definition to show the homotopy extension property defining cofibration holds (although I might be too optimistic). A reference in the literature to this result would be ideal although I would also be grateful for any counterexamples or explanations of why it isn't true. I've looked through the above mentioned references and skimmed through Hatcher but haven't been able to find it although it might follow from certain results on pushouts of cofibrations. I haven't been able to get my hands on Whitehead's big book yet. REPLY [7 votes]: There is a theorem of Dold to this effect: Dold, Albrecht Die Homotopieerweiterungseigenschaft (=HEP) ist eine lokale Eigenschaft. (German) Invent. Math. 6 1968 185–189.<|endoftext|> TITLE: ABC conjecture and Fermat's last theorem QUESTION [5 upvotes]: I have frequently read and heard that given the ABC-conjecture a number of important unsolved problems of number theory can be solved (with relatively simple proofs). Among them, the celebrated Fermat's Last theorem is frequently mentioned. So, my question is: Given that the $ABC$ conjecture is valid, can we prove that it implies Fermat's Last theorem ? P.S.: I can understand that ABC conjecture "easily" implies the asymptotic FLT (stating that: "the equ-ation $x^n+y^n=z^n$ can have solutions in positive integers only for $n< n_0$, where $n_0$ is some finite number"). This is outlined in Lang's Algebra (p.196, 1994 edition), see also here and here. REPLY [6 votes]: No, abc doesn't imply FLT. For all exponents $n > 3$, abc implies at most finitely many counterexamples to FLT, but it allows counterexamples to FLT. For exponent $n=3$ it allows infinitely many counterexamples and the Fermat-like equation $x^3+y^3=a z^3$ has infinitely many coprime solutions for some $a$ via the group law on the elliptic curve. Basically finite number of abc triples of arbitrary quality don't contradict abc, while a single counterexample contradicts FLT. Also, there is generalization of abc over number fields, widely believed to be true. Over number fields FLT fails, e.g. $1^n+1^n=(\sqrt[n]{2})^n$.<|endoftext|> TITLE: Volume of an ideal simplex in $\mathbb{H}^3$, idea/intuition behind result QUESTION [11 upvotes]: There is the following result on page 160 of Thurston's book "The Geometry and Topology of Three-Manifolds", as follows. The volume of an ideal simplex in $\mathbb{H}^3$ with dihedral angles $\alpha$, $\beta$, $\gamma$ is equal to$$\Lambda(\alpha) + \Lambda(\beta) + \Lambda(\gamma),$$where $\Lambda: \mathbb{R} \to \mathbb{R}$ is the Lobachevsky function$$\Lambda(\theta) = -\int_0^\theta \log|2\sin t|\,dt.$$ What is the idea/intuition behind the proof of this result, i.e. why does the volume of an orthoscheme have a formula in terms of the Lobachevsky function? I've read the proof in Thurston's book, but I'd still be interested in hearing what others have to say on this issue. REPLY [6 votes]: A variation of the other answer: first decompose an ideal tetrahedron into orthoschemes with two ideal vertices and (non-right) angles of the form $\alpha,\pi/2-\alpha, \alpha$. These orthoschemes have volume a function of $\alpha$. To compute the variation, truncate the ideal vertices by horospheres meeting the other two vertices. Then as we vary $\alpha$, only one edge contributes to Schlafli's formula, and a simple geometric computation shows that the edge length is $\log(2\sin\alpha)$. Integrating gives the Lobachevsky function. Note: The relevent version of Schlafli's formula was proved by Hodgson in his thesis. See the Remark on p. 125 of his thesis.<|endoftext|> TITLE: On a trace condition for positive definite $2\times 2$ block matrices QUESTION [6 upvotes]: Consider the following block matrix $$ X = \begin{bmatrix} A & C \\ C^\top & B\end{bmatrix}, $$ where $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{m\times m}$, and $C\in\mathbb{R}^{n\times m}$. To Prove (or disprove): If $X$ is positive definite, i.e. $X>0$, then the following trace inequality holds $$ \left[\mathrm{tr}(CC^\top)\right]^2< \mathrm{tr}(A^2)\mathrm{tr}(B^2). $$ Some comments. Based on Theorem 2.3 of Horn and Mathias. "Cauchy-Schwarz inequalities associated with positive semidefinite matrices." Linear Algebra and Its Applications 142 (1990): 63-82, I think it is possible to prove the previous fact if we replace strict inequalities with non-strict ones. REPLY [6 votes]: For any unitarily invariant norm it can be shown that \begin{equation*} \|X\| = \left\Vert \begin{bmatrix} A & C\\ C^* & B \end{bmatrix} \right\Vert \le \|A\| + \|B\|. \end{equation*} Thus, using the squared Frobenius norm on both sides and cancelling, we obtain \begin{equation*} \|C\|_F^2 \le \|A\|_F\|B\|_F, \end{equation*} as desired.<|endoftext|> TITLE: Topological Euler number of a singular variety QUESTION [5 upvotes]: Let $X$ be a projective variety over $\mathbb{C}$. Is there a way to define some number $\tilde{\chi}(X)\in \mathbb{Z}$ satisfying both of the following two properties? $\boldsymbol{(1)} \;$ When $X$ is smooth, $\tilde{\chi}(X)=\chi(X)$, where $\chi(X)$ is the usual Euler number of $X$ (as a topological space). $\boldsymbol{(2)} \;$ $\tilde{\chi}(X)$ is invariant under deformation, i.e. if we have a family of variety over a connected base $B$, then any two fibers have the same $\tilde{\chi}$. If I consider an affine variety (still over $\mathbb{C}$) instead of a projective variety, then what is the answer to the corresponding question? REPLY [13 votes]: The following example shows that the answer to abx's down-to-earth question Is the Euler number of the general fiber independent of the smoothing? is in general no. Take $X \subset \mathbb{P}^5$, the cone over a rational normal curve $C_4 \subset \mathbb{P}^4$. It is well-known that $X$ admits two differents smoothings: a $\mathbb{Q}$-Gorenstein smoothing whose general fibre isomorphic to $\mathbb{P}^2$ and a non-$\mathbb{Q}$-Gorenstein smoothing whose general fibre is isomorophic to $\mathbb{P}^1 \times \mathbb{P}^1$, see for instance this MathOverflow question and the corresponding answers. In the first case, the topological Euler number of the smooth fibre is $3$, whereas in the second case it is $4$. So, at least for varieties that are not $\mathbb{Q}$-Gorenstein, the topological Euler number of the general fibre actually depends on the smoothing we are choosing.<|endoftext|> TITLE: Clarifying the connection between 'etale locally' and 'formally locally' QUESTION [23 upvotes]: The following type of statements is often bandied about around the mathematical watering hole: "etale close is closer than formally close". Namely, if one imagines some sort of 'absolute topology' encompassing all the usual Grothendieck topologies on schemes (fppf, fpqc, etale, Nisnevich,...) that zooming in 'formally close' is a coarser zoom than zooming in 'etale close'. Can someone give meat to these statements? An actual comparison statement of some form? Also, perhaps more pressingly, are they 'intuitively true' (I've never really seen a rigorous formalization of these statements made)? Here is a sampler of the sort of things I'm curious about: Is there a natural Grothendieck topology on schemes whose 'local rings' are completions? Namely, in the etale topology the local ring of a scheme $X$ at a point $x$ is $\mathcal{O}_{X,x}^\text{sh}$. Is there a natural Grothendieck topology whose local ring of $X$ at $x$ is $\widehat{\mathcal{O}_{X,x}}$? Does this topology have a natural notion of fundamental groups? Is there some category of 'finite ____ covers" which forms a reasonable Galois category? How do the fundamental groups in this category compare? A literal question: Let $R$ be any local ring. How does, $\pi_1^{\acute{et}}(\text{Spec}(\widehat{R}))$ relate to $\pi_1^{\acute{e}t}(\text{Spec}(R^\ast))$ for $\ast\in\{\text{h},\text{sh}\}$? Similarly, how does $G_{\text{Frac}(\widehat{R})}$ (if $R$ is a domain) relate to $G_{\text{Frac}(R^\text{h})}$ and $G_{\text{Frac}(R^{\text{sh}})}$? What if $R$ is a DVR? I'm almost positive that Artin approximation enters this picture, but not totally sure how so. One reason I'm curious is that it's often easy to make nice intuitive relationships between geometry and the formal local geometry of a scheme (say over $\mathbb{C}$) but less obvious to me how to make analogies using henselizations. For example, when one wants to intuit $G_{\mathbb{Q}_p}$ as a 'group classifying the geometry of a punctured $\text{Spec}(\mathbb{Z}$) near $p$ ' one often times looks to $G_{\mathbb{C}(T)}$. There one sees that the analogue of $G_{\mathbb{Q}_p}$ is $G_{\mathbb{C}((T))}$ which one can imagine as being the (profinite) fundamental group of a small punctured disk around a point $p$ which makes total sense, and is beautiful. That said, it's a small punctured formal disk around the point whereas, for me, it would have made more sense to consider small etale disks (whatever this means) around the point. This is further backed up by the point that if $D^\ast$ is a small punctured disk around $p$ then $\widehat{\pi_1(D^\ast)}=\widehat{\mathbb{Z}}=G_{\mathbb{C}((T))}$. This should still be true if $\mathbb{C}((T))$ is replaced by $\mathbb{C}[T]_{(t)}^{\text{sh}}$, but it's then less clear that this equivalence of fundamental groups is not a mistake. Thanks! REPLY [19 votes]: For local noetherian rings, the henselization has much more direct algebro-geometric meaning than the completion since it is built from local-etale algebras. Hence, your statement that the completion provides more intuition than the henselization is just a matter of having less experience with henselizations. See Example 2 below for why there is no need in practice for anything like a "formal topology", which likely doesn't exist anyway. In particular, the answer #1 is likely "no", so #2 is then moot. But the real reason this is all somewhat of a wild goose chase is that you are correct about #4: for excellent local noetherian rings, the deep Artin-Popescu approximation is the ultimate answer to nearly all questions about passing between their completions and their henselizations. This theorem implies that for any such ring $R$ and any finite system of polynomial equations in several variables over $R$, any solution in the completion can be approximated arbitrarily closely (for the max-adic topology) by a solution in $R^{\rm{h}}$. (The real content, upon varying the system of equations, is that there is even a single solution in $R^{\rm{h}}$, let alone one that is "close" to the given solution in the completion.) However, that raw statement about systems of equations does not do justice to the significance, so let's illustrate with a couple of applications; see section 3 of Artin's IHES paper (http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1969__36_/PMIHES_1969__36__23_0/PMIHES_1969__36__23_0.pdf) for many more striking applications. Example 1. A pretty algebraic application of this approximation theorem is that if $R$ is an excellent local normal noetherian domain then $R^{\rm{h}}$ is the "algebraic closure" of $R$ in the normal local noetherian completion $\widehat{R}$. Even for $R$ the local ring at the origin of an affine space over a field, I am not aware of a proof of this application without invoking Artin approximation. Example 2. A geometric reason that Artin-Popescu approximation is so relevant is that it implies the following extremely useful fact: if $R$ is an exellent local noetherian ring and $A, B$ are two local $R$-algebras (with local structure map) essentially of finite type such that there exists an isomorphism $f:\widehat{A} \simeq \widehat{B}$ as $\widehat{R}$-algebras then $A$ and $B$ admit a common residually trivial local-etale neighborhood. (In particular, $A^{\rm{h}} \simeq B^{\rm{h}}$ over $R$; we definitely do not claim that such an isomorphism of henselizations can be found which induces $f$ on completions.) Even for $R$ a field, this is not at all obvious but is super-useful; e.g., it comes up when rigorously justifying the sufficiency of proving some facts about the geometry of specific singularities by studying analogous special cases with completions, such as in deJong's paper on alternations or various papers on semistable curves, etc. It remains to address #3, under reasonable hypotheses on $R$. We will see that the first question in #3 is quite easy, and the second generally has a negative answer when $\dim R > 1$ (whereas the case $\dim R = 1$ is affirmative under reasonable assumptions on $R$, a fact used all the time in number theory; more on this below). Now we come to the hypotheses that should have been made on $R$ for posing any questions of the sort being asked. Presumably you meant for $R$ to be noetherian (max-adic completion would not be appropriate otherwise), and also excellent (or else the completion could fail to be reduced when $R$ is reduced, etc.). Moreover, let's forget about strict henselization and only consider henselization, since the latter has the same completion and is the correct "algebraic" substitute for the completion. And since henselization of an excellent local ring is excellent (see the Remark after Prop. 1.21 in the book "Etale Cohomology and the Weil Conjectures" for a proof, which was overlooked in EGA even though preservation under strict henselization is done in EGA), we may as well assume $R$ is henselian. Since the category of finite etale algebras over a henselian local ring is naturally equivalent to that over its residue field, the invariance of etale fundamental group under passage to the completion for any henselian local noetherian ring is obvious. That is, the first question in #3 has an easy affirmative answer. I do not know what is the precise intended meaning of the 2nd to last paragraph in the posting (since equicharacteristic 0 is a poor guide to the rich arithmetic structure of mixed characteristic $(0,p)$), but one might contemplate trying to use the Artin-Popescu approximation theorem to affirmatively settle: Question. Let $R$ be an excellent henselian local domain that is normal, with fraction field $K$. Let $\widehat{K}$ denote the fraction field of the completion $\widehat{R}$. Do the Galois groups of $K$ and $\widehat{K}$ naturally coincide? Despite whatever intuition may be derived from Artin-Popescu approximation, the possibly surprising answer to the Question is generally negative when $\dim R > 1$, even when $R$ is regular. (I only realized that after failing to prove an affirmative answer beyond dimension 1 in this way.) Before discussing counterexamples, a few Remarks will be helpful to put things in context: Remark 1: That the local noetherian ring $\widehat{R}$ is actually a normal domain requires a proof. One way to see it is to use Serre's homological characterization of normality and the fact that the flat map ${\rm{Spec}}(\widehat{R}) \rightarrow {\rm{Spec}}(R)$ has regular fibers (since $R$ is excellent); see the Corollary to 23.9 in Matsumura's "Commutative Ring Theory". Note also that when $\dim R > 1$, there is no "topological field" structure on $K$ and so $\widehat{K}$ is not a "completion" for $K$ in some intrinsic sense not referring to $R$; it is just notation for the fraction field of the completion of $R$. Remark 2. In case $\dim R = 1$ the Question has an affirmative answer (even valid for rank-1 henselian valuation rings and valuation-theoretic completions thereof, without noetherian hypotheses; see 2.4.1--2.4.3 in Berkovich's paper in Publ. IHES 78). This does not require the approximation theorem, and is an instructive exercise in the use of Krasner's Lemma. The main task when $\dim R > 1$ is to try to replace Krasner's Lemma with an argument using the approximation theorem, bypassing the fact that $K$ (let alone its finite separable extensions) doesn't have a natural topological field structure when $\dim R > 1$; all such attempts are doomed to fail, as we'll see below. Remark 3. Equality of the Galois groups (through the evident natural map) would imply via Galois theory that every finite separable extension of $\widehat{K}$ arises by "completing" a finite separable extension of $K$ and more generally that $E \rightsquigarrow E \otimes_K \widehat{K}$ is an equivalence of categories of finite etale algebraic over $K$ and $\widehat{K}$ respectively. This equivalence is very useful in number theory for the case $\dim R = 1$ when the answer is affirmative. To see what goes wrong beyond dimension 1, consider a general $R$ as above, allowing $\dim R$ arbitrary for now and not yet assuming regularity. Let $K'/K$ be a finite separable extension, so by normality of $R$ the integral closure $R'$ of $R$ in $K'$ is module-finite over $R$. Since $R$ is henselian and $R'$ is a domain, it follows that $R'$ must be local (and henselian and excellent). Thus, $R' \otimes_R \widehat{R} = \widehat{R'}$ is a normal domain, and its fraction field $\widehat{K'}$ is clearly finite separable over $\widehat{K}$ with degree $[K':K]$. In case $K'/K$ is Galois with Galois group $G$, it follows that $G \subset {\rm{Aut}}(\widehat{K'}/\widehat{K})$ with $\#G = [\widehat{K'}:\widehat{K}]$, so $\widehat{K'}/\widehat{K}$ is Galois with Galois group $G$. In this way, we see (by exhausting a separable algebraic extension by finite separable subextensions) that $K_s \otimes_K \widehat{K}$ is a Galois extension of $\widehat{K}$ with Galois group ${\rm{Gal}}(K_s/K)$. The Question is exactly asking if this is a separable closure of $\widehat{K}$. By Galois theory over $\widehat{K}$, it is the same to ask that any given finite Galois extension $E/\widehat{K}$ is the "completion" (in the above sense based on integral closures of $R$) of some finite separable extension $K'/K$, necessarily of the same degree. Now let's see that already for quadratic extensions one has counterexamples with $R$ any countable henselian excellent regular local ring with $\dim R > 1$ and ${\rm{char}}(K) \ne 2$, such as the henselization at any closed point on a smooth connected scheme of dimension $>1$ over a field not of characteristic 2; e.g., $R = \mathbf{Q}[x_1,\dots,x_n]_{(x_1,\dots,x_n)}^{\rm{h}}$ is a counterexample for any $n > 1$. It is the same to show for such $R$ that the natural map $K^{\times}/(K^{\times})^2 \rightarrow \widehat{K}^{\times}/(\widehat{K}^{\times})^2$ is not surjective. Since $K^{\times}$ is countable, it suffices to show that the target is uncountable. This uncountability feels "obvious" when $\dim R > 1$, but we want to give a rigorous proof (e.g., to understand exactly how the condition $\dim R > 1$ and the completeness of $\widehat{R}$ are used). The regularity of $R$ implies that of $\widehat{R}$, so $\widehat{R}$ is a UFD. Thus, the height-1 primes of $\widehat{R}$ are principal, corresponding to irreducibles in $\widehat{R}$ up to unit multiple. Any two irreducibles that are not $\widehat{R}^{\times}$-multiples of each other give rise to distinct elements in $\widehat{K}^{\times}/(\widehat{K}^{\times})^2$, by the UFD property. Thus, it suffices to show that $\widehat{R}$ contains uncountably many height-1 primes. Complete local noetherian rings satisfy "countable prime avoidance": if an ideal $I$ is contained in a union of countably many prime ideals then it is contained in one of those primes. (This fact is originally a lemma of L. Burch. It can also be deduced easily from the Baire category theorem by using that such rings with their max-adic topology are complete metric spaces in which all ideals are closed, as noted in section 2 of the paper "Baire's category theorem and prime avoidance in complete local rings" by Sharp and Vamos in Arch. Math. 44 (1985), pp. 243-248.) By the UFD property, every non-unit in $\widehat{R}$ is divisible by an irreducible element and so lies in a height-1 prime. Thus, the maximal ideal $m$ of $\widehat{R}$ is contained in the union of all height-1 prime ideals. But $m$ is not contained in (equivalently, not equal to!) any of those height-1 primes since $\dim \widehat{R} = \dim R > 1$ (!), so by countable prime avoidance in $\widehat{R}$ it follows that $\widehat{R}$ must have uncountably many height-1 primes, as desired. (Related reasoning comes up in Example 2.4 in the paper of Sharp and Vamos mentioned above.) Remark: For any $R$ as in the above counterexamples, and $B'$ the module-finite integral closure of $\widehat{R}$ in a quadratic extension of $\widehat{K}$ not arising from a quadratic extension of $K$, there is no finite $R$-algebra $B$ satisfying $B \otimes_R \widehat{R} \simeq B'$. This may seem to contradict Theorem 3.11 in Artin's IHES paper linked above, but it does not (since there is no reason that the proper closed non-etale locus in the base for the generically etale ${\rm{Spec}}(B') \rightarrow {\rm{Spec}}(\widehat{R})$ is contained in a proper closed subset of ${\rm{Spec}}(\widehat{R})$ that is the preimage of one in ${\rm{Spec}}(R)$).<|endoftext|> TITLE: Locales as geometric objects QUESTION [17 upvotes]: There is the following analogy: $$\begin{array}{cc} \text{frames} & - & \text{commutative rings} \\ | && | \\\text{locales} & - & \text{affines schemes}\end{array}$$ Here, $\bigvee$ is analagous to $\sum$ and $\bigwedge$ is analogous to $\prod$. The category of locales is defined as the dual of the category of frames. We may also define the category of affines schemes as the dual of the category of commutative rings, this is done for example by Toen and Vaquié in relative algebraic geometry. But usually we define affine schemes geometrically as locally ringed spaces. Also, affine schemes are the local building blocks for schemes. This leads to the following two questions: Is there a geometric definition of locales? Are locales the local building blocks for some geometric objects? So, the second question is about completing the following analogy: $$\begin{array}{cc} \text{frames} & - & \text{commutative rings} \\ | && | \\\text{locales} & - & \text{affines schemes} \\ | && | \\ \text{?} & - & \text{schemes}\end{array}$$ After skimming through the nlab article on locales, I would guess that locales may be defined "geometrically" as localic toposes and that perhaps the glued geometric objects are toposes, too (which ones?). This is motivated by the fact that we have a fully faithful left adjoint functor $$\mathrm{Sh} : \mathsf{Loc} \to \mathsf{Topos}$$ from locales to toposes given by taking sheaves, and that toposes are sometimes regarded as geometric objects. (It is noted at the nlab that locales coincide with Grothendieck $(0,1)$-toposes. But this is just a reformulation of the definitions.) But this doesn't really look similar to the construction of the spectrum of a commutative ring. It rather looks like the fully faithful embedding from (affine) schemes into the category of cocomplete symmetric monoidal categories, given by taking quasi-coherent sheaves. And this is not how we usually define affine schemes geometrically. Perhaps there is an embedding from the category of locales into the category of locally ringed spaces? Of course, we cannot take the associated topological space of points, since that may turn out to be empty. Perhaps there is a different construction, though. REPLY [9 votes]: While Simon's answer is very good, I think one can also say something a little more along the lines of what you may be thinking. I haven't seen this written out in this way before, so I may have made some errors, but I think the general idea is valid. Note that the "space of points of a locale", when regarded as a functor on frames, is very similar to the maximal or prime spectrum of a ring. For a ring $R$, the points of $\mathrm{Spec}(R)$ are the prime ideals in $R$, and the open sets are generated by the sets $O_f$ of prime ideals not containing $f$ for $f\in R$. Similarly, for a frame $A$, the points of "$\mathrm{Spec}(A)$" are the completely prime filters in $A$ (filters and ideals being complementary in a lattice), and its open sets are generated by sets $O_u$ of filters containing $u$ for $u\in A$. Now, as we know, a ring $R$ induces a sheaf of rings on $\mathrm{Spec}(R)$, whose ring of sections over the open set $O_f$ generated by some $f\in R$ is the localization $R_f$. Similarly, from a frame $A$ we can try to induce a sheaf of frames on $\mathrm{Spec}(A)$, whose frame of sections over $O_u$ is the set $A_u$ of elements of $A$ that are $\le u$; this is the "localization" at $u$ in frames, since the only "unit" in a frame is the top element. The structure sheaf of a ring $R$ is a sheaf of local rings, in the sense that its stalks are local rings. A ring is local if $\sum_{i=1}^n a_i = 1$ implies there exists an $a_i$ that is invertible. Similarly, a frame is local if $\bigvee_i a_i = \top$ implies there exists an $a_i=\top$, i.e. the set $\{\top\}$ is a completely prime filter, or equivalently (in localic language) there is a point whose only neighborhood is the whole space (a "focal point"). The above sheaf of frames would be a local one; indeed, its stalk over a point $x\in \mathrm{Spec}(A)$ is the "localization of $A$ at $x$" in an appropriate sense (localically, it is the "germ at $x$", the intersection of all the open sublocales containing $x$). The problem, of course, is that unlike rings, frames may not have enough points. One manifestation of this is that a localization of a ring $R_f$ depends only on the radical of the principal ideal $(f)$, and hence only on the set of prime ideals containing (or not containing) $f$; thus the definition $O_{\mathrm{Spec(R)}}(O_f) = R_f$ is well-defined in that if $O_f = O_g$ then $R_f \cong R_g$. However, this is not true for frames: we can have $O_u=O_v$ but $A_u\ncong A_v$. The most drastic example is that there are nontrivial frames without any points at all, so that $O_u=O_v$ for any $u,v$. This means that our above attempted definition of a "structure sheaf of frames" on $\mathrm{Spec}(A)$ doesn't work. We can, I believe, fix it by taking colimits of $A_u$ over all appropriate $u$, and the stalks will still be as I claimed above. But the problem remains that the space $\mathrm{Spec}(A)$ itself could be completely empty even if $A$ is highly nontrivial, and so frames will not embed contravariantly into "locally framed spaces" in the same way that rings embed into locally ringed spaces. The solution, of course, is that we need to consider locally framed locales rather than spaces. Since this is obvious to locale theorists, they don't generally go through the above detour. But now the analogous construction looks very tautological: the locale version of the spectrum $\mathrm{Spec}_\ell(A)$ is just $A$ itself regarded as a locale. The "structure sheaf" is the subobject classifier of the sheaf topos $\mathrm{Sh}(A)$, which as is well-known is an internal frame, and the colimit-over-$u$ fix I described above is the pullback of this internal frame to the topos of sheaves over the topological space of points $\mathrm{Spec}(A)$.<|endoftext|> TITLE: Construction of irreps of path algebra of cyclic quiver, classification of all finite-dimensional irreps QUESTION [7 upvotes]: Originally posted here on Mathematics Stack Exchange. Let $Q$ be a quiver with vertex set $\{1, 2, \ldots, n\}$ such that $Q$ has a single edge $i \to i + 1$, for every $i = 1, 2, \ldots, n - 1$, one edge $n \to 1$, and no other edges. In other words, $Q$ is a cyclic quiver with $n > 1$ vertices. I am looking for a proof or a precise reference for the following. The construction of an irrep of $\mathbb{C}Q$ of (total) dimension $> 1$. The classification of all finite dimensional irreps of $\mathbb{C}Q$ up to isomorphism. Thanks in advance! REPLY [8 votes]: Every irreducible finite-dimensional representation is either $1$-dimensional (and there are exactly $n$ of them, corresponding to the vertices) or $n$-dimensional and they can be indexed by non-zero, complex numbers. Proof: Let $V$ be an irreducible representation, $V=\bigoplus_{i\in\mathbb{Z}/n} V_i$ its standard decomposition given by the vertices and $f_i : V_i\to V_{i+1}$ the linear maps given by the edges. Let $L:=f_{n-1}\ldots f_1 f_0$ be the endomorphism of $V_0$ given by the closed loop in the quiver. If $V$ is finite-dimensional, then there is a eigenvector $v_0$ of $L$ for the eigenvalue $\lambda$ (we can assume that there are eigenvectors at all by assuming $V_0\neq 0$ wlog). Define $v_1:=f_0(v_0), v_2:=f_1(v_1), \ldots, v_{n-1}:=f_{n-2}(v_{n-2})$. If $\lambda=0$, then $f_i(v_i)=0$ for some $i$ and $\mathbb{C} v_i$ is a one-dimensional submodule. By simplicity it is the whole of $V$ and we're in the first case. If $\lambda\neq 0$, then all $v_i$ are nonzero and $\sum_i \mathbb{C}v_i$ is a submodule. By simplicity it is all of $V$, $\{v_i\}$ is a basis and one can easily write down all matrices of the edge elements w.r.t. this basis. Therefore this is a complete set of pairwise non-isomorphic irreducibles. In fact it should be possible to describe all finite-dimensional modules in terms of a Jordan normal form of $L$. More abstractly there should be something close to a Morita equivalence between the quiver algebra $\mathbb{C}Q$ and $e_0 \mathbb{C}Q e_0$ which is just the polynomial ring $\mathbb{C}[L]$. Note that the irreducible finite-dimensional representations of $\mathbb{C}[L]$ are parametrized by complex numbers. There seems to be some sort of geometry here lurking in the background where $Irr(\mathbb{C}[L])$ is just the affine line and $Irr(\mathbb{C}Q)$ is the affine line with $n$-fold origin. Maybe some algebraic geometer can shed some light on this observation.<|endoftext|> TITLE: Periodic billiard paths in hyperbolic triangles QUESTION [10 upvotes]: It is a theorem of Masur that all rational triangles in the Euclidean plane posses a periodic billiard path, one obeying the reflection law that the angle of incidence equals the angle of reflection. A rational triangle has angles that are rational multiples of $\pi$. It remains an open question of whether all triangles have a periodic billiard path. Q. Is there some analogous theorem for hyperbolic triangles in the hyperbolic plane? A billiard path in a hyperbolic triangle consists of geodesics which reflect from the sides of the triangle by the same reflection law. Here is a possible start of a billiard path inside a triangle drawn in the Poincare half-plane model, where the geodesics are circular arcs and vertical line segments:           Is there some analog of the Euclidean rational angle condition that yields a theorem for hyperbolic triangles? REPLY [4 votes]: If the triangle is a fundamental domain for a discrete group, a periodic orbit corresponds to a closed geodesic in the quotient surface, which gives you boat-loads of periodic orbits (to be precise, $\sim \exp(L)/L$ of them of length bounded above by $L.$ I think that if all the angles are rational, the triangle is a fundamental domain for a cone-metric, and while a lot of the closed geodesics will go through the vertices, an entropy computation (that is, the set of geodesics passing through the vertices have smaller entropy than the entire geodesic flow) will give you a similar result.<|endoftext|> TITLE: Measure induced on [0, 1] by infinite tosses of biased coin QUESTION [19 upvotes]: It is well-known that one can get the Lebesgue measure on [0, 1] by tossing a fair coin infinitely (countably) many times and mapping each sequence to a real number written out in binary. I was trying to explore what happens if you follow the same procedure with a biased coin. I managed to prove that if the induced measure is absolutely continuous with respect to the Lebesgue measure, then the density must be discontinuous on at least a dense set (which includes all rational numbers in [0, 1] with a finite binary expansion). But I suspect it's much worse than that. When I try to visualize the cumulative distribution function, it seems to have a "fractal staircase" type of shape, so I suspect the measure is, in fact, singular, but I can't prove it. Q1. Are there any standard techniques for proving a measure is singular ? Q2. I strongly suspect that a natural question like this has already been resolved. Any link to a textbook and/or paper would be appreciated. REPLY [12 votes]: Just a side comment: if you pass from binary to trinary, you can still obtain the Lebesgue measure by choosing the digits $0$, $1$, $2$, with probabilities $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ (something one may call "a fair trinary coin"). Now, if you use the "biased" coin $(\frac{1}{2},0,\frac{1}{2})$, then you'll arrive exactly at the Cantor function, which puts mass $1$ to the Cantor set (which has null Lebesgue measure).<|endoftext|> TITLE: $G$ cocycle split to a coboundary in $J$, via a group extension QUESTION [7 upvotes]: Consider a generic nontrivial $d$-cocycle $\omega_d^G \in H^d(G,U(1))$ in the cohomology group of a group $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the $d$-cocycle $\omega_d^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega_d^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition $ \delta\omega_d^G=1$. question: We like to ask whether there always exists some Abelian group $N$ as a normal subgroup of some bigger group $J$, such that $G$ is the quotient group $$ \frac{J}{N}=G $$ and such that we can always trivialize (or split) the $d$-cocycle $\omega_d$ of $G$ into $d$-coboundary if we lift $G$ into a larger group $J$? Given that we know the group homomorphism $r$: $$J \overset{r}{\rightarrow} G.$$ Namely, $$ \omega_d^G(g_i,\dots)=\omega_d^G(r(j_i),\dots)= \delta \beta_{d-1}^J(j_i,\dots). $$ with $g \in G, j \in J$. You are welcome to comment or answer the partial case, for example, when $d=2$, $d=3$ or $d=4$, and when $G$ is a finite group. Partial comments or answers are still welcome! REPLY [6 votes]: In case d=1, the answer is always negative: 1-cocycles are homomorphisms, 1-coboundaries are always trivial, and inflation is injective. If you do not restrict yourself to the case where $N$ is abelian, the answer is positive: take a short exact sequence $1\to R\to F\to G\to 1$ where $F$ is free. For the abelian case we proceed as follows: you also have the short exact sequence $1\to R_{ab}\to F/[R,R]\to G\to 1$ and you have a morphism of short exact sequences from the first one to the second one. This also gives you a morphism on the corresponding Lyndon-Hochschild-Serre spectral sequences. In particular we have the following commutative diagram: $\require{AMScd}$\begin{CD}H^{d-2}(G,H^1(R_{ab}, U(1)))@> >> H^d(G,U(1))\\ @V V V @V V V \\ H^{d-2}(G,H^1(R,U(1)))@> >> H^d(G,U(1))\end{CD} For $d>2$ the lower horizontal morphism is an isomorphism since the spectral sequence converges to the cohomology of F, which is trivial. The right vertical arrow is the identity, and the left vertical arrow is an isomorphism, since all homomorphisms from $R$ to $U(1)$ factor through $R_{ab}$. But this implies that also the upper horizontal homomorphism is an isomorphism, and that the inflation from $G$ to $F/[R,R]$ is trivial. So this extension trivializes all cocycles at the same time. For $d=2$ we can similarly show that the upper horizontal morphism is surjective, and this is enough for the vanishing of the second cohomology. Another option: a two cocycle will give rise to an extension $$1\to U(1)\to J\to G\to 1.$$ The inflation of the cocycle to $J$ will be zero.<|endoftext|> TITLE: The fifth k-invariant of BSO(3) QUESTION [12 upvotes]: From work of Pontryagin and Whitney, as I understand it, the homotopy 4-type of $BSO(3)$ is $K(\mathbb{Z}/2,2) \times_{K(\mathbb{Z}/4,4)} K(\mathbb{Z},4)$, where the pullback is along the maps $\mathfrak{P}_2\colon K(\mathbb{Z}/2,2) \to K(\mathbb{Z}/4,4)$ (Pontryagin square) and the obvious $K(\mathbb{Z},4) \to K(\mathbb{Z}/4,4)$. In particular, the $k$-invariant $k_4$ is given by the composite $$ K(\mathbb{Z}/2,2)\to K(\mathbb{Z}/4,4)\to K(\mathbb{Z},5) $$ where the latter map arises from the short exact sequence $\mathbb{Z} \stackrel{\times 4}{\to}\mathbb{Z} \to \mathbb{Z}/4$ Has the next $k$-invariant been identified in the literature? This would be a map $$ K(\mathbb{Z}/2,2) \times_{K(\mathbb{Z}/4,4)} K(\mathbb{Z},4) \to K(\mathbb{Z}/2,6) $$ as $\pi_5(BSO(3)) \simeq \mathbb{Z}/2$. An obvious option would be (reduction mod 2 then) cup product. But there may be some other cohomology operation floating around. Edit: For $BSpin(3) = BSU(2)$ the analogous $k$-invariant is given by the map $$ K(\mathbb{Z},4) \to K(\mathbb{Z}/2,4) \xrightarrow{Sq^2} K(\mathbb{Z}/2,6) $$ REPLY [10 votes]: Recall that $SU(2)\cong Spin(3)$ and $PU(2)\cong SU(2)/\mathbb{Z}_2\cong SO(3)$. Originally Woodward calculated the lower stages of a Postnikov decomposition of $PU(n)$ in “The Classification of Principal PU(n)-bundles Over a 4-complex.” If I recall, his cacluations did not apply fully to the case $n=2$ and they were amended by Antieau and Williams in On the classification of oriented 3-plane bundles over a 6-complex [1]. I think the calculations they perform are as complete as exist in the literature. [1] Topology and its Applications, 173 (2014) pp 91-93, doi:10.1016/j.topol.2014.05.012, arXiv:1209.2219<|endoftext|> TITLE: On the distribution of roots modulo primes of an integral polynomial QUESTION [6 upvotes]: For motivation and related questions, see below. Rough sketch of the question. View $\bigsqcup_{p \text{ prime}} (\mathbb{Z}/p\mathbb{Z})$ as a ‘subset’ of the unit circle, via $a\pmod{p} \mapsto e^{2\pi i \cdot a/p}$. (This is not injective for $0 \pmod{p}$, but I wish to ignore that for the moment.) Let $f \in \mathbb{Z}[X]$ be a polynomial. If $a \pmod{p}$ is a root of multiplicity $m$ of the polynomial $f \pmod{p}$, then we attach a ‘weight’ $m$ to $e^{2\pi i \cdot a/p}$. Question 1: Are these weights uniformly distributed over the unit circle? Several remarks. I leave it as an exercise to the reader to make the question precise; using limits over circle segments, and primes going to infinity. If $\deg(f) \le 1$, the answer to question 1 is no. I have collected data for 1000 monic irreducible polynomials (with $2 \le \deg(f) \le 10$), and their roots modulo the first 10000 primes. Upon calculating some statistics I think there is some evidence for question 1. For almost all polynomials in my dataset, the first 3 moments of the samples are $< 3\%$ from the expected value for a uniform distribution. I am not an expert in statistics. But it might be the case that the correct distribution to look at is the circular uniform distribution. I have not yet calculated the circular moments for the dataset described above. Motivation. The main motivation comes from thinking about densities of places of finitely generated fields. Let $S$ be a subset of $\bigsqcup_{p \text{ prime}} \mathbb{F}_{p} = \{ \text{closed points of degree $1$ in $\mathbb{A}^{1}_{\mathbb{Z}}$}\}$ with positive Dirichlet density. Question 2: Is there a closed point of $\mathbb{A}^{1}_{\mathbb{Q}}$ whose closure in $\mathbb{A}^{1}_{\mathbb{Z}}$ intersects $S$ (a) infinitely often, or (b) with positive density? Such a closed point of $\mathbb{A}^{1}_{\mathbb{Q}}$ corresponds with a finite field extension of $\mathbb{Q}$, and thus with a polynomial in $\mathbb{Z}[X]$. Every circle segment of the unit circle gives rise to a positive density subset of $\bigsqcup_{p \text{ prime}} \mathbb{F}_{p}$. This provides a link between question 1 and question 2, although I do not see a logical implication from one to another. These questions are somewhat similar to statements of the Sato–Tate conjecture (or some generalisation of it). I do not see a direct link, but if someone sees how to connect this to statements about the distribution of eigenvalues of a certain Frobenius operator, please let me know. If I understand things correctly, the $0$-dimensional case of the Sato–Tate conjecture is Chebotarev's density theorem. Of course this describes at how many primes one expects a root of $f$, but it does not describe “where” the root in $\mathbb{F}_{p}$ will be. REPLY [9 votes]: In my paper, Polynomial congruences and density, Mathematics Magazine 80 (2007) 299-302, I cite the result of Christopher Hooley, On the distribution of roots of polynomial congruences, Mathematika 11 (1964) 39-49, MR 29 #1173, to the effect that if $f$ is an irreducible polynomial of degree at least 2, with integer coefficients, then the sequence formed by ordering $S_f$ by increasing denominator is uniformly distributed in $[0,1)$. Here, for positive integer $m$, $$S_f(m)=\{\,r/m:0\le r\le m-1,\,\gcd(r,m)=1,\,m\mid f(r)\,\}$$ and $S_f$ is the union of the sets $S_f(m)$ over all positive integers $m$. I then prove, without reference to Hooley's result, a theorem with weaker hypotheses, and a weaker conclusion: Let $f(t)=t^eg(t)$ where $e$ is a nonnegative integer, $g$ is a polynomial of degree at least 2 with integer coefficients, and $g(0)\ne0$. Define $T_f$ by $$T_f=\{\,r/m:\gcd(r,m)=1,\,m\mid f(r)\,\}$$ Then $T_f$ is dense in the reals.<|endoftext|> TITLE: Diophantine equation $3^n-1=2x^2$ QUESTION [14 upvotes]: How to solve a Diophantine equation like $$3^n-1=2x^2$$. One can easily see that the parity of $n$ and $x$ will be same and equation further can be seen taking if $$n\equiv0\pmod3\quad \text{then }x \equiv0\pmod{13} $$ but I don't know what to do further. REPLY [16 votes]: W. Ljunggren proved in 1 that the Diophantine equation $$\frac{x^{n}-1}{x-1} = y^{2}$$ doesn't admit solutions in integers $x>1, y>1, n>2$, except when $n=4, x=7$ and $n=5, x=3$. Since your equation can be rewritten as $$\frac{3^{n}-1}{3-1} = x^{2},$$ the aforementioned result of Ljunggren implies that its solutions in non-negative integers are $(n=0,x=0), (n=1,x=1), (n=2,x=2),$ and $(n=5,x=11)$. Alternatively, you can find a solution to this problem via Pell equations on page 243 of the March issue of vol. 110 (2003) of the American Mathematical Monthly... After noticing that this problem made it sometime to the problems & solutions department of the AMM, I couldn't help but recall what Léo Sauvé, former editor of Crux Mathematicorum, said on one occasion: "it seems like all problems were published in the Monthly once..." References W. Ljunggren, Some theorems on indeterminate equations of the form $\frac{x^{n}-1}{x-1} = y^{q}$ (In Norwegian). Norsk. Mat. Tidsskr. 25 (1943), pp. 17--20. On a result attributed to W. Ljunggren and T. Nagell, question 206645 in math-overflow.<|endoftext|> TITLE: Integral of the distance function to the boundary of a planar set QUESTION [5 upvotes]: I have been stuck for a few days in a seemingly harmless question. Given a simply connected open set $\Sigma\subset\mathbb{R}^2$, with smooth boundary $\partial\Sigma$, I am interested in estimating $$ \int_{\Sigma}d(x,\partial\Sigma)\;dx, $$ where $d(x,\partial\Sigma)=\inf_{y\in\partial\Sigma}|x-y|$. I would love to be able to prove that $$ \int_{\Sigma}d(x,\partial\Sigma)\;dx\ge\frac{2}{3} \frac{|\Sigma|^2}{|\partial\Sigma|}. $$ Equality holds for $\Sigma$ being a disk and violating the smoothness assumption for $\Sigma$ being a square. The only thing that seems to be doing on the right hand side of the inequality is using the Isoperimetric Inequality to say $$ \frac{2}{3} \frac{|\Sigma|^2}{|\partial\Sigma|}\le \frac{1}{3\sqrt{\pi}}|\Sigma|^{\frac{3}{2}}, $$ with equality, if and only if, $\Sigma$ is a disk. Ideas or counter-examples are very much welcomed. REPLY [5 votes]: The correct constant is $1/2$. Let $L(s)$ be the perimeter of the set of points whose distance to the boundary is $\ge s$. For simply-connected domains $L(s)$ is non-increasing, the integral of the distance to the boundary is $\int_0^\infty sL(s)\,ds$ and the area is $\int_0^\infty L(s)\,ds$. Now we want to minimize the first integral given the second one and $L(0)=\max_s L(s)$. Clearly, the worst case is when we push all the mass to the left as much as we can, so $L(s)=L(0)$ up to some point and then $0$. That gives you $1/2$ for the ratio in question and that scenario is almost realized by a long narrow rectangle, as Nate pointed out.<|endoftext|> TITLE: Eisenstein cohomology - explicit computation and relation to Franke's trace formula QUESTION [5 upvotes]: Let $G$ be a reductive group over $\mathbb{Q}_p$. Let $X_G$ be a locally symmetric space associated to the group $G$, and let $\partial X_G$ be the Borel-Serre boundary of $X_G$. The space $\partial X_G$ has a stratification by locally symmetric spaces $X_P$ associated to parabolic subgroups of $X_G$. Assuming I understand correctly, the cohomology of any local system (of sufficiently regular weight, with rational coefficients) on $\partial X_G$ should come from the cohomology of local systems on the $X_P$'s. Kostant's theorem further expresses the cohomology of local systems on $X_P$ in terms of the cohomology of local systems on $X_M$, where $M$ is the Levi subgroup of $P$. A cohomology class on $\partial X_G$ will appear in either the ordinary cohomology or the compactly supported cohomology of $X_G$. If $G$ has discrete series, then $X_G$ should only have cohomology in degree $\ge q$ and compactly supported cohomology in degree $\le q$, where $\dim X_G = 2q$. Given all of this information, if I have a cusp form on $X_M$, it seems that I should be able to write down very explicitly the Eisenstein classes associated to it. However, when I look at the trace formulas of Franke or Goresky-MacPherson, they do not look quite like I would expect. More specifically, the issue of whether a cohomology class of $X_P$ appears in the ordinary or compactly supported cohomology of $X_G$ does not seem to come up. Is this consideration implicitly there somehow, or is there something else going on? Does anything that I said in the first paragraph seem incorrect? More concretely, if I have a cusp form on some Levi subgroup of $G=Sp(4)$, what Eisenstein classes on $X_G$ does it contribute to? Can the answer be seen from any formulas in Franke's paper Harmonic Analysis on Weighted $L_2$ Spaces? REPLY [3 votes]: After rereading section 4 of On the Eisenstein Cohomology of Arithmetric Groups by Schwermer and Li, I think I understand this now. If $\lambda$ is sufficiently regular, then there is an isomorphism $$H^i(X_P,V_{\lambda}) \cong \bigoplus_{w \in W^M} H^{i+\dim \mathfrak{n}-\ell(w) }(X_M,V_{w(\lambda+\rho)-\rho})$$ where $V_{\lambda}$ denotes a local system of weight $\lambda$, $W^M$ is the set of elements $w$ of the Weyl group of $G$ such that $w^{-1}$ sends positive roots of $M$ to positive roots of $G$, and $\rho$ is half the sum of the positive roots. A class will show up in the cohomology (resp. compactly supported cohomology) of $X_G$ if the restriction of $w \lambda$ to the center of $M$ is positive (resp. negative). I believe that Franke uses the notation $\mathfrak{a}_{\mathcal{R}}^+$ for the set of weights whose restriction to the center of $M$ is positive. The relation between these criteria and the degree of the cohomology class can be found in Theorem 4.5 in Schwermer and Li's paper. For $G=Sp(4)$: If $P$ is either of the maximal parabolics, then cusp forms on $X_M$ occur in degree $1$ of the cohomology. They appear in degree $1,2,3,4$ of the cohomology of $X_P$. The degree $3,4$ classes occur in the same degree in the cohomology of $X_G$. The degree $1,2$ classes occur in one degree higher in the compactly supported cohomology of $X_G$. If $P$ is the minimal parabolic, then cusp forms on $X_M$ occur in degree $0$ of the cohomology. They appear in degree $0,1,1,2,2,3,3,4$ of the cohomology of $X_P$. The degree $4$ class occurs in the same degree in the cohomology of $X_G$, and the degree $0$ class occurs in the degree $2$ compactly supported cohomology of $X_G$. The other classes become zero in the cohomology of $\partial X_G$. One can check this as follows. If $P'$ is a maximal parabolic and $M'$ is the corresponding Levi, then Eisenstein series coming from cusp forms on $X_M$ appear in degree $1$ in $X_{M'}$ and in degrees $1,2,3,4$ in $X_{P'}$. Then we can apply Mayer-Vietoris. The degree $1,2,3$ classes on $X_P$ cancel with classes in one of the $X_P'$, and so they do not show up in the cohomology of $\partial X_G$. The degree $0$ class does not show up in either $X_{P'}$ and so it shows up in degree $1$ of the cohomology of $\partial X_G$. The degree $4$ class shows up in both of the $X_{P'}$, and so it shows up in degree $4$ of the cohomology of $\partial X_G$.<|endoftext|> TITLE: Transcendence of some number QUESTION [6 upvotes]: Everybody knows that $\sum_{k=0}^\infty{\frac{1}{2^{2^k}}}$ is transcendental. Is number $\sum_{k=0}^\infty{\frac{1}{2^{k^2}}}$ algebraic or not? REPLY [8 votes]: Yes, it is, and even the three numbers $\sum_{k \geq 0}{2^{-k^2}}$, $\sum_{k \geq 1}{k^22^{-k^2}}$ and $\sum_{k \geq 1}{k^42^{-k^2}}$ are algebraically independent. This results from algebraic independence results for theta functions, see Waldschmidt's excellent survey https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/SurveyTrdceEllipt2006.pdf, Corollary 52, and the reference given there: Nesterenko & Philippon, Introduction to algebraic independence theory, Lecture Notes in Mathematics, vol. 1752, Springer-Verlag, Berlin, 2001, Chapter 3. REPLY [4 votes]: I can't classify this question as a duplicate, but Is this number already known to be transcendental? Is there a survey about up-to-date trascendence results? asks the same thing. Introduction to algebraic Independence Theory, this number is transcendental.<|endoftext|> TITLE: Two different kinds of definitions of $C^k(\overline{\Omega})$ — extension and restriction QUESTION [5 upvotes]: This is cross-posted in MSE. I have seen two different kinds of definitions of the notation $C^k(\overline{\Omega})$ — by "extension" of functions on $\Omega$ or by "restriction" of functions on $\mathbb{R}^n$. I'm not sure about how different these two kinds of definitions could be. (I) Let the open set $\Omega\subset{\mathbb R}^n$, and $k$ be a positive integer. $C^k(\Omega)$ will denote the space of functions possessing continuous derivatives up to order $k$ on $\Omega$, and $C^k(\overline{\Omega})$ will denote the space of all $u\in C^k(\Omega)$ such that $\partial^{\alpha}u$ extends continuously to the closure $\overline{\Omega}$ for $0\leq|\alpha|\leq k$. The above definition defines $C^k(\overline{\Omega})$ as a subset of $C^k(\Omega)$. See for instance Folland's Introduction to Partial Differential Equations. On the other hand, one can also see in cited reference that $C^k(\overline{\Omega})$ is defined as restriction to $\overline{\Omega}$ of $C^k(\mathbb{R}^n)$ functions. (II) For instance, in Hermann Sohr's The Navier-Stokes Equations — An Elementary Functional Analytic Approach (page.23), $C^k(\overline\Omega)$ means the space of all restrictions $u|_{\overline\Omega}$ to $\overline\Omega$ of functions $u\in C^k(\mathbb{R}^n)$ such that $$ \sup_{|\alpha|\leq k,x\in\mathbb{R}^n}|D^\alpha u(x)|<\infty. $$ Here $|\alpha|\leq k$ is replaced by $|\alpha|<\infty$ if $k=\infty$. Here are my questions: If $u\in C^k(\overline{\Omega})$ as in (I), in general can $\partial^\alpha u$ extend continuously to $\mathbb{R}^n$ for $0\leq|\alpha|\leq k$? (*) [Added thanks to Pietro Majer's comment below] If $u\in C^k(\overline{\Omega})$ as in (I), does there exist $w\in C^k(\mathbb{R}^n)$ such that $w|_\Omega=u$? Are (I) and (II) essentially the same? REPLY [6 votes]: The standard reference for the extension problem is Whitney's extension theorem, which, has also many more modern presentations (a pretty good one is found in Stein's Singular Integrals and Differentiability Properties of Functions, Princeton University Press). It is worth paying attention, however, to the conditions that Whitney used in his extension theorem, which roughly states that at every point in the domain (up to the closure) the function is compatible with having the 'correct order' Taylor expansion. This condition is slightly stronger than mere $C^k(\bar{\Omega})$ in the sense of (I) of the question statement, as the example given in the link in this answer shows. In the case where you have some regularity conditions on the boundary $\partial\Omega$, however, you can use the regularity to show that $C^k(\bar{\Omega})$ in the sense of (I) automatically implies the condition that Whitney requires. This is why in many sources you will find Whitney's extension theorem stated only for closed sets with smooth boundary. In the context of PDE theory (such as Folland's book that you referred to for the definition (I)), domains usually have already nice boundaries (often you will need this for things besides just Whitney extension, such as boundary regularity estimates for elliptic equations), so in that setting the two definitions are essentially the same.<|endoftext|> TITLE: Dominant map from affine space implies unirationality QUESTION [8 upvotes]: Suppose $X$ is a variety of dimension $n$ over $k$ and there exists a dominant rational map $\mathbb{A}^N\dashrightarrow X$, where $N$ can be larger than $n$. Is it true that there is a dominant rational map $\mathbb{A}^n\dashrightarrow X$? If $k$ is characteristic 0, this is true because we can apply generic smoothness and look at the map of tangent spaces to pick out an $n$-dimensional linear subspace $\mathbb{A}^n$ of $\mathbb{A}^N$ that dominates $X$. In characteristic $p$, it seems unlikely, but one might worry that every $n$-dimensional linear subspace of $\mathbb{A}^N$ somehow fails to dominate $X$. REPLY [6 votes]: The proof in the case when $k$ is a finite field is given in Lemma 11 in Unirationality of Cubic Hypersurfaces by Kollár, published in Journal of the Institute of Mathematics of Jussieu, Volume 1, Issue 3 (2002). But you can also find the article in https://arxiv.org/abs/math/0005146 and Lemma 11 is given in page 5.<|endoftext|> TITLE: Why do we need random variables? QUESTION [48 upvotes]: In this MathStackExchange post the question in the title was asked without much outcome, I feel. Edit: As Douglas Zare kindly observes, there is one more answer in MathStackExchange now. I am not used to basic Probability, and I am trying to prepare a class that I need to teach this year. I feel I am unable to motivate the introduction of random variables. After spending some time speaking about Kolmogoroff's axioms I can explain that they allow to make the following sentence true and meaningful: The probability that, tossing a coin $N$ times, I get $n\leq N$ tails equals $$\tag{$\ast$}{N \choose n}\cdot\Big(\frac{1}{2}\Big)^N.$$ But now people (i.e. books I can find) introduce the "random variable $X\colon \Omega\to\mathbb{R}$ which takes values $X(\text{tails})=1$ and $X(\text{heads})=0$" and say that it follows the binomial rule. To do this, they need a probability space $\Omega$: but once one has it, one can prove statement $(\ast)$ above. So, what is the usefulness of this $X$ (and of random variables, in general)? Added: So far my question was admittedly too vague and I try to emend. Given a discrete random variable $X\colon\Omega\to\mathbb{R}$ taking values $\{x_1,\dots,x_n\}$ I can define $A_k=X^{-1}(\{x_k\})$ for all $1\leq k\leq n$. The study of the random variable becomes then the study of the values $p(A_k)$, $p$ being the probability on $\Omega$. Therefore, it seems to me that we have not gone one step further in the understanding of $\Omega$ (or of the problem modelled by $\Omega$) thanks to the introduction of $X$. Often I read that there is the possibility of having a family $X_1,\dots,X_n$ of random variables on the same space $\Omega$ and some results (like the CLT) say something about them. But then I know no example—and would be happy to discover—of a problem truly modelled by this, whereas in most examples that I read there is either a single random variable; or the understanding of $n$ of them requires the understanding of the power $\Omega^n$ of some previously-introduced measure space $\Omega$. It seems to me (but admit to have no rigourous proof) that given the above $n$ random variables on $\Omega$ there should exist a $\Omega'$, probably much bigger, with a single $X\colon\Omega'\to\mathbb{R}$ "encoding" the same information as $\{X_1,\dots,X_n\}$. In this case, we are back to using "only" indicator functions. I understand that this process breaks down if we want to make $n\to \infty$, but I also suspect that there might be a deeper reason for studying random variables. All in all, my doubts come from the fact that random variables still look to me as being a poorer object than a measure (or, probably, of a $\sigma$-algebra $\mathcal{F}$ and a measure whose generated $\sigma$-algebra is finer than $\mathcal{F}$, or something like this); though, they are introduced, studied, and look central in the theory. I wonder where I am wrong. Caveat: For some reason, many people in comments below objected that "throwing random variables away is ridiculous" or that I "should try to come out with something more clever, then, if I think they are not good". That was not my point. I am sure they must be useful, lest all textbooks would not introduce them. But I was unable to understand why: many useful and kind answers below helped much. REPLY [3 votes]: Practically everything we measure in real life (for instance the time taken for an apple to fall on Newton's head) is "random" in the sense that if we perform the experiment again, we will not get the same answer. So every measurement is a random variable $X$ whose probability of being within $x$ and $x+dx$ is usually $f(x)dx$ where $f(x)$ is the probability density function (or perhaps even better, the probability of $X$ being at least as large as $x$ is the cumulative distribution function).<|endoftext|> TITLE: Log canonical counterexample to Kawamata-Viehweg vanishing QUESTION [5 upvotes]: I found in the literature that, in characteristic 0, Kodaira vanishing holds for log-canonical pairs. On the other hand, the usual statement for Kawamata-Viehweg vanishing talks about a klt pair $(X,\Delta)$. Question: Are there (good) examples of failure of Kawamata-Viehweg vanishing for log-canonical pairs (even better, for $\Delta=0$)? REPLY [6 votes]: Here is a straightforward example with $\Delta \neq 0$. I think it can be adapted to give an example without a boundary. Let $X_0 = C \times C'$ be a product of elliptic curves viewed as an elliptic fibration $X_0 \to C$ with a fixed section $S_0$ and fiber $E_0$. Let $\mu : X \to X_0$ be the blowup of the intersection point $p = S_0 \cap E_0$. I'll denote the exceptional divisor by $A$ and the two strict tranforms by $S$ and $E$. Now consider the log canonical pair $(X, \Delta = S + F + E)$ where $F$ is the strict transform of a fiber disjoint from $p$. We have that $K_X \sim_\mathbb{Q} A$ and $$ L = K_X + \Delta = K_X + S + F + E $$ is big, nef and even semiample. $L$ is zero precisely on $E$ so the basepoint free linear series $|mL|$ for $m \gg 0$ induces a log canonical contraction $f : X \to Y$ that contracts $E$ to an elliptic singularity and $Y$ is the log canonical model of $(X, \Delta)$. Now we want to show that $\mathcal{O}_X(mL) = \omega_X(\Delta) \otimes \mathcal{O}_X((m-1)L)$ has nonzero higher cohomology. First note that $f_*\mathcal{O}_X(mL) = \mathcal{O}_Y(m(K_X + f_*\Delta))$ is sufficiently ample on $Y$ so $H^j(f_*\mathcal{O}_X(mL)) = 0$ for $j > 0$. Therefore $$ H^1(X, \mathcal{O}_X(mL)) = H^0(Y, R^1f_*\mathcal{O}_X(mL)) $$ from Leray spectral sequence. Now $R^1f_*\mathcal{O}_X(mL)$ is a skyscraper sheaf at $f(E)$ and we can compute this is nonzero by the theorem on formal functions using: $(1)$ $E$ is an elliptic curve, $(2)$ $mL|_E = \mathcal{O}_E$. Here the restriction of $mL$ to $E$ contributes to the cohomology since $E$ is a log canonical center of the pair. If I understand correctly, this is the only way that KV vanishing can fail for log canonical pairs: $\textbf{Theorem:}$ Let $(X, \Delta)$ be a projective log canonical pair. Let $L$ be a big and nef $\mathbb{Q}$-Cartier divisor such that $L$ remains big when restricted to each log canonical center of $(X,\Delta)$. Then $$ H^j(X, \mathcal{O}_X(K_X + \Delta + L)) = 0 $$ for $j > 0$. This is a special cause of Theorem 1.10 in this paper of Fujino. See also Theorem 1.7 in the paper linked in the question.<|endoftext|> TITLE: Selmer Group versus Selmer Variety QUESTION [5 upvotes]: For an abelian variety $A$, the $p$-adic Selmer group is defined to be the subset of $H^1(G_k,H_1^{et}(A;\mathbb{Q}_{p}))$ whose restriction to $G_{k_v}$ is in the image of $A(k_v)$ for all places $v$ of $k$. Minhyong Kim's paper states on its third page (p.91) that his Selmer varieties generalize $\mathbb{Q}_p$ Selmer groups of elliptic curves. As I understand it, his Selmer variety would refer to the subset of $H^1(G_k,H_1^{et}(A;\mathbb{Q}_{p}))$ that is unramified for $v \nmid p$ and crystalline at $v \mid p$ (assuming the abelian variety has good reduction there). Is it clear that these two definitions are equivalent? Why is this true? REPLY [9 votes]: You should read the Bloch--Kato paper in the Grothendieck Festschrift. This was, I believe, the first paper to consider Selmer groups of Galois representations defined by local conditions coming from p-adic Hodge theory (e.g. crystalline at p). The definition you quote from Kim is exactly the Bloch--Kato definition of the Selmer group; and Bloch and Kato prove in their paper that in the case of abelian varieties their definition recovers the classical Selmer group.<|endoftext|> TITLE: Dwork's proof of rationality of zeta function, crux of his generalization of a result of Borel along the way QUESTION [10 upvotes]: In this article by Katz and Tate here, there's a nice account of Dwork's argument for showing the rationality of the zeta function part of the Weil conjectures. Here is an excerpt. To recapitulate, we now know that the zeta function as power series has integer coefficients and that it is the ratio of two $p$-adically entire functions. We also know the zeta function has a nonzero radius of archimedean convergence (since we have the trivial archimedean bound $N_d \le (q^d - 1)^n$). Bernie's third new idea is to generalize a classical but largely forgotten result of E. Borel to show that any power series with these three properties is a rational function. Thus he proves the rationality of the zeta function. Can anybody give a sketch of the argument for this "third new idea"? What is the crux of the proof of Dwork's generalization of Borel's result that any power series with the aforementioned three properties is a rational function? What is the intuition behind the proof, what are the key steps that the proof boils down to? REPLY [6 votes]: Why don't you read Dwork's paper - it is quite clear (see p. 643)?<|endoftext|> TITLE: Etale cohomology can not be computed by Cech QUESTION [29 upvotes]: It can be proven that if in a quasicompact scheme $X$ any finite subset is contained in an affine open subset then for any sheaf $\mathcal{F}$ on $X$ its Cech cohomology $\hat{H_{et}^{\bullet}}(X,\mathcal{F})$ is naturally isomorphic to etale cohomology $H^{\bullet}_{et}(X,\mathcal{F})$. Is there a counterexaple for an arbitrary scheme? The assumptions of the theorem I cited fail, for instance for Hironaka's example of a threefold with two points not contained in any affine open subset but does it give a counterexample? REPLY [2 votes]: Čech cohomology fails for the plane with the doubled origin, $\mathbb A^2_{00}=\mathbb A^2\cup_{\mathbb A^2-0}\mathbb A^2$, with cohomology in the structure sheaf: $\check{H}^2(\mathbb A^2_{00};\mathcal O)=0\ne H^2(\mathbb A^2_{00};\mathcal O)$. Čech cohomology fails for both the Zariski and étale cohomologies. Mayer-Vietoris shows that the punctured plane $\mathbb A^2-0$ has nontrivial cohomology in degree 1. Similarly, if we take any open subset of $\mathbb A^2$ that contains the origin, the same group appears in its first cohomology. Mayer-Vietoris pushes that into degree 2: $H^2(\mathbb A^2_{00};\mathcal O)=H^1(\mathbb A^2-0;\mathcal O)$. (In other words, if we push forward from the punctured plane to the affine plane $j\colon \mathbb A^2-0\hookrightarrow A^2$, then $R^1j_*\mathcal O$ is nontrivial and concentrated on the origin.) The Čech cohomology for a particular cover amounts to pretending that the open sets of that cover and all of their $n$-fold intersections are acyclic for the sheaf. For a manifold and the constant sheaf, there do exist such “good covers” (eg, convex balls on a Riemannian manifold) and they are cofinal. For a separated scheme, affines are acyclic, so if intersections of affines are affine, then a single cover by affines computes quasi-coherent cohomology. This is why separated often appears as a hypothesis in cohomology (although it could be reduced to “semi-separated,” ie, affine diagonal). If $U\to X$ is a cover and $U^n$ is the $n$-fold fiber product over $X$, then the Čech complex is: $$F(U)\to F(U^2)\to F(U^3)\to\cdots$$ This defines $\check{H}^*(U\to X;F)$. For any cover, the derived version of this computes the full cohomology: $$RF(U)\to RF(U^2)\to RF(U^3)\to\cdots$$ This gives the Čech-to-cohomology spectral sequence: $E_1^{pq}=H^q(U^p;F)\Rightarrow H^{p+q}(X;F)$ and $E_2^{pq}=\check{H}^p(U\to X;R^qF)$). The edge map is the comparison from Čech cohomology to real cohomology and the groups off of the edge are obstructions to Čech cohomology working. Of course, we are not interested in a single cover, but in the limit over all covers. For any cohomology class there exists a cover such that the restriction to the cover kills the class. Refining by that cover moves that class off of $E_1^{p,0}=H^p(U;F)$ to at least $E_1^{(p-1),1}$. If $p=1$ then it has moved into Čech cohomology. That is why first Čech is always correct in the limit over all covers. But there is no guarantee that a cycle can be moved all the way to the Čech edge $E^{0,p}$. Indeed, this cannot be done in the example from the beginning: the second cohomology of the affine plane with the doubled origin with coefficients in the structure sheaf. Switching back to the language of ordinary spaces, our cover must have two open sets, each containing one of the two origins; their intersection is punctured in codimension 2 and thus has cohomology. Moreover, it is essentially the same cohomology, independent of the choice of cover. It cannot be killed by refining the cover, only by leaving the cover the same and introducing an auxiliary cover of the intersection, ie, Verdier’s theory of hypercovers, a small change to Čech cohomology that always works. I have identified a cycle in $E_1^{1,1}$ that I claim obstructs Čech cohomology from being correct in this example. I have shown that it survives the limit of all covers. I have not shown that it survives to $E_2$, let alone $E_\infty$. For proofs of such pathologies, see the literature on hypercovers.<|endoftext|> TITLE: How many different knot types can have the same shadow (projection) QUESTION [12 upvotes]: If we take a knot diagram and ignore the over/under-crossing information, we obtain a shadow (or projection). Thus a shadow is simply a plane embedding of a 4-regular graph. Clearly, two non-equivalent knots can give rise to the same shadow: in the figure below we have a knot diagram of the trefoil knot (left) and of the unknot (center), which yield the same shadow (right). Going in the opposite direction, a shadow with $n$ vertices induces $2^n$ knot diagrams, since we can assign to each vertex an overcrossing or an undercrossing. In this same forum, Joe O'Rourke asked if at least one of these $2^n$ knot diagrams corresponds to the unknot (the answer is yes). My question is: Start with a shadow of $n$ vertices. How many different knot types can be obtained from the $2^n$ knot diagrams induced by the shadow? For instance, for the shadow on the right hand side of the figure above, the answer is $3$: by assigning overcrossings or undercrossings to each vertex, one can only obtain the trefoil knot, its mirror image, and the unknot. The answer surely depends on the shadow. Andy Putman's answer highlights an important point (which was also observed by Lou Kauffman in an email message) : one should only consider shadows with no "cut vertices", since such vertices will become nugatory crossings in a knot diagram. If we don't impose this restriction, there are examples such as the one below (Andy's example). No matter what assignment of over/under-crossings this shadow receives, each crossing will be nugatory, and the corresponding knot will always be the unknot: So one should restrict the question to shadows with no cut vertices. Lou Kauffman has observed that in this case, the two alternating diagrams associated to the shadow correspond to non-trivial knots. This follows since that the number of crossings in an alternating, reduced diagram is an invariant of the alternating knot (for a proof, see http://homepages.math.uic.edu/~kauffman/Bracket.pdf). Thus a partial answer for a shadow with no cut vertices is: at least two of the diagrams it induces correspond to a non-trivial knot. My guess is that the answer is "exponentially many'' for "most" shadows. Or at least for some? More specifically: is it true that there exists a $c>1$ such that for all sufficiently large $n$, there exists a shadow whose $2^n$ associated diagrams yield at least $c^n$ different knot types? I have spent quite a bit of time searching for pretty much any result in this direction, without any success. REPLY [2 votes]: Here is a way to get the desired exponential growth, at least for some shadows. (This is morally similar to Ian's comment, but I thought of this method before I read that (ha!), and my proof is different.) Fix $\Gamma$ a four-valent planar graph without cut vertices, cut pairs (of edges), or "Conway spheres". For example, a large-ish, square-ish region of the square tiling of $\mathbb{R}^2$, plus some extra edges along the boundary, will work. I also want that $\Gamma$ has no symmetries. (I'll add an additional hypothesis below). Fix $k < \ell - 2$ both "large enough" and independent of $\Gamma$. We now replace all of the vertices of $\Gamma$ with twist regions having somewhere between $k$ and $\ell - 2$ twists. (Some fussing about is needed to ensure that the result is the diagram of a knot, not a link.) We choose $\Gamma$, $k$, $\ell$, and the sizes of the twist regions so that the resulting knot is hyperbolic and so that each clasp (about each twist region) is a very short geodesic in the hyperbolic metric. Thus, hyperbolic symmetries of the knot complement preserve the set of clasps, so preserve the diagram $\Gamma$, and thus are trivial. We can now add (or not) one more full twist to each twist region. We count - the number of crossings is (at most) $\ell \cdot |V(\Gamma)|$ while the number of knots produced is more than $2^{|V(\Gamma)|}$. Since $\ell$ is independent of $\Gamma$, we are done.<|endoftext|> TITLE: Extending the discussion on "super Catalan" numbers QUESTION [9 upvotes]: Ira Gessel "dubbed" the name super Catalan to $$S(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}$$ and offers a combinatorial proof in his paper Note. The numbers $\frac12S(m,n)$ are also integers. I would like to extend the discussion by asking for a proof that the following numbers, which I call super super Catalan numbers type 1, are integers $$S(x,y,z)=\frac13\frac{(3x)!(3y)!(3z)!}{x!^2y!^2z!^2(x+y+z)!}$$ and the same question of integrality about the numbers, which I call super super Catalan numbers of type 2, $$T(x,y,z)=\frac{x}3\frac{(3x)!(3y)!(3z)!}{x!^3y!^3z!^3(x+y+z)}$$ provided that $x, y, z$ are non-negative integers (plus $x+y+z>0$ for the latter). Remark. It is evident that $\frac{(3x)!(3y)!(3z)!}{x!^3y!^3z!^3}$ are integral; the extra factor $x$ in the numerator of $T(x,y,z)$ is necessary, although $y$ or $z$ would do. I don't have a proof to these claims, but I am convinced of their truth. Even a generating function method is acceptable, yet a combinatorial proof is much desirable. REPLY [7 votes]: I have a somewhat more algebraic proof than Jan-Christoph Schlage-Puchta's carry-based proof. I believe the idea goes back to Landau. Let's start with your super super Catalan numbers type 1. A rational number $a$ is integer iff $v_p(a)\ge 0$ for all primes $p$. Hence, to prove integrality of those numbers, it suffices to show that $$(*) \forall p: v_p( \frac13\frac{(3x)!(3y)!(3z)!}{x!^2y!^2z!^2(x+y+z)!} ) \ge 0 .$$ Define the following function $f:\mathbb{R}^3 \to \mathbb{Z}$: $$ f(x_1,x_2,x_3) = \lfloor 3x_1 \rfloor + \lfloor 3x_2 \rfloor + \lfloor 3x_3 \rfloor - 2\left( \lfloor x_1 \rfloor + \lfloor x_2 \rfloor + \lfloor x_3 \rfloor \right) - \lfloor x_1+x_2+x_3 \rfloor.$$ Legendre has shown that $$v_p(n!) = \sum_{ k \ge1} \lfloor \frac{n}{p^k} \rfloor,$$ which implies that $$(**) v_p( \frac13\frac{(3x)!(3y)!(3z)!}{x!^2y!^2z!^2(x+y+z)!} ) = \sum_{k \ge 1} f(\frac{x}{p^k}, \frac{y}{p^k}, \frac{z}{p^k}) - \delta_{p=3}.$$ Hence, to establish $(*)$, at least for $p \neq 3$, it suffices to show that $f$ is always non-negative. Note that $f$ has period 1 in each of its variables. Hence, WLOG we may assume that $0\le x_i <1$, in which case $$(***) f(x_1,x_2,x_3) = \lfloor 3x_1 \rfloor + \lfloor 3x_2 \rfloor + \lfloor 3x_3 \rfloor - \lfloor x_1+x_2+x_3 \rfloor.$$ Write $x_i$ as $\frac{a_i}{3}+r_i$, where $a_i \in \{0,1,2\}$ and $0 \le r_i < \frac{1}{3}$. Then $(***)$ becomes $$f(x_1,x_2,x_3) = a_1+a_2+a_3 - \lfloor \frac{a_1+a_2+a_3}{3} + (r_1+r_2+r_3) \rfloor,$$ which is non-negative since $$\frac{a_1+a_2+a_3}{3} + (r_1+r_2+r_3) < \frac{a_1+a_2+a_3}{3} + 1 \le a_1+a_2+a_3 + 1.$$ What about $p=3$? Because of the extra term $-\delta_{p=3}=-1$ in $(**)$, we must show that at least one of the summands in $(**)$ is positive. If we let $k$ equal the largest integer such that $3^{k-1} \mid x,y,z$, and write the fractional values of $\frac{x}{3^k}, \frac{y}{3^k}, \frac{z}{3^k}$ as $\frac{a_1}{3} + r_1, \frac{a_2}{3}+r_2, \frac{a_3}{3} + r_3$, then: $$r_1=r_2=r_3=0, a_i \in \{0,1,2\}, a_1+a_2+a_3 \ge 1.$$ Hence, the calculations above imply that $$f(\frac{x}{3^k}, \frac{y}{3^k}, \frac{z}{3^k}) = a_1+a_2+a_3 -\lfloor \frac{a_1 + a_2+a_3}{3} \rfloor \ge 1, $$ which implies that the $k$'th summand in $(**)$ is positive. As for the super super Catalan numbers of type 2, note that $$T(x,y,z) = S(x,y,z) \binom{x+y+z-1}{x-1,y,z},$$ so it is integer as a product of two integers. Remark: The same method shows that $C_m(a_1,\cdots,a_n)=\frac{\prod_{i=1}^{n} (m \cdot a_i)!}{\prod_{i=1}^{n} a_i!^{m-1} (\sum a_i)!}$ is integer for all positive integers $a_1,\cdots,a_n,m$, and that $\frac{C_m}{m}$ is integer whenever $m$ is prime. Although I'm not sure that this is the "right" generalization of the super Catalan numbers. REPLY [5 votes]: Let $p\neq 3$ be a prime. Then \begin{eqnarray*} \nu_p\left(\frac{(3x)!}{x!^3}\right) & = & \sum_k \left[\frac{3x}{p^k}\right]-3\left[\frac{x}{p^k}\right]\\ & = & \sum_k 3\left\{\frac{x}{p^k}\right\} - \left\{\frac{3x}{p^k}\right\}\\ & = & \frac{3s_p(x)-s_p(3x)}{p-1}, \end{eqnarray*} where $s_p$ denotes sum of digits to base $p$. The last quantity equals the number of carries in the multiplication $3\cdot x$ and is certainly non-negative. We have to show that if $p^k|\frac{x+y+z}{(x,y,z)}$, then there are at least $k$ carries in the multiplications $3\cdot x$, $3\cdot y$, $3\cdot z$. We may assume that $(x,y,z)=1$. Now $x+y+z\equiv 0\pmod{p}$, but not all of $x,y,z$ are divisible by $p$, hence $x+y+z=\nu p$, $1\leq\nu\leq 2$, thus one of them is $>\nu p/3$, which implies there are at least $\nu$ carries at the last position. If $k=1$, we are done, otherwise let $x_2,y_2,z_2$ be the second to last digits of $x,y,z$. We have $x_2+y_2+z_2+\nu\equiv 0\pmod{p}$, thus one of $x_2,y_2,z_2$ is larger than $\frac{\mu p-\nu}{3}$, $1\leq\mu\leq 2$, together with the contribution $\nu$ coming from the carry in the last position we get $\mu$ carries here. Continuing in this way we find that the number of carries is at least $k$, hence $\nu_p\left(\frac{(x,y,z)(3x)!(3y)!(3z)!}{x!^3y!^3z!^3(x+y+z)}\right)$ is non-negative. One could probably deal with $p=3$ in a similar way using the fact that to base 3, $3\cdot x$ automatically has a carry at the last position, although the details would become a little lengthy.<|endoftext|> TITLE: Sum over Permutations and Possible Relations with the Polya Enumeration Theorem QUESTION [7 upvotes]: A specialization of the Polya enumeration theorem can give rise to the following identity: $\frac{1}{k!} \sum_{\pi \in S_k} n^{cyc(\pi)} = {n+k-1 \choose k}, $ where $S_k$ is the permutation group of degree $k$ and $cyc(\pi)$ is the number of cycles in the permutation $\pi$. Now, I am interested in the following summation $\sum_{\pi \in S_k} n_1^{cyc(\pi)} n_2^{cyc(\pi \xi )} n_3^{cyc(\pi \xi^{-1})}, $ where $\xi$ is the cyclic permutation of the cycle type $(123....k)$. Is there an expression in closed form for this summation just like the previous summation? Is this summation related to the Polya enumeration theorem in some way？ We might need to separately discuss the two cases with even $k$ and odd $k$. The naive reasoning is that, for example, if we look at the term with the highest power in, say $n_2$, we get $n_1n_2^k n_3$ for odd $k$ and $n_1n_2^k n_3^2$ for even $k$. REPLY [2 votes]: This is not an answer but stated here for lack of margins in the "comments" area. First, I do not believe there will be a nice structural description for the sum, let alone a closed form. A couple of things are apparent: (a) the sum is symmetric in $n_2$ and $n_3$ but not in $n_1$; (b) while from above, $$\sum_{\pi\in S_k}x^{cyc(\pi)}=(x)^{(k)}.$$ Here is a contrasting analogue $$-\sum_{\pi\in S_k}x^{cyc(\pi)}(-1)^{cyc(\pi\xi)}=(x)_{(k)}.$$ Notations. $(x)^{(k)}$ and $(x)_{(k)}$ stand for rising and falling factorials, respectively.<|endoftext|> TITLE: Knot invariants from skein relations QUESTION [5 upvotes]: Let $k$ be a fixed field. A skein relation defined on $k$ is a local relation on $k$-linear combination of oriented planar link diagrams saying that one can trade an overcrossing for an undercrossing by paying a price in the form of an error term which contains no crossings, and vice versa. More formally, it is a linear relation of the form $$ overcrossing = a\, undercrossing + b\, noncrossing $$ with $a,b\in k$, with $a\neq 0$. This last condition is what is needed to get the other way round relation $$ undercrossing = a^{-1}\, overcrossing - a^{-1}b\, noncrossing. $$ Moreover, if $b\neq 0$, which is the generic case, we can also write $$ noncrossing = b^{-1}overcrossing - ab^{-1}\, undercrossing. $$ The first two relations and Reidemeister relations allow one to reduce any closed link diagram to a $k$-linear combination of disjoint copies of unknot diagrams. Moreover, the third relation shows that two disjoint unknot diagrams are equivalent (under Reidemeister) to a certain scalar multiple of a single unknot, and so inductively that an arbitrary $k$-linear combination of disjoint copies of unknot diagrams is ultimately equivalent to some scalar multiple of a single unknot diagram. Then, adding a single relation of the form $$ unknot = c\, \emptyset $$ for some $c\in k$, we see that every closed link diagram is equivalent to a scalar multiple of the empty link diagram. Clearly if $c=0$ then this scalar multiple is zero, so in order to get a nontrivial result, let us assume $c\neq 0$. This shows that there exists at most a unique link invariant $$ P: \text{closed links in $\mathbb{R}^3$}/\text{isotopy} \to k $$ with $P(skein)=0$, $P(unknot)=c$ and $P(\emptyset)=1$. However the skein relation imposes a constraint on $c$ in order for $P$ to possibly exist. Namely, from $$ noncrossing = b^{-1}overcrossing - ab^{-1}\, undercrossing $$ we see that the disjoint union of two unknots is equivalent to $b^{-1}(1-a)$ times a single unknot, so we get the equation $$ c^2=cb^{-1}(1-a), $$ which, since we are assuming $c\neq 0$, determines $c$ as $$ c=\frac{1-a}{b} $$ For instance, for the Jones polynomial, $k=\mathbb{Q}(q)$, $a=q^4$ and $b=q-q^3$ so that $c=q+q^{-1}$ in this case (there are other possible conventions/normalizations for the Jones polynomial, this is the one I use). The question is the following: is the necessary condition $c=(1-a)b^{-1}$ also sufficient to have a nontrivial closed link invariant $P$ with $P(skein)=0$, $P(unknot)=c$ and $P(\emptyset)=1$? In other words, can it be shown that the reduction from a closed link diagram to a scalar multiple of an unknot diagram done using a skein relation and Reidmeister moves is always independent of the particular reduction process? (i.e., that two reductions necessarily lead to the same scalar multiple of the the value of the invariant on the unknot diagram?) REPLY [6 votes]: Yes, this condition is sufficient. The $b=0$ case is trivial. For the $b\ne 0$ case, embed $k$ in a larger field $K$ where $-a$ has a square root; then your knot invariant can be derived from the HOMFLY-PT knot invariant $P_H(\ell,m)$ (which lives in $K[\ell,\ell^{-1},m]$), which satisfies the skein relation $$ \ell(\text{over})+\ell^{-1}(\text{under}) + m(\text{none}) = 0 $$ and $P_H(\text{unknot})=1$ (with no defined value on the empty link). Then your invariant $P$ is just $cP_H\left(\frac{1}{\sqrt{-a}},\frac{-b}{\sqrt{-a}}\right)$ with the extra value $1$ for the empty link. The paper definining HOMFLY-PT is only 8 pages and includes the outlines of four distinct proofs of the well-definedness. P. Freyd, D. Yetter, J. Hoste, W. B. R. Lickorish, K. Millett, and A. Ocneanu, MR 776477 A new polynomial invariant of knots and links, Bull. Amer. Math. Soc. (N.S.) 12 (1985), no. 2, 239--246.<|endoftext|> TITLE: Is the ring of meromorphic modular forms on a fine modular curve generated in degree 1? QUESTION [6 upvotes]: Consider a torsion-free congruence subgroup $\Gamma\le SL_2(\mathbb{Z})$ (for example, $\Gamma(N)$ for $N\ge 3$, or $\Gamma_1(N)$ for $N\ge 4$). By a meromorphic modular form for $\Gamma$ of weight $k$, I mean a holomorphic function $f$ on $\mathcal{H}$ with $f(\gamma z) = (cz+d)^kf(z)$ for $\gamma\in\Gamma$, which is meromorphic at the cusps. Let $M_k(\Gamma)$ be the $\mathbb{C}$-vector space of meromorphic modular forms for $\Gamma$ of weight $k$. Is the graded ring $\bigoplus_{k\ge 0} M_k(\Gamma)$ generated in degree 1? I believe this follows from the geometric interpretation of modular forms as sections of line bundles on moduli stacks (at least when the moduli stack is representable), but I just wanted to check that this is also known from an analytic perspective as well. Also, if the above is true, then can someone explain roughly why this fails if $\Gamma$ is not torsion-free? (ie, are line bundles on stacks, viewed as the Spec of the symmetric algebra of the corresponding invertible module not necessarily generated in degree 1? Perhaps I don't have the right definition of a line bundle on a stack) REPLY [3 votes]: Tyler basically answered the question in the comments, but I might as well fill in an answer. Your definition of "meromorphic modular form" is often called "weakly holomorphic modular form" in the literature, to distinguish such forms from those that may have poles in the interior of the half-plane. Weakly holomorphic modular forms of weight $k$ are algebraic sections of some line bundle $\Omega^{k/2}$ on the affine quotient $Y = \Gamma \backslash \mathfrak{H}$, and by affineness, the space of sections $\Gamma(Y, \Omega^{k/2}) \cong M_k(\Gamma)$ is given by the $k$th tensor power of $\Gamma(Y, \Omega^{1/2})$ as a $\Gamma(Y, \mathcal{O})$-module. Multiplication then yields a ring isomorphism $\operatorname{Sym}^*_{\Gamma(Y, \mathcal{O})} \Gamma(Y, \Omega^{1/2}) \cong \bigoplus_{k \geq 0} M_k(\Gamma)$. When $\Gamma$ has torsion, the stack quotient is no longer an affine curve, and the line bundle $\Omega^{1/2}$ is nontrivial. We can see this concretely as follows: if $\Gamma$ has an element $\left( \begin{smallmatrix} a&b \\ c & d \end{smallmatrix} \right)$ of finite order $n > 1$, then it has a fixed point $x \in \mathfrak{H}$, and $(cx + d)^1$ is not equal to 1. Thus, all elements of $M_1(\Gamma)$ vanish at $x$. For example, if $\Gamma$ contains $\left(\begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$, then any weight 1 form satisfies $$f(\tau) = f\left(\frac{-\tau}{-1}\right)= (-1)^1 f(\tau)=-f(\tau),$$ at all $\tau$, so weight 1 forms for $\Gamma$ necessarily vanish everywhere. Since there are positive weight level 1 forms like $\Delta$ that vanish nowhere, there must exist generators whose weight is greater than 1.<|endoftext|> TITLE: Coend of $\mathscr{D}(F(\bullet), G(\bullet))$ QUESTION [8 upvotes]: (Crosspost from stack) Given categories $\mathscr{C}$ and $\mathscr{D}$ and functors $F,G: \mathscr{C} \to \mathscr{D}$, we can form a bifunctor $$\mathscr{D}(F(\bullet), G(\bullet)): \mathscr{C}^\text{op} \times \mathscr{C} \to \mathsf{Set}$$ and the end of this functor is the set of natural transformations from $F \Rightarrow G$. (I guess we need $\mathscr{C}$ to be small in order to guarantee there is a set of such natural transformations, in general.) Can I say anything about the coend? Is it some familiar thing? REPLY [4 votes]: I'm not sure I have much to add beyond my comment, but I might add a point of view (which could in turn provide some search terms). I'd situate this construction within the bicategory of (small) categories, profunctors/bimodules, and transformations between them. Recall that a profunctor $R: C \nrightarrow D$ is a functor of the form $C^{op} \times D \to Set$ (conventions may differ), composed much in the way relations are, via the formula $$(C \stackrel{R}{\nrightarrow} D \stackrel{S}{\nrightarrow} E)(c, e) = \int^{d: D} R(c, d) \times S(d, e).$$ (If you like, you can consider the bicategory of profunctors as biequivalent to a strict 2-category whose objects are small categories and whose morphisms $C \to D$ are given by cocontinuous functors $Set^C \to Set^D$.) This bicategory is compact closed in an evident bicategorical sense: we have a symmetric monoidal bicategory whose tensor at the object level is given by cartesian product of small categories, and each object $C$ has a monoidal dual given by the opposite category $C^{op}$. For each $C$, the unit $\eta_C: 1 \nrightarrow C^{op} \times C$ is given by $\hom_C: C^{op} \times C \to Set$. (In the cocontinuous functor picture, it's the unique (up to isomorphism) cocontinuous functor $Set \to Set^{C^{op} \times C}$ that takes the terminal object $1$ to $\hom_C$.) The counit $\epsilon_C: C \times C^{op} \nrightarrow 1$ may also be described by a hom-functor, but it is probably more illuminating to think of it in terms of the cocontinuous functor picture, given by taking the coend $\int^C: Set^{C^{op} \times C} \to Set$. Since we are working in a compact closed (bi)category, we can expect certain resonances with constructions in other compact closed categories, such as the category of finite-dimensional vector spaces. The construction in question is a profunctor composite $$1 \stackrel{\eta_C}{\nrightarrow} C^{op} \times C \stackrel{F^{op} \times G}{\nrightarrow} D^{op} \times D \stackrel{\epsilon_{C^{op}}}{\nrightarrow} 1$$ which is certainly akin to trace operations in linear algebra. Thus, in linear algebra over a field $k$, we have the notion of trace of an endomorphism $f: V \to V$, which we can form categorically as the composite: $$\text{Tr}(f) = \left(k \stackrel{\eta_V}{\to} V^\ast \otimes V \stackrel{1 \otimes f}{\to} V^\ast \otimes V \stackrel{eval_V}{\to} k \right)$$ where the first map $\eta_V$ takes $1 \in k$ to $\sum_{i = 1}^n f^i \otimes e_i$ (here $\{e_1, \ldots, e_n\}$ is a basis of $V$ and $f^i$ is the dual basis; the expression $\sum_{i=1}^n f^i \otimes e_i$ is independent of basis). Similarly, we speak of the trace of an endoprofunctor $B: C \nrightarrow C$; after a brief Yoneda-lemma type calculation, one finds that the composite $$1 \stackrel{\eta_C}{\nrightarrow} C^{op} \times C \stackrel{1 \otimes B}{\nrightarrow} C^{op} \times C \stackrel{\epsilon_{C^{op}}}{\nrightarrow} 1$$ is the profunctor $1 = 1^{op} \times 1 \to Set$ taking the unique object of $1$ to $\text{Tr}(B) = \int^{c: C} B(c, c)$. Thus we could also describe your construction as the trace of the endoprofunctor or endobimodule $B: C \nrightarrow C$ defined by $B(c, d) = \hom_D(Fc, Gd)$. Possibly this gives a useful search term. Usually such traces are challenging to calculate explicitly (for example, determining the trace of an identity functor can be nontrivial). Among the properties of trace formally deducible from compact closed structure is $\text{Tr}(B \circ B') \cong \text{Tr}(B' \circ B)$.<|endoftext|> TITLE: Spin 4-manifold bounded by a mapping torus of tori QUESTION [10 upvotes]: Consider a smooth torus endowed with the non-bounding spin structure. Pick a basis of its first homology and a diffeomorphism inducing the S-transformation $\left(\begin{array}{cc} 0 & 1 \\-1 & 0\end{array}\right)$ in the chosen basis. Construct the mapping torus $U$ of this diffeomorphism. Take the non-bounding spin structure on the $S^1$ base. This induces a spin structure on $U$. As the 3rd spin cobordism group vanishes, there is a smooth spin 4-manifold $W$ whose boundary is $U$. The question is: How can one concretely construct such a $W$? More generally, when given a manifold that is known to bound, what are the tools available to construct a concrete bordism to the empty manifold? Update: I explain why Marco's answer below does not provide 4-manifolds bounding the mapping torus above with the required spin structure. It is not difficult to see that $H^1(U;\mathbb{Z}_2) = \mathbb{Z}_2^2$, with one generator associated to the base circle and one generator associated to the cycle on the torus fiber left invariant under the S transformation. The required spin structure restricts to the non-bounding spin structure on representative loops in both classes. This means that these loops cannot become homotopically trivial in the 4-manifold. As the 4-manifolds Marco constructs are all simply connected, they cannot carry a spin structure restricting to the one I am interested in on the boundary. REPLY [4 votes]: Marco's answer above comes close to a solution, but as he mentions, one still has to figure out which of the manifolds bounds the mapping torus with the specified spin structure. I want to record this part of the reasoning in this answer. The possible four manifolds are described by the following diagrams: Marco's second diagram above. A diagram with a trivalent node labeled 0, linked to a -4 node, a -2 node and a string of three 2 nodes, obtained from the characteristic sublink given by one of the -4 nodes in Marco's first diagram. The same diagram as above, obtained from the other -4 node. An affine E7 Dynkin diagram, obtained from the characteristic sublink given by the three nodes -2, -4, -4 in Marco's first diagram. The Rokhlin invariant of a spin 3-manifold is given by the signature mod 16 of a spin 4-manifold admitting it as a boundary. When the 4-manifold is specified by a plumbing diagram as above, this is just the signature of the adjacency matrix of the diagram (where the links indicate an off-diagonal value of -1). For reasons that will become clear, we only need Rokhlin invariants mod 8. The Rokhlin invariants mod 8 of the four 4-manifolds above are -1,1,1,-1. Now we need a way of computing the Rokhlin invariant directly on the mapping torus to compare with the values above. A method for computing the Rokhlin invariant mod 8 in this way has been described by Taylor in Taylor, L. R, Relative Rochlin invariants, Topology and its Applications, 1984, 18, 259 - 280 The Rokhlin invariant is the Arf invariant of a quadratic refinement of the linking pairing on the torsion part of the first integral homology. The value of the quadratic refinement on a given homology class is determined by the restriction of the spin structure of the 3-manifold to a representing cycle. Getting all the sign conventions straight to compare with the values of the Rokhlin invariant computed above looks like a nightmare, but fortunately we can reason as follows. On the one hand, the first torsion homology can be represented by one of the cycle generating the homology of the fiber of $U$. This means that switching the spin structure on the base circle will not change the Rokhlin invariant. On the other hand, Johannes' solution above corresponds to the affine E7 diagram, as it is the only one that can accomodate the degree 2 cohomology classes that will necessarily appear when resolving the orbifold singularities. As I mentioned in a comment of his solution, he has the right spin structure on the fibers, but not on the base. This means that the 4-manifold that I am after is the one that has the same Rokhlin invariant, namely the one originally proposed by Marco.<|endoftext|> TITLE: Divisibility labeling on a boolean lattice and positive Euler totient QUESTION [6 upvotes]: Let $B_n$ be the rank $n$ boolean lattice (i.e. the subset lattice of $\{1,2, \dots , n \}$). Let $\hat{0}$ and $\hat{1}$ be the minimum and the maximum of $B_n$. Let $f: B_n \to \mathbb{N}$ be a labeling satisfying: $f(\hat{0}) = 1$ $a < b \Rightarrow f(a) < f(b)$ and $f(a) \mid f(b)$ $ f(a)f(b) \le f(a \vee b)f(a \wedge b)$, $\forall a,b \in B_n$ Let $\varphi(f) := (-1)^n\sum_{a \in B_n} (-1)^{|a|}f(a)$, with $|a|$ the cardinal of $a$ as subset of $\{1,2, \dots , n \}$. Example: For $B_3$ labeled as below we have $\varphi(f) = 30-6-10-15+2+3+5-1 = 8$. Question: Is it true in general that $\varphi(f) > 0$ ? Examples: the divisor lattice of a square-free integer is a labeled boolean lattice satisfying the properties above, and $\varphi(f)$ is exactly the Euler totient of this integer, so that $\varphi(f) > 0$. If we label a boolean interval of finite groups $[H,G]$ with $f(K) = |K:H|$, the labeling satisfies the properties above using the product formula, and $\varphi(f)$ is exactly the number of cosets $Hg$ such that $\langle Hg \rangle = G$, so that $\varphi(f)>0$ by Ore's theorem. Checking: We have checked $B_3$ by SAGE for $f(\hat{1})<10^6$. Remark: it is immediate for $B_1$, and for $B_2$, take $x_1 = f(\{1,2\})$, $x_2 = f(\{1\})$, $x_3 = f(\{2\})$ and $x_4 = f(\emptyset)$, then $x_2 + x_3 \le \frac{x_1}{2} + \frac{x_1}{2} < x_1+x_4$. REPLY [2 votes]: It's false in general, there is a counter-example of rank $6$. Consider the labeling $f$ on a boolean lattice of rank $n$ such that for every maximal chain $$\hat{0} = a_0 < a_1< \cdots TITLE: Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$ QUESTION [11 upvotes]: Let $q = e^{2\pi i\,z}$. I. 24th power The Ramanujan tau function $\tau(n)$ is given by the expansion of the Dedekind eta function $\eta(z)$'s $\text{24th}$ power. Then $$\begin{aligned}\eta(z)^{24} &= \sum_{n=1}^\infty\tau(n)q^n\\&=q - 24q^2 + 252q^3 - 1472q^4 + 4830q^5 - 6048q^6 - 16744q^7 + \dots\end{aligned}$$ Ramanujan observed that $$\tau(n)\equiv\sigma_{11}(n)\ \bmod\ 691\tag1$$ II. 12th power Assume the rho function $\rho(n)$ as, $$\begin{aligned}\eta(2z)^{12} &= \sum_{n=1}^\infty\rho(n)q^n\\&=q - 12q^3 + 54q^5 - 88q^7 -99q^9 +540q^{11} - 418q^{13} -648q^{15} + \dots\end{aligned}$$ Note the odd powers. Is it true that $$\rho(n)\equiv\sigma_{5}(n)\ \bmod\ 2^8\tag2$$ analogous to $(1)$? P.S. It's true for the first 10000 coefficients in OEIS A000735. REPLY [4 votes]: To keep the discussion alive and local, I add here further manifestations of the above behavior. Let $q = e^{2\pi i z}$, III. 8th power Define the numbers $a(n)$ according to $$\begin{aligned}\eta(3z)^8 &= \sum_{n=1}^\infty a(n)q^n\\ &=q - 8q^4 + 20q^7 - 70q^{13}+64q^{16} +56q^{19} - 125q^{25} -160q^{28} + \dots\end{aligned}$$ Then I claim that $$a(n)\equiv \sigma_3(n) \mod 81.$$ IV. 6th power Define the numbers $b(n)$ according to $$\begin{aligned}\eta(4z)^6 &= \sum_{n=1}^\infty b(n)q^n\\ &=q - 6q^5 + 9q^9+10q^{13}-30q^{17} +11q^{25}+42q^{29}-70q^{37} + \dots\end{aligned}$$ Then I claim that $$b(n)\equiv \sigma_3(n) \mod 4.$$<|endoftext|> TITLE: Reference to "bounds of Weil and Deligne" QUESTION [5 upvotes]: In the this paper by Friedlander and Iwaniec, it is said that they are "able to avoid much of the high-powered technology frequently used in modern analytic number theory such as the bounds of Weil and Deligne." What bounds are being referred to here? REPLY [4 votes]: The first of these bounds was explicitly stated by Weil in the short paper André Weil, "On some exponential sums" (1948) It depends on the Riemann hypothesis for curves over finite fields, which he had proved back in the early 40s, and shows that $$|S(m,n;p)|\leq 2 \sqrt{p}$$ where $S(m,n;p)$ is the classic Kloosterman sum $$S(m,n;p)=\sum_{\substack{{x (\mathrm{mod}\, p)} \\ (x,p)=1}} e\left(\frac{mx+nx^{-1}}{p}\right)$$ The second bound is due to Deligne, and holds for the more general hyper-Kloosterman sum, that is, sums of the form $$K_n(p)=\sum_{\substack{{x_1,...,x_n (\mathrm{mod}\, p)} \\ x_1 \cdots x_n=1}} e\left(\frac{x_1+\cdots+x_n}{p}\right)$$ and the precise estimate is $$|K_n(p)|\leq np^{(n-1)/2}$$ This one depends on the Riemann hypothesis for functions fields, which Deligne had proved on the famous paper Pierre Deligne, "La conjecture de Weil: I" (1974) The applications to hyper-Kloosterman sum were worked out in detail in Pierre Deligne, "SGA 4½: Cohomologie étale" (1977) particularly the Exposé 6, "Applications de la formule des traces aux sommes trigonométriques", section 7, "Sommes de Kloosterman généralisées". It also important to mention the work of Katz on exponential sums, building directly on that of Deligne, particuarly the book Nicholas Katz "Gauss Sums, Kloosterman Sums, and Monodromy Groups" (1988)<|endoftext|> TITLE: Atiyah on the "Galois group of the octonions" and Physics QUESTION [10 upvotes]: Apparently Atiyah was talking about the "Galois group of the octonions" and the unification of the forces of physics at the Heidelberg Forum. Unfortunately not on the stage -- it didn't make its way into his talk. Can anyone who has heard Atiyah talk about this (at the Heidelberg Forum or elsewhere) say what he has said? (If Atiyah is ok with that being posted here, of course). "Listening in on some of his conversations, I don’t think I’m violating any confidentiality by reporting that he’s quite taken with the idea that if one could make sense of the “Galois group of the octonions” one would find that it lies at the heart of unification of the forces of physics. I can’t do justice to his arguments for this, but, if you can find him, I’m sure you’ll get an enthusiastic explanation." (Via http://scilogs.spektrum.de/hlf/sir-michael-atiyah-unity-mathematics-physics/ ) REPLY [2 votes]: The video of his lecture is at the HLF website. I was unable to find it myself but thanks to Lashi Bandara we have it: http://www.heidelberg-laureate-forum.org/blog/video/lecture-monday-september-19-2016-sir-michael-atiyah/<|endoftext|> TITLE: Why do we say fano varieties have rich geometry? QUESTION [9 upvotes]: Fano varieties are defined by the ampleness of $-K_X$, and a rough statement of a step in the Mori program is to check whether a variety is a Fano-fibered one. By that reason, Fano ones are important in classification problems. From somewhere, I have read the phrase "Fano varieties have rich geometry". What is geometry in this context? How does their geometry help us to classify the Fano varieties? REPLY [7 votes]: Let $X$ be an smooth Fano variety of dimension $n$, defined over $\mathbb{C}$ to simplify. Here is a non exhaustive list of some nice properties of Fano varieties, which of course can be complemented by the experts here in mathoverflow. From Kodaira vanishing theorem, $\operatorname{H}^i(X,\mathcal{O}_X)=0$ for $i>0$. In particular, $\operatorname{Pic}(X)\cong \operatorname{H}^2(X,\mathbb{Z})$. In particular the Picard number of $X$ equals to the second Betti number and it is a topological invariant. Moreover, for every $n$ there only finitely many families of Fano manifolds. The Minimal Model Program for a variety with negative Kodaira dimension is expected to end up with a fiber type morphism whose general fibers are (singular) Fano varieties. Moreover, the Cone Theorem easily implies that Fano manifolds has rational polyhedral Mori cone. Even better, they are Mori Dream Spaces. A lot of examples can be explicitly constructed by using toric varieties. They are combinatorial and explicit computations can be done. Also, since general elements of the anticanonical complete linear system $|-K_X|$ are Calabi-Yau varieties, Fano varieties appear in Mirror Symmetry and explicit constructions can be done in the toric case by looking at polytopes. Fano varieties are rationally connected, so there are plenty of rational curves on $X$, their first fundamental group is trivial, there are not non-trivial étale coverings, etc. The singular analogs of these facts are very deep. Maybe you can look at the recent article of C. Birkar "Singularities of linear systems and boundedness of Fano varieties" for (1), BCHM article "Existence of minimal models for varieties of log general type" for (2), D. Cox and S. Katz wrote "Mirror Symmetry and Algebraic Geometry" which talks about (3) at some point, and Kollar's article "Which are the simplest algebraic varieties?" is related to (4). Some lists to look for examples/classifications are the Graded Ring Database (for toric Fano 3-folds and toric Del Pezzo surfaces), the original works of Iskovskikh and Mori/Mukai (see also this article), and also there is a nice list of Fano 4-folds of Fano index $r>1$ here (see also the references therein). I recently assisted to a talk of T. Coates about a joint work with E. Kalashnikov where they found 138 new Fano 4-folds, he also explained how this fits into a program (joint work with Corti, Galkin, Golyshev, Kasprzyk, and others) to find and classify Fano manifolds using mirror symmetry.<|endoftext|> TITLE: Convex representation of a measure QUESTION [7 upvotes]: Let $\mathcal P(X)$ denote the space of all probability measure defined on a measurable space $X$. We canonically endow the former with its own measurability structure, generated by evaluation maps. Let $P \subseteq \mathcal P([0,1])$ be a measurable subset of probability measures, and let $\hat p\notin P$ be such that for every bounded measurable $f:[0,1] \to \Bbb R$ there exists $p_f\in P$ satisfying $$ \int_{[0,1]} f(x) \hat p(\mathrm dx) = \int_{[0,1]} f(x) p_f(\mathrm dx), $$ or $\hat p f = p_f f$ in a short form. Does it necessarily means that there exists a probability measure $\nu \in \mathcal P(\mathcal P([0,1]))$ such that $\nu(P) = 1$ and $\hat p = \int_P p\,\nu(\mathrm dp)$? I think this result is quite easy to show for finite $X$ where $\mathcal P(X)$ is just a subset of $\Bbb R^n$, however I am not sure whether it still holds true in my more general case. Obviously, from $[0,1]$ it would generalize to any Borel space. REPLY [3 votes]: The answer is no. I understand the condition on $p_f$ as belonging to the closure $\bar{P}$ for the weak-$\star$ topologie. And one can then ask whether $[\int p d\nu(p),\nu\in \mathcal{P}(\mathcal{P}([0,1]))]$ contain this closure. Consider the following counter example $$P=[\lambda \delta_0+(1-\lambda)\delta_1:0<\lambda<1]$$ Then $\delta_0 \in \bar{P}$ but not to in the convex set. Here we have $p_ff=\hat{p}f$ which is a stronger condition than $\inf_{p\in P}|pf-\hat{p}f|=0$. We therefore have to work a bit harder to construct the counter example. Let $\hat{p}=\frac{1}{2}\delta_0+\frac{1}{2}1_{[0,1]}$ and choose $$P=\{q_x :x\geq \frac{1}{2}\}\cup[p\in \mathcal{P}([0,1]):p1_{\{0\}}<\frac{1}{2}]$$ with $q_x=\frac{1}{2}\delta_0+\frac{1}{2}\delta_x$ We see that if $\hat{p}=\int_P pd\nu(p)$ then $$\hat{p}1_{\{0\}}=\frac{1}{2}=\frac{1}{2}\nu(\{q_x : x\geq \frac{1}{2}\})+\int_{P-\{q\}} p1_{\{0\}}d\nu(p)$$ and therefore the support of $\nu$ is a subset of $\{q_x :x\geq \frac{1}{2}\}$ which is impossible. We now check that $P$ satisfies the condition. Let $f$ a bounded (measurable) function. If $f(0)=a$ and $f(t)=b$ for all $t\in E\subset [0,1]$ a set of Lebesgue measure 1. Then there exist $x\geq \frac{1}{2}$ such that $f(x)=b$ and we have $\hat{p}f=\frac{1}{2}(a+b)=q_x f$. If $f$ is not constant on a set of Lebesgue measure 1. then there exists $\epsilon >0$ and $\tilde{p}$ with $\tilde{p}1_{\{0\}}=0$ and $$\tilde{p}f>(1+\epsilon)\int f(t)dt $$ Suppose $f(0)\geq \int f(t)dt$. Then by continuity we can find $\lambda<\frac{1}{2}$ such that $\lambda f(0)+(1-\lambda)\tilde{p}f=\hat{p}f$. Suppose $f(0)\leq \int f(t)dt$ then we use the same argument but with $$\tilde{p}f<(1-\epsilon)\int f(t)dt $$<|endoftext|> TITLE: Modifying a Cohen generic QUESTION [7 upvotes]: Let $M$ be a transitive model of ZFC (set- or class-sized) and let $\kappa\in M$ be a regular cardinal (in $V$). Let $G$ be $M$-generic for $\operatorname{Add}(\kappa,1)$. Now suppose that there is an $X\subseteq\kappa$ of size $\kappa$ which has no size $\kappa$ subsets in $M$ (for example, we could take $X=G$). My question concerns which "patterns" can be realized in an $M$-generic on the coordinates in $X$. Given a subset $Y\subseteq X$, is there an $M$-generic $G'$ for $\operatorname{Add}(\kappa,1)$ such that $G'\cap X=Y$? In other words, if we see $Y$ as a pattern of 0s and 1s, is there a generic $G'$ whose pattern on $X$ matches $Y$? The requirement that $X$ has no large subsets in $M$ is clearly necessary: if $X'\subseteq X$ is in $M$ and has size $\kappa$, a density argument shows that $G'\cap X'\neq \emptyset$ for any generic $G'$, preventing us from realizing the all 0 pattern $Y=\emptyset$. Certain patterns can be easy to achieve. For example, if we take $X=G$, then it is simple to realize the all 1 pattern $Y=G$: just take $G'=G$. Similarly, the all 0 pattern $Y=\emptyset$ is realized by $G'=\kappa\setminus G$. This last example also shows that it does not suffice, in general, to simply perform surgery on $G$ on the coordinates in $X$ and that global changes to $G$ might be necessary. I am particularly interested in the special case when $M$ is the ultrapower by a measure on a measurable cardinal $\lambda$, we have $\kappa=\lambda^+$ and GCH holds. In this situation there is a variety of $M$-generics in $V$, which might make the problem easier. REPLY [3 votes]: In some cases, it is possible to construct a set $X$ for which every pattern $Y\subseteq X$ is realized in the way you describe. For example, consider the very natural case where $M$ is a countable transitive model of ZFC and $\kappa=\omega$. So we are talking about adding $M$-generic Cohen reals here. Enumerate the dense subsets $D_0,D_1,\ldots$ of $M$, and let us build a generic $G$ by diagonalization, while also constructing $X$ for which any $Y\subseteq X$ is realized. Construct a descending sequence of conditions $p_0\geq p_1\geq\dots$ as follows. Let $p_0$ be any element of $D_0$. Now, let the first element $i_0$ of $X$ be the first unspecified bit of $p_0$. Let $\bar p_1$ extend $p_0$ into $D_1$, and then flip the bit at $i_0$ in $\bar p_1$ and further extend to $p_1$ in $D_1$. So $p_1\in D_1$, extending $p_0$, and flipping the bit at $i_0$ keeps the condition in $D_1$. Continue similarly. At stage $n$, let $i_n$ be the first unspecified bit of $p_n$; first extend to $\bar p_{n+1}$ into $D_{n+1}$, and then flip the bits at all the $i_k$ for $k\leq n$ in any desired pattern, and extend into $D_{n+1}$ again, repeating for each pattern. At the end, let $p_{n+1}$ be the resulting extension of $p_n$, which has all its finite flips on the $i_k$'s in $D_{n+1}$. Let $G$ be the filter generated by the conditions $p_n$, which is $M$-generic since we have met every dense set, and let $X$ be the set of all $i_n$. By design, for any bit-flipping pattern on $X$, the construction exactly arranges that the resulting filter obtained by surgery will also be $M$-generic. Thus, every pattern $Y\subseteq X$ is realized by some $M$-generic surgical modification of $G$. Update. Let me explain how a similar idea can work higher up, such as in the case of your ultrapower, with some additional assumptions. Specifically, let's assume that $\kappa$ is a regular cardinal and $M$ is a transitive set or class model of ZFC, with $M^{<\kappa}\subset M$ and $|P(\kappa)^M|=\kappa$, which is true for your ultrapower example at the end of the question, if we have the GCH at your measurable cardinal. These assumptions are enough to construct $M$-generic Cohen sets $G\subset\kappa$ by the usual diagonalization procedure. You enumerate in $V$ the dense sets of $M$, and then construct a descending $\kappa$-sequence that gets inside them. But let me now assume additionally that the $\Diamond_\kappa$ principle holds in $V$. In this case, I shall construct a set $X\subseteq\kappa$ such that every pattern $Y\subseteq X$ is realized in the way you request. Fix a $\Diamond_\kappa$-sequence $\langle A_\alpha\mid\alpha<\kappa\rangle$. Enumerate the dense sets of $M$ as $D_\alpha$, for $\alpha<\kappa$. At stage $\alpha$ of the construction, we have constructed a descending sequence of conditions $p_\beta$ for $\beta<\alpha$, and we have identified the first $\alpha$ many elements of $X$. Let $\bar p_\alpha$ be the union of the prior conditions, which is a condition in $M$ because $M^{<\kappa}\subset M$. We only do something special if the domain of $\bar p_\alpha$ is $\alpha$ itself, and if furthermore $A_\alpha$ is coding an ordinal $\beta<\kappa$ and a subset $Y$ of the first $\alpha$ members of $X$. In this case, we first extend $\bar p_\alpha$ into $D_\beta$, and then we further extend to a condition $p_\alpha$, so that after flipping the bits on $Y$, the condition remains in $D_\beta$. After this, let the $\alpha^{th}$ member of $X$ simply be the first bit unspecified by $p_\alpha$. By design, the filter $G$ generated by the descending sequence of conditions $p_\alpha$ is an $M$-generic filter. If $Y\subseteq X$ is any set, then let $G_Y$ be the filter obtained by flipping the bits of $G$ on the coordinates of $Y$. This filter is also $M$-generic, since for any dense set $D_\beta$, there is a stationary set of stages $\alpha$ for which the domain of the conditions up to $\alpha$ is $\alpha$ itself (since that occurs on a club) and where $A_\alpha$ codes $\beta$ and the set $Y\cap\alpha$. In this case, the condition $p_\alpha$ had the property that flipping the bits on $Y$ remained in $D_\beta$, and so $G_Y$ meets $D_\beta$. Thus, any desired pattern can be arranged on the digits of $X$, by choosing $Y$ suitably.<|endoftext|> TITLE: What can the extremely large cardinals tell us about small sets? QUESTION [11 upvotes]: Are there any applications of the largest large cardinals to consistency results concerning, say, cardinals below $\aleph_{\aleph_\omega}$? Or perhaps to prove results in descriptive set theory? I am thinking of ZFC + Axiom I0, ZF + Reinhardt, ZF + Berkeley. See here for definitions. The largest cardinal I know of to be used for a consistency result is a 2-huge cardinal to prove the consistency of $(\aleph_3,\aleph_2,\aleph_1) \twoheadrightarrow (\aleph_2,\aleph_1,\aleph_0)$. I would be remiss not to mention the results of Laver about left-distributive algebras proved using Axiom I3. Some of the results have not been brought down to ZFC. This is remarkable and I suppose answers the question as posed, but my motivation is to see if there are forcing constructions aimed at small cardinals that use these very large cardinals. REPLY [9 votes]: The paper Generic $I_0$ at $\aleph_\omega$ by Vincenzo Dimonte might be of interest to you. It introduces the notion of being generic $I_0$ at $\aleph_\omega$ (Def. 3.1 of the paper), and proves several consequences of it. In particular it is shown that if eneric $I_0$ holds at $\aleph_\omega$, then $\aleph_\omega$ is Jonsson. Of course it is open if generic $I_0$ at $\aleph_\omega$ can be consistent!.<|endoftext|> TITLE: are the congruence subgroups $\Gamma(n)$ characteristic inside $\mathrm{SL}_2(\mathbb{Z})$? QUESTION [12 upvotes]: For which $n$ is the "principal congruence subgroup" $\Gamma(n)\le \mathrm{SL}_2(\mathbb{Z})$, the subgroup consisting of matrices congruent to the identity modulo $n$, characteristic? I.e., for which $n$ is $\Gamma(n)$ stable (as a set) under all automorphisms of $\mathrm{SL}_2(\mathbb{Z})$? REPLY [19 votes]: No for $n$ odd, yes for $n$ even. By a result of Hua and Reiner, the automorphism group of $SL_2(\mathbb{Z})$ is generated by (1) conjugation by $GL_2(\mathbb{Z})$ (the matrices with determinant $\pm 1$) and (2) the map $X \mapsto \epsilon(X) \cdot X$ where $\epsilon$ is the unique nontrivial character $SL_2(\mathbb{Z}) \to \pm 1$. The maps of the form (1) clearly preserve $\Gamma(n)$, so we just need to check the single map $X \mapsto \epsilon(X) \cdot X$. We can describe $\epsilon$ as follows: $SL_2(\mathbb{Z})$ acts on the three points of $\mathbb{P}^1(\mathbb{F}_2)$, giving a map $SL_2(\mathbb{Z}) \to S_3$. Compose this with the sign character $S_3 \to \pm 1$. If $n$ is odd, then $\left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) \equiv \left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right) \bmod 2$, so $\epsilon \left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) = -1$. So $\left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right)$ is in $\Gamma(n)$ but $\epsilon \left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) \cdot \left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) = \left( \begin{smallmatrix} -1 & -n \\ 0 & -1 \end{smallmatrix} \right)$ is not. On the other hand, if $n$ is even then $X \equiv \mathrm{Id} \bmod n$ shows that $X$ maps to $e$ in $S_3$, so $\epsilon(X)=1$ and $\epsilon(X) X = X$ for all $X \in \Gamma(n)$.<|endoftext|> TITLE: For which $n$ is it true that all surjections $SL_2(\mathbb{Z})\rightarrow SL_2(\mathbb{Z}/n\mathbb{Z})$ have kernel $\Gamma(n)$? QUESTION [22 upvotes]: For which integers $n$ does every surjection $SL_2(\mathbb{Z})\twoheadrightarrow SL_2(\mathbb{Z}/n\mathbb{Z})$ have kernel $\Gamma(n)$? (this is the usual kernel, ie, the subgroup of matrices congruent to 1 mod $n$). A positive answer (for some $n$) would certainly imply that $\Gamma(n)$ is characteristic inside $SL_2(\mathbb{Z})$, which is true for even $n$ by David Speyer's beautiful answer (pointing to a result of Hua and Reiner) here: are the congruence subgroups $\Gamma(n)$ characteristic inside $\mathrm{SL}_2(\mathbb{Z})$? From his answer, it's clear that our statement is false for odd $n$. However, even for $n\equiv 0\mod 2$, what's strange is that after doing some computations, the statement is true for $n = 2,4,6,\ldots, 24$, is false for $n = 26$, is true for $n = 28,30,32$, and is false again for $n = 34$. Has there been any work on this? (I always find questions about $SL_2(\mathbb{Z})$ hard to google, due to the difficulty of making queries involving symbols). EDIT: So far, after computing examples for even $n$ ranging from 2 through 58, the even $n$ for which the statement is false are: $$26,\ldots, 34,\ldots, 38,\ldots, 46,\ldots, 52, 54,\ldots, 58$$ where "$\ldots$" indicate gaps. REPLY [9 votes]: The following mostly answers the question: Claim 1: If every surjection $SL_{2}(\mathbb{Z}) \to SL_{2}(\mathbb{Z}/n\mathbb{Z})$ has kernel $\Gamma(n)$, then all prime factors $p | n$ have $p \leq 11$. To prove Claim 1, I need another claim. Claim 2: If $p > 11$ is prime, there are elements $x, y \in PSL_{2}(\mathbb{F}_{p})$ so that $\langle x, y \rangle = PSL_{2}(\mathbb{F}_{p})$, $x$ has order $2$, $y$ has order $3$, and $xy$ does not have order $p$. Proof of Claim 1 (using Claim 2): Assume that $p^{k} \parallel n$ with $p > 11$. Choose elements $X, Y \in SL_{2}(\mathbb{Z}/n\mathbb{Z})$ so that $\bullet$ the images of $X$ and $Y$ in $PSL_{2}(\mathbb{F}_{p})$ have orders $2$ and $3$, respectively, $XY$ does not have order $p$, and $X$ and $Y$ generate $PSL_{2}(\mathbb{F}_{p})$, and $\bullet$ replacing $X$ and $Y$ by $p$th powers of themselves (if necessary), assume that $X \bmod p^{k}$ and $Y \bmod p^{k}$ have order $4$ and $6$, respectively, in $SL_{2}(\mathbb{Z}/p^{k} \mathbb{Z})$, and $\bullet$ take $X \equiv \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \pmod{n/p^{k}}$ and $Y \equiv \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} \pmod{n/p^{k}}$. It then follows that $X^{2} \equiv Y^{3} \equiv \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \pmod{n}$ and $X^{4} \equiv I \pmod{n}$. Note that $\langle X, Y \rangle = SL_{2}(\mathbb{Z}/n\mathbb{Z})$ because the map $\langle X, Y \rangle \to SL_{2}(\mathbb{Z}/(n/p^{k})\mathbb{Z})$ is surjective and the map $\langle X, Y \rangle \to SL_{2}(\mathbb{Z}/p^{k} \mathbb{Z})$ is also surjective (because $PSL_2(\mathbb{F}_{p})$ is a proper quotient of the image mod $p^{k}$, and every proper subgroup of $SL_{2}(\mathbb{Z}/p^{k} \mathbb{Z})$ has all composition factors abelian or isomorphic to $A_{5}$). Since $\langle x, y | x^{2} = y^{3}, x^{4} = 1 \rangle$ is a presentation for $SL_{2}(\mathbb{Z})$, there is a homomorphism $\phi : SL_{2}(\mathbb{Z}) \to SL_{2}(\mathbb{Z}/n\mathbb{Z})$ so that $\phi\left(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\right) = X$ and $\phi\left(\begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}\right) = Y$. Now, $$\left(\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}^{3} \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}\right) = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, $$ has order $n$ in $SL_{2}(\mathbb{Z}/n\mathbb{Z})$, but the power of $p$ dividing the order of $X^{3} Y$ is at most $p^{k-1}$, because $(X^{3} Y)^{r} \equiv 1 \pmod{p}$ for some $r$ with $p \nmid r$ and $A^{r} \equiv 1 \pmod{p}$ implies that $A^{rp^{k-1}} \equiv 1 \pmod{p^{k}}$. Thus $(x^{3} y)^{rp^{k-1}} \in \ker \phi$, but $(x^{3} y)^{rp^{k-1}} \not\in \Gamma(n)$. QED Claim 1 Proving Claim 2 should not be too difficult, but it is a bit tedious. This requires carefully counting pairs $x, y$ (of orders $2$ and $3$ respectively) so that $xy$ has order $p$, and also counting pairs $x,y$ that both lie in one of the proper (Borel, dihedral, $A_{4}$, $S_{4}$ or $A_{5}$) subgroups of $PSL_{2}(\mathbb{F}_{p})$. Note that there are some unusual things that happen for $n$ all of whose prime factors are $\leq 11$. In particular, there are examples with $n = 54$, $n = 98$, and $n = 108$. It seems plausible to me that there are only finitely many $n$ for which the kernel must be $\Gamma(n)$.<|endoftext|> TITLE: When the boundary of any subset is compact? QUESTION [8 upvotes]: Let $X$ be a Tychonoff space with no isolated points such that the boundary of any subset of $X$ is compact. Does it mean that $X$ is compact ? (If $X$ is a resolvable space then it is clearly compact.) REPLY [10 votes]: Your property implies in particular: "Every nowhere dense closed subset of $X$ is compact." (A nowhere dense closed subset is the boundary of its complement.) This in turn is equivalent to compactness for $T_1$ spaces with no isolated points, as shown by Katetov in 1947. Interestingly, the result holds as well for Lindelöf spaces, or more generally for $[\lambda,\kappa]$-compact spaces (a space is $[\lambda,\kappa]$-compact if every cover of it by $\le\kappa$ open sets has a subcover of cardinality $<\lambda$). This is due to Mills and Wattel, a very nice short proof can be found in Blair's "Some nowhere densely generated topological properties" (easily found online). Completely unaware of these results, I actually recently published a paper whose main result is weaker than Mills and Wattel's, and with a more complicated proof than Blair's.<|endoftext|> TITLE: How Would an Intuitionist Prove This? QUESTION [32 upvotes]: My question concerns the proof of the following: Let $a,b,n \in \mathbb{N}$. If $n \neq 1$ and $n$ divides both $a$ and $b$, then $b$ is a composite number or $b$ divides $a$. My proof: Suppose $b$ is not composite. Then $b$ is prime. Since $n \neq 1$ and $n$ divides $b$, we must have $n = b$. Thus $b$ divides $a$. This proof has the form $(P \wedge \neg R) \to Q$. If we were using normal first order logic, I could conclude $P \to (Q \vee R)$. But, using intuitionistic logic, I cannot conclude that which was to be proven. How would you carry out this proof in intuitionistic logic? REPLY [19 votes]: You ask how a constructive mathematician would prove the theorem. Just like you did! We only need to verify that every number is either composite or prime, see below. But Carl Mummert gave the "proof from the book", which is constructive: Theorem: If $n \neq 1$ and $n$ divides both $a$ and $b$, then $b$ is composite or $b$ divides $a$. Constructive proof (Carl Mummert). Either $n = b$ or $n \neq b$. If $n = b$ then $b$ divides $a$. If $n \neq b$ then $b$ has a non-trivial divisor and is composite. QED. Here is your proof, worked out in a bit more detail. Constructive proof. If $n$ is not composite then it is prime by the Lemma below. Because $n$ divides $b$ it follows that $b = n$, hence $b$ divides $a$. QED. Lemma: A number $n > 1$ is either composite or prime. Constructive proof. Let us be careful about the meaning of words here. By composite we mean "a product of two numbers, each of which is different from $1$". By prime we mean "a number $p > 1$ whose only divisors are $1$ and $p$". Since every number is dividisble by $1$ and itself, primality is equivalent to "a number $p > 1$ such that it has no divisor between $2$ and $p-1$." Suppose $\phi$ is a decidable predicate on natural numbers, i.e., we have $\forall k \in \mathbb{N} \,.\, \phi(k) \lor \lnot\phi(k)$. Then also the predicates $\forall k \leq n \,.\, \phi(k)$ and $\exists k \leq n \,.\, \phi(k)$ are decidable. (Exercise, prove by induction on $n$.) Using this we can prove: Given $n$ and $k$, it is decidable whether $k$ divides $n$. Given $n$ it is decidable whether there is $k$ such that $2 \leq k < n$ and $k$ divides $n$. But the second statement says that it is decidable whether $n$ is composite. To finish the proof we need to show that a number $n > 1$ which is not composite is prime. Suppose $n > 1$ is not composite. Consider any $k$ such that $2 < k < n$. Either $k$ divides $n$ or it does not. But it cannot divide $n$, or else $n$ would be composite. Therefore $n$ is prime. QED.<|endoftext|> TITLE: Automatic proof in Euclidean Geometry using Theory of Groebner Bases QUESTION [5 upvotes]: I've done the same question in math.stackexchange here "https://math.stackexchange.com/questions/1938261/automatic-proof-in-euclidean-geometry-using-theory-of-groebner-bases?noredirect=1#comment3978891_1938261" with meager responses, so for curiosity's sake I posted here in case of some member knows! Recently, whilst I was reading something in the Theory of Groebner bases, came across the aforementioned question. In the book of Cox, Litlle, O'Shea "Ideals, Varieties and Algorithms" there is a section dealing with that question and some examples, remarks and definitions. Although, I understand the whole concept and the idea behind all this beautiful application of Groebner bases the general question still is wandering around my head. Is it feasible to prove everyevery Theorem arises in Euclidean geometry using Groebner bases, in other words, can we produce an algorithm which provides a way to encode everyevery statement in Euclidean Geomtry in terms of polynomials and conclude to the result of hypotheses we've done by answering the ideal membership question for a polynomial? Also, if the latter isn't true, under which circumstances can we have it and what's the general problem? Plays any particular role the number of hypotheses that we have and the number of dependent variables we use to encode the problem? Because seems to be a matter of concern for the writers under the phrase: "...the number of hypotheses and the number of dependent variables are the same. This is typical of properly posed geometric hypotheses." P.S. Please if you can give me some more information for the above questions like definitions, remarks or references (apart from the aforementioned) would be really helpful REPLY [6 votes]: Not every theorem in Euclidean geometry can be proven by Gröbner basis methods, because the connection between Gröbner bases and geometry only goes through for algebraic closed fields, such as the complex numbers. Euclidean plane geometry is defined over the real numbers, so you need a technique that works over the reals. There such a technique, called quantifier elimination. You can find some details on Wikipedia here. In general, both methods can be very slow. Gröbner bases are known to require doubly exponential time, and quantifier elimination is slower still. REPLY [4 votes]: Although I cannot answer your question of whether every Euclidean geometry theorem can be proved via Gröbner bases (I doubt it), there is literature on the topic. Here is one relatively recent source: Petrovic, Danijela. "Automated Proving in Geometry using Gröbner Bases in Isabelle/HOL." 2011. (PDF download.)         And here is an older, highly cited paper: Kapur, Deepak. "Using Gröbner bases to reason about geometry problems." Journal of Symbolic Computation 2.4 (1986): 399-408. (Journal link.) Abstract excerpt: "Two kinds of geometry problems are considered: (i) Given a finite set of geometry relations expressed as polynomial equations, in conjunction with a finite set of subsidiary conditions stated as negations of polynomial equations to rule out certain degenerate eases, check whether another geometry relation expressed as a polynomial equation and given as a conclusion, holds. (ii) Given a finite set of geometry relations expressed as polynomial equations, find a finite set of subsidiary conditions, if any, stated as negations of polynomial equations which rule out certain values of variables, such that another geometry relation expressed as a polynomial equation and given as a conclusion, holds under these conditions."<|endoftext|> TITLE: are there finite nonabelian characteristic quotients $G$ of $F_2$ inducing a surjection $Aut(F_2)\twoheadrightarrow Aut(G)$? QUESTION [7 upvotes]: Let $F_2$ be the free group of rank 2. Let $K\le F_2$ be a characteristic subgroup, such that $G := F_2/K$ is finite. Do there exist examples of such nonabelian $G$ such that the induced map $$Aut(F_2)\rightarrow Aut(G)$$ is surjective? For example, if $C_n$ is the cyclic group of order $n$, then for every $n$, $G = C_n\times C_n$ is an abelian example of such a group. EDIT: As Derek Holt pointed out in the comments, $C_n\times C_n$ is only an example in the cases $n = 1,2,3,4,6$. REPLY [7 votes]: The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal. There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization. The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms. Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has one only Nielsen equivalence class of generating pairs for $n \in \{2, 3, 4, 6\}$. The group $\text{Heisenberg}_2(\mathbf{Z}/2\mathbf{Z})$ is the dihedral group of order $8$, which is not a quotient of $F_2$ by a characteristic subgroup. But $B(2, 3) = \text{Heisenberg}_2(\mathbf{Z}/3\mathbf{Z})$, is certainly another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by a characteristic subgroup. [1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.<|endoftext|> TITLE: Identity involving a sum over all partitions of $n$ QUESTION [7 upvotes]: In some work I've been doing on the cohomology of the moduli space of curves, the following identity has come up: $$\prod_{i=1}^n \frac{x^{i-1}}{x^i-1} = \sum_{(a_1^{r_1},\ldots,a_{\ell}^{r_{\ell}}) \vdash n} \left(\prod_{j=1}^{\ell} \frac{1}{a_i^{r_i} (x^{a_i}-1)^{r_i} (r_i)!}\right).$$ Here $x$ is a formal variable and the sum on the RHS is over all partitions of $n$. By $(a_1^{r_1},\ldots,a_{\ell}^{r_{\ell}}) \vdash n$, I mean a partition of the form $$r_1 a_1 + r_2 a_2 + \cdots + r_{\ell} a_{\ell} = n$$ with $r_1,\ldots,r_{\ell} \geq 1$ and $a_1>a_2>\cdots>a_{\ell} \geq 1$. I have verified this identity with Mathematica for $1 \leq n \leq 20$. However, I cannot figure out how to prove that it is always true. Can anyone help me? This reminds me a little bit of the identity in this question, and I've tried without success to use the tools discussed in the answers to that question to solve it. EDIT: In case anyone is interested, a version of this identity now appears as Lemma 5.2 in my paper "The high dimensional cohomology of the moduli space of curves with level structures" (joint w/ Neil Fullarton), which can be downloaded from my webpage here. Thanks to Lucia for telling me how to prove it! REPLY [15 votes]: Here's a quick sketch (since I'm pressed for time). Multiply both sides of the identity by $t^n$ and sum over $n$ from $0$ to infinity. From the cycle decomposition identity (Polya's formula) the right side becomes $$ \exp \Big( \sum_{i=1}^{\infty} \frac{t^i}{i (x^i-1)} \Big)= \exp\Big( -\sum_{i=1}^{\infty} \frac{t^i}{i} \sum_{j=0}^{\infty} x^{ji} \Big) = \exp\Big( \sum_{j=0}^{\infty} \log (1-t x^j) \Big) = \prod_{j=0}^{\infty} (1-tx^j). $$ The RHS is also known as a Pochhammer symbol (see the Wikipedia article linked in my comment): it is $(t;x)_{\infty}$. The wikipedia article already describes the combinatorial identity (simple partition relation) $$ (t;x)_{\infty} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n(n-1)/2} }{(x;x)_n} t^n, $$ where $$ (x;x)_n = (1-x) (1-x^2) \cdots (1-x^n). $$ This matches what you get from multiplying your LHS by $t^n$ and summing.<|endoftext|> TITLE: Counter-example that "almost preserving" is not preserved under countable support iteration QUESTION [8 upvotes]: Let $\sqsubset=\bigcup_n \sqsubset_n$ be a relation on $\omega^\omega$ where each $\sqsubset_n$ is arithmetic and $\{f: f\sqsubset_n g\}$ is closed for each $g\in \omega^\omega, n\in \omega, i.e. \Pi_1^0(g)$, a typical example is domination past $n$. In the following fix $\chi $ large enough regular cardinal. Definition [Goldstern]. $\mathbb{Q}$ almost preserves $\sqsubset$ (or is almost preserving) if for all countable $N\prec H(\chi)$ containing $\mathbb{Q}, p\in \mathbb{Q}\cap N, g\in \omega^\omega$ and $N$ is $\sqsubset$-covered by $g$ (any $f\in N, f\sqsubset g$). There exists $q\leq p$ $(N,\mathbb{Q})$-generic such that $q\Vdash \forall f\in N[G] f\sqsubset g$. We can check that this notion is preserved under iterations of finite length, more precisely, if $P$ is almost preserving and $1\Vdash_P \dot{Q}$ is almost preserving, then $P*\dot{Q}$ is almost preserving. My question is: is there any known example (references) that this notion is not preserved at limit stages (countable support iteration of proper forcings)? REPLY [7 votes]: We can do something like setting $f\sqsubset_n g$ if and only if $f(m)=g(n^m)$ for all $m$ (so $f$ looks like $g$ restricted to the powers of $n$). For each $g$, the set $\{f:f\sqsubset_n g\}$ is a singleton, hence closed. Also, given countably many $f$ we can find a single $g$ covering them all using powers of primes. Now we use the fact that there are countable support iterations of length $\omega$ where no new reals are added at each finite stage, but new reals are added in the limit. If $g$ $\sqsubset$-covers $N$ in the ground model and forcing with $P$ adds a new real, then $g$ no longer covers $N$ in the extension.<|endoftext|> TITLE: Does there exist a Penalized Conditional Expectation? QUESTION [6 upvotes]: In my recent work I've become interested in working with the minimizer of $$ \mathbb{E}[(Y-Z)^2] + \lambda P(Z), $$ $Y$ is an observed random variable, $P$ is a positive-convex penalty function, $Z$ is a measurable random variable with respect to the $\sigma$-algebra generated by $Y$ and $\lambda\geq 0$. If $\lambda=0$ this is the MSE and is minimized by the conditional expectation. My question is, is there a well developed theory of "penalized conditional expectation"? That is a theory studying the above equation's minimizer? So far I have found nothing really. All help is greatly appreciated. REPLY [3 votes]: This is a bit different and doesn't address the question, but hopefully close enough to be useful: we know some things about $\mathbb{E} L(Z,Y)$ for other loss functions $L$. If and only if $L$ is a Bregman divergence, the mean is the minimizer. This is attributed to Banerjee et al. 2005; try Mark Reid's blog post and its references. By phrasing things differently and asking to maximize an expected score rather than minimize an expected loss (though there is no formal difference), the setting starts to look like proper scoring rules (e.g. Savage 1971, Gneiting and Raftery 2007). These are loss functions for eliciting distributions and they are well-understood; again, essentially just Bregman divergences. A generalization of this question which is just receiving study recently is property elicitation: other loss functions and how they connect to the "property" of the distribution that solves the minimization problem. For example, if the loss is $|Y-Z|$, the minimizer is the median; if it is $\mathbf{1}_{Y=Z}$ then it is the mode, etc. For references there I would just point to the Information Elicitation page created by Rafael Frongillo and I for this purpose, which has some slides as well. I think the paper to look at first there would be Lambert et al. 2008. ... (1971) Leonard J. Savage. "Elicitation of personal probabilities and expectations". (2005) Banerjee et al. "On the Optimality of Conditional Expectation as a Bregman Predictor". (2007) Tilman Gneiting and Adrian E. Raftery. "Strictly proper scoring rules, prediction, and estimation". (2008) Nicolas S. Lambert, David M. Pennock, and Yoav Shoham. "Eliciting properties of probability distributions".<|endoftext|> TITLE: Abstract definition of differential operators QUESTION [6 upvotes]: Let $E,F$ be complex vector bundles over some closed manifold $M$. We investigate operators $T:C^{\infty}(M,E) \to C^{\infty}(M,F)$ between smooth sections of these bundles. We say that such operator is order zero differential operator if $[T,f]=0$ this means that $T$ is a bundle endomorphism. If we know what is order $k$ operator we define $T$ to be order $k+1$ if $[T,f]$ is order $k$. $f$ denotes the smooth function. I would like to understand why this definition implies that every such operator is really a honest differential operator (in local coordinates it has to be ordinary differential operator). I would be grateful if anyone could help me with this. REPLY [5 votes]: First of all, you can show that operators defined by your recursive rule are local operators, i.e., if a section $s$ has support in an open set $U$ then $T(s)$ has support in $U.$ By using the partition of unity it remains to do the work locally. As already mentioned by Simon Henry in his comment, when the order $k=1$, you can directly check that your operator is a first order operator in the sense that it can be written locally as $$(\sum_{i=1}^n A_i\nabla_{X_i})+B$$ for a connection $\nabla$ and suitable homomorphisms $A_i,B\in Hom(E;F).$ This can be generalized by introducing the symbol map for $k$th order differential operators: It can be shown that for functions $f_j,$ $j=1,..k$ the iterated commutator evaluated at $p\in M$ $$[..[T,f_1],..,f_k]$$ is a homomorphism which only depends on $d_pf_1,..,d_pf_k$ in a multilinear symmetric way. Up to a scaling, this commutator is the so called symbol $\sigma$, and it measures the highest order differential operator terms. By subtracting these highest order terms, i.e., something like $$T-constant\sum \sigma(X_{{i_1}}^*,..,X_{{i_k}}^*)\nabla_{X_{i_1}\circ..\circ \nabla_{X_{i_k}}}$$ for a local basis of vector fields and its dual basis, you obtain that the symbol of the difference is $0$, hence it is a differential operator of order $k-1.$ A good reference for this is chapter 2.1. in the book "Heat kernels,.." by Berline, Getzler, Vergne.<|endoftext|> TITLE: Definition of "Lagrangian skeleton" QUESTION [7 upvotes]: I'm looking for the precise definition of "Lagrangian skeleton", as I'm eventually going to give a talk on this topic. As I asked a professor in my university about references on Lagrangian skeleta, he listed the following: Lagrangian Non-intersection by Paul Biran http://arxiv.org/pdf/math/0412110v2.pdf From Stein to Weinstein and Back Symplectic Geometry of Affine Complex Manifolds by Kai Cieliebak, Yakov Eliashberg http://www.mathematik.uni-muenchen.de/~kai/research/stein.pdf As I googled "Lagrangian skeleton", the only useful information I've found is this post of MO Has anything precise been written about the Fukaya category and Lagrangian skeletons? in which Ben Webster said the following: As I understand it, a Lagrangian skeleton is a union of Lagrangian submanifolds which a symplectic manifold retracts to. Two references above never used the term "Lagrangian skeleton", but I have often seen "skeleton of a Morse function", which is the union of all stable manifolds. Is this what Lagrangian skeleton is? Or is Webster's definition the right/standard one? In fact, I haven't seen Ben's definition in the papers. If the latter is the case, which concept in the above papers correspond to Lagrangian skeleton? I'm 100% confident that it is in the papers, especially the latter one, which is, according to the professor, the canonical text on the topic. REPLY [11 votes]: Indeed the Cieliebak-Eliashberg reference is canonical. I summarize some things here: The notion of skeleton in symplectic geometry is generally used in the exact setting, i.e. the symplectic form is $\omega = d\lambda$. Note that in this case the symplectic manifold $M$ must be noncompact. Having fixed the primitive $\lambda$, one forms the Liouville vector field $Z$ by demanding $\omega(Z, \cdot) = \lambda$. In this case one says that the skeleton is the locus in $M$ which does not escape every compact set under the flow of $Z$. One generally restricts further to the ``Liouville'' setting where there is a convex contact hypersurface $V$ whose flow under $Z$ exhausts the non-compactness of the symplectic manifold; in this case the flow gives a retraction from $M$ to the skeleton. The relation to the Morse theoretic notion is that one often asks moreover for a ``Weinstein structure'', i.e. a Morse function, convex with respect to some compatible almost-complex structure, with respect to which $Z$ is gradient-like. In this case one can show (easily) that the skeleton is necessarily isotropic. One reason for some of the confusion regarding the definitions in this area is the surprisingly large number of open (to my knowledge) questions about really basic things. Here are some examples. In the first two I take $M$ to be a Liouville manifold, i.e. I have fixed a primitive for the symplectic form and the manifold is asymptotically a cylinder on a contact manifold. (1) Suppose the skeleton is Lagrangian. Does there exist a Weinstein function with respect to which the Liouville flow is gradient like? (2) Suppose given a 1-parameter family $L_t$ of Lagrangian subvarieties of M, with $L_0$ the skeleton determined by $\lambda$. When does there exist a family $\lambda_t$ of 1-forms for which these are the skeleta? (3) Now let $(M, \omega)$ be any symplectic manifold, and let $L \subset M$ be a singular Lagrangian. When is it true that $L$ admits a neighborhood $W$ such that $(W, \omega)$ admits a Weinstein structure for which L is the skeleton? [Mathematical pun: an answer to the 3rd question above would be a ``Weinstein neighborhood theorem'']<|endoftext|> TITLE: Vandermonde's remarkably clever notation for determinants QUESTION [21 upvotes]: The entry on Alexandre-Théophile Vandermonde at the MacTutor History of Mathematics archive ends with the description of the contents of Vandermonde's fourth and last mathematical paper, concluding with the sentence Finally he gave a remarkably clever notation for determinants which has not survived. Sadly, it does not tell us what this remarkably clever notation was! I've searched a bit for the actual paper, Mémoire sur l'élimination, with no success. Maybe someone knows what this notation was? REPLY [26 votes]: The history of the Vandermonde notation is described, in the context of the Vandermonde determinant, in section 2.1 of A case of mathematical eponymy: the Vandermonde determinant (2010). It seems Lebesgue didn't like it because it could have induced a mix-up between indices and exponents, and that may be a reason it did not survive. Leibniz used a similar notation. The following quotation from his 1772 paper shows how Vandermonde used this positional notation for coefficients to construct determinants: Here is the original in French, from Mémoire sur l'Élimination, M. Vandermonde, Histoire de L'Academie Royale des Sciences, 1772, part two, pages 516-532.<|endoftext|> TITLE: Literature on Kripke models QUESTION [5 upvotes]: Which is the best introduction to Kripke-models for modal logics? I am a M.Sc in mathematics and know predicatlogic. REPLY [3 votes]: Brian F. Chellas: Modal Logic: An Introduction, 1980. Starts very basic but covers in detail the beautiful completeness theorem proofs for the basic systems like $S5$, $S4$, $K$.<|endoftext|> TITLE: Cocontinuous product-preserving functor between Grothendieck toposes QUESTION [9 upvotes]: What is an example of a functor $$F : \mathcal{C} \to \mathcal{D}$$ between two Grothendieck toposes which preserves colimits and finite products, but is not left exact (i.e., does not preserve pullbacks)? I just assume that there is such an example, since otherwise the notion of an algebraic morphism between toposes would probably not include left-exactness. But I am not experienced enough in topos theory to see such an example. My question is essentially why the forgetful strict $2$-functor from (toposes with algebraic morphisms) to (cocomplete symmetric monoidal categories with cocontinuous symmetric monoidal functors) is not fully faithful. I already asked this on math.SE. REPLY [15 votes]: For any small category $J$, the colimit functor $\mathsf{Set}^J \to \mathsf{Set}$ preserves colimits. It preserves finite limits if and only if $J$ is filtered and it preserves finite products if and only if $J$ is sifted. So we only need an example of a sifted but non-filtered category. One such example is $\Delta^{\mathrm{op}}$ in which case the colimit functor reduces to $\pi_0 \colon \mathsf{sSet} \to \mathsf{Set}$.<|endoftext|> TITLE: Introduction to representation theory of algebraic groups QUESTION [10 upvotes]: This is a very basic and most definitively a naive question but coming from a student it is probably OK. I am trying to learn representation theory of (linear) algebraic groups and looking for a an easy resource. The books and notes which I have come across start with extensive knowledge of algebraic geometry (which is essential for a comprehensive treatment) but coming from a weak background of mathematics by the time you absorb all that you are exhausted. I am wondering if there is another way to get into representation theory of algebraic groups without worrying too much about algebraic geometry part. We have tried to go through the famous books (Humphreys, Springer, Borel), although excellent but too much for a beginner only interested in representation theory part to start with. Any help will be highly appreciated. REPLY [3 votes]: I know close to nothing about algebraic groups, but the little I know comes from this article. Basically, it assumes that you know what an algebraic variety is, and develops from scratch both the elementary properties of these groups and their actions on varieties (which I think is what motivates their introduction in the first place).<|endoftext|> TITLE: Section of universal curve QUESTION [7 upvotes]: Let $\mathcal{M}_{g,1}\to \mathcal{M}_g$ be the universal genus $g$ curve, let $K_g$ denote the funciton field of $\mathcal{M}_g$. Take the generic fiber $C\to \mathrm{Spec}{(K_g)}$, after some finite extension $L/K_g$, $C_L$ will admit a rational point. Is there a good choice for $L$? If there is, do we know $\mathrm{Gal}(L/K_g)$? REPLY [7 votes]: Let me assume $g>2$, so that there is a curve $C$ over $K_g$ of genus $g$ (the generic curve) -- the genus condition assures that the generic curve has no automorphisms. It is known but not obvious that $C$ has no $K_g$-rational points (this is due to Hain-Matsumoto in characteristic zero and Watanabe in positive characterstic, if I remember correctly). If you choose any closed point of $C$ with residue field $L$, the base change to $L$ will have a rational point. There is a reasonably canonical choice of $L$. Namely, let $W_g$ be the moduli space of curves with a Weierstrass point. Then $W_g$ is connected and $W_g\to M_g$ is not Galois, but it is generically etale with monodromy group is $S_{g(g^2-1)}$ by this paper of Eisenbud-Harris. Thus letting $L_g$ be the function field of $W_g$, the generic curve obtains a rational point over $L_g$ (namely, the generic Weierstrass point). $L_g/K_g$ is not Galois, but its Galois closure has Galois group $S_{g(g^2-1)}$. Of course this works in genus $<2$ as well, as long as you consider the generic point of $\mathcal{M}_g$ to be a stacky point rather than Spec of a field.<|endoftext|> TITLE: Showing that a certain matrix is not positive definite QUESTION [18 upvotes]: Let $J_k$ be a $k \times k$ all ones matrix and $B$ any $k \times k$ binary matrix - that is $B$ only has entries from $\{0,1\}$. I would like to show that the matrix $$X_B = (J_k -I) - B (J_k - I)^{-1} B^T\,,$$ is not positive-definite. In other words, I'd like to show that At least one eigenvalue of $X_B$ is non-positive. I can show that for certain specific matrices $B$ but don't see how to prove the more general statement. Does anybody know why this property seem to hold in general? REPLY [17 votes]: Counterexample: let $k=7$, and let $B$ be the circulant matrix with $B_{ij}=1$ iff $i-j \in \{1,2,4\} \bmod 7$. Then $X_B$ is $I + \frac12 J$, with characteristic polynomial $(x-1)^6 (x-\frac92)$. Or use $B+I$ instead to get $I + \frac13 J$, with characteristic polynomial $(x-1)^6 (x-\frac{10}{3})$.<|endoftext|> TITLE: A simpler proof that compact sets have cardinality continuum? QUESTION [6 upvotes]: Is there a simple reason why uncountable compact sets of real numbers have cardinality continuum? I know that this is immediate from the Cantor-Bendixon Theorem, but I wonder whether this consequence in its own can be proved in a simpler/shorter manner. Perhaps building a Cantor set inside the compact set is one answer to my question, but even this is not so trivial if you did not see similar things before (many small assertions need to be proved in such a case). I wonder if there are better approaches. The question is relevant for a survey paper I write, where going into the CB Theorem (just for the mentioned purpose) may distract the attention of the reader off the main thread of the paper. REPLY [4 votes]: I think "the" conceptual reason why an uncountable compact subset of $K\subset \mathbb{R}$ has the cardinality of the continuum is that it has a "canonical" map onto a space homeomorphic to $[0,1]$. The complement of $K$ is a union of open intervals: a left ray, a right ray and at most countably many bounded ones. The canonical map is the quotient map to the space obtained by gluing the end points of these bounded intervals in the complement (it is easy to see that the equivalence classes of the equivalence relation thus obtained are at most countable). This answers the question in the sense that this is a "simple reason why...". I am not saying that this is the simplest proof, as one still has to argue that the quotient space is homeomorphic to $[0,1]$ (a choice of such a homeo is certainly not canonical). There are various ways to do this. Edit: below I replace a previous argument (for $K/{\sim}\simeq [0,1]$) by a more conceptual one. Recall that $K$ is assumed uncountable and that the equivalence relation $\sim$ we defined on it has countable fibers. Denote $X=K/{\sim}$. Endow it with the quotient topology and quotient order. The following is easy. Observation: $X$ is a connected, separable, compact linearly ordered space which is not a singleton. (separablity and compactness are inherited from $K$, connectedness follows from the definition of $\sim$ and $X$ is not a singleton is by the fact that the equivalence classes are countable - this is where we use that $K$ is uncountable). We are left to prove the following. Proposition: Every space $X$ saisfying the properties above is homeomorphic to $[0,1]$. Fix a countable dense subset $A\subset X$. Assume as you may that $\min X$ and $\max X$ are not in $A$. The proof of the proposition consists of the combination of the following facts: $X$ is an order completion of $A$. $A$ is order isomorphic to $\mathbb{Q}$. The order completion of $\mathbb{Q}$ is the two points compactification of $\mathbb{R}$. Facts 1 and 3 are easy (details could be found in https://en.m.wikipedia.org/wiki/Dedekind-MacNeille_completion). In fact, fact 3 could be regarded as the definition of $\mathbb{R}$. For fact 2, observe that $A$ is a countable dense linear order with no min and max. It is a classical fact (due to Cantor) that every two models of this theory are isomorphic, see https://en.m.wikipedia.org/wiki/Dense_order. The proof is given in https://en.m.wikipedia.org/wiki/Back-and-forth_method. It is a first-course-in-set-theory-exercise (but don't give it in the exam).<|endoftext|> TITLE: $\pi_1$ of 4-manifolds that "look like" disk bundles QUESTION [5 upvotes]: Let $X$ be a smooth compact oriented 4-manifold with $\partial X=L(p,1)$, $H_2(X;\Bbb Z)=\Bbb Z$, $H_3(X; \Bbb Z)=0$ and the induced map $\pi_1(L(p,1)) \to X$ surjective. What are the possibilities for $\pi_1(X)$? In particular, are there examples where $\pi_1 \ne 0$? REPLY [3 votes]: Let's take your $p$ to be prime. Then $X$ has to be simply connected, even without all of the hypotheses. Here is the argument. From the map on $\pi_1(L) \to Z_p$ you get a map $L \to BZ_p$. This map is clearly $0$ in $H_3$, since it factors through the inclusion of $L$ into the $4$-manifold $X$. On the other hand, it's well-known that $L$ generates $H_3(BZ_p) \cong Z_p$. (A quick explanation: you can build $BZ_p$ by attaching a 4-cell by a map of degree $p$ (the universal covering $S^3 \to L$) and then higher cells.) I think a similar proof works if $p=mn$ with $m$ and $n$ relatively prime, and $\pi_1(X) = Z_n$. You have to know a little more about the map $H_3(BZ_p) \to H_3(BZ_n)$ induced by a surjection. I'm pretty sure it's given by multiplication by $m^2$.<|endoftext|> TITLE: What is this analogy between manifolds and bundles (or schemes and locally free sheaves)? QUESTION [9 upvotes]: There's a kind of analogy between the way manifolds work and the way bundles work. Let me try to give some examples of the analogy (although there may be better ones). I'll stick to smooth manifolds and vector bundles for definiteness. (the main idea) A manifold is $M$ a space with an open cover $\{U_i\}$ by copies of the model space $\mathbb{R}^n$, with transition functions coming from the diffeomorphism group. A vector bundle $V \to X$ is a map where the base $X$ has an open cover $\{U_i\}$ on which the map pulls back to copies of the standard map, which is a projection from $\mathbb{R}^n \times U_i \to U_i$, with transition functions required to be fiberwise in the general linear group. An orientation on a (compact) manifold $M$ is a choice of a fundamental class, i.e. $[M] \in H_n(M)$ which restricts to a generator on the local homology group $H_n(M,M-x)$ at each point $x \in M$. An orientation on a vector bundels $V \to X$ is a choice of Thom class, i.e. $v \in H^n(V,V-0)$ (where $0$ is the zero section of $V$) which restricts to a generator on the local cohomology group $H^n(V_x, V_x - 0)$ for each point $x \in X$, where $V_x$ is the fiber at $x$. Manifolds have a Pontrjagin-Thom construction, while vector bundles have a Thom construction. The idea that you can "collapse at infinity" is something in common. This connection may be me grasping at straws -- anyway, the two constructions are already closely related: the Thom construction of the normal bundle receives the map from the manifold in the Pontrjagin-Thom construction. (attempt at (1) in a more Cech-like language:) A manifold is some kind of simplicial object of model spaces. A vector bundle is some kind of simplicial object of maps to a structure group (satisfying a cocycle condition). (attempt at (1) in a more sheaf-like language:) I think a manifold $M$ is some kind of locally free sheaf on the site of Euclidean spaces and $C^\infty$-functions. The sheaf of sections of a vector bundle $V \to X$ is a locally free module over the sheaf $C^0(X)$ of continuous functions on $X$. Point (1), in particular, leads me to want to view bundles as a "relative version of manifolds" (although vector bundles in particular happen to have a much smaller structure group). I suppose all of this could be considered in other types of geometry, too -- e.g. the same sort of analogy holds between schemes and certain kinds of sheaves over schemes. But I'd like to be able to state this in some precise way. I'd also like to understand whether the analogy stops there, or whether there's a whole hierarchy of notions of which these are just 0th-order and 1st-order notions. REPLY [2 votes]: I think you really want to think of a vector bundle as a family of vector spaces, parametrized over a manifold and not so much like a manifold itself. I'll just point out some of the ways I see an essential distinction in your proposed analogies. It is true that vector bundle are equipped with locally trivializing patches, but these patches must respect the linear structure on the fibers of the bundle. The fundamental pieces by which smooth vector bundles (with fiber $V$) are built are projections ${\rm pr}:V\times \mathbb{R}^n\to \mathbb{R}^n$. Manifolds, by contrast, locally look like $\mathbb{R}^n$, which is an object and not a type of morphism. An orientation on a smooth manifold is equivalent to an orientation on its tangent bundle. Fix a Riemannian metric $g$ on $M$ and consider the induced map classifying the cotangent bundle of of $M$: $$g:M\to B{\rm O}(n)\;.$$ Post composing with the determinant map gives a map $M\to B\mathbb{Z}/2$, classifying the determinant line bundle $\Lambda^n(T^*M)$ over $M$. If this bundle is trivilizable, then we can choose a global nonvanishing section which is a choice of global volume form on $M$. This volume form can be used explicitly to build a Thom class for the Tangent bundle of $M$ (see for example Bott and Tu). If you look at things this way, then you see that orientability of a smooth vector bundle is really a generalization of the concept of orientability of a smooth manifold via the tangent bundle. The Pontryagin-Thom construction is basically Whitney embedding theorem + Thom construction of the normal bundle. PT and the Thom space construction are not two separate constructions that are like each other, but rather the former crucially uses the latter. The Cech and sheaf like language can be combined in a nice way in the $\infty$-topos of smooth stacks on the site of cartesian spaces (i.e. objects look like copies of $\mathbb{R}^n$). Dugger, Hollander and Isaksen show that a cofibrant replacement of a smooth manifold in Jardines model structure on simplicial presheaves is given by resolving a smooth manifold by the simplicial object given by the Cech nerve of a cover. The smooth stack which classifies locally trivial vector bundles of rank $n$ is the homotopy orbit stack $U:=\mathbb{R}^n/\!/{\rm O}(n)$, where ${\rm O}(n)$ acts as usual, by matrix multiplication. In fact, after making some convenient choices for homotopy colimits, a map $$M\overset{\simeq}{\leftarrow} C(\{U_{\alpha}\})\to U\;,$$ where $C(\{U_{\alpha}\})$ is a homotopy colimit over the Cech nerve diagram for a cover of $M$, gives you exactly the Cech cocycles neede to define a locally trivial vector bundle on $M$. There is a canonical map $U\to \mathbf{B}{\rm O}(n)$ to the moduli stack of principal orthogonal bundles on $M$ (this map simply projects our $\mathbb{R}^n$). A map to $\mathbf{B}{\rm O}(n)$ classifies a principal bundle on $M$ and pulling back by the map $U\to \mathbf{B}{\rm O}(n)$ gives you a locally trivial vector bundle over $M$, equipped with a resolution $C(\{U_{\alpha}\times \mathbb{R}^n\})$. The sections of this object recover exactly the sheaf version of vector bundle over $M$. In this abstract language, the relationship between manifold and vector bundle is made manifest by the fact that the local trivializing patches of a manifold prescribe the base space over which the vector bundle ought to trivialize. Still the two notions are quite distinct from one another. A vector bundle is really a map which looks like a projection over trivializing patches of $M$. The trivializing patches of $M$ are prescribed by the site of (cartesian spaces in this case).<|endoftext|> TITLE: Large cardinals ordered by cardinality of least instance QUESTION [6 upvotes]: This question is an extension of something I asked earlier here: Ordering of large cardinals by cardinality I have seen large cardinals ordered by consistency strength in several places but no ordering by cardinality of the least instance, although this is probably common knowledge among experts. Based on information scattered in various articles and the answer to my prior question from Dr. Hamkins, here is what I have gathered so far (I have deliberately avoided "identity crisis" cases):- Strongly inaccessible < Mahlo < Weakly Compact < Totally Indescribable < Measurable < Huge (1-Huge upwards) < Rank-into-Rank < Supercompact < Extendible My questions are: 1) Is this ordering correct ? Maybe < should be replaced by $\leq$ in some cases ? 2) Any important types I have left out which can be placed in the ordering without ambiguity ? 3) If a Reinhardt cardinal existed (maybe in ZF), would it be at the top of this size hierarchy ? My intuition is that it should, since a Reinhardt cardinal is an "ultimate extendible cardinal" in a sense, but I find my intuition isn't terribly good on these issues. REPLY [8 votes]: I added some information elsewhere, but it was to long to put in one post: http://metaordinals.azurewebsites.net/?p=261 Part 1: Your ordering is correct, but the size of the gaps are much more complex than just $LC_1\kappa$, and we can call a cardinal totally-indescribable if it is $\lambda$-indescribable for every cardinal $\lambda$. Under the typical definition the totally indescribable cardinals would be the $\omega$-indescribable cardinals. The $(\lambda+1)$-indescribable cardinals are stationarily transcendent over the $\lambda$-indescribable cardinals, and every critical point of a non-trivial elementary embedding $j:M\rightarrow M$ is totally indescribable. If I am not mistaken, every weakly measurable cardinals is $\Sigma^2_1$-indescribable. Theorem: Every measurable cardinal is a measurable limit of totally indescribable cardinals. Proof. Consider the normal measure given by $D=\{X\subseteq \kappa|\kappa\in j(X)\}$. Then $D$ is $\kappa$-complete normal ultrafilter. Note that for every $\alpha$th order formula $\phi$, its revitalization $\phi^{V_\kappa}$ is first order and moreover there is a truth predicate for $V_\kappa$. Also note that if $V_{j(\kappa)}\vDash \phi(S)$, then $V_\kappa\vDash \phi(S\cap V_\kappa)$ and so $j(\kappa)$ is totally indescribable. Moreover, the statement $\kappa$ is totally indescribable is $\Sigma_1$ and $M$ is an inner model and so $M\prec_{\Sigma_1}V$ and so $M\vDash(\text{$j(\kappa)$ is totally indescribable})$. Therefore $U\in D$, where $U=\{\lambda<\kappa\mid \text{$\lambda$ is totally indescribable}\}$.■ Beyond that, however, the gaps are much larger, being instead measurable gaps. The $(n+1)$- are measurably transcendent over the $n$-huge cardinals (By a simple alteration to the standard proof that they are measurable limits of $n$-huge cardinals). Theorem: The rank-into-rank cardinals are measurably transcendent over the $\omega$-huge cardinals. Proof. Consider the filter generated by sets of the from $C\cap U_\omega$, where $U_\omega=\{\lambda<\kappa\mid \text{$\lambda$ is $\omega$-huge}\}$, and $C$ is club in $\kappa_n$. It is clearly a normal filter, assuming $U$ is stationary, and similarly if $U$ is stationary closed under diagonal intersection. Now suppose we have a non-trivial elementary embedding $j: V_\lambda\rightarrow V_\lambda$, then $\kappa_n$ is $\omega$-huge$^1$ and so it remains to show $U$ is stationary in $\kappa_n$. Consider the normal measure, $D=\{X\subseteq \kappa_n|\kappa_n\in j(X)\}$. Then $\kappa_{n+1}$ is $\omega$-huge and so $U_\omega\in D$, where $U_\omega=\{\lambda<\kappa\mid\text{$\lambda$ is $\omega$-huge}\}$. To see that it is closed under $o(X)$, observe $o(C\cap U)=o(C)\cap o(U)$, and by a similar argument such $o(U)=C\cap U$, where $C$ is the class of measurable limits of $U$, and it is clear $o(C)$ is closed. To see that it is unbounded, first observe that as $C$ is club, for every normal measure $D$ we have $C\in D$. Consider the normal measure $D=\{X\subseteq \kappa_n\mid \kappa_n\in j(X)\}$. As $M\vDash\text{There is a normal measure on $\kappa_{n+1}$ that contains $j(C)$}$, we have $o(C)\in D$.■ The gap between supercompact and rank-into-rank cardinals is also quite large. Theorem: The supercompact cardinals are measurably transcendent over the rank-into-rank cardinals. Proof. Every supercompact cardinal $\kappa$ is $\Sigma_2$-reflecting, and the property "$\kappa_n$ is rank-into-rank," "There exists a "rank-into-rank embedding," are both $\Sigma_2{}^1$. Moreover, it is easy to see that the supercompact cardinals are measurably transcendent over the rank-into-rank cardinals.■ Lemma: The extendible cardinals are measurably transcendent over the rank-into-rank cardinals. Proof. Simply replace "There exists a rank-into-rank cardinal" with "There exists a supercompact cardinal" and use the fact that extendible cardinals are $\Sigma_3$-reflecting.■ So in conclusion, this list is technically correct, yet very misleading, as it gives the impression that the size gaps are around uniform. Part 2: Yes, you omitted several very important cardinals. First, I am not sure if you omitted these because you considered them identity crises, but the weakly measurable cardinals are above the totally indescribable cardinals but below (Or equal, assuming the $\mathsf{GCH}$) the measurable cardinals. Also the weakly indescribable cardinals are rather interesting; if $\kappa$ is weakly indescribable $L_\kappa\vDash \mathsf{ZFC}+V=L$. In fact, this fact doesn't even require Choice. The most tragic omission is definitely the worldly cardinals. A cardinal $\kappa$ is worldly if and only if $V_\kappa\vDash \mathsf{ZFC}$. The worldly cardinals are the smallest type of large cardinal; if $\kappa$ is inaccessible there is a $\kappa$-complete filter $F$ on $\kappa$ closed under the worldly limit point operator (By a simple alteration to $^2$). The pseudo-uplifting cardinals are weaker than the Mahlo cardinals, yet every pseudo-uplifting cardinal is a limit of inaccessibles, and the pseudo-uplifting cardinals are stationarily transcendent over the $\Sigma_3$-reflecting cardinals. Therefore, assuming the existence of a supercompact cardinal, the pseudo-uplifting cardinals are stationarily transcendent over the supercompact cardinals, and the uplifting cardinals are limits of pseudo-uplifting cardinals. I am not sure if you omitted because you considered them identity crises, yet they are still interesting. Beyond the totally indescribable cardinals are the strongly unfoldable cardinals. Theorem: The strongly unfoldable cardinals are stationarily transcendent over the totally indescribable cardinals. Proof. Consider the filter generated by sets of the form $C\cap U$, where $U=\{\lambda<\kappa\mid\text{$\lambda$ is totally indescribable}\}$, and $C$ is club in $\kappa$. It is clearly a normal filter, assuming $U$ is stationary, and similarly if $U$ is stationary closed under diagonal intersection. Now suppose we have a non-trivial elementary embedding $j: M\rightarrow N$, then $(V_\kappa,\in,S)\prec(V_{j(\kappa)},\in,j(S))$. Then $j(S)\cap V_\kappa=S$ and so it remains to show $U$ is stationary in $j(\kappa)$. By the same logic as before, $j(\kappa)$ is totally indescribable, and moreover, if $T$ is the higher order truth predicate for $V_{j(\kappa)}$ then $V_\kappa\prec _T V_{j(\kappa)}$. Then let $C$ be club. Then there exists some $\alpha\max\{\alpha,\lambda\}$, such that $V_\alpha\subset N$ and $N\vDash\text{ ($\kappa$ is strongly unfoldable of degree }\beta)$ for every $\beta<\alpha$. A cardinal is totally unfoldable if and only if it is strongly unfoldable of degree $\alpha$ for every ordinal $\alpha$. Theorem: The unfoldable with degree $\alpha+1$ cardinals are stationarily transcendental over the unfoldable with degree $\alpha$ cardinals. The superstrongly unfoldable cardinals are stationarily transcendent over the totally unfoldable cardinals. Proof. Every cardinal $\kappa$ that is unfoldable with degree $\alpha+1$ is clearly unfoldable with degree $\alpha$. Consider the filter generated by sets of the from $C\cap U$, where $U=\{\lambda<\kappa\mid\text{$\lambda$ is totally indescribable}\}$, and $C$ is club in $\kappa$. It is clearly a normal filter, assuming $U$ is stationary, and similarly if $U$ is stationary closed under diagonal intersection. Now suppose we have a non-trivial elementary embedding $j: M\rightarrow N$, then $(V_\kappa,\in,S\cap V_\kappa)\prec(V_{j(\kappa)},\in,S)$ and so it remains to show $U$ is stationary in $j(\kappa)$. Then $j(\kappa)$ is unfoldable with degree $\alpha$. Let $D=\{X\subseteq \kappa\mid\kappa\in j(X)\}$. Then if $S\in D$, $S$ is stationary, and $N\vDash (\text{$j(\kappa)$ is unfoldable with degree }\alpha)$ and so $U\in D$. To see that it is closed under $M(X)$, observe $M(C\cap U)=M(C)\cap M(U)$, and by a similar argument such $M(U)=C\cap U$, where $C$ is the class of stationary limits of $U$, and it is clear $M(C)$ is club. For the rest, see $^3$.■ The supercompact cardinals are more complex than simply being smaller then extendibles. Theorem: If $\eta$ is not an infinite limit ordinal with $\operatorname{cf}\eta<\kappa$, and $\kappa$ is $(\eta+1)$-extendible then $\kappa$ is $\beth_{\kappa+\eta}$-supercompact. If $\kappa$ is $(\eta+2)$-extendible then $\kappa$ is $\beth_{\kappa+\eta}$-supercompact. If $\eta<\kappa$ and $\kappa$ is $\kappa$ is $\beth_{\kappa+\eta}$-supercompact, then $\kappa$ is measurably transcendent over the $\eta$-extendible cardinals. Moreover, the $1$-extendible supercompacts are measurably transcendent over the supercompact cardinals. Proof. For the first part, see $^4$. For the second part, suppose $\eta$ is a limit ordinal with $\operatorname{cf}\eta<\kappa$. Then $\xi=\eta+1$ is not a limit ordinal and so if $\kappa$ is $\xi$-extendible it is $\beth_{\kappa+\xi}$-supercompact. For the third part, let $j: V\rightarrow M$ witness the $|\beth_{\kappa+\eta}|$-supercompactness of $\kappa$, and let $j`=j\restriction V_{\kappa+\eta}$. Then $j': V_{\kappa+\eta}\rightarrow V_{j(\kappa+\eta)}$ is a non trivial elementary embedding witness the $\eta$-extendibility of $\kappa$. The rest follows simply. For the fourth part, let $j: V_{\kappa+1}\rightarrow V_{j(\kappa)+1}$ witness the $1$-extendibility of $\kappa$ and $D=\{X\subseteq \kappa|\kappa\in j(X)\}$. Then $V_{j(\kappa)}\vDash (\kappa\text{ is $\lambda$-supercompact for $\lambda>\kappa$ and }\lambda TITLE: Geodesics for non differentiable riemannian metric QUESTION [6 upvotes]: Let $M$ be a differentiable manifold of dimension $n>2$ with a Riemannian metric $g=\sum_{i,j=1}^ng_{ij}dx_idx_j$ such that in some points on $M$ its coefficients $g_{ij}$ are not differentiable (so $g_{ij}$ are just continuous on $M$). Call $d_g$ the metric induced on $M$ by $g$ (by the infimum of the lengths of the curves). Is $(M,d_g)$ a geodesic space (by "geodesic space" I mean that for every point $p\in M$ there is a neighborhood $U$ of $p$ such that for every couple of points $x,y\in U$ there exists a geodesic for $d_g$ from $x$ to $y$) although $g$ has coefficients which are not differentiable at some points? Clearly the problem is that in the geodesic equations $\displaystyle{\frac{d^2 x^a}{ds^2}+\Gamma^a_{bc}\frac{d x^b}{ds}\frac{dx^c}{ds}=0}$ there are Christoffel symbols $\Gamma^a_{bc}$ in which the derivatives of the $g_{ij}$ appear. REPLY [7 votes]: This is a length-metric, that is any two points $x$ and $y$ can be joined by a path with length arbitrary close to the distance from $x$ to $y$. Further, your metric space is locally bi-Lipschitz to the Euclidean space, in particular it is locally compact. If your space is complete then by Hopf–Rinow theorem it is geodesic. For sure your space is locally complete. Then the same proof imply that it is locally geodesic as you asked.<|endoftext|> TITLE: Alexander's theorem (on stellar moves) QUESTION [6 upvotes]: A widely used theorem these days says that given (abstract) simplicial complexes $K$ and $K'$, a polyhedron $P\subset R^n$ and homeomorphisms $|K|\to P$ and $|K'| \to P$ which are linear (better to say affine) in each simplex then $K$ and $K'$ are stellar equivalent as abstract simplicial complexes. Therefore we have a sequence: $$K \to K_1 \to K_2 \to K_3 \to \dots \to K_n\to K' $$ of abstract simplicial complexes where $K_{i+1}$ is obtained from $K_i$ by an isomorphism ("vertex name change"), a stellar subdivision or a welding (inverse stellar subdivision). Question: can you do this in order that each $K_i$ is linearly (as opposed to piecewise linearly) embedded in $P$? My version of the answer. Looking at the classical references of the proof of Alexander's theorems it does appear that the proofs do not imply this stronger statement: the problem being that sometimes $K_i$ may very well contain the star of a non-convex cell (which my be not "star-shaped") in $P$ and some welds may be not-realizable in $P$ without modifying $P$, slightly, to an isomorphic complex: c.f. L. C. Glaser, Geometrical combinatorial topology volume 1, pages 23 and 29. PS: Great texts on Alexander's theorem are: a) The first two chapters of L. C. Glaser, Geometrical combinatorial topology volume 1. b) Alexander's original paper: Alexander, James W. The combinatorial theory of complexes. Ann. of Math. (2) 31 (1930), no. 2, 292–320. c) Lickorish, W. B. R.: Simplicial moves on complexes and manifolds. (English summary) Proceedings of the Kirbyfest (Berkeley, CA, 1998), 299–320 (electronic), Geom. Topol. Monogr., 2, Geom. Topol. Publ., Coventry, 1999. REPLY [3 votes]: Yes, two triangulations of a polyhedron can be connected by a sequence of geometric stellar moves (subdivisions and weldings). And you are right, this does not follow from the Alexander theorem, which deals with combinatorial stellar moves. This was proved by [Morelli, The birational geometry of toric varieties J. Algebraic Geom. 5(1996), no. 4, 751–782.] and [Wlodarczyk, Decomposition of birational toric maps in blow-ups & blow-downs, Trans. Amer. Math. Soc. 349 (1997), no. 1, 373–411.] (They proved also more: if the triangulation is unimodular, then there is a path through unimodular triangulation.) An outline of Morelli's proof can be found in our article with Jean-Marc Schlenker: On the infinitesimal rigidity of polyhedra with vertices in convex position, here is the arXiv version. Morelli-Wlodarczyk theorem deals with convex polyhedra. It is not completely clear to me how to extend the theorem to non-convex polyhedra. If we modify the triangulations of the cells in some order, there can appear problems on the boundaries between the cells. Maybe one can do the following: extend some stellar subdivisions of $K$ and $K'$ to triangulations $\overline{K}$ and $\overline{K'}$ of the convex hull of $P$. A sequence of stellar moves from $\overline{K}$ to $\overline{K'}$ induces a sequence of stellar moves between the stellar subdivisions of $K$ and $K'$. Hence we also have a sequence of moves joining $K$ and $K'$.<|endoftext|> TITLE: Cohomology of the mapping class group of a non-orientable surface? QUESTION [7 upvotes]: What is the low degree cohomology of the mapping class group of a non-orientable surface? More specifically, what is the universal central extension of the mapping class group of a non-orientable surface? (I've done some googling, but so far have only found presentations of the MCG.) (I would be happy to have the answer in just the negative Euler characteristic case.) REPLY [12 votes]: Let me write $\mathcal{N}_g$ for the mapping class group of the connect sum of $g$ projective planes. Nathalie Wahl proved that these groups enjoy homological stability, and in O. Randal-Williams, The homology of the stable non-orientable mapping class group, Algebraic & Geometric Topology 8 (2008) 1811-1832. I calculated the stable (co)homology. Using the best currently available general homological stability ranges, one has $$H_1(\mathcal{N}_g ; \mathbb{Z}) = \mathbb{Z}/2 \quad \text{for} \quad g \geq 7$$ and $$H_2(\mathcal{N}_g ; \mathbb{Z}) = (\mathbb{Z}/2)^2 \quad \text{for} \quad g \geq 10.$$ In fact, by M. Korkmaz, First homology group of mapping class groups of nonorientable surfaces, Math. Proc. Cambridge Philos. Soc. 123 (1998), 487-499. the abelianisations of all $\mathcal{N}_g$ are known: starting at $g=1$ they are $$0, (\mathbb{Z}/2)^2, (\mathbb{Z}/2)^2, (\mathbb{Z}/2)^3, (\mathbb{Z}/2)^2, (\mathbb{Z}/2)^2, \mathbb{Z}/2, \mathbb{Z}/2, \ldots.$$ As the abelianisation does not vanish for $g > 1$ these groups are not perfect, so they do not have a universal central extension.<|endoftext|> TITLE: A variant of Ramsey numbers QUESTION [5 upvotes]: The well known Ramsey number $R(k)$ is the least integer $n$ so that every 2-edge coloring of $K_n$ contains a monochromatic $K_k.$ Another interpretation of the above definition is that every graph on $R(k)$ vertices has a $K_k$ or $\overline{K_k}$ as a (induced) subgraph. There are many generalizations of Ramsey numbers and I am curious to see what happens if we push this in the direction of complete multipartite graphs. Let $\widetilde{R}(k)$ be the least integer $n$ so that every graph on $n$ vertices must contain an induced complete multipartite graph on $k$ vertices. Since $K_k$ and $\overline{K}_k$ are complete multipartite graphs we have $\widetilde{R}(k) \leq R(k)$. What I am wondering is whether allowing other complete multipartite graph reduces the order of $\widetilde{R}(k)$ significantly. More precisely Is it true that $\widetilde{R}(k) = o(R(k))$? I am still looking at the available literature so if anyone is aware of results in this direction that is also appreciated. In particular is anybody aware of non-obvious bounds for $\widetilde{R}(k)$? REPLY [7 votes]: There is a conjecture of Erdős, Fajtlowicz and Staton which is closely related to your question. Define $\hat{R}(k)$ to be the smallest positive integer $n$ such that any graph on $n$ vertices contains an induced regular subgraph on $k$ vertices. Note that this class includes complete graphs, independent sets and complete balanced multipartite graphs (this balancedness requirement makes it a little different to your question as stated, but you can easily get a large balanced multipartite graph from an unbalanced one). The conjecture then states that $\hat{R}(k) = 2^{o(k)}$. However, to date, the best known bounds are just $$k^{2 - o(1)} \leq \hat{R}(k) \leq R(k).$$ To my mind, this points to your problem being difficult.<|endoftext|> TITLE: Plane curves through a given collection of points QUESTION [5 upvotes]: Let $d$ be a positive integer and $0 \leq k \leq 3d-2$ another integer. Question 1. If $p_1,...,p_k$ are $k$ general points in ${\mathbb P}^2$, is it known in general whether the space parametrizing degree $d$ plane rational curves passing through $p_1,...,p_k$ is irreducible? I think it's clear that the answer is yes, but I don't know if this is supposed to be easy to prove or not. Of course, if $k=0$, this is trivial, but I'm more curios about $k$ at the opposite end of the spectrum. I was imprecise about the specific type of compactification we're using (e.g. it could be a closed subvariety of the Severi variety, or a geometric fiber of the evaluation map $ev:\overline{\mathcal M}_{0,k}({\mathbb P}^2,d) \to ({\mathbb P}^2)^k$), since this seems rather irrelevant. Some thoughts. If we think in terms of the evaluation map -- since the source is smooth (at least as a stack) -- it looks like probably it should be enough to prove that the fibers are connected. If, on the other hand, we think in terms of Severi varieties, then imposing "incidence with points" conditions amounts to taking hyperplane sections, but I'm not sure if that's useful, since (closures of) Severi varieties are quite singular and these hyperplane sections are not general. Question 2 (Generalization). What if instead we take $k \leq 3d+g-2$ and ask the same question about curves of geometric genus $g$? Does this follow somehow from the irreducibility of Severi varieties? EDIT: Marked as solved, thanks to Jason Starr's answer below. However, comments on the second question are still appreciated! REPLY [3 votes]: Yes, for every integer $k$ with $0\leq k \leq 3d-2$, the following evaluation morphism is surjective and the generic fiber is geometrically irreducible, $$\text{ev}_{1,2,\dots,k}:\overline{\mathcal{M}}_{0,k}(\mathbb{P}^2,d)\to (\mathbb{P}^2)^k.$$ One proof uses the version of Bertini's connectedness theorem as formulated by Cristian Minoccheri in Proposition 3.1 of "On the Arithmetic of Weighted Complete Intersections of Low Degree". Bertini's Connectedness Theorem [Cristian Minoccheri] Let $h:M\to X$ be a projective morphism of $k$-schemes with $X$ a smooth variety that is algebraically simply connected, and with $M$ a normal, quasi-projective variety. If the closed subscheme of $M$ where $h$ is not smooth has codimension at least $2$, then $h$ is surjective and the generic fiber is geometrically irreducible. In this theorem, it suffices for $M$ to be normal and "pure" in the sense of SGA2. In particular, this is true for the coarse moduli space of a smooth Deligne-Mumford stack, as with $\overline{\mathcal{M}}_{0,k}(\mathbb{P}^2,d)$. Also, we may as well assume that $k$ equals $3d-2$, since the case of $k\leq 3d-2$ follows from the case of $k=3d-2$. It is convenient to begin with a result about the branching behavor of the evaluation morphism when $k$ equals $3d-1$. In this case, the evaluation morphism is surjective and generically etale. By the Purity Theorem from SGA2, the non-smooth locus of the evaluation morphism in $\overline{\mathcal{M}}_{0,3d-1}(\mathbb{P}^2,d)$ is a Cartier divisor. We know the divisor class group of this moduli space. In particular, the restriction map to the boundary locus $\Delta_{(3,\{1,\dots,8\})(d-3,\{9,\dots,3d-1\}}$ is injective. Thus, by analyzing the case that $d=3$ and $3d-1=8$, the non-smooth locus has multiplicity $1$ (i.e., simple branching) and every generic point parameterizes a stable map with one cusp. By Minoccheri's version of Bertini's theorem, to prove that the generic fiber of $\text{ev}_{1,2,\dots,k}$ is geometrically irreducible, it suffices to prove that over codimension one points of $(\mathbb{P}^2)^k$, the fiber is generically nonreduced. Because $k<3d-1$, every fiber of the evaluation morphism has dimension $\geq 1$. Because $k$ equals $3d-2$, which is $\geq 2$, by the famous Bend-and-Break theorem, every irreducible component of every fiber of the evaluation morphism has nonempty intersection with the boundary divisor $\Delta$ of $\overline{\mathcal{M}}_{0,k}(\mathbb{P}^2_k,d)$. Thus, it suffices to prove that over codimension $1$ points of $(\mathbb{P}^2)^{3d-2}$, the intersection of the fiber with the boundary $\Delta$ is generically nonreduced. The boundary divisors are indexed by partitions $d=d'+d''$ and a corresponding partitions of $P'\sqcup P'' = \{1,\dots,3d-2\}$. In order for that boundary divisor to dominate a divisor in $(\mathbb{P}^2)^{3d-2}$, the sizes of the partition sets must be either $(3d',3d''-2)$, $(3d'-1,3d''-1)$, or $(3d'-2,3d'')$. The first and last are symmetric: one component has degree $e$ and contains $3e$ of the marked points. For degree $e$, genus $0$ stable maps, for $k=3e$, the evaluation morphism $ev_{3e}$ is generically an embedding to a Cartier divisor $D$ in $(\mathbb{P}^2)^{3e}$. For a point of $(\mathbb{P}^2)^{3d-2}$, it is already a codimension one condition for the image point in $(\mathbb{P}^2)^{3e}$ to be contained in $D$. Thus, the remaining $3d-3e-2$ points are unconstrained. So the fiber is a product of the reduced fiber of $\text{ev}_{3e}$ (just one reduced point) and the reduced fiber of $ev_{1,\dots,3(d-e)-2}$ for the component of degree $d-e$. The main case to consider is when the two partition sets have sizes $3d'-1$ and $3d''-1$. In this case, in order for a stable map in the boundary to lie over a codimension one point of $(\mathbb{P}^2)^{3d-2}$ and also to be a nonsmooth point of $\text{ev}_{1,\dots,3d-2}$, the restriction of the stable map to precisely one of the two components of the domain, say the component with degree $d'$, must be a ramification point for the evaluation morphism $\text{ev}_{1,\dots,3d'-1}$. But we know what are these ramification points. In particular, so long as the other irreducible component is generic, then the union is a smooth point of $\text{ev}_{1,\dots,3d-2}$. This is the same deformation theory analysis as in my paper with Graber and Harris about section of rationally connected fibrations. So long as the stable map is unramified at the node where the two components of the domain intersect, then the normal sheaf of the stable map restricted to each component equals the normal sheaf of that component "twisted up" by the degree $1$ invertible sheaf of the node as a reduced Cartier divisor on that component. Since the normal sheaf of the component was just shy of being $(3d'-1)$-globally generated (by the analysis of ramification points), twisting up by a degree $1$ invertible sheaf is enough to kill the $H^1$.<|endoftext|> TITLE: Applications of Morley's Categoricity Theorem QUESTION [9 upvotes]: I just attended a lecture by Rami Grossberg and he mentioned that he is not aware of any applications of Morley's Categoricity Theorem. This is exactly my question. Question: Do you know of any applications of Morley's Categoricity Theorem outside of Logic? Morley's Categoricity Theorem If $T$ is a first-order theory in a countable vocabulary and $T$ is categorical in one uncountable cardinal, then it is categorical in all uncountable cardinals. REPLY [3 votes]: If you want to apply the theorem without using the structure theory arising from the proof, then I claim that there cannot be an application. This is since the ordinary mathematician is not interested in comparing uncountable structures of different cardinality. In this sense Morley's Categoricity Theorem is a negative result, i.e. an uncountable structure (which is uncountably categorical) cannot be elementarily equivalent to some uncountable structure with exotic properties. Of course if you take the structure theory arising from the proof into account then the picture is completely different.<|endoftext|> TITLE: Reference for Chang's Conjecture at $\aleph_{\omega}$ QUESTION [6 upvotes]: The following theorem is well known: Theorem: $(\aleph_{\omega + 1}, \aleph_{\omega}) \not\twoheadrightarrow (\aleph_{n + 1}, \aleph_n)$ for every $n \geq 3$. Under CH, $(\aleph_{\omega + 1}, \aleph_{\omega}) \not\twoheadrightarrow (\aleph_{n + 1}, \aleph_n)$ for every $n > 0$. where $(\kappa, \lambda)\twoheadrightarrow (\mu, \nu)$ stands for Chang's Conjecture between the pair of cardinals $(\kappa, \lambda)$ and the pair $(\mu, \nu)$. Who was the first to prove this statement? Was it published somewhere? REPLY [4 votes]: Levinski+Magidor+Shelah, Chang's conjecture for $\aleph_ \omega$ -- Israel J Math 69 (1990) 161-172. See also "Some consequences of reflection on the approachability ideal" by Matteo Vilale, Assaf Sharon, Transactions of the American Mathematical Society 362, 4201-4212, 2009. In particular fact 4.2 and remarks after it.<|endoftext|> TITLE: All maximal subgroups have odd index QUESTION [10 upvotes]: The maximal subgroups of $A_{7}$ are $A_{6}$ (index 7), $S_{5}$ (index 21), $(A_{4} \times A_{3}):2$ (index 35), and $GL_{3}(2)$ (index 15). If $G$ is a nonabelian finite simple group and all maximal subgroups of $G$ have odd index, does it follow that $G \cong A_{7}$? I don't know the Lie type groups well enough to answer this for myself, but here I'll prove it for the alternating groups (and a case-by-case check works for the sporadic groups): Assume that $n \geq 5$. If $n+1$ is not a power of 2, then choose an $r < \frac{n}{2}$ such that $\binom{n}{r}$ is even; then the stabilizer of an $r$-element subset of $\{ 1, \ldots, n \}$ in $A_{n}$ is a maximal subgroup of even index. Otherwise, write $n = 2^{k}-1$. Since we are assuming $n \neq 7$, we say that $k \geq 4$. Then $GL_{k}(2)$, in its action on vectors of $\mathbb{F}_{2}^{k} \setminus \{ 0 \}$, is a subgroup of $A_{n}$. Since $GL_{k}(2)$ is a proper subgroup of $A_{n}$, consider a maximal subgroup $M$ containing $GL_{k}(2)$. Claim. $[A_{n}:M]$ is even. Proof of Claim. $GL_{k}(2)$ acts doubly transitively on the indices, so $M$ does too. Therefore $M$ acts primitively on the indices. If $[A_{n}:M]$ were odd, then every 2-element of $A_{n}$ would be conjugate to an element of $M$; in particular, $M$ would contain permutations of cycle shape $(2,2)$. But $n = 2^{k}-1 \geq 15$ and, on more than $8$ points, the only primitive permutation groups containing a permutation of cycle shape $(2,2)$ are the symmetric and alternating groups. This is a contradiction because we assumed $M < A_{n}$. REPLY [15 votes]: There is an answer to the more general question of which finite (almost) simple groups have any maximal subgroup of odd index in the paper M. Liebeck and J. Saxl, The Primitive Permutation Groups of Odd Degree J. London Math. Soc. (1985) s2-31 (2): 250-264. They give a complete list of all groups of Lie type with this property. There are not so many examples of groups of Lie type that have some subgroup of odd index, and and there are no examples of groups of Lie type in which all maximal subgroups have odd index. They refer to a different paper for the sporadic groups, but you could check them in the ATLAS.<|endoftext|> TITLE: Construction of an optimal electron cage QUESTION [17 upvotes]: I will describe the question first in 2D, but my interest is in $\mathbb{R}^3$. An electron $x$ will shoot from the origin along an initial vector $v$. You know the speed $|v|$ but not the direction. Your task is to arrange a minimum number of point-charge electrons at fixed locations on an origin-centered unit circle so that, no matter in which direction $v$ is pointing, $x$ cannot escape the disk. All the charges are equal, and repel each other via inverse-square Coulomb force. Below there are four charges on the circle. The left electron escapes, the right, which is shot along the same direction but with a lesser speed, does not. (Of course aiming $v$ along a diagonal is the best escape strategy in this configuration.)     Now here is my question. Consider the same problem in $\mathbb{R}^3$. Given $|v|$, arrange fixed electrons on a sphere to cage-in the electron $x$ starting from the origin with velocity $v$ in any direction. It seems natural to think that an optimal arrangement is a type of disk-packing on a sphere, for example, the solutions to the Thompson problem or the Tammes problem. (Here I am mining an earlier MO question.)                     (Image from Paul Sutcliffe.) Q. Is an optimal electron cage for a given speed $|v|$ a configuration that minimizes electrostatic potential? Or maximizes the minimum distance between any pair of electrons, i.e., an optimal disk packing on the sphere? Answered by Robert Israel: No such cage is possible in $\mathbb{R}^3$! ($\mathbb{R}^2$ is rather different and not a reliable guide to $\mathbb{R}^3$.) REPLY [3 votes]: Here is Robert Israel's animation.<|endoftext|> TITLE: Are there textbooks on logic where the references to set theory appear only after the construction of set theory? QUESTION [18 upvotes]: This is cross posted from MathStackExchange. Since this is a reference request, I believe there will not be duplications of efforts in answers. This is also related to the question here. In textbooks on logic I see a tradition of using notions of set and map from the very beginning of the exposition, long before the set theory is formally constructed. This causes multiple misunderstandings for the reader (I remember these difficulties when I was a student, and even now I have questions, as you can see), since formally this is a violation of the principle that a mathematician can't use a term before giving its accurate definition. That is why I believe that there must be texts where this mistake is repaired. Can anybody advise me a textbook on logic with a "linear structure", without these "circles in definitions"? I already told this to people at MSE, I don't see serious technical obstacles for such a book to exist: the author can just formulate the axioms of predicate calculus + axioms of set theory, then consider the corollaries (i.e. construct a first-order theory of sets), and only after that discuss what they call "semantics of logic" (i.e. everything connected to interpretations of first order languages). That is strange, I can't find such a textbook. I would appreciate very much if somebody could give a reference. P.S. I have a feeling that I must specify the question, because people seem to do not quite understand what I ask. Look at these examples. Elliott Mendelson in his Introduction to Mathematical Logic formulates and proves in Chapter 2 (Corollary 2.14) the Gödel completeness theorem. Of course, for formulating and proving this he needs the notion of logically valid formula. He gives this definition a little bit earlier (also in Chapter 2), and of course, he uses the notion of interpretation, which he defines with the help of the notions of set and function. These are not sets and functions in some "trivial, everyday sense" -- these are "true sets" and "true functions" in the sense of Set theory. But he builds Set theory (and gives accurate definitions of sets and functions) only in Chapter 4, long after Gödel's theorem. The same happens in the book by Joseph R. Shoenfield Mathematical logic. He also defines valid formulas, with the help of sets and functions, and also proves Gödel's completeness theorem (in his Chapter 4), and he does this long before constructing Set theory (which appears only in Chapter 9). And this is everywhere, in all the textbooks on Logic that I know. The only difference is that sometimes the author does not build Set theory at all (like P.S.Novikov in his Elements of Mathematical Logic). So my question is Is there a book on logic where sets and functions are mentioned only after constructing ("true", axiomatic) Set theory? I understand that the word "set" can be used in non-mathematical, everyday sense, but as I wrote above, Gödel completeness theorem is not that case. I also understand the desire to have a similar notion for simpler, "everyday" situations, say, for describing syntax of your language. But this desire does not imply the necessity to use the word "set", which appears in your text later as a term of a rigoruos, axiomatic theory (where it has a much more sophisticated meaning than in everyday life). The English language is rich enough (as well as other languages, Russian, French, etc.), it is always possible to find a better solution, which does not provoke misunderstandings, accusations and controversy. Finally, it is not nice to arrange the course of a mathematical discipline in that way, because students (and readers of your book) perceive this as a mockery. A natural way to overcome this, I believe, is what I wrote before: to construct a first-order theory of sets (with all axioms, including the axioms of predicate logic), and only after that to speak about things like "interpretation", "completeness", etc. Is it possible that nobody did this up to now? P.P.S. To people who vote to close: I hope you have convincing explanations of your motives. Edit 24.02.2019. I would like to draw the attention of those who might be interested, to my own attempt to solve this problem. This is chapter 1 in my unfinished course of undergraduate mathematics. This text is designed for my students, that is why it is in Russian, unfortunately. My problem is that I am not an expert in this field and because of that I have to spend too much time on working with the details. And I actually do not have this time, I do this sporadically. This problem with Gödel's completeness theorems (theorems 1.1.22 and 1.1.23 in the text, separately for theories with finite and infinite systems of axioms) is now the only blank spot, if it were resolved, the textbook could be considered finished (I already asked this question at MO before, it is here). The best solution for me would be if someone published an article with accurate proof of these statements in their “highly formalized form” as they are presented in my text, so that I could just refer to his article. (But I must say that a part of the problem is that Gödel's theorem for a theory with an infinite system of axioms is even not accurately formulated in my text, since I was trying to avoid the standard trick of "embedding the given first order theory into arithmetics". I believe it can be replaced by an equivalent trick of "embedding into set theory", but the details are not well written in my text, because I don't see how to "translate this".) So if somebody could give an advise or a reference I would appreciate this very much. Edit 17.11.2020. It seems to me, I did it. I would be grateful if some logicians could have a look at this text (chapter 2) and share critical comments. If need be I can translate this into English. REPLY [6 votes]: Bourbaki's book on set theory is kind of terrible, but it does develop set theory in this super-formalistic way that you're looking for. I suggest the French second edition instead of the English first edition, since the second edition is a lot easier to follow (and has fewer typos). I think it's worth adding the warning that actually reading this book is almost totally unrewarding, unless it's something you'd like to see done once. Logicians do not think this way, nor do they really seem to like this book (see for example the two polemics against it by ARD Mathias).<|endoftext|> TITLE: Principal ideal ring, does there exist an invertible matrix such that certain matrix is upper triangular? QUESTION [7 upvotes]: I asked here on Math Stack Exchange the following question. Let $R$ be a principal ideal ring. If $A$ is any $p \times q$ matrix over $R$, then does there exist an invertible matrix $U$ in $\text{M}_p(R)$ such that the $p \times q$ matrix $UA$ is upper triangular? rschwieb gave the following answer. Of course you can try row reduction: you should probably go look at Smith normal form. Of course that is written for principal ideal domains, but I don't remember the domain part being the important condition to execute row operations. I think you just need GCDs. (I should check, maybe there is something tricky.) Part of the algorithm is the process of performing row operations to put the matrix in upper triangular form. The process continues with aim invertible matrix on the right to get the matrix into a diagonal form. My question is, can anybody here beef up what he wrote here into a more complete (but hopefully concise) proof? I still don't really see why what he suggests works. REPLY [2 votes]: An elementary matrix is a $p\times p$ matrix of the form $$ E^a_{i,j} = I + a e_{i,j}, $$ where $I$ is the identity matrix, $a$ is any element in your ring, $i\neq j$, and $e_{i,j}$ is the standard matrix unit. It is easy to see that $E^a_{i,j}$ is always invertible and that its inverse is $E^{-a}_{i,j}$. The other crucial property of this matrix is that, for every $p\times q$ matrix $A$, the product $E^a_{i,j}A$ coincides with the matrix you get by multiplying the $j^{th}$ row of $A$ by $a$, and adding it to the $i^{th}$ row. This is therefore one of the elementary operations used to perform row reduction. The other important operation used in row reduction, namely exchange of rows, is also easily seen to be obtained by left multiplying $A$ by a suitable invertible matrix. Thus, if you can transform $A$ into an upper triangular matrix using a finite number of elementary operations, you may also left multiply $A$ by the same number of invertible matrices, obtaining an upper triangular matrix, say $$ U_1U_2\ldots U_nA=T, $$ from where your conclusion follows easily.<|endoftext|> TITLE: Is a paraconsistent and provably non-trivial foundation for math possible? QUESTION [8 upvotes]: Would it be possible to use a paraconsistent logic and axioms similar to ZFC to create a formal sytem, that can be proven to be non-trivial (so that there are some statements which can´t be proven in the system), and which can serve as a foundation for mathematics? Because it´s possible that the current ZFC + first order logic foundation is inconsistent, and if it´s inconsistent then it would be completely trivial and worthless. And according to my understanding of gödel´s second incompleteness theorem it´s impossible to show that ZFC is consistent (at least without using a formal system that´s even more questionable). This means it´s impossible to prove that the current ZFC foundations are non-trivial, and the same holds for any other foundation that uses first order logic and contains basic arithmetic. Naturally, I think it would be more desirable to have a foundation that is demonstrably non-trivial, than a foundation where we can´t possibly prove non-triviality. That´s why I wonder whether a paraconsistent foundation could be demonstrably non-trivial. So basically I have two questions: Is it possible to have a non-trivial paraconsistent formal system containing basic arithmetic, whose non-triviality can be proven in "weaker" formal systems? If such a system exist, could it possibly be a reasonable foundation for mathematics? REPLY [5 votes]: One should note that as regards your question 1, systems of relevant arithmetic with inconsistent models such as $R{\sharp}$, $R{\sharp}{\sharp}$, and the systems $RM3^{i}$ can prove (with finitary proofs) their own non-triviality (see Friedman's and Meyer's paper "Whither Relevant Arithmetic",The Journal of Symbolic Logic, Vol. 57, No. 3(Sep., 1992), pp. 824-831; and Meyer's and Mortensen's paper, "Inconsistent Models for Relevant Arithmetics", Journal of Symbolic Logic, Vol. 49, No. 3 (Sep.,1984)). Note also that (for example) that $$ PRA+(\text{Quantifier-free Transfinite Induction up to }\epsilon_0) $$ can prove the consistency of $PA$, but are incomparable in logical strength. Why then is it necessary that the nontrivial paraconsistent system that can serve as a foundation for mathematics have its non-triviality proven in a weaker formal system? If one drops the criterion mentioned in question 1 (the proof of non-triviality of a paraconsistent system in a weaker formal system), there is a paraconsistent system in which one can develop standard set theory (and in so doing produces a paraconsistent foundation for mathematics). This is the system Hyper-Frege ($HF$)+ 'There is an infinite well-founded set' ($HF_{\infty}$). $HF$ first appears in Thierry Libert's paper "$ZF$ and the Axiom of Choice in Some Paraconsistent Set Theories" (Logic and Logical Philosophy, Vol. 11 (2003), 91-114; and $HF_{\infty}$ appears in Olivier Esser's paper "A Strong Model of Paraconsistent Logic", Notre Dame Journal of Formal Logic, Vol 44, No. 3 (2003), pp. 149-156. If one considers Esser's Theorem 3.2 (my comments will be in square brackets), The theory $HF_{\infty}$ is mutually interpretable with $GPK^{+}_{\infty}$ which is also [equiconsistent with] $KM$ [Kelly-Morse class theory] + 'On is weakly compact'. The theory $HF$ is mutually interpretable with $PA_2$ [second-order arithmetic--practitioners of Reverse Mathematics please take note]. one sees that $HF_{\infty}$ can serve as a foundation for most (if not all) modern mathematics, and that most ordinary mathematics can be interpreted and developed in $HF$. It also should be noted that Esser's construction of a model for $HF_{\infty}$ shows that $HF_{\infty}$ is non-trivial. It is unknown to me whether $HF$ or $HF_{\infty}$ can prove their own non-triviality.<|endoftext|> TITLE: Group of units of a ring spectrum vs of its connective cover QUESTION [5 upvotes]: Let $R$ be a commutative ring spectrum (interpret this as you will; as an $E_\infty$-ring or as a commutative $S$-algebra etc.) and $\operatorname{GL}_1(R)$ as usual denote its space of units. If $\tilde R$ is the connective cover of $R,$ is there a simple relationship between $\operatorname{GL}_1(\tilde R)$ and $\operatorname{GL}_1(R)$? REPLY [10 votes]: For symmetric ring spectra $R$ there is also a definition of the graded group of units, $GL_1^J(R)$, which retains information about the negative homotopy groups of $R$. See Sagave-Schlichtkrull, "Diagram spaces and symmetric spectra", Advances in Mathematics (2012). Here $J = \Sigma^{-1} \Sigma$ is a specific category, and the graded units can be viewed as spaces over $BJ \simeq QS^0$. For instance, $GL_1^J(ku)$ is different from $GL_1^J(KU)$, where $ku$ is the connective cover of the periodic complex $K$-theory spectrum $KU$.<|endoftext|> TITLE: Ramsey Theorem for the class ORD QUESTION [12 upvotes]: Is it true that given a (definable) 2-coloring of the ORD (class of ordinals), $\chi:[ORD]^{2}\rightarrow\lbrace 0,1\rbrace$, there exists an unbounded $H\subseteq ORD$ which is homogenous, i.e., $\chi:\upharpoonright [H]^2$ is constantly 0 or 1? ` In the above $[X]^2$ is the collection of unordered pairs from $X$. REPLY [12 votes]: Ali Enayat and I have proved that with respect to definable classes, Ord is NOT weakly compact. In particular, we show, in every model of ZFC, there is a definable Ord-tree with no definable cofinal branch. there is a definable 2-coloring of a definable proper class, with no definable homogeneous proper class. there is a definable set-satisfiable $L_{\text{Ord},\omega}$-theory, which has no definable class model. This result surprised me very much, since it shows that with respect to definable classes, we can prove that Ord fails to have a large cardinal property that reasonable people might have expected to hold true. The article is now available: A. Enayat and J. D. Hamkins, ZFC proves that the class of ordinals is not weakly compact for definable classes, manuscript under review. (arχiv, blog post)<|endoftext|> TITLE: Evolution of $W_+$ and $W_-$ under the Ricci flow QUESTION [5 upvotes]: In dimension $4$ the Weyl operator $W$ splits in two parts $$W_+:\Lambda^{2}_{+} \to \Lambda^{2}_{+}$$ and $$W_-:\Lambda^{2}_{-} \to \Lambda^{2}_{-}.$$ (a) Has there been a study of the evolution under the Ricci flow of these two endomorphisms? (b) If so, is there a source where these are written explicitly? REPLY [4 votes]: You can find a computation for the evolution of $A=\tfrac{S}{6}I_++W_+$ in Hamilton's "Four-manifolds with positive curvature operator", it is right before Lemma 3.1. What is found is : $\partial_tA=\Delta A + A^2+2A^\#+ BB^T$ Where $A^\#$ is the cofactor matrix of the $3\times 3$ matrix of $A$ in an orthonormal frame and $B$ is the matrix of $\tfrac{1}{2}Ric_0\wedge id:\Lambda^2_-\to\Lambda^2_+$. This can be proved just by "projecting" the evolution of the full curvature tensor on the $S^2(\Lambda^2_+)$ component. To get the evolution of $W_+$ alone, just subtract to it the evolution of $S$. Remark : All computations are done assuming Uhlenbeck's trick.<|endoftext|> TITLE: Intuition behind Harmonic Analysis in Analytic Number Theory QUESTION [37 upvotes]: As far as I know, in analytic number theory, harmonic analysis appears often. The thing is that I would see the proof of some results where they use harmonic analysis, and I can follow the argument of the proof and it makes sense, but I have no intuition behind why one would consider using harmonic analysis there (other than that using it works...). For example, maybe in a proof one has to estimate a sum of the form $\sum f(n)$ and so they would take the Fourier transform and use Poisson summation formula or something and it works. I would understand the proof, but I just have no idea why it was the "right" thing to do or why it was a good thing to try (other than of course that it worked out). I know my question is rather vague, but I would appreciate some explanations if possible! Also I would try to modify the question in a better way if anyone has any suggestion. Thank you very much! REPLY [5 votes]: Consider the Real numbers modulo the integers. You then get a circle which is of course the group $\mathbb{R}/\mathbb{Z}$, this group is compact and we know that that the dual space is a discrete infinite cyclic group and thus isomorphic to $\mathbb{Z}$. Consider a continuous function $h:\mathbb{R}/\mathbb{Z} \longrightarrow \mathbb{C}$. We have by the Plancherel theorem $$h\left(\bar{0}\right) = \sum_{n\in \mathbb{Z}} \hat{h}(n)$$ Now, there is a canonical map from $C_c(\mathbb{R}) \longrightarrow C_c(\mathbb{R}/\mathbb{Z})$, where $C_C(X)$ is the space of continuous functions of compact support on $X$ with values in the complex numbers. Indeed the map is given by $f\mapsto \bar{f}$, where $$\bar{f}(\bar{i})= \sum_{n\in\mathbb{Z}} f(n+i)$$ in particular $$\sum_{n\in\mathbb{Z}} f(n)=\bar{f}\left(\bar{0}\right) = \sum_{n\in \mathbb{Z}} \hat{\bar{f}}(n)$$ it is not very difficult to show that $$\hat{\bar{f}}(n)= \hat{f}(n)$$ where the $\hat{}$ means their respective fourier transfoms i.e. one for the group $\mathbb{R}/\mathbb{Z}$ and one for $\mathbb{R}$. So this is essentially a short proof of poisson summation formula. The question is how can one have an intuition of why the Fourier transfom will be useful in this kind of things. The reason in this case is because the sum $\sum_{n\in\mathbb{Z}} f(n)$ turn out to be a trace and thus can be decomposed into a spectral sum. In more general terms, when can you expect this kind of relationship. I would say every time you work with automorphoic forms. An automorphic form is in general a function that lives in $L^2$ of some topological group say $G$ modulo some discrete subgroup $Q$. In number theory we are interested in the case where $$G(\mathbb{Q})\backslash G(\mathbb{A})$$, where $\mathbb{A}$ are the adeles and $\mathbb{Q}$ are the rationals. In this case we can decompose the space $$L^2(G(\mathbb{Q})\backslash G(\mathbb{A}))$$ into factors related to irreducible automorphic representations. This decomposition relies on a measure and as one would expect a "Fourier transform" with respect to this measure. So in short it is natural to think on Fourier transforms every time you see something like functions on $Q\backslash G$ and in the particular case of number theory this is the study of automorphic forms.<|endoftext|> TITLE: Non-homogeneous forcing and HOD QUESTION [7 upvotes]: Is there a separative forcing notion $\mathbb{P}$ such that: 1) For any $p \in\mathbb{P}, \mathbb{P}/p = \{q \in \mathbb{P}: q \leq p \}$ is not forcing isomorphic to any homogeneous forcing notion, 2) For all $G$, $\mathbb{P}$-generic over $V$, $HOD^{V[G]} \subseteq V$. Here by homogeneity I mean cone homogeneity, i.e. for any p, q in $\mathbb{P}$, there are $p' \leq p, q' \leq q$ and an isomorphism from $\mathbb{P}/ p' $ onto $\mathbb{P}/q'.$ REPLY [8 votes]: It is consistent that the answer is positive and it is consistent that the answer is negative. Claim: There is a generic extension, $V[G]$ by a weakly homogeneous forcing notion in which there is a rigid forcing notion $\mathbb{P}$ such that for every generic filter $H \subseteq \mathbb{P}$, $$HOD^V = HOD^{V[G]} = HOD^{V[G][H]}.$$ Proof: Let $\kappa$ be a regular uncountable cardinal. Recall that $\square(\kappa)$ sequence is a sequence $\mathcal{C} = \langle C_\alpha \mid \alpha < \kappa\rangle$ such that $C_\alpha \subseteq \alpha$ is a club, for every accumulation point $\beta \in \text{acc }C_\alpha$, $C_\beta = C_\alpha \cap \beta$ and there is no club $D \subseteq \kappa$ such that for every $\beta \in \text{acc }D$, $D \cap \beta = C_\beta$. Let $\mathbb{S}$ be the forcing notion for adding a $\square(\kappa)$ sequence using bounded approximations (or successor ordinal length). Let $\mathcal{C}$ be the generic $\square(\kappa)$ sequence. Let $\mathbb{P}$ be the forcing notion $\mathcal{C}$, where $C_\alpha$ is stronger than $C_\beta$ in the order of $\mathbb{P}$ if $C_\alpha$ is an end extension of $C_\beta$. $\mathbb{S}$ is weakly homogeneous and $\mathbb{S} \ast \mathbb{P}$ has a dense subset isomorphic to the Cohen forcing $\text{Add}(\kappa, 1)$. Therefore the Boolean completion of $\mathbb{S} \ast \mathbb{P}$ is weakly homogeneous. On the other hand $\mathbb{P}$ is rigid. In fact, if $V[G] \subseteq W$ is a model of $ZFC$ and $H_1, H_2\in W$ are two distinct $V[G]$-generic filters for $\mathbb{P}$ then $W \models \text{cf }\kappa = \omega$. By density arguments (on $\mathbb{S}$ and $\mathbb{P}$), $D_1 = \bigcup H_1,\ D_2 = \bigcup H_2$ are both threads, namely, for $i = 1, 2$ and every accumulation point $\beta\in \text{acc }D_i$, $D_i\cap \beta = C_\beta$. But if $\text{cf }\kappa$ is uncountable, then $\text{acc }D_1 \cap \text{acc }D_2$ is unbounded at $\kappa$ and therefore $D_1 = D_2$ and thus $H_1 = H_2$. This implies that there is no automorphism for the Boolean completion of $\mathbb{P}$ which is non-trivial on the generic filter, since this automorphism sends the generic filter to a different generic filter. Since $\mathbb{P}$ preserves the regularity of $\kappa$ - this is impossible. Claim: Assume $V = L$. Then for every complete Boolean algebra $\mathbb{B}$, $\Vdash_{\mathbb{B}} HOD^{V[G]} = L$ iff for densely many $b\in \mathbb{B}$, $\mathbb{B} \restriction b$ is weakly homogeneous. Proof: Let $$\mathbb{C} = \{ x \in \mathbb{B} \mid \forall \sigma \in \text{Aut}({\mathbb{B}}),\ \sigma(x) = x\}.$$ Where we take the set of all automorphisms that respect also arbitrary $\inf$ and $\sup$. $\mathbb{C}$ is a complete subalgebra of $\mathbb{B}$. Let $G \subseteq \mathbb{B}$ be $L$-generic and let us claim that the generic filter for $\mathbb{C}$ belongs to $HOD^{V[G]}$. Indeed, $b \in G\cap \mathbb{C}$ iff for every $L$-generic filter $H\subseteq \mathbb{B}$, $b\in H$. This follows from a theorem of Vopenka and Hajek: Theorem (Vopenka and Hajek): If $H_1, H_2\subseteq \mathbb{B}$ are $V$-generic and $V[H_1] = V[H_2]$ then there is an automotphism $\sigma\in \text{Aut}(\mathbb{B})$ such that $\sigma '' H_1 = H_2$. Thus, we conclude that $\Vdash \dot{G} \cap \mathbb{C}\in L$, and in particular, in $\mathbb{C}$ there is a dense set of atoms. Otherwise, there is $c\in \mathbb{C}$ such that there are no atoms below $c$ in $\mathbb{C}$. Therefore, $G \cap \mathbb{C} \notin L$ for every generic filter $G$ that contains $c$, since it is a generic filter for non-atomic forcing. Let $a$ be an atom of $\mathbb{C}$. The forcing $\mathbb{B} \restriction a$ is weakly homogeneous, as for every $b \leq a$, the orbit of $b$, $\{\sigma(b) \mid \sigma \in \text{Aut}(\mathbb{B})\}$ is below $a$, and therefore $\bigvee_{\sigma \in \text{Aut}(\mathbb{B})} \sigma(b) = a$. Therefore, for every $c \leq a$, there is $\sigma$ such that $\sigma(b) \wedge c \neq 0$.<|endoftext|> TITLE: Non-residually-finite finitely-presented sofic group with all finitely generated subgroups Hopfian QUESTION [7 upvotes]: Is there a finitely presented sofic group which is not residually finite, but all of its finitely generated subgroups are Hopf groups? It seems like the Baumslag Solitar groups $BS(m,n)$ don't work (i.e. for $|m|=1$ or $|n|=1$ or $|m|=|n|$ they are residually finite, and otherwise they contain a non-Hopf finitely generated subgroup). Note: Thanks to YCor for pointing out that in the initial formulation I said "subgroups" instead of "finitely generated subgroups". The latter is my intention. REPLY [8 votes]: Houghton's group $H_3$ (see Section 5.3 here for a definition) is (locally finite)-by-$\mathbf{Z}^2$, which easily implies that all its finitely generated subgroups are Hopfian. (Not all its subgroups are Hopfian: it admits an isomorphic copy of $F^{(\mathbf{N})}$ as a subgroup for every finite group $F$.) It is finitely presented (K. Brown 1987, reference at the above link). It is not residually finite because it has the finitary symmetric group $S_\infty$ as a subgroup. It is amenable hence sofic. Edit (Jan. 12 '20): the claim that finitely generated (locally finite)-by-$\mathbf{Z}^d$ groups are Hopfian is false, even for $d=1$ (one can construct a counterexample as central extension of a lamplighter group). Therefore I retract my claim that all finitely generated subgroups of $H_3$ are Hopfian; I don't know if it's true. Yet $H_3$ is Hopfian as well as its finite index subgroups, using that the locally finite kernel is virtually simple). (As long as this is not fixed the answer should probably be unaccepted.)<|endoftext|> TITLE: Modular curve X(2) QUESTION [7 upvotes]: Let $\mathfrak{M}(2)$ be the algberaic stack over $\mathbb{Z}[1/2]$ which classifies the elliptic curves with the two level structure and let $X(2)$ be the coarse moduli space of $\mathfrak{M}(2)$ ($X(2)$ exists since $\mathfrak{M}(2)$ is smooth and proper). It is easy to see that $X(2)(\mathbb{C})$ is the modular curve of level $2$ and it is known that the modular lambda function rises to an isomorphism $X(2)(\mathbb{C}) \simeq \mathbb{P}^1_{\mathbb{C}}$. Is there a meromorphic function on $X(2)$ or on $X(2)_{\mathbb{Q}}$ compatible with the modular lambda function on the $\mathbb{C}$-points of $X(2)$? REPLY [6 votes]: $y^2 =x (x-1)(x-\lambda)$ is a family of elliptic curves with level two structure on $\mathbb P^1_{\mathbb Q}$. Hence by the definition of coarse moduli space, it defines a map from $\mathbb P^1$ to the coarse moduli space $X(2)$. It is sufficient to check that this map is an isomorphism and that it send $\lambda$ to the modular $\lambda$. The second claim implies the first, and the second claim can be checked easily from the definition of the modular $\lambda$ as the cross-ratio of the $x$-coordinates of the $2$-torsion points.<|endoftext|> TITLE: Is there a bound on the growth of homotopy groups of spheres? QUESTION [7 upvotes]: That is to say, can one find a good bound on $|\pi_i(S^n)|$? Let us assume that $i\ge 2n$ to avoid all infinite quantities. Particularly, I am interested to see if there is a bound of exponential type. I do not see a way to do this though. REPLY [5 votes]: There are relevant estimates in the 1986 paper by Hans-Werner Henn. Hans-Werner Henn, MR 850372 On the growth of homotopy groups, Manuscripta Math. 56 (1986), no. 2, 235--245.<|endoftext|> TITLE: When is the Jacobian a product? QUESTION [6 upvotes]: When is the Jacobian of a hyperelliptic curve $$y^2=x(x-1)(x-a)(x-b)(x-c)$$ a product of two elliptic curves? (This is a sort of reverse to When is a product of elliptic curves isogenous to the Jacobian of a hyperelliptic curve?). Obviously, it implies some algebraic relations between $a,b,c$; the question is: which ones? P.S. This must be classical, but I am having some trouble figuring it out or finding it in the literature. There is an old example due to Jacobi, but is it all? EDIT. As pointed out by abx, the set of parameters $(a,b,c)$ for which the Jacobian is isogenous to a product is actually dense (e.g. in $\mathbb{C}^3$ with complex topology). For this reason, the question in this form does not have a reasonable answer. (In hindsight, it is obvious, but I missed it.) REPLY [5 votes]: A genus two curve $C$ has its jacobian isogenous to a product of elliptic curves if and only if there is a nonconstant map from $C$ to an elliptic curve $E$. For each degree of map $C \to E$ there is a hypersurface in the moduli space of genus two curves of such jacobians. So you have a countable union of hypersurfaces, each of which gives you an algebraic relation. Probably one can make these conditions on $a,b,c$ explicit for small degrees. The easiest ones to characterize are the bielliptic ones: in this case, the genus two curve must have a Weierstrass equation of the form $y^2 = f(x^2)$, and the bielliptic involution is $x \mapsto -x$.<|endoftext|> TITLE: Weil's book L'intégration dans les groupes topologiques et ses applications QUESTION [12 upvotes]: The book L'intégration dans les groupes topologiques et ses applications published by André Weil in 1940 is regarded as one of the classical references for harmonic analysis on topological groups. Unfortunately I am not fluent in French, so reading the book in all details is simply impossible. However, the reason why I am interested in reading this book is because it seems that Weil's treatment is different from than the now "standard treatment" of so-called abstract harmonic analysis, which is in general characterised by its extensive use of Gelfand theory. This book by Weil was never translated into English, although there are Russian and Japanese editions, as noted in the comments. Therefore, I was curious if someone could point out references to treatments of harmonic analysis which are similar to Weil's, but are available in English. I am aware of the book Classical Harmonic Analysis and Locally Compact Groups by Hans Reiter, which was a student of Weil and which seems to have a similar approach to harmonic analysis as Weil had. However, the proofs of classical results such as the existence and uniqueness of the Haar measure and Pontryagin's Duality Theorem are all omitted in this text. Other books where I expect a similar approach as in Weil's book are, of course, the books by Bourbaki. However, I do not think that, for example, Pontryagin's Duality Theorem is proved in any of these books. I am aware that the uniqueness and existence of the Haar measure is proved in the book on Integration though. Any reference is highly appreciated. REPLY [3 votes]: I really like Cartan and Godement's Théorie de la dualité et analyse harmonique dans les groupes abéliens localement compacts (1947). It isn't quite what you hope for, since it is again untranslated. But it is very much in Weil's spirit — improved to avoid the structure theory, but unlike Bourbaki's Théories spectrales, still eschewing Gelfand theory (replaced by use of the Kreĭn-Milman theorem). The French is more accessible than Weil's, and it's much shorter: just 20 pages to progress through the theorems Riemann-Lebesgue, Bochner, Fourier inversion, Plancherel, and Pontryagin duality. Add two pages for Godement's generalized Stone's theorem (1944), and all the bases are covered.<|endoftext|> TITLE: Existence of periodic orbits in rational billiards QUESTION [9 upvotes]: Recently I've got interested in dynamical billiards. Some results in this field are obtained by elementary methods. For instance, see George W. Tokarsky's Polygonal Rooms Not Illuminable from Every Point or Andrew M. Baxter and Ron Umble's Periodic Orbits of Billiards on an Equilateral Triangle. Then I stumbled across this Every rational billiard has periodic orbits. I tried to find a proof which was comprehensible to a freshman as I am, but I couldn't. Almost every article I had a look at somehow brought me to Howard Masur and Serge Tabachnikov's Rational billiards and flat structures, which is mostly beyond my knowledge. My question is, is it possible to prove that theorem without Teichmuller spaces, quadratic differentials, ergodicity,...? If not, what background is needed to deal with this and related problems? REPLY [8 votes]: Schwartz, Richard Evan. Mostly surfaces. Vol. 60. American Mathematical Society, 2011. On p.219ff of Schwartz's book, he sketches "an elementary proof, due to Boshernitsyn, that every rational polygon has at least one periodic billiard path." (He doesn't cite an explicit reference for Boshernitsyn's proof.)                    In answer to "what background is needed...?": The first 218 pages of Schwartz's book. :-) Less flippantly, you could essentially read backwards from the proof sketch to the concepts that precede it.<|endoftext|> TITLE: Has anyone catalogued the "first generation" proof of the classification of finite simple groups? QUESTION [27 upvotes]: It has been estimated that the original proof of the CFSG spans around 15,000 journal pages written by hundreds of authors over most of the 20th century. The GLS project attempted to simplify this original proof, with a target length around 5000 pages, but this involved a variety of changes to the original proof style, and from what I understand this project has stalled (the last volume was published in 2004 and five more are planned), possibly due to the mounting difficulties of changing the structure of such a large proof. Has anyone attempted to present an argument for the full CFSG, encoding the high level structure of the proof using journal references as necessary to establish the various subcases? That is, the part of the proof that was happening in the minds of specialists who felt comfortable declaring the problem "solved" after the last journal article had been published (Aschbacher-Smith Quasithin theorem, 1996). R. Solomon's article above is near to such a project, with a very comprehensive reference list, but it lacks the rigorous presentation of the proof itself, and the division into subcases except in an illustrative sense. REPLY [42 votes]: There are two books which together have the purpose of answering this question. D. Gorenstein, The Classification of Finite Simple Groups. Volume 1: The Noncharacteristic 2 Type Case. Plenum Press, 1983. (Gorenstein died without writing Volume 2.) M. Aschbacher, R. Lyons, S.D. Smith, R. Solomon, The Classification of Finite Simple Groups: Groups of Charateristic 2 Type. A.M.S. Mathematical Surveys and Monographs 172, 2011.<|endoftext|> TITLE: Which unimodular matrices are integer combinations of permutations? QUESTION [7 upvotes]: Consider the set of all integer linear combinations of permutation matrices of some fixed dimension. Is there a description of the set of unimodular matrices in this lattice? REPLY [5 votes]: The $\Bbb{Z}$-span $T_n$ of the order $n$ permutation matrices consists of integer $n\times n$ matrices with the row and column sums equal to $k$ for some integer $k$. Thus it can be characterized as the subset (in fact, subring) of $M_n(\Bbb{Z})$ consisting of matrices $A$ such that $Av=kv$ and $A^t v=kv$ for some $k\in\Bbb{Z}$, where $v=(1,\dots,1)^t$ is the column with all entries $1$. Let $L\subset\Bbb{Z}^n$ be the sublattice of integer vectors with sum of the entries $0$, $L=\{u: u^t v=0\}$. Note that $\Bbb{Z}^n=L\oplus \Bbb{Z}v$ and $A\in T_n$ preserves this decomposition. Clearly, $A\in T_n$ is invertible only if $k=\pm 1$ and the restriction of $A$ to $L$ is invertible. Conversely, any automorphism of $L$ can be extended in exactly two ways to an invertible matrix acting by $\pm 1$ on $v$. Since $L$ has rank $n-1$, it follows that the group in question is the direct product ${\rm GL}_{n-1}(\Bbb{Z})\times\{\pm 1\}$, with the first factor represented by the automorphisms of $L$ and the second factor represented by the scalar matrices $\pm I_n$.<|endoftext|> TITLE: Is there a non-metrizable topological space for which any countably compact subset is compact? QUESTION [8 upvotes]: The title is the question : Is there a non-metrizable topological space for which any countably compact subset is compact ? EDIT : non-metrizable and Hausdorff REPLY [7 votes]: In addition to the above examples, it might be of interest that functional analysis is awash with such spaces. For example, the space of distributions on a compact interval or that of functions analytic on a closed domain in the complex plane, both with their natural topologies. More generally, any Silva space has this property. A further ubiquitous source of examples is the dual of a separable Banach space with the bounded weak $\ast$ topology. A suitable reference would be Köthe's classic on topological linear spaces.<|endoftext|> TITLE: $GCH$ and special Aronszajn trees QUESTION [15 upvotes]: Question. Does $\text{GCH}$ imply the existence of a non-special $\aleph_2$-Aronszajn tree ? Remark 1. By a result of Jensen, it is consistent that $\text{GCH}$ holds and all $\aleph_1$-Aronszajn trees are special. Remark 2. The above question is related to the famous question ``does $\text{GCH}$ imply the existence of an $\aleph_2$-Souslin tree ?''. REPLY [5 votes]: It turned out the answer to the question is no. David Aspero and I, have proved the following: Theorem. Assuming the existence of a weakly compact cardinal, there exists a generic extension of the universe in which $GCH$ holds and all $\aleph_2$-Aronszajn trees are special. See The special Aronszajn tree property at $\aleph_2$ and $GCH$. Comments and suggestions on the paper are welcome.<|endoftext|> TITLE: When $k = \mathbb{F}_q$ finite field, $X$ always has $k$-rational point, and so $A \simeq X$? QUESTION [11 upvotes]: Let $k$ be an arbitrary field. Let $(A, e)$ be an abelian variety over $k$, and let $X$ be a torsor for $A$, i.e. $X$ is a proper smooth $k$-variety, and there is an $A$-action acting $:A \times X \to X$ such that for any $k$-scheme $L$ and a point $x \in X(L)$, the induced "orbit" map $A_L \to X_L$ given by $a \mapsto a + x$ is an isomorphism. When $k = \mathbb{F}_q$ is a finite field, how do I see that $X$ always has a $k$-rational point, and thus $A \simeq X$? REPLY [14 votes]: A bit of overkill, but it follows from the Weil conjectures. The structure of cohomology ($H^i = \wedge^i H^1$) is computed over the algebraic closure and it follows that the number of points is $\prod(\alpha_i-1)$ where the $\alpha_i$ are the eigenvalues of Frobenius on $H^1$ so $|\alpha_i| = q^{1/2}$ and the product is therefore not zero.<|endoftext|> TITLE: Simply connected manifolds with dense geodesics on the tangent bundle QUESTION [5 upvotes]: A unit speed geodesic $\gamma:I\to M$ on a Riemannian manifold $M$ can be lifted to a curve $\sigma=(\gamma,\gamma')$ on $SM$, the unit sphere bundle (the unit tangent bundle) of $M$. Let us say that the geodesic $\gamma$ is dense on the tangent bundle if the trace $\sigma(I)$ of $\sigma$ is dense on $SM$. The existence of a dense geodesic in this sense can be regarded as a form of ergodicity (or mixing) of the geodesic flow. Are there simply connected Riemannian manifolds for which a geodesic is dense on the tangent bundle? Does the answer depend on dimension? I know that such geodesics exist on all compact Anosov manifolds. This includes negatively curved closed surfaces, but they have genus $\geq2$ and are not simply connected. I failed to find an example or proof of non-existence. REPLY [8 votes]: Burns and Donnay proved that every surface (including a sphere) admits a Riemannian metric that makes the geodesic flow ergodic with respect to Liouville measure, and hence topologically transitive (there is some $v\in SM$ whose orbit under the geodesic flow is dense in $SM$, in other words, the corresponding geodesic is dense in $SM$ in the sense you describe in the question): see "Embedded surfaces with ergodic geodesic flows", Internat. J. Bifur. Chaos Appl. Sci. Engrg. 7 (1997), no. 7, 1509–1527. The abstract to that paper reads as follows: Following ideas of Osserman, Ballmann and Katok, we construct smooth surfaces with ergodic, and indeed Bernoulli, geodesic flow that are isometrically embedded in $\mathbb{R}^3$. These surfaces can have arbitrary genus and can be made analytic. As Misha's comment points out, this is related to an earlier question of Joseph O'Rourke, although that one seems to be about density on the surface, rather than in the unit tangent bundle. Misha's answer to that question references a 2004 paper by Donnay and Pugh that builds on the Burns-Donnay construction from the paper above.<|endoftext|> TITLE: Homogeneous polynomials, mixed determinants, positive definiteness QUESTION [10 upvotes]: Are there $n\times n$ real matrices $A_{1}, \ldots, A_{n}$ such that the $n$-homogeneous polynomial $$ f(x_{1}, \ldots, x_{n}) = \det(x_{1} A_{1}+\cdots +x_{n} A_{n}) $$ never vanishes on $\mathbb{R}^{n}\setminus\{0\}$? I was listening to a seminar of a student, and a certain problem boils down to this linear algebraic question. I know that if $n$ is odd then the answer is negative; also if $n=2^{k}$ then the answer is positive. I do not quite see right now what happens for an arbitrary even $n$. The first interesting case is $n=6$. I believe this should be something well-known. REPLY [12 votes]: I doubt about powers of 2, it looks that the answer is $n=1,2,4,8$. Without loss of generality $A_1=I$ (else replace $A_i$ to $A_iA_1^{-1}$ for all $i$). Then for any $x\in \mathbb{S}^{n-1}$ the vectors $x=A_1 x,A_2x,\dots,A_nx$ should be linearly independent (else $x$ belongs to a kernel of a certain linear combination of $A_i$'s). Projecting $A_2x,\dots,A_nx$ onto the hyperplane $x^{\perp}$ we get for any $x$ an $(n-1)$-tuple of linearly independent vectors orthogonal to $x$, and they of course are continuous in $x$. That is, the sphere $\mathbb{S}^{n-1}$ is parallelizable, this is only the case for $n=1,2,4,8$. For these values of $n$, we take $n$-dimensional associative real division algebra (of real, complex, quaternionic or octavic numbers) generated by the lements $g_1,\dots,g_n$ and let $A_i$ be the operator of right multiplication by $g_i$.<|endoftext|> TITLE: Is it decidable to check if an element has finite order or not? QUESTION [36 upvotes]: Suppose we have a finitely presented group $G$ with decidable word problem. Is it decidable to check whether a given element $x\in G$ has finite order or infinite? REPLY [21 votes]: The decidability of the word problem does not imply the decidability of the order problem, and in fact the following more general result holds. Theorem. Let $\mathbf{a}, \, \mathbf{b}, \, \mathbf{c}$ be three recursively enumerable degrees of unsolvability (i.e., Turing degrees) with $\mathbf{a} \leq \mathbf{b}$ and $\mathbf{a} \leq \mathbf{c}$. Then there exists a finitely presented group $L$ such that the word problem for $L$ is of degree $\mathbf{a};$ the power problem for $L$ is of degree $\mathbf{b};$ the order problem for $L$ is of degree $\mathbf{c}.$ See D. J. Collins: The word, power and order problems in finitely presented groups, in "Word Problems, Decision Problems and the Burnside Problem in Group Theory", Studies in Logic and Fundations of Mathematics 71 (1973).<|endoftext|> TITLE: Polynomials with more than one common root QUESTION [5 upvotes]: Given two polynomials $p_1 (x) , p_2 (x) \in \mathbb{C} [x]$, we know they have a common root iff their resultant vanishes. What is the condition ensuring $p_1$ and $p_2$ share more than one common root? Many thanks in advance. REPLY [6 votes]: The real condition is that the greatest common divisor of the two polynomials has degree greater than one. Computationally, that's about as good as you are going to get, but if you like resultants, the vanishing of the resultant is just a way of restating that the system of linear equations on the coefficients of $a, b$ given by $$a p_1 + b p_2 = 1$$ has no solutions (meaning that the vector $(1, 0, \dots)$ is not in the image). If the gcd has degree greater than one, the span of $(1, 0, \dots)$ and $(0, 1, \dots)$ is not in the image, so the resultant has co-rank at least $2,$ as suggested by Fedor.<|endoftext|> TITLE: Topology and the 2016 Nobel Prize in Physics QUESTION [119 upvotes]: I was very happy to learn that the work which led to the award of the 2016 Nobel Prize in Physics (shared between David J. Thouless, F. Duncan M. Haldane and J. Michael Kosterlitz) uses Topology. In particular, the prize was awarded "for theoretical discoveries of topological phase transitions and topological phases of matter". Having read both the popular and advanced versions of the scientific background found on the website linked above, I'm left with the impression that the most advance topological concepts used are winding numbers, vector fields and the Poincaré-Hopf index theorem. I wonder if anyone who reads MO is sufficiently familiar with this work to explain to me why this impression is correct or incorrect. To ask a precise question: Which topological concepts and results are involved in the work which led to the award of the 2016 Nobel Prize in Physics? And a follow-up question: Where should a topologist go to read about topological phases of matter and topological phase transitions? REPLY [16 votes]: The involved topology are 1) $\pi_1(S^1)$ in TK transition (Thouless-Kosterlitz) 2) First Chern class and Chern number in TKNN relation to Hall conductance (Thouless) and in Haldane model for quantum anomalous Hall effect (Haldane) 3) $\pi_2(S^2)$ in Haldane phase ($\theta$-topological term) (Haldane) Now some physics. 1) TK transition is a transition between non-topological phases induced by topological defects (vortexes). Thouless and Kosterlitz revealed the importance of the topological defects in phase transition. TK transition is also the first example of a continuous transition that does not change the symmetry. 2) TKNN related the Chern number to quantized Hall conductance in integer quantum Hall effect on lattice. Haldane showed that integer quantum Hall effect can appear on a lattice if we break the time reversal and parity without the need of external magnetic field. 3) Haldane showed that spin-1/2 chain has a $\theta$-topological term with $\theta=\pi$, and spin-1 chain has a $\theta$-topological term with $\theta=2\pi \sim 0$. Thus he concluded that spin-1/2 chain is gapless (agreeing with previously known results) while spin-1 chain is gapped (this was a new result and a surprise. The gapped phase is named Haldane phase). Later, algebraic topology enter the study of topological phases, such as group coohomology or cobordism (classify SPT phases, including Haldane phase), unitary fusion category (classify 2+1D topological orders with gappable boundary), unitary modular tensor category (classify 2+1D topological orders), $G$-crossed unitary modular tensor category or unitary braided fusion category over a symmetry fusion category (classify 2+1D bosonic/fermionic topological orders with symmetry), etc. Are those "quantum topology"?<|endoftext|> TITLE: Boundary triangulation induces triangulation QUESTION [8 upvotes]: In $R^n$ (the real space) we have an open connected set $D$, such that $\partial D$ is triangulable. Can we prove the closure $\bar{D}$ is triangulable or any counterexample? Furthermore, the $\partial D$ are piecewise algebraic in the question I am considering, I do not know whether this would be helpful for the above statement. Thanks for any help. REPLY [4 votes]: In many categories, the answer is known to be yes, see Emil Saucan, MR 2184196 Note on a theorem of Munkres, Mediterr. J. Math. 2 (2005), no. 2, 215--229.<|endoftext|> TITLE: Concrete bounds on the discriminant of a number field QUESTION [5 upvotes]: There exists a well known concrete bound on the discriminant of a number field by Minkowski. Are there any concrete (completely explicit) improvements of this bound? I know of a bound by Odlyzko, but it is only asymptotic, and thus not explicit/concrete enough. In particular, I do not know how to apply this bound to a number field of degree (say) $7$. That is, I am looking for a bound on the discriminant as a function of the degree $n$ of the number field over $\mathbb{Q}$ (improving Minkowski's bound even if $n$ is not too large). I am most interested in the case of at most two real embeddings. REPLY [8 votes]: Theorem 2.4 of the paper Local corrections of discriminant bounds and small degree extensions of quadratic base fields by Brueggeman and Doud gives two lower bounds for the discriminant of a number field of arbitrary signature. The formula is a bit complicated, but can easily be computed with a computer algebra system like SAGE. Note that the formula also makes use of Poitou's refinement of the Odlyzko bounds in that the bound incorporates knowledge about the existence of primes of small norm. (For instance, the bound implies that a degree 10 number field with 2 real embeddings has root discriminant at least 5.834, whereas a degree 10 number field with 2 real embeddings in which 2 splits completely has root discriminant at least 28.951.)<|endoftext|> TITLE: how to show the existence of root for a system of polynomial equations? QUESTION [6 upvotes]: I come across a system of polynomial equations. Although I can solve it by Mathematica (numerically), I wonder whether it is possible to prove this root indeed exists. Any general method to do it? For example, $f(x_1,x_2)=0$ ; $g(x_1,x_2)=0$ ; $f$ and $g$ are polynomials of $x_1$ and $x_2$, and I compute an approximate root $(x_1^*,x_2^*)$ by some numerical method. How can I show there indeed exists a root around this approximate calculated root? I just wonder whether there exist some general theories in doing so. REPLY [8 votes]: The magic words are: Cylindrical Algebraic Decomposition.<|endoftext|> TITLE: Eliminating constant in Rado graph QUESTION [11 upvotes]: Let $R$ denote the Rado graph, and let $c$ be a fixed vertex. Question 1. Is the structure obtained by extending $R$ by the constant $c$ interpretable in $R$ without parameters? By interpretable I mean first-order interpretable; see below for an equivalent formalism. A related group-theoretic question is the following. Let $\text{Aut}(R)$ denote the group of automorphisms of the Rado graph. Question 2. Is there an action of $\text{Aut}(R)$ on $R$ by automorphisms fixing $c$ which has finitely many orbits? Edit (11 Oct'16): Here's a purely group-theoretic weakening of Question 2: Question 2'. Is the group $\text{Aut}(R)$ isomorphic to a subgroup of $\text{Aut}(R,c)$, the group of automorphisms of $R$ fixing $c$? Edit: Question 2' has been answered positively by 'Rado the explorator' below. Unfortunately, this still doesn't answer Questions 1 and 2, since the induced action has infinitely many orbits. Here is some background to this question, and some minor observations. For a relational structure $\mathbb A$ and its element $c$, let $(\mathbb A,c)$ denote the structure obtained from $\mathbb A$ by adding the constant $c$. If $\mathbb A$ has the property that for every constant $c$, $(\mathbb A,c)$ interprets without constants in $\mathbb A$, then let us say that $\mathbb A$ has elimination of parameters. Equivalently, this means that any relational structure $S$ which interprets in $R$ with parameters also interprets in $S$ without parameters. In a comment to this question: ω-categorical, ω-stable structure with trivial geometry not definable in the pure set I remarked that $(\mathbb N,=)$ has elimination of parameters, and stated that this holds for any $\omega$-categorical structure, but Alex Kruckman pointed to an error in my reasoning, but we figured out a proof for $(\mathbb N,=)$. Here is an even simpler proof. Fact. $(\mathbb N,=)$ has elimination of parameters. In the proof, I allow myself to use a more convenient syntax for defining interpretations in a structure $\mathbb A$, which I call definable sets. The syntax of sets definable over $\mathbb A$ allows using set-builder expressions with variables ranging $\mathbb A$, and which can be constrained using first-order formulas in the language of $\mathbb A$; additionally, we can take finite unions or tuples of such expressions, and nest these operations. For instance, the set $\{(x,y):x,y\in\mathbb N, x\neq y\}\cup\{x:x\in\mathbb N, x\neq 5\}\cup \{\{x,y\}:x,y\in\mathbb N\}$ is definable over $(\mathbb N,=)$, using the parameter $5$. A relational structure $\mathbb B=(B,R_1,\ldots,R_n)$ is definable over $\mathbb A$ if $B$ and each relation $R_i$ is a set which is definable over $\mathbb A$. Up to isomorphism, structures definable over $\mathbb A$ are the same as structures which interpret in $\mathbb A$, using first-order interpretations (this correspondence preserves the used parameters), but sometimes using definable sets makes constructions easier, as e.g. below. When the definition does not involve parameters, we say that the structure is 0-definable. Proof. We show that $(\mathbb N,=,c)$ is isomorphic to a structure which is definable over $(\mathbb N,=)$: Indeed, the structure $(\mathbb N\cup\{\emptyset\},R_=,c)$, where $c$ is the constant interpreted as $\emptyset\in\mathbb N\cup\{\emptyset\}$ and $R_==\{(m,m):m\in\mathbb N\}\cup\{(\emptyset,\emptyset)\}$ is clearly 0-definable over $(\mathbb N,=)$ and isomorphic to $(\mathbb N,=)$. (It is also easy to construct a two-dimensional interpretation in $(\mathbb N,=)$ without parameters). $\square$ Fact. $(\mathbb Q,\le)$ has elimination of parameters. Proof. Here the construction is slightly more interesting. A constant $c\in\mathbb Q$ splits $\mathbb Q$ into two parts: the rationals smaller than $c$ and the rationals larger than $c$, and the interaction between the two parts is very simple. The structure $(\mathbb Q,\le, c)$ is isomorphic to the following structure. It's domain is $X=\{(q,L):q\in\mathbb Q\}\cup\{(q,R):q\in\mathbb Q\}\cup\{C\}$, where $L,R,C$ are symbols, e.g. implemented as $\emptyset,\{\emptyset\},\{\{\emptyset\}\}$ which are 0-definable, and so is $X$. We now define the relation $\le$ on $X$ by $R_\le=\Big\{((q,L),(q',L)):q,q'\in \mathbb Q, q TITLE: References for the minimal model program QUESTION [8 upvotes]: What are some references for a beginning graduate student in algebraic geometry to learn about the minimal model program? I'm not thinking about entering this field, but rather I just want to know what it means "to run the MMP". Related link: Training towards research on birational geometry/minimal model program (the OP had a different goal in mind here) REPLY [6 votes]: A very light introduction is contained in A first glimpse at the minimal model program. (Also available here.) I would also second Simone's suggestion: The first two chapters of Kollár-Mori are illuminating.<|endoftext|> TITLE: Dirichlet's unit theorem for reductive schemes QUESTION [15 upvotes]: Let $O_{K,S}$ be the ring of $S$-integers in a number field $K$. Dirichlet's unit theorem implies that the group of units in $O_{K,S}$ is a finitely generated group. In other words, the group $\mathbb G_{m}(O_{K,S})$ is finitely generated. In particular, if $T$ is a torus over $K$, then $T(O_{K,S})$ is finitely generated. My question is whether this statement generalizes to other group schemes. [Edit: Daniel Litt's comment makes it clear that unipotent groups don't satisfy this finiteness property.] Question. Let $G$ be a reductive group scheme over $O_{K,S}$. Is $G(O_{K,S})$ finitely generated? The analogous question for proper group schemes has a positive answer by the Mordell-Weil theorem. What if we restrict to semi-simple group schemes? REPLY [4 votes]: nfdc23 answered the question: For connected reductive $G$ over a number field $K$ and a finite set $S$ of places of $K$ containing the archimedean places, any affine flat finite type ,$O_{K,S}$-group scheme $\mathcal G$ with $K$-fiber $G$ is the schematic-closure in $\mathrm{GL}_{n,O_{K,S}}$ of some $G \to \mathrm{GL}_{n,K}$, since $O_{K,S}$ is a Dedekind domain. Therefore, $\mathcal G(O_{K,S})$ is an $S$-arithmetic subgroup of $G(K)$. All $S$-arithmetic groups are finitely presented, by Theorem 5.11 in the book of Platanov-Rapinchuk. Over global function fields some hypotheses are needed, see H. Behr's 1998 paper in Crelle 495.<|endoftext|> TITLE: "Role" of cohomology of coherent sheaves in SGA 4.5, étale cohomology QUESTION [5 upvotes]: As the question title suggests, what is the role cohomology of coherent sheaves plays for SGA 4.5, étale cohomology? Why are they so important for the construction and establishing properties of étale cohomology? What are some of the main essences, intuitions, and theorems of the cohomology of coherent sheaves that are exploited or get "carried over" from the classical setting, in some sense? REPLY [9 votes]: You ask two different questions, I believe. 1) When Grothendieck invented étale cohomology, cohomology of coherent sheaves was already well known, thanks to Oka/Cartan's theorems A and B, Serre's FAC, Gaga, and the Serre-Grothendieck duality theorems). It was also well known (Weil, already) that the invention of some cohomology theory closer to singular cohomology was one path towards the Weil conjecture. In some sense, cohomology of coherent sheaves was an insufficient prototype. 2) Coherent sheaves naturally furnish examples of étale sheaves and Grothendieck's descent theorems imply that nothing is gained by viewing them in the étale framework. This implies in particular that étale line bundles or étale vector bundles are the same objects than line bundles or vector bundles. This gives at least some étale sheaves, such as $(\mathbf G_m)_X$ or $\mathrm{GL}(n)_X$, whose cohomology can be computed; for example: $$ H^1_{\rm ét}(X,\mathbf G_m)=H^1_{\rm Zar}(X,\mathbf G_m)=\mathop{\rm Pic}(X).$$ For $n$ invertible on $X$, the Kummer exact sequence $$ 1\to (\mu_n)_X \to (\mathbf G_m)_X\xrightarrow{x\mapsto x^n}(\mathbf G_m)_X\to 1 $$ gives rise to a long exact sequence which is the starting path to the computation of étale cohomology groups with coefficients in $\mu_n$. When $X$ is an $\mathbf F_p$-scheme, the Artin-Schreier exact sequence $$ 0 \to (\mathbf Z/p\mathbf Z)_X \to (\mathbf G_a)_X \xrightarrow{x\mapsto x^p-x} (\mathbf G_a)_X\to 0 $$ allows to compute the étale cohomology groups of $X$ with coefficients in $\mathbf Z/p\mathbf Z$. One sees in particular that they vanish beyond the dimension of $X$, hence are insufficient for the Weil conjectures.<|endoftext|> TITLE: What is integration along the fibers in D-module theory? QUESTION [9 upvotes]: In Hotta, Takeuchi, Tanisaki's book on "D-modules, Perverse Sheaves, and Representation theory", for a morphism of smooth algebraic varieties $f:X \to Y$, they use the notation $$ \int_f:D^b(D_X^{op}) \to D^b(D_Y^{op}) $$ for the derived pushforward of right $D$-modules. What motivates this notation with the integral? I have hear this has something to do with "integrating along the fibers", but wikipedia gives an article about partially integrating differential forms. REPLY [7 votes]: I think, for a locally constant $D$-module it basically cashes out to Defining a formal solution to the differential equation. Defining a formal integral along the fibers. Writing down all the differential relations satisfied by the formal integral that follow by integration by parts from the differential equation applied to the formal solution. Converting this to a differential equation satisfied by the formal solution. More precisely we are taking the de Rham cohomology, so there may be more than one integral over more than one cycle, all of which satisfy the same differential equation. To calculatee this you take a formal ensemble of integrals of an arbitrary differential operator applied to $f$ (which is just the D-module as a module) and find all the formal differential relations between them. Unfortunately in your example you have a single PDE in two variables, so the solutions are not a holonomic D-module. I don't know if this still works in that case. If it does, you can observe that $\int \partial_x^2f dx =0$ (by integration by parts) forces $\partial_y^2\int f df dx = \int \partial_y^2 f dx =0$ (by the differential equation). Let's do a holonomic example instead. If we have the equations $\partial f/ \partial x=2xy f$, $\partial f/\partial y=x^2 f$, which is satisfied by $e^{x^2 y}$, whose integral is proportional to $\sqrt{y}$, we can find the differntial equation satisfied by $\sqrt{y}$ by observing $$ y \partial_y \int f dx = \int y \partial_y f dy = \int y x^2 f dx= \int x \partial_x f dx/2 = \int \partial_x ( xf)dx/2 - \int fdx/2 dx= 0 - \int f dx/2$$ which is the correct differential equation, and I think you can check that this algebra is exactly what the definition of D-module cohomology by de Rham cohomology is doing.<|endoftext|> TITLE: Vorticial ground states for the O(2) rotor model QUESTION [5 upvotes]: Is there a sensible notion of a ground state for the classical $O(2)$ rotor model "frustrated at infinity by a single unit of counterclockwise vorticity"? Here is a picture of the kind of thing I mean, taken from Brian Skinner's blog: I'm not sure the phrase in quotation marks is standard terminology, so let me clarify the question. The model I have in mind assigns to each site $i$ in the lattice $\Lambda = {\bf Z}^2$ a a unit-length vector ${\bf s}_i \in {\bf R}^2$ that we call a spin. The interaction energy between two adjacent sites $i,j$ is $- {\bf s}_i \cdot {\bf s}_j$. We naively define the energy of a spin-configuration as the sum of these interaction energies (summed over all pairs of adjacent sites in $\Lambda$), with the understanding that since this sum almost always diverges, we need to regularize it in some principled way. More on regularization later. If we didn't bring some frustration into the situation, it's pretty clear what the ground-states of the model would have to be, regardless of the regularization scheme one chooses: spin-configurations in which all the spins line up with each other. However, I want to impose a couple of conditions. The first is that the configuration should be "slowly varying near infinity": for all $\epsilon > 0$, there should be a radius $R$ such that if $i$ and $j$ are adjacent sites at distance $>R$ from the origin, the angle between ${\bf s}_i$ and ${\bf s}_j$ is less than $\epsilon$. (Note that I stipulated adjacency between $i$ and $j$ in this condition; if $i$ and $j$ are far apart, I don't require ${\bf s}_i$ and ${\bf s}_j$ to be close.) The second is that the configuration should be "vorticial": if one looks at the spins along any closed lattice path sufficiently far away from the origin, the (slowly-changing) spins should make a single counterclockwise turn. I would like to know whether within this class of "vorticial" spin-configurations there are ones that minimize energy, and if so, what those configurations look like. Note that the class of vorticial configurations (unlike the product space it sits inside) is not compact, so it's not guaranteed that an actual minimum exists (even leaving aside the issue of what "energy-minimizing" means when the naively-defined energy diverges). One way to regularize the sum is to change the global guage so that two vectors being parallel counts as energy 0 rather than energy $-1$. However, I think that even with this change, all the candidates for "vortex ground-states" have energy that diverges like the harmonic series. One could sidestep divergence by replacing $\Lambda$ by a large finite patch of $\Lambda$ and imposing "vorticial" conditions along the boundary, and then look for limiting behavior of the resulting energy-minimizing configuration as the patch grows to exhaust $\Lambda$. One might work directly with $\Lambda$ but mitigate divergence issues by looking not at energies of configurations but energy-differences between two configurations. Indeed, in this vein one might replace the search for globally energy-minimizing configurations by a search for locally energy-minimizing configurations, having the property that changing the spins at any finite number of sites results in higher energy (as measured by an increase in the sum of the finitely-many interaction energies that involve the affected sites). Another way to regularize would be to use a summable multiplier as described in Regularizing divergent sums over lattices. Assuming that a minimum does exist, or a family of minima, one might seek an exact solution. If this is old hat for condensed matter theorists, where can I learn about it? (Condensed matter is not my field, so if I am misusing terminology or laboring under misconceptions I hope readers of this forum will set me straight.) REPLY [3 votes]: I think your question is interesting and reasonable. But I don't know the answer, so I'll say something else. Kosterlitz and Thouless won the physics Nobel prize this week for their work on this subject. According to them, the 'vortex' configuration in Brian Skinner's picture has infinite energy - at least for an ideal thin film of magnetic material that's infinite in spatial extent. So, they didn't try to find an energy-minimizing configuration of this sort: indeed they used the infinite energy to argue that we'd never see a single vortex by itself. Instead, they noticed that a 'vortex-antivortex' pair can have finite energy - and they showed that as you heat up a thin film starting from absolute zero, a 'phase transition' occurs when the concentration of vortex-antivortex pairs suddenly increases. I wrote a blog article about this today: John Baez, Kosterlitz-Thouless transition, Azimuth, 7 October 2016.<|endoftext|> TITLE: Can we define exterior derivatives using pushforwards and connections? QUESTION [7 upvotes]: Let $\alpha$ be a differential form on a smooth manifold $M$. For simplicity, let's suppose that it is a $1$-form. Then we can think of $\alpha$ as a smooth map from $M$ to $T^* M$, the cotangent bundle. The exterior derivative $d\alpha$ is a $2$-form on $M$ that somehow "differentiates" $\alpha$. On the other hand, a $2$-form is a special kind of (i.e. alternating) bundle map from $TM$ to $T^* M$. Thinking of $\alpha$ as a smooth map, we obtain a map $D \alpha$ from $TM$ to $TT^* M$. Given a linear Ehresmann connection on the vector bundle $T^* M$, there is a linear map $\phi$ from $TT^* M$ to $T^* M$, with the latter viewed as the sub-bundle of $TT^* M$ which projects to $0$ in $M$. So, $\phi \circ D \alpha$ is a bundle map from $TM$ to $T^* M$, also known as a section of $T^*M \otimes T^* M$. This has a canonical projection $P$ to its skew-symmetrization $T^* M \wedge T^* M$, so $P \circ \Phi \circ D \alpha$ gives a section of $T^* M \wedge T^* M$, also known as a $2$-form. Perhaps more transparently, we can write this in terms of the covariant derivative associated to the Ehresmann connection. This is an operator $\nabla$ which maps sections $\alpha$ of $T^*M$ to a "$T^*M$-valued $1$-form" $\nabla \alpha: X \mapsto \nabla_X \alpha$, where $\nabla_X \alpha$ is a $1$-form, such that the map is tensorial in $X$ and follows the Leibniz rule in $\alpha$. This is related to $\phi$ precisely by the formula $\phi \circ D \alpha(X) = \nabla_X \alpha$, so this is the same as the previous definition, up to the skew-symmetrization. This should replace the contravariant $2$-tensor $\nabla \alpha(X, Y) = (\nabla_X \alpha)(Y)$ with the $2$-form $\omega_\alpha(X, Y) = (\nabla_X \alpha)(Y) - (\nabla_Y \alpha)(X)$. By the Leibniz rule for $\nabla$, we have \begin{align*}\omega_{f\alpha}(X, Y) &= \nabla_X (f\alpha)(Y) - \nabla_Y (f\alpha)(X)\\ &= f \omega_\alpha(X, Y) + X(f)\alpha(Y) - Y(f) \alpha(X)\\ &= f\omega_\alpha(X, Y) + (df \wedge \alpha)(X, Y) \end{align*} So the map $\alpha \mapsto \omega_\alpha$ satisfies the usual derivation property of the exterior derivative and produces an honest $2$-form. However, it uses the entirely non-intrinsic Ehresmann connection $\nabla$. What gives? Does something like this work for higher $k$-forms? Replacing the bundle $T^* M$ with $\wedge^k T^* M$ everywhere in the arguments gives an analogous definition of an "exterior derivative" of an arbitrary $k$-form, but it is not so clear that this now satisfies $d(\omega_1 \wedge \omega_2) = d\omega_1 \wedge \omega_2 + (-1)^k \omega_1 \wedge d \omega_2$. EDIT I realized this was confusing - when I say an Ehresmann connection, I'm referring to one that is linear, so it is equivalent to the usual notion of covariant derivative for a vector bundle. However, I wanted the definition to include connections that aren't necessarily determined by an affine connection on $TM$. REPLY [6 votes]: An affine connection induces the exterior derivative by taking the ant symmetrization of the covariant derivative if and only if the torsion of the connection vanishes. This can be computed directly.<|endoftext|> TITLE: Arakelov divisor on $\operatorname{Spec } O_F$: places or embeddings? QUESTION [8 upvotes]: Let $F$ be a number field such that $[F:\mathbb{Q}]=n$ and with ring of integers $O_F$. Let's put $B=\operatorname{Spec } O_F$, then an Arakelov divisor is an element of: $$Div(X)\times \bigoplus_\sigma \mathbb R[\sigma]$$ namely it can be written as $$\bigg(\sum_{\text{$\mathfrak p$ prime $\neq 0$}}n_{\mathfrak p}[\mathfrak p],\sum_{\sigma}x_{\sigma}[\sigma]\bigg)\quad \text{with $x_\sigma\in\mathbb R$, $n_{\mathfrak p}\in\mathbb Z$}$$ Now: some references say that $\sigma$ runs among the archimedean places of $F$ other say that $\sigma$ runs among the complex embeddings of $F$. There is a substantial difference in the two cases, because in the former the dimension of the vector space of "divisors at infinity" is $r_1+r_2$, whereas in the latter it is $n=r_1+2r_2$ (as usual $r_1$ and $2r_2$ are the number of real and complex embeddings of $F$ ). Which is the most natural approach? In the first case the degree of an Arakelov divisor is $\sum_{\mathfrak p} n_{\mathfrak p}\log(\#(O_F/\mathfrak p))+\sum_\sigma x_\sigma$ whereas in the second case I think that we should put $$\sum_{\mathfrak p} n_{\mathfrak p}\log(\#(O_F/\mathfrak p))+\sum_{\text{$\sigma$ real }} x_\sigma+\frac{1}{2}\sum_{\text{$\sigma$ complex }} x_\sigma$$ Is it correct? Edit: books/papers using places: Van De Geer,Schoof- Effectivity of Arakelov divisors of a number field,Neukirch ANT. books/papers using embeddings: Moriwaki's book, De Jong thesis. Moreover I'd like to understand what happens when you consider Arakelov theory on arithmetic surfaces $X\to B$. Here I think that would be meaningful to consider all the embeddings. REPLY [5 votes]: You would usually want the principal Arakelov divisors, i.e. those of the form $(\sum_{\mathfrak{p}}{\rm ord}_{\mathfrak{p}}(a), \sum_\sigma -\log|\sigma(a)|)$ for $a\in F^\times$, to be cocompact in the group of degree $0$ divisors — the volume of the quotient should be the familiar product ${\rm Reg}(F)\cdot{\rm h}(F)$, where ${\rm h}$ denotes the class number. For that, you want $\sigma$ to only run over places, not over all embeddings. Incidentally, you also want to normalise your degree function carefully: the degree of a real place should be $1$, while the degree of a complex place should be $2$. REPLY [2 votes]: I think the natural definition is the one using archimedean places (or equivalently, all embeddings, with the pairs of complex conjugates considered as a single embedding), and I'm very curious of the context where you have seen the definition using complex embeddings only. Consider for example the basic result that the kernel of the mapping $\mathrm{div}:K^*\to \mathcal{Div}(\overline{\mathcal{O}})$ is $\mu(K)$. This follows directly from the exact sequence $$1\longrightarrow \mu(K) \longrightarrow \mathcal{O}^*\longrightarrow \Gamma\longrightarrow 0$$ and from the equality (up to a sign) of the composition $$K^*\longrightarrow\mathcal{Div}(\overline{\mathcal{O}}) \longrightarrow \prod_{\mathfrak{p}|\infty}\mathbb{R}$$ and the logarithmic Minkowski map $$\lambda: f \mapsto (...,\log |f|_\mathfrak{p},...)$$ This shows quite explicitly why the definition uses necessarily all the embeddings. For reference, this is also consistent with the use in Moriwaki's and Lang's textbooks on Arakelov theory.<|endoftext|> TITLE: Optimal sphere packings ==> Thinnest ball coverings? QUESTION [29 upvotes]: It was proved by Kershner long ago that the thinnest (least density) covering of the plane by congruent disks can be obtained by enlarging the radii of the optimal circle packing to just cover the gaps in that packing:           My question is: Q. Is it known that, in the dimensions for which the optimal sphere packing is known $(2, 3, 8, 24)$, is it also known that a thinnest ball covering can be achieved by enlarging the spheres to just cover the interstices in the packing? For example, now that the Kepler problem has been settled by Hales, does it follow that a thinnest ball packing of $\mathbb{R}^3$ can be obtained from the cannonball packing by enlarging the spheres to just cover the gaps?           Kershner, Richard. "The number of circles covering a set." American Journal of Mathematics 61.3 (1939): 665-671. REPLY [32 votes]: Nope, this is false in three dimensions (where the body-centered cubic beats the face-centered cubic for covering) and eight dimensions (where $E_8$ is not even locally optimal). The Leech lattice is locally optimal and may be the best covering in twenty-four dimensions, but that would be a special fact rather than an instance of a general pattern. Overall sphere covering is more complicated than sphere packing, and there seems to be no close relationship between the optimal solutions.<|endoftext|> TITLE: Core model with $\omega$ Woodin cardinals QUESTION [6 upvotes]: In [KwoM] it is proved that the the core model K exists under the assumption that there is no inner model having a Woodin cardinal and satisfying ZFC. Furthermore, they also mention that the result is the "best possible", in the sense that we cannot weaken this anti-large-cardinal hypothesis without strengthening the remainder of the hypothesis. From what I understand, using the core model induction (which I must admit I don't know the details of yet), K can be constructed below any finite number of Woodin cardinals. So my question is then whether or not K has been constructed below $\omega$ many Woodins and if so, what the extra hypothesis is? If this is indeed the case, I would greatly appreciate a reference to the given paper. [KwoM] Jensen and Steel: "K without the measurable" Edit: In the introduction to Schindler and Steel's "The Core Model Induction", they write that if we assume that there is no proper class model with $n+1$ Woodins and over every set there is a proper class model with $n$ Woodins, then $K$ exists (in the sense of the above [KwoM]). Can this be extended to: if there is no proper class model with $\omega$ Woodins and over every set there is a proper class model with $n$ Woodins, for every $n<\omega$, then $K$ exists? REPLY [3 votes]: Dan, the answer is "no." You might be in a situation where Kc is a mouse (i.e., its ctble. substructures are ω1+1 iterable), but Kc is not fully iterable. E.g. it could be that Kc has M2F,#, where F is the operator simultaneously closing under all Mn#, n<ω, but for some θ, V = LF(Vθ) (same F, V could even be an extender model): the Q-structures needed to fully iterate M2F,# would be given by the M1F,# operator which is not available thruout the universe. If Kc is not fully iterable, one can't use it to isolate K, and in such a situation usually there is no inner model which has enough of the relevant properties that it might be called "the core model." See e.g. the first two pages of https://ivv5hpp.uni-muenster.de/u/rds/K.pdf for a list of such properties. (There are exceptions to this, though, see e.g. https://ivv5hpp.uni-muenster.de/u/rds/CMPW.pdf and also https://ivv5hpp.uni-muenster.de/u/rds/RalfGrigor_version_Sept11_2017.pdf .)<|endoftext|> TITLE: Curvature of maximum of two riemannian metrics QUESTION [11 upvotes]: Consider $g_1$ and $g_2$ two Riemannian metrics on a differentiable manifold $M$ of dimension $n\ge 4$. Suppose locally $g_i=f_i\sum_{j=1}^ndx_j^2$, where $f_i:M\rightarrow \mathbb{R}$ are non negative functions. Suppose that both $g_1$ and $g_2$ have non positive sectional curvature. Define the metric $\hat g$ on $M$ as $\hat g:=max\{f_1,f_2\}\sum_{j=1}^ndx_j^2$. Short question: Will $(M,\hat g)$ have non positive curvature? More elaborate question: A priori $\hat g$ has just continuous coefficients: we can ask if its intrinsic induced metric $d_{\hat g}$ has non positive Alexandrov curvature (it means that every $p\in M$ has an open neighborhood $U$ such that for every $x,y,z\in U$ $d_{\hat g}(z,m)^2\le \frac 1 2(d_{\hat{g}}(z,x)^2+d_{\hat{g}}(z,y)^2)-\frac 1 4d_{\hat{g}}(x,y)^2$, where $m$ is the midpoint between $x$ and $y$, definition taken from http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/NPC0606.pdf) Clearly in the open sets $U_1:=\{p\in M|f_1(p)>f_2(p)\}$ or $U_2:=\{p\in M|f_2(p)>f_1(p)\}$ this is true, because the metrics $g_1$ and $g_2$ have negative sectional curvature. But I don't know what happens in points where $f_1=f_2$. Say $p\in M$ is such that $f_1(p)=f_2(p)$. Then every open neighborhood $U$ of $p$ will intersect both $U_1$ and $U_2$ and evaluating directly if the preceding inequality is true will be quite difficult. Is there a more straightforward way (than direct computation) to conclude that $d_{\hat{g}}$ has non positive Alexandrov curvature? REPLY [8 votes]: Dimension 2 : The answer to your question is yes in dimension 2. I am not sure about higher dimension. Set $f_i=e^{2u_i}$, so that $f=e^{2u}$ with $u=\max (u_1,u_2)$. The Gauss curvature of a metric $e^{2v}(dx^2+dy^2)$ is given by $-e^{-2v}\Delta v$, so that each of the $u_i$ is subharmonic, hence $u$ it self is subharmonic as the maximum of two subharmonic functions. There is a theory of metrics on surfaces of the form $e^{2u}(dx^2+dy^2)$ where $u$ is subharmonic (or a difference of subharmonic functions) which has been developed by Reshetnyak in the 50s. Basically, there is notion of curvature measure, which is $-\Delta_d u$, the opposite of the distributional laplacian of $u$. Since $u$ is subharmonic, this curvature measure will be nonpositive. From there you can show that $e^{2u}(dx^2+dy^2)$ has nonpositive curvature in the sense of Alexandrov. Reshetnyak's theory is exposed in "Two-Dimensional Manifolds of Bounded Curvature", Yu. G. Reshetnyak, in Encyclopaedia of Mathematical Sciences, Volume 70 1993, Geometry IV. Higher Dimension : Unfortunately the formula for the conformal change of curvature in higher dimension is not so nice and to my knowledge bothered to make this approach work in higher dimensions. The $(4,0)$ curvature tensor of $e^{2u}g$ where $g$ is flat and $u$ is smooth is given by : $$R_u-e^{2u}g\wedge(\nabla^2u-du\otimes du+\tfrac{1}{2}|du|^2g).$$ Nonnegativity of the sectional curvature is then equivalent to the sum of the two lowest eigenvalues of $S_u=\nabla^2u-du\otimes du+\tfrac{1}{2}|du|^2g$ being nonnegative (I'll say that $S_u$ is $2$-nonnegative). If $u=\max(u_1,u_2)$ where $u_1$ and $u_2$ are smooth it should be possible to make sense of $S_u$ at least as a symmetric tensor with measure coefficients. A starting point would be to see if the $2$-nonnegativity of $S_{u}$ (in some weak sense) can be inferred from the $2$-nonnegativity of $S_{u_1}$ and $S_{u_2}$. If $S_u$ was just $\nabla^2u$, it would boil down to a statement whose proof would be roughly the same as "sup of convex is convex" or "sup of subharmonic is subharmonic" (this is what was going on in 2d). However the occurence of the gradient terms in a nonlinear fashion makes this not as straight forward. Then comes the real question in some sense : does the negativity of this $R_u$ in a weak sense have anything to do with the metric geometry of $(M,e^{2u}g)$ ? Maybe smoothing could help there. PS: I've been thinking about this kind of problems for quite some time, feel free to directly email me if you have further questions.<|endoftext|> TITLE: On a characterization of the recursively inaccessible ordinals QUESTION [5 upvotes]: For a given set of numbers $A$, let $O^A$ be the hyperjump of $A$. It is possible to iterate inductively the hyperjump of a set, through the computable ordinals, in a way that the $\alpha$-th hyperjump is computably stronger than any $\beta$-th hyperjump for $\beta < \alpha$. It is even possible to keep iterating it through the ordinals which becomes computable in a previous iterations of the hyperjump. Formally, let us do the following inductive definition of an order $<$ on the natural numbers, together with sets $J_a^A$ for any $a$ in the field of the order $<$ (these elements $a$ are meant to be notations for ordinals) : Let $J_1^A$ be the set $A$ and $J_2^A$ be the set $O^A$. Define $1 < 2$. If $a$ is in the field of $<$, let $J_{2^a}^A = O^{J_a^A}$. Define $a < 2^a$ and $b < 2^a$ for any $b < a$. If $a$ is in the field of $<$ and if $e$ is the code of a computable functional such that $\varphi_e(0, J^A_a) = a$ and $\varphi_e(n, J^A_a) < \varphi_{e}(n+1, J^A_a)$, then $J^A_{3^e5^a}$ is the disjoint union of the sets $J^A_{m_i}$ for $m_i = \varphi_e(i, J^A_a)$. We define $b < 3^e5^a$ for any $b < \varphi_e(i, J^A_a)$ for some $i$. There is a smallest ordinal $\lambda^A$ which does not have a corresponding code in the field of $<$. It seems clear (even though I haven't gone through the details carefully) that $\lambda^A$ should be recursively inaccessible. Sacks showed that for any countable admissible ordinal $\alpha$, there is a set $A$ such that $\alpha$ is the smallest ordinal non-computable ordinal in $A$. My question is : Is it known/true that any countable recursively inaccessible ordinal is of the form $\lambda^A$ for some set $A$ ? (with $\lambda^A$ defined as above). REPLY [2 votes]: Well assuming that $\lambda^A$ is always the first recursively admissible which is bigger than $\omega_1^A$, which I think should be true, I think my question is after all not so interesting: Either there is a largest recursively inaccessible smaller than $\omega_1^A$, in which case $\lambda^A$ is a successor in the recursively inaccessible ordinals, or $\omega_1^A$ is a limit of recursively inaccessible. In this case it is itself also recursively inaccessible (as an admissible being a limit of admissible). Thus $\lambda^A$ is also a successor in the recursively inaccessible ordinals. It would follow that $\lambda^A$ is never meta-recursively inaccessible : a recursively inaccessible being a limit of recursively inaccessible.<|endoftext|> TITLE: Proper discontinuity and existence of a fundamental domain QUESTION [12 upvotes]: I am currently teaching a topics course where I talk about some discrete groups acting properly. A student asked a very basic question that stumped me: what is the precise relationship between proper discontinuity and existence of a fundamental domain? Namely, let $G$ be a discrete group acting on a topological space $X$. Recall that the action is properly discontinuous if for every compact subset $K \subset X$, the set $$\{g \in G | gK \cap K \neq \emptyset \}$$ is finite. A fundamental domain is a subset $U \subset X$ such that its images by all elements of $G$ are disjoint and cover $X$ (possibly up to some "thin" set, e.g. the boundary of $U$). Is there any theorem saying that the action is proper if and only if there exists a "sufficiently nice" fundamental domain on $X$, possibly with some additional conditions? This seems to be taken for granted by everyone, and I often used it as an explanation of what a "properly discontinuous action" means intuitively. But to my extreme surprise, a search for precise references brought up almost nothing. I only found one statement of this sort: Theorem 3.6 in T. Drumm and V. Charette, "Complete Lorentzian 3-manifolds" (in Contemporary Mathematics, vol. 639 "Geometry, Groups and Dynamics", pp. 43--72). This statement seems to be actually wrong, as the well-known counterexample of the cyclic group generated by $\left( \begin{smallmatrix} 2 & 0 \\ 0 & 1/2 \end{smallmatrix} \right)$ acting on a closed quadrant (or on $\mathbb{R}^2 \setminus \{0\}$) seems to contradict it (unless I missed some additional assumptions they made somewhere). Possible restrictions needed to make it work: Clearly we need to impose some regularity condition on $U$, to prevent people from just picking a point in every orbit at random and pretending that the resulting mess is a fundamental domain. One that is often used is being regular in the topological sense, i.e. open with $\overset{\circ}{\overline{U}} = U$, or (which is closely related) closed with $\overline{\overset{\circ}{U}} = U$. We need a local finiteness condition (e.g. every compact set meets only finitely many tiles), to exclude the counterexample above. Points with nontrivial stabilizers are annoying, as pointed out in the comments. We must either restrict our attention to free actions, or modify the definitions somehow. (It would still be nice to be able to deal with orbifolds.) Maybe things are simpler for groups acting on smooth manifolds by diffeomorphisms? Or if we have some sort of measure? Or something else? I tried to think a little about this on my own. The "if" part seems more or less doable by hand. For the "only if" part however, I don't really know how to proceed. There is indeed a well-known theorem that says that if you have a proper action, then you have a nice (Hausdorff) quotient. But then you would have to somehow prove that by "cutting open" this quotient in a suitable way, you can flatten it and lift it to a fundamental domain - which seems really hard to do in general. Case in point: there is one example of proper actions from my field (namely the Margulis spacetimes - see for example "Properly Discontinuous Groups of Affine Transformations: A Survey" by H. Abels) which was found in 1983, but the description of the fundamental domain (the "crooked planes" construction) was not done until 1992. REPLY [12 votes]: I will assume that you are interested in group actions on connected manifolds: In the case of more general spaces it is not even completely clear what a fundamental domain means since an element of finite order can fix a nonempty open subset. Definition. Let $M$ be a manifold (in any category you like, DIFF, PL or TOP). Let $G\times M\to M$ be a proper action of a discrete group. An open subset $D\subset M$ is a fundamental domain for the action if: The interior of the closure of $D$ equals $D$. $G$-orbit of $cl(D)$ equals $M$. $g(D)\cap D=\emptyset$ for all $g\in G -\{1\}$. The collection of subsets $\{g cl(D): g\in G\}$ is locally finite. I will prove that every smooth proper effective action on a smooth connected manifold admits a fundamental domain. First, you construct a $G$-invariant Riemannian metric $g$ on $M$; let $d$ be the corresponding distance function on $M$. This is done using a partition of unity on the quotient orbifold $M/G$. Now, since $M$ is connected, $G$ is countable and the action is effective, there is a point $x\in M$ not fixed by any $g\in G$. Now, define $D$ to be the Dirichlet fundamental domain: $$ D= \{y\in M: d(y,x)< d(y, gx) \forall g\in G - \{1\} \}. $$ I will leave you to verify that $D$ has the right properties (this is very standard). Suppose now that $M$ is a connected topological n-manifold and $G\times M\to M$ be an effective proper action of a discrete group. In order to show that there is a fundamental domain you can use a theorem of Morton Brown or, rather its generaization by Berlanga. Consider $X$ which is the quotient $M/G$ minus the projection of the union of fixed point sets of nontrivial elements of $G$. Then $X$ is open and dense in $M$. By Brown's theorem in the case when $X$ is compact and Berlanga's theorem in the general case, $X$ contains a dense open subset $U$ homeomorphic to $R^n$. Now, take $D$ to be a component of the preimage of $U$ in $M$. This will satisfy conditions 1--3 but I am not at all sure about 4 even in the case when $M/G$ is compact. Proving 4 would require some serious work. You probably do not want to unload any of these on your students. in this case, it is best to work in the simplicial category with triangulated manifolds. You can find a proof in this case in the book by Seifert and Threlfall "A textbook on topology". The construction is quite simple and easy to explain even to undergraduate students taking an introductory topology class. Namely, Let $T$ denote the quotient-triangulation of $M/G$. Let $\alpha$ denote the 1-skeleton of the dual triangulation (vertices of $\alpha$ are facets and the edges of $\alpha$ are codimension 1 faces, called panels). Let $\tau\subset \alpha$ be a maximal subtree. Now, take the union $U$ of all open facets of $T$ with all the open panels. This union is simply-connected. Now, take $D$ to be a connected component of the preimage of $U$ in $M$. This will be a fundamental domain: Part 4 works because of the combinatorial nature of the construction since the triangulation of $M$ is locally finite, of course.<|endoftext|> TITLE: Is $x^2+x+1$ ever a perfect power? QUESTION [14 upvotes]: Using completing the square and factoring method I could show that the Diophantine equation $x^2+x+1=y^n$, where $x,y$ are odd positive and $n$ is even positive integers, does not have solution, but I could not show that for odd positive $x,y$ and odd $n>1$ the equation does (does not) have solution. I already asked the above question of some expert and I received good information about such equation, but how it will be if we assume $x,y$ were "odd prime" numbers? Thank you for your contribution. P.S. I already put it here, but did not get some useful suggestion. REPLY [39 votes]: Your equation can be rewritten as $$\frac{x^{3}-1}{x-1} = y^{N}.$$ As Gerhard Paseman commented above, the Diophantine equation $$ \frac{x^{n} − 1}{x-1} = y^{q} \quad x > 1, \quad y>1, \quad n>2 \quad q \geq 2 \quad \mbox{ (*) }$$ was the subject matter of a couple of papers of T. Nagell from the 1920's. Some twenty-odd years later, W. Ljunggren clarified some points in Nagell’s arguments and completed the proof of the following result. Theorem. Apart from the solutions $$\frac{3^{5}-1}{3-1}=11^{2}, \quad \frac{7^{4}-1}{7-1}=20^{2}, \quad \frac{18^{3}-1}{18-1} = 7^{3},$$ the equation in $(*)$ has no other solution $(x, y, n, q)$ if either one of the following conditions is satisfied: $q = 2$, $3$ divides $n$, $4$ divides $n$, $q = 3$ and $n$ is not congruent with $5$ modulo $6$. Clearly enough, this theorem implies that there are only two solutions to the equation you are considering: $(x=1, y=3, N=1)$ and $(x=18, y=7, N=3)$.<|endoftext|> TITLE: Numerical coincidence? Why is sum(x^(k^2)) = sum(x^((k+1/2)^2)) for x = 0.8? QUESTION [8 upvotes]: A well-known formula for the logarithm is given by $$\log x = -\frac{\pi}{AGM(a^2,b^2)}, \qquad x < 1$$ where AGM is the arithmetic-geometric mean, and $a$ and $b$ are given by $$a = \sum_{k\in\mathbb{Z}}x^{k^2}, \qquad b = \sum_{k\in\mathbb{Z}}x^{(k+1/2)^2};$$ Or, equivalently, $$a = 1 + 2x + 2x^4 + 2x^9 + 2x^{16} + 2x^{25} + \dots$$ and $$b = 2x^{1/4} + 2x^{9/4} + 2x^{25/4} + 2x^{49/4} + \dots$$ Why does it happen that $a = b$ occurs when $x = 0.8$? REPLY [18 votes]: $$\sum_{k \in \mathbb Z} x^{k^2} = \theta_3(0,x)$$ while $$\sum_{k \in \mathbb Z} x^{(k+1/2)^2} = \theta_2(0,x)$$ where $\theta_2$ and $\theta_3$ are Jacobi theta functions. The difference $$\theta_2(0,0.8) - \theta_3(0,0.8) \approx 9.280378636257491074676461535977 \times 10^{-19}$$ according to Maple. EDIT: The difference $$\theta_3(0,x) - \theta_2(0,x) = \sum_{j \in \mathbb Z} (-1)^j x^{(j/2)^2} =\theta_3(\pi/2, x^{1/4}) $$ The Poisson summation formula gives us the identity $$ \theta_3(\pi/2, e^{-t^2}) = \frac{\sqrt{\pi}}{t} \theta_2(0, e^{-\pi^2/t^2})$$ and for $t \to 0+$, this goes to $0$ very rapidly: $$\theta_3(\pi/2, e^{-t^2}) \sim \frac{2 \sqrt{\pi}}{t} e^{-\pi^2/(4 t^2)}$$ i.e. $$ \theta_3(0,x) - \theta_2(0,x) \sim \frac{4 \sqrt{\pi}}{\sqrt{\ln(1/x)}} \exp\left(-\frac{\pi^2}{\ln(1/x)}\right) $$ For $x = 0.8$, the right side above is extremely close to the value I gave for $\theta_2(0,0.8) - \theta_3(0,0.8)$ (all digits shown match). REPLY [5 votes]: The two values $a(0.8)$ and $b(0.8)$ appear "deceivingly" equal, but they actually are not! Other "near-miss" values include $$0<\theta_3(0,0.9)-\theta_2(0,0.9)<0.5\times 10^{-39}.$$ Let $f(x)=\theta_3(0,x)-\theta_2(0,x)$. The graph of $f(x)$, for values $00$, still) and also a local maximum for some $0.9 TITLE: Reverse Minkowski (and related) Determinant Inequalities QUESTION [10 upvotes]: For positive semidefinite matrices $A,B,C \in \mathbb{R}^{n\times n}$, the following inequalities are well known: $$(\det(A+B))^{1/n} \geq (\det A)^{1/n} + (\det B)^{1/n} $$ and $$\det(A+B+C) + \det(C) \geq \det(A+C) + \det(B+C).$$ The first is, of course, Minkowski's determinant inequality. I'm not sure whether the second has a name, but it can be found, e.g., here and here. I would like to know whether anyone knows of any reverse forms of these inequalities. To be more specific, I note that the Minkowski determinant inequality can be viewed as a consequence of the Entropy Power Inequality applied to Gaussian random vectors with covariance matrices $A$ and $B$. Applying similar reasoning to the entropy power inequality appearing in Theorem 4 of this paper, we may conclude that the following reverse counterparts to the above inequalities hold: $$(\det(A+B))^{1/n} \leq (\det A)^{1/n}\left(\frac{\frac{1}{n}\operatorname{Tr}(B^{-1})}{(\det B^{-1})^{1/n}} \right) + (\det B)^{1/n}\left(\frac{\frac{1}{n}\operatorname{Tr}(A^{-1})}{(\det A^{-1})^{1/n}} \right),~~~~~(\star) $$ and $$(\det(A+B+C) \det(C) )^{1/n} + (\det(A) \det(B) )^{1/n} \leq (\det(A+C) \det(B+C) )^{1/n} .~~~(\star\star)$$ The first may be obtained from the second by taking $C = \varepsilon I$ and letting $\varepsilon \downarrow 0$. I view $(\star)$ as a reverse counterpart to Minkowski's determinant inequality, since the ratio $$\frac{\frac{1}{n}\operatorname{Tr}(B^{-1})}{(\det B^{-1})^{1/n}}\geq 1$$ by the AM-GM inequality applied to the eigenvalues of $B^{-1}$ (and similarly for the ratio involving $A^{-1}$), with near equality if the matrix $B$ has eigenvalues all roughly the same. Has anyone seen these inequalities before, or does anyone have a direct proof of $(\star\star)$? REPLY [7 votes]: Inequality ($\star\star$) essentially follows from the original Minkowski plus an implication of Lidkskii's inequality (Fiedler's inequality, noted below). $\newcommand{\da}{\downarrow} \newcommand{\ua}{\uparrow}$ Assume $C$ is strictly positive definite (otherwise, $\det C=0$ rendering ($\star\star$) trivial), and let $C^{1/2}$ denote the positive square root of $C$. Renaming the matrices $A \leftarrow C^{-1/2}A C^{-1/2}$ and $B \leftarrow C^{-1/2}B C^{-1/2}$, the original inequality reduces to showing that \begin{equation*} \det(A+B+I)^{1/n} + [\det A\det B]^{1/n} \le [\det(I+A)\det(I+B)]^{1/n}. \end{equation*} Let $a^\da$ denote the eigenvalues of $A$ in decreasing order; similarly, define $b^\ua$ for $B$. Then, we have \begin{equation*} \det(I+A)\det(I+B) = \prod_i(1+a_i^\da)\prod_i(1+b_i^\ua) = \prod_i(1+a_i^\da + b_i^\ua + a_i^\da b_i^\ua), \end{equation*} whence by the Minkowski inequality we get \begin{equation*} [\det(I+A)\det(I+B)]^{1/n} \ge \prod_i(1+a_i^\da+b_i^\ua)^{1/n} + \prod_i(a_i^\da b_i^\ua)^{1/n}. \end{equation*} Now from Fiedler's (1971) determinantal inequality we know that \begin{equation*} \prod_i(1+a_i^\da+b_i^\ua) \ge \det(I+A+B), \end{equation*} while clearly $\prod_i(a_i^\da b_i^\ua) = \det(A)\det(B)$. Thus, the inequality ($\star\star$) follows.<|endoftext|> TITLE: Extremal functions for the 'packing density in dimension one' QUESTION [5 upvotes]: The $n = 1$ case of Theorem 3.1 of Cohn and Elkies's paper New upper bounds on sphere packings I amounts to the inequality $f(0) \geq 1$ for all ('admissible') functions $f$ on $\mathbb{R}$ satisfying $\widehat{f}(0) = 1$; $f \leq 0$ outside of $[-1,1]$; and $\widehat{f} \geq 0$ everywhere. I am interested in the functions which attain the equality: $f(0) = \widehat{f}(0) = 1$. Can anything (everything?) be said about those functions, or even more particularly about the ones among them having the second condition strengthened to $\mathrm{supp}(f) \subset [-1,1]$? Cohn and Elkies give $f_1(t) = \Big( \frac{\sin{\pi t}}{\pi t} \Big)^2$ and $f_2(t) = \frac{1}{1-t^2}\Big( \frac{\sin{\pi t}}{\pi t} \Big)^2$ as examples of extremal functions. Are there additional examples besides the convex hull of $f_1, \widehat{f_1}$ and $f_2$? REPLY [3 votes]: You didn't make the precise assumptions on $f$ explicit, but even in a fairly general situation, your $\widehat{f}_1$ is the only example that is supported by $[-1,1]$. Given an $f$ with the stated properties, we can let $\widehat{g}=\widehat{f}^{1/2}$, so $f=g*g$, and if $\textrm{supp}\, f\subseteq [-1,1]$, then Titchmarsh's convolution theorem shows that $g$ is supported by $[-1/2,1/2]$. Since we're dealing with real valued functions $f,\widehat{f}$ here, they are even, and thus $$ 1=f(0) = \int_{-1/2}^{1/2} g(x)g(-x)\, dx = \int_{-1/2}^{1/2} g(x)^2\, dx \ge \left(\int_{-1/2}^{1/2} g(x)\, dx\right)^2 = \widehat{g}(0)^2=1 . $$ Equality in the Cauchy-Schwarz inequality means that the functions $1,g$ are linearly dependent, so $g=1$ (or $=-1$), and this gives us the triangular function $f(x)=1-|x|$ as the only possible example supported by $[-1,1]$.<|endoftext|> TITLE: covering convex sets by round balls QUESTION [9 upvotes]: Let $X=\{x_1,...,x_k\}\subset E^n$ be a finite subset in the Euclidean $n$-space, $r>0$ and $B(x_i,r)$ are open balls of radius $r$ centered at the points $x_i\in X, i=1,...,k$. Suppose that $$ \bigcap_{i=1}^k B(x_i,r)\ne \emptyset. $$ Is it true that the convex hull of $X$ is contained in $$ \bigcup_{i=1}^k B(x_i,r) ? $$ I needed a version of this in some Riemannian geometry argument and while I found a way around, the proof would have been a bit cleaner if this statement were true. REPLY [11 votes]: Yes. Any point $y$ in the convex hull of $x$'s is a barycenter of some non-negative masses $m_i$ in $x_i$, $\sum m_i=1$, $y=\sum m_i x_i$. Point $y$ minimizes the moment of inertia $I(p)=\sum m_i |p-x_i|^2$, $p\in E^n$, just because $I(p)=I(y)+|p-y|^2$. In particular, it can not happen that all distances $|y-x_i|$ are at least $r$, but for some other point $p$ all distances $|y-x_i|$ are less than $r$.<|endoftext|> TITLE: Pseudo-Anosov maps with same dilatation. QUESTION [8 upvotes]: Let $S$ be a hyperbolic surface. Suppose $\mathcal{T}$ denotes the Teichmuller space of $S$ and $Mod(S)$ denotes the mapping class group of $S$. Given any pseudo-Anosov element $f\in Mod(S)$, suppose $\lambda_f$ denote the stretch factor (or dilatation) of $f$. Q) Does there exist $f,g\in Mod(S)$ such that $f\neq g$ and $\lambda_f=\lambda_g.$ REPLY [3 votes]: Here is a general source of examples. Suppose $f$ is a pseudo-Anosov map whose centralizer is not cyclic. Say $\rho$ is finite order and is in the centralizer of $f$. Note that $f$ and $\rho f$ have the same dilatation, because they have a common power. However, $\rho$ can change the way the singularities (and the separatrices) are permuted, and this permutation data is also a conjugacy invariant. To be concrete, suppose that $f$ is any pseudo-Anosov map on the twice-punctured torus and that $\tau$ is the hyperelliptic element. Then $f$ and $\tau f$ are not conjugate. With a bit of care you can play similar games in hyper-elliptic strata in any genus. There are more subtle obstructions as well -- see the paper Polynomial invariants of pseudo-Anosov maps by Birman, Brinkmann, and Kawamuro. Near the end of the paper they say: "It seems to be an open question to describe all the ways to construct all pA maps having a fixed dilatation." [Edit] Here is another construction. Suppose that $f$ is an Anosov map on the torus. Take a power to ensure that $f$ has at least two fixed points $x$ and $y$. Now form $S_g$ by taking a $g$--fold branched cover over $x$ and $y$. So $S_g$ is a surface of genus $g$. We can take a further power of $f$ to ensure that it lifts to a map on $S_g$. Note that $f$, acting on $S_g$, has exactly two singularities. Ok. Now form the surface $T$ by taking a unbranched double cover of $S_2$. So $T$ has genus three and is homeomorphic to $S_3$. There is again a power of $f$ that lifts to $T$, and this lift is a pA map with four singularities. Taking the correct powers of $f$ equips $S_3$ and $T$ with pA maps of the same dilatation, but with different numbers of singularities.<|endoftext|> TITLE: Vinogradov's method for sums of more than three primes QUESTION [6 upvotes]: In Hardy-Littlewood's 1923 paper "Some problems of 'Partitio Numerorum' III" it is proven, assuming a weak version of GRH (namely that there is $\varepsilon>0$ s.t. all zeroes of $L(s,\chi)$ have $\Re(s)<3/4-\varepsilon$), that for all $k\geqslant 3$, when $n\to\infty$ through the integers with same parity than $k$: $$ r_{k}(n) \sim \frac{2C_k}{(k-1)!}\frac{n^{k-1}}{\log(n)^k}\prod_{\substack{p\mid n \\ p\geqslant 3}} \left(\frac{(p-1)^k+(-1)^k(p-1)}{(p-1)^k -(-1)^k} \right), $$ where $r_{k}(n):= \{(p_1,\ldots,p_k)\in\mathbb{P}^k:\sum p_i = n \}$ and $C_k$ is a constant given by: $$ C_k := \prod_{p\geqslant 3} \left(1- \frac{(-1)^k}{(p-1)^k}\right). $$ Roughly 15 years later Vinogradov introduced his influencial technique that allowed him to prove this estimate unconditionally for $k=3$. One thing that bothers me is: what about the other $k$? The wikipedia article on Goldbach's conjecture has (as of today [08/Oct/2016]) the following passage: This formula has been rigorously proven to be asymptotically valid for $k \geqslant 3$ from the work of Vinogradov, but is still only a conjecture when $k=2$.$^{\text{[citation needed]}}$ I gave a quick look at K. F. Roth & Anne Davenport's translation of Vinogradov's The Method of Trigonometrical Sums in the Theory of Numbers, and in Chapter X. "Goldbach's Problem" it has the following passage at the start: In the present chapter I give a solution of Goldbach's problem concerning the representability of every sufficiently large odd number $N$ as the sum of three primes, and I establish an asymptotic formula for the number of representations. The method used here enables one also to solve more general additive problems involving primes, for example the question of representability of large numbers $N$ in the form $$ N = p_1^n + \ldots + p_s^n $$ (Waring's problem for primes). But I do not consider these more general questions here. Every other source I consulted only talks about the case $k=3$. Does Vinogradov's method for $k=3$ implies the other cases as a corollary? Where can I find more (historical) information about the other $k$ in this Hardy & Littlewood's estimate? Thanks in advance! REPLY [3 votes]: Your question is concerned with the so-called Waring-Goldbach problem. A classic in this topic is Hua Lo Keng's book, Additive theory of prime numbers (Translations of Mathematical Monographs, 13, American Mathematical Society, Providence, R.I. 1965), which focuses on how large the number of summands $s$ should be in terms of $n$ for an asymptotic formula to hold. It is a characteristic feature of the method that if it works for some value of $s$, then it also works for all larger values of $s$. Hua himself derived an asymptotic formula for $s>cn^2\log(2n)$, with some absolute constant $c>0$, and a more concrete lower bound is available for individual $n$'s. The method has been refined in may ways since, but this book is certainly a good starting point. For later developments, and a good overview in general, I recommend warmly the survey by Kumchev-Tolev: see especially Theorem 3 and the subsequent comments there.<|endoftext|> TITLE: How much should the average mathematician know about foundations? QUESTION [28 upvotes]: How much should an average mathematician not working in an area like logic, set theory, or foundations know about the foundations of mathematics? The thread Why should we believe in the axiom of regularity? suggests that many might not know about the cumulative hierarchy, which is the intended ontology that ZFC describes. (Q1) But is this something everyone should know or does the naive approach of set theory suffice, even for some mathematical research (not in the areas listed above)? EDIT: Let me add another aspect. As Andreas Blass has explained in his answer to the MO-question linked above, when set theorists talk about "sets", they mean an entity generated in the following transfinite process: Begin with some non-set entities called atoms ("some" could be "none" if you want a world consisting exclusively of sets), then form all sets of these, then all sets whose elements are atoms or sets of atoms, etc. This "etc." means to build more and more levels of sets, where a set at any level has elements only from earlier levels (and the atoms constitute the lowest level). This iterative construction can be continued transfinitely, through arbitrarily long well-ordered sequences of levels. Now, for foundational purposes it may be prudent to take the "none atoms" (pure) version of set theory. But in mathematical practice, one is working with many different types of mathematical objects, and many of them are not sets (even though one might often encode them as such). But even if one adds some atoms/non-sets – for example the real numbers –, one does not get a universe which is satisfactory for the practice of mathematics. This is because notions like "functions" or "ordered tuples" are from a conceptual perspective not sets; but we can't take them as our atoms for the cumulative hierarchy – the set of all "functions" or "ordered pairs" ... leads to paradoxes (Russell). Now I wonder: (Q2) What should mathematicians not working in an area like logic, set theory, or foundations understand by "set"? Note that there is also the idea of structural set theory (see the nlab), and systems such as ETCS or SEAR (try to) solve these "encoding problems" and remove the problem with "junk theorems" in material set theories. One can argue that these structural set theories match (more than material set theories do) with the mathematical practice. But my personal problem with them is that they have no clear ontology. So this approach doesn't answer Q2, I think. Or, (Q3) do mathematicians not working in foundational subjects not need to have a particular picture in mind of how their set-theoretic universe should look like? This would be very unsatisfactory for me since this would imply that one had to use axioms that work – but one doesn't understand why they work (or why they are consistent). The only argument I can think of that they are consistent would be "one hasn't found a contradiction yet". REPLY [4 votes]: I agree with the previous answer: it is difficult to state what an average mathematician "should" know. (I dare to conjecture that there is no such thing!) Each university has its own opinion on this: look at the graduate programs, and notice which courses are mandatory. In some countries this is decided by the state. I can share my own experience of an "average mathematician". As student I has a 1-semester course of logic (it was required), the text was E. Mendelson, Introduction to mathematical logic, but we did not cover the whole book. (In my country the curriculum was established by the state and a course of logic was required). At the same time I read Lyndon, Notes on logic, and I liked this small and very clear book. After my first year, I decided that I will do something else, not logic, and never read anything else on the subject.<|endoftext|> TITLE: Invariants of higher genus curves QUESTION [8 upvotes]: Are there analogues of the $j$-invariant for higher genus curves? Of course for genus $g \ge 2$ there will have to be at least $3g-3$ invariants. Are there "canonical choices" for these invariants, viewed as functions on the coarse moduli schemes? (ideally with relatively nice arithmetic properties) I would appreciate any references on this. Googling seemed to only yield results for genus 2 curves. REPLY [14 votes]: For $g > 2$ you're not going to be able to write such invariants as global functions on the moduli space. In general for an algebraic variety $X$, there is a canonical morphism $a \colon X \to \mathrm{Spec}(H^0(X,\mathcal O_X))$, the affinization of $X$. Points $x$ and $y$ of $X$ are separated by some global regular function if only if $a(x) \neq a(y)$. So your question can only have a positive answer if the affinization morphism is injective (so $M_g$ is affine or quasi-affine). Now $M_2$ is affine, but $M_g$ for $g >2$ contains a complete curve. Since a regular function restricted to a complete subvariety is constant, this means that $a$ has positive dimensional fibers. In fact one can prove using the Satake compactification that through any two points of $M_g$ passes a complete curve, if $g>2$. So the affinization is in fact a point. This is also why in the $g=3$ case of David Lehavi's answer there are only such choices of invariants on the hyperelliptic locus and on the non-hyperelliptic locus separately. Both loci are affine, but their union is not. It also seems relevant to mention that $M_g$ is of general type for $g \gg 0$, so that there will be no rational parametrization of the set of genus $g$ curves. And it also seems relevant to mention Fenchel--Nielsen coordinates: these will give you local real-analytic coordinates around any point of the moduli stack in any genus. REPLY [5 votes]: AFAIK, these are know only up to genus 3. Genus 2: Igusa (classical). Hyperelliptic genus 3: Shioda (classical). Non hyperelliptic genus 3: a decade ago by Dixmier & Ohno - see https://www.win.tue.nl/~aeb/math/ternary_quartic.html<|endoftext|> TITLE: Iterations of $2^{n-1}+5$: the strong law of small numbers, or something bigger? QUESTION [65 upvotes]: I've discovered what I believe is a quite remarkable sequence (A318970), defined by $$n_1 = 3,\qquad n_{k+1} = 2^{n_k-1}+5\quad(k\geq 1).$$ Here are the first four terms with their prime factorizations: $$ \begin{split} n_1 &= 3,\\ n_2 &= 2^2 + 5 = 9 = 3^2,\\ n_3 &= 2^8 + 5 = 261 = 3^2\cdot 29, \\ n_4 &= 2^{260} + 5 = 3^2\cdot 29\cdot p_{76}, \end{split} $$ where $p_{76}$ is a 76-digit prime. The remarkable property of this sequence is that $n_k\mid n_{k+1}$ for $k=1,2,3,4$. And I'm truly puzzled if this divisibility holds for all positive $k$ (in other words, if this sequence is a subsequence of A245594). Here are some properties that I can prove: $(n_k-n_{k-1})\mid (n_{k+1}-n_k)$ for all $k\geq 2$; If $n_k\mid n_{k+1}$, then $n_k\mid n_{k+t}$ for all integer $t\geq 0$. If $n_k\mid n_{k+1}$ and $\frac{n_{k+1}}{n_k}$ is a prime, then $n_{k+1}\mid n_{k+2}$. The last property seems to explain why we have the observed divisibility: $\frac{n_2}{n_1} = 3$, $\frac{n_3}{n_2}=29$, and $\frac{n_4}{n_3} = p_{76}$ are all primes. However, it is still possible that something bigger is going on, and I've not yet convinced myself in favor of either of the two possible outcomes: Outcome #1. It's an instance of the strong law of small numbers, and we just got lucky to get the divisibility for $k\leq 4$. To disprove it for larger $k$, we need simply to find a prime factor of $n_5$ that does not divide $n_6$ (more generally, we need a prime factor of $n_k$ that does not divide $n_{k+1}$). The trouble here is that $n_5$ is really huge, and we can only hope to discover its small factors. Outcome #2. The divisibility holds for all $k$ because this sequence is very special and satisfies some additional (yet unknown) property. Here is an analogy with numbers $m$ such that $m\mid (2^m+2)$ (see A006517). In general, iterations of $2^x+2$ do not preserve this property (e.g., $946$ belongs to A006517, but $2^{946}+2$ does not). However, if we additionaly require that $(m-1)\mid (2^m + 1)$, then $2^x+2$ starts to "magically" preserve both properties, and thus delivers an infinite subsequence A219037 of A006517. Does anybody have a clue what's going on here? UPDATE #1 (Oct 11 2016). Seeing here and there erroneous attempts to compute $n_k$ modulo primes for large $k$, I'd like to make some remarks. To compute $n_k\bmod m$, one needs to represent $m=2^s\cdot t$ with $t$ being odd, compute $n_k\bmod 2^s$ and $n_k\bmod t$, and combine these residues using CRT. For the former, we have $n_k\bmod 2^s = 5$ provided that $n_{k-1}>s$ (for $k\geq 4$, this holds for all $m<2^{260}$), while for $k\leq 4$ we can take as an answer $n_k$ itself. Computing $n_k\bmod t$ reduces to computing $n_{k-1}$ modulo Euler totient $\varphi(t)$: $$n_k \equiv 2^{(n_{k-1}-1)\bmod \varphi(t)} + 5 \pmod{t}.$$ These observations allow one to compute $n_k\bmod m$ recursively. In the described recursive process, we can start rolling back as soon as we reach modulus $1$, modulo which the answer is always $0$. Reaching 1 happens within A227944$(m) \leq \log_2(m)$ iterations. An implication of this observation is that the sequence $n_k\bmod m$ stabilizes within first $\log_2(m)$ terms. Here is an important consequence: if a prime $p$ does not divide any of the first $\log_2(p)$ terms, then it never divides any terms. This partially explains why it is so hard to find prime factors of the terms of this sequence. UPDATE #2 (Nov 04 2016). Let me state some statements that are equivalent to $n_k\mid n_{k+1}$: $n_k\mid n_{k+1}$ $2^{n_k-1}\equiv -5\pmod{n_k}$ or $2^{n_k}\equiv -10\pmod{n_k}$ (both congruences hold modulo $n_{k+1}$ by definition) $2^{n_k}\equiv 2^{n_{k-1}}\pmod{n_k}$ or $2^{n_k-n_{k-1}}\equiv 1\pmod{n_k}$ (both congruences hold modulo $n_{k+1}-n_k$) $n_k\mid (n_{k+1}-n_k)$ $10^{\frac{n_{k}-n_{k-1}}{n_{k-1}}}\equiv 1\pmod{n_k}$, provided that $n_{k-1}\mid n_k$ (via Pietro Majer) UPDATE #3 (Jan 24 2018). As explained in UPDATE #1 above, for every prime $p$, one can compute the set of indices $k$ such that $p\mid n_k$. This set is either finite (possibly empty), or includes all large enough integers. Recently I've tested all primes below $10^{13}$ and found that the only primes in this range that ever divide $n_k$ are $3$, $29$, and $31821709567$. The last prime divides $n_k$ for all $k\geq 8$, and so it does not help to resolve the question in favor of either outcome. UPDATE #4 (Oct 25 2018). I've extended the search for prime factors to $10^{14}$ and found only one more prime 28480625878963, which divides $n_k$ for all $k\geq 11$. The known prime factors are now listed in the sequence A318971. REPLY [5 votes]: I spent some time working this problem and discovered the following generalization. There's no new information here about the $2^{x-1}+5$ problem, so this is not much of an answer to that specifically. But we can say some similar things about some similar functions. Let $F(x)$ be a composition of functions $x$, $c$, $c^\square$, $\square + \square$, $\square \cdot \square$, and $\square!$. For example, we might have $F(x) = 2^{6^x + x^2} + (x !)^2 + 3^x \cdot x - 3$, but not $F(x) = x^x$. Let $F^k(x)$ denote the $k^\text{th}$ iterate of $F$, so for example $F^2(x) = F(F(x))$. Lemma: $F^k(x) ~\text{mod}~ m$ is eventually periodic in $k$. Proof: First re-write $F(x)$ so that all the bases are factored into primes, for example $F(x) = 2^{6^x} = 2^{2^x \cdot 3^x}$. Now with $m = p^a \cdot b$ and $(p,b) = 1$, define $g_p(m) = \text{ord}_b(p)$, and observe that $p^{a+x} \equiv p^{(a + x) ~\text{mod}~ g_p(m)} ~\text{mod}~ m$. Assume as an inductive hypothesis that $F^k(x) ~\text{mod}~ n$ is eventually periodic in $k$ for all $1 \leq n < m$. By taking all the exponents $\text{mod}$ an appropriate composition of $g$ functions we get a function $G$ such that for sufficiently large $x$, $F(x) \equiv G(x) ~\text{mod}~ m$ — for example, if $F(x) = 2^{3^x+x}+x^7$, consider $G(x) = 2^{3^{x ~\text{mod}~ g_3(g_2(m))} + x ~\text{mod}~ g_2(m)} + x^7$ — then for sufficiently large $k$ we have $F^{k+1}(x) \equiv G(F^k(x)) ~\text{mod}~ m$ and combined with the inductive hypothesis and $g_p(m) < m$ this implies $F^k(x) ~\text{mod}~ m$ is an eventually periodic function of $k$. Factorials are allowed too since they are eventually equal to $0 ~\text{mod}~ m$ and we can remove them from the expression, but I'm not sure how to handle general $\square^\square$ power compositions or primorials. $\square$ Corollary: If $x$ never occurs outside of an exponent in the expression defining $F(x)$, like $F(x) = 2^{x-1} + 5$ and $F(x) = 2^{7^x+x}\cdot 5^x +3$, but not like $F(x) = 2^x + x$ or $F(x) = 2^x \cdot x$, call it restricted. Then $F^k(x) ~\text{mod}~ m$ is eventually fixed (periodic with period $1$) for all $m$ if and only if $F$ is restricted. However, this doesn't really explain why it should be that $F^k(x) ~\vert~ F^{k+1}(x)$ as requested. It's simply a generalization of the observations in Update #1, but it applies to a much wider class of functions than I originally suspected. Corollary: For all $x, m \in \mathbb{N}$, there exists a $q \in \mathbb{Q}$ such that for all $k \in \mathbb{N}$, $F^k(x) \equiv \lfloor q \cdot m^k \rfloor ~\text{mod}~ m$. If $F$ is restricted then $q$ has a denominator of the form $m^z \cdot (m-1)$. An idea I have is to compute some of these rationals for various $F, m, x$ and find cases with the same $m$ and two different functions $F$ and $G$ where the corresponding rationals have some of the same base-$m$ digits at the same positions. Since it is impossible to evaluate iterates of $F$ and $G$ beyond small arguments, if there is no obvious reason for this relationship, then determining for exactly which $k$ is it the case that $m ~\vert~ F^k(x) - G^k(y)$ may turn out to be a good puzzle problem requiring essentially the argument above plus calculations.<|endoftext|> TITLE: Linear independence of p-adic logarithms (analog of Baker's theorem) QUESTION [5 upvotes]: We have the following theorem of Baker: Theorem 1. Let $\alpha_1, \ldots, \alpha_m \in \mathbb{C}$ be algebraic numbers $\neq 0, 1$ such that $\log \alpha_1, \ldots, \log \alpha_m$ are linearly independent over $\mathbb{Q}$. Then $\log \alpha_1, \ldots, \log \alpha_m$ are linearly independent over the algebraic numbers. Here I read that this theorem has an analog in the $p$-adic setting, so I guess it holds something like: Theorem 2 (?). Let $p$ be a prime number and $\alpha_1, \ldots, \alpha_m \in \{x \in \mathbb{C}_p : |x - 1|_p < p^{-1/(p-1)}\}$ be algebraic numbers such that $\log_p \alpha_1, \ldots, \log_p \alpha_m$ (where $\log_p$ is the $p$-adic logarithm) are linearly independent over $\mathbb{Q}$. Then $\log_p \alpha_1, \ldots, \log_p \alpha_m$ are linearly independent over the algebraic numbers. My question is: Is Theorem 2 true? Where I can find a reference? Thanks. REPLY [7 votes]: Yes. For linear independence, the result goes back to A. Brumer in connection with the Leopoldt conjecture. Brumer, A. "On the units of algebraic number fields", Mathematika 14 (1967) 121–124 For lower bounds on linear forms, there was an early version due to van der Poorten that I think had some problems and then a correct version was proved by Kunrui Yu. You should be able to find Yu's papers through Mathscinet. I am sure there are even more recent developments for linear forms.<|endoftext|> TITLE: Are inverse eigenvalue problems (IEPs) hopeless and not a fruitful area of research? QUESTION [5 upvotes]: I've been studying IEPs, in particular, the Nonnegative Inverse Eigenvalue Problem, some basic theoretical framework, the many open questions that IEPs have, and now sort of realize the computational difficulty of reconstructing matrices from prescribed spectral data (e.g., prescribed spectrum). Are these problems sort of dead right now, like hopeless? Or perhaps IEPs is actively being researched today? Most of the literature that I have found on IEPs with the most partial progress seems to be very old - late 70s to mid 80s. (Newer papers seem to just reiterate old results.) For the nonnegative inverse eigenvalue problem, the problem is open for $n\ge 5$. Thanks in advance for any advice or suggestions. REPLY [3 votes]: Although I work exclusively on the nonnegative inverse eigenvalue problem (NIEP), I can assure you that IEPs are far from dead, both on the theoretical side and the applied side (there has been a flurry of activity in recent years, along with a special session at the 2017 Meeting of the International Linear Algebra Society). I would like to point out that in the paper by Charlie Johnson and I [MR3452738; *Linear Algebra Appl.* 493 (2016), 281–300], connections between the real NIEP to Bose-Mesner algebras and association schemes were established. For more details on current developments on the NIEP, see the recent survey paper (which will appear later this year in Operator Theory: Advances and Applications by Johnson et al (it is worth noting that this survey already dated needs updating).<|endoftext|> TITLE: Uniquely complemented but not Boolean QUESTION [6 upvotes]: What is an example of a lattice $(L,\leq)$ that is uniquely complemented, but not Boolean? REPLY [9 votes]: As mentioned in the comments, there are and this follows from Dilworth's construction of free uniquely complemented lattices. A simpler construction was later given by Chen and Grätzer [On the construction of complemented lattices, J. Algebra 11 (1969), 56–63]. There is a nice overview by George Grätzer in the AMS Notices in 2007: "Two Problems That Shaped a Century of Lattice Theory". Here is an excerpt that indicates that there were no other examples of non-distributive uniquely complemented lattices in 2007; I don't know if any new examples have been found in the last decade. All known examples of nondistributive uniquely complemented lattices are freely generated, one way or another. Is there a construction of a nondistributive uniquely complemented lattice that is different? In the same vein, is there a “natural” example of a nondistributive uniquely complemented lattice from geometry, topology, or whatever else?<|endoftext|> TITLE: Correspondence between $SBT (n)$ and $W(B_n)$ QUESTION [6 upvotes]: Let $W(B_n)$ be a Weyl group of type $B_n$ and $SBT (n)$ the set of standard bitableaux of size $n$. Similar to Robinson-Schensted correspondences, I know that there exists a map $W(B_n) \to SBT (n) \times SBT (n)$ such that the image of $W(B_n)$ is the set of the same shape pairs of standard bitableaux. But, I don't know what the map $W(B_n) \to SBT (n) \times SBT (n)$ is. In particular, $W(B_2) \to SBT (2) \times SBT (2)$? Here \begin{align*} W(B_2) \cong & \left\{ \begin{pmatrix} 1 & 2 & -1 & -2 \\ 1 & 2 & -1 & -2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & -1 & -2 \\ 2 & 1 & -2 & -1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & -1 & -2 \\ -1 & 2 & 1 & -2 \end{pmatrix}, \right. \\ & \left. \begin{pmatrix} 1 & 2 & -1 & -2 \\ 1 & -2 & -1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & -1 & -2 \\ -2 & 1 & 2 & -1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & -1 & -2 \\ 2 & -1 & -2 & 1 \end{pmatrix}, \right. \\ & \left. \begin{pmatrix} 1 & 2 & -1 & -2 \\ -1 & -2 & 1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & -1 & -2 \\ -2 & -1 & 2 & 1 \end{pmatrix}\right\}. \end{align*} REPLY [3 votes]: Before getting into my answer, let me explain what I understand by a bitableau of size $n$ (I hope it is the same as what you mean). This is a pair of tableaux $(P, P')$, where each of the integers $1,\dotsc,n$ occurs exactly once, and moreover, the numbers increase along rows and columns in both $P$ and $P'$. The shape of the bitableaux is $(\mu, \mu')$ where $\mu$ and $\mu'$ are the shapes of $P$ and $P'$. Bitableaux of shape $(\mu, \mu')$ correspond to maximal chains in the Bratelli diagram of the sequence of groups $W(B_n)$ which begin at $(\emptyset, \emptyset)$ and end at $(\mu, \mu')$. The Robinson-Schensted correspondence has a generalization - the RSK correspondence. This correspondence is between (nonnegetive) integer matrices and pairs of semistandard Young tableaux of the same shape (see Knuth's 1970 paper in PJM, or my book). If the rows of the integer matrix sum to $\lambda = (\lambda_1,\dots, \lambda_l)$, and the columns sum to $\mu = (\mu_1,\dotsc, \mu_m)$, then the correspondence associates to $A$ a pair $(P, Q)$ of semistandard young tableaux, $P$ having content $\lambda$, and $Q$ having content $\mu$. The matrix $A$ is said to be a partial permutation if all the row and column sums are at most $1$. Now, an element $w\in W(B_n)$ may be regarded as a pair of partial permutations, $(A, B)$ such that $A + B$ is a permutation matrix as follows: $A_{ij} = 1$ if $w:j\mapsto i$, and is $0$ otherwise. $B_{ij} = 1$ if $w:j\mapsto -i$, and is $0$ otherwise. Now apply the RSK correspondence to $A$ and to $B$, getting $(P_A, Q_A)$ and $(P_B, Q_B)$. Then $(P_A, P_B)$ and $(Q_A, Q_B)$ are both bitableaux of the same shape. Consider for example, the element (in your notation) $$w = \begin{pmatrix} 1 & 2 & 3 & -1 & -2 & -3\\ 2 & -1 & -3 & -2 & 1 & 3 \end{pmatrix}$$ corresponds to the matrices: $$ A = \begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1\end{pmatrix} $$ Under the RSK correspondence $A$ maps to $P=(1)$, $Q = (2)$ (of shape $(1)$), while $B$ corresponds to $(2, 3)$ and $(1, 3)$ (of shape $(2)$). So the bitableaux pair for $w$ is $(1), (2, 3)$ and $(2), (1, 3)$. The relation to domino tableaux is rather interesting. Given a bipartition $(\alpha, \beta)$, let $\lambda$ denote the unique partition whose $2$-core is trivial and whose $2$-quotient is $(\alpha, \beta)$. Then adding a box to either $\alpha$ or $\beta$ corresponds to adding a domino to $\lambda$ (see the book by James and Kerber, or the paper of Frame, Robinson and Thrall for details). The size of $\lambda$ will be twice the sum of the sizes of $\alpha$ and $\beta$. Thus we may regard a bitableau of size $n$ as a domino-tableau of size $2n$. Thus the correspondence that I described above may also be regarded as a bijection from $W(B_n)$ onto the set of pairs of domino tableaux of the same shape. In the running example, the associated domino tableau are $P = \begin{array}{cccc}1&2&3&3\\1&2&&\end{array}$ and $Q = \begin{array}{cccc}1&1&3&3\\2&2&&\end{array}$. Here is some Sage code for the computation: from sage.combinat.tableau import from_chain def matrices(w): n = len(w) A = Matrix(ZZ, n, n, lambda i,j: int(w[j] == i+1)) B = Matrix(ZZ, n, n, lambda i,j: int(w[j] == -i-1)) return A, B def domino_rsk(w, dominoes=True): n = len(w) A, B = matrices(w) P1, Q1 = tuple(RSK(A)) P2, Q2 = tuple(RSK(B)) if dominoes: ch_P1 = P1.to_chain(max_entry=n) ch_P2 = P2.to_chain(max_entry=n) ch_Q1 = Q1.to_chain(max_entry=n) ch_Q2 = Q2.to_chain(max_entry=n) P = from_chain([Partition(core=[], quotient=[ch_P1[i], ch_P2[i]]) for i in range(n+1)]) Q = from_chain([Partition(core=[], quotient=[ch_Q1[i], ch_Q2[i]]) for i in range(n+1)]) return P, Q else: return [P1,P2], [Q1, Q2] For the elements of $W(B_2)$, I got (in terms of bitableaux): [1, 2] ([[[1, 2]], []], [[[1, 2]], []]) [1, -2] ([[[1]], [[2]]], [[[1]], [[2]]]) [-1, 2] ([[[2]], [[1]]], [[[2]], [[1]]]) [-1, -2] ([[], [[1, 2]]], [[], [[1, 2]]]) [2, 1] ([[[1], [2]], []], [[[1], [2]], []]) [2, -1] ([[[1]], [[2]]], [[[2]], [[1]]]) [-2, 1] ([[[2]], [[1]]], [[[1]], [[2]]]) [-2, -1] ([[], [[1], [2]]], [[], [[1], [2]]]) and in terms of domino tableaux: [1, 2] ([[1, 2, 2], [1]], [[1, 2, 2], [1]]) [1, -2] ([[1, 2], [1, 2]], [[1, 2], [1, 2]]) [-1, 2] ([[1, 1], [2, 2]], [[1, 1], [2, 2]]) [-1, -2] ([[1, 1, 2, 2]], [[1, 1, 2, 2]]) [2, 1] ([[1], [1], [2], [2]], [[1], [1], [2], [2]]) [2, -1] ([[1, 2], [1, 2]], [[1, 1], [2, 2]]) [-2, 1] ([[1, 1], [2, 2]], [[1, 2], [1, 2]]) [-2, -1] ([[1, 1], [2], [2]], [[1, 1], [2], [2]]) This is not the same as the answer obtained by Brian Hopkins, whose dominoes have an extra $0$, and I must admit that I have not looked carefully the references he gave as yet (and am not convinced that this question should need one to read four papers in Compositio). However, if I tweak my code so that $\lambda$ has core $(1)$ and quotient $(\alpha,\beta)$, then I do seem to get his answer.<|endoftext|> TITLE: What is the largest subgroup of $GL^{+}(7,\mathbb{R})$ which smoothly retracts onto $G_2$? QUESTION [5 upvotes]: There is a nice smooth retraction from $\operatorname{GL}(n,\mathbb{C})$ onto $\operatorname{U}(n)$, which can be explained using polar decomposition. There is an analogous one from $\operatorname{GL}(n, \mathbb{R})$ onto $\operatorname{O}(n)$. Both are "natural" from a Lie theoretic point of view (in a sense that shall not be made precise for now). I am wondering what is the largest subgroup of $\operatorname{GL}^{+}(7,\mathbb{R})$ ('+' here just means with positive determinant) which smoothly retracts onto $\operatorname{G}_2$? I have a strange feeling it could be $\operatorname{G}_2$ itself, because I don't think that one can split a $G_2$ structure into 2 independent pieces, one being a Riemannian inner product, and something else. For instance if one uses the usual $3$-form as one of the pieces, then the subgroup of $\operatorname{GL}^{+}(7,\mathbb{R})$ which preserves that $3$-form is $G_2$ itself (so that preserving that $3$-form implies preserving an inner product). REPLY [12 votes]: There is no retraction of $\mathrm{SO}(7)$ onto $\mathrm{G}_2$. If such a retraction $\rho:\mathrm{SO}(7) \to \mathrm{G}_2$ existed, then the composition $$ \mathrm{G}_2 \hookrightarrow \mathrm{SO}(7)\ {\buildrel{\rho}\over{\rightarrow}} \ \mathrm{G}_2\,, $$ would induce a composition of the homotopy group homomorphisms $$ \pi_6\bigl(\mathrm{G}_2\bigr) \rightarrow \pi_6\bigl(\mathrm{SO}(7)\bigr)\ {\buildrel{\rho_*}\over{\rightarrow}} \ \pi_6\bigl(\mathrm{G}_2\bigr) $$ that was the identity. However, according to M. Mimura, The homotopy groups of Lie groups of low rank, J. Math. Kyoto Univ. 6 (1967), 131–176 (see the tables on page 132), $$ \pi_6\bigl(\mathrm{G}_2\bigr)\simeq \mathbb{Z}_3 \quad\text{while}\quad \pi_6\bigl(\mathrm{SO}(7)\bigr)\simeq 0\,, $$ so this is impossible. Beyond this, though, I feel that the OP's given motivation for asking the question suggests that the issue of concern is not really whether $\mathrm{SO}(7)$ retracts onto $\mathrm{G}_2$. The usual reason that one wants to retract $\mathrm{GL}(n,\mathbb{C})$ onto $\mathrm{U}(n)$ and $\mathrm{GL}(n,\mathbb{R})$ onto $\mathrm{O}(n)$ is that these are the fundamental steps in the proof that a complex vector bundle over a paracompact base can always be given a unitary structure and a real vector bundle over a paracompact base can always be given an Euclidean structure (i.e., a positive definite inner product field). The analogous question concerning $\mathrm{SO}(7)$ and its subgroup $\mathrm{G}_2$ would be whether every oriented $7$-plane bundle $E\to M$ endowed with an Euclidean structure can be given a $\mathrm{G}_2$-structure, i.e., whether there exists a 'definite' (i.e., a $\mathrm{G}_2$-defining) section of $\Lambda^3(E)\to M$. Now, the answer to this question is a definite 'no'. For example, let $M^7$ be any orientable-but-not-spinnable $7$-manifold, say $M^7 = \mathbb{CP}^2\times S^3$. Then $E=TM\to M$ certainly has a $\mathrm{SO}(7)$-structure (defined by the orientation and any Riemannian metric on $M$), but, because $M$ is not spinnable (because $\mathbb{CP}^2$ is not), its tangent bundle does not allow a structure reduction to $\mathrm{G}_2$. (This is because $\mathrm{G}_2$ is simply-connected, so that any $7$-manifold whose tangent bundle has a reduction of structure to $\mathrm{G}_2$ would necessarily be spinnable.) Thus, this particular problem does not motivate the retraction question. The OP's other remark expressing doubts about whether a $\mathrm{G}_2$-structure can be split into two 'independent pieces' seems to be asking for an analog of what happens in complex and symplectic geometry, where one has the intersections $$ \mathrm{U}(n) = \mathrm{GL}(n,\mathbb{C})\cap \mathrm{Sp}(n,\mathbb{R}) = \mathrm{GL}(n,\mathbb{C})\cap \mathrm{O}(n) = \mathrm{Sp}(n,\mathbb{R})\cap \mathrm{O}(n), $$ so that one can think of a unitary structure as a combination of two 'independent' (but compatible) weaker structures. The natural analog of this in the present context would be whether $\mathrm{G}_2$ can be expressed as the intersection of two (or more?) Lie subgroups of $\mathrm{GL}(7,\mathbb{R})$ that properly contain $\mathrm{G}_2$. The answer to this is a(n uninteresting) 'yes'. There are only three connected Lie groups that lie strictly between $\mathrm{G}_2$ and $\mathrm{GL}^+(7,\mathbb{R})$, and they are $$ \mathrm{R}^+{\cdot}\mathrm{G}_2,\quad \mathrm{SO}(7),\quad\text{and} \quad \mathrm{R}^+{\cdot}\mathrm{SO}(7). $$ Thus, the only nontrival way to do this is the uninteresting intersection $$ \mathrm{G}_2 = \mathrm{R}^+{\cdot}\mathrm{G}_2 \cap \mathrm{SO}(7). $$ Thus, the largest (connected) subgroup of $\mathrm{GL}^+(7,\mathbb{R})$ that retracts onto $\mathrm{G}_2$ is $\mathrm{R}^+{\cdot}\mathrm{G}_2$. As a final argument as to why the retraction question seems to be irrelevant to the OP's apparent concerns, consider the analogous question of whether $\mathrm{SO}(2n)$ can be retracted onto $\mathrm{U}(n)$ for $n>1$. It cannot, because, such a retraction $\rho$, if it existed in the form the OP desired, i.e., $$ \mathrm{U}(n)\hookrightarrow \mathrm{SO}(2n)\ {\buildrel{\rho}\over{\rightarrow}} \ \mathrm{U}(n), $$ where the composition is the identity on $\mathrm{U}(n)$, then the composition of the induced map on fundamental groups $$ \mathbb{Z}\to \mathbb{Z}_2\to \mathbb{Z} $$ would be an isomorphism, which is manifestly impossible.<|endoftext|> TITLE: Entropy of chain decompositions of posets QUESTION [7 upvotes]: Let $P$ be a finite poset and let $\,\mathcal{C}=(C_1,\ldots,C_\ell)\,$ be its decomposition into chains. We can define $$ f(\mathcal{C}) = |C_1|! \, \cdots \,|C_\ell|! $$ and ask for what $\mathcal{C}$ we have $f(\mathcal{C})$ maximal. Roughly speaking, the smaller is $\ell$ and the less evenly distributed are $|C_i|$, the better. Specifically, I am interested in the following. Question: Does $f(\mathcal{C})$ maximize for the number $\ell$ of chains equal to the size of the maximal antichain in $P$? Motivation: If true, this would be an unusual extension of Dilworth's theorem. I am really interested in bounds on the number $e(P)$ of linear extensions, and it is easy to see that $e(P)\le |P|!/f(\mathcal{C})$ for all chain decompositions $\mathcal{C}$. The connection to entopy is also standard: take $\log e(P)$ in the last equation and think of $|C_i|/|P|$ as a probability distribution. Refs disclaimer: There is a large number of papers bounding $e(P)$ in terms of the entropy, see e.g. here, here and here. They are not completely unrelated, but do not seem concerned with this type of questions. REPLY [3 votes]: The answer is no. Take a chain $a_1>\dots>a_{100}$ and add two elements $b>a_{51}$ and $c TITLE: Closed convex bounded sets are weakly compact for which spaces? QUESTION [6 upvotes]: It is known that for all reflexive Banach spaces, closed convex bounded sets are weakly compact (compact for the weak topology). What is the general class of topological vector spaces for which this is true ? For example, is it true that for all reflexive complete locally convex topological vector spaces, closed convex bounded sets are weakly compact ? REPLY [6 votes]: It is well-known that a Hausdorff locally convex space is semi-reflexive (i.e., the canonical map into its bidual is surjective) if and only if every weakly closed bounded set is weakly compact. This is proposition 23.18 in Introduction to Functional Analysis of Meise and Vogt. (It is mainly a consequence of Alaoglu's theorem.) REPLY [4 votes]: The following is from Shaefer's "Topological Vector Spaces", sections 5.5 and 5.6. For a locally convex (Hausdorff) $E$, the injection into its bidual is surjective iff every bounded set in $E$ is weakly-relatively-compact. In this case $E$ is said to be semi-refelexive. The map into the bidual need not be continuous. This bijection is continuous iff $E$ is also barreled. In this case this bijection is in fact an isomorphism of topological vector spaces and $E$ is said to be reflexive. A general topological vector space which injects into its bidual, when taken with the weak topology, is clearly locally convex Hausdorff, so we can safely restrict ourselves to this class. So, the answer to the first question is "the general class of locally convex Hausdorff topological vector spaces for which this is true is called semi-reflexive spaces" and the answer to the second is "yes, in particular it is true for reflexive spaces".<|endoftext|> TITLE: Klein's curve (algebraic geometry) QUESTION [9 upvotes]: I can't find any information about the canonical ring of Klein's quartic curve (the one with 168 automorphisms). I would imagine there is a lot known about the structure of this ring. Can anybody help me please? Thank you! REPLY [11 votes]: Klein quartic, $X$, is a smooth degree $4$ plane curve, given by the equation $F(x,y,z) = x^3 y + y^3 z + z^3 x = 0$. Doesn't this mean that the canonical ring is $k[x,y,z]/(x^3 y + y^3 z + z^3 x)$? Details: This embedding is the canonical embedding, since we compute by adjunction that $\mathcal{O}(1)$ is the canonical bundle. Thus, we have an injection $k[x,y,z]/F \hookrightarrow \bigoplus_{n=0}^{\infty} H^0(X, K^{\otimes n})$. The Hilbert series of the left hand side is $(1-t^4)/(1-t)^3 = 1+3t + \sum_{n \geq 2} (4n-2) t^n$. The Hilbert series of the canonical ring is $1+gt + \sum_{n \geq 2} ((2g-2)n-(g-1)) t^n$. These match, so the map is an isomorphism. (Actually, I think I recall that the canonical ring of a nonhyperelliptic curve is always generated in degree $1$, but it seemed easier to check the Hilbert series in this case.)<|endoftext|> TITLE: Are bounded sets always weakly metrizable in reflexive separable spaces? QUESTION [5 upvotes]: It is known that if a Banach space is reflexive and separable, its unit ball is weakly metrizable. My question is about the generalization of this property : 1) Is it true that for all reflexive separable locally convex space, bounded sets are weakly metrizable? 2) If that's true, is there a way to explicitly construct a distance for the weak-topology on any bounded set? REPLY [6 votes]: No. Let $I$ be an index set with the cardinality of the continuum. Endow $X=\mathbb R^I$ with the product topology. According to (a particular case of) the Hewitt-Marczewski-Pondiczery theorem (which is 2.3.15 in Engelking's General Topology) $X$ is separable. Moreover, it is semi-reflexive (by Tychonov) and barrelled (because barrelledness is stable w.r.t. products, proposition 4.2.5 in Barrelled Locally Convex Spaces of Bonet and Perez Carreras). Therefore, $X$ is reflexive. The set $B=[-1,1]^I$ is bounded. Assume that it is weakly metrizable (by the way, $X$ carries its weak topology). Then it has a countable basis of $0$-neighbourhoods and since every $0$-neighbourhood in $X$ contains one of the form $\lbrace x\in X: |\varphi_i(x)|<\varepsilon, i=1,\ldots,n \rbrace$ for some $n\in \mathbb N$, $\varphi_1,\ldots,\varphi_n \in X'$, and $\varepsilon>0$, we find a sequence of $\phi_n\in X'$ such that $$ B\cap \bigcap_{n\in\mathbb N} \text{kern}(\phi_n) = \lbrace 0 \rbrace. $$ But each $\phi_n$ only depends on finitely many coordinates, i.e., it is of the form $\phi_n((x_i)_{i\in I})= \sum_{i\in E_n} a_{n,i} x_i$ for some finite $E_n \subseteq I$, and it is enough to consider $j\in I\setminus \bigcup_{n\in\mathbb N} E_n$ and $b=(\delta_{i,j})_{i\in I} \in B$ to get a contradiction.<|endoftext|> TITLE: Question on the Sato-Tate conjecture QUESTION [6 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Q}$ without CM. For a good prime $p$, define $\theta_{E}(p)$ by $$\cos\theta_{E}(p)=\frac{p+1-N_{p}(E)}{2\sqrt{p}}\quad (0\leq \theta_{E}(p)\leq \pi).$$ I wonder if the function $\theta_{E}$ is injective? REPLY [12 votes]: No. If $E_p$ is a supersingular elliptic curve and $p>3$ then trace of Frobenius on $E_p$ is zero, so $\theta_E(p)=\pi/2$. By a result of Elkies any elliptic curve over $\mathbb{Q}$ has supersingular reduction in infinitely many primes, so for infinitely many $p$ this function takes the same value.<|endoftext|> TITLE: Restrictions on $\pi_1(X)$ of geometric origin (Kähler groups as example) QUESTION [6 upvotes]: There's and old and extensively studied question about characterisation of fundamental groups of smooth compact Kähler manifolds. Restrictions imposed by Kählerness are somewhat fragile, and if we drop some properties and go aside a bit, every finitely presented group can be realized as a $\pi_1$ of 1) symplectic complex threefold — no relation between $I$ and $w$, however (an old result by Gompf); 2) projective surface with mild singularities (M. Kapovich paper https://www.math.ucdavis.edu/~kapovich/EPR/tiling.pdf); 3) 6-dimensional symplectic manifold with trivial $c_1$ (A. Petrunin afair). Let dimensions of our manifolds to be $\geq 4$ to omit purely dimensional properties of $\pi_1$. Are there some geometric structures (say, defined by collection of forms, e. g. quaternionic Kähler, hermitian almost complex, CR manifolds with additional properties — you name it) or "locally defined" constraints of different kind (Chern class relations, comparison morphisms between cohomology of natural sheaves) for which fundamental groups of smooth manifolds of that kind have non-obvious nice properties such as superrigidity in symmetric spaces of some kind? More hand-wavy version: is there some differential-geometric description for coastline of country where manifolds with restricted fundamental group live? REPLY [13 votes]: Some properties of Kähler groups are retained by other classes of closely related groups, such as fundamental groups of smooth, quasi-projective varieties, or fundamental groups of Sasakian manifolds. For instance, if $M$ is a smooth, quasi-projective variety, then the first characteristic variety $V^1_1(M)$, i.e., the set of characters $\rho\colon \pi_1(M)\to \mathbb{C}^*$ for which $H^1(M, \mathbb{C}_{\rho})\ne 0$, is a finite union of torsion-translated subtori of the character group $\operatorname{Hom}(\pi_1(M), \mathbb{C}^*)$. As another example, if $M$ is a compact Sasakian manifold, then the same conclusion holds for the irreducible components of $V^1_1(M)$ passing through the identity. Moreover, if $\dim(M)\ge 5$, then $\pi_1(M)$ is $1$-formal, i.e., its Malcev Lie algebra is quadratic.<|endoftext|> TITLE: Does a compact nonflat surface without conjugate points have ergodic geodesic flow? QUESTION [8 upvotes]: I read this as a conjecture in the paper by Ballmann-Brin-Burns, titled "On Surfaces with No Conjugate points" JDG 25(249-273), 1987. What is current status of this conjecture? REPLY [4 votes]: This is an open question. Actually it is still not known if a compact non-flat surface with non-positive curvature has an ergodic geodesic flow (with respect to the Liouville measure).<|endoftext|> TITLE: Symmetric functions arising from continuous-time Markov chains QUESTION [6 upvotes]: I'm interested in the following symmetric functions $s_k: \mathbb{R}_+^{k}\mapsto\mathbb{R_+}$: \begin{align*} s_k(x_1,x_2,\dots, x_k)&= \int\limits_{0=t_0 TITLE: Is there a row vector $x$ with integer entries such that no entry of $xM$ is $0 \text{ (mod }p\text{)}$? QUESTION [13 upvotes]: Let $p$ be a prime and let $M$ be an $n \times m$ matrix with integer entries such that $M\vec{v} \not\equiv \vec{0} \text{ (mod }p\text{)}$ for any column vector $\vec{v} \neq \vec{0}$ whose entries are $0$ or $1$. Is there a row vector $\vec{x}$ with integer entries such that no entry of $\vec{x}M$ is $0 \text{ (mod }p\text{)}$? REPLY [10 votes]: It seems that my previous answer was completely wrong, and that the fact is true for all nonzero integer (not necessarily prime) $p$. Let $M=[m_{ij}]_{i\leq n,\;j\leq m}$. If there is no $x$ satisfying the requirements (we regard all entries as integers, not residues!), then one of the sums of the form $\sum_{i\leq n} m_{ij}x_i$, $j\leq m$, is divisible by $p$ for every integer vector $x$. Equivalently, $$ \prod_{j\leq m}\left(1-\exp\left(\frac{2\pi i}p\sum_{i\leq n} m_{ij}x_i\right)\right)=0 $$ for every integral vector $x$. After expanding all the brackets, we get the equality of the form $$ \sum_k c_k\xi_{1k}^{x_1}\dots \xi_{nk}^{x_n}=0, $$ where $c_i$ are some (integer) numbers, and $\xi_{ik}$ are some nonzero complex numbers; after collecting terms, we may assume that all the $n$-tuples $(\xi_{1k},\dots,\xi_{nk})$ are distinct. Due to the usual Vandermonde argument, this may happen only if $c_k=0$ for all $k$. This means, in particulat, that the term $1^m$ cancels with some other term, which has the form $$ (-1)^t\prod_{j\in T}\exp\left(\frac{2\pi i}p\sum_{i\leq n} m_{ij}x_i\right) =(-1)^t\prod_{i\leq n}\left(\exp\left(\frac{2\pi i}p\sum_{j\in T}m_{ij}\right)\right)^{x_i}. $$ for some $\varnothing\neq T\subseteq \{1,2,\dots,m\}$ with $t=|T|$. This yields that $\sum_{j\in T}m_{ij}$ is divisible by $p$ for all $i\leq n$, so the characteristic column of $T$ (taken as $v$) violates the condition imposed on $M$. Moreover, one may notice that there is such $T$ with odd cardinality; so we have proved that $v$ may be chosen with odd number of ones in it. ADDENDUM. The `usual Vandermonde argument' I meant consists in the following. Assume that $(\xi_{1,k},\dots,\xi_{n,k})$, $k=1,\dots,K$, are distinct tuples of complex numbers. Then the arrangements $(\xi_{1,k}^{x_1}\cdots\xi_{n,k}^{x_n})_{x_i\in\mathbb Z}$ are linearly independent. The proof goes by the induction on $n$; the base case $n=1$ is classical Vandermonde. For the step of induction, assume the linear dependence, collect the terms with identical $\xi_{1,k}$ and apply first the hypothesis (to prove that one of the coefficients of $\xi_{1,k}^{x_i}$ does not vanish for some choice of $x_2,\dots,x_n$) and then the base to finish the step.<|endoftext|> TITLE: Subgroups of the tensor product $A\otimes A$ QUESTION [7 upvotes]: I have this problem about subgroups of the tensor product of an abelian group $A$ with itself which arises from a complete different setting. I fell into this question studying quandles and quandle coverings, but the question can be stated just in terms of abelian groups. Let $A$ be an Abelian group and $\alpha$ an automorphism of $A$ such that $1-\alpha$ is invertible. I need to compute the following subgroups in $A\otimes A$: $B=\langle x\otimes y-\alpha(y)\otimes x,\quad x,y\in A\rangle$ The goal is to find some condition on $A$ and $\alpha$ in order to have that $B=A\otimes A$. For instance, if $A$ is a cyclic group then $B=A\otimes A$. Moreover, I know that $B$ is the whole group when $A$ is elementary abelian different from $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\alpha$ has order $|A|-1$ (it follows by an equivalent property in quandles setting). A counterexample is given when $\alpha=-1$. Since $B=\langle x\otimes y+y\otimes x,\quad x,y\in A\rangle$ is a subgroup of the element fixed by the flip $\tau$ ($\tau(x\otimes y)=y\otimes x$) and then if $A$ is not cyclic, this is a proper subgroup. Thank you. M. P.S. I forgot to say that $A$ is finite. REPLY [6 votes]: Here's a full answer. First I consider a general linear algebra problem, then I apply it to the specific given problem. Let $K$ be a field, $V$ a finite-dimensional vector space, $T$ an operator of $V$. Consider the endomorphism $b_T$ of $V\otimes V$ (all $\otimes$ are over $K$) given by $$x\otimes y\mapsto b_T(x\otimes y)=x\otimes y-Ty\otimes x.$$ Then I claim that $b_T$ is an automorphism (or equivalently surjective) if and only if all the following hold: (a) 1 is not eigenvalue of $T$; (b) $-1$ is not double eigenvalue of $T$ [that is, $\dim(\mathrm{Ker}(1+T))\le 1$]; (c) $T$ has no pair of distinct inverse eigenvalues (in any extension field of $K$). Since both the problem and the characterization ignore field extension, it is enough to prove this when $K$ is algebraically closed; we do it by induction on $d=\dim(V)$; the case $d=0$ being trivial, assume $d\ge 1$. Write $b_T=b_{V,T}$. Fix an eigenline $E\subset V$ of $T$. It yields a $b_T$-stable $(2d-1)$-dimensional subspace $W=(E\otimes V+V\otimes E)$ of $V\otimes V$. So $\det(b_T)=\det((b_T)|_{W})\det(b_{V/E,T})$, because the endomorphism of $(V\otimes V)/W$ induced by $T$ can be identified with $b_{V/E,T}$. So we have to compute this first determinant $\det((b_T)|_{W})$. Let $(e_1,\dots,e_d)$ be a basis of $V$ with $e_1\in E$ and $Te_1=\lambda e_1$. Then a basis of $W$ is given by $$(e_1\otimes e_1,e_2\otimes e_1,\dots e_d\otimes e_1,e_1\otimes e_2,\dots e_1\otimes e_d).$$ Writing matrices by blocks $1+(d-1)+(d-1)$, we see that the matrix of $(b_T)|_{W}$ in this basis is $$\begin{pmatrix}1-\lambda & 0 & *\\ 0 & I & -T'\\0 & -\lambda I & I\end{pmatrix},$$ where $T'$ is the matrix of the endomorphism induced by $T$ on $V/E$ on the basis $(e_2,\dots,e_d)$. The determinant of $(b_T)|_{W}$ is therefore $(1-\lambda)\det(1-\lambda T')$. Thus the determinant of $b_T$ is $$\det(b_T)=(1-\lambda)\det(1-\lambda T')\det(b_{V/E,T}).$$ From this formula, we can deduce the result. First, for the easy implication: if $1$ is eigenvalue as in (a), pick $\lambda=1$ and then $\det(b_T)=0$. If there are two inverse eigenvalues with independent eigenvectors (this covers (b) and (c)), then pick $\lambda$ to be one of them, and then since $\lambda^{-1}$ is an eigenvalue of $T'$, we have $\det(1-\lambda T')=0$ and hence $\det(b_T)=0$. Conversely, assume that none of (i),(ii),(iii) holds. By induction and using (i)-(ii)-(iii), $\det(b_{V/E,T})\neq 0$. Also $\det(1-\lambda T')\neq 0$: this is clear if $\lambda=0$, and if $\lambda\neq 0$ we know from (ii),(iii) that $\lambda^{-1}$ is not eigenvalue of $T'$. Finally $\lambda\neq 1$ by (i) and hence $\det(b_T)\neq 0$. Now consider the analogous problem in a finite abelian group $A$. An endomorphism of a finite abelian group $Q$ is surjective if and only if the induced endomorphism of $Q/pQ$ is surjective for every prime $p$. Noting that $(A/pA\otimes A/pA)=(A\otimes A)/p(A\otimes A))$ and that this "commutes" with defining the operator $b_T$, we thus reduce all the problem to the case when $A$ is a $p$-elementary abelian group, in which case it is solved by the above linear algebra problem (with $K=\mathbf{Z}/p\mathbf{Z}$). To summarize, given an endomorphism $T$ of $A$, the endomorphism of $A\otimes A$ given by $b_T(x\otimes y)=x\otimes y-Ty\otimes x$ is surjective if and only if the three following condition hold, denoting $T_p$ the endomorphism of $T$ induced on $A/pA$: (a') for every prime $p$, 1 is not eigenvalue of $T_p$; (b') for every prime $p$, $-1$ is not double eigenvalue of $T_p$ [that is, $\dim_{\mathbf{Z}/p\mathbf{Z}}(\mathrm{Ker}(1-T_p))\le 1$]; (c') for every prime $p$, $T_p$ has no pair of distinct inverse eigenvalues (in any extension field of $\mathbf{Z}/p\mathbf{Z}$). These conditions can sometimes be restated: (a') just means that $1-T$ is invertible (which was one of your assumptions). Actually, in the linear setting, given (a), (b)-(c) means that 1 is not an eigenvalue of the endomorphism $T\wedge T$ of the exterior product $V\wedge V$. So probably given (a'), (b')-(c') means that the $1-T\wedge T$ is an invertible endomorphism of $\Lambda^2 A=(A\otimes A)/\langle a\otimes a:a\in A\rangle$. Note that in spite of the latter characterization being shorter, it is much more convenient in practice to reduce modulo $p$.<|endoftext|> TITLE: smooth functional to detect whether a function has a zero QUESTION [11 upvotes]: Does there exist a function $F : C^\infty(\mathbb{R}, [0, \infty)) \to \mathbb{R}$ with the following properties: $F(f) = 0$ if and only if there exists an $x \in [0,1]$ such that $f(x) = 0$. $F$ is smooth in the following sense: if $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0,\infty))$ and $F$ is applied to $f$ in the $x$ variable, the function of $t$ that results is smooth ($C^\infty$). The second question is whether the following candidate satisfies the smoothness property. For $f \in C^\infty(\mathbb{R}, [0, \infty))$, define $G(f)$ to be $$ \left( \int_0^1 \frac{1}{f(x)} \, dx \right)^{-2} $$ if $f$ has no zeros in $[0,1]$, and $0$ otherwise. Then $G$ satisfies the first condition by definition, and one can show that when $f$ is a function of $x$ and $t$ as above, $G(f)$ is (continuous and) differentiable as a function of $t$. Is it smooth? Edit: Willie Wong answered the second question in the negative. So let's instead define $G(f)$ to be $$ \exp \left( -\int_0^1 \frac{1}{f(x)} \, dx \right) $$ if $f$ has no zeros in $[0,1]$, and $0$ otherwise. Is this $G$ smooth? REPLY [4 votes]: A functional that works is $$ F(f)=\begin{cases} \exp(-\exp(\int_0^1 \frac{1}{f(x)}dx)), & \textrm{if $f(x)>0$ for all $x \in [0,1]$} \\ 0, & \textrm{otherwise}. \end{cases} $$ The idea to use this formula and a sketch of the proof that it works are both due to Chengjie Yu. The proof is in the appendix of a paper I wrote with Enxin Wu, Smooth classifying spaces, http://arxiv.org/abs/1709.10517 I don't know whether the functional described in the question, $G(f) = \exp(-\int_0^1 \frac{1}{f(x)} dx)$, works.<|endoftext|> TITLE: Number of all different $n\times n$ matrices where sum of rows and columns is $3$ QUESTION [8 upvotes]: For a given positive integer $n$, I need to learn the number of $n\times n$ matrices of nonnegative integers with the following restrictions: The sum of each row and column is equal to $3$. Two matrices are considered equal if one can be obtained by permuting rows and/or columns. For example, for $n=2$ there are two different matrices as follows: $$M_1=\begin{pmatrix} 1&2\\2&1 \end{pmatrix},\qquad M_2=\begin{pmatrix} 3&0\\0&3 \end{pmatrix}.$$ For $n=3$ there are five different matrices as follows: $$\begin{pmatrix} 1&1&1\\1&1&1\\1&1&1 \end{pmatrix},\quad \begin{pmatrix} 0&1&2\\1&2&0\\2&0&1 \end{pmatrix}, \quad \begin{pmatrix} 0&1&2\\1&1&1\\2&1&0 \end{pmatrix}, \quad \begin{pmatrix} 1&2&0\\2&1&0\\0&0&3 \end{pmatrix}, \quad \begin{pmatrix} 3&0&0\\0&3&0\\0&0&3 \end{pmatrix}.$$ REPLY [3 votes]: http://oeis.org/A001501 has an enumeration of binary matrices with row and column sums equal to 3. If n is the order of the matrix, the count grows like n^{3n}. You want classes up to row and column permutations, so your number will grow like n^n. There are some details to handle, but most of your cases reduce to the following. Such a matrix will be equivalent to one with t 3's on the diagonal, s 2x2 diagonal blocks with 2's and 1's, r more 2's on the diagonal with parallel offset diagonals having 1's, and a remaining block of size qxq with q=n-(t+2s+r), and this block is like one of your matrices but it is binary, and unless r is small you can arrange r zeros on the upper diagonal. When you flip those r diagonal entries from 0 to 1, the qxq block is like one of the matrices enumerated by the OEIS sequence above. So an upper bound will be of order n^{n+3} or smaller. A professional combinatorialist can tighten up the bound for you. Gerhard "Doesn't Need To Be Professional" Paseman, 2016.10.11.<|endoftext|> TITLE: How bogus is the glitzy proof of Borsuk-Ulam? QUESTION [19 upvotes]: Suppose $f: S^2 \rightarrow {\bf R}^2$ is continuous; let $A$ be the set of points $u \in S^2$ such that $f(u)-f(-u) \in {\bf R} \times \{0\}$ (where $-u$ denotes the antipode of $u$). Given $u,-u \in A$, must there exist a path in $A$ joining $u$ to $-u$? The appealing argument presented at https://www.youtube.com/watch?v=csInNn6pfT4 (which I have also seen elsewhere), which purports to prove the existence of a $u \in S^2$ with $f(u)-f(-u) = (0,0)$, hinges on this unstated assumption. Can anyone devise an $f$ for which $A$ is something like a topologist's sine curve? ADDED LATER: I like Ilya Bogdanov's example, and I wonder whether we can improve it. Is there an $f$ such that no point in $A$ can be joined to its antipode by a continuous curve? REPLY [18 votes]: Let $S^2$ be the unit sphere in $\mathbb R^3$, $S^2_+=\{(x,y,z)\in S^2\mid z\geq0\}$ the upper hemisphere, $S^2_-$ the lower hemisphere and $B^2$ the closed unit disk in $\mathbb R^2$. Define $p:S^2\to B^2$ by $p(x,y,z)=(x,y)$. Then the restrictions of $p_+:S^2_+\to B^2$ and $p_-:S^2_-\to B^2$ of $p$ are homeomorphisms. Define a copy of the topologist's sine wave in $B^2$ by $$W=\overline{\{(x,y)\in B^2\mid x\neq 0\text{ and } y = \tfrac12\sin\tfrac{\pi}x\}}.$$ Then, $K=p_+^{-1}(W)\cup p_-^{-1}(W)$ is a closed subset of $S^2$ with the following properties: $S^2\setminus K$ has two components $U_+$ and $U_-$ and the antipode of each point in $U_+$ lies in $U_-$ and vice versa. Antipodes of points in $K$ lie in $K$, but there is no path in $K$ between any such pair of antipodes. Now consider $f:S^2\to\mathbb R^2$ defined by $$f(x,y,z)=(0,\pm d((x,y,z),K)),$$ where $d$ is e.g. the Euclidean metric on the sphere and the sign is $+$ for $(x,y,z)\in U_+$ and $-$ for $(x,y,z)\in U_-$. This function $f$ has all the desired properties: it is continuous and the set of points $u=(x,y,z)$ such that $f(u)-f(-u)\in\mathbb R\times\{0\}$ is precisely $K$, because for any $u\in S^2\setminus K$, the second components of $f(u)$ and $f(-u)$ have different signs. REPLY [15 votes]: Let $S^2=\{(x,y,z)\colon x^2+y^2+z^2=1\}$, and set $f(x,y,z)=(x,\sin 1000x)$. Then the set $A$ consists of many components, and only one of them contains opposite points. You may choose whatever more interesing function instead of $\sin 1000x$...<|endoftext|> TITLE: Ultrafilters preserved by $\mathbb{P}$ but not by products? QUESTION [8 upvotes]: Let $U\in V$ be an ultrafilter on $\omega$. We say $U$ is preserved under forcing with $\mathbb{P}$ if $\Vdash \forall x\subset \omega \ \exists Z\in U \ Z\subset x \vee Z\subset x^c$. In other words, $U$ generates an ultrafilter in $V[G]$. Familiar examples of such ultrafilters: Ramsey ultrafilters, P-points are preserved under Sacks forcing and Miller forcing (also their products) P-points/Ramsey ultrafilters are also preserved under iterations of Sacks forcing with countable support There are also combinatorial characterizations of ultrafilters preserved by Sacks forcing, related to Halpern-Lauchli theorem, see https://www.math.wisc.edu/~miller/res/ultra-s.pdf My question is: is it known that an ultrafilter preserved by Sacks forcing necessarily needs to be preserved by side-by-side products of Sacks forcing (finite/countable support)? There are probably more ad-hoc examples of forcing $\mathbb{P}$ that preserves an ultrafilter $U$ but not the product $\mathbb{P}\times\mathbb{P}$ (is there any)? Maybe there is something easy that I overlook. REPLY [5 votes]: Let $\mathbb M$ be Miller forcing. Then $\mathbb M$ adds an unbounded real, so $\mathbb M\times \mathbb M\times\mathbb M$ adds a Cohen real (Velickovic; also an unpublished result of Shelah). Hence this forcing destroys every ultrafilter from the ground model. So for every ultrafilter $U$ in the ground model, one of the following is true: $\mathbb M$ destroys $U$. $\mathbb M$ preserves $U$, but $M\times M$ destroys $U$ $P:=\mathbb M\times\mathbb M$ preserves $U$, but $P\times P$ destroys $U$. The first alternative cannot happen for P-points, but I am not sure which of the other two may hold.<|endoftext|> TITLE: Does GAGA hold over other topological fields? QUESTION [10 upvotes]: If k is a non-discrete topological field, we can define an analytic space over k just like complex analytic spaces over $\mathbb{C}$. If you replace "complex analytic space" and "complex algebraic variety" with "analytic space over $k$" and "algebraic variety over $k$", respectively, under what conditions on $k$ does GAGA, or weaker similar results, hold? Presumably $k$ must be algebraically closed, but I'm wondering whether this is enough, or whether more conditions must be added, or whether this really only works for $k=\mathbb{C}$. REPLY [8 votes]: If $k$ is a field that is complete with respect to some ultrametric valuation, then there is the "GAGR" (i.e. géométrie algébrique et géométrie rigide) theorem. A succinct explanation (in French, without proof) is: Jarraud, Pierre. À propos de G.A.G.R.. Groupe de travail d'analyse ultramétrique 11 (1983-1984): 1-4. The original source for the result, listed as $[4]$ in Jarraud's note, seems to be: Köpf, Ursula. Über eigentliche Familien algebraischer Varietäten über affinoïden Räumen, Schriftenreihe Univ. Münster, 2. Serie, Heft 7 (1974).<|endoftext|> TITLE: Algebraic vs. homological equivalence for curves on a smooth complex projective surface QUESTION [6 upvotes]: Let $X$ be a smooth projective surface over $\mathbb{C}$. Then there is the exponential sheaf sequence: $$ 0 \rightarrow \mathbb{Z} \rightarrow \mathscr{O}_X \rightarrow \mathscr{O}_X^\times \rightarrow 0 $$ with the map from $\mathscr{O}_X$ to $\mathscr{O}_X^\times$ given by $f \mapsto \exp(f)$. Since the sequence is exact on global sections (which are just constant functions), taking cohomology gives: $$ 0 \rightarrow H^1(X, \mathscr{O}_X)/H^1(X, \mathbb{Z}) \rightarrow H^1(X, \mathscr{O}_X^\times) = \text{Pic}(X) \rightarrow H^2(X, \mathbb{Z}) \rightarrow H^2(\mathscr{O}_X) $$ In terms of this sequence, how can I understand the various notions of equivalence of divisors on $X$? I'm willing to believe that the map from $\text{Pic}(X)$ to $H^2(X, \mathbb{Z})$ is the map taking a divisor to the Poincaré dual of its homology class (although if someone knows a place where this is explained very clearly, that would be great! The only reference I know for this is Griffiths and Harris, which I find to be very challenging), which explains homologically trivial divisors are numerically trivial. Similarly, the Lefschetz $(1,1)$ theorem and the stability of the Hodge decomposition under cup product show that numerically trivial divisors are represented by torsion cohomology classes. However, what I don't understand is how to see that homologically trivial divisors are the same thing as algebraically trivial divisors here. It seems plausible that algebraically equivalent divisors are homologous, but I don't even have a guess as to why the converse is true. REPLY [11 votes]: Super vast generalisation: for divisors on a smooth projective variety over an algebraically closed field of any characteristic, the notions of algebraic, homological (for any Weil cohomology theory), and numerical equivalence agree (up to torsion). Remark. Recall: the group of cycles $\alpha \sim_{\text{alg}} 0$ is generated by cycles of the form $$\pi_{1,*} (\pi_2^* ([t_1] - [t_2]))$$ for $C$ a curve, $t_1, t_2 \in C$, and $V \subseteq X \times C$ a subvariety flat over $C$, with projections $\pi_1 \colon V \to X$, $\pi_2 \colon V \to C$. To show that $\operatorname{cl}(\alpha) = 0$, one reduces to the case $[t_1] - [t_2]$ on a curve. On the other hand, if $\operatorname{cl}(\alpha) = 0$, then compatibility between the intersection product and the cup product shows that $\alpha \sim_{\text{num}} 0$. Lemma. If $D$ is a divisor with $D \sim_{\text{num}} 0$, then $nD \sim_{\text{alg}} 0$ for some $n \in \mathbb Z_{>0}$. Remark. If $k = \mathbb C$ and we use singular cohomology, then we can actually say a little bit more: If $D \sim_{\text{num}} 0$, then $nD \sim_{\text{hom}} 0$ for some $n$, If $D \sim_{\text{hom}} 0$ (integrally), then $D \sim_{\text{alg}} 0$. Fulton's Lemma 19.3.1 states both, but only proves the second assertion (it seems). I will prove both. Remark. Since in general we don't have access to an integral Weil cohomology theory (the best we can do is $\mathbb Z_\ell$), there is no analogue of the second statement for other Weil cohomology theories. Proof 1. (Over $k = \mathbb C$ for singular cohomology) Let $H$ be an ample class, and recall that $\operatorname{NS}(X)_\mathbb Q$ is the intersection $H^{1,1}(X) \cap H^2(X,\mathbb Q)$, by the Lefschetz $(1,1)$-theorem. By hard Lefschetz and Poincaré duality (and since $- \cup H$ is compatible with both the Hodge decomposition and the $\mathbb Q$-structure), the pairing \begin{align*} \operatorname{NS}(X)_\mathbb Q \times \operatorname{NS}(X)_\mathbb Q &\to H^{2n}(X,\mathbb Q) \cong \mathbb Q\\ (\alpha, \beta) &\mapsto \alpha \cup H^{n-2} \cup \beta \end{align*} is a perfect pairing. Hence, $\operatorname{cl}(D) = 0 \in H^2(X,\mathbb Q)$ if and only if $D \cdot C = 0$ for all curves $C$. (The summary is that hard Lefschetz and Lefschetz $(1,1)$ give us enough curves to cup with.) We have the exact sequence $$0 \to H^1(X, \mathbb Z) \to H^1(X, \mathcal O_X) \to \operatorname{Pic}(X) \to H^2(X,\mathbb Z) \to H^2(X,\mathcal O_X) \to \ldots.$$ The map $\operatorname{Pic}(X) \to H^2(X,\mathbb Z)$ is the Chern class map (which is the same thing as the cycle class map in this case). Thus, any element in the kernel will come from $H^1(X, \mathcal O_X)/H^1(X, \mathbb Z)$. But that's exactly the identity component $\operatorname{Pic}^0_{X/\mathbb C}$. Write $D = D_1 - D_2$ as a difference of effective divisors. Then $D_1$ and $D_2$ live in the same component of $\operatorname{Pic}_{X/\mathbb C}$. Choose a curve connecting them. The universal family restricted to this curve is now a family of effective divisors from $D_1$ to $D_2$, showing that $D \sim_{\text{alg}} 0$. $\square$ Proof 2. (Over any (algebraically closed?) field for any Weil cohomology theory) By Riemann–Roch, we have $$\chi(X,\mathcal O(D)) = \int_X \operatorname{ch}(\mathcal O(D)) \cdot \operatorname{td}(T_X),$$ and similarly for $\mathcal O$. The point is that the right hand side is defined purely in terms of intrinsic geometry of $X$ and the intersection behaviour of $D$. Indeed, for a line bundle $\mathcal L$, we have $$\operatorname{ch}(\mathcal L) = 1 + c_1(\mathcal L) + \frac{c_1(\mathcal L)^2}{2} + \ldots + \frac{c_1(\mathcal L)^n}{n!};$$ we cup it with some fixed thing $\operatorname{td}(T_X)$, and we integrate (i.e. take the degree of the component in dimension $0$). This only depends on the intersection behaviour of $c_1(\mathcal L)$. We conclude that $$\chi(X,\mathcal O(D)) = \chi(X,\mathcal O),$$ since the intersections of $D$ and $0$ with any curve agree. Edit: An argument is missing here; I thought I fixed it but that created a different gap. In order to deploy the theory of Quot schemes, we have to write $\mathcal O$ and $\mathcal O(D)$ as quotient of the same vector bundle. For this, some boundedness argument is needed. This is carried out in characteristic $0$ in Lazarsfeld's Positivity in Algebraic Geometry I, Prop 1.4.37 (using Fujita's vanishing theorem), or in general in FGA Explained, Lemma 9.6.6 (using a variant of Mumford's argument for boundedness of the Hilbert scheme, cf. Lectures on curves on an algebraic surface, Lecture 14, Theorem). Now by the theory of Quot schemes, when we fix the Hilbert polynomial, the Quot scheme is projective. In particular, it has finitely many components, so a multiple of $\mathcal O(D)$ has to land in the identity component $\operatorname{Pic}^0_{X/k}$. Now write $D = D_1 - D_2$ with $D_1, D_2$ effective; then $D_1$ and $D_2$ lie in the same component of the Picard scheme. Choosing a curve connecting these two points in $\operatorname{Pic}^0_{X/k}$ gives a family of effective divisors from $nD_1$ to $nD_2$, showing that $nD \sim_{\text{alg}} 0$. $\square$ Remark. If you want to see this Riemann–Roch thing in action: just look at the formula of the Euler characteristic of a line bundle on a surface. The point is that the formula only depends on certain intersection numbers; we don't even need to know which intersection numbers. Remark. Why does this approach fail for higher codimension subvarieties? For instance, there are examples of irreducible subvarieties $Z_1, Z_2 \subseteq X$ that are numerically equivalent, but their Hilbert polynomials do not agree. In fact, you can even do this (exercise!) for smooth curves in $\mathbb P^3$, for which numerical equivalence coincides with algebraic (or even rational) equivalence! (If you are confused about this last example, note that cycles are parametrised by Chow schemes, not Hilbert schemes. So the fact that algebraically equivalent subvarieties can have different Hilbert polynomials is not a contradiction.) Although in this example the conclusion does hold, it shows that the method above cannot be generalised. Another reason is of course that the result $\sim_{\text{alg}} = \sim_{\text{num}}$ is not true in general. However, the equality $\sim_{\text{hom}, \mathbb Q} = \sim_{\text{num}}$ is not known; this is one of Grothendieck's standard conjectures on algebraic cycles. Remark. In the first proof, we need that the complex torus $\operatorname{Pic}^0_{X/\mathbb C} = H^1(X,\mathcal O_X)/H^1(X,\mathbb Z)$ is actually algebraic when $X$ is smooth projective. Proof 1 (analytic). Recall that a complex torus $V/U$ is projective (hence algebraic) if and only if there is a positive definite hermitian metric $h$ on $V$ such that $\operatorname{im} h$ is integral on $U \times U$ (see for example Mumford's book on abelian varieties, p. 33). To construct such a form for $V = H^1(X,\mathcal O_X)$ and $U = H^1(X,\mathbb Z)$, let $H$ be a hyperplane class, and set \begin{align*} h \colon V \times V &\to \mathbb C\\ (v,w) &\mapsto \frac{1}{4\pi^2 i} \cdot v \cup \bar w \cup H^{n-1}. \end{align*} Here, we view elements in $H^1(X,\mathcal O_X)$ as living in $H^1(X,\mathbb C) = H^1(X, \mathcal O_X) \oplus H^0(X,\Omega^1_{X/\mathbb C})$. Remember that $H^1(X,\mathbb Z)$ for our purposes maps to $H^1(X,\mathbb C)$ by multiplication by $2 \pi i$; this is different from the natural inclusion $H^1(X, \mathbb Z) \subseteq H^1(X, \mathbb C)$. Note that \begin{align*} h(w,v) &= \frac{1}{4 \pi^2 i} \cdot w \cup \bar v \cup H^{n-1} \\ &= \frac{1}{4 \pi^2 i} \cdot (-1) \cdot \bar v \cup w \cup H^{n-1} = \overline{h(v,w)}, \end{align*} since the cup product is graded commutative. Thus $h$ is hermitian. Moreover, for $v, w \in H^1(X,\mathbb Z)$, we have \begin{align*} h(v,w) &= \frac{1}{4\pi^2 i} \cdot (2\pi i v) \cup (-2\pi i w) \cup H^{n-1}\\ &= -i \cdot v \cup w \cup H^{n-1}. \end{align*} This is a purely imaginary algebraic integer, since $v \cup w \cup H^{n-1}$ is an integer. Finally, we need to check that $h$ is positive definite. This is an immediate consequence of the Hodge index theorem. $\square$ Proof 2 (algebraic, sketch). Construct the Picard scheme $\operatorname{Pic}_{X/\mathbb C}$ (a lot of work!), and prove that $\operatorname{Pic}^0_{X/\mathbb C}$ is projective, so it is a complex torus. Prove that it coincides with the complex torus $H^1(X,\mathcal O_X)/H^1(X,\mathbb Z)$. $\square$<|endoftext|> TITLE: How to find a permutation of [n] so that $\sum\{\min(i-l[i],r[i]-i)\}$ is maximized? QUESTION [7 upvotes]: Given a sequence $a_1, a_2,\dots,a_n$, define the two sequences $$l_i=\max_{1 \leq j < i, a_j \geq a_i} j$$ or $0$ if it does not exist; and $$r_i=\min_{i < j \leq n, a_j > a_i} j$$ or $n+1$ if does not exist. I want to find a permutation of $1, 2, ..., n$ so that $$\sum_{1 \leq i \leq n} \min(i-l_i,r_i -i)$$ is maximized. How can this be done? REPLY [4 votes]: Denote the maximum of your sum by $f(n)$, agree also that $f(0)=0$. If we fix $k$ such that $a_k=n$, then the maximal possible value is $\min(k,n+1-k)+f(k-1)+f(n-k)$. So, we get a recursive formula $$f(n)=\max_{1\leqslant k\leqslant (n+1)/2} k+f(k-1)+f(n-k).$$ Now we may forget about permutations and study this recurrence. Assume that $f(n)$ grows as $c(n+1)\ln(n+1)$ for some yet unknown $c$ and we try to prove that $f(n)\leqslant c(n+1)\ln(n+1)$ by induction (base $n=0$ is ok). Denote $k=\alpha(n+1)$, $\alpha\leqslant 1/2$, this reduces to $\alpha-cH(\alpha)\leqslant 0$ where $H(x)=-x\log(x)-(1-x)\log(1-x)$ is the entropy function. $H$ is concave, thus inequality $H(\alpha)\geqslant \alpha/c$ holds iff it holds for $\alpha=1/2$, $c=\frac1{2\log2}$, so $f(n)\leqslant (n+1)\log_2(n+1)$ and we should insert the maximal value $n$ to the middle of the permutation (if you do so, you get almost equality.)<|endoftext|> TITLE: Is there a perfect set of ground model reals in the Cohen extension? QUESTION [6 upvotes]: This question is motivated by the "interesting tidbit" in Hamkins' response here: https://mathoverflow.net/a/99025/10671, in which he demonstrates that, after Cohen forcing, there is a perfect set consisting entirely of mutually generic Cohen reals. In particular, this demonstrates that there is a perfect set in the extension containing no ground model reals. I'm interested in a related question and, while I feel it must have been answered, I can't find a reference. QUESTION. Consider the reals in the extension $V[c]$ obtained by forcing to add a Cohen real. Does there exist a perfect set in $V[c]$ consisting entirely of ground model reals? Or does every perfect set of reals in $V[c]$ contain a "new" real (added by the forcing)? REPLY [9 votes]: There is a more general result proved by Groszek and Slaman which says that if there is a nonconstructible real, then every perfect set has a nonconstructible element. The constructibility can be replaced with any inner models.<|endoftext|> TITLE: What are the $j$-invariants of the genus 1 modular curves? QUESTION [8 upvotes]: I believe there are only finitely many congruence subgroups $\Gamma\le SL_2(\mathbb{Z})$ such that the compactification of $\mathcal{H}/\Gamma$ is genus 1. Is there somewhere I can find a list of these genus 1 modular curves and look at their $j$-invariants (and ideally also reduction types over their field of definition). Here, I'm referring to the moduli-theoretic model of the elliptic curve, which is necessarily a quotient of some $X(n)$ (which is defined over $\mathbb{Q}(\zeta_n)$) and hence the $j$-invariant will lie in $\mathbb{Q}(\zeta_n)$. In particular I'm interested in whether or not you can say anything about the integrality of the $j$-invariants of these genus 1 modular curves, especially in the case when the congruence subgroup is torsion-free. For example, are the $j$-invariants all integral? Do they all have additive/multiplicative reduction? Do they have CM? REPLY [5 votes]: First, there is no comprehensive list of all models of modular curves of genus $1$. (There is a list of congruence subgroups of $SL_{2}(\mathbb{Z})$ here.) Many cases have been computed, including the $X_{0}(n)$, $X_{1}(n)$, $X(n)$, as well as many cases that have prime power level. (With David Zureick-Brown, I did all of the cases whose level is a power of $2$ that might have rational points. See our paper here). Another really handy tool is the code that David Zywina has written (see the paper here) that will count points on any modular curve $X_{G}$ (provided $G \subseteq GL_{2}(\mathbb{Z}/n\mathbb{Z})$ has surjective determinant). This will allow one to determine the isogeny class of ${\rm Jac}~X_{G}$, but this is not quite enough to determine the $j$-invariant. Based on the examples that are known, it is not clear to me there is a relationship between the integrality of the $j$-invariant and the torsion structure. Here are some examples: $X_{0}(11): y^{2} + y = x^{3} - x^{2} - 10x - 20$, $j$-invariant is $-122023936/161051$ is not integral, but the subgroup $\Gamma_{0}(11)$ is torsion-free (well, its image in ${\rm PSL}_{2}(\mathbb{Z})$ is). $X_{0}(17): y^{2} + xy + y = x^{3} - x^{2} - x - 14$, $j$-invariant is $-35937/83521$ is not integral, and the subgroup $\Gamma_{0}(17)$ is not torsion-free (because it has elliptic elements of order $2$). $X_{0}(27): y^{2} + y = x^{3} - 7$, $j$-invariant is zero (and the curve is isomorphic to the Fermat cubic). The subgroup is torsion-free. $X_{0}(49): y^{2} + xy = x^{3} - x^{2} - 2x - 1$, $j$-invariant is $-3375$ (and has CM by the ring of integers in $\mathbb{Q}(\sqrt{-7})$). The subgroup is not torsion-free, as it contains elliptic elements of order $3$. Many of these curves have CM. ($X_{0}(32)$, $X_{0}(36)$, $X_{0}(49)$, $X_{{\rm ns}}^{+}(11)$, $X(6)$, most, but not all of the $2$-power level genus $1$ curves.)<|endoftext|> TITLE: Is the Gromov conjecture still open? QUESTION [16 upvotes]: Today I read about Gromov's definition of minimal volume for smooth manifolds. $$\min {\rm Vol}(M):=\inf_{|K_g|\leq1}\{{\rm Vol}(M,g)\}.$$ Gromov's conjecture states that for every closed simply connected odd-dimensional manifold $\min {\rm Vol}(M)=0$. Is the Gromov conjecture still open? Can anybody give an example for this conjecture? Update: My teacher told to me that this conjecture can be solved by the Ricci flow method but I don't know how to use it. Can anybody give me an explanation to this? and how it works? Thanks. REPLY [11 votes]: According to wikipedia, the conjecture is still open. I do not see at all how RF can help here. (Apart from dimension 3 when the statement is of course a corollary of Perelman's geometrization theorem.) There is no positivity assumption on curvature(s) of $M$ in this conjecture and the Ricci Flow in higher dimensions is mostly a mystery without some positivity assumptions. From what you wrote, it sounds like your teacher has no idea how to approach this problem via Ricci Flow either; I suggest you work on something else, more doable. One more thing: The conjecture is only in odd-dimensional case. For even dimensional manifolds Gromov noted in his paper that there is a bound $MinVol(M^n)\ge c_n|\chi(M)|$, $c_n>0$. Hence, all even-dimensional spheres have positive minimal volume. Edited: (4 September 2017) An Example of vanishing minimal volume which is due to Gromov is three-sphere. (see Gromove paper,section 0.4)<|endoftext|> TITLE: Properties of the petit Zariski topos QUESTION [10 upvotes]: What are some (intrinsically formulated) properties of the locally ringed topos $(\mathbf{Sh}(X),\mathcal{O}_X)$ for some scheme $X$, which do not hold for arbitrary locally ringed toposes? Is there, perhaps, even an intrinstic characterization of those locally ringed toposes which are equivalent to $(\mathbf{Sh}(X),\mathcal{O}_X)$ for some scheme $X$? REPLY [11 votes]: Unfortunately I don't know an interesting intrinsically formulated sufficient criterion for a locally ringed topos to be the little Zariski topos of a scheme. This is an extremely interesting question! There are necessary conditions, for instance (formulated in the internal language of the topos): For any element $f : \mathcal{O}_X$: If $f$ is not invertible, then $f$ is nilpotent. While this condition does exclude some locally ringed toposes, it doesn't exclude the locally ringed topos given by a smooth manifold (if a smooth function has the property that the only open subset on which it is (multiplicately) invertible is the empty set, then it is zero). On a reduced scheme, where it holds that $\mathcal{O}_X$ is internally a reduced ring, a consequence of this condition is that $\mathcal{O}_X$ is "$\neg\neg$-separated": For any $f : \mathcal{O}_X$, if $\neg\neg(f = 0)$, then in fact $f = 0$. Update: There is a further, but more convoluted, property which is enjoyed by the little Zariski topos of any scheme but not in general by the topos given by a smooth manifold: For any element $f : \mathcal{O}_X$ the localized module $\mathcal{O}_X[f^{-1}]$ is a sheaf with respect to the internal modality $\Box$, where $\Box\varphi :\equiv (\text{$f$ invertible} \Rightarrow \varphi)$. This condition can be rephrased so as to not refer to internal modalities: For any element $f : \mathcal{O}_X$ it holds that: For any $s : \mathcal{O}_X$ such that $\text{$f$ invertible} \Rightarrow s = 0$, there is a natural number $n$ such that $f^n s = 0$. For any subset $K \subseteq \mathcal{O}_X$ such that $\text{$f$ invertible} \Rightarrow \text{$K$ is a singleton}$, there is a natural number $n$ and an element $s : \mathcal{O}_X$ such that $\text{$f$ invertible} \Rightarrow f^{-n} s \in K$. The first part is satisfied by (the topos of sheaves over a) manifold, but the second part is not. A counterexample is $X = \mathbb{R}^1$, $f(x) = x$, $K = \text{"$e^{1/x}$ on $x \neq 0$"}$. In fact a manifold satisfies this condition if and only if it is empty. The big Zariski topos of a scheme $X$, that is approximately the topos of sheaves on the Zariski site $\mathrm{Sch}/X$ with structure sheaf $\mathbb{A}^1 : T \mapsto \mathcal{O}_T(T)$, has the following special properties, which are more helpful in distinguishing it from other locally ringed toposes: For any element $f : \mathbb{A}^1$: If $f$ is not zero, then $f$ is invertible. For any finitely presented $\mathbb{A}^1$-algebra $A$, the canonical map $A \to \mathrm{Hom}(\mathrm{Hom}_{\mathbb{A}^1\mathrm{-Alg}}(A, \mathbb{A}^1), \mathbb{A}^1)$ is an isomorphism of $\mathbb{A}^1$-algebras. In the special case that $A = \mathbb{A}^1[T]$, the second condition yields Any map $\mathbb{A}^1 \to \mathbb{A}^1$ is given by a unique polynomial in $\mathbb{A}^1[T]$. which somehow expresses the vague intution that "in algebraic geometry, every morphism is a polynomial".<|endoftext|> TITLE: Cartan formula for Steenrod squares on the cochain level QUESTION [8 upvotes]: Steenrod originally defined his squares using explicit cochain-level formulas for simplicial mod-2 cochains. To this end, he introduced higher cup products, which control the failure of the usual cup product to be supercommutative on the cochain level. On the other hand, H. Cartan formula tells us how Steenrod squares relate to the usual cup product of mod-2 cohomology classes. Questions. (1) How does one show this using Steenrod's definition of Steenrod squares? (2) Is there some cochain-level relation between cup products of various degrees which implies Cartan's formula? REPLY [8 votes]: I think that what you are searching for is explained in P. May's paper "A general algebraic approach to Steenrod operations". In this paper you will find a very general treatment of cochain level cup-i product operations. In particular P. May introduces the concept of Cartan object and shows that the cohomology of Cartan object comes equipped with the action of Steenrod operations that satisfy the Cartan relations (see proposition 2.6). All the computations are made at the cochain level. The algebra of singular cochains of a topological space is a typical example of a Cartan object. Moreover, we have an operadic interpretation: take an algebra $A$ over an $E_{\infty}$-algebra, then you have evaluation products $$\mu_r:E_{\infty}(r)\otimes A^{\otimes r}\rightarrow A$$ where each $E_{\infty}(r)$ is a projective resolution of the symmetric group $\Sigma_r$. The evaluation product $\mu_2$ is risponsible for the action of cup-i products. If you take for $E_{\infty}$ the Baratt-Eclles operad $E$ it is a Hopf operad you have a coassiociative diagonal of operads $$\Delta:E\rightarrow E\otimes E.$$ Cartan relations come from the Hopf operad structure AND the fact that $E$ is the resolution of the commutative operad $Com$. Cartan relations can be detected in the complex $E(4)$.<|endoftext|> TITLE: $\Theta$-Sets and Higher-QuasiCategories QUESTION [9 upvotes]: In his well-known paper "Disks, Duality and $\Theta$-Categories, Joyal defines at the very end a $\Theta$-Category to be a cellular set suitably "fibrant". I was wondering whether someone has worked on this proposed model for weak $\omega$-Category, and its relationship with Dimitri Ara's Higher-QuasiCategories and Harry Gindi's attempt of putting a model structure on cellular sets. A question could be, for instance: is it the case that the localizer for higher-quasicategories produces the same fibrant objects, namely those with the inner-horns filling property? Any idea or comment is welcome! REPLY [7 votes]: I think my preprint answers this question in the appendices when I compare the horizontal Joyal model structure on $\Theta[C]$ sets with the Rezk model model structure for $\operatorname{Se}_C\cup \operatorname{Cpt}_C$-local Reedy-fibrant simplicial presheaves. If you unwind what the vertical model structure localizer is, you can extract from this that all of the Oury inner horn inclusions are local in Ara's model structure (by induction on $C=\Theta_n$). It's somewhat easy to see that you still need the higher dimensional "completeness maps". Yuki Maehara said he has full results showing that even "bigger" horns than the Oury horns generate the inner anodynes as well, specifically the horns suggested by Joyal and Berger (omitting exactly one codimension 1 subobject), but I don't think he has made his preprint available yet. A more useful heuristic argument to see why Joyal's proposed model structure doesn't work is that since all strict $\omega$-categories are fibrant in that model structure (they all admit unique inner horn fillers) and the embedding of strict $\omega$-categories is full in cellular sets, it would mean that every pseudofunctor between strict $\omega$-categories is naturally (not pseudonaturally) isomorphic to a strict one, which is trivially false. Edit: Here is Yuki's paper, which came out a bit after I gave this answer. It only covers the case n=2, but I can't imagine that this is too hard to generalize to higher n.<|endoftext|> TITLE: Higgs bundles and stable vector bundle QUESTION [7 upvotes]: Let $\mathcal M_X(r,0,K_X)$ be tha moduli space of semistable Higgs bundles over a smooth irreducible algebraic curve over $\mathbb C$. And let $E$ be a stable vector bundle of rank $r$ and degree $0$. Then clearly we have an injection $$H^0(X,E\otimes E^*\otimes K_X)\rightarrow \mathcal M_X(r,0,K_X)$$ Is this map a closed immersion? Is it proper? Thanks REPLY [3 votes]: This is a partial answer to my question! We can deduce easly that if this map is closed immersion, then $E$ is very stable vector bundle (because in this case, the Hitchin map is finite). Hence if $E$ is stable non very stable vector bundle (there exist of course such vector bundles) then we deduce that the above map couldn't be closed immersion.<|endoftext|> TITLE: Early Papers and Books on Math Logic QUESTION [5 upvotes]: I would like to go back to the beginnings of math logic and read selected papers throughout the years connecting the beginnings to today. My first stop was Wikipedia where it claims that math logic basically begins with De Morgan and Boole. I have gotten Boole's "The Mathematical Analysis of Logic" to read. It seems De Morgan's book is either expensive on Amazon or is a scan of an old enough edition that I can't read it. I have three questions: First, can I ignore De Morgan as a starting place? Second, should I be looking into anyone earlier or contemporary to Boole? Third, though my main is to study the development of mathematical logic, I would also like to learn the philosophical traditions a little bit. Is there anything that would be good to read that maybe isn't quite mathematical in nature? REPLY [4 votes]: It seems that the references mentioned in the comments collectively answer the question, so I'm compiling them into an actual (community wiki) answer so that the system recognizes that the question has been answered. Handbook of the History of Logic. There are several anthologies of key original papers: From Frege to Gödel and two volumes of From Kant to Hilbert, and From Brouwer to Hilbert also. Two secondary sources are worth consulting: Kneale and Kneale's The Development of Logic (Oxford: Clarendon-Oxford University Press, 1962) and Grattan-Guinness's The Search for Mathematical Roots, 1870-1940: Logics, Set Theories and the Foundations of Mathematics from Cantor through Russell to Gödel (Princeton, New Jersey: Princeton University Press, 2000). Both contain helpful pointers to primary texts. You would do worse than to consult references like the Handbook of Mathematical Logic, which should have other references that cover the early history. If you are looking for elementary treatments, you might enjoy Lewis Carroll's Game of Logic. Symbolic Logic may also be a good read.<|endoftext|> TITLE: How do I evaluate this sum :$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n} $? QUESTION [9 upvotes]: I have tried evaluating this series $$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n} $$ using some methods but it's seems to me that it is very hard. However, I noticed that the series converges faster than the Riemann series. My question here is: Is there some mathematical technique for evaluating the above series? Note1: Here, $H_n$ denotes the harmonic numbers. Edit : I have a wrong type I meant in the denomenator $2^n$ Thank you for any help. REPLY [7 votes]: Let $g(a)=\frac{a}{1-a}, f(a)=\log(1-a)$ and $h(a)=\log^2(1-a)$. Let $\square^m$ denote a $m$-dimensional unit hypercube. The following is an application or reformulation of the above infinite series evaluation. $$\int_0^1\int_0^1\frac{h(\frac12xz)-xh(\frac12z)+xzh(\frac12)-zh(\frac12x)}{x(1-x)z(1-z)}dxdz= \frac{\log^42}{12}-\frac{\pi^4}{144}+\zeta(2)\log^22+\zeta(3)\log2.$$ Proof outline: Since $\frac1k=\int_0^1x^{k-1}dx$, we may write \begin{align} \sum_{n\geq1}\frac{t^nH_n^3}{n+1}&=\sum_{n\geq1}t^n\int_0^1w^ndw\sum_{i,j,k=1}^n\int_{\square^3}x^{i-1}y^{j-1}z^{k-1}dxdydz \\ &=\int_{\square^4}dxdydzdw\sum_{n\geq1}(tw)^n\sum_{i,j,k=1}^nx^{i-1}y^{j-1}z^{k-1}\\ &=\int_{\square^4}\frac{dxdydzdw}{(1-x)(1-y)(1-z)}\sum_{n\geq1}(tw)^n(1-x^n)(1-y^n)(1-z^n)\\ &=\int_{\square^4}dxdydzdq\frac{g(tw)-g(twy)-g(twx)+g(twxy)-g(twz)+g(twyz)+g(twxz)-g(twxyz)}{(1-x)(1-y)(1-z)}. \end{align} Next, make a repeated use of the integral evaluations $$\int_0^1\frac{g(tw)-g(twy)}{1-y}dy=-\frac{f(tw)}{1-tw}$$ and $$\int_0^1\left(\frac{f(twx)}{1-twx}-\frac{f(tw)}{1-tw}\right)\frac{dw}{1-w}= \frac{xh(t)-h(tx)}{2tx}$$ to arrival at the double integral upon replacing $t=\frac12$. The rest follows from the series evaluation algorithm that Julian Rosen alluded to. $\square$ Remark. This approach actually could prove (directly) the easier-looking $$\sum_{n\geq1}\frac{H_n^2}{(n+1)2^n}=\frac13\log^32+\zeta(2)\log2-\frac12\zeta(3).$$<|endoftext|> TITLE: Subconvexity bounds and zero-free regions QUESTION [5 upvotes]: When I see results in analytic number theory, I often have trouble seeing how they relate and their relative strength. Here is a specific question that should help me (and hopefully others too). Here is my understanding, please correct anything that is wrong here. An $L$ function $L(s)$ comes with an analytic conductor $q$. I will also write $s = \sigma + it$ ($\sigma,t\in\mathbb{R}$). We believe that it satisfies a Riemann hypothesis: $L(s)\neq 0$ whenever $\sigma>\frac12$ and a Lindelöf hypothesis: $|L(1/2+it)|\ll |qt|^\epsilon$. Ignoring real zeroes, we have bounds towards the Riemann hypothesis, i.e. zero-free regions of the form: $L(s)\neq 0$ whenever $\sigma > 1-\frac{c}{\log(q(1+|t|))}$ and $t\neq 0$, and sometimes improved zero-free regions $L(s)\neq 0$ whenever $\sigma > 1-\frac{c}{\log(q(1+|t|))^a}$ and $t\neq 0$, for some $a<1$. We also have bounds towards the Lindelöf hypothesis, i.e. convexity bounds $L(1/2+it) \ll |qt|^{1/4}$, and sometimes subconvexity bounds, i.e. $L(1/2+it) \ll |qt|^{1/4-\delta}$ for some $\delta>0$. Both zero-free regions and subconvexity bounds are related to bounding the values of the $L$-function in the critical strip and to bounding certain sums (character/exponential/...). However, I don't see very well how they relate: for instance, I found out that in the $q$-aspect for Dirichlet $L$-functions, subconvexity bounds are known (Burgess) but improved zero-free regions are not (according to the answer to this question). Can you clarify the relations between: bounds of various sums subconvexity bounds zero-free regions? Feel free to separate the $t$-aspect from the $q$-aspect if you think that it will clarify the answer. REPLY [11 votes]: Let's just discuss the $t$-aspect, i.e. bounds for the zeta function and its zeroes. Let $T$ be a large ordinate, and let $H$ be a medium-sized quantity (much larger than $1$, but much less than $T$). Roughly speaking, subconvexity bounds at ordinates $t = T + O(H)$ measure the average failure of the Riemann hypothesis in this region, whilst zero-free regions at these ordinates measure the worst-case failure of the Riemann hypothesis here. Indeed, consider a rectangular contour with corners $\frac{1}{2}+\varepsilon+iT$, $\frac{1}{2}+\varepsilon+i(T+H)$, $2 + i(T+H)$, and $2+iT$ for some very small $\varepsilon>0$. Suppose first that there are no zeroes of $\zeta$ in this rectangle, so we have a holomorphic branch of $\log \zeta$ in this rectangle. By Cauchy's theorem, the integral of $\log \zeta$ on this rectangle is zero. For medium-sized $H$ (and perturbing $T$ and $H$ slightly if needed), the contribution of the short horizontal edges of the rectangle are small, and from elementary bounds the contribution of the right vertical edge is $O(H)$. Sending $\varepsilon \to 0$, and taking real parts, we conclude (heuristically at least) that $$ \frac{1}{H} \int_T^{T+H} \log |\zeta(\frac{1}{2}+it)|\ dt = o( \log T )$$ (in fact we can get a better error term here than $o(\log T)$). This is a cousin of the well known fact that the Riemann hypothesis implies the Lindelof hypothesis. Now suppose we have some zeroes $\sigma_j + i t_j$ in the rectangle. To maintain a holomorphic branch of $\log \zeta$ in the rectangle, we need to cut out some slits connecting $\frac{1}{2}+it_j$ to $\sigma_j + it_j$. If one does so and performs the usual contour integration calculations, one eventually arrives (heuristically at least) at the Jensen-type formula $$ \frac{1}{H} \int_T^{T+H} \log |\zeta(\frac{1}{2}+it)|\ dt = \frac{2\pi }{H} \sum_{T \leq t_j \leq T+H} (\sigma_j-\frac{1}{2})_+ + o( \log T ).$$ Recall from the Riemann-von Mangoldt formula that there are about $\frac{H}{2\pi} \log T$ zeroes $\sigma_j+it_j$ with $T \leq t_j \leq T+H$; by the functional equation, they are symmetric around the critical line. The expression $ \frac{2\pi }{H} \sum_{T \leq t_j \leq T+H} (\sigma_j-\frac{1}{2})_+$ is thus at worst about $\frac{1}{4} \log T$, which is attained when about half of the zeroes have abscissas $\sigma_j$ close to 1, and the other half have abscissas close to zero. This is basically a slightly averaged form of the convexity bound $\zeta(\frac{1}{2}+iT) \ll T^{1/4+o(1)}$. More generally, a subconvexity bound of the form $\zeta(\frac{1}{2}+iT) \ll T^{\alpha+o(1)}$ is basically asserting that near the ordinate $T$, the quantities $(\sigma_j-\frac{1}{2})_+$ have average value at most $\alpha+o(1)$, thus as mentioned previously subconvexity controls the average failure of the Riemann hypothesis. In particular, if the Lindelof conjecture is true, $\alpha=0$ and most of the zeroes at ordinates near $T$ are within $o(1)$ of the critical line (this observation is I believe due to Backlund). Meanwhile, zero-free regions are upper bounds on $(\sigma_j-\frac{1}{2})_+$, that is to say worst-case failure of the Riemann hypothesis near a given ordinate $T$. The worst-case bounds the average-case, so a really good zero-free region does give some subconvexity bound, but this is extremely inefficient (in particular, the best zero-free region currently available unconditionally does not come anywhere close to implying the best subconvexity bounds currently known). I like to formalise the above discussion using limiting profiles of the log-zeta function: https://terrytao.wordpress.com/2015/03/01/254a-supplement-7-normalised-limit-profiles-of-the-log-magnitude-of-the-riemann-zeta-function-optional/ . I think Titchmarsh's book also has some discussion of these topics. Finally, $\zeta(\frac{1}{2}+it)$ can be related to partial sums $\sum_{n \leq N} \frac{1}{n^{1/2+it}}$ by elementary arguments (or the approximate functional equation), and by dyadic decomposition and summation by parts these partial sums can in turn be related to exponential sums such as $\frac{1}{N^{1/2}} \sum_{N \leq n \leq 2N} \frac{1}{n^{it}} = \frac{1}{N^{1/2}} \sum_{N \leq n \leq 2N} e( - \frac{t}{2\pi} \log n )$. A subconvexity bound $\zeta(\frac{1}{2}+iT) \ll T^{\alpha+o(1)}$ is basically the same thing as an exponential sum bound $\sum_{N \leq n \leq 2N} e( - \frac{t}{2\pi} \log n ) \ll N^{1/2} T^{\alpha+o(1)}$ (see e.g. these notes of mine for some more rigorous statements of this form).<|endoftext|> TITLE: An equivariant social choice in Mathematical economics QUESTION [10 upvotes]: Motivated by this paper and its economics motivations, we recall that a social choice among $n$ objects is a continuous function $$f:\overbrace{M\times M\times\cdots\times M}^{\text{$n$ times}}\to M$$ which satisfy the following conditions: 1) $f(x,x,\ldots,x)=x$ 2) $f$ is unchanged under all permutations $\sigma \in S_{n},$ the symmetric group on $n$ elements. Perhaps the following could be considered as a social model for the above mathematical problem. A client has to choose one item among $n$ items $(x_{1},x_{2},\ldots,x_{n})$. His preference function is denoted by the above $f$. Now it is natural that we assume that the client faces with n different items, so $x_{i} \neq x_{j}$, $\forall i \neq j$. So we consider the ordered configuration space $$F_{n}(M)=\{(x_{1},x_{2},\ldots,x_{n})\in M^{n}\mid x_{i} \neq x_{j},\;\forall i \neq j\} $$. There is an obvious action of the symmetric group $S_{n}$ on $F_{n}(M)$. Now the following question can be counted as an equivariant analogy of the social choice problem: Assume that $M$ is a manifold which is acted by the symmetric group $S_{n}$. Is there always an equivariant continuous map $f:F_{n}(M)\to M$? If not, for what type of manifolds the answer is affirmative? What type of algebraic topological obstructions would appear? Can we find an economics interpretation for this equivariant version? Example: For $M=S^{1},\; n=2$, the symmetric group $S_{2}$ can act on $S^{1}$ by complex conjugation $z\mapsto \bar{z}$ Then the function $f(z,w)=z\bar{w}$ satisfies $f(z,w)=\overline{f(w,z)}$. In this example let's replace the conjugate action by antipodal action. Then $f(z,w)= (z-w)/|z-w|$ is a map $f:F_{2}(S^{1})\to S^{1}$ which satisfy $f(z,w)=-f(w,z)$? REPLY [4 votes]: I'll attempt an answer to the mathematical question, without discussing the motivation. As I understand it, $M$ is a manifold with $S_n$-action, and we are asking whether there exists an $S_n$-equivariant map $f:F_n(M)\to M$, where the action on $F_n(M)$ is by permutation of coordinates (and in particular has nothing to do with the action on $M$). As a first observation, note that if $x\in M$ is a fixed point $x$, then such an $f$ exists; just map everything to $x$. A general necessary condition for the existence of a $G$-map $f:X\to Y$ between $G$-spaces is given by the Faddell-Husseini index, as decribed for example in this paper: Pavle V. M. Blagojević, Wolfgang Lück, Günter M. Ziegler, Equivariant Topology of Configuration Spaces, J. Topology 8 (2015) pp 377–413, doi:10.1112/jtopol/jtu029, arXiv:1207.2852. Given a $G$-space $X$ and a commutative ring $R$, the Faddell-Husseini index of $X$ is the ideal in $H^*(BG;R)$ defined by $$ \operatorname{Index}_G(X;R):=\ker(p^*:H^*(BG;R)\to H^*(EG\times_G X;R)), $$ where $p:EG\times_G X\to BG$ is the projection of the Borel fibration. Then it is easy to see that if there exists a $G$ map $f:X\to Y$ then $$ \operatorname{Index}_G(Y;R)\subseteq \operatorname{Index}_G(X;R) $$ must hold. You can sometimes rule out existence of equivariant maps using this property. The index of the configuration space $F_n(M)$ is probably difficult to compute in general, but is done in the linked paper for $M=\mathbb{R}^d$.<|endoftext|> TITLE: Piecewise definition of a functor QUESTION [9 upvotes]: I am in the following situation: I have a category $A$ and an increasing chain of full subcategories $A_0\subseteq A_1\subseteq\, ...\subseteq A_n\subseteq \, ...$ inside $A$, such that any object of $A$ belongs to some $A_n$. Roughly speaking, $A$ is the union of the $A_n$. I would like to construct a functor from $A$ to a category $B$, and what I can do is the following: for any $n\in \mathbb N$ I can construct a functor $F_n\colon A_n\to B$ and prove that the restriction of $F_{n+1}$ to $A_n$ is naturally isomorphic to $F_n$. The question is, can we use these functors $F_n$ to construct a functor $F\colon A\to B$ in some canonical way? Of course one basic requirement would be that $F(a)\cong F_n(a)$ if $a\in A_n$. I think I can give a construction, but it is quite tedious and definitely it is not nice to give this construction in the context I am writing. So, is there any reference I can quote? Or, do you have any one-line argument for the existence of $F$? REPLY [8 votes]: Basically you want to prove that $A = \mathrm{colim}_n A_n$ in the $2$-categorical sense. There is a general construction of $2$-colimits; we may use this and check that it is equivalent to $A$. But I think it is a good idea to just wrote down the functor. Let us denote by $\theta_n : F_n \to F_{n+1}|{A_n}$ the given isomorphisms. These may be composed to compatible isomorphisms $\theta_{n,m} : F_n \to F_m |{A_n}$ for $n \leq m$. Define $t : \mathrm{Ob}(A) \to \mathbb{N}$ by $t(a) = \min\{n \in \mathbb{N} : a \in A_n\}$. Define $t(a,b) := \max(t(a),t(b))$. The functor $F : A \to B$ is defined as follows: An object $a \in \mathrm{Ob}(A)$ is mapped to $F(a):=F_{t(a)}(a)$. A morphism $f:a \to b$ is mapped to the composition $$F(f) : \quad F_{t(a)}(a) \xrightarrow{~\large \theta_{t(a),t(a,b)}(a)~} F_{t(a,b)}(a) \xrightarrow{~\large F_{ t(a,b)}(f)~} F_{t(a,b)}(b) \xrightarrow{~\large \theta_{ t(a),t(a,b)}^{-1}(b)~} F_{t(b)}(b).$$ It is easy to check that $F$ is, in fact, a functor. If $a \in A_n$, then $t(a) \leq n$. Hence $\theta_{t(a),n} : F(a)=F_{t(a)}(a) \to F_n(a)$ is an isomorphism. It is natural in $a$; this follows from the construction of $F(f)$. Hence, there is an isomorphism $\vartheta_n : F|{A_n} \cong F_n$. Moreover, these isomorphisms are compatible with the $\theta_n$: The diagram $$\begin{array}{cc} F|A_n & \xrightarrow{\large \vartheta_n} & F_n\\ \mathrm{id}\downarrow ~~ && ~~\downarrow \theta_n \\ F|A_{n+1}|A_n & \xrightarrow{\large \vartheta_{n+1}|A_n} & F_{n+1}|A_n\end{array}$$ commutes.<|endoftext|> TITLE: A proof without derivatives that a real polynomial of degree $n$ has at most $n-1$ local extrema QUESTION [10 upvotes]: This question is about math education and is not research level, so do not hesitate to delete it if it feels inappropriate. I already asked it here a year ago: https://math.stackexchange.com/questions/1401938/a-real-polynomial-of-degree-n-cannot-have-more-than-n-1-local-extrema-a-p but although Michael Hardy had some ideas about it, I think someone might have a more general argument. I teach elementary mathematics to engineers, and I'd like to prove the theorem in the title: a real polynomial of degree $n$ has at most $n-1$ local extrema. If you use derivatives, it's basically a one liner, but since I did not introduce them, I'd like to know if there is a slick proof that uses only elementary facts about polynomials. (Subtleties such as the intermediate value theorem can be swept under the rug.) Actually, I found one proof that uses derivatives only in disguise, but it is not really satisfying as it feels very 'ad hoc'. Let me give it so that it does not come up in an answer. First, define a formal derivative $p'(x)$ of a given polynomial $p(x)$ by its action on the monomials, that is: $a\cdot x^k$ becomes $ka\cdot x^{k−1}$. (It seems that it is exactly what Rolle did when proving the first version of Rolle's Theorem in 1691, which predates calculus and was done only for polynomials.) Then, it is easy to show that this formal derivative obeys the usual product rule and that $p'(x)=q'(x)$ iff $q(x) = p(x) + b$ with $b\in\mathbb{R}$. Now, $a\in\mathbb{R}$ is a local extremum of $p(x)$ iff $a$ is a root of even multiplicity of $p(x) - p(a)$, so $p(x)-p(a) = (x-a)^2\cdot q(x)$ for some $q(x)$. By the (formal) product rule, $p'(a)=0$. Since $p'(x)$ has degree $n-1$, there are no more than $n-1$ local extrema. So, does anyone has a proof without derivatives, disguised or not ? REPLY [16 votes]: From your post it seems you are permitted to use the following: $a\in\mathbb{R}$ is a root of $p(x)$ (i.e. $p(a) = 0$) iff $p(x) = (x-a)q(x)$ for some polynomial $q(x)$. $a\in\mathbb{R}$ is a local extremum of $p(x)$ iff (thanks to Ilya Bogdanov; consider e.g. $p(x) = x^3$) only if $p(x)-p(a) = (x-a)^2q(x)$ for some polynomial $q(x)$. A proof using these two facts is as follows: write $p(x) = \sum_{k=1}^n c_kx^k$. Then you can write $p(x) - p(a) = (x-a)Q(x,a)$ for some polynomial $Q$ in two variables. But then by those two facts, if $a\in\mathbb{R}$ is a local extremum, then $q(a) = 0$, where $q(x) := Q(x,x)$. You can explicitly show (e.g. by computing the $(n-1)$-degree term of $q$) that it is a non-zero polynomial of degree $n-1$. This can also serve as a motivation to introduce/define the derivative.<|endoftext|> TITLE: $q$-Eulerian type B enjoy symmetry QUESTION [7 upvotes]: Let $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q;q)_0:=1$. Define a $q$-exponential by $$e(z;q)=\sum_{n\geq0}\frac{z^n}{(q;q)_n}.$$ There is a notion of $q$-Eulerian polynomials, see the reference. I like to introduce $q$-Eulerian polynomial of type B via the generating function $$\sum_{n\geq1}B_n(t,q)\frac{z^n}{(q;q)_n}=\frac{(e(z;q)-e(tz;q))(e(tz;q)+te(z;q))}{e(2tz;q)-te(2z;q)}.$$ Now, expand $B_n(t,q)$ as a polynomial $$B_n(t,q)=\sum_{k=0}^nB_{n,k}(q)t^k$$ and call $B_{n,k}(q)$ $q$-Eulerian numbers type B. Here are the first few terms: \begin{align} B_1(t,q)&=1+t, \\ B_2(t,q)&=1+(2q+4)t+t^2, \\ B_3(t,q)&=1+(7q^2+7q+9)t+(7q^2+7q+9)t^2+t^3. \end{align} Claim. if $a, b\in\Bbb{N}$ and $\alpha=a+b+1$, then the symmetric relation holds: $$\binom{\alpha}a_q+\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,b}(q)= \binom{\alpha}b_q +\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,a}(q).$$ QUESTIONS: (a) I don't have a proof for my claim which seems very true though. Do you? (b) Is there a combinatorial interpretation for these polynomials $B_n(t,q)$ or the Eulerian numbers $B_{n,k}(q)$? You might be inspired by the reference. REPLY [3 votes]: $\bf{Step~1}.$ $B_{n,a}(q)=B_{n,n-a}(q)$. $\it{Proof}$. Write $$ \sum_{n\geq1}\dfrac{B_n(t,q)}{t^{n/2}}\frac{z^n}{(q;q)_n}=\frac{e(z/\sqrt{t};q)-e(z\sqrt{t};q)}{\dfrac{e(2z\sqrt{t};q)}{\sqrt{t}}-\sqrt{t}~e(2z/\sqrt{t};q)}\left(\dfrac{e(z\sqrt{t};q)}{\sqrt{t}}+\sqrt{t}~e(z/\sqrt{t};q)\right). $$ The rhs is symmetric under $t\to 1/t$, so $\dfrac{B_n(t,q)}{t^{n/2}}$ is also symmetric: $$ \dfrac{B_n(t,q)}{t^{n/2}}=t^{n/2}{B_n(1/t,q)}.\tag{1} $$ We know that $B_n(t,q)=B_{n,0}t^0+...$ starts with the power $t^0$. Due to $(1)$ this means that the highest power of $t$ in the polynomial $B_n(t,q)$ is $n$, and $B_{n,a}(q)=B_{n,n-a}(q)$ as required. $\bf{Step~2}.$ $\displaystyle{\binom{\alpha}a_q+\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,b}(q)= \binom{\alpha}b_q +\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,a}(q)},~ a,b,\in \mathbb{N},~a+b+1=\alpha.$ $\it{Proof}$. According to q-binomial theorem $e(z;q)=\sum_{n\geq0}\frac{z^n}{(q;q)_n}=\frac{1}{(z;q)_\infty}$. Now we equate the coefficients of $z^n$ in both sides of \begin{align} \left(\sum_{l\geq0}\frac{(2tz)^l}{(q;q)_l}-t\sum_{l\geq0}\frac{(2z)^l}{(q;q)_l}\right)\cdot\sum_{k\geq1}B_k(t,q)\frac{z^k}{(q;q)_k}={(e(z;q)-e(tz;q))(e(tz;q)+te(z;q))}\\ =\left(\sum_{l\geq0}\frac{z^l}{(q;q)_l}-\sum_{l\geq0}\frac{(tz)^l}{(q;q)_l}\right)\left(\sum_{l\geq0}\frac{(tz)^l}{(q;q)_l}+t\sum_{l\geq0}\frac{z^l}{(q;q)_l}\right) \end{align} and obtain $$ \sum _{k=1}^n \binom{n}{k}_q\frac{B_k(t,q) \left(t^{n-k}-t\right) }{2^{k-n}}-\sum _{k=1}^n \left(1-t^k\right) \left(t^{n-k}+t\right) \binom{n}{k}_q=0. $$ The coefficient of $t^{a+1}$ in the above formula is (note that for $a\in \mathbb{N}$ we have $a+1\neq1$ and for $b=n-a-1\in \mathbb{N}$ we have $a+1\neq n$, so the 'boundary' terms with $t$ and $t^n$ don't make any contribution in the formula below) $$ -\sum _{k=1}^n\binom{n}{k}_q \frac{B_{k,a}(q) }{2^{k-n}}+\sum _{k=1}^n \binom{n}{k}_q\frac{B_{k,a+1+k-n}(q)}{2^{k-n}}+\binom{n}{a}_q-\binom{n}{b}_q=0. $$ Finally, taking into account $B_{k,a+1+k-n}(q)=B_{k,k-b}(q)=B_{k,b}(q)$ we have $$ \binom{n}{b}_q+\sum _{k=1}^n\binom{n}{k}_q \frac{B_{k,a}(q) }{2^{k-n}}=\binom{n}{a}_q+\sum _{k=1}^n \binom{n}{k}_q\frac{B_{k,b}(q)}{2^{k-n}}. $$<|endoftext|> TITLE: What is a Fermi surface? QUESTION [6 upvotes]: I posted this question on the physics site, but then received immediately the sense that I won't be able to understand answers even if they come. So I hope it's all right if I post it here, since there appears to be mathematical interest as evidenced by this question: Topology and the 2016 Nobel Prize in Physics Allow me to explain my confusion. Given a solid, I believe I have some feeling for the Fermi level. I can understand it, for example, as the characteristic parameter $\mu$ in the Fermi-Dirac distribution of energy levels for the electrons in the system: $$f(\epsilon)=\frac{1}{e^{(\epsilon-\mu)/kT}+1}$$ ignoring for the moment other physical interpretations. Thus, it is the unique energy level that has probability 1/2 of being occupied. The definition of the Fermi surface, on the other hand, is usually given as 'the iso-surface of states with energy equal to the Fermi level' in the three-dimensional space of wave-vectors $k$, for example in this Wikipedia article: https://en.wikipedia.org/wiki/Electronic_band_structure In other words, it is defined to be those $k$ such that $$E(k)=\mu.$$ So far, so good. The problem is, I don't quite understand what $E(k)$ is. One situation seems to be straightforward, namely a Fermi gas of identical particles. Then $$E(k)=\frac{k^2}{2m}$$ and the Fermi surface is a sphere. However, if we are in an infinite periodic potential, the usual idealised model for Bloch theory, then the solutions to the Schroedinger equation come out in the form $$\psi_{kn}(r)=e^{ik\cdot r}u_{kn}(r),$$ where $u_{kn}$ is a periodic function and $n$ is a discrete index for energy levels. In other words, for each wave vector $k$, there are many energy levels $E_n(k)$. So the equation for the Fermi surface would actually look like $$E_n(k)=\mu.$$My question, therefore, is which energy level is the $E(k)$ that occurs in the definition of the Fermi surface? Perhaps there is one Fermi surface for every level $n$? (Assuming that the levels vary continuously over the momentum space, enabling us to consistently index the levels for varying $k$.) If I could elaborate on my confusion a little bit more, I don't quite understand the definition in this answer to this question: https://physics.stackexchange.com/q/5739/ It is stated that 'The Fermi surface is simply the surface in momentum space where, in the limit of zero interactions, all fermion states with (crystal) momentum $|k|<|k_F|$ are occupied, and all higher momentum states are empty. ' For one thing, as mentioned above, for any momentum $k$, there is an infinite sequence of fermion states. The other problem is that I'm not sure that the statement above defines a unique surface, even if I were able to somehow pick out a fermion state $\psi(k)$ for each $k$ that the statement refers to. (I would need to draw a picture to explain this point, which I don't have the competence to do.) REPLY [2 votes]: Quoting p. 142 of the trusty Ashcroft-Mermin (who write $\mathcal E_F$ for your $\mu$): For each partially filled band there will be a surface in $k$-space separating the occupied from the unoccupied levels. The set of all such surfaces is known as the Fermi surface, and is the generalization to Bloch electrons of the free electron Fermi sphere. The parts of the Fermi surface arising from individual partially filled bands are known as the branches of the Fermi surface. We shall see (Chapter 12) that a solid has metallic properties provided that a Fermi surface exists. Analytically, the branch of the Fermi surface in the $n$th band is that surface in $k$-space (if there is one) determined by $$\mathcal E_n(\mathbf k)=\mathcal E_F.$$ As to "what $\mathcal E_n(\mathbf k)$ is": it arises from looking for eigenfunctions of a "single-electron" hamiltonian $$ \left(-\frac{\hbar^2}{2m}\nabla^2+U(\mathbf r)\right)\psi(\mathbf r)=E\psi(\mathbf r) $$ in the "Bloch" form: $\psi(\mathbf r)=e^{i\mathbf k\cdot\mathbf r}u(\mathbf r)$ with (Bravais lattice) periodic boundary conditions on $u$, i.e. $$ \left(\frac{\hbar^2}{2m}\left(\frac1i\nabla + \mathbf k\right)^2+U(\mathbf r)\right)u_{\mathbf k}(\mathbf r)=\mathcal E(\mathbf k)u_{\mathbf k}(\mathbf r). $$ Because of the periodic boundary condition we can regard [this] as a Hermitian eigenvalue problem restricted to a single primitive cell of the crystal. Because the eigenvalue problem is set in a fixed finite volume, we expect on general grounds to find an infinite family of solutions with discretely spaced eigenvalues, which we label with the band index $n$.<|endoftext|> TITLE: Is Teichmüller distance bigger than Weil-Petersson distance on Teichmüller space? QUESTION [8 upvotes]: It is known that Teichmüller distance ($d_{Teich}$) on Teichmüller space is complete, whereas Weil-Petersson distance ($d_{WP}$) is not complete. See for example the article Wolpert, Scott. Noncompleteness of the Weil-Petersson metric for Teichmüller space. Pacific J. Math. 61 (1975), no. 2, 573--577 , in which Wolpert constructs an the explicit path on the Teichmüller space, which diverges in $d_{Teich}$, but converges in $d_{WP}$. My question is : is it true in general that there is a positive constant $C$ such that $d_{WP} < C d_{Teich}$? REPLY [12 votes]: Yes, this is a result of Michele Linch, 1974.<|endoftext|> TITLE: What is forcing indescribability? QUESTION [7 upvotes]: Suppose $m, n\in\omega$ and $\kappa$ is a cardinal. Then $\kappa$ is $\Pi^m_n$-indescribable if every $\Pi^m_n$-sentence true about $\kappa$ is true about some $\lambda<\kappa$; formally, if for every $\Pi_n$-sentence $\varphi$ in the language of set theory with a unary predicate and every $A\subseteq V_\kappa$, there is some $\lambda<\kappa$ such that $$(V_{\kappa+m}, \in, A)\models\varphi\implies (V_{\lambda+m}, \in, A\cap V_\lambda)\models\varphi.$$ For example, the $\Pi^1_1$-indescribable cardinals are exactly the weakly compact cardinals. (Note that we can go beyond $\Pi^m_n$-describability - quite a ways, even - but let's ignore that for now.) I'm interested in a version of indescribability where we pay attention to what can happen in generic extensions of $V$ (I don't think this really qualifies as a "generic" version indescribability, but there's a vague connection). Specifically, for an ordinal $\alpha$ and a set $C\subseteq V_\alpha$, write "$(V_\alpha, \in, C)\models_f\varphi$" if $((V_\alpha)^{V[G]}, \in, C)\models\varphi$ for every $G$ which is set-generic over $V$ (note that the forcing for which $G$ is generic need not be an element of $V_\alpha$). Now say that a cardinal $\kappa$ is $\Pi^m_n$-forcing indescribable if for every $\Pi_n$-sentence $\varphi$ in the language of set theory with a unary predicate and every $A\subseteq V_\kappa$, there is some $\lambda<\kappa$ such that $$(V_{\kappa+m}, \in, A)\models_f\varphi\implies (V_{\lambda+m}, \in, A\cap V_\lambda)\models_f\varphi.$$ Note that $A$ is a fixed set in the ground model. My question is, roughly how "big" are the forcing-indescribable cardinals? For example, it is not clear to me what the relationship is between $\Pi^1_1$-indescribable (= weakly compact) cardinals and $\Pi^1_1$-forcing indescribable cardinals. I believe that in $L$, $\Gamma$-forcing indescribability implies $\Gamma$-indescribability (since we can replace $\varphi$ with "$\varphi$ holds in $L$"), but I don't see the converse, or how this holds in general $V$. REPLY [3 votes]: Let me start things off by providing an upper bound. The bound is very large, however, and I expect that it can be improved, perhaps dramatically. But at least it shows the consistency of your large cardinal relative to some other well-studied large cardinals. Theorem. If $\kappa$ is $1$-$C^{(2)}$-extendible, then it is forcing $\Pi^m_n$-indescribable for every $m,n$. Definition. A cardinal $\kappa$ is $1$-$C^{(2)}$-extendible, if there is an elementary embedding $j:V_{\kappa+1}\to V_{\theta+1}$, with critical point $\kappa$, such that the target $j(\kappa)=\theta$ is $\Sigma_2$-correct in $V$, meaning $V_\theta\prec_{\Sigma_2} V$. This is a fairly strong large cardinal notion, far stronger than the totally indescribable cardinals you mention in your question. For example, every 1-extendible cardinal is superstrong and much more. But these cardinals are weaker than Vopěnka's principle. Proof. Assume that $\kappa$ is $1$-$C^{(2)}$-extendible. So there is an elementary embedding $j:V_{\kappa+1}\to V_{\theta+1}$ with critical point $\kappa$ and $j(\kappa)=\theta$ is $\Sigma_2$-correct in $V$. Suppose now that $A\subset V_\kappa$ and $\langle V_{\kappa+m},\in,A\rangle\models_f\varphi$, which means that $\langle V[G]_{\kappa+m},\in,A\rangle\models\varphi$ for every forcing extension $V[G]$. This is a $\Sigma_2$ property about $A$ and $\kappa$, since any violation of it would be revealed inside some large enough $V_\eta$, using forcing inside that $V_\eta$. Thus, by $\Sigma_2$-correctness, we see that $V_\theta$ agrees that $\langle V_{\kappa+m},\in,A\rangle\models_f\varphi$. Since $A=j(A)\cap\kappa$, we may pull this back by elementarity to conclude that there is some $\lambda<\kappa$ with $\langle V_{\lambda+m},\in,A\cap V_{\lambda+m}\rangle\models_f\varphi$ inside $V_\kappa$. But $\kappa$ itself must also be $\Sigma_2$-correct, and so actually $\langle V_{\lambda+m},\in,A\cap V_{\lambda+m}\rangle\models_f\varphi$ in $V$, as desired. QED I'll think some more about lower bounds and about pulling down the strength of the hypothesis. Update. I've realized that we can improve the upper bound as follows. We don't really need the "$+1$", since that actually provided a uniform version of the phenomenon, with the same embedding working for every $A$. Thomas Johnstone and I defined that a cardinal $\kappa$ is uplifting, if it is inaccessible and $V_\kappa\prec V_\theta$ for cofinally many inaccessible cardinals $\theta$. (J. D. Hamkins, T. Johnstone, Resurrection axioms and uplifting cardinals) A boldface version is that $\kappa$ is strongly uplifting, if for every $A\subset V_\kappa$ there are cofinally many inaccessible cardinals $\theta$ for which $\langle V_\kappa,\in,A\rangle\prec\langle V_\theta,\in,A^*\rangle$ for some $A^*\subset V_\theta$. (J. D. Hamkins, T. Johnstone, Strongly uplifting cardinals and boldface resurrection) These have diverse equivalent formulations, as I mention on the linked blog post, connected with strengthenings of the strongly unfoldable cardinals. Let me now strengthen this a little more, for the present application, with the following new large cardinal concept. Definition. A cardinal $\kappa$ is strongly $C^{(n)}$-uplifting, if for every $A\subset V_\kappa$ there is a $\Sigma_n$-correct cardinal $\theta$ and $A^*\subset V_\theta$ with $\langle V_\kappa,\in,A\rangle\prec\langle V_\theta,\in,A^*\rangle$. This is what we really needed in the theorem above. Theorem. If $\kappa$ is strongly $C^{(2)}$-uplifting, then it is forcing $\Pi^m_n$-indescribable for every $m$ and $n$. Proof. Argue as in the first theorem above, but now we have only $\langle V_\kappa,\in,A\rangle\prec\langle V_\theta,\in,A^*\rangle$, instead of $j$. If $\langle V_\kappa,\in,A\rangle+m\models_f\varphi$, then this will be true inside $V_\theta$ since it is $\Sigma_2$-correct, and so $V_\theta$ thinks that this holds on an initial segment of $A^*$, and so we get $\lambda<\kappa$ with $\langle V_\lambda,\in,A\cap V_\lambda\rangle\models_f\varphi$ inside $V_\kappa$, which is right about this since $\kappa$ is itself $\Sigma_2$-correct. QED I think the strongly $C^{(2)}$-uplifting cardinals are comparatively weak, and absolute to $L$, but I'll think more about it.<|endoftext|> TITLE: relativistic dipole problem QUESTION [5 upvotes]: I am trying to understand something in a relativistic dipole problem. The system is described by the following Hamiltonian function $$H=\sqrt{p_r^2+\frac{p_\varphi^2}{r^2}+m^2}+\gamma\frac{\cos\varphi}{r^2},$$ where $m, \gamma$ are positive constants and $r, \varphi$ are the polar coordinates in the plane. *QUESTIONS. (a) I am interested in solutions for values near $r\to 0$. (b) What about asymptotic-in-time behavior of those solutions? It seems that those solutions must form smooth manifolds in the phase-space. Actually I can not see any thread to pull to unravel this problem. Please do let me know any references, at least to similar problems. Anyway, I appreciate any prompt comments or references. REPLY [3 votes]: Qualitatively, there are two types of solutions to this problem: trajectories are either attracted (resp. repelled) to the negative (resp. positive) end of the dipole. Which type of solution depends on the initial angle of the particle and its initial radius. This conclusion is supported by numerical evidence (samples are given below) obtained by solving this Hamiltonian system using a fourth order accurate Runge-Kutta method. Note that it's a bit tricky to apply a symplectic integrator in this context because the Hamiltonian of this system is not separable, but otherwise this problem is trivial to numerically solve because the time intervals of simulation are not that long. Before delving into these numerical results, let me just say that it does not seem that this conclusion depends strongly on whether the kinetic energy is relativistic (i.e., strictly convex in momentum) or classical (i.e., strongly convex in momentum). It is simply a consequence of the potential energy lacking coercivity, and its not clear how either type of kinetic energy can compensate for this. For $m=\gamma=1$, here is a sample output showing pieces of trajectories in the $(r,\varphi)$-plane starting (resp. ending) at the point marked with a circle (resp. hexagram). The initial radius for all trajectories is $1/2$ and the initial angles vary as shown. In the background, a few filled contours of the dipole potential energy $U(r,\varphi) = \gamma \cos(\varphi)/r^2$ are shown. Note that for any $r>0$, this potential energy is positive when $\varphi \in [0, \pi/2)$ and negative when $\varphi \in (\pi/2,\pi]$. Here is a plot of the total energy along the trajectory starting at $\varphi=0$, which is near the positively charged end of the dipole). It shows the RK integrator preserving total energy (black) quite nicely, and the potential energy (blue) being converted to kinetic energy (red) as the distance to the dipole increases. Here is a plot of the total energy along the trajectory starting at $\varphi=\pi$, which is near the negatively charged end of the dipole. In this case, while the total energy (black) is preserved the kinetic and potential energies become unbounded as the distance to the dipole decreases.<|endoftext|> TITLE: Self-contained proof that finite index subgroup in which $g$ is one element of a basis? QUESTION [6 upvotes]: Let $g$ be a nontrivial element of a finitely generated free group $G$. Is there a finite index subgroup $H \subset G$ in which $g$ is one element of a basis? Andy Putman says this in his answer below that this is an immediate corollary of Marshall Hall's theorem. However, I'm wondering if there is a more "elementary"/"self-contained" way of seeing that such a finite index subgroup exists. REPLY [22 votes]: Yes, this is a corollary of the following famous theorem of Marshall Hall: Theorem: If $A$ is a finitely generated subgroup of a free group $G$, then there exists a finite-index subgroup $H$ of $G$ containing $A$ such that $A$ is a free factor of $H$, i.e. such that $H = A \ast A'$ for some subgroup $A'$ of $G$. Just apply this to the cyclic subgroup $A = \langle g \rangle$. For a proof of Marshall Hall's theorem, see e.g. this blog post.<|endoftext|> TITLE: Abelian varieties with p-rank zero QUESTION [6 upvotes]: Let $X$ be an abelian variety over a finite field of characteristic $p$ such that the $X[p]=0$. In other words, none of the Newton slopes are $0,1$. QUESTIONS. (a) Is it possible for the endomorphism algebra $End(X)$ to be commutative? (b) If so, are there examples that are fairly easy to construct? REPLY [7 votes]: (a) Oh, yes. This is proven in a paper of Hendrik Lenstra and Frans Oort http://www.sciencedirect.com/science/article/pii/0022404974900292 . (b) A construction of the corresponding CM-field of endomorphisms is described in the paper mentioned above.<|endoftext|> TITLE: Applications of linear programming duality in combinatorics QUESTION [19 upvotes]: So, I know that one can apply the strong LP duality theorem to specific instances of maximum flow problems to recover some nontrivial theorems in combinatorics, such as Hall's theorem, Koenig's theorem and Menger's theorem. It's neat that these theorems, all equivalent to one another, can be seen to follow from a single framework. I was wondering if there are other such interesting applications of LP duality (or more general kinds of duality) in combinatorics, perhaps of a different flavor? Is there a good reference for these topics? The more surprising the applications, the better. REPLY [5 votes]: Here is a fairly recent example. Davies, Jenssen, Perkins and Roberts use linear programming duality to obtain a tight upper bound on the number of independent sets and matchings in a $d$-regular graph on $n$ vertices. In fact, more generally, their proof actually gives a bound on the so called "independence polynomial" and the "matching polynomial" of $d$-regular graphs on $n$ vertices. Basically, (I think) their proof uses linear programming duality to provide bounds on the logarithmic derivative of this polynomial and translates this to give bounds on the polynomial itself. In the case of independent sets, their result strengthens a result of Zhao (which, itself, extended a result of Galvin and Tetali, which strengthened a result of Kahn). For matchings, their result proves the asymptotic Upper Matching Conjecture of Friedland, Krop, Lundow and Markström. Yufei Zhao has written a blog post and a survey paper about this topic.<|endoftext|> TITLE: Bernstein–Zelevinsky classification for classical groups QUESTION [8 upvotes]: Bernstein and Zelevinsky classifies the irreducible complex smooth representations of a general linear group over a local field in terms of cuspidal representations. The irreducible modules are parametrized by multi-segments. Are there some references about parametrizations of irreducible representations of classical groups (types A, B, C, D) using something like multi-segments? Thank you very much. REPLY [3 votes]: You have such a classification for discrete series representations of classical groups by Moeglin and Moeglin-Tadic (and probably for tempered representations by Chris Jantzen) but it is not only using multisegment.<|endoftext|> TITLE: Review paper/book on Finite Difference Methods for PDEs QUESTION [8 upvotes]: I am looking for a good, relatively modern, review paper/book on Finite Difference Methods for PDEs with a theoretical emphasis in mind. By theoretical emphasis I mean that I care about theorems (i.e. with proofs) of convergence (and rate of convergence, if available) to an actual solution. A non-modern (late 1950s) example of the sort of review I'm looking for is O. Ladyzenskaja's "The Method of Finite Differences in the theory of partial differential equations". Any help finding such papers/books is very well appreciated. More broadly, modern reviews on other discrete schemes for solving PDEs are also welcome, provided the reviews are theoretical in nature (in the sense defined above). REPLY [12 votes]: There are many well-written books/notes on this topic including: J. C. Strikwerda, Finite difference schemes and partial differential equations, SIAM, 2004. R. J. LeVeque, Finite difference methods for ordinary and partial differential equations: steady-state and time-dependent problems, vol. 98, SIAM, 2007. B. Gustafsson, H.-O. Kreiss, and J. Oliger, Time-Dependent Problems and Difference Methods, 2nd Edition, vol. 123, John Wiley & Sons, 2013. Thomas, James W. Numerical partial differential equations: finite difference methods. Vol. 22. Springer Science & Business Media, 2013. Arnold, Douglas N. Lecture notes on Numerical Analysis of Partial Differential Equations, 2012. 2014-2015 version is available at http://www.math.umn.edu/~arnold/8445/notes.pdf Lloyd N. Trefethen, Finite Difference and Spectral Methods for Ordinary and Partial Differential Equations, unpublished text, 1996, available at https://people.maths.ox.ac.uk/trefethen/pdetext.html ADD The above references are primarily aimed at graduate students in applied mathematics and are an excellent intro to basic/general concepts like consistency, stability, and convergence of finite difference methods. However, they may not adequately cover topics like Higher Order Accuracy, Singularly Perturbed PDEs, Long-Time Simulation and related to this Structure-Preserving Discretizations, Adaptive Mesh Refinement, the Method of Lines (e.g., for asset pricing), Meshless Finite Difference Methods for irregularly shaped regions, and Monte-Carlo Methods for local solutions to high-dimensional PDEs. These works are a bit more specialized. To conclude, let me mention that the above numerical PDE references may not be sufficient and one might need to complement them with a suitable book/reference on analytical methods for PDE. (What is suitable really depends on the specific PDE problem one is dealing with.) Let me expand on this point just a bit: there are typically two main ingredients to proving accuracy of a numerical method for a PDE (be it finite difference, volume or element) the quantity to be computed is sufficiently regular; and the approximation is stable and accurate in approximating this quantity. Local accuracy is typically straightforward to verify with finite-difference approximations both in the interior and on the boundary of the PDE problem: it just requires doing a Taylor expansion. (If the problem has internal discontinuities, then this verification is not so straightforward.) However, deriving an estimate of the global error of the approximation requires more work. Typically this error is expressed in terms of the quantity you wish to compute evaluated at the numerical solution. Thus, in order to estimate this global error, one needs both some notion of stability of the numerical solution and a priori estimates on the quantity being approximated. This latter point emphasizes that understanding the PDE you wish to approximate is essential for the design and analysis of good schemes.<|endoftext|> TITLE: Counting higher-dimensional partitions with symmetric function theory QUESTION [13 upvotes]: My coauthors and I are writing a (mostly expository) paper in which we construct the Specht module. Our proof that the Specht module is irreducible in characteristic zero implies the following stronger fact: The number of monomial ideals of $\mathbb{Q}[x_1, \ldots, x_k]$ of codimension $n$ is equal to the number of $k$-tuples of partitions, $(\lambda^1, \ldots, \lambda^k)$, $\lambda^i \vdash n$ , for which $\langle h_{\lambda^1} \ast \cdots \ast h_{\lambda^k} \;,\; e_n \rangle = 1$. Here, $\ast$ denotes Kronecker product, $h$ denotes the homogeneous symmetric function, $e$ denotes the elementary symmetric function, and $\langle \;\;, \;\; \rangle$ denotes the Hall inner product. It seems likely that this fact is known. However, the literature on planar partitions (and higher-dimensional partitions) focuses on finding a generating function, which is a different, much harder question. Does anyone know a reference for the easier fact above? To use Sage to compute these numbers, start with the following lines: h = SymmetricFunctions(QQ).homogeneous() e = SymmetricFunctions(QQ).elementary() Now, the number of monomial ideals in three variables up to codimension 7 may be computed using for n in range(7): print sum([1 if h[a].itensor(h[b].itensor(h[c])).scalar(e[n]) == 1 else 0\ for a in Partitions(n)\ for b in Partitions(n)\ for c in Partitions(n)]) which lists the numbers $1, 1, 3, 6, 13, 24, 48,$ an initial segment of OEIS A000219. Similarly, monomial ideals in four variables may be counted for n in range(5): print sum([1 if h[a].itensor(h[b].itensor(h[c].itensor(h[d])))\ .scalar(e[n]) == 1 else 0\ for a in Partitions(n)\ for b in Partitions(n)\ for c in Partitions(n)\ for d in Partitions(n)]) The resulting sequence, $1, 1, 4, 10, 26$, is an initial segment of OEIS A000293. It is a famous open problem in combinatorics to find a generating function for this sequence. EDIT for the curious, the paper has been posted: https://arxiv.org/abs/1701.05277 REPLY [5 votes]: This conjecture is false. It has to be, because it is asymtotically wrong: The number of plane partitions of size $n$ grows like $\exp(n^{2/3- \epsilon})$ and the number of triples of partitions of $n$ (with no condition on them) grows like only $\exp(n^{1/2+\epsilon})$. (citation) As I'll explain below, there is a relationship between the two sides. Using that relationship, I can predict that the result will fail for $n=13$, $k=3$. I don't know if this is the first failure. Details: Let's consider the condition that $\langle h_{\lambda_1} \ast \cdots \ast h_{\lambda_k}, e_n \rangle = 1$. As is well known, $h_{\lambda_1} \ast \cdots \ast h_{\lambda_k}$ expands positively in the $h$'s, and the coefficient of $h_{\mu}$ is the number of $k$-dimensional tables of nonnegative integers where summing up the $k-1$ dimensional slices gives the partitions $\lambda_1$, ..., $\lambda_k$ and just listing all the entries in sorted order gives $\mu$. (We pad partitions with infinitely many zeroes at the end.) Also, $\langle h_{\mu}, e_{n} \rangle$ is $0$ unless $\mu = 1^n$, in which case it is $1$. So you are counting $k$-tuples of partitions $(\lambda_1, \ldots, \lambda_k)$ such that there is a unique way to fill a $k$-dimensional table with $0$'s and $1$'s so that the $(k-1)$-dimensional slices total $(\lambda_1, \ldots, \lambda_k)$. Let $T$ be a $k \times k$ array of $0$'s and $1$'s, and let $\sigma(T) = (\sigma_1(T), \ldots, \sigma_k(T))$ be its sums along slices. I'll call $T$ "ordered" if each of the $\sigma_j(T)$ is a partition. I'll call $T$ an "ideal" if the set of $1$'s of $T$ is an order ideal. And I'll call $T$ "singleton" if no other $T'$ has $\sigma(T) = \sigma(T')$. It is clear that, if $T$ is an ideal, then $T$ is ordered. Your conjecture is that the number of $(k,n)$ ideals is equal to the number of $(k,n)$ ordered singletons. Here are the true statements: Claim 1 Every ordered singleton is an ideal. Proof Suppose that $T$ is ordered, but not an ideal. Then there must be some $(i_1, \ldots, i_k)$ and some index $j$ for which $T(i_1, \ldots, i_k)=1$ but $T(i_1, \ldots, i_j-1, \ldots, i_k)=0$. Since $\sigma_j(T)_{i_j} \leq \sigma_j(T)_{i_j-1}$, there must be some other $(i'_1, \ldots, i'_k)$ with $i_j=i'_j$, $T(i'_1, \ldots, i'_k)=0$ but $T(i'_1, \ldots, i'_j-1, \ldots, i'_k)=1$. But then define $T'$ by switching the values of $T$ in these $4$ places, and $\sigma(T) = \sigma(T')$, a contradiction. $\square$ Claim 2 There are ideals of type $(3,13)$ which are not singleton. Proof Let $I$ be the order ideal which contains the $10$ triples $(i_1, i_2, i_3)$ with $i_1+i_2+i_3 \leq 2$ and also $\{ (2,1,0), (1,0,2), (0,2,1) \}$. Let $I'$ be the same with $\{ (0,1,2), (1,2,0), (2,0,1) \}$. Let $T$ and $T'$ be the corresponding indicator functions. Then $\sigma(T) = \sigma(T') = (7,4,2)^3$. $\square$<|endoftext|> TITLE: How few $k$-dimensional subspaces of $V$ are enough to have a complement to each $n-k$-dimensional subspace? QUESTION [14 upvotes]: Let $n$ and $k$ be nonnegative integers such that $k\leq n$. Let $F$ be a field, and let $V$ be an $n$-dimensional $F$-vector space. A set $\mathcal{S}$ of $k$-dimensional subspaces of $V$ is said to be a complement repository if for every $n-k$-dimensional subspace $U$ of $V$, there exists some $P \in \mathcal{S}$ such that $V = U \oplus P$ (internal direct sum). (Of course, $V = U \oplus P$ is equivalent to $U \cap P = 0$, since $\dim U + \dim P = n = \dim V$.) Question: What is the smallest size of a complement repository (for given $n$ and $k$) ? Let me denote this smallest size by $d\left(n,k\right)$. It is easy to see that there exists a complement repository of cardinality $\dbinom{n}{k}$. Namely, fix a basis $\left(e_1,e_2,\ldots,e_n\right)$ of the $F$-vector space $V$. For each subset $S$ of $\left\{1,2,\ldots,n\right\}$, let $E_S$ be the $F$-vector subspace of $V$ spanned by the $e_s$ with $s \in S$. Then, the set $\left\{E_S \mid S \subseteq \left\{1,2,\ldots,n\right\};\ \left|S\right| = k\right\}$ is a complement repository of cardinality $\dbinom{n}{k}$. (Here is a proof that this set is a complement repository: Set $\mathcal{S} = \left\{E_S \mid S \subseteq \left\{1,2,\ldots,n\right\};\ \left|S\right| = k\right\}$. We need to show that $\mathcal{S}$ is a complement repository. Let $U$ be an $n-k$-dimensional subspace of $V$. Let $\left(u_1,u_2,\ldots,u_{n-k}\right)$ be a basis of $U$. Thus, the list $\left(u_1,u_2,\ldots,u_{n-k}\right)$ is linearly independent, and spans $U$. By the Steinitz exchange lemma (applied to the linearly independent list $\left(u_1,u_2,\ldots,u_{n-k}\right)$ of vectors, and the basis $\left(e_1,e_2,\ldots,e_n\right)$ of $V$), we have $n-k \leq n$ (which is no surprise), and possibly after reordering the basis $\left(e_1,e_2,\ldots,e_n\right)$, the list $\left(u_1,u_2,\ldots,u_{n-k},e_{n-k+1},e_{n-k+2},\ldots,e_n\right)$ spans $V$. We can WLOG assume that the basis $\left(e_1,e_2,\ldots,e_n\right)$ is already reordered in such a way that the list $\left(u_1,u_2,\ldots,u_{n-k},e_{n-k+1},e_{n-k+2},\ldots,e_n\right)$ spans $V$ (because reordering the basis $\left(e_1,e_2,\ldots,e_n\right)$ does not change the set $\mathcal{S}$). The list $\left(u_1,u_2,\ldots,u_{n-k},e_{n-k+1},e_{n-k+2},\ldots,e_n\right)$ spans $V$, and thus is a basis of $V$ (since it has length $n = \dim V$). Thus, the span of the first $n-k$ entries of this list and the span of the last $k$ entries of this list are complementary subspaces of $V$. But since the former span is $U$ (because the list $\left(u_1,u_2,\ldots,u_{n-k}\right)$ spans $U$), while the latter span is $E_{\left\{n-k+1,n-k+2,\ldots,n\right\}}$ (because $E_{\left\{n-k+1,n-k+2,\ldots,n\right\}}$ is defined as the span of $\left(e_{n-k+1},e_{n-k+2},\ldots,e_n\right)$), this rewrites as follows: The spaces $U$ and $E_{\left\{n-k+1,n-k+2,\ldots,n\right\}}$ are complementary subspaces of $V$. Thus, $V = U \oplus E_{\left\{n-k+1,n-k+2,\ldots,n\right\}}$. Hence, there exists some $P \in \mathcal{S}$ such that $V = U \oplus P$ (namely, $P = E_{\left\{n-k+1,n-k+2,\ldots,n\right\}}$). This shows that $\mathcal{S}$ is a complement repository.) This gives an upper bound on the smallest size of a complement repository: namely, $d\left(n,k\right) \leq \dbinom{n}{k}$. For a lower bound, I so far can only see $d\left(n,k\right) \geq k\left(n-k\right)+1$ for $F = \mathbb{C}$, and I am not sure of that either. (An old MO answer claims that a smooth projective variety of dimension $d$ over $\mathbb{C}$ cannot covered by less than $d+1$ affine open subsets. Applying this to the Grassmannian $\operatorname{Gr}\left(V,n-k\right)$, which I hope is smooth and has dimension $k\left(n-k\right)$, we conclude that $\operatorname{Gr}\left(V,n-k\right)$ cannot be covered by less than $d+1$ affine open subsets. But if $P$ is a $k$-dimensional subspace of $V$, then the complements of $P$ form an affine open subvariety of $\operatorname{Gr}\left(V,n-k\right)$. Hence, a complement repository of cardinality $g$ would induce a covering of $\operatorname{Gr}\left(V,n-k\right)$ by $g$ affine open subsets, and as we know this is impossible for $g < k\left(n-k\right)+1$. This should at least take care of the case $F = \mathbb{C}$; but I doubt that the bound thus obtained is anywhere near optimal.) It is also clear that $d\left(n,1\right) = d\left(n,n-1\right) = n$. Moreover, a simple argument using orthogonal subspaces in dual spaces shows that $d\left(n,k\right) = d\left(n,n-k\right)$. REPLY [5 votes]: See these papers: Covering by Complements of Subspaces, II., W. Edwin Clark and Boris Shekhtman, Proc. Amer. Math. Soc.125 (1997), no. 1, 251--254. (link here, unrestriced access; MR review) Covering by Complements of Subspaces, W. Edwin Clark and Boris Shekhtman, Linear and Multilin. Algebra, Vol 49,1995, pp. 1--13. (link here, restricted accessMR review)<|endoftext|> TITLE: Relation between representations of p-adic groups and affine Hecke algebras QUESTION [8 upvotes]: Let $R_n$ be the category of complex-valued smooth finite-length representations of the group $GL_n(F)$, where $F$ is a local field. By the result of Borel, the subcategory of $R_n$ consisting of representations that admit a Iwahori-invariant vector is equivalent to the category of finite-dimensional representations of the affine Hecke algebra $H(n, F)$ attached to $GL_n(F)$. The equivalence translates parabolic induction to module induction from $H(n_1, F) \otimes H(n_2, F)$ to $H(n_1 + n_2, F)$. This also follows from the results of the paper and the paper of Bushnell and Kutzko by type theory and the results by Heiermann. Therefore to study the subcategory of $R_n$ consisting of representations that admit a Iwahori-invariant vector is equivalent to study finite dimensional representations of affine Hecke algebras. This is very useful since we can apply results of representations of affine Hecke algebras to representations of complex-valued smooth finite-length representations of the group $GL_n(F)$. Are there some references about similar results of the relation between classical groups of other types and some algebras similar to affine Hecke algebras. Thank you very much. REPLY [3 votes]: For the inner forms of $\mathrm{SL}_n$ it was done by Aubert, Baum, Plymen and Solleveld in "Hecke algebras for inner forms of p-adic special linear groups". Moreover, the equivalence between the category of representations in one Bernstein block and the category of modules over an affine Hecke algebra (may be extended by a finite group in the case of classical group), is true for all Bernstein blocks, not only for the Iwahori block.<|endoftext|> TITLE: Extending Sacks forcing QUESTION [7 upvotes]: Sacks forcing allows us to build a model $V[G]$, such that there is no "intermediate model" between $V$ and $V[G]$, meaning if $V \subseteq W \subseteq V[G]$ is a model of ZFC then either $W = V$ or $W = V[G]$. My question is: Whether we know how to force a model to have exactly $1$ intermediate model? Assuming the answer is Yes, do we know how to extend this result arbitrarily long? Meaning, do we know how to force a model where all the inner models (i.e standard, transitive classes that contain all the ordinals) are well ordered and have a $1-1$ correspondence with the ordinals? This would necessate to start with $V_0 = L$, and then possibly take $V_1 = V_0[G]$ (where this would be a regular Sacks forcing) and somehow continue to create $V_\alpha$ for all $\alpha \in ORD$, without 'accidentally' creating any more inner models. REPLY [5 votes]: Yes, there is a whole literature on this kind of thing. One of the main methods is to perform iterations and products of Sacks forcing, so as to realize a given (set-sized) partial order of inner models. Marcia Groszek has been very active in this area. For example, you could begin with the following. Marcia J. Groszek, Applications of iterated perfect set forcing, Annals of Pure and Applied Logic Volume 39, Issue 1, July 1988, Pages 19-53. Although the methods are extremely flexible for achieving set-sized orders for the structure of inner models, and in particular, any ordinal can be realized. But to have actually exactly Ord many inner models, you cannot use Sacks forcing alone, since there would be too many reals. So I am less sure about question 2.<|endoftext|> TITLE: Is every group an ideal class group of a number field? QUESTION [11 upvotes]: The inverse Galois problem asks whether every finite group appears as the Galois group of some finite extension of $\mathbb Q$. I was wondering to what extent the analogous problem for ideal class groups has been investigated. More precisely, consider the following question: Is every finite abelian group the ideal class group of some number field (finite extension of $\mathbb Q$)? I'd be interested to hear about any partial results, as I suppose this question is still open. I'd be also interested in any results about a weaker problem: Is every positive integer the ideal class number of some number field? Again, any reference, even to a partial result, will be appreciated. REPLY [9 votes]: The recent paper by Homlin, Jones, Kurlberg, McLeman, and Petersen (Experimental Math., to appear) is devoted to these questions especially in the context of imaginary quadratic fields. One should expect that every natural number arises as the class number of an imaginary quadratic field. Refining an earlier conjecture of Soundararajan, in this paper a precise asymptotic is formulated for the number of imaginary quadratic fields with a given class number. They also formulate conjectures on what kind of groups can arise as class groups of imaginary quadratic fields -- for example, one expects that for any odd prime $p$, $({\Bbb Z}/p{\Bbb Z})^3$ is not the class group of any imaginary quadratic field. The paper gives much data on such questions together with many related references.<|endoftext|> TITLE: On J.G.Thompson's conjecture about conjugacy classes of finite simple groups QUESTION [5 upvotes]: This conjecture in "Unsolved problems in group theory" No.18: 9.24: Conjecture: every finite simple non-abelian group $G$ can be represented in the form $G=CC$, where $C$ is some conjugacy class of $G$. I want to prove this conjecture is right for simple group $A_n$ in $S_n$ ($n>4$), but I have no idea how to deal with it. I hope someone could give me some advice, or some old results about this. REPLY [5 votes]: This was proved in this paper: MR0183763 (32 #1241) Reviewed Xu Cheng-hao The commutators of the alternating group. Sci. Sinica 14 1965 339–342. 20.20 My institution does not have an electronic copy, so there you are on your own.<|endoftext|> TITLE: Four manifold without point homotopy equivalent to wedge of two-spheres? QUESTION [9 upvotes]: Let $X$ be a closed, simply-connected four-manifold. Let $X'$ be obtained from $X$ by removing a point. Is $X'$ homotopy equivalent to a wedge of $S^2$s? REPLY [6 votes]: Here is a hopefully better answer which is copied from page 104 of Milnor-Husemoller's book on symmetric bilinear forms. $X^\prime$ is also simply connected (Seifert-van Kampen reversed) and thus has torsion-free $H_2={\mathbb Z}^r$ (otherwise the torsion would contribute nontrivially to $H^3$ via the universal coefficient theorem) and $\pi_2=H_2$ (Hurewicz). So there is a map $S^2\vee\ldots\vee S^2\to X^\prime$ that induces a homology isomorphism, hence (by Hurewicz) a weak homotopy equivalence, hence (because $X^\prime$ is an ANR) a homotopy equivalence.<|endoftext|> TITLE: Cohomology of function spaces QUESTION [6 upvotes]: Let $M,N$ be smooth manifolds and $C^\infty(M,N)$ be the function space with Whitney topology. If we know the cohomology groups of $M,N$, can we calculate the cohomology groups of $C^\infty(M,N)$? REPLY [8 votes]: It seems there exists such a thing: There is a spectral sequence by Anderson that computes the homology of the mapping space out of the cohomology of $M$ and the homology of $N$. Here is a link to the announcement: http://www.ams.org/journals/bull/1972-78-05/S0002-9904-1972-13034-9/S0002-9904-1972-13034-9.pdf I have not been able to find the subsequent paper with the proof.<|endoftext|> TITLE: Relations among Pontryagin numbers QUESTION [15 upvotes]: The Hattori-Stong theorem describes the image of the morphism: $$\tau:\Omega^{SO}_*\rightarrow H_*(BSO;\mathbb{Q})$$ that associates to any closed, smooth, oriented manifold $M$ the homology class $\phi_*([M])$ where $\phi$ is the classifying map $M\stackrel{\phi}{\rightarrow}BSO$ of the tangent bundle and $[M]$ is the fundamental class of $M$. This theorem describes all the relations satisfied by the Pontryagin numbers of smooth manifolds in terms of integrality conditions. Question: do we know the topological analogue of the Hattori-Stong theorem, i.e. do we know all the relations that Pontryagin numbers of topological manifolds should satisfy? Edit: as we have an isomorphism $\Omega^{Top}_*/Tors\cong \Omega^{PL}/Tors$, this question also amounts to knowing all the relations that Pontryagin numbers of $PL$-manifolds should satisfy. REPLY [2 votes]: Check this out: Madsen-Milgram, The Classifying Spaces for Surgery and Cobordism of Manifolds, Corollary 11.26.<|endoftext|> TITLE: Inner product over finite fields QUESTION [9 upvotes]: Let $F$ be a finite field, For every $c \in F$, let $X_1, X_2,..., X_9, Y_1,..., Y_9$ be independent non-zero random variables over $F$. Denote $X=(X_1,...,X_9)$, $Y=(Y_1,...,Y_9)$, also let $\langle X,Y \rangle =\sum_{i}X_i Y_i$. Question: Show that $$\sum_{x,y: \langle x,y\rangle =c} (P[(X,Y)=(x,y) ])^{17/18} \leq 1.$$ Remarks: -feel free to swap $17/18$ for any other positive constant smaller then $1$. -I can prove this for flat distributions (using min-entropy). -I can prove it for dimension $2$ (with constant $1/2$ instead of $17/18$), that is for $X_1, Y_1$ instead of $18$ random variables. I'd be super grateful for the proof or sketch or idea that actually works :-). Best regards, Maciej REPLY [7 votes]: This follows from Young's inequality for convolutions which claims that $\|f*g\|_r\leq \|f\|_p\|g\|_q$ whenever $+\infty\geq p,q,r\geq 1$ and $\frac1p+\frac1q=\frac1r+1$. It can be generalized by the induction as $$ \|f_1*\dots*f_n\|_r\leq \prod_{i=1}^n \|f_i\|_{p_i}, $$ whenever $\sum_{i=1}^n\frac1{p_i}=\frac1r+(n-1)$ and $+\infty\geq p_1,p_2,\dots,p_n,r\geq 1$. We regard $X_i$, $Y_i$ as functions $f\colon F\to[0,1]$ such that $\sum_{a\in F} f(a)=1$. For such functions, we may define two different convolutions: $$ (f*_\times g)(a)=\sum_{xy=a}f(x)g(y) \quad \text{and} \quad (f*_+ g)(a)=\sum_{x+y=a}f(x)g(y). $$ The first one works well on the functions vanishing at zero (and produces also a function vanishing at zero); the second one works well on all functions. For every $i$ we have $$ 1=\|X_i^\alpha\|_{1/\alpha}\|\cdot \|Y_i^\alpha\|_{1/\alpha}\geq \bigl\|X_i^\alpha*_\times Y_i^\alpha\bigr\|_{1/(2\alpha-1)}, $$ as soon as $1/2\leq \alpha\leq 1$. Next, we have $$ 1\geq \prod_{i=1}^n\bigl\|X_i^\alpha*_\times Y_i^\alpha\bigr\|_{1/(2\alpha-1)}\\ \geq \bigl\|(X_1^\alpha*_\times Y_1^\alpha)*_+\dots*_+(X_n^\alpha*_\times Y_n^\alpha)\bigr\|_{1/(2n\alpha-(2n-1))}, $$ whenever $\frac{2n-1}{2n}\leq \alpha\leq 1$. Finally, substitute $\alpha=\frac{2n-1}{2n}$. We get exactly what we need, and even a bit more: $$ 1\geq \max_{a\in F}\sum_{x,y\in F^n\colon \langle x,y\rangle=a}\prod_{i=1}^n X_i(x_i)^\alpha Y_i(y_i)^\alpha. $$<|endoftext|> TITLE: Relation between $h^1(\mathcal{O}_X)$, $h^0(\Omega_X)$, and first betti number for general complex manifold? QUESTION [5 upvotes]: I was wondering in the world of non-Kähler compact complex manifold, what is the relationship between $h^1(\mathcal{O}_X)$, $h^0(\Omega_X)$, and first betti number? The only thing I know is that for surfaces, we have $h^1(\mathcal{O}_X) \geq \frac{b_1}{2} \geq h^0(\Omega_X)$, the reason being that every holomorphic 1-form on a compact surface must be closed. Does this inequality hold true for general compact complex manifold? I'm mainly interested in the relation between $h^1(O_X)$ and $h^0(\Omega_X)$. Another somewhat related question is: does there exist a compact complex manifold with non-closed holomorphic 1-forms? Thanks to an anonymous friend's comment. On page 444 of Griffiths and Harris' book Principles of Algebraic Geometry, they gave an example (Iwasawa manifold) of a non-closed holomorphic 1-form. REPLY [2 votes]: It feels awkward to answer the question posed by myself, but I was told it's okay to do so. There are few points I want to say here. First of all, regarding Remy's question. On a compact complex surface $X$, every holomorphic 1-form is closed, hence $h^0(\Omega_X) \leq h^1(O_X)$ holds true in this case (by David Speyer's answer). So suppose $\eta$ is a holomorphic 1-form on $X$, then $\int_X d\eta \wedge \bar{d\eta}=0$ by Stokes. Then one expands $d\eta \wedge \bar{d\eta}$ in local coordinate to see that this would imply $d\eta=0$. More generally on a compact complex n-fold, every holomorphic (n-1)-form is closed by the same argument. Secondly, my main question is actually whether the inequality $h^0(\Omega_X) \leq h^1(O_X)$ holds in general. Now I knew a counterexample to this inequality which is Iwasawa manifold. One can just consult this and that. Probably I should describe Iwasawa manifold here in case someone forgot the definition. Let $G=\{\begin{bmatrix} 1 &a & b \\ 0 &1 & c \\ 0 &0 & 0 \end{bmatrix}|a, b, c \in \mathbb{C}\}$ be the Heisenberg group of complex coefficient and let $\Gamma$ be the subgroup with entries in $\mathbb{Z}[i]$, then we form a quotient $X=\Gamma \backslash G$. By projecting to $(a,c)$ coordinates, we exhibit $X$ as a holomorphic fibration with fiber $E$ and base $E \times E$, where $E$ is the beautiful elliptic with j-invariant 1728. One compute directly that $\Gamma^{ab}=\mathbb{Z}[i]^2$, hence $h^1(X,\mathbb{C})=4$. One sees (arguably not so obvious) that $da$, $dc$, and $db-adc$ form a basis of $H^0(X,\Omega^1_X)$. Applying Leray spectral sequence to the projection $X \to E \times E$, one can actually show that $h^1(O_X)=2$ (the point being that the differential $d_2: H^0(R^1\pi_*(O_X)=O_{E \times E}) \to H^2(O_{E\times E})$ is nonzero, and it sends 1 to $\bar{da} \wedge \bar{dc}$ if one normalizes everything appropriately). Thirdly, Hopf surface would also give an example where the inequality $h^{01} \geq z^{10}$ in Davide Speyer's answer is not an equality ($z^{10}=h^{10}=0$ in that case). Lastly, I was also wondering if there's an example where the Hodge-deRham doesn't degenerate at $E_2$ page. I was informed by my friend yesterday that such examples (nilmanifold) do exist, but I know nothing about it so perhaps I should stop here.<|endoftext|> TITLE: A representation of the Bernoulli numbers QUESTION [8 upvotes]: Let \begin{equation} \ell_{m,p}:=\sum_{j=1}^m\gamma_{m,j}\sigma_{p,j}, \end{equation} where \begin{equation} \gamma_{m,j}:=\frac{2 (-1)^{j-1} }{j }\binom{2 m}{m+j}\Big/\binom{2 m}{m},\quad \sigma_{p,j}:=\sum _{i=0}^{j-1} \left(\frac{j}{2}-i-1\right)^p, \end{equation} $p$ and $m$ are natural numbers, and \begin{equation} m\ge m_p:=\left\lceil \frac{p+1}{2}\right\rceil. \end{equation} Problem 1 (the more important one to me): Show that $\ell_{m,p}$ does not depend on $m$ (as long as $m\ge m_p$). Problem 2 (a more specific and difficult (?) form of Problem 1): Show that (again for $m\ge m_p$) \begin{equation} \ell_{m,p}=B_p, \tag{1} \end{equation} where $B_p$ is the $p$th Bernoulli number -- see e.g. \url{http://mathworld.wolfram.com/BernoulliNumber.html} and/or \url{https://en.wikipedia.org/wiki/Bernoulli_number}. Formula $(1)$ has been verified for $p=1,\dots,20$ and $m=m_p,\dots,m_p+5$. I think I know how to show that the restriction $m\ge m_p$ cannot be relaxed, for any given natural $p$. These problems arise in a certain work in approximation theory. REPLY [3 votes]: I'll sketch the idea behind your claims, starting with Problem 1. Step 1: Convince yourself that if we denote $$F_p(j):=\frac1j\sum_{i=0}^{j-1}\left(\frac{j}2-i-1\right)^p$$ then $F_p(j)$ is always an even polynomial in $j$; a polynomial in $j^2$. For example, when $p$ is odd, we get $F_p(j)=-\left(\frac{j}2\right)^{p-1}$. Step 2: Therefore, from Step 1, it suffices to consider even powers of $j$ instead of $F_p(j)$. We shall also disregard $\frac{-2}{\binom{2m}m}$ and focus on the sum $$G_p(m):=\sum_{j=1}^m(-1)^j\binom{2m}{m-j}j^{e}$$ where $e\geq0$ is a fixed even integer. Observe that $$\sum_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e =\begin{cases} 2G_p(m) \qquad \qquad \text{if $e\neq0$} \\ 2G_p(m)+1 \qquad \,\,\text{if $e=0$}.\end{cases}$$ Step 3: After re-indexing $k=m-j$, $$\sum_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e=(-1)^m\sum_{k=0}^{2m}(-1)^k\binom{2m}k(m-k)^e.$$ It's well-known (actually the crux of the matter here) that if $n>a$ then we've the vanishing of $$\sum_{k=0}^n(-1)^k\binom{n}k k^a=0.\tag{0}$$ The reason why you get a persistent value (for $m\geq m_p$) is caused by the one single term when $e=0$ which happens for $p$ even. At any rate, we gather that if $e$ is even and $m\geq \frac{e}2$ then $$G_p(m)=\begin{cases} \,\,\,\,\,0 \qquad \,\text{when $e\neq0$} \\ -\frac12 \qquad \text{when $e=0$}. \end{cases}$$ To understand Problem 2, it remains to identify the constant term in $F_p(j)$ w.r.t. the variable $j$. To this end, the Binomial Theorem furnishes \begin{align} F_p(j) =\frac1j\sum_{i=1}^j\sum_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-k}(-1)^ki^k =\sum_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-1-k}(-1)^k\sum_{i=1}^ji^k \tag{1} \end{align} revealing that the only way to encounter the sought-after constant term is provided that $k=p$. Keeping in mind that $\sum_{i=1}^ji^k$ is divisible by $j$, we just look at the quantity (using know expressions for Bernoulli numbers) $$\frac{(-1)^p}j\sum_{i=1}^ji^p=(-1)^p\left[j^{p-1}+\frac1{p+1}\sum_{r=0}^p\binom{p+1}rB_rj^{p-r}\right].\tag{2}$$ Thus, the term we seek is $(-1)^pB_p=B_p$ (remember: $p$ is even) leading to \begin{align} \frac2{\binom{2m}m}\sum_{j=1}^m(-1)^{j-1}\binom{2m}{m+j}F_p(j) &=\frac{2B_p}{\binom{2m}m}\sum_{j=1}^m(-1)^{j-1}\binom{2m}{m+j} \\ &=\frac{2B_p}{\binom{2m}m}\frac12\binom{2m}m \\ &=B_p \end{align} which is exactly what we want to arrive at. We've already known from above the $B_p=0$ when $p$ is odd. The proof is complete for both problems.<|endoftext|> TITLE: What kind of compactness does "expanding $\mathbb{R}$ by constants" have? QUESTION [10 upvotes]: EDIT: in retrospect this question should have been split up; I've accepted Joel's answer to the first part below, and asked the second part here. This question is crossposted at MSE; however, it has not gotten any responses there despite a bounty, so I'm asking it here. Say that a theory $T$ in the language of ordered fields + constants is $\mathbb{R}$-satisfiable if it has a model whose ordered field part is $\mathbb{R}$, with the usual ordered field structure. I'm interested in what compactness-like properties $\mathbb{R}$-satisfiability has. To begin with, we can easily knock off compactness itself: since $\mathbb{R}$ is Archimedean, $\mathbb{R}$-satisfiability is not compact. But things get murkier when we consider “compactness at higher cardinalities.” For cardinals $\kappa<\lambda$, say that $\mathbb{R}$-satisfiability is $(\kappa, \lambda)$-compact if whenever $\Gamma$ is a set of sentences of cardinality $<\lambda$, and every subset of cardinality $<\kappa$ is $\mathbb{R}$-satisfiable, then $\Gamma$ is $\mathbb{R}$-satisfiable. (So usual compactness is $(\omega, \infty)$-compactness, and countable compactness is $(\omega,\omega_1$)-compactness.) It’s easy to show that $\mathbb{R}$-satisfiability is not $(\omega_1,\omega_2)$-compact: any countable linear order embeds into $\mathbb{R}$, but $\omega_1$ does not. And it follows from Easton’s theorem that for every cardinal $\kappa$ with $cf(\kappa)>\omega$, it is consistent with ZFC that $\mathbb{R}$-satisfiability is not $(\kappa^+, \kappa^{++})$-compact (look at a theory asserting the existence of $\kappa^+$-many distinct reals assuming $\mathfrak{c}=\kappa$). This suggests two natural questions: Can ZFC prove that $\mathbb{R}$-satisfiability is $(\omega_{\omega+1}, \omega_{\omega+2})$-compact? Is it consistent with ZFC that $\mathbb{R}$-satisfiability is $(\omega_2, \omega_3)$-compact? I believe the answer to the second question should be a relatively easy "yes", while the first question should be "no" but might require some work. However, I don't immediately see how to resolve either piece. REPLY [7 votes]: The answer to the first question is no, by an easy argument. What I claim is that ZFC proves that $\newcommand\R{\mathbb{R}}\R$-satisfiability is not $(\mathfrak{c},\mathfrak{c}^+)$-compact. (This improves upon your observation about $(\kappa^+,\kappa^{++})$.) To see this, let $T$ be the elementary diagram of $\R$, with a constant for every real, plus another constant, asserted to be none of them. Every size-less-than-continuum subtheory of this theory is $\R$-satisfiable. But for the whole theory, the only way to interpret the $\R$-constants in an $\R$-structure is as themselves, since the rational constants will be interpreted as themselves and the other constants are then constrained by the type of their cuts in the rationals. So there is nobody left for the one extra constant. In particular, if $2^\omega=\aleph_{\omega+1}$, then this shows $\R$-satisfiability needn't be $(\omega_{\omega+1},\omega_{\omega+2})$-compact.<|endoftext|> TITLE: Laws characterizing the trivial group QUESTION [13 upvotes]: What are laws characterizing the trivial group? I mean all group words $w$ such that if the identity $w=1$ holds in a group $G$, then $G=1$. For example, it can be easily verified that if a group word $w=x_{i_1}^{\alpha_1}x_{i_2}^{\alpha_2}\cdots x_{i_t}^{\alpha_t}$ has the following property, then it characterizes the trivial group: $$\exists J\subseteq \{x_{i_1},\ldots, x_{i_t}\}:\ gcd(deg_{x_{i_j}}(w): x_{i_j}\in J)=1.$$ It seems that such a problem must be already studied. Anybody knows a reference? P.S. Note that a given non-trivial group cannot be characterized by a law, because the identities of any group $A$ and its powers are the same. REPLY [3 votes]: Every word $w$ on free generators $x_1,\dots,x_n$ can be written as $$w=x_1^{\alpha_1} \cdots x_n^{\alpha_n} c(x_1,\dots,x_n),$$ where $c$ is a word in the commutator subgroup of $\langle x_1,\dots,x_n \rangle$ and $\alpha_1,\dots, \alpha_n$ are some integers. Note that the integer $\alpha_i$ is the same as the sum of the exponents of letter $x_i$ in the word $w$. You have denoted $\alpha_i$ by $\deg_{x_i}(w)$. Now we can state our characterization of such words: A word $w$ is charaterizing the trivial group if and only if $\gcd (\deg_{x_1}(w),\dots,\deg_{x_n}(w) )=1$. This is because if $d:=\gcd(\alpha_1,\dots,\alpha_n)\neq 1$, then the cyclic group of order $d\neq 1$ satisfies $w$. Now consider $w(1,\dots,a,\dots,1)$, where $a$ is an arbitrary element of the group satisfying the law $w$ and $a$ is in the possition $i$. It follows that $a^{\alpha_i}=1$ for all $i=1,\dots,n$ (note that $c(1,\dots,a,\dots,1)=1$). Thus $a^d=1$ and so $a=1$.<|endoftext|> TITLE: Current state of Straus's illumination problem QUESTION [8 upvotes]: In George W. Tokarsky's Polygonal Rooms Not Illuminable from Every Point (1995) it is stated that the problem Is a polygonal region illuminable from at least one point in the region? was still open at the time. What is its current state? Has it been settled or is it still open? REPLY [10 votes]: As far as I know, the specific question you ask (entirely illuminable from at least one point) remains open. (Tokarsky showed that placing a light in some spots can leave some points dark.) But you may be interested to know that an old conjecture has been settled, as I reported in this earlier question: Lelièvre, Monteil and Weiss have shown that if $P$ is a rational polygon, for every $x$ there are at most finitely many $y$ not illuminated by $x$. Lelievre, Samuel, Thierry Monteil, and Barak Weiss. "Everything is illuminated." arXiv:1407.2975 (2014). REPLY [6 votes]: This started as a comment, but became too long. It might be worth mentioning that from a physics point of view, this problem has a certain ambiguity that somewhat diminishes its interest: when the walls are mirrors it is clear what happens when a light ray hits the wall (specular reflection), but what happens when a light ray hits a vertex? The answer to the "illumination problem" depends crucially on how one treats the vertices, because the "dark points" (points inside the region that cannot be illuminated by a point source of light) may appear only if one assumes that a light ray that hits a vertex is extinguished. (Diffuse reflection, rather than specular reflection, might seem a more natural assumption from the physics point of view.) A variation on this problem that avoids this ambiguity, is to exclude dark points of measure zero, and then ask whether a point source can illuminate the entire interior up to points of measure zero. This is listed as an open problem in the 2008 collection of open problems in computational geometry, by Eric Demaine and our own Joseph O'Rourke.<|endoftext|> TITLE: Transitive actions of finite subgroups of ${\rm GL}(n,\Bbb Z)$ on projective geometries QUESTION [16 upvotes]: For any $n$, the group ${\rm GL}(n,\Bbb Z)$ has a natural action on $\Bbb Z^n$. Modding out a prime $p$ yields an action on the vector space $F_p^n$, where $F_p$ is the finite field with $p$ elements. Projectivising gives an action on the finite projective geometry $PG_{n-1}(p)$. Therefore, every subgroup of ${\rm GL}(n,\Bbb Z)$ has a natural action on $PG_{n-1}(p)$, for every prime $p$. Question: Do there exist arbitrarily large primes $p$ such that, for some $n>1$, there is a finite subgroup of ${\rm GL}(n,\Bbb Z)$ whose natural action on the points of $PG_{n-1}(p)$ is transitive? Motivation: in https://arxiv.org/abs/1502.06114, Corollary 6.6, I defined a function $k_n$; in my proof of its existence, this function grows with $n$. The referee has asked the reasonable question of whether or not the actual function is unbounded. I believe this is equivalent to the question I'm asking here, but I don't know the answer. REPLY [14 votes]: Probably final revision: I am indebted to Dave Witte-Morris, who added a reference to a refinement of Zsigmondy's Theorem by W. Feit, of which I was unaware, and pointed out that consequently, a complete answer to the question implicitly followed from what was previously written. In fact, going beyond Dave Witte-Morris's original suggestion (but still making use of Feit's result in its more precise statement as his Theorem A), we may obtain the sharper conclusion that if there is a transitive action of the form requested by the OP, then $p \leq 3$, so I now incorporate this observation. The order of a finite subgroup $X$ of ${\rm PGL}(n,\mathbb{Z})$ is always a divisor of $\frac{(2n)!}{2}$( by the result of Blichfeldt or Minkowski mentioned below, applied to the preimage $G$ of $X$ in ${\rm GL})$. In particular, the order of a finite subgroup of ${\rm PGL}(2,\mathbb{Z})$ divides $12.$ However, we note that ${\rm GL}(2,\mathbb{Z})$ has no quaternion subgroup of order $8,$ from which it easily follows that a finite subgroup of ${\rm GL}(2,\mathbb{Z})$ has a normal $2$-complement (necessarily of order dividing $3$), and also that ${\rm PGL}(2,\mathbb{Z})$ has no subgroup of order $12.$ Now ${\rm GL}(6,\mathbb{Z})$ contains no element of order $31,$ so we need not concern ourselves with the case $p =5, n = 6.$ A finite group $H$ which acts transitively on a set of $\frac{p^{n}-1}{p-1}$ points with $n >1$ has order at least $p+1$ (and if $n =2$, then such a group has order divisible by $p+1$). Hence if there is a prime $p$ as asked for in the question, we certainly have $2p < (2n)!$. If $n = 2,$ we can exclude $p = 7,$ since $p+1 =8$ does not divide the order of ${\rm PGL}(2,\mathbb{Z}).$ If $X$ is a finite subgroup of ${\rm PGL}(2,\mathbb{Z})$ of order $6,$ then $X$ is the image of a subgroup $G \cong ( \mathbb{Z}/2 \mathbb{Z}) \times S_{3}$ in which every non-central involution has the eigenvalue $1$. Thus $X$ does not act regularly on ${\rm PG}_{1}(5),$ and hence does not act transitively as $|X| = 6 =|{\rm PG}_{1}(5)|.$ We can also exclude the possibility $p =11$ when $n =2,$ since no subgroup of ${\rm PGL}(2,\mathbb{Z})$ has order $12.$ Hence we may ( and do, from now) assume that $n >2$ and $p \geq 5.$ Furthermore, if $ p = 5,$ we may suppose that $n \neq 6,$ so we do. A strengthening of Zsigmondy's theorem proved by Feit as his Theorem A shows (as $p \geq 5$ and $n >2,$ with $n \neq 6$ if $p=5$), that there is a "large" Zsigmondy prime divisor $q$ of $\frac{p^{n}-1}{p-1}$, which means we can assume either that $q > n + 1$ or else that $p^n - 1$ is divisible by $q^2$. Then $q$ divides $\frac{p^{n}-1}{p-1}$ and $p$ has multiplicative order $n$ in $\mathbb{Z}/q\mathbb{Z},$ so that $q \equiv 1$ (mod $n$). However, by Cauchy's theorem, $G$ contains an element, say $x,$ of order $q.$ But $\langle x \rangle$ has no non-trivial irreducible representation of degree less than $q-1$ over $\mathbb{Q},$ so we must have $n = q-1$ (this also implicitly shows that $G$ contains no element of order $q^{2}).$ Thus $x$ has trace $-1$ in the given representation, and a Lemma of Blichfeldt (or maybe Minkowski, as J-P. Serre attributes it) shows that $\langle x \rangle$ must be a Sylow $q$-subgroup of $G,$ so that, in particular, $\frac{p^{n}-1}{p-1}$ is not divisible by $q^{2}.$ Since $q$ is a Zsigmondy prime, we know that $p - 1$ is not divisible by $q$, so this implies that $p^{n}-1$ is not divisible by $q^{2}.$ This contradicts the fact that $q$ is a "large" Zsigmondy prime. ( For the sake of completeness, I outline the argument of Blichfeldt used, or an alternative using basic character theory: Let $Q$ be a Sylow $q$-subgroup of $G,$ and let $\chi$ denote the character of $G$ afforded by the given representation. Then $Q$ has exponent $q,$ and we have see that $\chi(x) = -1$ for each non-identity element of $Q$. Now the (necessarily integral) multiplicity of the trivial character in ${\rm Res}^{G}_{Q}(\chi)$ is given by $\frac{1}{|Q|} ( (q-1) - (|Q|-1)).$ Hence $|Q|$ divides $|Q|-q$ which forces $|Q| = q.$ The result of Blichfeldt alluded to is the observation that if $\mu$ is a faithful complex character of a finite group $X$ and $c_{1},c_{2}, \ldots, c_{r}$ are all the distinct values assumed by $\mu$ on non-identity elements of $X$ then $|X|$ divides $\prod_{i=1}^{r}( \mu(1) - c_{i})).$<|endoftext|> TITLE: Geometrical meaning of admissible hermitian metric on a line bundle QUESTION [7 upvotes]: Let $(X,\Omega)$ be a complex compact Kahler manifold, where $\Omega$ is the fundamental $(1,1)$-form. Moreover let $L$ be a holomorphic line bundle on $X$. A (smooth) hermitian metric $h$ on $X$ is said admissible with respect to $\Omega$ if $$c_1(L,h)=a\Omega$$ for $a\in\mathbb R$. Here $c_1(L,h)$ is the curvature form or first Chern form of the hermitian line bundle $(L,h)$. I'd like to visualize geometrically (in some informal way) what is the meaning of the condition $c_1(L,h)=a\Omega$. Admissible hermitian metrics are unique up to a multiplication by a smooth function, so they seem to be very particular hermtian metrics. What is their peculiarity? REPLY [2 votes]: I don't know if this answers to your question, and very likely you already know it, but one easy fact is the following. First of all an obvious necessary condition is that at the level of cohomology classes we have $$ c_1(L)=a[\Omega]. $$ Once you have this, you put any smooth hermitian metric on $L$, say $h_0$, and look at its Chern curvature $c_1(L,h_0)=i/2\pi\,\Theta(L,h_0)$. Since it defines the same cohomology class as $a\Omega$, and since $X$ is Kähler, the $\partial\bar\partial$-lemma tells you that you can find a smooth function $f\colon X\to\mathbb R$ such that $$ c_1(L,h_0)-a\Omega=\frac i{2\pi}\partial\bar\partial f. $$ Now, set $h=e^{f}h_0$. Then, $$ c_1(L,e^fh_0)=-\frac i{2\pi}\partial\bar\partial\log(e^fh_0)=c_1(L,h_0)-\frac i{2\pi}\partial\bar\partial f=a\Omega. $$<|endoftext|> TITLE: Is there any set theory $T$ such that $T$ plus true arithmetic is complete with respect to statements in set theory? QUESTION [9 upvotes]: Is there an effective set theory $T$ such that $T + $$TA$ is consistient and complete. It should at least prove all theorems of $ZF$ true, so that it is a "standard" set theory. In particular, the axiom of infinity is required, since otherwise finite set theory would work, making the answer trivial. By complete, I mean every statement in the language of set theory is either provable or disprovable. Another way of phrasing this question is what are a set of axioms of $T$ (which is either finite or such that its contents can be listed by a Turing machine), such that for any statement in the language of set theory, it can be decided by $T$ and statements in arithmetic (which are true of the standard integers). If it fails, perhaps the requirement that $T$ be effective could be replaced by a requirement that $T$ is an Arithmetical set. Also, perhaps $TA$ could be replaced with some other consistent and complete theory of arithmetic (although this wouldn't be as useful as with $TA$). The nice thing about such a theory is that, once $T$ is finalized, selecting new axioms would be metaphysically simpler, you could say. Instead of arguing over whether a given axiom is actually true of sets or not, the problem would be reduced to arguing about whether statements in arithmetic are true of integers or not (after $T$ has been selected). tl;dr. Can you have a complete set theory, modulo arithmetic. REPLY [12 votes]: There can be no such theory $T$, even if you weaken the requirement to $T$ being merely arithmetically definable, rather than insisting it must be effective. To see this, consider the theory T+TA, which should be consistent and complete. Let $X$ be the set of Gödel codes of assertions in this theory, and let $A$ and $B$ be $X$-computably inseparable sets, which means that they are disjoint and with an oracle for $X$, we can enumerate each of them, but there is no $X$-computable set containing $A$ and disjoint from $B$. Fix specific programs for enumerating $A$ and $B$ from an oracle for $X$. Now, let $C$ be the set of numbers $n$ such that T+TA proves that they are enumerated into $A$ before they are enumerated into $B$, using the fixed programs and using what the theory thinks is TA. Because every member of $A$ is enumerated into $A$ by that program on account of only finitely much of the oracle, the theory T+TA will agree on that element being enumerated into $A$, and similarly with $B$. Thus, the set $C$ contains every actual element of $A$ and no actual element of $B$. But $C$ is computable from $X$, since we need only search for proofs from T+TA. Contradiction. So there is no such theory T.<|endoftext|> TITLE: Relationship between motivic Galois groups and Langlands program QUESTION [5 upvotes]: I would like to know if there is any relationship between the motivic Galois groups and the Langlands program. Many thanks. REPLY [4 votes]: Take $K=\mathbb{Q}$ for simplicity, but this applies to any number field $K$, and let $L$ be the Langlands group and $\mathcal{G}$ the motivic Galois group of $\mathbb{Q}$. Then the conjectural relation between automorphic forms and motives (the Langlands program) implies that there is an homomorphism (up to a certain conjugation) such that the diagram $$\require{AMScd}\begin{CD} L @>{}>> \mathcal{G};\\ @VVV @VVV \\ \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) @>{}>>\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}); \end{CD}$$ is commutative. To get a better idea of both sides of this picture it is useful to use some classic (and better understood) intermediate groups. Let $W$ be the Weil group, $\mathcal{S}$ the Serre group and $\mathcal{T}$ the Taniyama group, then we have something like this $$L \longrightarrow W \longrightarrow \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(automorphic side)}$$ $$\mathcal{G} \longrightarrow \mathcal{S} \longrightarrow \mathcal{T} \longrightarrow \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(motivic side)}$$ Most of this goes back to Robert Langlands, "Automorphic representations, Shimura varieties and motives" (1977) Some other relevant references are James Milne, Kuang-yen Shih, "Langlands's Construction of the Taniyama Group" (1982) Norbert Schappacher, "CM motives and the Taniyama group" (1994) James Arthur, "A note on the automorphic Langlands group" (2002)<|endoftext|> TITLE: If $X$ is a genus $g\geq 2$ curve over a number field $K$, then is there a bound on $X_L(L)$ for $L/K$ that depends only on $X$ and $[L:K]$? QUESTION [6 upvotes]: If $X$ is a genus $g\geq 2$ curve over a number field $K$, then $X(K)$ is finite by Falting's Theorem. My question is how does $X_L(L)$ behave for finite field extensions $L/K$? In particular, is there a bound on $X_L(L)$ for $L/K$ that depends only on $X$ and $[L:K]$? REPLY [11 votes]: Assuming the conjecture that varieties of general type cannot have a Zariski dense set of points over a number field (which is a consequence of the more precise Vojta conjectures, for example), Patricia Pacelli [1] proved the following, which is much stronger than what you're asking, since the constant only depends on the genus of $X$: Theorem Fix $d\ge1$ and $g\ge2$. There is a constant $C=C(g,d)$ such that for all number fields $L/\mathbb Q$ of degree at most $d$ and all smooth projective curves $X/L$ of genus at most $g$, one has $\#X(L)\le C(g,d)$. Pacelli's result is a generalization of an earlier result by Caparaso, Harris, and Mazur [2] that wasn't quite as uniform, but that really set in motion the use of this conjecture to prove uniformity result of this sort. But if you want unconditional results, I don't think there's anything known in general, although one might(?) be able to get something like $$ \#X(L) \le C(X,[L:K])^{1+\text{rank Jac}_X(L)}. $$ [1] Patricia L. Pacelli, MR 1448017 Uniform boundedness for rational points, Duke Math. J. 88 (1997), no. 1, 77--102. [2] Lucia Caporaso, Joe Harris, and Barry Mazur, MR 1325796 Uniformity of rational points, J. Amer. Math. Soc. 10 (1997), no. 1, 1--35.<|endoftext|> TITLE: Where can I find details of Elie Cartan's thesis? QUESTION [12 upvotes]: I am interested in the details of Elie Cartan's thesis, and, more specifically the explicit construction of the exceptional Lie groups as groups of symmetries of some specific homogeneous polynomials (according to what I have read in many places). I am interested in the details. For instance, what does one such a polynomial look like? What do all such polynomials (with groups of symmetry a fixed exceptional Lie group) look like? Is the set of all such polynomials dense in the relevant polynomial space? Also, I am interested in different realizations of the exceptional Lie groups: for instance, $G_2$ is the group of symmetries of a skew-symmetric trilinear form on $\mathbb{C}^7$. And in this case, the space of such skew-symmetric trilinear forms is dense in the set of all skew-symmetric trilinear forms on $\mathbb{C}^7$. So I guess in general, I am interested in concrete and explicit realizations of the exceptional Lie groups as groups of symmetries of a specific algebraic object. Also what is the set of all such objects with this property? And, is the set of all such objects dense in the relevant vector space? I hope my questions are sufficiently clear! REPLY [3 votes]: All details about Cartan's thesis can be found in the thesis itself: https://archive.org/details/surlastructured00bourgoog There is also a German translation for those who do not read French: Ueber die einfachen Transformationsgruppen. (German) JFM 25.0638.01 Leipz. Ber. XLV. 395-420 (1893).<|endoftext|> TITLE: Uncertainty principle QUESTION [14 upvotes]: A version of the uncertainty principle says that a function and its Fourier transform cannot be both with compact support: it is not difficult to prove since a compactly supported distribution has an entire Fourier transform. Another version is that $$ \left\Vert{\frac{du}{dx}}\right\Vert_{L^2(\mathbb R)} \left\Vert{xu}\right\Vert_{L^2(\mathbb R)}\ge \frac12\left\Vert{u}\right\Vert_{L^2(\mathbb R)}^2, $$ and many other quantitative versions are available for functions bounded above by Gaussians functions as well as their Fourier transform. All this seems to be compatible with the existence of a function $u$ in $L^2(\mathbb R)$ such that $$ \text{support u}\subset \mathbb R_+,\quad \text{support $\hat u$}\subset \mathbb R_+. $$ My question: is there an "explicit" example of such a function? REPLY [23 votes]: Yes, there is one such example: $u \equiv 0$. The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications). By Titchmarsh's theorem, if $u\in L^2(\mathbb{R})$ and its Fourier support is on the positive real line, $u$ must be equal to the trace of some holomorphic function $F$ defined on the upper half plane. If $u$ itself further vanishes on the left half line, which has positive measure, by the Luzin-Privalov Theorem the function $F$ must vanish identically. Hence the only function satisfying your condition is identically zero.<|endoftext|> TITLE: Expected Size of Independent Set QUESTION [5 upvotes]: Q. Let $G = (V, E)$ be a graph with $V = \{v_1, \cdots, v_n\}$ and $E = \{(v_i, v_{i+1}) \mid 1 \leq i < n\}$. If we repeatedly remove vertices from $G$ uniformly randomly until the set of vertices $V'$ remained constitute an independent set of the original $G$ (i.e., $\forall u, v \in V'$, $(u, v)\notin E$), what is the expected size of the independent set $V'$? REPLY [3 votes]: [Previous answer was completely wrong.] Denote by $p_k$ the probability that at least $k$ vertices remain. Then the expectation is of course $\sum p_i$. I claim that $p_k=\frac{(n-k+1)! (n-k)!}{(n-2k+1)! n!}$, it is about $e^{-k^2/n}$ and thus the asymptotics is the same as for $\sum e^{-k^2/n}\sim \int_0^\infty e^{-x^2/n}dx=\frac12 \sqrt{\pi n}$. I do not know whether the sum $\sum p_k$ may be simplified. Now the claim. We need to prove that the probability that exactly $k$ vertices remain equals $p_k-p_{k+1}$. It is convenient to replace the graph to the cycle with $n+1$ vertices (it becomes a path after the first move). Fix the last removed vertex $v_0$, and the remained set $A$, $|A|=k$, $A$ should be independent and contain at least one neighbor of $v_0$. The probability of this event (fixed $v_0$, $A$) equals $\frac1{\binom{n+1}{k+1}(k+1)}$. It has to be multiplied by the number of ways to choose $v_0$ and $A$. There are $n+1$ ways for $v_0$ and $\binom{n-k+1}{k}-\binom{n-k-1}{k}$ ways to choose $A$ after that. It remains to note that $$\frac{\binom{n-k+1}{k}}{\binom{n+1}{k+1}(k+1)}=p_k,\,\, \frac{\binom{n-k-1}{k}}{\binom{n+1}{k+1}(k+1)}=p_{k+1}. $$<|endoftext|> TITLE: Teichmüller space on non-orientable closed surfaces QUESTION [8 upvotes]: It is known that any closed orientable surface of genus $g \geq 2$ admits a hyperbolic metric, and the Teichmüller space of such metrics has dimension $6g - 6$. I was wondering if there is a corresponding statement for non-orientable closed surfaces of negative Euler characteristic. By the Gauss-Bonnet theorem and Ricci flow theory, for example, I can see that such surfaces admit many hyperbolic metrics, but I am not sure about the structure of the space of such metrics. Also, on the Teichmüller space for a genus $g \geq 2$ closed orientable surface, there is a concept of distance (see, for example, Definition 6.4.1 of Buser's book "Geometry and spectra of compact Riemann surfaces"). Is there a similar concept in the non-orientable case? Thanks in advance! REPLY [2 votes]: A very nice paper about the Teichm\"uller space of non-orientable surfaces, Fenchel-Nielsen coordinates and other generalizations of Thurston's theory to non-orientable surfaces is this one by Papadopoulos and Penner: https://arxiv.org/pdf/1309.0383.pdf<|endoftext|> TITLE: The Freedman Dichotomies QUESTION [12 upvotes]: Definition: A categorical dichotomy is said to be a “Freedman Dichotomy” if the sole evidence for its non-vacuity is the Disk Theorem (Theorem 1.1 of (F82)). This claims that a Casson handle is homeomorphic to a standard 2-handle (rather than just homotopic to a standard 2-handle, as originally proven by Casson (C86)). Question: What are all of the known Freedman Dichotomies? Let me illustrate by presenting examples I have found. Definition: A smooth manifold is called “exotic” if it is homeomorphic to some standard smooth manifold, but not diffeomorphic to it. Examples of exotic spheres in dimension 7 and above were famously constructed by Milnor in 1959 (M59). No examples of exotic manifolds of lower dimension are known, other than in dimension 4, where examples arise due to Freedman’s work. Answer 1: 4-manifold exoticness is a Freedman Dichotomy. In a previous MO question I sought, without success, to find any evidence for the Disk Theorem other than that contained in (F82). A follow-up MO question sought to quarantine the implications of the Disk Theorem with regard to exotic 4-manifolds. The result of these questions is that, without the Disk Theorem there would be no evidence for exoticness in 4 dimensions. In fact, this can be improved: Answer 2: The distinction between “small” and “large” exotic 4-manifolds is also a Freedman (sub)Dichotomy. Small exotic structures arise from the contradiction between the Disk Theorem and the smooth h-cobordism theorem, while large exotic arise from the contradiction between the Disk Theorem and smooth connected-sum-splitting. A good overview of this is contained in (Sc05). Moving on, we have the following. Definition: A knot in the 3-sphere (the boundary of a 4-ball) is “smoothly slice” if it bounds a proper smoothly embedded disk in the 4-ball. A knot is “topologically slice” if it bounds a proper continuously embedded disk $D$ which can be extended to a continuous embedding of $D\times D$ into a normal neighbourhood of the disk. Answer 3: The distinction between topological sliceness and smooth sliceness for knots in dimension 3 is a Freedman Dichotomy. The reason is that the existence of a topological slice disk for knots with Alexander polynomial equal to 1 follows solely from the Disk Theorem, while the non-existence of smooth slice disks comes from smooth knot theory – often allied to gauge theory or smooth invariants of knots. The E8 lattice is a symmetric $8\times 8$ matrix of integers that appears in the study of Lie groups. As it is unimodular, the question arises as to whether E8 is the intersection form of a closed simply connected 4-manifold (which must be unimodular). Since the E8 lattice has signature 8, a theorem of Rohlin says that it cannot be the intersection form of a smooth closed simply connected 4-manifold (which must have signature divisible by 16). Answer 4: The existence/non-existence of a (non-smooth) topological 4-manifold with E8 intersection form is a Freedman Dichotomy. Such a 4-manifold exists by the classification of topological 4-manifolds, which once again, follows from the Disk Theorem. Definition: A “triangulation” of a topological manifold is a homeomorphism to a locally finite simplicial complex. Topological manifolds of dimension greater than 4 which do not admit a triangulation have been known for decades (KS77). Answer 5: The triangulability/non-triangulability of 4-manifolds is a Freedman Dichotomy. Unpublished work of Casson relates his invariant to the triangulability or otherwise of topological 4-manifolds, whose existence is guaranteed by their classification - see the exposition (AMcC14). So what other Freedman Dichotomies exist? And, of course, any corrections or comments on the above five examples are welcome. (AMcC14) Akbulut, Selman, and John D. McCarthy. Casson's Invariant for Oriented Homology Three-Spheres: An Exposition.(MN-36). Princeton University Press, 2014. (C86) Casson, A. J. (1986), "Three lectures on new infinite constructions in 4-dimensional manifolds", A la recherche de la topologie perdue, Progr. Math., 62, Boston, MA: Birkhauser Boston, 201-244. (F82) Freedman, Michael Hartley. "The topology of four-dimensional manifolds." J. Differential Geom 17.3 (1982): 357-453. (KS77) Kirby, Robion C., and Laurence Siebenmann. Foundational essays on topological manifolds, smoothings, and triangulations. No. 88. Princeton University Press, 1977. (M59) Milnor, John. "Differentiable structures on spheres." American Journal of Mathematics 81.4 (1959): 962-972. (Sc05) Scorpan, Alexandru. The wild world of 4-manifolds. American Mathematical Soc., 2005. REPLY [6 votes]: One of Freedman's results is that a homology $3$-sphere admits a tame topological embedding into $\mathbb R^4$. So here is an odd fact: Let $M$ be the Poincare dodecahedral space. There is an open subset of $\mathbb R^4$ homeomorphic to $M \times \mathbb R$. But there is no open subset of $\mathbb R^4$ diffeomorphic to $M \times \mathbb R$. The obstruction to the smooth embedding of $M$ comes from the Rochlin invariant of the smooth $4$ manifold $M$ bounds in $\mathbb R^4$. One additional twist to this is that $M \setminus \{*\}$, i.e. the once-punctured Poincare Dodecahedral Space admits a beautiful smooth embedding in $\mathbb R^4$. The problem with this embedding with respect to the above is that the sphere surrounding the puncture point is knotted, i.e. the punctured Poincare dodecahedral space is a Seifert Surface for a non-trivially embedded $S^2$ in $\mathbb R^4$. So there's no clear way of how to modify this embedding to get an embedded $M$. These embeddings of Freedman's use the standardness of Casson 2-handles. I have not seen a construction of any such embedding by any other means. edit: I suppose one reason the above could be deemed hard to accept is you would think you could `smooth' a topological embedding $M \to \mathbb R^4$. Smooth maps are dense, so every topological embedding can be approximated by a smooth map, but that smooth map (I presume) would have degeneracies. I think it would be useful to complete this picture to describe the degeneracies of such smooth approximations, in some combinatorial language.<|endoftext|> TITLE: Smoothness of the branch divisor and ramification on surfaces QUESTION [6 upvotes]: Let $f \colon X \longrightarrow Y$ be a finite morphism of degree $n \geq 2$ between smooth compact, complex surfaces. Let $B \subset Y$ be the branch divisor of $f$ and assume that the corresponding branching order is $2$, namely $$f^*B = 2R + R_0,$$ where $R \subset X$ is the ramification divisor and $R_0$ is the residual curve (sometimes this condition is expressed by saying that $f$ is a generic cover). Question. If the branch locus $B$ is a smooth divisor, is it true that $RR_0=0$? I know that the answer is yes for $n=2$ (trivially, because $R_0$ is empty in that case) and for $n=3$ (as a consequence of the general theory of triple covers developed by R. Miranda), but what happens for general $n$? REPLY [6 votes]: It seems to me that the intersection is zero in general, i.e. the answer is YES. Let us prove that $f^{-1}(B)$ is a smooth curve in $X$. This clearly implies the desired result. The proof of smoothness of $f^{-1}(B)$ relies one the following definition and a lemma. Definition. Let $C$ be a complex curve in a complex surface $S$ and $x\in C$. Let us call the local fundamental group of $S\setminus C$ at $x$ to be $\pi_1(U_x\setminus (C\cap U_x))$ where $U_x$ is a sufficiently small neighborhood of $x$. I think that the following lemma is correct. Lemma. A complex analytic curve $C$ in $X$ is smooth at $x\in C$ iff the local fundamental group of $C$ at $x$ is $\mathbb Z$. If this lemma holds, then it is easy to see indeed that in your question the total perimage of $B$ is smooth, because for any point $x\in B$ and a sufficiently small neighborhood of $U$ of $x$ the map from $f^{-1}(U\setminus (U\cap C))$ to $U\setminus (U\cap C)$ is etale. I.e. all connected components of $f^{-1}(U\setminus (U\cap C))$ have $\pi=\mathbb Z$. Intuition for the lemma. I think that the lemma can be obtained as a corollary of Mumford's result stating that a point $x$ on a surface is smooth iff its link is the 3-sphere $S^3$. Here is why. Indeed, suppose that $\pi_1(U_x\setminus (C\cap U_x))=\mathbb Z$. This means that the intersection of $C$ with the boundary of $U_x$ is an unknot. Indeed, the only knot is $S^3$ whose complement has fundamental group $\mathbb Z$ is the unknot. Now, let us take a double cover of $U_x\setminus (C\cap U_x)$ and take its analytic completion. This is just a double cover of $U_x$ and its boundary is again $S^3$. So this double cover is smooth. Finally we see that $U_x$ is a smooth quotient of a ball by a smooth involution, hence its branching set (i.e., $C$) is a smooth curve.<|endoftext|> TITLE: Taut foliations and closed leaves QUESTION [8 upvotes]: EDITED A codimension-1 foliation $\mathcal{F}$ of a (closed, connected, oriented) 3-manifold is taut if there exists a simple closed curve $\gamma$ that intersects each leaf of $\mathcal{F}$ transversely. Using work of Thurston, you know that if a taut foliation has closed leaves, they must be homologically non-trivial. But how do you know when a taut foliation admits closed leaves at all? Is there some way to tell, or some class of manifolds which you know support a taut foliations with closed leaves? Thanks! REPLY [4 votes]: The existence of compact leaves of codimension 1 foliations defined on 3-manifolds is characterized by Novikov see: https://en.wikipedia.org/wiki/Novikov%27s_compact_leaf_theorem If you read Novikov paper: (Theorem 6.1) and (Theorem 7.1) imply that if the image of $t(A)\rightarrow \pi_1(M)$ has a non trivial kernel, then the foliation has a compact leaf, here $t(A)$ is the semi-group of homotopy class of tranversal through a point $x$ the leaf $A$. In particular, for a taut foliation one can define the class of the closed transversal in $t(A)$ for any leaf and apply the previous theorems if this class is a non trivial element of the kernel $t(A)\rightarrow \pi_1(M)$. http://www.mi.ras.ru/~snovikov/23.pdf<|endoftext|> TITLE: Degree of irrationality and hyperelliptic curves QUESTION [5 upvotes]: For a variety $V$ of dimension $n$, let $Irr(V)$ denote the minimal degree of a dominant rational map $V\to \mathbb{P}^n$. Suppose that a curve $X$ admits a dominant map from a variety $V$ with $Irr(V)=2$. Does it follow that $X$ is hyperelliptic? REPLY [5 votes]: Yes, because then there is a nonconstant map from projective space to the symmetric square of $C$.<|endoftext|> TITLE: Reference result: proof of theorem of Kazhdan-Margulis on monodromy group of a Lefschetz pencil of odd fiber dimenion is "as big as possible" QUESTION [7 upvotes]: In Deligne's paper on his first proof of the Weil conjectures, we have the following result. Theorem 5.10 (Kazhdan-Margulis). L'image de $\rho: \pi_1(U, u) \to \text{Sp}(E/(E \cap E^\perp), \psi)$ est ouverte. This theorem says the monodromy group of a Lefschetz pencil of odd fiber dimension is "as big as possible". My question is, where has this proof been written down in the literature? I'm finding the discussion in Deligne to be quite terse, and I searched the papers of Kazhdan and Marguilis and I couldn't find a paper which establishes this result. REPLY [3 votes]: The proof of this result is worked out in detail in page 250 (theorem 7.5) of the book Eberhard Freitag, Reinhardt Kiehl "Etale Cohomology and the Weil Conjecture" You can preview that page in particular here.<|endoftext|> TITLE: Gromov's pseudogroups and Tao's approximate groups QUESTION [6 upvotes]: In his article Gromov, M. Almost flat manifolds. J. Differential Geom. 13 (1978), no. 2, 231–241 Gromov exploited a notion of a pseudogroup. In his book Tao, Terence. Hilbert's fifth problem and related topics. Graduate Studies in Mathematics, 153. American Mathematical Society, Providence, RI, 2014 Tao systematically developed the notion of an approximate group and gave several applications. Both of these notions are in the subject that has come to be called geometric group theory. Thus, Gromov builds his pseudogroup out of parts of the fundamental group with a suitable operation called Gromov product (by Buser and Karcher) which approximates composition of loops. What is the precise relationship between the notion of pseudogroup and that of an approximate group? REPLY [7 votes]: After some thinking, I no longer believe in any deep connection between the two notions. Their definitions sound similar and can indeed be used to establish some of the same results, yet the analogy seems superficial. Let me record some references that might help those who want to think more. A pseudogroup seems more of a technical device similar to thinking a space as the quotient of its universal cover by the deck-transformation action. In the collapsing theory of manifolds with two sided curvature bounds a pseudogroup is a technical device allowing to desingularise the Gromov-Hausdorff convergence. To see how it is done it is most efficient to glance through the following: pp. 33-34 of Fukaya's book "Metric Riemannian geometry" https://www.math.kyoto-u.ac.jp/preprint/2004/16fukaya.pdf. p.494 of Lott's paper "Dimensional reduction and the long-time behavior of Ricci flow", https://math.berkeley.edu/~lott/2010-85-03-01.pdf. Let me informally summarize what happens in the above references: a collapsing sequence of metric balls can be (under some curvature assumptions) realized as quotients of balls in the tangent spaces by action of pseudogroups. The actions subcoverge to an isometric action of a local Lie group on some Riemannian manifold. This is a convenient way to think about local collapse developed by Gromov and Fukaya. (There are other ways to think about collapse that do not use the language of pseudogroups.) Thus in this construction a pseudogroup approximates a local Lie group, whose action gives a description of the collapse. On the other hand, an approximate group sits, by definition, in a local group, and sometimes captures the algebraic properties of the associated global group, such as virtual nilpotence. To see how this applies to geometric problems look at the proof of Corollaries 11.13 in Breuillard-Green-Tao's "The structure of approximate groups", http://front.math.ucdavis.edu/1110.5008. Here the geometric input is minimal (namely, the Bishop-Gromov volume comparison which gives a packing condition on the orbit of the fundamental group action in the universal cover). The notion of an approximate group allows to decouple group theory and geometry, which is quite striking.<|endoftext|> TITLE: Are small $\varepsilon$-balls convex in geodesic metric spaces? QUESTION [12 upvotes]: Let $(M,d)$ be a complete, separable, compact metric space. Assume $M$ is geodesic, that is for any $x,y \in M$ there exists a distance realizing geodesic between $x$ and $y$ (not necessarily unique). A set $U \subseteq M$ is convex if for any $x,y \in U$ there exists a geodesic between $x$ and $y$ that lies entirely in $U$. Question: Is it true that for any $x \in M$ there exists an $\varepsilon' >0$ such that all $\varepsilon$-balls around $x$ for $\varepsilon < \varepsilon'$ are convex? Two examples to illustrate the question: Let $M= \{(x_1,x_2) \in \mathbb{R}^2 | x_2 \ge 0, x_1^2+x_2^2 \le 1 \} / \sim$ where $\sim$ is the equivalence relation $(x,0) \sim (-x,0)$, that is $M$ is a flat cone with the standard metric induced from $\mathbb{R}^2$. Then for any $\varepsilon >0$, the $\varepsilon$-ball around the point $(0,\varepsilon/2)$ is not convex. In particular, arbitrarily small non-convex balls exists. However, for every point sufficiently small balls around it are convex. Let $M= \{(x_1,x_2) \in \mathbb{R}^2 | x_1^2+x_2^2 \le 1 \}$ be the unit disk with the Manhattan metric $d((x_1,x_2),(y_1,y_2)):=|x_1-y_1|+|x_2-y_2|$. Then the $\varepsilon$-balls are diamond shaped and are convex. However, distance realizing geodesics are not unique, in general there exist uncountably many and not all of them lie inside the $\varepsilon$-balls. Consider the radius $1/2$ ball around $(0,0)$ and the points $(2/5,0)$ and $(0,2/5)$. Then the geodesic through the point $(0,0)$ lie inside the ball but the geodesic through $(2/5,2/5)$ does not. Hence the formulation 'there exists a geodesic inside the ball' is crucial, the same statement with 'all geodesics lie inside the ball' is false. A theorem that shows this under some additional assumptions on the metric space would also be interesting. REPLY [7 votes]: There is actually a much more striking result than non-convexity of balls in the Heisenberg group discussed in the answer of Nate Eldredge. While the definition of the sub-Riemannian metric on $\mathbb{H}^3$ has been explained by Nate Eldredge, I will skip that part and just formulate the theorem. Theorem. Let $A=\{(x,y,z_1),(x,y,z_2)\}\subset\mathbb{H}^3$, where $z_1\neq z_2$ be a set consisting of two points in the Heisenberg group. Then the smallest geodesically convex set containing $A$ is $\mathbb{H}^3$. That means there are very few convex sets and in particular the smallest geodesically convex set containing a ball must be $\mathbb{H}^3$. This is a remarakable result because, the Heisenberg group is one of the nicest non-Euclidean metric spaces that allows one to carry out most of the analysis in Euclidean spaces. It also serves as an ulitmate example for testing new conjectures in analysis on metric spaces. The above result is Theorem 1.1 in: Monti, R. Rickly, M.: Geodetically convex sets in the Heisenberg group. J. Convex Anal. 12 (2005), 187-196. Remark: The space $\mathbb{H}^3$ considered above is often denoted by $\mathbb{H}^1$. Although, as a linear space it is $\mathbb{R}^3$, it is the first Heisenberg group. Moreover the sub-Riemannian metric discussed above is often called the Carnot-Caratheodory metric.<|endoftext|> TITLE: Cofinality of $j(\kappa)$ for a measurability embedding $j:V\to M$ with critical point $\kappa$ QUESTION [13 upvotes]: Suppose $\kappa$ is a measurable cardinal and $j:V\to M$ is the ultrapower by a normal measure on $\kappa$. Let's say, for instance, that $2^\kappa=\kappa^{++}$ (note that this assumption has consistency strength greater than just a measurable cardinal). It is easy to see that $j(\kappa)$ has size $\kappa^{++}$ in $V$. So the cofinality of $j(\kappa)$ in $V$ can be either $\kappa^+$ or $\kappa^{++}$ (it cannot be $\kappa$ because $M$ is closed under $\kappa$-sequences). Can each of these possibilities be realized in some model of set theory? More generally, what can be said about the cofinality of $j(\kappa)$ under various GCH assumptions? As Asaf Karagila notes in the comments below, we can reformulate the question to ask what are the possible cofinalities of the order $\kappa^\kappa/U$ for a normal measure $U$ on $\kappa$. This is equivalent because elements of $j(\kappa)$ are precisely the equivalence classes of functions $f:\kappa\to\kappa$. REPLY [10 votes]: As Mohammad Golshani remarked, it is possible to control the cofinality of $j(\kappa)$ by iterating the forcing that adds a function $f\colon \kappa \to \kappa$ which is eventually larger than any ground model function. The conditions of the forcing notion are pairs of the form $(s, g)$ where $s{\in} ^{<\kappa}\kappa$ and $g\colon \kappa \to \kappa$, where $(s, g)$ is stronger than $(t, h)$ if $s\supseteq t$, $g \geq h$ everywhere and $\forall \alpha \in \text{dom } s \setminus \text{dom }t$, $s(\alpha) \geq h(\alpha)$. Assuming $\kappa^{<\kappa} = \kappa$, this forcing is $\kappa$-centred and $\kappa$-directed closed. Let $\mathbb{P}_\alpha$ be the iteration of adding dominating function for $\alpha$ many steps with support ${<}\kappa$. Using the $\kappa$-closure of iteration and standard $\Delta$-system arguments - this iteration is $\kappa^{+}$-c.c. Therefore, it doesn't collapse cardinals. Let $\langle f_i \mid i < \alpha\rangle$ be the sequence of the generic dominating functions. Lemma: If $\text{cf }\alpha \geq \kappa^{+}$ then the true cofinality of $\kappa^{\kappa} / J^{bd}$ is $\text{cf }\alpha$. Namely, there is a cofinal, increasing sequence of functions in $\kappa^{\kappa}$ of order type $\text{cf }\alpha$. Proof: Let $\lambda = \text{cf }\alpha$. Let $\langle \gamma_i \mid i < \lambda\rangle$ be a cofinal sequence. The sequence of functions $g_i = f_{\gamma_i}$ is increasing (modulo bounded error) and by the chain condition of $\mathbb{P}_\alpha$, every function in the generic extension is bounded by one of them. $\square$ Let $\kappa$ be a measurable cardinal. If $\text{tcf } \kappa^{\kappa} / J^{bd} = \lambda$, then for every $\kappa$-complete measure $\mathcal{U}$ on $\kappa$, $\text{cf }j_{\mathcal{U}}(\kappa) = \lambda$. Therefore, in order to construct a model in which $\text{cf }j(\kappa) = \lambda$, $2^\kappa = \mu$ where $\kappa < \lambda \leq \mu$, $\lambda$ regular, $\text{cf } \mu \geq \kappa^{+}$, we may start with indestructible supercompact $\kappa$ such that $2^\kappa = \kappa^{+}$ and force with $\mathbb{P}_{\mu + \lambda}$.<|endoftext|> TITLE: The fundamental group of the complement of an analytic subset of codimension at least $2$ QUESTION [12 upvotes]: Let $X$ be a compact complex manifold and $V$ an analytic subset of $X$ of codimension $\ge 2$. By intuition, il might be true that the fundamental group of $X$ equals that of $X \backslash V.$ I would like to ask if the last claim is true or not and in the affirmative case, how do we prove it? Thank you in advance. REPLY [11 votes]: The answer is yes. Just triangulate $X$ in such a way that $V$ is a subpolyhedron (this can be done by Lojasiewicz's theorem). Now it is well-known that removing a subpolyhedron of (real) codimension at least $3$ does not affect the fundamental group for transversality reasons. Edit. Actually, Lojasiewicz's theorem is not necessary. In fact, the analytic subset $V$ provides a closed real submanifold of codimension at least $4$ of the underlying real manifold of $X$, so we can use the following result, whose proof can be found for instance in C. Godbillon, Eléments de Topologie Algebrique, Théorème 2.3 page 146: Theorem. Let $X$ be a smooth, connected real manifold without boundary, $V \subset X$ a closed submanifold, $x$ a point of $X\setminus V$ and $i \colon X \setminus V \longrightarrow X$ the inclusion map. Then if the codimension of $V$ is at least $2$, the group homomorphism $$i_* \colon \, \pi_1(X \setminus V, \, x) \longrightarrow \pi_1(X, \, x)$$ is surjective; if the codimension of $V$ is at least $3$, the group homomorphism $$i_* \colon \, \pi_1(X \setminus V, \, x) \longrightarrow \pi_1(X, \, x)$$ is an isomorphism.<|endoftext|> TITLE: Lawvere's 'Categories of space and of quantity" - the projection formula QUESTION [6 upvotes]: I'm trying to read bits and pieces of W. Lawvere's Categories of Space and of Quantity, and, as usual have lots of questions. Page numbers will refer to the number printed on the corners of the pages. The article is pretty dense (for me), so I won't paraphrase much. An extensive quantity is, broadly speaking, a "quantity of space", and an intensive quantity is a "ratio" between extensive ones. For instance, mass and volume are extensive (measures), while density is intensive (function). [See last paragraph of page 18.] For a variety of reasons, "ratios" are harder to formalize than "multiplications", so the author posits the relationship between intensive and extensive quantities should be seen as an action of intensive quantities on extensive ones which produce more extensive ones. For instance, density acting on volume to give mass. An excerpt from the bottom of page 21 and the top of page 22 (boldface is added by me): $\;\;\;$The common spatial base of extensive and intensive quantities also supports the relation between the two, which is that the intensives act on the extensives. For example, a particular density function acts on a particular volume distribution to produce a resulting mass distribution. Thus it should be possible to "multiply" a given extensive quantity on a certain space by an intensive quantity (of appropriate type) on the same space to produce another extensive quantity on the same space. The definite integral of the intensive quantity "with respect to" the first extensive quantity is defined to be the total of this second resulting extensive quantity. This action (or "multiplication") of the contravariant on the covariant satisfies bilinearity and also satisfies, with respect to the multiplicative structure within the intensive quantities and along any inducing spatial map, an extremely important strong homogeneity condition which so far has carried different names in different fields: in algebraic topology this homogeneity is called the "projection formula", in group representation theory it lies at the base of "Frobenius reciprocity", in quantum mechanics it is called "covariance" or the "canonical commutation relation", while in subjective logic it is often submerged into a side condition on variables for the validity of the rule of inference for existential quantification when applied to a conjunction. In particular this brings to mind these two questions on (the origins of) projection formulas. My question is simple - can anyone explain, illustrate, simplify, and/or give details about this? (The concept of "total" is discussed on page 19.) In particular: In what sense is the projection formula a "homogeneity" condition? What is the intuitive geometric meaning of the projection formula at this great and conceptual generality? Of course all additional stories, references, details, examples, anecdotes, warnings, etc are welcome! REPLY [3 votes]: One way to interpret the projection formula as a "homogeneity" condition is by thinking of it as saying that the push-forward is a module map, i.e. it preserves an action. I don't know if this is what is meant by Lawvere, however. I will be vague and general. We start with some category of things (maybe spaces or groups or whatever). To each object $X$ we associate a thing $M(X)$ which might be a category or an abelian group or a vector space (maybe a category of sheaves, or the representation ring, or cohomology) and $M(X)$ has a product $\times$ making it into a monoid in a suitable sense (a monoidal category or a ring or an algebra). Given a morphism $f:X\to Y$ we get a pull-back $f^\ast\colon M(Y) \to M(X)$ and this respects the products in $M(X)$ and $M(Y)$, so $f^\ast$ is a morphism of monoids: $$f^\ast(y_1\times y_2) = f^\ast y_1 \times f^\ast y_2.$$ Using this fact we can make $M(X)$ into an $M(Y)$-module: $y\cdot x:= (f^\ast y) \times x$. On the other hand, being a monoid $M(Y)$ is also an $M(Y)$ module using the regular action: $y\cdot y' := y\times y'$. Associated to $f\colon X\to Y$ we also have the push-forward $f_\ast\colon M(X)\to M(Y)$. The projection formula says $$f_\ast(f^\ast y \times x)= y \times f_\ast x$$ in other words the projection formula says that $f_\ast$ is a morphism of $M(Y)$-modules: $$f_\ast( y \cdot x)= y \cdot f_\ast x.$$<|endoftext|> TITLE: About Morley congruence QUESTION [5 upvotes]: Let $p>3$ be an odd prime and $a$ be a positive integer, is the following congruence true? $$\binom{p^a-1}{\frac{p^a-1}{2}}\equiv(-1)^{\frac{p^a-1}{2}}4^{p^a-1}\pmod{p^3}.$$ When $a=1$, this is Morley's congruence. REPLY [6 votes]: The answer is yes. See Corollary 3 of "A congruence involving the quotients of Euler and its applications (I)" by Tianxin Cai (Acta Arithmetica, 2002). Namely, Cai shows that $$(-1)^{\frac{p-1}{2}} \binom{p^k-1}{\frac{p^k-1}{2}}/\binom{p^{k-1}-1}{\frac{p^{k-1}-1}{2}} \equiv 4^{\phi(p^k)} \bmod {p^{3k}}.$$ In particular, for any $k\ge 1$, $$(-1)^{\frac{p-1}{2}} \binom{p^k-1}{\frac{p^k-1}{2}}/\binom{p^{k-1}-1}{\frac{p^{k-1}-1}{2}} \equiv 4^{\phi(p^k)} \bmod {p^{3}}.$$ By applying the above for all $1 \le k \le a$, he deduces that $$(−1)^{\frac{(p-1)a}{2}} \binom{p^a-1}{\frac{p^a-1}{2}} \equiv 4^{p^a-1} \bmod {p^3}.$$ To see that this is the same as your congruence, note that $(p-1)a \equiv p^a-1 \bmod 4$ by considering two cases: $p \equiv 1 \bmod 4$ and $p \equiv -1 \bmod 4$.<|endoftext|> TITLE: The flat support of a module QUESTION [5 upvotes]: Let $R$ be a Noetherian commutative ring and $M$ a finitely generated $R$-module. What is known about the following subset of $ \mathrm {Spec}(R)$: $$\mathrm{supp}_{fl}(M)=\{P\in \mathrm {Spec}(R):\ \mathrm{Tor}^R_1(M, R/P)= 0\}?$$ In particular, Is $\mathrm{supp}_{fl}(M)$ open in $\mathrm {Spec}(R)$? REPLY [7 votes]: I suspect that Graham Denham's argument only shows that the weaker condition $\mathrm{Tor}_1(M,\kappa(P))=0$ is open, where $\kappa(P)$ is the residue field of $P$. With the problem as stated, here is a counterexample: for a field $k$, take $R=k[x,y]$ and $M=R/(x,y)$. Then it is easy to check that $\mathrm{supp}_{fl}(M)$ consists of: $\bullet$ the generic point of $\mathrm{Spec}(R)$, $\bullet$ all closed points except the origin $(x,y)$, and $\bullet$ all principal primes $(f)$ where $f$ is irreducible and not in $(x,y)$, i.e. all generic points of curves not containing the origin. This is not open in $\mathrm{Spec}(R)$.<|endoftext|> TITLE: The Gray tensor product as a Kan extension QUESTION [7 upvotes]: $\require{AMScd}$I've been told that the Gray tensor product of two 2-categories can be obtained with a Kan extension (forgive me for my carelessness of foundational issues): the product $\boldsymbol\otimes \colon 2\textbf{-Cat} \times 2\textbf{-Cat} \to 2\textbf{-Cat}$ can be obtained with a Kan extension of $\Phi$ along $J$: $$ \begin{CD} \{\text{some 2-category}\} @>\Phi>> 2\textbf{-Cat}\\ @VJVV \\ 2\textbf{-Cat} \times 2\textbf{-Cat} \end{CD} $$ Is this true? If yes, can you provide a reference with an explicit proof, or provide a proof yourself? Thanks! REPLY [7 votes]: The nice answer given by Alexander can be consider as a particular case of the general framework developed in AM16, specifically in its appendix A. The objects of the parity complex of cubes in $\infty$-$\mathcal{C}at$ can be defined inductively as the tensor products $\Delta_1 \otimes \dots \otimes \Delta_1$ and they are free objects (in the sense of polygraph/computads), see remark A.16. So it is more natural to define at first the general Gray tensor product for $\infty$-$\mathcal{C}at$ and then truncate to get a Gray tensor product in any degree $n$. Let $\mathcal{C}_{\text{da}}$ be the category of directed augmented complex and $\mathcal{S}t_{\text f}$ its full subcategory of Steiner complexes, see ch. 2 of AM16. There exists a functor $\nu \colon \mathcal{C}_{\text{da}} \to \infty$-$\mathcal{C}at$ which by a theorem of Steiner is fully faithful when restricted to $\mathcal{S}t_{\text f}$, so that we call the objects of the essential (restricted) image Steiner $\infty$-categories (these are all free in the sense of polygraphs). Another result by Steiner ensures that Steiner complexes are closed under tensor product of directed augmented complexes, see proposition A.4. Moreover, the orientals, the cubes and $\Theta$ are Steiner complexes/categories. Theorem A.14 tells us that the Gray tensor product on $\infty$-$\mathcal{C}at$ is precisely the left Kan extension $$ \require{AMScd} \begin{CD} \Theta \times \Theta @>\nu\times \nu>> \infty\text{-}\mathcal {C}at\times\infty\text{-}\mathcal{C}at\\ @V\otimes VV @VV\otimes_{\text{Gray}} V\\ \mathcal{S}t_f @>>\nu> \infty\text{-}\mathcal{C}at \end{CD}$$ Finally, by lemma A.25 and proposition A.26, we get the 2-categorical Gray tensor product as the following left Kan extension: $$ \require{AMScd} \begin{CD} \Theta_2 \times \Theta_2 @>\nu\times \nu>> 2\text{-}\mathcal {C}at\times 2\text{-}\mathcal{C}at\\ @V\otimes VV @VV\otimes_{\text{Gray}} V\\ \mathcal{S}t_f @>>\tau^\text{i}_{\leqslant 2}\circ \nu> 2\text{-}\mathcal{C}at \end{CD},$$ where $\tau^\text{i}_{\leqslant 2}$ is left adjoint to the inclusion functor $2\text{-}\mathcal{C}at \to \infty\text{-}\mathcal{C}at$. The role of $\Theta$ and $\Theta_2$ is that they are small dense sub-categories satisfying a bunch of nice properties, see theorem 6.3. Role which can be played by the 2-trucation of the cubes $\mathbf{Cu}$ for the second case of $2\text{-}\mathcal{C}at$. I don't know whether the full subcategory of cubes is dense in $\infty$-$\mathcal{C}at$, though.<|endoftext|> TITLE: Real world example of use of Monte Carlo method for high dimensional integrals QUESTION [5 upvotes]: The Monte Carlo method for numerical integration is usually presented as a method invented to efficiently compute high dimensional integrals numerically. However, I haven't found any source which has a believable real world example of when it may be applied. All the sources I have provide "toy problems" which are simple integrals like say the 3D integral exp( -x^2 - y^2 - z^2 ) over a box, without any motivation. I would like a real world example, that is hopefully also accessible. For instance, for what specific problem did Ulam and Neumann originally invent the method? REPLY [5 votes]: Some buzzwords that should lead to some non-textbook examples: In the fields of uncertainty quantification, statistical inverse problems or Bayesian inference one wants, for example, compute conditional expectations for posterior distributions. The domain of integration has as many dimensions as the the quantity of interest has degrees of freedom, and this can be a distributed parameter which, after discretization, may well be in the ten thousands up to millions. Before Monte Carlo methods can be applied one needs to think carefully, how to generate samples from the distribution and one often uses Markov chains to do so (leading to Markov chain Monte Carlo methods). To be a bit more concrete: Consider an inference problem where the quantity of interest $u$ is observed through an operator $A$. Moreover, take into account that the measurement $Au$ is not exact, i.e. we observe $v^\delta = Au + \eta$ with some noise $\eta$ and also consider the case where $A$ has no continuous inverse (or, discretized, a large condition number). Assuming that the distribution of the noise is known, we can write down the probability $$ p(v^\delta\mid u) = p(\eta). $$ If we further assume prior knowledge about the real $u$, formulated in a prior distribution $p(u)$ (could be Gaussian or something else - whatever makes sense in the application), then the distribution that is really of interest is the posterior $$ p(u\mid v^\delta) \propto p(v^\delta\mid u)p(u) $$ i.e. the one that answers the question "What is a probable $u$, now that we have seen the data $v^\delta$?". It's not easy to deal with the posterior, since it is a distribution on the space where $u$ comes from, so if the problem is a discretized problem and you want $u$ with $n$ degrees of freedom, then, it is a distribution over $\mathbb{R}^n$ - $n$ in millions. One quantity of interest for the posterior is the conditional mean, i.e. something like "the expected solution $u_{CM}$, given the current data $v^\delta$" and this is the integral $$ u_{CM} = \int u \, p(u\mid v^\delta) du $$ and this is an integral over $\mathbb{R}^n$. Here people (and especially companies) indeed use Monte Carlo integration. Even more concrete: $u$ is oil distribution under ground, $A$ takes this distribution and gives back the surface measurements under seismic stimulation and $v^\delta$ is the measurement you take. Oil companies invest a lot of time and money to infer as much as possible about the oil reservoirs and indeed use techniques as described above. They want a good resolution of the oil distribution and hence, $n$ as large as possible (but to have at least a crude picture of how it looks like under ground, you need at least some ten or hundred thousand…) This is probably not what Ulam and von Neumann had in mind but I think you can't get more real world than that. A starting point for the general idea is Kaipio, Jari, and Erkki Somersalo. Statistical and computational inverse problems. Vol. 160. Springer Science & Business Media, 2006. and there are also some work on the concrete application of oil reservoirs like Iske, Armin, and Trygve Randen. Methods and Modelling in Hydrocarbon Exploration and Production. Springer, 2005. (but I have to admit that I do not know the latter book well).<|endoftext|> TITLE: "Surprising" examples of Markov chains QUESTION [59 upvotes]: I am looking for examples of Markov Chains which are surprising in the following sense: a stochastic process $X_1,X_2,...$ which is "natural" but for which the Markov property is not obvious at first glance. For example, it could be that the natural definition of the process is in terms of some process $Y_1,Y_2,...$ which is Markovian and where $X_i=f(Y_i)$ for some function $f$. Let me give you an example which is slightly surprising, but not surprising enough for my taste. Suppose we have $n$ bins that are initially empty, and at each time step $t$ we throw a ball into one of the bins selected uniformly at random (and independently of previous time steps). Let $X_t$ be the number of empty bins at time $t$. Then $X_1,X_2,...$ form a Markov chain. Are there good, more surprising examples? I apologize if this question is vague. REPLY [3 votes]: The Dyson Brownian motion. Take a random matrix whose entries perform independent standard Brownian motions, subject to the condition that the matrix stays Hermitian, so that entries above diagonal perform Brownian motions in the complex plane, the diagonal entries perform a BM on the real line, and the entries below the diagonal are complex conjugates of the ones above that diagonal. Then, the eigenvalue process $(\lambda_1(t),\dots,\lambda_n(t))$ of this random matrix is Markov. This is closely related to the example mentioned in Leonid Petrov's answer, but I think the reasons for this "surprising Markov property" are rather different in the two cases.<|endoftext|> TITLE: certain smoothness of principal eigenvalue of Dirichlet Laplacian on polygons QUESTION [5 upvotes]: For a given polygon $P_N$, with side lengths $x_1,\cdots,x_N$ and interior angles $\theta_1,\cdots,\theta_N$ let $\lambda(x_1,\cdots,x_N,\theta_1,\cdots,\theta_N)$ denote the least eigenvalue of Dirichlet Laplacian on $P_N$. Question. Is $\lambda$ as a function of $x_1,\cdots,x_N,\theta_1,\cdots,\theta_N$ smooth in each variable?if not, is it at least twice continuously differentiable? Edit 1. The exitence and continuity of the first derivative follows from Theorem 2.5.1 of A. Henrot's book. REPLY [5 votes]: $P_N$ seems to be real analytic. Part L of the main theorem of Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. (pdf) shows that the eigenvalues can be chosen real analytic after a local blow up of the coordinates. Since eigenvalues can cross each other, the smallest one may change. In each single parameter, the eigenvalues are real analytic.<|endoftext|> TITLE: All fiber bundles over $S^2$ extendable to $\mathbb{C}P^\infty$? QUESTION [24 upvotes]: I ran into the following sanity check. Is the following statement true? Every smooth fiber bundle (with compact fiber) over $S^2$ can be extended to a smooth fiber bundle over $\mathbb{C}P^\infty$ (or just $\mathbb{C}P^n$). For the case of $\mathbb{C}P^2$, if the bundle is a principal bundle, then it can be extended, since $\pi_3(BG)=\pi_2(G)=0$. So I thought possible counterexamples might come from a fiber $F$, with $\pi_2(\text{Diff}(F))\ne 0$, for example $\text{Diff}(S^1\times S^2)\simeq O(2)\times O(3)\times \Omega SO(3)$, but seems like the fiber bundle can still be extended, since $S^3\rightarrow S^2\rightarrow B(\text{Diff}(S^1\times S^2))$ is still nullhomotopic. Update: The counterexample is provided in Oscar's post, when the fiber is $(D^7,\partial D^7)$. Now my question is, if the fiber is a closed manifold, is the following statment true? Every smooth fiber bundle(with closed fiber) over $S^2$ can be extended to a smooth fiber bundle over $\mathbb{C}P^\infty$ (or just $\mathbb{C}P^n$). In fact, this should be my original question, but somehow I wrote "compact" but with "closed" in mind. I am really sorry about that. REPLY [37 votes]: No it isn't, but I had to dig quite deep to get a counterexample. Let us look at smooth $(D^7, \partial D^7)$-bundles over $S^2$, i.e $D^7 \to E \overset{\pi} \to S^2$ with an identification $\partial E \cong S^2 \times \partial D^7$. These are classified by a map $$f : S^2 \to BDiff_\partial(D^7)$$ to the classifying space of the group of diffeomorphisms of $D^7$ relative to the boundary. I will show that there exists an $f$ which does not extend to $\mathbb{CP}^2$. Smoothing theory gives an identification $$BDiff_\partial(D^7) \simeq \Omega^7\left(\frac{TOP(7)}{O(7)}\right)$$ and by page 246 of Kirby--Siebenmann the map $$\Omega^7\left(\frac{TOP(7)}{O(7)}\right) \to \Omega^7\left(\frac{TOP}{O}\right)$$ is 2-connected, so there is a surjection $$\pi_2(BDiff_\partial(D^7)) \to \pi_9(\tfrac{TOP}{O}) \quad \quad(*).$$ To find an example of such a bundle which does not extend over $\mathbb{CP}^2$, it is therefore enough to show that the map $$- \circ \eta : \pi_9(\tfrac{TOP}{O}) \to \pi_{10}(\tfrac{TOP}{O}) \quad \quad(**)$$ given by precomposition with the (suitably suspended) Hopf map is not trivial. By surgery theory $\pi_n(\tfrac{TOP}{O}) \cong \Theta_n$, the group of homotopy $n$-spheres, and a homotopy $n$-sphere has a unique Spin structure, determining a $\pi_n(\tfrac{TOP}{O}) \cong \Theta_n \to \Omega^{Spin}_n$, which we can further compose with the Atiyah--Bott--Shapiro map $$\alpha : \Omega^{Spin}_n \to KO_n.$$ Applying this construction to both sides, the map $(**)$ above becomes the map $$- \circ \eta : KO_9 = \mathbb{Z}/2 \to KO_{10} = \mathbb{Z}/2,$$ which is an isomorphism. Finally, it is well-known that $$\pi_n(\tfrac{TOP}{O}) \cong \Theta_n \to \Omega^{Spin}_n \to KO_n$$ is surjective for $n \equiv 1,2 \mod 8$. So choose a $f' \in \pi_9(\tfrac{TOP}{O}) \cong \Theta_9$ which is non-trivial under the Atiyah--Bott--Shapiro map, and use that $(*)$ is surjective to lift it to a $f \in \pi_2(BDiff_\partial(D^7))$. This has the desired property. Addendum The following gambit can be used to get a closed fibre example. The product of the natural maps $SO(n+1) \to Diff^+(S^n)$ and $Diff_\partial(D^n) \to Diff^+(S^n)$ gives a map $$SO(n+1) \times Diff_\partial(D^n) \to Diff^+(S^n)$$ which is a weak homotopy equivalence (though not a homomorphism). Thus the map $$BDiff_\partial(D^7) \to BDiff^+(S^7)$$ is (split) injective on all homotopy groups, so the example above also gives an oriented $S^7$-bundle over $S^2$ which cannot extend to $\mathbb{CP}^2$.<|endoftext|> TITLE: When does Scott topology generated by specialization order induced by a sober space (X,$\tau$) equal the initial topology $\tau$? QUESTION [7 upvotes]: Let X be a $T_{0}$ space. The specialization order ≤ on X is that if x is contained in cl{y}, then we call "x≤y". Obviously (X,≤) is a partially ordered set. A sober space is a topological space such that every irreducible closed subset of X is the closure of exactly one point of X: that is, this closed subset has a unique generic point. It is not difficult to see that specialization order induced by a sober space is a directed complete partial order (dcpo). A subset O of a dcpo P is called Scott-open if it is an upper set and if it is inaccessible by directed joins, i.e. if all directed sets D with supremum in O have non-empty intersection with O. The Scott-open subsets of a dcpo P form a topology on P, the Scott topology. (https://en.wikipedia.org/wiki/Scott_continuity) Let (X,$\tau$) be a sober space. The specialization order ≤ induced by a sober space makes (X,≤) a dcpo. Generally, the Scott topology on (X,≤) is finer than $\tau$. But when do these two topologies coincide with each other? Two examples (sober c space and sober locally finitely compact space) are given in a book "Non-Hausdorff topology and Domain theory". (http://www.cambridge.org/ca/academic/subjects/mathematics/geometry-and-topology/non-hausdorff-topology-and-domain-theory-selected-topics-point-set-topology) Can you find more examples like this? or Can you characterize this kind of sober space by simple property? REPLY [4 votes]: Here is a partial answer. In P. T. Johnstone, "Stone Spaces", pp. 292, 294, it is shown that Scott topologies of continuous posets are precisely all completely distributive complete lattices. Recall that a poset $P$ is continuous iff it is a dcpo and moreover the join map $\operatorname{Ideals}(P)\to P$ has a left adjoint. I don't know a nice characterization of spaces whose specialization posets are continuous, but for those the answer is thus that their topology coincides with the Scott topology of the specialization iff open sets form a completely distributive lattice. And moreover, any completely distributive complete lattice is the topology of a sober space which is at the same time the Scott topology of its specialization order.<|endoftext|> TITLE: References for properties of Atiyah-Hirzebruch Spectral Sequence for a spectrum $X$ and generalised homology theory $MSpin_*$ QUESTION [6 upvotes]: Currently I'm working on the following version of the AHSS $$ E^2_{pq}\cong H_p(M\eta; MSpin_q(\ast))\Rightarrow MSpin_{p+q}(M\eta)$$ where $\eta \colon B \to BSO$ is a stable vector bundle, and $M\eta$ denotes its Thom Spectrum. My experience with the AHSS is "graduate level", therefore most properties and, let's say, tricks I'm aware of it is for the simple version with $E^2_{pq}(X)\Rightarrow h_*(X)$ where $X$ is a CW space, NOT a spectrum. So these are the questions I'm looking for an answer (and a reference for it) Third differentials for $ E^2_{pq}\cong H_p(M\eta; MSpin_q(\ast))\Rightarrow MSpin_{p+q}(M\eta)$. I was able to find a reference for the second differentials (https://math.berkeley.edu/~teichner/Papers/Signature.pdf) Prop 1 page $750$, but nothing for the third differentials. In my specific case, there is a possibly non-trivial $d_3\colon E^3_{5,0}\to E^3_{2,2}$. Asking around it seems that it's general knowledge that the $d_3$ starting from the $0$th row is a secondary cohomology operation associated to $Sq^2Sq^2=0$. I'm not able to find any reference for this, and my little knowledge in secondary cohomology operations prevents me from proving it on my own. Edge homomorphisms: Are there some characterisations for the edge homomorphisms in this case? I'm aware that the characterisation can't be something easy, but if there is something it might prove useful sometimes. Is there some multiplicative structure on the cohomological version? I'm referring to something on the lines of Kochman page 34 def. $2.2.1$, in particular differentials are derivations. I think there should be, but I'm looking for some opinion from someone more experienced than me. Is there a pairing between the AHSS's for $MSpin_*(M\eta)$ and $MSpin^*(M\eta)$ like in the case for a finite CW complex $X$? I'm referring to a pairing analogue to the one explained in Kochman page 129 Proposition $4.2.10$. In particular I'd like to have some kind of relationship between the differential in the homology version and in the cohomology version, since my idea is to try figure out the cohomological one (maybe using some kind of multiplicative structure) and infer something about the homological ones. Thanks in advance for your help. REPLY [7 votes]: Answer to Q2: By the Thom isomorphism, $E_{n,0}=H_n(B;\mathbb Z)$. The bordism group you are interested in is isomorphic to a twisted spin bordism group, i.e. elements are represented by $f:M^n\to B$ together with a spin structure on $TM\oplus f^*\eta$, up to bordism. This follows by a Pontryagin-Thom construction. The edge homomorphism sends such an element to $f_*([M])$, which follows from the Thom isomorphism. For the study of the edge homomorphism, you may want to use surgery below the middle dimension, which reduces to the case of manifolds $M$ of a particular form. Maybe you can use some classification result of manifolds in dimensions 4 and 5 with given fundamental group. More could be said if you would specify the stable bundle. Another thought: If one is interested in calculating the low dimensional spin bordism groups of $M\eta$, then an Adams spectral sequence might help. Note that $MSpin$ is isomorphic to $ko$ in dimensions up to 7, so that at the prime 2 one will only need to know $Sq^1$ and $Sq^2$ on $M\eta$.<|endoftext|> TITLE: Do real analytic functions on $\mathbb{C}\mathbb{P}^n$ form a Noetherian ring? QUESTION [5 upvotes]: Question: Is the ring of real-analytic functions on $\mathbb{C}\mathbb{P}^n$ (real valued) a Noetherian ring? References or counterexamples are welcome. I know that the ring of germs of holomorphic functions on $\mathbb{C}^n$ is Noetherian, but the ring of holomorphic functions on $\mathbb{C}^n$ is not ! I also know how to verify that the ring of polynomials is Noetherian. REPLY [8 votes]: Since I found the answer to my question I will write it here, hoping that it will be helpful to someone else. Answer: The ring of global real analytic functions on a compact real analytic Stein manifold is noetherian due to Theorem I.9. [J.Frisch.: Points de platitude d'un morphisme d'espaces analytiques complexes. Invent. Math., 4:118-138, 1967.], see also last paragraph in [J. M. Ruiz. On Hilbert's 17th problem and real Nullstellensatz for global analytic functions. Math. Z., 190(3):447-454, 1985]<|endoftext|> TITLE: Direct limits of a matrix and its transpose QUESTION [5 upvotes]: Let $A \in M_n(\mathbb Z)$ and $A^T$ denote the transpose of $A$. Define the direct limits $$H_1 = \mathrm{colim} (\mathbb Z^n \xrightarrow{A} \mathbb Z^n \xrightarrow{A} \mathbb Z^n \xrightarrow{A} \dots)$$ and $$H_2 = \mathrm{colim} (\mathbb Z^n \xrightarrow{A^T} \mathbb Z^n \xrightarrow{A^T} \mathbb Z^n \xrightarrow{A^T} \dots).$$ QUESTION. Is it true that $H_1 \cong H_2$? If not, are there nice counterexamples? Remarks. If $z = \det A = \det A^T$ then $H_1, H_2 \subseteq \mathbb Z[\frac{1}{z}]^n$. Therefore, a counterexample would need the have strict inclusions. Also, in the examples we have looked at, if the eigenvalues $\lambda_1, \ldots, \lambda_n \in \mathbb C$ of $A$ are integers then we find $H_1 \cong \mathbb Z[\frac{1}{\lambda_1}] \oplus \dots \oplus \mathbb Z[\frac{1}{\lambda_n}] \cong H_2$ (though we don't claim this to be true in general). Further remarks. Thank you to David Handelman for the answer and the example. The matrices we are interested in arise in the context of substitutional tilings. In most cases we were able to calculate the direct limits $H_1$ and $H_2$ using ad hoc methods. We did however find a matrix for which we were unable to calculate $H_1$: That matrix is $A = \left( \begin{array}{ccccc} 5 & -2 & 2 & -4 & -2 \\ 2 & 0 & 2 & -6 & -5 \\ 5 & -3 & 5 & -11 & -9 \\ -2 & 1 & 1 & -3 & -3 \\ 3 & -1 & 0 & 0 & 0 \\ \end{array} \right)$. We tried to guess that $H_1$ is isomorphic to $\mathbb Z^2 \oplus \mathbb Z[\frac{ 1 }{ 6 }]^3$ but we couldn't show this (it might be false). Unfortunately we're not well versed in algebraic number theory, so we are not sure how to decide whether $H_1$ is isomorphic to $H_2$ for this particular matrix. REPLY [4 votes]: No. Define the matrix [nasty comment about LaTeX deleted] $$ A =\pmatrix {3 & 7 \\ 2 & 1 \\}. $$ A left eigenvector for the eigenvalue $c = 2 + \sqrt{15}$ is $(2,\sqrt{15}-3)$, whose entries generate the maximal ideal sitting over $2$; it is non-principal in $R = \mathbf{Z}[\sqrt {15}]$ (since $a^2- 15b^2 = \pm 2$ is unsolvable in $\mathbf{Z}$), and a little more is true. A right eigenvector is $(7, \sqrt{15}-3)^T$, whose entries generate a principal ideal. The basic idea is to use the fact that the eigenvalues are $c= 2+ \sqrt 15$ and $d =2- \sqrt{15}$, either one generates $R = \mathbf{Z}[\sqrt{15}]$, the ring of integers in $\mathbf{Q}[\sqrt 15]$; from the right eigenvector, thus $\mathbf{Z}^2$ as a left $R$-module corresponds to a principal ideal; on the other hand, $\mathbf{Z}^2$ as right $R$-module is not principal. Moreover, it remains non-principal in $R[c^{-1}] = \mathbf{Z}[c^{-1}]$. Finally, the limiting groups are strongly indecomposable, so that isomorphism as abelian groups entails isomorphism as $R$-modules. Set $G$ to be the stationary direct limit determined by $A$ and $G’$ that determined by $A^T$. We need a few (interesting) results, which are sketched below. (a) $G$ contains no elements of infinite height. Follows easily from the fact that the content of each column in all powers of the matrix is one. (b) $G$ is strongly indecomposable. If not, there would exist a subgroup $H = C \oplus D$ of finite index. Since the rank of $G$ is two, each of $C$ and $D$ is isomorphic to a subgroup of the rationals. Since neither can include elements of infinite height, each must be free. But then $G$ would be free, which it clearly isn’t (since the determinant of $A$ is nonzero but not $\pm1$). [There are more general sufficient conditions for a stationary limit group to be strongly indecomposable; e.g., for size two, determinant not zero, one, or minus one, and the trace and determinant be relatively prime.] (c) Let $E$ denote the endomorphism ring of $G$. Let $\alpha$ denote the automorphism of $G$ induced by $A$ (explcitly, $[v,m] \mapsto [Av,m]$; its inverse is $[v,m] \to [v,m+1]$). Then every endomorphism of $G$ commutes with $\alpha$. To see this, note that $\mathbf{Q}[A]$ is a two-dimensional field inside $M_2 \mathbf{Q}$ (isomorphic to $\mathbf{Q}[\sqrt{15}]$), hence is a maximal $\mathbf{Q}$-subalgebra thereof. Thus if $e \in E$ does not commute with $\alpha$, there exists an element $f$ of $E \otimes \mathbf{Q}$ that is idempotent. Hence there exists an integer $N$ such that $Nf \in E$, and then the range of $Nf$ is rank one, and so is that of $N(1-f)$. This yields a direct sum of finite index in $G$, contradicting strong indecomposability. (d) The centralizer in $M_2 \mathbf{Z}$ of $A$ includes $\mathbf{Z}[A]$, and since this is isomorphic to an integrally closed ring, $\mathbf{Z}[\sqrt {15}]$, we must have that the centralizer is $\mathbf{Z}[A]$. Then it is easy to check that $E = \mathbf{Z}[c^{-1}]$. (e) We also see that if $d = 2 - \sqrt {15}$, then $d^{-1} \not\in E$ (elementary). (f) Any (group) isomorphism between the strongly indecomposable groups $G$ and $G’$ induces an isomorphism of their endomorphism rings. Hence if $G$ and $G’$ were isomorphic, then they would be isomorphic up to an automorphism of $E$. But the only candidate for an automorphism is the Galois automorphism, which is not defined on $E$, because $d^{-1} \not\in E$. Hence any group isomorphism would be an $E$-module isomorphism. (g) Since the right eigenvector corresponds to a principal ideal, $G$ is isomorphic the principal ideal, $E$ itself. On the other hand, the left eigenvector of $A$ yields the ideal $I= (2,\sqrt{15}-3)$ of $R$, and this remains non-principal in $E$. Thus $G’$ is isomorphic to $I\cdot E$, which is not a free $E$-module, so not isomorphic to $G$ as $E$-modules, and therefore by (f), is not isomorphic to $G$ as abelian groups. Historical comment In the context of partially ordered direct limits (where the matrix $A$ is primitive, that is, all the entries are nonnegative, and in some power, they are all strictly positive), this type of construction has appeared in connection with classification up to shift equivalence of shifts of finite type (dynamical systems appearing as subshifts of full shifts, in topological dynamics) since the 70s. Here, the invariant includes the induced action of $A$, and it was well-known that there are size two matrices $A$ for which $A$ is not shift equivalent to its transpose. We cannot just adapt these examples directly, since the order data (the positivity) is crucial; for example, we could have in this context $|det\ A|= 1$, so that the underlying group of the limit is free. Instead, we use strong indecomposability to restrict considerably the endomorphisms (in the ordered case, the natural set of endomorphisms are those that are continuous).<|endoftext|> TITLE: Hard to compute real numbers QUESTION [10 upvotes]: Turing proved that not all real numbers are effectively computable. In the sense that no algorithm exists to compute some real numbers. Here is Turing's definition: A real number is computable if its digit sequence can be produced by some algorithm or Turing machine. The algorithm takes an integer $ n \geq 1$ as input and produces the $n$-th digit of the real number's decimal expansion as output. I am interested in what is known about the efficiency of representing real numbers and decoding speed. Specifically, I want to know the relationship between the compactness of the algorithm (that encodes a real number) and the speed of producing the digits of a real number (decoding speed as a function of $n$). A real number is efficiently computable if the number of steps needed to compute digit $n$ is bounded by a polynomial in $n$. It is hard to compute if no such polynomial bound exists. Are there any real numbers that are hard to compute (not necessarily computable one)? What are the references? I have no formal definitions in mind. REPLY [6 votes]: Noah Schweber has answered the question, but here is a simple example that may help drive the point home more intuitively. List all graphs. Let $r$ the be real number whose $n$th binary digit is 1 if and only if the $n$th graph is Hamiltonian. In other words, any computational problem with a yes/no answer can be encoded in the binary expansion of a real number. Since there exist hard computational problems, there exist hard-to-compute real numbers.<|endoftext|> TITLE: Sketch of Weil's proof of the Riemann hypothesis for curves QUESTION [6 upvotes]: I was wondering if anybody could provide a sketch of Weil's proof of the Riemann hypothesis for curves that uses the Jacobian $\text{Pic}^0(X)$ and a bit of the intersection theory on $X \times X$ and the underlying intuitions. The previous version of this question was asking for a "proof without words", which I realize now was a silly thing to ask for. REPLY [5 votes]: Alexander Beilinson gave four lectures on this topic (probably, during the same REU). Unfortunately, right now the link is not working, but maybe someone could provide another link? Also, in Manin’s “Cubic Forms: Algebra, Geometry, Arithmetic” one can find beautiful and self-contained exposition of the Weil conjectures for del Pezzo surfaces (and some arithmetic applications).<|endoftext|> TITLE: Is the coordinate ring of a variety a finitely generated algebra? QUESTION [6 upvotes]: Let $k$ be an algebraically closed field and let $X$ be a (not necessarily affine) variety over $k$. Is the coordinate ring of $X$ ($k[X]$ or $O_{X}(X)$) always a finitely generated $k$-algebra? REPLY [12 votes]: Feed google with "variety whose ring of global sections is not finitely generated". One gets An example of a nice variety whose ring of global sections is not finitely generated by Ravi Vakil.<|endoftext|> TITLE: Geodesics (with the same limit point) in a surface group of genus two QUESTION [5 upvotes]: Consider a discrete Gromov-hyperbolic group $\Gamma$ (and its Cayley graph $\mathcal{G}$ w.r.t. some generating set). The notion of Gromov-boundary, indicated with $\partial\Gamma$, is naturally associated with the group. It is well known that two geodesics starting at the same point (assume a vertex) in the graph $\mathcal{G}$ and having the same limit point at infinity (i.e., on the boundary) have to stay at a bounded distance: if $\gamma_1$, $\gamma_2$ are s.t. $\gamma_1(0)=\gamma_2(0)$ and $\gamma_1(\infty)=\gamma_2(\infty)$, then, there exists $K>0$ s.t. $\sup_{t\in [0,\infty)}\mathrm{dist}(\gamma_1(t),\gamma_2(t)) TITLE: On the divisibility of $(x+y)^k - 1$ by $xy$ QUESTION [5 upvotes]: For a fixed $k \geq 2$, are there infinitely many non-trivial coprime integer pairs $(x,y)$ for which $xy$ divides $(x+y)^k-1$? By trivial I mean parametrized pairs $(x,y)$ of the form $(-1,-1),(x,1),(1,y),(x,1-x),(x,1 - x^n),(1-y^n,y),(x,p(x)),(p(y),y)$, where $p(x)$ is a polynomial factor of $\frac{x^k-1}{x-1}$. When $k$ is even there are a few more trivial pairs, namely $(x,-1),(-1,y),(x,-1-x),(x,x^n-1),(y^n-1,y)$; where $n>0$ is a multiple of $k$. REPLY [3 votes]: After doing some research online I found out that there's a lot of theory behind this subject and it was studied by several authors (see the references below). As it was pointed out by @Geoff Robinson, the question essentially reduces to finding all coprime integer pairs $(x,y)$ such that $x \mid 1-y^k$ and $y \mid 1-x^k$. It turns out that the solutions have a recursive pattern, and this phenomenon was observed by @Aaron Meyerowitz in one of the answers given above. From a single solution one can produce an infinitude of solutions, which form a so-called 1 chain. A 1 chain is a recursive sequence $u_n = (1-u_{n-1}^k)/u_{n-2}$, with non-zero $u_0, u_1$ being given (note that this definition of a 1 chain differs from [4]). The following construction is a slight variation of Mohanty's argument [4]: Suppose that $x$ and $y$ are coprime integers which satisfy $x \mid 1-y^k$ and $y \mid 1-x^k$. Then there exists a unique $r \in \mathbb Z$ such that $xr=1-y^k$. But then $$xr \equiv 1 \pmod{y},$$ $$(xr)^k \equiv 1 \pmod{y},$$ and since $x^k \equiv 1 \pmod{y}$ we have $$(xr)^k-x^k\equiv 1 - 1 \pmod{y},$$ which results in $x^k(r^k - 1) \equiv 0 \pmod{y}$. Since $\gcd(x,y)=1$, we conclude that $y \mid r^k - 1$. Since $r \mid y^k - 1$, we conclude that $(y,r)$ is also a solution. By proceeding in the same fashion we can generate an infinitude of solutions (see [4] for more details). Mohanty himself studied a slightly different problem, namely he looked for coprime solutions $(x,y)$ such that $x \mid y^3+1$ and $y \mid x^3 + 1$. One can see that $(x,y)=(1,1)$ is a solution, and the sequence $u_0=1, u_1=1, u_n=(u_{n-1}^3+1)/u_{n-2}$ looks as follows: $$1,1,2,9,365,5403014, \ldots$$ This is a sequence A003818 in OEIS. According to Benoit Cloitre, this sequence "is asymptotic to $\exp(F_{2n}\cdot \log c)$ where $F_n$ is the $n$-th Fibonacci's number and $c\approx1.1137$". In 1], Bier studies the polynomial system of congruences of the form $x \mid g(y)$ and $y \mid f(x)$, where $f(t), g(t) \in \mathbb Z[t]$. He fully classifies for which pairs $(\deg f, \deg g)$ it is possible to have only finitely many solutions. It turns out that there are only finitely many such pairs, and the pairs $(k,k)$ for $k \geq 2$ are not among them. Any suggestions on how to obtain an asymptotic formula $A(M)$ for the number of solutions satisfying $\max(|x|,|y|)\leq M$ would be greatly appreciated. References: 1] T. Bier, Finite or infinite number of solutions of polynomial congruences in two positive integer variables, Springer Proceedings in Mathematics & Statistics, pp. 11 - 26, 2014. [2] J. Brzeziński, W. Holsztyński, P. Kurlberg, On the congruence $ax+by\equiv 1$ modulo $xy$, Experimental Mathematics 14 (4), pp. 391-401, 2005. [3] E. Dofs, Solutions of $x^3+y^3+z^3=nxyz$, Acta Arithmetica 73 (3), pp. 201 - 213, 1995. [4] S. P. Mohanty, A system of cubic Diophantine equations, Journal of Number Theory 9, pp. 153 - 159, 1977. [5] L. J. Mordell, The congruence $ax^3+by^3+c\equiv 0 \pmod{xy}$, and integer solutions of cubic equations in three variables, Acta Mathematica 88, pp. 77 - 83, 1952.<|endoftext|> TITLE: What are some applications of Sperner style theorems? QUESTION [11 upvotes]: I'm currently working through "Combinatorics of Finite Sets" by Ian Anderson, mostly to improve at a style of mathematics that I've historically been quite bad at, and I find myself wondering why this sort of work actually matters outside of intrinsic interest - the two books I have on this subject are this and Bollobas, and neither of them seem to have so much of a hint at the motivations for studying this other than because we can. In particular things like Sperner's Theorem (the one about counting antichains, not the one about triangulations of simplices) are neat but I can't see any really interesting consequences of it, either in combinatorics, other areas of mathematics or outside of mathematics altogether. Are there? I'm aware of the Littlewood-Offord problem, but it doesn't seem any more applicable than Sperner's theorem itself and it's more or less the only application I've found. I'm not wedded to examples just about Sperner's theorem. I'd be happy with anything "related" - any of the generalisations of it, the Lubell–Yamamoto–Meshalkin inequality, etc. REPLY [2 votes]: To give an idea of how real-life applications may arise outside mathematics, let's consider a "testing" problem of some sort. I'll set up a problem to solve first, so Sperner's theorem doesn't appear until much later in this post. What I describe is the basic theory behind Intel's X-compact. If you wish, a more engineer-friendly description can be found in S. Mitra and K. S. Kim, X-compact: An efficient response compaction technique, IEEE Trans. Comput.-Aided Design Integr. Circuits Syst., 23 (2004) 421-432. So, let's say you've got a bunch of $n$-bit vectors and want to know if they're all exactly the ones you actually want. More formally, you are given two ordered (multi-) sets $S = (\boldsymbol{s}_0,\dots,\boldsymbol{s}_{x-1})$, $T = (\boldsymbol{t}_0,\dots,\boldsymbol{t}_{x-1})$ $\subseteq \mathbb{F}_2^n$ of $n$-dimensional binary vectors over the finite field $\mathbb{F}_2$ of order $2$, where their sizes $\vert S\vert = \vert T\vert = x$ are the same. Your task is to check if for any $0 \leq i \leq x-1$, it holds that $\boldsymbol{s}_i = \boldsymbol{t}_i$, i.e., $S$ and $T$ are completely the same list of vectors. You may simply assume that $S$ is the list of vectors you asked for and $T$ is what you are actually given. It may sound trivial to check whether a pair of lists of binary vectors are the same. However, if the number $x$ of vectors in $S$ (and also in $T$) is huge and/or if the number $n$ of bits in each vector is humongous, it can be hopeless to check each and every bit one by one. To make this hypothetical problem more concrete, you may think of a brand-new CPU you, as an engineer, designed. You want to make sure that your newly designed CPU behaves as expected no matter what input you throw at it. Let's say, your CPU has $1,000,000$ input pins and $1,000,000$ output pins. So, there are $2^{1000000}$ possible input patters that may get into the input pins, and you want to make sure that your CPU responds correctly to any possible input. In this example, $S$ is the list of expected responses to the $2^{1000000}$ possible input patters. And $T$ is the corresponding actual behaviors of your CPU, i.e., the actual output bits from the output pins. So $T$ is also the list of $1000000$-dimensional vectors of cardinality $2^{1000000}$. Since we have only $10^{80}$ or so atoms in our entire universe, I think it is now clear how absurdly impossible to first make the golden list $S$ of correct responses and then compare actual output $T$ against it bit by bit; if you want to be practical, the size of your golden list better be at the very most in the order of the number of atoms on Earth, not gazillions of times the number of atoms in our galaxy. The first thing you should do is to reduce the insane number $2^{1000000}$ to something more sober. This may be done by, for example, limiting the input patters to really important ones you definitely, absolutely want to check. Also, because you are the CPU designer who devised the product, you know how wires and stuff are (supposed to be) connected in the CPU, so you might be able to come up with a clever, small set of input patterns to test which reveals all practically important design flaws and manufacturing faults. So, after all those efforts, you now have a reduced golden list $S'$ of sane but still very large size. (This was already in the order of gigabytes for an ordinary CPU in an run-of-the-mill PC about a decade ago.) So, $T$ is also automatically reduced in size, and now you have $T'$ of very large but not crazy large size. The next task is to reduce the "$1,000,000$," i.e., the dimension of each vector in $S'$ (and $T'$). This is where Sperner's theorem comes in handy (at least in the simplest case). In an abstract language, what we want to do is devise a good map $h: \mathbb{F}_2^n \rightarrow \mathbb{F}_2^m$ (where $m$ is much smaller than $n$) and hash each vector $\boldsymbol{s}_i \in S'$ and corresponding actual output vector $\boldsymbol{t}_i \in T'$ to shorter vectors $h(\boldsymbol{s}_i), h(\boldsymbol{t}_i)$ in such a way that if $\boldsymbol{s}_i \not= \boldsymbol{t}_i$, then $h(\boldsymbol{s}_i) \not= h(\boldsymbol{t}_i)$. This way, we only need to compare the hashed vectors, and hence the golden list $S'$ can be made smaller. Of course, because $n > m$, this is asking the impossible. So, we settle for "being able to detect pretty much all discrepancies that can practically happen." To make this condition more mathematics-friendly, let's consider the situation where your CPU is well designed and very stable against potential fault, so that if there is any unexpected behavior (i.e., $\boldsymbol{s}_i \not= \boldsymbol{t}_i$ for some $i$), the discrepancy between the correct and actual responses is just $1$ bit (or equivalently, if $\boldsymbol{s}_i \not= \boldsymbol{t}_i$, then only one of the $n$ pairs of bits is different). If this were all there to the problem of testing a CPU's behavior, you could simply use a linear map $h$ from $\mathbb{F}_2^n$ to $\mathbb{F}_2^m$ defined by an $m \times n$ matrix $H$ over $\mathbb{F}_2$ in which no columns are the $m$-dimensional zero vector $\boldsymbol{0}$. Indeed, it is easy to see that for any $\boldsymbol{s}, \boldsymbol{e} \in \mathbb{F}_2^n$ with $\operatorname{wt}(\boldsymbol{e}) = 1$, we have $$H(\boldsymbol{s}+\boldsymbol{e})^T \not= H\boldsymbol{s}^T.$$ So, our hashing would simply be "multiplying $H$," which shrinks $n$-dimensional vectors into $m$-dimensional vectors. This is very nice because a linear hashing circuit is simple hardware-wise and runs very fast. Now, the thing is that because modern CPUs are very complicated, you can't always perfectly predict its behavior even if you have its exact blueprint. So, for some input patterns, the correct, expected output from certain output pins are "not determined," represented by symbol $X$. In other words, some bits of some vectors in $S'$ are $X$ rather than $0$ or $1$. Let's see how this affects our linear hashing. For the sake of simplicity, let $n=7$ and $m=3$. Take the following linear map $H$ defined by $$H = \left(\begin{array}{ccccccc} 1001101\\ 0101011\\ 0010111 \end{array}\right).$$ Assume that the expected behavior is $\boldsymbol{s}_i = (1,1,0,0,X,0,1)$. Since the sums and multiplications involving unknown behavior $X$ are again unknown (except when we take the product of $X$ and $0$, which would surely be $0$). So, our hashed vector $h(\boldsymbol{s}_i)$ is $$H\boldsymbol{s}_i^T = (X,0,X)^T.$$ Obviously, because the $X$'s mask two bits in the resulting hash value, we have to detect any discrepancy between $\boldsymbol{s}_i$ and the actual, corresponding behavior $\boldsymbol{t}_i$ by looking at the single surviving bit. Of course, this is impossible because, for example, $\boldsymbol{t}_i = (0,1,0,0,X,0,1)$ gives the exact same hash value as $h(\boldsymbol{s}_i)$. How do we design a linear hash function that shrinks $n$-dimensional vectors as much as possible while ensuring detection of at most one-bit discrepancy under the presence of unknown bits $X$? Let's focus on the simplest case where the blueprint of your CPU is simple enough so that there is at most one $X$ in each expected response $\boldsymbol{s}_i \in S'$. In this case, an optimal hashing circuit $H$ is given by Sperner's theorem. To see this, recall that Sperner's theorem gives the size of a largest possible family $A = \{A_0,\dots,A_{n-1}\}$ of finite sets $A_i \subseteq \{0,1,\dots,m-1\}$ none of which contains any other sets in $A$, i.e., $A_i \not\subseteq A_j$ for any $i \not= j$. Such a family is called a Sperner family. The theorem says that for any Sperner family $A$, we have $$\vert A \vert = n \leq \binom{m}{\lfloor m/2 \rfloor},$$ and that the equality can be achieved by taking all distinct $\lfloor m/2 \rfloor$-subsets of $\{0,1,\dots,m-1\}$. Now, given an $m \times n$ binary matrix $H = (h_{i,j})$, construct the family $B = \{B_0,\dots,B_{n-1}\}$ of subsets of $\{0,1,\dots,m-1\}$ by defining $i \in B_j$ if and only if $h_{i,j} = 1$. It is straightforward to see that, with this set representation of a binary matrix, the kind of linear hash function we want for detecting discrepancies under the presence of at most $1$ unknown $X$ is exactly a Sperner family. Indeed, the condition that no set covers any other corresponds to the requirement that propagated $X$'s do not mask any possible $1$-bit-wise discrepancy. So, the condition that the linear map should compress data as much as possible corresponds to the requirement that for given $m$, the number $n$ of elements in a Sperner family should be as large as possible. Or if $n$ is given, $m$ should be as small as possible. Either way, we know the bound and how to achieve it thanks to Sperner's theorem. Of course, if your CPU gives multiple-bit-wise discrepancies or if there are more than one unknown bit, you need more sophisticated extremal set theory. But the point is that the kind of research related to Sperner's theorem gives a solid foundation of practical problems outside of mathematics like this.<|endoftext|> TITLE: Why does the proof of Myers and Steenrod fail in the Lorentzian case? QUESTION [18 upvotes]: This is my first question on this site. I hope it is not inappropriate on MO. Myers and Steenrod proved 1939 that the isometry group of a Riemannian manifold is a lie group. I add a picture where Kobayashi describes the base idea of this proof. Now I read in the book "Recent trends about lorentzian geometry" the following: (The reference [6] is Myers and Steenrod). Sadly the author doesn't give more details. Can anyone tell my why this particular proof of M&S does not work for Pseudo-Riemannian manifolds? EDIT: Maybe this will clear things up a bit. (1) My questions is NOT about why theorem 1 (on page 278) fails for Lorentzian manifolds. (Theorem 1 is: If $(M,g)$ is a compact Riemannian manifold, then $Iso(M,g)$ is compact.) (2) I know that the theorems provided by Kobayashi in his book "Transformation groups in differential geometry" can be applied to Lorentzian manifolds as well (e.g. Thm 5.1). So I wonder why this specific proof of M&S only works for Riemannian manifolds. REPLY [6 votes]: I think the main reason is basically Myers and Steenrod use properties of Riemannian manifolds related to their other theorem of differential geometry by the same name in the same 1939 paper (but not involving Lie groups) for distance function and isometries. Since this other theorem doesn't generalize to Lorentzian manifolds for obvious reasons, their proof of their theorem for Lie groups using those arguments is not generalizable to Lorentzian manifolds. The usual modern proofs(by Sternberg and Kobayashi for instance) avoid arguments related to distance functions by using local flows and orthonormal basis bundles.<|endoftext|> TITLE: References for differential cohomology and differential characters QUESTION [9 upvotes]: I am interested in learning differential cohomology and differential characters, and am currently studying these lecture notes on the subject. I sometimes feel it would be great if I could keep some more good references besides the lecture notes, as it could greatly help me speed up when I get stuck (which is happening often with me). Could someone please advise some alternate references which might supplement my study ? I am familiar with elementary algebraic topology (singular homology & cohomology theories, basic homotopy theory) and differential geometry (connections, curvature, de-Rham cohomology, Chern-Weil theory). Thanks ! Note : I earlier asked this question on m.se and offered a bounty but I did not get any answers there, I am taking the liberty of asking here. REPLY [4 votes]: The following book has recently appeared: Araminta Amabel, Arun Debray, Peter J. Haine (eds.), Differential Cohomology: Categories, Characteristic Classes, and Connections. Based on Fall 2019 talks at MIT's Juvitop seminar by: A. Amabel, D. Chua, A. Debray, S. Devalapurkar, D. Freed, P. Haine, M. Hopkins, G. Parker, C. Reid, and A. Zhang. (arXiv:2109.12250)<|endoftext|> TITLE: kth finite difference always positive when kth derivative is? QUESTION [6 upvotes]: Let $f : \mathbb{R} \to \mathbb{R}$ such that the $k^{\rm th}$ derivative of $f$ is strictly positive for every $x \in \mathbb{R}$. Define the forward difference operator to be: $$\Delta(g,h) = g(x+h) - g(x),$$ and for $h_1, \ldots , h_k > 0$, $$\Delta(g, h_1, \ldots , h_k) = \Delta( \Delta(g , h_1 , \ldots , h_{k-1}), h_k).$$ Is it true that for all $x \in \mathbb{R}$ $$\Delta(f , h_1 , \ldots , h_k)(x) > 0?$$ For $k = 2$ this is true by convexity. I would like to have a reliable reference for this if it is true. If it is not true, is it true when $h_1 = \ldots = h_k$? Any relevant information would be much appreciated, even if you feel it is only indirectly related, please leave a comment! REPLY [6 votes]: Yes, this is true. For the function $\Delta(f,h)$ its $(k-1)$-st derivative is strictly positive, since by Lagrange theorem it equals $$f^{(k-1)}(x+h)-f^{(k-1)}(x)=hf^{(k)}(x+\theta h)>0,$$ for some $\theta\in (0,1)$. Then induct on $k$.<|endoftext|> TITLE: Stable homotopy groups of $QX$ QUESTION [5 upvotes]: If $X$ is a space, we can form $QX=\varinjlim \Omega^n\Sigma^nX$ which is an infinite loop space with homotopy groups $\pi_i(QX)=\pi^{s}_i(X)$ the stable homotopy groups of $X.$ But these are the unstable homotopy groups of $QX.$ Q: Is there any similar expression for the stable homotopy groups of $QX$? Just to get some feeling, I would already be very happy to know just what $\pi_0^s(QS^0)$ is. If I understand things correctly, it should be the same as $\pi_0^s(B\Sigma_\infty^+),$ but I still don't know how to compute it. I guess $\pi_i^s(QX)$ seems like some sort of "secondary stable homotopy group" of $X$. But there is no reason to stop there. How about groups $\pi_i(Q^nX)$ which should be "$n$-ary stable homotopy groups" of $X$ - what sort of information do they carry? Of course then we could start forming other very silly objects such as about some sort of stabilizations $\pi_{i+n}(Q^nX)$ as $n\to\infty$ (does that make sense and/or produce anything sensible?) and so on, but at that point I am burried so deep in things that I don't understand that it's a bit pointless. REPLY [10 votes]: The "Snaith splitting" gives the following spectrum level statement: for a pointed connected space $X$, there is a weak equivalence: $$ \Sigma^\infty_+ (\Omega^\infty \Sigma^\infty X) \simeq \bigvee_{n \geq 0} \Sigma^\infty (X^{\wedge n})_{h\Sigma_n} $$ There is also an unbased version: $$ \Sigma^\infty (\Omega^\infty \Sigma^\infty X) \simeq \bigvee_{n \geq 1} \Sigma^\infty (X^{\wedge n})_{h\Sigma_n} $$ This gives an isomorphism on the level of stable homotopy groups: $$ \pi_*^{s}((QX)_+) = \pi_*(\Sigma^\infty_+ \Omega^\infty \Sigma^\infty X) \cong \bigoplus_{n \geq 0} \pi_*^s((X^{\wedge n})_{h\Sigma_n}) $$ When the space $X$ is not connected, one needs to add a group-completion to get the correct statement. The net effect is that on $\pi_0$ we get an isomorphism between $\pi_0^s((QX)_+)$ and the group algebra on the free abelian group on $\pi_0(X)$: it is freely generated as a commutative ring by generators coming from $\pi_0(X)$ and their inverses. In the case of $QS^0$ this becomes the Laurent polynomial ring $\Bbb Z[t^{\pm 1}]$.<|endoftext|> TITLE: Can an irreducible representation have a zero character? QUESTION [15 upvotes]: I asked the following question on Stackexchange, https://math.stackexchange.com/questions/1978407/can-an-irreducible-representation-have-a-zero-character but it got no answer, so I ask it here. Is there an example of the following situation : $F$ is a field, $G$ is a finite group, $\rho$ is an irreducible $F$-representation of $G$ (with finite degree) and the character of this representation takes the zero value at every element of $G$ ? If I'm not wrong, $F$ cannot be algebraically closed (Robinson, A Course in the Theory of Groups, 8.1.9, p. 220) and must have a nonzero characteristic $p$ such that the degree of the representation is divisible by $p$ and such that $G$ is not a $p$-group (Robinson, exerc. 8.1.5, p. 222). But that doesn't solve the problem. Thanks in advance for the answers. REPLY [16 votes]: Well, it seems that the character of an irreducible representation of a finite group over whatever field is always nonzero. I find this statement in I.M. Isaacs, Character Theory of Finite Groups, coroll 9.22, Dover, p. 155. REPLY [11 votes]: Representations of semigroups whose characters are identically zero have been classified in A. Freedman, R. N. Gupta, and R. M. Guralnick, Shirshov’s theorem and representations of semigroups, Pacific J. Math. (1997), no. 181, 159--176. Their result is the following, see Theorem 2.5 in the aforementioned paper: Theorem. Let $\rho \colon S \to \mathrm{End}(V )$ be a finite irreducible representation of the semigroup $S$ and let $A$ be the $F$-algebra generated by $\rho(S) \subseteq \mathrm{End}(V )$. Then the character $\chi$ of $\rho$ is identically $0$ if and only if $A \simeq M_r(D)$, where $D$ is division ring such that either $Z = Z(D)$ is not separable over $F$ or $\mathrm{dim}_Z(D)$ is a multiple of the characteristic. In particular, if $F$ is perfect, the character in nonzero. ${}$ Edit. As explained by B. Steinberg in the comments below, none of the conditions in the previous Theorem can occur when $S$ is a finite semigroup. In particular, when $G$ is a finite group over any field $F$, the character of any irreducible $F$-representation of $G$ is non-zero. This agrees with the result in Isaac's book quoted in Panurge'a answer.<|endoftext|> TITLE: Large deviation/concentration inequality for submartingale QUESTION [7 upvotes]: Let $S_t = M_t + D_t$ be the sum of a martingale $\left(M_t\right)_{t=1,2,\ldots}$ and a predictable process $(D_t)_{t=1,2,\ldots}$ such that the variance of the increments of $M$ is uniformly bounded $$\mathbb{E}\left[\left(M_{t+1}-M_t\right)^2 | \mathcal{F}_t\right] \leq v$$ and the increments of $D$ are uniformly bounded from below. i.e. there exists $\Delta>0$ such that $$ D_{t+1} - D_t \geq \Delta > 0 \,. $$ I want to show that for all $\alpha<\Delta$ $$ \lim_{t \to \infty} \mathbb{P}[S_t \leq \alpha \, t ] = 0\,. $$ REPLY [5 votes]: This looks like a weak law of large numbers, and in fact a strong law holds: I claim that $\liminf_{t \to \infty} \frac{S_t}{t} \ge \Delta$ almost surely, which implies the desired result. The key is to show that $\frac{M_t}{t} \to 0$ almost surely. Then we have $\frac{S_t}{t} = \frac{M_t}{t} + \frac{D_t}{t} \ge \frac{M_t}{t} + \Delta$ and can take the liminf of both sides. Let $\newcommand{\qm}{\langle M \rangle}\qm_t$ be the quadratic variation of $M_t$. The given assumption implies that $\qm_t \le vt$. We know $\qm_t$ is an increasing process so its limit $\qm_\infty$ exists almost surely (but could take the value $+\infty$). We need the following two results which can be found in http://math.ucsd.edu/~pfitz/downloads/courses/winter05/math280b/martslln.pdf. (I think they should also be in Williams's Probability with Martingales.) (Theorem 2(a) from link) Almost surely, if $\qm_\infty < \infty$ then $M_t$ converges to a finite limit. So in this case we have $\frac{M_t}{t} \to 0$. (Theorem 3 from link) Almost surely, if $\qm_\infty = \infty$ then $\frac{M_t}{\qm_t} \to 0$. Since $\frac{\qm_t}{t}$ is bounded between $0$ and $v$, again we have $\frac{M_t}{t} \to 0$. REPLY [3 votes]: For convenience, suppose that $D_0 = 0$ and $M_0=0$. The lower bound $D_{t+1} - D_t \ge \Delta$ implies that $D_{t} \ge \Delta t$ a.s., i.e., $D_t$ grows at least linearly with $t$. Thus, for any $t \in \mathbb{N}$ we have that $$ \{ S_t = M_t + D_t \le \alpha t \} \subset \{ M_t + \Delta t \le \alpha t \} \tag{$\star$} $$ Also, note that $$ \mathbb{E} \left\{ (M_{t+1}-M_t)^2 \mid \mathcal{F}_t \right\} \le \nu \implies \mathbb{E} M_{t}^2 \le \nu t \tag{$\diamond$} $$ uniformly in $t$. We apply these two simple facts below. Let $\tilde M_t = -M_t$ and $\tilde M_t^* = \max_{1 \le s \le t} \tilde M_s$. Then, ($\star$) implies that \begin{align*} \mathbb{P}( S_t \le \alpha t) &\le \mathbb{P}( M_t \le - (\Delta - \alpha) t ) \\ &\le \mathbb{P}( \tilde M_t \ge (\Delta - \alpha) t ) \\ &\le \mathbb{P}( \tilde M_t^* \ge (\Delta - \alpha) t) \\ &\le \frac{1}{(\Delta - \alpha) t} \mathbb{E} \tilde M_t^+ \\ &\le \frac{1}{(\Delta - \alpha) t} \frac{\sqrt{\mathbb{E}M_t^2 }}{2} \\ &\le \frac{\sqrt{\nu}}{2 (\Delta - \alpha) \sqrt{t}} \end{align*} where we used a discrete-time martingale inequality in terms of $(a)^+ = \max(a,0) = (a+|a|)/2$, Jensen's inequality to bound $\mathbb{E} |M_t|$, and the bound ($\diamond$) on the second moment of $M_t$. Pass to the limit as $t \to \infty$ in the final inequality to obtain the desired asymptotic result.<|endoftext|> TITLE: Upper bounds for regulators of real quadratic fields QUESTION [11 upvotes]: We have an elementary sharp lower bound for the regulator of a real quadratic field as a function of the discriminant $$R\geq \tfrac{1}{2}(\sqrt{d-4}+\sqrt{d})$$ It is sharp because the equality holds infinitely often for $d=x^2+4$. The problem of finding a good upper seems much more complicated, but there's still is a very nice (and relatively easy) bound for $\mathbb{Q}(\sqrt{D})$ depending only on $D$. Loo-Keng Hua proved that $L(1,\chi)<1+\tfrac{1}{2} \log D$, so using the trivial estimate $h\geq 1$ and substituting on Dirichlet's class number formula we get $$R<\sqrt{D}(\tfrac{1}{2}\log D+1)$$ The way this bound is presented (indirectly) in this survey suggests that it might be the best one currently known for all real quadratic fields (well, for $D>5$). Given how old Hua result is (1942), this seems unlikely, but I haven't been able to find a better one so far. I am aware of much better estimates which work for sufficiently large $D$. For example it follows from work of Lavrik that $$R<(0.263+\mathcal{o}(1))\sqrt{D}\log D$$ What is the best known bound for $R$ which holds for all real quadratic fields and depends only on $D$? (or for $D>k$ with the $k$ explicit and "small") I'm also interested in what the true bound is expected to be. REPLY [11 votes]: Stephane Louboutin has several papers on getting explicit bounds for $L(1,\chi)$, for $\chi$ a character $\pmod q$. They're all of the strength of $1/2 \log q + $ an explicit constant. Some of his results include information on $\chi(2)$ which is sometimes helpful. The best theoretical upper bound is due to P.J. Stephens: it gives $$ L(1,\chi) \le \frac{1}{4} \Big( 2- \frac{2}{\sqrt{e}} + o(1)\Big) \log q = (0.1967\ldots +o(1)) \log q, $$ for a quadratic character $\chi \pmod q$ (see for example Upper bounds for $|L(1,\chi)|$). Here the $1/4$ is from Burgess and the $2-2/\sqrt{e}$ comes from Vinogradov's trick for the least quadratic non-residue. To go beyond $1/2$ in the bound for $L(1,\chi)$ explicitly, one would have to work with either explicit Burgess type bounds or with explicit versions of Vinogradov's trick. I don't think anyone has carried that out -- nothing new here, but just needs elbow grease. On GRH Theorem 1.5 from Lamzouri-Li-Soundararajan will give you explicit bounds for $L(1,\chi)$. These show that the regulator is bounded by something of size $\sqrt{d} \log \log d$, which is likely the correct order of magnitude (but this is unknown).<|endoftext|> TITLE: Where can I find explicit descriptions of principal $SL(2,\mathbb{C})$s? QUESTION [7 upvotes]: I am interested in an explicit description of the principal homomorphism from $SL(2,\mathbb{C})$ to $G$, for each complex semisimple Lie group $G$. Does any one have specific references please? Kostant's original paper is of course great, and contains a lot, but I do not think it contains explicit descriptions for each specific example (which is what I really want at the moment). REPLY [2 votes]: I would like to add to the suggested references, one more: A.L. Onishchik, E.B. Vinberg, Lie Groups and Algebraic Groups, Springer, 1990. Exercise 4.2.28.<|endoftext|> TITLE: What is the definition of computational content? QUESTION [9 upvotes]: I am interested in type theory and proof theory. I have read a lot of papers and books that use the term "computational content" (For example: https://scholar.google.com/scholar?hl=en&q=%22computational+content+of%22&btnG=&as_sdt=1%2C33&as_sdtp=) and I have developed an intuitive sense of the meaning, but I have never seen the term formally defined. Is anyone aware of a paper or book which defines this term? REPLY [6 votes]: Let me respond to your comments to Bjorn and Matt's answers: Let's say I have a sentence of the form $\forall x\exists y\theta(x, y)$, where "$\theta(x, y)$" is "simple" (say, only bounded quantifiers) - I'm not interested in how hard it is to evaluate $\theta(a, b)$ for a given $a, b$. Then a computational interpretation begins with Skolemization: we look at the sentence $$\exists F\forall x(\theta(x, F(x))).$$ We're interested in what sorts of $F$ work here. One reasonable question to ask is, Can such an $F$ be given by an algorithm? This yields a distinction between "computably true" sentences and "not computably true" sentences. But as you correctly observe, this is a very coarse distinction in many cases. We often care about the runtime of an algorithm, or its complexity in some other sense (e.g. space used). So we may ask: "Is there an $F$ which works which is in $\mathcal{C}$?" for some nice complexity class $\mathcal{C}$. We may also be interested in provable correctness. Maybe there's a polynomial-time computable $F$, but proving that this $F$ works requires large cardinals :P. It would be quite reasonable to say that this is pretty bonkers. So now we're interested in the theories within which statements like "There is a polytime-computable solution" are proved. And, finally, we may also run into the opposite problem: given two non-computably true sentences, we may want to argue that one is "more computably true" than the other. To do this we now need to compare noncomputable functions, e.g. via Turing reducibility. So we can also go to a "larger" context, and work there. (See e.g. this paper, and also reverse mathematics.) All of these concerns - the complexity of Skolem functions, and the difficulty of proving such complexity results - are broadly part of one intuition: that our desire to prove a theorem does not stop only at determining its correctness. I would call all the nuances of a theorem, in this sense, its "computational content" - and since this is so broad, I would not hope for a formal definition of this term. (You could create one, but I think it would necessarily lose some of the aspects of the term. Not everything has to be formalized, and I say this as a formalist!)<|endoftext|> TITLE: Action of the degree 2 map on $\pi_8(S^4)$ QUESTION [24 upvotes]: I am currently reading Sullivan's Geometric Topology: Localization, Periodicity, and Galois Symmetry, on page 34 Sullivan claims that the degree 2 map $2:S^4 \to S^4$ induces the map $\left(\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\right)$ on $\pi_8(S^4) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2$. I don't see how one would perform this computation. Sullivan attributes this contribution to David Frank, I skimmed a few of his papers and saw no mention of this computation. I have a few ideas about how I might try to attack this, but none of them seem very palatable, I thought that I could look at the induced map on postnikov towers and see what happens in cohomology, but that doesn't seem to be very easy to work with. I would appreciate it if anyone had a direct method to attack this computation, or if someone had a reference about how similar computations are performed. I'm waiting on my library to retrieve a copy of Toda's "Composition Methods..." to see if this has any input on this computation. REPLY [18 votes]: Use Theorem 8.9 from Whitehead's "Elements of Homotopy Theory" on page 537. Take $k=2$, $\beta=\iota_4$ and $\alpha\in\pi_8S^4$. Use $[\iota_4,\iota_4]=\Sigma\nu'-2\nu_4$, $h_0(\nu_4\circ\eta_7)=h_0(\nu_4)\circ\eta_8=\iota_7\circ\eta_8=\eta_8$ and $h_0(\Sigma\nu'\circ\eta_7)=h_0(\Sigma(\nu\circ\eta_6))=0$. The triple product disappears and you are left with David Frank's answer. Just a reminder that $\pi_8S^4=\mathbb{Z}_2\{\Sigma\nu'\circ\nu_7\}\oplus\mathbb{Z}_2\{\nu_4\circ\eta_7\}$. The Hopf invariants and whitehead products (and some formulas to compute them) are available in Oguchi's paper "Generators of 2-Primary Components of Homotopy Groups of Spheres, Unitary Groups and Symplectic Groups".<|endoftext|> TITLE: Whether a total order set of size $n$ has the fewest endomorphisms among posets of size $n$ QUESTION [10 upvotes]: A function $f: P \to P$ is an endomorphism iff for any $x \le y$ in the poset $P$ , $f(x) \le f(y)$. So among posets of size $n$, whether the total order set $[n]$ (with the usual ordering) has the fewest endomorphisms? REPLY [11 votes]: No. The zig-zag poset on 4 elements has only 31 endomorphism, whereas the total order has 35 endomorphisms. I added the number of automorphisms and endomorphisms of a poset to http://www.findstat.org, should be visible shortly. Update: To make things a little bit clearer, note that adding more covering relations may increase the number of endomorphisms. Consider the two posets below, then the map sending $[0,1,2,3]$ to $[0,0,0,2]$ is not a poset endomorphism of the fence, but it is a poset endomorphism of the second poset.<|endoftext|> TITLE: Connections between Complexity Theory & Set Theory QUESTION [12 upvotes]: Inspired by Joshua Grochow and Iddo Tzameret's answers in a post on http://cstheory.stackexchange.com , I would like to get more references on possible connections between complexity theory and set theory here in this post. My question simply is: What are examples of those points where set theoretical axioms, concepts and tools like large cardinals and forcing appear in the complexity theory? As a matter of fact note that the statement $\mathsf P=\mathsf{NP}$ (and other similar assertions) remains unaffected by most set theoretic methods and assumptions. So this may help you restrict the searching domain for finding possible connections between set theory and complexity theory through excluding straightforward independence results. REPLY [6 votes]: As far as I'm aware, large cardinals have not found any application in complexity theory. The main concepts from set theory that have found some application in complexity theory are forcing and generic filters. In addition to the references mentioned by Mohammad Golshani, there is also the earlier book by Krajíček, Bounded Arithmetic, Propositional Logic and Complexity Theory, as well as Gaisi Takeuti's papers, e.g., Forcing on bounded arithmetic (with Yasumoto). It is not clear, however, how useful these connections are for addressing the fundamental questions in complexity theory. In this MO question I raised this question and learned that there are some serious obstacles to using forcing to construct models of arithmetic that satisfy (or refute) arithmetic sentences of interest. In a different direction, Joel Friedman has attempted to phrase certain problems about circuit lower bounds in the language of sheaves on a Grothendieck topology. I don't know if this counts as "set theory" in your book, but if you're interested, I commented briefly on Friedman's work at cstheory.stackexchange.com.<|endoftext|> TITLE: Determinant of arbitrary sum of positive semidefinite hermitian matrices QUESTION [11 upvotes]: Suppose that $A_i$, $i=1,\ldots,m$ are positive semidefinite Hermitian $n\times n$ matrices, with $a_i^{(j)}$ being the $j$-th eigenvalue of $A_i$. Let $A_0=I$. QUESTION. Can we extend the result of M. Fiedler for $m=2$ (Bounds for the Determinant of the Sum of Hermitian Matrices, Proc. Amer. Math. Soc. 30 (1971) 27-31) to establish the bound $$ \det \sum_{i=0}^mA_i \le\max_{\sigma_0,\dots,\sigma_m\in S_n}\prod_{j=1}^n\sum_{i=0}^m a_i^{(\sigma_i(j))}\,\,? $$ ($S_n$ is the group of permutations of $n$ elements) Edit: it should be noted $A_0$ is superfluous, since the statement would hold for singular matrices too. I didn't want the answers below to have to consider the additional case. REPLY [8 votes]: The previous answer was incorrect due to an incorrect use of majorization. I decided to replace it by a correct version below for completeness, and to remedy the embarrassing error (though now the answer becomes essentially equivalent to the OPs own answer, so feel free to ignore!). $\newcommand{\da}{\downarrow} \newcommand{\ua}{\uparrow} \newcommand{\nlsum}{\sum\nolimits}$ Recall that for Hermitian matrices $A$ and $B$, Lidskii's eigenvalue majorization inequality implies that there exist doubly stochastic matrices $D_1$ and $D_2$ such that $\lambda(A+B)=D_1\lambda(A)+D_2\lambda(B)$. Introduce the shorthand $a_{m+1}=\lambda(A_1+\cdots+A_m)$, and $a_j = \lambda(A_j)$ for $1\le j\le m$. Then, applying Lidskii's result repeatedly and noting that doubly stochastic matrices are closed under multiplication, it follows that there exist doubly stochastic matrices $D_1,\ldots,D_m$ such that \begin{equation*} a_{m+1} = D_1a_1+\cdots+D_ma_m. \end{equation*} We wish to upper bound $\det(A_{m+1}):=\prod_i a_{m+1,i}$. Consider the function \begin{equation*} f(D_1,\ldots,D_m) := \prod_i (D_1a_1+\cdots+D_ma_m)_i. % \le \left(\frac{\nlsum_i e_i^T(D_1a_1+\cdots+D_ma_m)}{n} \right)^n. \end{equation*} We now show that $f$ is maximized over permutation matrices. First, we make the change of variables $(D_l)_{ij}=D_{l,ij} = e^{u_{l,ij}}$. Moreover, we relax the doubly-stochastic constraints on each $D_l$ to \begin{equation*} \Omega_n := \left\lbrace(U_1,\ldots,U_m) \in \mathbb{R}^{m\times n \times n}\mid \nlsum_i e^{u_{l,ij}} \le 1,\ \forall i,\quad \nlsum_j e^{u_{l,ij}} \le 1,\forall j, \quad\text{for}\ 1 \le l \le m\right\rbrace. \end{equation*} Then, we consider the optimization problem \begin{equation*} \sup_{(U_1,\ldots,U_m) \in \Omega_n}\quad g(U_1,\ldots,U_m) := \nlsum_{i=1}^n \log\left(\nlsum_{lj} a_{l,j}e^{u_{l,ij}}\right). \end{equation*} Since all the matrices $A_1,\ldots,A_m$ are positive definite, each $a_{l,j}\ge0$. Thus, $g$ is a convex function of the $U_i$. Hence its supremum will be at an extreme point of $\Omega_n$. These points correspond to the permutation matrices. Thus, mapping back to the $D_i$ space, we observe that $f$ will be maximized at permutation matrices. This reasoning immediately yields the desired inequality: \begin{equation*} \det(A_{m+1})=\det(A_1+\cdots+A_m) \le \sup_{\sigma_1,\ldots,\sigma_m \in S_n}\prod_{i=1}^n\left(\nlsum_{l=1}^m a_{l,\sigma_l(i)} \right). \end{equation*}<|endoftext|> TITLE: When did the abuse of notation $y=y(x)$ start? QUESTION [14 upvotes]: It's quite common nowadays to name a function and the application of the function to its input with the same letter. (Possibly more so in applied areas. Certainly many calculus textbooks do this.) When did this practice start? In particular, did any of the old masters like Newton, Leibniz, Euler etc. ever write something like $y=y(x)$? Clarification: The question is really about the history of this practice. With whom did it start? I didn't want to discuss merits or demerits of this notation. If you want to provide a non standard interpretation of $y=y(x)$, please also back it up with historical references. REPLY [16 votes]: Regarding the original question of who started literally writing $y=y(x)$ or something like it, which I understand Jacobi didn't do in the quoted 1840 paper: Cayley (1859, p. 3), where $\Omega$ is regarded as a function of $r,v,y,$ or (as this may be expressed) where $\Omega = \Omega(r, v, y)$ sounds like an early example, in that he feels the need to explain the notation.<|endoftext|> TITLE: Is this diffeomorphism on standard two sphere an isometry? QUESTION [10 upvotes]: Suppose $f$ is a diffeomorphism from the standard two sphere $S^2$ to itself. Given any three points $a,b,c$ on a geodesic curve on the sphere $S^2$ and $b$ is the middle points of the geodesic arc $\widehat{ac}$, $f$ satisfies the following two conditions: (1) $f(a), f(b), f(c)$ are also on a geodesic. (2) $f(b)$ is the middle points of the geodesic arc $\widehat{f(a)f(c)}$; I want to know whether $f$ is an isometry on $S^2$ or not. REPLY [11 votes]: Yes, $f$ is an isometry; you only need to know that $f$ is a homeomorpphism, diffeomorphism is not needed. You first show that (by continuity of $f$) that $f$ sends geodesics to geodesics. From this, you conclude that $f$ is the lift of a projective transformation $g: RP^2\to RP^2$. ($f$ preserves anipodality since antipodal points are intersections of two great circles; then use von Staudt's theorem about characterization of elements of $PGL(3,R)$.) By composing with a random isometry of $S^2$, you can assume that $g$ has three fixed points $x, y, z$ in general position on $RP^2$. If $f$ is not an isometry, you get a contradiction by taking $a, c$ to be lifts of two of the fixed points, say, $x, z$, and $b$ the midpoint of $a, c$ and observing that $$ \lim_{n\to\infty} g^n(p)\in \{x, z\} $$ for every $p$ on the projective line $xz$. Edit. Here are the details. Let $SL_{\pm}(n,R)$ denote the subgroup of $GL(n,R)$ consisting to matrices with determinant $\pm 1$. This subgroup projects onto $PGL(n,R)$. Definition. An element $g\in SL_\pm(n,R)$ is a transvection if it is diagonalizable with real positive eigenvalues. A transvection is nontrivial if it is different from the identity matrix. Examples of transvections used in the proof will come from the following. Let $sl(n,R)= o(n) \oplus {\mathfrak p}$ be the Cartan decomposition of the Lie algebra of $SL_{\pm}(n,R)$; the subspace ${\mathfrak p}$ consists of traceless symmetric matrices. Then $exp({\mathfrak p})$ consists of transactions. (Actually, every transvection is conjugate to one of these.) Proposition. If $G< SL_\pm(n,R)$ is a closed subgroup strictly containing $O(n)$ then $G$ contains a nontrivial transvection. (Working more one can show that $G=SL_\pm(n,R)$ but I will not need this.) Proof. Lemma. $O(n)$ preserves unique up to scalar multiple positive definite quadratic form, namely the form $q_0$ with the identity Gramm matrix. Corollary. The normalizer of $O(n)$ in $SL_\pm(n,R)$ equals $O(n)$. Proof. If $g\in SL_\pm(n,R)$ normalizes $O(n)$, it sends $q_0$ to an $O(n)$-invariant quadratic form, hence, $g^*(q_0)=q_0$, hence $g\in O(n)$. qed Let $g\in G - O(n)$. Then $g O(n) g^{-1}\ne O(n)$ and is contained in $G$. The subset $g O(n) g^{-1}$is connected and passes through $1\in G$. Hence, the identity component of $1$ in $G$ is strictly larger than $SO(n)$. Since $G$ is a closed subgroup of $SL_\pm(n,R)$, it is a Lie subgroup (by Chevaliey’s theorem). Thus, the Lie algebra ${\mathfrak g}$ of $G$ is strictly larger than $o(n)$, hence, it has nonzero intersection with ${\mathfrak p}$. Taking a nonzero element $\xi$ of this intersection, $\exp(\xi)\in G$ is a nontrivial transvection. qed Lemma. If $g\in SL(n,R)$ is a transvection, it cannot send midpoints to midpoints in $RP^{n-1}$. Proof. Let $v_1,…,v_n$ be a basis of eigenvectors of $g$. There exist two vectors in this basis, say, $v_1, v_2$ which correspond to distinct eigenvalues $\lambda_1, \lambda_2$, say, $\lambda_1> \lambda_2$. Let $V\subset R^n$ be the span of $v_1, v_2$ and let $L\subset RP^{n-1}$ be the projectivization of $V$, a projective line in $RP^{n-1}$. Considering the action of $g$ on the projective space, we see that it preserves the line $L$ and has exactly two fixed points in this line, $p_1, p_2$, which are projections of the eigenvectors $v_1, v_2$. Since $\lambda_1> \lambda_2$, the fixed point $p_1$ is attractive and $p_2$ is the reclusive fixed point for the action of $g$ on $L$. Let $p\in L$ denote the midpoint of $p_1, p_2$. Since $g$ is not the identity on $L$, $g(p)\ne p$ (otherwise $g$ has three fixed points in $L$). But this implies that $g$ does not preserve midpoints in $RP^{n-1}$. A contradiction. qed We conclude that $O(n)$ is the maximal subgroup of $SL_\pm(n,R)$ which sends midpoint to midpoints in $RP^{n-1}$.<|endoftext|> TITLE: Is there a known construction of Cuspidal representations of GL(3) isomorphic to their own twist? QUESTION [10 upvotes]: Take $F$ a number field, $\pi$ a cuspidal automorphic representation of $GL(3, \mathbb{A}_F).$ Suppose $\pi \cong \pi\otimes \chi.$ Comparing central characters we see that $\chi$ must be cubic. Now suppose $\chi$ is a cubic character of $\mathbb{A}_F^\times$. Is it known how to construct $\pi$ such that $\pi \cong \pi\otimes \chi$ (or whether it is possible to do so)? REPLY [10 votes]: Let $E / F$ be the cyclic cubic extension corresponding to $\chi$ by class field theory. Let $\sigma$ be a generator of $\operatorname{Gal}(E / F)$, and let $\psi$ be a character of $E^\times \backslash \mathbb{A}_E^\times$ such that $\psi \ne \psi^\sigma$. Then we can consider the automorphic induction of $\psi$ to an automorphic representation $\pi$ o $GL(3, \mathbb{A}_F)$. This exists, by Example 1.1 of Clozel's 1986 ICM survey; it is cuspidal, because of our assumption on $\psi$ (Theorem 2.4(iv) of op.cit.); and it satisfies $\pi \cong \pi \otimes \chi$. (If I understand Clozel's article correctly, these are actually the only examples of cuspidal $\pi$ such that $\pi \cong \pi \otimes \chi$.)<|endoftext|> TITLE: Is there a topologically mixing and minimal homeomorphism on the circle (or on $\mathbb S^2$)? QUESTION [6 upvotes]: The irrational rotation on the circle is both a homeomorphism and minimal but is not topologically mixing. The argument-doubling transformation on the circle is topologically mixing but is neither a homeomorphism nor is it minimal. Is there a topologically mixing and minimal homeomorphism on the circle (or on $\mathbb S^2$)? REPLY [4 votes]: By Lefschetz fixed point theorem, any continuous map $f\colon\mathbb{S}^2\to\mathbb{S}^2$ has a fixed point. So, there is no hope of getting any minimal homeomorphism on $\mathbb{S}^2$. However, there exist minimal mixing homeomorphisms on the $2$-torus. The first examples were constructed by A. Kochergin in http://iopscience.iop.org/article/10.1070/SM2002v193n03ABEH000636 and more recently Artur Avila has announced the existence of smooth minimal and mixing diffeomorphisms.<|endoftext|> TITLE: Can we parametrize the Fermat curve by $g(t)^n+g'(t)^n=1$ for already named $g(t)$? QUESTION [5 upvotes]: Basically the Fermat curve is $x^n+y^n=1$. For $n > 2$, can we parametrize it by $x=g(t),y=g'(t)$ for already named function $g(t)$ of complex argument? i.e., $g(t)^n+g'(t)^n=1$. Some of what I know: For $n=2$, solution is $g(t)=\sin(t)$. For $n=4$, Wolfram alpha appears to give functional equation. For $n \ge 3$, SAGE gives functional equation with a lot of constants: #n = 3 x == 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-t^3 + 1)^(1/3)/t - 1)) + _C + 1/3*log((-t^3 + 1)^(1/3)/t + 1) - 1/6*log(-(-t^3 + 1)^(1/3)/t + (-t^3 + 1)^(2/3)/t^2 + 1), g(x) == (-t^3 + 1)^(1/3) #n = 5 x == 1/5*sqrt(5)*(sqrt(5) + 1)*arctan((sqrt(5) + 4*(-t^5 + 1)^(1/5)/t - 1)/sqrt(2*sqrt(5) + 10))/sqrt(2*sqrt(5) + 10) + 1/5*sqrt(5)*(sqrt(5) - 1)*arctan(-(sqrt(5) - 4*(-t^5 + 1)^(1/5)/t + 1)/sqrt(-2*sqrt(5) + 10))/sqrt(-2*sqrt(5) + 10) + _C - 1/10*(sqrt(5) + 3)*log(-(-t^5 + 1)^(1/5)*(sqrt(5) + 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) + 1) - 1/10*(sqrt(5) - 3)*log((-t^5 + 1)^(1/5)*(sqrt(5) - 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) - 1) + 1/5*log((-t^5 + 1)^(1/5)/t + 1), g(x) == (-t^5 + 1)^(1/5) REPLY [3 votes]: Suppose $u,v:\mathbb{C} \to \mathbb{C}$ are entire functions such that $u^{n}+v^{n}=1$ (disregarding relations like $u^{'}=v$ momentarily). Then $\frac{u}{v}$ is a meromorphic function which never attains a value among the $n$ solutions to $z^{n}=−1$. Picard's Little Theorem says that a meromorphic function omitting 3 values from its range is constant. Then $\frac{u}{v}$ is constant so $u$ and $v$ are themselves constant, or $n \leq 2$.<|endoftext|> TITLE: Steepest descent/gradient descent as dynamical system QUESTION [7 upvotes]: In my research on optimization research, I thought about attempts to bridge gradient descent (deterministic standard) via a dynamical system of sorts, meaning if I look at the iterations of Gradient Descent as the discrete orbit of some dynamical system in the phase space (maybe a Poincare map here...) So I was wondering if someone could possibly tell me if it is possible to use the lense of dynamical systems to characterize steepest descent in such a way that Hamiltonian dynamics and possibly perturbative analysis works? Analysis of the stable and unstable manifolds here? Maybe asymptotic analysis and orbits in the phase space like attractors and the center focus problem? I thank all helpers. REPLY [13 votes]: This topic has long history. Here are some references: Bloch, Anthony M. "Steepest descent, linear programming and Hamiltonian flows." Contemp. Math. AMS 114 (1990): 77-88. Brockett, Roger W. Dynamical systems that sort lists, diagonalize matrices and solve linear programming problems. Decision and Control, 1988., Proceedings of the 27th IEEE Conference on. IEEE, 1988. Helmke, Uwe, and John B. Moore. Optimization and Dynamical Systems. Springer Science & Business Media, 2012. Also, there are plenty of physically relevant PDEs which can be seen as implementing gradient descent in some Banach space. For example, see Ambrosio, Luigi, Nicola Gigli, and Giuseppe Savaré. Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media, 2008. Terry Tao, The Euler-Arnold equation, June 2010.<|endoftext|> TITLE: Is there a way to "puncture" a topos? QUESTION [9 upvotes]: Let $E$ be a (Grothendieck) topos, e.g. $E = \text{Sh}(X)$ for a topological space $X$. And let $p = (p^*, p_*):\text{Set}\to E$ be a point of $E$, is there a way to "puncture" $E$ in some sense? By "puncture" I mean a way to remove the point $p$ and still get a topos. e.g. one may expect to get $\text{Sh}(X\backslash\{p\})$ if $E = \text{Sh}(X)$ for a topological space $X$ and $p \in X$ is a usual point. But it's not obvious to see how to do this at the level of topos. I'm asking this question because I want to know if topoi are "unpuncturable" in some sense. REPLY [5 votes]: A more general question is how do you form the image of a geometric morphism f : E -> F, as a subtopos of F, and how do you form its complement? Recall that if U is a subobject of 1 in F then it defines an open subtopos (given by the operator U => ? on subobjects of 1) and its closed complement F-U is given by the operator of union with U on subobjects of 1. The closure of the image of f has to be the complement of the suprema of all the subobjects U of 1 in F for which f*(U) is 0. I have memories of a nice paper by Anders Kock about glueing which may yield more.<|endoftext|> TITLE: Why study Higher Sheaf Cohomology? QUESTION [34 upvotes]: The classical lore is that $H^1(X,\mathcal F)$ is obstruction to lifting local data to global data. However I don't understand why one would want to compute $H^3(X,\mathcal F), H^4(X,\mathcal F), \cdots$. For complex manifold $X$, $H^1(X,\mathcal O),H^{0,1}(X)$ both represent obstruction to local-to-global lifting of holomorphic functions. This in particular allows one to determine whether Mittag-Leffler problem can be solved. $H^1(X,\mathcal O)=0$ implies local solutions can be modified to identify a global solution, and $H^{0,1}(X)=0$ implies that local solutions can be multiplied by a smooth bump function, after which $\bar\partial$-exactness kicks in to save the day. However, what does $H^q(X,\Omega^p)$ or $H^{p,q}(X)$ mean (which are the same by Dolbeault theorem)? $H^1(X,\Omega^p)$ or $H^{p,1}(X)$ should represent the same local-to-global problem for holomorphic $p$-forms, but what about, say, $H^3(X,\Omega^p)$? Sure, one can talk about Cech cohomology for a good cover; $H^3(X,\mathcal F)$, for example, is about lifting sections defined on quadruple-intersections to triple-intersections. That's fine and all, but that doesn't sound very compelling to me. It seems to me that lifting sections on $q$-fold intersections to sections on $(q-1)$-fold intersections doesn't really solve any natural, interesting problems that arise independently of the formalisms introduced (for example, Lefschetz fixed point formula solves the problem of counting fixed points, and this is defined independently of singular homology and therefore I'd consider this a very compelling reason to study singular homology groups). Similar situation appears when we study Chern class and line bundles. The Bockstein morphism for exponential sheaf sequence $H^1(X,\mathcal O^\times) \rightarrow H^2(X,\mathbb Z)$ is precisely taking Chern class for line bundle, so it helps to know things like $H^1(X,\mathcal O)=H^2(X,\mathcal O)=0$, which allow us to classify line bundles for manifolds like $\mathbb {CP}^n$. However, there does not seem to be a reason to care about $H^3(X,\mathcal O)$, etc. Note: I have checked out a similar question. Here, one of the answers point out that for sheaf $\mathcal F$, if we can find an acyclic sheaf $\mathcal A$ such that $\mathcal F$ is a subsheaf of $\mathcal A$, then $$0\rightarrow \mathcal F \rightarrow \mathcal A \rightarrow \mathcal A/\mathcal F \rightarrow 0 $$ is exact and therefore by long exact sequence coming from this, $$ H^{p}(X,\mathcal F) \cong H^{p-1}(X,\mathcal A / \mathcal F)$$ and therefore higher cohomology groups can be understood as obstruction ($H^1$), and actually even global section ($H^0$) of $\mathcal A_1/(\mathcal A_2 / \cdots (\mathcal A_p /\mathcal F)\cdots )$. This just mystifies the issue further for me, somewhat, largely because I can't think of a canonical choice of such acyclic $\mathcal A$ and therefore I can't interpret the meaning of local-to-global lifting of iterated-quotient sheaves. REPLY [7 votes]: In addition to the other great answers, there are some proofs in algebraic geometry where we establish a result about $H^q(X, \mathcal{E})$ by descending induction on $q$: The base case is $q = \dim X$ and then we show Case $q$ implies Case $q-1$. Even if you only care about the result for $H^0$, this may be the best route. The big example is Grothendieck's proof that proper pushforwards of coherent sheaves are coherent. See the discussion before Cor 18.1.15 in Vakil, and related examples throughout Chapter 18.<|endoftext|> TITLE: Approximation of convex hull in high dimension QUESTION [6 upvotes]: What are efficient methods (polytime) to compute an approximation of the convex hull in high dimension (say, $30000$) for a given set of points? Edit: I am looking for an algorithm for getting the hyperplanes that define the convex hull, not the extreme points that are the vertices of the convex hull. REPLY [5 votes]: This recent paper Sartipizadeh, Hossein, and Tyrone L. Vincent. "Computing the Approximate Convex Hull in High Dimensions." arXiv:1603.04422 (2016). includes a summary of previous work on approximate convex hulls. The time complexity of their algorithm is independent of the dimension, and quadratic in the number of points. "The proposed algorithm uses a greedy method to attempt to find the best approximation to the convex hull for a given number of vertices."<|endoftext|> TITLE: Does every equivalence of operads in the category of small categories have a weak inverse? QUESTION [7 upvotes]: Call a map of operads $\mathcal{O}\rightarrow \mathcal{U}$ in the category of small categories an equivalence, if each functor $\mathcal{O}(n)\rightarrow \mathcal{U}(n)$ is an equivalence of categories. Arbitrary choices of inverses $\mathcal{U}(n)\rightarrow \mathcal{O}(n)$ won't assemble to a map of operads. Is there always a choice which does? (assuming a sufficiently strong version of the axiom of choice) If not, what would be a counterexample? Does the situation get any better, if I restrict to the category of small groupoids? REPLY [7 votes]: No, not even when you restrict to groupoids. In fact there's a counterexample in the book of Fresse that you cite (and it's explicitly said in §I.5.2.2 that the arity-wise inverse don't always assemble to an operad morphism, in fact). Consider the equivalence $\mathsf{PaB} \to \mathsf{CoB}$ from the operad of parenthesized braids to the operad of colored braids (which are both operads in groupoids). It doesn't admit a weak inverse, as in fact there does not even exist any morphism in the reverse direction. This can be seen on the level of objects: there's no way to send the only element of arity 2 in $\mathsf{CoB}$ to an element of $\mathsf{PaB}(2)$ and get an operad morphism (because $m \in \mathsf{CoB}(2)$ satisfies $m \circ_1 m = m \circ_2 m$, i.e. it is associative, but no element of $\mathsf{PaB}(2)$ satisfies this relation). Indeed, these "categorical equivalences" are equivalences as in "weak equivalences of spaces", you can only invert them under some conditions (eg cofibrant and fibrant objects). The point of these categorical equivalences is that they induce equivalences of simplicial operads after applying the classifying space functor, see the volume II of the book mentioned above.<|endoftext|> TITLE: Vertex deletion in knotless and linkless graphs QUESTION [5 upvotes]: If $G$ is a graph and $G-v$ is linkless for some vertex $v$, is $G$ necessarily knotless? Of course, one can assume that $v$ is adjacent to every vertex in $G-v$. Here, a graph is linkless if it has an embedding in 3-space with no two linked cycles. And a graph is knotless if it has an embedding in 3-space such that every cycle is an unknot. REPLY [4 votes]: No, this is false. Your question was posed by Adams in 1994, and was disproved by Foisy in 2003.<|endoftext|> TITLE: Approximate intermediate value theorem in pure constructive mathematics QUESTION [20 upvotes]: The ordinary intermediate value theorem (IVT) is not provable in constructive mathematics. To show this, one can construct a Brouwerian "weak counterexample" and also promote it to a precise countermodel: the basic idea is that the root may not depend continuously or computably on the function, since a small perturbation in a function's value may cause a root to appear or disappear. There are, however, many variants of the IVT that are constructively provable. This question is about the approximate IVT, which says that if $f(a)<00$ there is a point $x$ with $|f(x)|<\epsilon$. It seems to be well-known that the approximate IVT can be proven assuming either (1) countable (or maybe dependent) choice, or (2) that $f$ is uniformly continuous. This paper contains many versions of approximate IVT using somewhat weaker hypotheses such as "strong continuity" of $f$. But I would like to know: Can the approximate IVT be proven constructively about an arbitrary (pointwise) continuous function $f$, without using any form of choice or excluded middle (e.g. in the mathematics valid in any elementary topos with NNO)? If the answer is no, I would like to see at least a weak counterexample, or even better a specific countermodel (e.g. a topos in which it fails). Edit: To clarify, the functions in question are from the real numbers (or some interval therein) to the real numbers. I'll accept an answer that defines "the real numbers" either as equivalence classes of Cauchy sequences or as Dedekind cuts (but not as a "setoid" of Cauchy sequences). REPLY [3 votes]: [I just started here and do not have enough reputation to comment...so I´m kind of forced to give an answer.] I believe there is a different way to eliminate countable choice in the proof of aIVT (approximate intermediate value theorem). I've described this way on the constructive-news forum https://groups.google.com/forum/#!topic/constructivenews/e3JfKk_W2jI I'm not sure if that description is too specialist to post here. But what it boils down to is this: in Matt's first proof, using bisection, there is a hidden use of countable choice. Because to evaluate the real number $f(c_n)$, one has to pick a representative of its equivalence class. (This is brilliantly avoided in Matt's second proof). My first attempt to avoid this involved looking at recursive mathematics. If we take $a,b$ and $f$ to be recursive, then if we pick (recursive) representatives $a', b'$ we can apply the bisection method without using countable choice. Because taking the mean is a recursive function $m(a,b)=\frac{(a+b)}{2}$, we see that every iterative construction and evaluation of $f(c_n)$ is recursive in $a', b'$. This approach can be characterized as: `we only need countable choice because the objects that we are working with have been insufficiently specified beforehand'. In other words: we have left too much choice in $a,b$ AND $f$ However I believe there is a more general way to avoid choice in the construction of $a,b$ AND $f$. By describing R and continuous real functions in a different way, namely R as a #-quotient and continuous real functions as #-morphisms on Baire space, we can apply the bisection method without using countable choice. The reason for this is comparable to the recursive situation. If we pick #-representatives $a'',b''$ of $a,b$ AND #-representatives $m'', f''$ of the functions $m, f$, there is no choice left in the bisection procedure using $a'', b'', m'', f''$, because composition of #-morphisms is completely deterministic. In CLASS, INT and RUSS I believe we can prove that every continuous real function can be represented by such a #-morphism. This proof can be found in my book Natural Topology [but it really should be checked by some more people]. A preprint which I submitted to LMCS three years ago (!) is still under reviewer's consideration. So I believe that using #-morphisms in BISH is a general way to avoid countable choice in theorems comparable to aIVT. Since #-quotient representations can be found for any Polish space, this goes a lot further than just real functions.<|endoftext|> TITLE: Geometry of the complex quadric QUESTION [5 upvotes]: The complex orthogonal group $O(n+1, \mathbb{C})$ acts transitively on the complex quadric $$ Q_{n-1} := \{[z_0:z_1: \cdots :z_n] : z_0^2 + \cdots z_n^2 = 0 \} \subset \mathbb{CP}^n. $$ What is known about the geometry of curves, surfaces, invariant tensors, etc. in this geometry? If I'm not mistaken, the case $n = 2$, where the quadric is a conic in the projective plane, is just conformal geometry. REPLY [3 votes]: Consider a $(n+2)$-dimensional vector space $V$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $V'$ be the set of all non-zero vectors. Then $V'$ is an open (real or complex) submanifold of $V$. The projective space $\mathbb{P}:=V'/\sim$ (where $\sim$ is the equivalence relation on $V'$ denied by $v\sim w$ iff there exists some $\lambda\in\mathbb{K}$ such that $w=\lambda v$), is a compact manifold of dimension $n+1$ (over $\mathbb{K}$), such that $V'\to\mathbb{P}$ is a submersion. In particular, $\mathbb{P}$ consists of all the lines $[v]=\{\lambda v :\lambda\in\mathbb{K}\}$ through $v\in V'$. Now, $T\mathbb{P}$ is identified with $TV'/\sim$ where we appropriate extend $\sim$ in $TV'\cong V'\times V$. In particular, if $E=\mathbb{P}\times V\to \mathbb{P}$ is the trivial bundle over $\mathbb{P}$ and $$F:=\{([v], x) : x\in[v]\}$$ denotes the canonical line bundle over $\mathbb{P}$, then one can show that there is a natural isomorphism of vector bundles $T\mathbb{P}\cong {\rm Hom}(F, E/F)$. Let us pass now to the definition of the quadric. Assume $n\geq 1$ and consider the light cone in $V'$, i.e. the $(n+1)$-submanifold (hypersurface) $\cal{C}$ of $V'$ defined by $$ {\cal{C}}=\{v\in V' : g(v, v)=0\}, $$ where $g$ is a scalar product on $V$ satisfying specific properties (e.g. for $\mathbb{K}=\mathbb{R}$ the signature of $g$ must be $(p+1, q+1)$ with $p, q\geq 0$). Then, the quadric $Q$ is a $n$-dimensional submanifold of $\mathbb{P}$ defined by $$ Q=\{[v]\in\mathbb{P} : v\in\cal{C}\subset V'\}. $$ In particular, the map $V'\to\mathbb{P}$ restricts to a submersion ${\cal{C}}\to Q$. Also the vector bundle $E\to\mathbb{P}$ restricts to a vector bundle over $Q$ and similarly for the line bundle $F\to\mathbb{P}$. Now, the Lie group ${\rm SO}(g)$ acts transitively on $Q$ with kernel $A=\{\rm Id \}$ if $n$ is odd, and $A=\{{\rm Id}, -{\rm Id}\}$ if $n$ is even. The quotient ${\rm SO}(g)/A$ is the (effective) Möbius group of transformations on $Q$. Exercise: Show that $TQ\cong {\rm Hom}(F, F^{\perp}/F)$, where $F^{\perp}$ is the orthogonal subbundle of $F$ relative to the fibre metric $g$ on $E$. Mention that although $g$ induces a scalar product in the fibres of the bundle $F^{\perp}/F\to Q$, the quadric $Q$ inherits only a weaker conformal structure, which is preserved by the Möbius group. Example: The complex quadric, say of (complex) dimension n, $$ Q_{n}=\{[z]\in\mathbb{C}P^{n+1} : (z,z)=0\} $$ is diffeomorphic to the Grassmannian ${\rm Gr}_{+}(2, n):={\rm SO}(n+2)/({\rm SO}(n)\times{\rm SO}(2))$ of oriented two planes in $\mathbb{R}^{n+2}$. The complex quadric admits a holomorphic conformal structure induced by the quadratic form $(z, z)$ on $\mathbb{C}^{n+2}$. It admits also a Kähler structure, induced by the hermitian form $(z, \bar{z})$, which can be thought of as induced by the embedding of $Q_{n}$ in $\mathbb{C}P^{n+1}$. A complex quadric doesn't admit any smooth complex-bilinear Riemannian metric, although it admits a Kähler metric as we said above. Further details on complex quadrics can be found in the books of Kobayashi and Nomizu.<|endoftext|> TITLE: Elementary Proof of Riemann-Roch for Compact Riemann Surfaces QUESTION [28 upvotes]: I am supposed to give a talk about the Riemann-Roch theorem to a seminar of first and second year graduate students. I want to do Riemann-Roch for compact Riemann surfaces, but I am open to perhaps doing the version for projective curves. I assume the crowd knows little algebraic geometry. I assume however, knowledge of elementary differential geometry, sheaves, cohomology and complex analysis. I am looking for an elementary proof (especially because I want time to be able to work out some examples). Any references? REPLY [2 votes]: Here is one that is not elementary. Joe Polchinski mentions a derivation of Riemann-Roch via Feynman Path integrals. Unfortunately most people don't know what "ghosts" are and try to do String Theory without them. https://physics.stackexchange.com/questions/110322/number-of-zero-modes-on-the-sphere It is worked out in Chapter 5 of Polchinski's String Theory Vol 1 - he is trying to talk about the Moduli space of Riemann Surfaces<|endoftext|> TITLE: A $\mathbb{R}^{n}$ -fiber bundle which do not admit a n-dimensional vector bundle structure QUESTION [5 upvotes]: Is there a fiber bundle $(E,B, \mathbb{R}^{n})$, with typical fiber $\mathbb{R}^{n}$, such that there is no any $n$-dimensional vector bundle structure on the pair $(E,B)$? That is there is no a continuous map $p:E \to B$ such that the triple $(E,B,P)$ would be a $n$ dimensional vector bundle. REPLY [6 votes]: Yes, there a lots of them. The topological tangent bundle $\tau$ of any topological non-smoothable manifold $M$ gives you an example since a vector bundle reduction of $\tau$ is by smoothing theory equivalent to a smoothing of $M$.<|endoftext|> TITLE: Why does it suffice to study modular forms for $\Gamma_0(N)$? QUESTION [7 upvotes]: Every reference I've seen on modular forms seems to jump from the general definition of a modular form for congruence subgroups to studying modular forms just for $\Gamma_0(N)$. Once upon a time, I saw something like $$M_k(\Gamma(N)) \cong\bigoplus_{\chi\mod N} M_k(N^2,\chi)$$ where $M_k(\Gamma(N))$ is the vector space of all weight $k$ modular forms for $\Gamma(N)$, and on the right side $\chi$ ranges over all dirichlet characters mod $N$, and $M_k(N^2,\chi)$ is the vector space of weight $k$ modular forms $f$ satisfying $$f(\gamma z) = \chi(d)(cz+d)^kf(z)$$ for all $\gamma = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\in \Gamma_0(N^2)$. However, I don't see how this makes sense, since $\Gamma(N)$ is not a subgroup of $\Gamma_0(N^2)$. REPLY [5 votes]: Conjugating $\Gamma(N)$ by $\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix}$ $$\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix} \displaystyle\Gamma(N)\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & 1/N \end{bmatrix} \scriptstyle \ \ = \ \ \underbrace{\left\{\ \scriptstyle\begin{bmatrix} aN+1 & b \\ cN^2 & dN+1\end{bmatrix} \in SL_2(\mathbb{Z})\right\}}_{\displaystyle\tilde{\Gamma}_1(N^2)}\displaystyle \ \supset \Gamma_1(N^2)$$ Define the linear operator $T : M_k(\Gamma(N)) \to M_k(\tilde{\Gamma}_1(N^2)), \ \ T f(\tau) = f (N\tau)$, and the usual linear operators for showing $M_k(\Gamma_1(N^2)) =\displaystyle \bigoplus_{\chi \bmod N^2}M_k(\Gamma_0(N^2),\chi)$ For $gcd(d,N^2)=1$, let $\langle d \rangle : M_k(\Gamma_1(N^2)) \to M_k(\Gamma_1(N^2)) $, $\ \ \langle d \rangle g = g|_k\gamma, \quad \gamma \in \Gamma_0(N^2), \quad\gamma_d \equiv d \bmod N^2$ (which is well-defined, not depending on the chosen $\gamma$). Note that $\langle d d' \rangle = \langle d \rangle\langle d' \rangle$ And for a $\chi \bmod N^2$ : $$ \pi_\chi g= \frac{1}{\varphi(N^2)}\sum_{\begin{array}{l}d \bmod N^2\\gcd(d,N^2) =1\end{array}} \overline{\chi(d) }\langle d \rangle g$$ $\pi_\chi$ is an orthogonal projection $M_k(\Gamma_1(N^2)) \to M_k(\Gamma_0(N^2),\chi)$, and $\displaystyle\sum_{\chi \bmod N^2} \pi_\chi g=\langle 1\rangle g= g$ and for any $\chi \ne \chi'$ : $\pi_\chi \pi_{\chi'} = 0$ Finally, $Tf = \langle dN+1\rangle T f$, so that $\langle dN+d'\rangle T f= \langle d'\rangle T f$ and hence $\pi_\chi T f = 0$ whenever $\chi$ isn't a character $\bmod N$. Thus $\sum_{\chi \bmod N} \pi_\chi Tf = Tf$, and together with $M_k(\Gamma_0(N^2),\chi) \subset M_k(\tilde{\Gamma}_1(N^2))$ for any $\chi \bmod N$, it means that $$M_k(\Gamma(N)) \simeq M_k(\tilde{\Gamma}_1(N^2)) =\bigoplus_{\chi \bmod N}M_k(\Gamma_0(N^2),\chi)$$<|endoftext|> TITLE: The notions of $H^0(\widehat{ D})$ and $h^0(\widehat{D})$ are not satisfactory QUESTION [5 upvotes]: Let $K$ be a number field with ring of integers $O_K$. Moreover consider an Arakelov divisor $\widehat{D}\in\overline{\operatorname{Div }(\operatorname {Spec }O_K)}$, namely $$D=\sum_{\mathfrak p\;\text{prime}}n_\mathfrak p\mathfrak p+\sum_\sigma\lambda_\sigma \sigma$$ where $\sigma$ is an archimedian place i.e. a complex embedding of $K$ up to conjugation. Clearly we have the proper notion of principal Arakelov divisor $\widehat{(f)}$ for any $f\in K^\ast$. The usual (and most natural) definition of $H^0(\widehat{D})$ is the following: $$H^0(\widehat{D})=\{f\in K^\ast:v_\mathfrak p(f)\ge -n_\mathfrak p \;\forall \mathfrak p,\; v_\sigma(f)\ge-\lambda_\sigma\; \forall \sigma\}$$ where $v_\sigma$ is the adequate normalized real valuation at archimedian places. The above definition of $H^0(\widehat{D})$ in my opinion emulates perfectly the notion of $H^0(X,\mathcal O_X(D))$ in the usual geometric setting. One can show that $H^0(\widehat{D})$ is a finite set and its "dimension" is defined simply as (a multiple of): $$h^0(\widehat{D})=\log(\#H^0(\widehat{D}))$$ Now in this paper (van der Geer, Schoof- Effectivity of Arakelov Divisors and the Theta Divisor of a Number Field), the authors claim that the above notions of $H^0(\widehat{D})$ and $h^0(\widehat{D})$ are not satisfactory, and they suggest new definitions which lead to a "Riemann-Roch theorem". The point is that they don't explain why the usual objects $H^0(\widehat{D})$ and $h^0(\widehat{D})$ don't work, so I'd like to understand the motivation of their strong claim. On the contrary it seems to me that the above objects are very useful to get the usual Riemann-Roch theorem for number fields which is due to Lang. REPLY [6 votes]: The Arakelov-theoretical analogue of Riemann's inequality is Minkowski's theorem. As Felipe Voloch's answer indicates, Tate's approach makes the Riemann-Roch theorem a consequence of the Poisson summation formula, and the self duality of the characteristic functions of (ultrametric) balls is crucial. Over archimedean fields, the characteristic functions of balls are no longer self-dual, but the Gaussian kernel is. It is thus natural, in the Arakelov-theory framework, to consider a mollified version of the $H^0$, as has been done by van der Geer and Schoof and, apparently independently, by Damian Rössler; see also Groenwegen. This has also various advantages: A better Minkowski-type lower bound; A clean version of Serre duality; Subadditivity properties in exact sequences. This is amply explained in recent preprint of J-B. Bost, Theta invariants of euclidean lattices and infinite-dimensional hermitian vector bundles over arithmetic curves, https://arxiv.org/abs/1512.08946, where he uses them to study infinite dimensional analogues (that will crucially appear in forthcoming work).<|endoftext|> TITLE: Do set theorists work in T? QUESTION [20 upvotes]: In the thread Set theories without "junk" theorems?, Blass describes the theory T in which mathematicians generally reason as follows: Mathematicians generally reason in a theory T which (up to possible minor variations between individual mathematicians) can be described as follows. It is a many-sorted first-order theory. The sorts include numbers (natural, real, complex), sets, ordered pairs and other tuples, functions, manifolds, projective spaces, Hilbert spaces, and whatnot. There are axioms asserting the basic properties of these and the relations between them. For example, there are axioms saying that the real numbers form a complete ordered field, that any formula determines the set of those reals that satisfy it (and similarly with other sorts in place of the reals), that two tuples are equal iff they have the same length and equal components in all positions, etc. There are no axioms that attempt to reduce one sort to another. In particular, nothing says, for example, that natural numbers or real numbers are sets of any kind. (Different mathematicians may disagree as to whether, say, the real numbers are a subset of the complex ones or whether they are a separate sort with a canonical embedding into the complex numbers. Such issues will not affect the general idea that I'm trying to explain.) So mathematicians usually do not say that the reals are Dedekind cuts (or any other kind of sets), unless they're teaching a course in foundations and therefore feel compelled (by outside forces?) to say such things. Question: If set theorists just want to do set theory and not worry about foundations (and encodings of mathematical objects as sets), do they also work in the theory T? Or are they always regarding every object as a set? Also, do I understand it correctly that it's hard to actually formalize the syntax of the theory T, because of the many types and connotations of natural language involved? But then, what's "first-order" about T, if T is communicated through natural language? REPLY [11 votes]: You address your question to set theorists, and so let me answer as a set theorist that, yes, when I think purely as a set theorist, then indeed the idea that every object is a set just goes without saying — it is a basic elementary part of the ZFC conceptual framework. One needn't ever even remark on this in an argument made to other set theorists. For example, the cumulative $V_\alpha$ hierarchy provides an extremely rich structural background for the set-theoretic universe, which is especially informative and helpful in the analysis of various set-theoretic axioms, especially the large cardinal axioms that reach very high into this hierarchy. The picture of all objects existing as sets in the cumulative hierarchy is basically pervasive in set-theoretic arguments, and is definitely a fundamental part of the set-theoretic understanding of the mathematical universe. One may freely carry out an argument by $\in$-induction, for example, concluding that every object $x$ has a certain property, although really what one has proved is that every well-founded set has the property. In this sense, I would say that when set theorists are operating as set theorists amongst set theorists, they are not usually operating in the theory T described by Andreas, but rather in something much closer to ZFC, in a language expanded by concepts that have been defined in set theory. (As David mentioned, esentially no mathematician, including set theorists, works in a purely formal system.) Meanwhile, however, this doesn't mean that set-theorists don't make use of type-theoretic concepts. For example, set theorists have diverse concepts of what counts as a real number, and one can commonly find various real-number concepts used in set theory, including: an element of Cantor space $r\in 2^\omega$; a subset of the natural numbers $r\subseteq\omega$; an element of Baire space $r\in \omega^\omega$; and so on. Often the algebraic field-theoretic structure of the reals is less important or relevant to set-theoretic concerns, and set theorists typically care about real numbers as: a package containing countably much information. In many set-theoretic contexts, it doesn't matter which particular real-number concept one is using, and in this sense talk about the reals reduces essentially to the use of a real-number type. Finally, however, when set theorists communicate with other mathematicians, then of course they are naturally able to communicate in something much closer to the theory $T$ that you mentioned.<|endoftext|> TITLE: Which quartic fields contain the 4th roots of unity in their Galois closure? QUESTION [5 upvotes]: Let $K/\mathbb{Q}$ be a degree $4$ number field. Is it known how to determine whether the Galois closure of $K$, say $K'$, contains the 4-th roots of unity? In the cubic case, there is a succinct answer: the Galois closure $E'$ of a cubic field $E/\mathbb{Q}$ contains the third roots of unity if and only if $E$ is a pure cubic field: that is, $E = \mathbb{Q}(\sqrt[3]{n})$ for some cube-free integer $n$. Such a simple characterization doesn't seem to work for the quartic case, since for example the field $\mathbb{Q}(\sqrt{a}, \sqrt{-1})$ with a square-free integer $a \ne \pm 1$ contains the $4$-th roots of unity but is not of the form $\mathbb{Q}(\sqrt[4]{n})$ for a $4$-free integer $n$. If this characterization is known classically, can someone give a reference? REPLY [7 votes]: As explained in the comments, the only non-trivial case is where $K/\mathbb{Q}$ has Galois closure $K'/\mathbb{Q}$ with $\operatorname{Gal}(K'/\mathbb{Q})$ isomorphic to $D_4$. I will do this case here since it is too long for a comment. In that case, since all proper subgroups of $D_4$ are contained in a subgroup of index $2$, we must have a quadratic subextension $F/\mathbb{Q}$, so write $F = \mathbb{Q}(\sqrt{d})$. Also write $K = F(\sqrt{\delta})$, with $\delta = a + b \sqrt{d}$, where $a,b$ are in $\mathbb{Q}$. Since $D_4$ has three subgroups of index $2$, $K'$ has three quadratic subextensions, of which only one can be contained in $K$, since otherwise we would have had the Klein $4$-group as our Galois group. Now $K'$ is generated over $K$ by an element $\sqrt{\delta'}$ where $\delta' = a - b \sqrt{d}$. Claim. We have that $\operatorname{Nm}_{F/\mathbb{Q}}(\delta)=\delta\delta'=a^2-db^2$ is not contained in $F^{\times 2} \cap \mathbb{Q}^{\times}$, which by Kummer theory is the subgroup of $\mathbb{Q}^{\times}$ generated by $\mathbb{Q}^{\times 2}$ and $d$. Proof. This says precisely that $K/\mathbb{Q}$ is not normal: if $\operatorname{Nm}_{F/\mathbb{Q}}(\delta)=\epsilon^2$ with $\epsilon \in F$, then $(\epsilon/\sqrt{\delta})^2 = \delta'$, implying that all conjugates of $\delta$ are in $K$, contradiction. Now, $K'$ contains $\sqrt{\delta \delta'}$, which is a square root of the rational number $a^2-db^2$, so generates a quadratic subextension $F'$, which differs from $F$ precisely because of the claim. So $\sqrt{-1} \in K'$ is equivalent to $\overline{-1} \in \left\langle \overline{d}, \overline{a^2-db^2} \right\rangle \subset \mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$.<|endoftext|> TITLE: convergence of 2nd eigenvalue QUESTION [9 upvotes]: Fix $0 0$ for sufficiently large $n$, while if $\lambda > c/m_{11}(\infty)$, $P_n(\lambda) < 0$ for sufficiently large $n$. Thus any interval $c/m_{11}(\infty)-\epsilon, c/m_{11}(\infty)+\epsilon)$ will contain an eigenvalue for sufficiently large $n$. We conclude that the second eigenvalue does converge, to $c/m_{11}(\infty)$.<|endoftext|> TITLE: Abstract definition of convex set QUESTION [39 upvotes]: I'd like to formulate an abstract definition of convex sets: a set $K$ is convex if it is endowed with a ternary operation $K\times[0,1]\times K\to K$, written $(x:t:y)$, satisfying axioms $(x:0:y)=(x:t:x)=x$ $(x:t:y)=(y:1-t:x)$ $(x:t:(y:\frac ut:z))=((x:\frac{t-u}{1-u}:y):u:z)$ The axioms imply that, for every $x_1,\dots,x_n\in K$ and $t_i\ge0$ with $\sum t_i=1$ the convex combination $\sum t_i x_i\in K$ is well defined. Examples are the usual convex subsets of real vector spaces, with $(x:t:y)=(1-t)x+ty$, but also trees with $(x:t:y)$ the unique point at ratio $t$ on the geodesic from $x$ to $y$, and more generally $\mathbb R$-trees (geodesic metric spaces in which all triangles are isometric to tripods). I'm sure this has already been explored, and I'd rather not reinvent the wheel (and I possibly missed some useful axioms), but I couldn't find any reference to such notions. Natural questions that come to mind are: define an affine map between convex sets $K,L$ as a map $f\colon K\to L$ with $f(x:t:y)=(f(x):t:f(y))$. Topologize then convex sets by making all affine maps to $\mathbb R$ continuous. What can be said of these topological spaces? can every convex set be represented in a vector space? I have in mind the map $K\to\ell^1(K)/\{\delta_{(x:t:y)}=(1-t)\delta_x+t\delta_y\forall x,t,y\}$, though the topologies will probably not match (and I'm not sure I want to close the space I quotient $\ell^1$ by). Thanks to all! Any kinds of references or comments are welcome. REPLY [5 votes]: A good (but probably not up to date) reference on several abstractions of the notion of convexity is Singer, Ivan. Abstract convex analysis. Vol. 20. John Wiley & Sons, 1997. There you'll find notions like "order convexity" in posets, "metric convexity" for metric spaces, abstract convex combinations, but also notions based on approximation from the outside or based on abstraction of the "convex hull" operation. I am not sure if something along th lines of your ideas is in there, but it is a good read anyway.<|endoftext|> TITLE: Related Forms for the Riemann Hypothesis over Function Fields QUESTION [9 upvotes]: There are several formulations and consequences of the Riemann Hypothesis over Function Fields (RH, from now on). I am interested in the logical implications between those, and in proofs\references for those implications (which are surely known to the experts). Personally, I would be most satisfied with proofs which are explicit and which only use the knowledge that was available to Weil at the time of his proof of the RH. I will state below the various formulations\consequences. Let $\mathbb{F}_q$ be the finite field with $q$ elements. Curves: Let $C/\overline{\mathbb{F}_q}$ be a smooth, projective algebraic curve defined over $\mathbb{F}_q$. Let $\zeta_C(s)$ be the zeta function of $C$, defined as $$\zeta_C(s) = \exp(\sum_{n \ge 1} \frac{N_m}{m}q^{-ms}),$$ where $N_m$ is the number of points of $C$ defined over the degree $m$ extension $\mathbb{F}_{q^m}$ of $\mathbb{F}_q$. RH: All the zeros of $\zeta_C(s)$ lie on the line $\Re(s)=\frac{1}{2}$. Field Extensions: Let $K/\mathbb{F}_q$ be a function field with constant field $\mathbb{F}_q$. Let $\zeta_K(s)$ be the zeta function of $K$, defined as follows: $$\zeta_K(s) = \sum_{A \ge 0} (NA)^{-s},$$ where the summation is over all effective divisors $A$ of $K$, and $NA=q^{\deg A}$. RH implies: All the zeros of $\zeta_K(s)$ lie on the line $\Re(s)=\frac{1}{2}$. Rings of Integers (Dedekind zeta functions): Let $K/\mathbb{F}_q(T)$ be a field extension of finite degree. Let $O_K$ be the integral closure of $\mathbb{F}_q[T]$ in $K$. Let $\zeta_{O_K}(s)$ be the zeta function of $O_K$, defined as follows: $$\zeta_{O_K}(s) = \sum_{I \ge 0} (NI)^{-s},$$ where the summation is over all ideals $I$ of $O_K$, and $NI=|O_K/I|$. RH implies: All the zeros of $\zeta_{O_K}(s)$ lie on the line $\Re(s)=\frac{1}{2}$. Characters (L-functions): Let $\chi:\mathbb{F}_q[T] \to \mathbb{C}$ be a Dirichlet character. Let $$L(s,\chi) = \sum_{f \in \mathbb{F}_q[T], f \text{ monic}} \chi(f) |f|^{-s}$$ be its L-function, where $|f|=|\mathbb{F}_q[T]/f|$. RH implies: All the zeros of $L(s,\chi)$ lie on the line $\Re(s) = \frac{1}{2}$. I believe the implications are $1 \leftrightarrow 2 \leftrightarrow 3 \implies 4$, although I cannot show this rigorously. Are these the only implications, and are they correct? Below are some hand-waving arguments that partially explain the implications. They are not proofs, and I am not satisfied with them. I am most comfortable with Variant 2, mostly because there is an elementary proof for it, due to Stepanov-Bombieri, found in the appendix "Number Theory in Function Fields" by Michael Rosen. Morally, Variant 1 and Variant 2 are equivalent, since one can associate a function field to any curve, and vice versa. The only difference between Variant 2 and Variant 3 seems to be the contribution of the prime at infinity, which only contributes a pole ($s=1$). Variant 3 implies Variant 4, as follows: Note $L(s,\chi)$ is a polynomial. Construct an abelian extension $K$ of $\mathbb{F}_q(T)$ whose set $S$ of "associated Dirichlet characters" contains $\chi$. In that case, $\zeta_{O_K}(s) = \prod_{\chi' \in S} L(s,\chi')$. $RH$ for $\zeta_{O_K}(s)$ implies $RH$ for $L(s,\chi)$. The hard part, which I am not sure how to do, is the construction of the abelian extension - Is there a simple procedure for producing such an extension? (Cross-posted from MSE. Got no answer there in the duration of 8 months.) REPLY [9 votes]: 1 is not just morally equivalent with 2 - the equivalence is really easy. As you say, there is a correspondence between smooth, projective, geometrically connected algebraic curves over $\mathbb F_q$ and function fields with constant field $\mathbb F_q$. The equivalence sends the set effective divisors on the curve (=formal sums of closed points) to the set of effective divisors of the field. By the definition of effective divisors, one writes the zeta function as a product over the closed points $P$ of $1/(1- q^{-d_P s})$ where $P$ is the degree of $d$. Because a closed point of degree $d$ corresponds to $d$ $\mathbb F_{q^m}$ points when $m$ is a multiple of $d$ and $0$ otherwise, these zeta functions are identical. 3 as stated is wrong, because there may be extra primes at $\infty$, which create zeroes off the line, and the prime at $\infty$ may be inert, creating the same effect. It is only valid when the extension is totally ramified at $\infty$. 4 is also wrong as stated when $\chi$ is trivial on $\mathbb F_q^\times$. Indeed one can see this explicitly, as by the polynomial description, $L(0,\chi)$ is equal to the sum over monic polynomials $f$ of bounded degree of $\chi(f)$, thus $1/(q-1)$ times the sum over all polynomials of bounded degree of $\chi(f)$, which vanishes when the bound is large enough. The abelian extension may be constructed by class field theory. This can be done explicitly. If $\chi$ is a character modulo a polynomial $f(T)$ of degree $d$, consider the extension generated by variables $a_0,\dots,a_{d-1}$ satisfying the system of equations $$(a_0^q+ a_1^q X + \dots + a_{d-1}^q X^{d-1}) \equiv (a_0+ a_1 X + \dots + a_{d-1} X^{d-1}) (T-X) \mod f(X)$$ where $X$ is a formal variable. The left side and right side, after applying Euclids algorithm to mod out by $f(X)$, are polynomials of degree $d-1$ in $X$, so this is really $d$ equations. Also we should demand $(a_0+ a_1 X + \dots + a_{d-1} X^{d-1})$ be invertible modulo $f(X)$. Now the group $(\mathbb F_q[X]/f(X))^\times$ acts on this equation via its action by multiplication on $(a_0+ a_1 X + \dots + a_{d-1} X^{d-1})$ modulo $f(X)$. (Because the coefficients are $\mathbb F_q$, they are equal to their own $q$th powers, and cancel on both sides). In fact these equations define a field extension with Galois group $(\mathbb F_q[X]/f(X))^\times$. I don't remember how easy it is to prove this. The claim is that this extension's zeta function has the factorisation you seek. The idea is that it is sufficient to calculate how the Frobenius elements of various primes lie in the Galois group, and this can be done easily because, by the equation we have written down, the Frobenius acts on $(a_0+ a_1 X + \dots + a_{d-1} X^{d-1})$ by multiplication by $X-T$. If you do this calculation you will see an extra factor appearing at $\infty$ when $\chi$ is trivial on $\mathbb F_q^\times$. Finally, you are correct that 4 does not imply 3 because not all function fields are of this form, or are subfields of fields of this form. Because it came up in Keith's and Ofir's comments, let me explain how this relates to the Carlitz module. If we let $\mathbb F_q[Y]$ act on $\overline{\mathbb F_q[T]}$ in the unique linear way such that $Y (g) = g^p + Tg$, then $\overline{\mathbb F_q[T]}$ is called the Carlitz module and the Carlitz torsion - the elements sent to $0$ by the action of $f(Y)$ - generate an abelian extension, which I claim is equivalent to this one (modulo sign changes). Indeed let me adjust this slightly so $Y(g) = T - g^p$, which should be the same thing. The system of equations $$(a_0^q+ a_1^q X + \dots + a_{d-1}^q X^{d-1}) \equiv (a_0+ a_1 X + \dots + a_{d-1} X^{d-1}) (T-X) \mod f(X)$$ may be rewritten $$ X (a_0+ a_1 X + \dots + a_{d-1} X^{d-1}) \equiv (Ta_0-a_0^q+ (Ta_1-a_1^q) X + \dots + (Ta_{d-1}-a_{d-1}^q) X^{d-1}) \mod f(X)$$ So the action of $X$ by multiplication is the same as the action of $Y$. Since the action of X on the left side clearly satisfies $f(X)=0$, it follows that the action of $Y$ satisfies $f(Y)(a_i)=0$. So the elements $a_0, \dots, a_{d-1}$ are all elements of the Carlitz $f$-torsion.<|endoftext|> TITLE: A strong relationship between $\mathrm{ad}(X)$ and $1-\mathrm{Ad}_g$ when $\mathrm{Ad}_gX=X$ QUESTION [5 upvotes]: Let $G$ be a linear algebraic semisimple group over $\mathbb{C}$ with Lie algebra $\mathfrak{g}$, and let $(g,X)\in G\times\mathfrak{g}^{\operatorname{reg}}$ be such that $\mathrm{Ad}_gX=X$ (where $X\in\mathfrak{g}^{\operatorname{reg}}$ means that $Z_{\mathfrak{g}}(X)$ has minimal dimension). Conjecture. We have $$\operatorname{im}(1-\mathrm{Ad}_g)\subseteq\operatorname{im}\mathrm{ad}(X), \quad\text{and}\quad\ker\mathrm{ad}(X)\subseteq\ker(1-\mathrm{Ad}_g).$$ I already know that the second identity is true. Indeed, using the accepted answer to this question, $Z_G(X)$ is abelian and hence $Z_G(X)\subseteq Z_G(g)$. Passing to Lie algebras gives that $\ker\mathrm{ad}(X)\subseteq\ker(1-\mathrm{Ad}_g)$. I believe the first identity is also true. At least, it is true when $G=\mathrm{SL}(2,\mathbb{C})$, as I verified it by brut force. It also seems true for $G=\mathrm{SL}(n,\mathbb{C})$, as I verified the identity in many cases using a symbolic mathematical computation program. But I don't have any conceptual reason for why this happens. I would be interested to know if the conjecture is true, or if we can determine a subclass of the linear algebraic semisimple groups for which it holds (for all pairs $(g,X)$). REPLY [6 votes]: Consider the Killing form on $\mathfrak g$. Since $Ad_g$ is orthogonal, it is easy to see that $im(1-Ad_g)^\perp=ker(1-Ad_g)$. Since $ad_g$ is anti-selfadjoint, $im(ad(x))^\perp=ker(ad(x))$. Therefore, your two statements are equivalent. Since you know the second holds, so does the first.<|endoftext|> TITLE: Non-phantom maps from $\mathbb CP^{\infty}$ to a finite complex QUESTION [13 upvotes]: It is not true that every map from $\mathbb CP^{\infty}$ to a finite complex is null-homotopic. For example it is a result due to Brayton-Gray that there are non-null maps $\mathbb CP^{\infty} \to S^3$, this story is told in P. May's "More Concise Algebraic Topology" (see cor. 2.4.3 on page 40). However, the examples constructed there are phantom maps, i.e. they become trivial when restricted to any finite subcomplex $\mathbb CP^k \subset \mathbb CP^{\infty}$. (A way to think about this is that it is just not possible to find a compatible choice of null-homotopies.) So my question is: Does there exist a finite CW-complex $X$ and a non-phantom map $\mathbb CP^{\infty} \to X$? Here are some thoughts: (1) One can show that any such map has to be trivial on cohomology (with $\mathbb Z$ coefficients, using the cup product structure of $\mathbb CP^{\infty}$) and consequently also on homology (here we have to use that $X$ is finite). Lifting the map to the universal cover $\tilde{X}$ of $X$ and using Hurewicz, one can also see that the map is zero on all homotopy groups ($\mathbb CP^{\infty}$ only has one). However, there are plenty of non-trivial maps inducing zero on all homotopy and homology groups, the easiest example is maybe given by the composition $T^3 \to S^3 \to S^2$ where the first map collapes the complement of a ball and the second one is the Hopf map. (2) Using the Sullivan conjecture/Miller's theorem one can probably show that the $p$-completion of any such map is null. (3) Going in the opposite direction, maybe one can start with a non-trivial map $\mathbb CP^{k} \to X$ which induces zero on all homotopy and homology groups and then somehow prove that it extends to $\mathbb CP^{\infty}$. However I do not see how one could show in practice that extending the map is possible... REPLY [11 votes]: All maps of $\mathbf CP^\infty$ to a finite complex are phantom maps; see A. Zabrodsky, On phantom maps and a theorem of H. Miller, Israel J. Math 58 (1987) 129-143. For maps out of classifying spaces of Lie groups, see Friedlander-Mislin, Locally finite approximations of Lie groups, I, Invent. Math. 83 (1986), 425-436.<|endoftext|> TITLE: Are metric isometries smooth at the boundary? QUESTION [6 upvotes]: Let $M,N$ be smooth Riemannian manifolds with boundary (In particular, we assume the boundaries are smooth). Suppose we have a map $\phi:M \to N$ which satisfies the following properties: $$(1) \, \, \phi:M \to N \, \, \text{is a bijection}$$ $$ (2) \, \, \phi(\operatorname{int}M)=\operatorname{int}N,\phi(\partial M)=\partial N $$ $$ (3) \, \, \phi:M \to N \, \,\text{is a metric isometry}$$ By the Myers-steenrod theorem, applied to $\phi|_{\operatorname{int}M} :\operatorname{int} M \to \operatorname{int}N $, $\phi$ is a diffeomorphism between $\operatorname{int} M , \operatorname{int}N$. Question: Is $\phi$ necessarily smooth as a map $M \to N$? When looking at the proof of Myers-steenrod theorem here, the problem seems to be that initial conditions do not determine a unique geodesic, if the starting point is on the boundary. The basic idea of the proof is to express the map $\phi$ in exponential coordinates, then show this representation is linear, hence smooth. However, constructing this representation relies on the uniqueness of geodesics. I suspect there might be a counter example where singularity occurs at the boundary, but I could not fine one. REPLY [3 votes]: Yes, $\phi$ is smooth. Indeed, fix any $x\in\partial M$ and take any nieghbourhood $x\in U\subseteq\partial M$ with compact closure. Calling $\nu:\partial M\to TM$ the (unit) inward-pointing normal vector, we can find some $\epsilon>0$ so small that the map $\alpha:U\times [0,\epsilon]\to M$, $\alpha(y,t):=\exp(t\nu(y))$ is well-defined and is a diffeomorphism onto its image; $\text{dist}(\alpha(y,t),\partial M)=t$ for all $(y,t)\in U\times [0,\epsilon]$. For any $y\in U$, the curve $t\mapsto\phi\circ\alpha(y,t)$ is a unit-speed geodesic: this is true on the interval $(0,\epsilon]$ by interior smoothness of $\phi$, so it is true on $[0,\epsilon]$ by continuity. Since $\text{dist}(\phi\circ\alpha(y,t),\partial N)=t$, we deduce that it is a minimizing geodesic from $\partial N$ to $\phi\circ\alpha(y,\epsilon)$, implying that $\frac{d}{dt}(\phi\circ\alpha)|_{t=0}\perp\partial M$, i.e. $$ \phi\circ\alpha(y,t)=\exp(t\nu(\phi(y)))\qquad (*)$$ (now $\nu$ denotes the inward-pointing normal in $N$). Up to shrinking $\epsilon$, we can assume that $(z,t)\mapsto\exp(t\nu(z))$ gives a diffeomorphism from $\phi(U)\times [0,\epsilon]$ onto its image, as well. We call $(\beta,\tau)$ its smooth inverse. Finally, $\phi|_U$ is smooth (as $\phi(y)=\beta\circ\phi\circ\alpha(y,\epsilon)$ and as $\phi$ is smooth on $\text{int}(M)$), so also $\phi|_{\alpha(U\times[0,\epsilon])}$ (by $(*)$).<|endoftext|> TITLE: Constructing real forms of the Tits-Freudenthal magic square for (Rosenfeld) projective planes QUESTION [5 upvotes]: If $\mathbb{K},\mathbb{L} \in \{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}$ then the Rosenfeld projective ("elliptic"?) plane $\mathbb{P}^2(\mathbb{K}\otimes\mathbb{L})$ is "the" compact Riemannian symmetric space associated with the non-compact Lie algebra $\tilde L_3(\mathbb{K},\mathbb{L})$ given by the following version of the Tits-Freudenthal magic square: $$\begin{array}{r|c|c|c|c|} &\mathbb{R}&\mathbb{C}&\mathbb{H}&\mathbb{O}\\\hline \mathbb{R}&\mathfrak{so}(1,2)&\mathfrak{su}(1,2)&\mathfrak{sp}(1,2)&\mathfrak{f}_4(\mathfrak{so}_9)\\\hline \mathbb{C}&\mathfrak{su}(1,2)&\mathfrak{su}(1,2) \oplus \mathfrak{su}(1,2)&\mathfrak{su}(2,4)&\mathfrak{e}_6(\mathbb{R}\oplus\mathfrak{so}_{10})\\\hline \mathbb{H}&\mathfrak{sp}(1,2)&\mathfrak{su}(2,4)&\mathfrak{so}(4,8)&\mathfrak{e}_7(\mathfrak{su}_{2}\oplus\mathfrak{so}_{12})\\\hline \mathbb{O}&\mathfrak{f}_4(\mathfrak{so}_9)&\mathfrak{e}_6(\mathbb{R}\oplus\mathfrak{so}_{10})&\mathfrak{e}_7(\mathfrak{su}_{2}\oplus\mathfrak{so}_{12})&\mathfrak{e}_8(\mathfrak{so}_{16})\\\hline \end{array}$$ (Notation is hopefully understandable: in the case of exceptional Lie algebras, the real form is indicated by putting their maximal compact subalgebra in parentheses. For example, $\mathbb{P}^2(\mathbb{H}\otimes\mathbb{H}) = \mathit{SO}_{12}/\mathit{S}(\mathit{O}_4\times \mathit{O}_8)$ and $\mathbb{P}^2(\mathbb{C}\otimes\mathbb{O}) = \mathit{E}_6/(\mathit{U}_1\cdot\mathit{Spin}_{10})$. I am unsure about finite coverings, which is why I only put Lie algebras in this table and why I put quotation marks around "the" above; but this is irrelevant for my question.) This table is found, e.g., in Hans Freudenthal's 1964 report "Lie Groups in the Foundations of Geometry" (Advances in Math. 1, 145–190), ¶4.16 on p. 172. It is also found in Robert Goss's 2015 Ph.D. thesis The topology of the higher projective planes in §2.3 but with an incorrect claim as to its construction. When the set of $3\times 3$ Hermitian matrices with coordinates in $\mathbb{K}\otimes\mathbb{L}$ is a Jordan algebra, then $\mathbb{P}^2(\mathbb{K}\otimes\mathbb{L})$ can be defined as the set of trace $1$ idempotents of that algebra. But as far as I understand, there is no such construction of $\mathbb{P}^2(\mathbb{K}\otimes\mathbb{L})$ from $\mathbb{K}\otimes\mathbb{L}$ in the case of $\mathbb{H}\otimes\mathbb{O}$ and $\mathbb{O}\otimes\mathbb{O}$. (Rosenfeld's book Geometry of Lie Groups seems to suggest that there is one, but I find it very difficult to understand what this book actually says.) Instead, the way we know which particular real form of the magic square to use is that it gives a symmetric space of the desired real dimension $2pq$ (with $p := \dim_{\mathbb{R}}(\mathbb{K})$ and $q := \dim_{\mathbb{R}}(\mathbb{L})$) and desired real rank $\min(p,q)$. Anyway, my question is: How can we construct the real form $\tilde L_3(\mathbb{K},\mathbb{L})$ given in the above table from $\mathbb{K}$ and $\mathbb{L}$ using an algebraic construction? I emphasize that the difficult part is getting the correct real form: the complexification is the usual magic square of Lie algebras. There are a number of constructions given in Barton & Sudbery's article "Magic squares and matrix models of Lie algebras", but unless I missed something, they do not produce the above tables (for $\mathbb{K},\mathbb{L} \in \{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}$, their $L_3(\mathbb{K},\mathbb{L})$ defined in theorem 4.4 gives a compact real form; and if we replace $\mathbb{K}$ and $\mathbb{L}$ by the split form of the algebra, the result, shown near the end of section 7 of their paper, produces a different table from the one above). So, is there a way to produce the above table without resorting to filtering real forms by their real rank? REPLY [2 votes]: Alberto Elduque obtained all real forms of exceptional Lie algebras in his variant of Freudenthal square in A new look at Freudenthal's magic square, Non-associative algebra and its applications, 149–165, Lect. Notes Pure Appl. Math., 246. There is also a PhD thesis (pdf link) of Yongdong Huang that shows how to get all Riemannian symmetric spaces out of a magic square construction.<|endoftext|> TITLE: Reason for putting log weight for exponential sums over primes? QUESTION [5 upvotes]: In analytic number theory, for example as in ternary Goldbach problem via circle method, when one has to deal with exponential sums over primes often people use von Mangoldt function or log weight. In other words, instead of $$ \sum_{p < X}e^{2 \pi i f(p)} $$ one considers $$ \sum_{n < X} \Lambda(n) e^{2 \pi i f(n)} $$ or $$ \sum_{n < X} \log (n) 1_{\mathbb{P}}(n) e^{2 \pi i f(n)} $$ where $1_{\mathbb{P}}$ is the characteristic function on the primes and $f$ is a polynomial with integer coefficients. Is the reason for introducing these log weights only so that certain computation becomes less messy? or are there cases where introducing such weights was actually crucial? [are there even some philosophy behind this maneuver?] Thank you very much! REPLY [10 votes]: Ultimately, it is because of the fundamental theorem of arithmetic, which expresses each natural number $n$ as a product of the primes dividing it: $$ n = p_1^{a_1} \dots p_k^{a_k}.$$ Taking logarithms to make this multiplicative formula additive, we conclude that $$ \log n = \sum_{p,j: p^j|n} \log p $$ or in terms of the von Mangoldt function $$ \log n = \sum_{d|n} \Lambda(d).$$ This formula is the starting point for many computations involving primes, and so we see that the logarithmic weighting enters in at the very beginning.<|endoftext|> TITLE: Covering systems QUESTION [6 upvotes]: Suppose we have a minimal covering system. If $k$ is the maximal positive integer such that the $k$-th power of a prime $p$ divides some modulus, then the $k$-th power of $p$ is a divisor of at least $p$ moduli. Is this true or is there a counterexample? REPLY [6 votes]: A covering system (or, simply, a cover) is a finite collection of congruences $x\equiv a_i\pmod{m_i}$ with distinct moduli, each modulus exceeding 1, such that every integer satisfies at least one of the congruences. We suppose a cover is minimal, in the sense that for each congruence there is an integer satisfying that congruence and no other congruence. Let $p$ be a prime, and suppose $p^k$ ($k\ge1$) divides at least one of the moduli, say, $p^k\mid m_1$, but $p^{k+1}$ doesn't divide any of the moduli. We want to prove that $p^k$ divides at least $p$ of the moduli. Let $n$ be an integer satisfying $n\equiv a_1\bmod{m_1}$, but not satisfying any other congruence. Let $L$ be the least common multiple of all the moduli that are not divisible by $p^k$. Then the integers $n,n+L,n+2L,\dots,n+(p-1)L$ don't satisfy any congruence to a modulus not divisible by $p^k$ (since $n$ doesn't, and since they are all congruent modulo $L$, and thus to each modulus not divisible by $p^k$), and they lie in different congruence classes modulo $p^k$ (since $p^k$ doesn't divide $L$). Thus, these $p$ numbers must be covered by different congruences to moduli divisible by $p^k$, so there must be at least $p$ such moduli.<|endoftext|> TITLE: In what sense can we "describe" the rational points on a unirational surface? QUESTION [5 upvotes]: This is not a completely precise question, but I hope someone can offer an interesting perspective on my problem. In the field of Diophantine geometry, an important question is deciding whether a geometrically rational surface $X$ defined over a number field $k$ admits a dominant rational map $\mathbb{P}^2_k \dashrightarrow X$; if it does, we say $X$ is unirational (or unirational over $k$, but the ground field is often understood to be part of the data of $X$ so really we don't need to mention it explicitly). So for example, if $X$ is a del Pezzo surface of degree $\geq 3$, and $X(k) \neq \emptyset$, we know $X$ is unirational. On the other hand, the question is open in general for del Pezzo surfaces of degrees $1$ and $2$. However, I would like to ask: What good does it do us to know that $X$ is unirational? To be a little more specific, I am wondering how much help it is to know that $X$ is unirational, if what we're really interested in is as good a description of the set $X(k)$ of rational points on $X$ as we can get. Of course, if $X$ is actually rational, i.e. there exists a birational map $\mathbb{P}^2 \dashrightarrow X$, we have as good a description of $X(k)$ as we could wish. But there are plenty of examples of geometrically rational $X$ that admit a unirational parametrization, but not a birational one. An example. Let $X/\mathbb{Q}$ be degree $4$ del Pezzo surface given by $$ xy + x + y - 6 = u^2, ~~~ xy - x - y + 6 = v^2. $$ Projection to the $y$-coordinate gives $X$ a conic bundle structure $\pi : X \rightarrow \mathbb{P}^1$; however, $\pi$ does not have a section (as can be verified by checking that $\pi^{-1}(2)$ does not have any $2$-adic points). But if $y$ has the form $(t^2+1)/(t^2-1)$, then $X_y := \pi^{-1}(y)$ takes the form $t^2u^2-v^2 = P_6(t)$, for some degree $6$ polynomial $P_6$ with rational coefficients, and this clearly admits a parametrization. This shows that if we pull back $\pi:X \rightarrow\mathbb{P}^1$ along the map $t \mapsto (t^2 +1)/(t^2-1)$, and we denote the result by $\pi':X'\rightarrow \mathbb{P}^1$, then $\pi'$ has a section, which shows that $X'$ is birational to $\mathbb{P}^2$. It follows that there exists a degree $2$ dominant rational map $$\phi:\mathbb{P}^2 \stackrel{2:1}{\dashrightarrow} X.$$ We note that, since $t^2=(y+1)/(y-1)$, the corresponding extension of function fields is obtained by adjoining a square root of $(y+1)/(y-1)$. On the other hand, since the Brauer group of $X$ is $\mathbb{Z}/2\mathbb{Z}$ (if I have made no mistakes), there does not exist a birational map $\mathbb{P}^2 \dashrightarrow X$. The problem is of course that $\phi$ does not give all rational points on $X$. Indeed, it only gives those rational points $(x_0,y_0,u_0,v_0)$ where $y_0^2-1 = t_0^2$ is the square of a rational number. So for example, the fiber $X_3$ has the point $(x,u,v)=(3,3,3)$. So while $\phi$ does give infinitely many rational points on $X$, and even a Zariski dense set of them, it certainly does not give a complete description of $X(\mathbb{Q})$. [One half-baked idea that did occur to me at this point, is that one could consider "twists" of $\phi$. Indeed, since $\phi$ is $2$-to-$1$, we can define quadratic twists of it in the following way: for each $c \in \mathbb{Q}^{\times}$, consider the field extension $K_c:=\mathbb{Q}(X)[t]/((y-1)t^2-c(y+1))$ of the function field $\mathbb{Q}(X)$ of $X$. This extension corresponds to a dominant rational map $\phi_c:X'_c \dashrightarrow X$, where $X'_c$ is some geometrically rational surface with function field $K_c$, which may be chosen to be smooth and projective, defined over $\mathbb{Q}$. So we get a family of rational maps $\{ \phi_c : X'_c \dashrightarrow X \}$, which together give all rational points on $X$. Unfortunately, I have no indication that the arithmetic of the $X'_c$ should be any easier to analyze than that of $X$ itself. Moreover, there doesn't even seem to be a good reason why the set of all $c$ such that $X'_c(\mathbb{Q})\neq \emptyset$ (independent of the choice of $X'_c$ by Lang-Nishmura) should admit of an easy description.] Is there any way to get around this? It feels disconcerting to me that, even in the case of varieties that are so agreeable as to be unirational, it seems a hard problem to actually describe the set of rational points in any non-trivial manner. But of course, this might just be one of those cases where life doesn't turn out to be as pleasant as one might have hoped... REPLY [4 votes]: As requested, I'm upgrading my comments to an answer. In fact one can even show something stronger for your surface: there is no finite collection of rational maps $\phi_i:\mathbb{P}^2 \to X$ such that $X(ℚ)$ lies in the image of the union of the $\phi_i(\mathbb{ℙ}^2(ℚ))$. To see this, we note that $X$ satisfies so-called weak weak approximation. This follows from the fact that the Brauer group of $X$ modulo constants is finite, and that the Brauer-Manin obstruction is only obstruction to weak approximation for $X$ (this being a theorem of Skorobogatov and Salberger) The union of the $\phi_i(\mathbb{ℙ}^2(ℚ))$ is a thin set, by definition, but the rational points on X are not thin as X satisfies weak weak approximation by a theorem of Ekedahl, cf. the discussion in https://en.wikipedia.org/wiki/Thin_set_(Serre) This property should hold more generally for any (geometrically) rationally connected non-rational smooth projective variety $X$ over a number field (the same argument applies in this case, assuming Colliot-Thélène's conjecture that the Brauer-Manin obstruction is the only one to weak approximation).<|endoftext|> TITLE: Direct decompositions of torsion-free abelian groups QUESTION [6 upvotes]: In his paper A. L. S. Corner, A note on rank and direct decompositions of torsion-free abelian groups, Camb. Phil. Soc. Proc. 57, (1961), 230--233, in which the late Prof. Corner proves various criteria for a finite rank torsion--free abelian group to have non--isomorphic indecomposable decompositions, he writes It can be shown quite readily that an equation such as $1 + 1 + 2 = 1 + 3$ cannot be realized in terms of direct sums of indecomposable groups. By this he means that there is no abelian group $A=B_1\oplus B_2\oplus B_3=C_1\oplus C_2$ with indecomposable summands of rank $(B_1)=$rank $(B_2)=$ rank $(C_1)=1$, rank $(B_3)=2$ and rank $(C_2)=3$. This result does not follow from anything else in the paper nor any of the references to the paper and I am unable to prove it either readily or otherwise. Can anyone suggest a suitable approach? REPLY [2 votes]: I switch notation to make it intuitive (at least for me): assume by contradiction that $G$ is a torsion abelian free group of rank 4 (i.e., isomorphic to a subgroup of $\mathbf{Q}^4$ containing $\mathbf{Z}^4$) and $$A_1\oplus A_2\oplus B=G=A\oplus C,$$ with $A_1,A_2,A$ subgroups of rank 1, $B$ of rank 2, $C$ of rank 3 and $B,C$ indecomposable. Then $A$ cannot be contained in $A_2\oplus B$ because otherwise we would have $G/A\simeq A_1\oplus (A_2\oplus B)/A\simeq C$, contradicting that $C$ is indecomposable, and similarly $A$ is not contained in any of $A_1\oplus B$, $A_1\oplus A_2$. Again similarly, $A_1\oplus A_2$ is not contained in $C$ since this would imply $G/(A_1\oplus A_2)\simeq B\simeq A\oplus C/(A_1\oplus A_2)$, contradicting that $B$ is indecomposable. Write $\{1,2\}=\{i,j\}$. Then $A$ being a subgroup of $A_i\oplus (A_j\oplus B)$ not contained in $A_j\oplus B$, it is isomorphic to a subgroup of $A_i$. So $A$ is isomorphic to a subgroup of both $A_1$ and $A_2$. Also, $A_1\oplus A_2$ being not contained in $C$, there exists $i$ in $\{1,2\}$ such that $A_i$ is not contained in $C$; we assume $i=2$ up to switching notation. Since $A_2\subset G=A\oplus C$, we deduce that $A_2$ is isomorphic to a subgroup of $A$. If $U,V$ are subgroups of $\mathbf{Q}$, $U$ and $V$ are isomorphic if and only if $U\cap V$ has finite index in both $U$ and $V$ (simple consequence of classification, or easy exercise). In particular, if $U$ is isomorphic to a subgroup of $V$ and $V$ is isomorphic to a subgroup of $V$, then $U,V$ are isomorphic. So $A$ and $A_2$ are isomorphic, and are isomorphic to a subgroup of $A_1$. We now distinguish two cases. (a) $A_1$ is contained in $C$. Then $C=A_1\oplus ((A_2\oplus B)\cap C)$: indeed, first it's clear that the sum is direct, and if $c\in C$, write it as $a_1+x$ with $a_1\in A_1$ and $x\in A_2\oplus B$: since $a_1\in C$, we deduce that $x\in C$. So we contradict that $C$ is indecomposable. (b) $A_1$ is also not contained in $C$. Then, for the same reason as previously (when we checked that $A$ and $A_2$ are isomorphic), $A$ and $A_1$ are isomorphic. Therefore the group $\mathrm{GL}_2(\mathbf{Z})$ acts on $A_1\oplus A_2\simeq A\oplus A$ in the standard way, and this action is transitive on the set of maximal rank 1 subgroups, and also it extends to an action on $G$ (because $A_1\oplus A_2$ is a direct factor in $G$). Therefore, if $A'$ is the projection of $A$ on $A_1\oplus A_2$ modulo $B$, we can use such an automorphism and change $A_1$ and $A_2$ so as to assume that $A'\subset A_1$. This means that $A\subset A_1\oplus B$, and we have shown that this cannot be the case.<|endoftext|> TITLE: Continuous deformation of soap films QUESTION [5 upvotes]: Let $S$ be a soap film bounded by an unknotted wireframe cycle (in $R^3$). Why is it the case that as we deform the wireframe in $R^3$, $S$ deforms continuously? REPLY [14 votes]: You are asking whether a least area surface depends continuously on its boundary curve, or in mathematical terms: whether for every curve $\gamma$ there exists a least area surface $S$ with $\partial S=\gamma$ such that for any neighborhood of $S$ in the space of surfaces there exists a neighborhood of $\gamma$ in the space of curves such that for any curve $\gamma^\prime$ in that neighborhood one will find a least area surface $S^\prime$ with $\partial S^\prime=\gamma^\prime$ in the first neighborhood. This is however not always the case, see Beeson: Non-continuous dependence of surfaces of least area on the boundary curve. Pacific J. Math. 70 (1977), no. 1, 11–17.<|endoftext|> TITLE: Algebras whose subalgebras are finitely generated QUESTION [9 upvotes]: Let $k$ be a commutative ring. Is there a name for those commutative $k$-algebras with the property that every subalgebra is finitely generated? (Equivalently, the partial order of subalgebras is Noetherian.) Can we say something about their structure? Where can I read more about them? I am particularly interested in the cases $k=\mathbb{Z}$ and when $k$ is field. Notice that if $k$ is a field then $k[x,y]$ does not have the property, but $k[x]$ does. One can also check directly that the subrings of $\mathbb{Z} \times \mathbb{Z}$ are finitely generated because they are given by $\mathbb{Z}[(n,0)] = \mathbb{Z} \times_{\mathbb{Z}/n} \mathbb{Z}$, where $n \geq 0$. I wonder if there is a more conceptual reason for this. (Usually an object is called Noetherian if its partial order of subobjects is Noetherian, and this applies for example to (non-abelian) groups and modules. But a ring is usually called Noetherian if its partial order of quotients is Noetherian. This is somewhat confusing. Therefore, although it would be consistent to call a ring Noetherian if its partial order of subrings is Noetherian, this would contradict the usual terminology.) Edit. If $k$ is a field, then a necessary condition is that the algebra is finitely generated of Krull dimension $\leq 1$. This follows directly from Noether normalization and the fact, already mentioned above, that $k[x,y]$ does not have the property. But this condition is not sufficient, as the example $k[x,y]/(x^2)$ shows. Summary of the answers so far. Keith Kearnes suggests the terms "supernoetherian" (if $k$ is Noetherian) and "hereditary finitely generated" (which sounds very good). YCor has reduced the general classification to the case of finitely generated $k$-domains (if $k$ is Noetherian). The classification in this case is still open. If $k$ is a field, is it equivalent to Krull dimension $\leq 1$? Is there a characterization if $k$ is not a field? YCor has also generalized my observation on $\mathbb{Z} \times \mathbb{Z}$: The $k$-algebra $k \times k$ is hereditarily finitely generated if and only if $k$ is Noetherian, because in fact there is an isomorphism of partial orders between ideals of $k$ and subalgebras of $k \times k$ given by $I \mapsto k \cdot (1,1) + I \times \{0\}$. REPLY [3 votes]: I'll assume that $k$ is noetherian. I'll just write $k$-ACC, or ACC if no ambiguity, to mean the ascending chain condition on $k$-subalgebra. (For $k$ arbitrary, $A=k$ is the only $k$-subalgebra of itself so satisfies the property, so can be arbitrarily bad.) Here's an equivalence which then boils down to the case of a domain: A $k$-algebra $A$ ($k$ is noetherian) has ACC iff the following 3 condition hold: (i) $A$ is noetherian (ii) the nilradical $N_A$ of $A$ is a finitely generated $k$-module (iii) $A/P$ has ACC for every (minimal) prime ideal $P$ of $A$. Indeed suppose that $A$ has ACC. (iii) immediately follows. Let $(I_n)$ be an increasing sequence of ideals; so $(I_n\cap k1_A)$ is an ascending sequence of ideals of $k1_A$ and hence is stationary, say for $n\ge n_0$. Also $(k1_A+I_n)$ is an ascending sequence of $k$-subalgebras of $A$. So there exists $n_1$ (say $\ge n_0$) such that for every $n\ge n_1$ and $x\in I_n$, one can write $x=x'+t1_A$ with $x'\in I_{n_1}$ and $t\in k$. Then $x'-x\in I_n\cap k1_A$ and hence $x'-x\in I_{n_0}$. Thus $x\in I_{n_1}$; whence $I_n=I_{n_1}$ for all $n\ge n_1$. This proves (i). To prove (ii), use that $A$ is noetherian to write a nested sequence of submodules $0\le N_1\le \dots N_k=N_A$ with each $N_i$ isomorphic as an $A$-module to $A/P_i$ for some prime $i$. Suppose by contradiction that some $N_i$ is an infinitely generated $k$-module. Since ACC passes to quotients, we can suppose that $i=1$. Since $P_i$ annihilates $N_1$ and contains the nilradical, we see that $xy=0$ for all $x,y\in N_1$. Therefore, for every $k$-submodule $V$ of $N_1$, the $k$-subalgebra generated by $V$ is reduced to $k1_A+V$. So if $(V_n)$ is an increasing sequence of submodules, from ACC we deduce that for large $n$, $k1_A+V_n=k1_A+V_{n+1}$ (in other words, the canonical map $V_n/(k1_A\cap V_n)\to V_{n+1}/(k1_A\cap V_{n+1})$ is an isomorphism). Since $k$ is noetherian, $(k1_A\cap V_n)$, as an ascending sequence of $k$-submodule of $k1_A$, is also stationary, say with union $W$. Hence the above canonical map is the inclusion $V_n/W\to V_{n+1}/W$; since it is an isomorphism it means that $V_{n+1}=V_n$. Conversely suppose that (i),(ii),(iii) hold. It is easy to check (see below) that a finite direct product of $k$-algebras with ACC has ACC. By (i), $A$ has finitely many minimal primes $P_i$, so $A/N_A$ embeds as subalgebra in the finite product $\prod A/P_i$, which has ACC using (iii), hence $A/N_A$ has ACC. Also it is immediate that if an algebra $A$ has an ideal $I$ that is a f.g. $k$-module and $A/I$ has ACC then $A$ has ACC. Then using (ii) $A$ has ACC. Fact: ($k$ noetherian) the $k$-ACC condition passes to finite direct products. Proof: it's enough to do it for a product $A\times B$. Let $(H_n)$ be an ascending sequence of subalgebras of $A\times B$. The projections being stationary, we can suppose that the projections of $H_n$ on both $A$ and $B$ are surjective. It follows that the intersection $H_n\cap (A\times\{0\})$ is an ideal in $A$. Since $A$ is noetherian (as I first checked: this didn't use this finite product claim), this intersection is stationary and we're done.<|endoftext|> TITLE: How should one call and use categories that are not locally small? QUESTION [5 upvotes]: In the paper I am now writing most of the categories are locally small (so, object form a class and $Mor(X,Y)$ is always a set). Yet they are not small, and sometimes I have to mention something like the category of additive functors from $C$ into abelian group. What can I say about the "size" of this functor category (to note that it is not locally small in contrast to other categories used in the paper)? What can one say about the functor category if $C$ is essentially small but not small? I appears that to treat categories of functors "properly" one has to use Grothendieck universes. Yet the papers (and a book) on triangulated categories that I cite in my text (almost) ignore this matter, and I don't want to pay much attention to it either. So I wonder which terminology can I use and which "precautions" should be taken so that my simple arguments concerning functor categories will be mathematically correct. In particular, if the restriction of certain functors to a small subcategory of $C$ gives an equivalence of functor categories then can I say that the original category of functors is locally small? Do you know any text that introduces "the most conventional" terminology for these matters? I would not like to mention universes, and I don't like terms like "small set". REPLY [3 votes]: As indicated by user337830, understanding your questions, hence their answers, depends on the foundational setting you use for mathematics. Beyond a few remarks in MacLane's Category Theory for the Working Mathematician (Chapter I, Section 6) and the initial setting of Kashiwara/Schapira's Categories and sheaves, the standard litterature is rather poor on this matter, possibly reflecting the absence of interest of “working mathematicians” about foundations. Grothendieck's SGA 4 gave slightly more comments in Expose 1 (2 sections on universes, and an Appendix by Bourbaki). More positively, this deficiency may reflect the fact that for “usual mathematicas”, the precise foundational setting is rather innocuous. For example, Kashiwara and Schapira introduce universes, but forget them right away, as does Grothendieck. (In K-S's book, this leads to unprecise expressions, such as “for a sufficiently large universe”, and some change of universes in the middle of a proof I can't understand the meaning of...) The most naive point of view consists of having all categories and functors viewed as formulas in the language of set theory. It is OK for a lot of things such as Grothendieck's Galois equivalence of categories (but not for categories of functors, I would guess). Then come universes or classes. The point of view of universes has the advantage of lying within classical (ZFC) set theory, while introducing classes requires another foundational system, Gödel-Bernays's (GB), say, something mathematicians may be reluctant to do. As MacLane writes: if you have an universe $U$, you can call an element of $U$ a small set, and a subset of $U$ a class (or a large set), so that a model of ZFCU furnishes a model of GB. For more complicated constructions, you'll need higher classes, or a (possibly infinite) sequence of nested universes. For example, categories of functors (categories of presheaves, for example) depend on the chosen universes; the fact that they are (or not) locally small is a theorem. I try never to forget that the fpqc-sheaf associated with a presheaf depends on the chosen universe (an example is given in Bosch/Lütkebohmert/Raynaud's book, Néron models). Two papers sum up this with great clarity, I believe, and explain the merits/demerits of various approaches. One is due to Andreas Blass, The interaction between category theory and set theory, Contemp. Math. 30 (1984). I particularly like the (not so) recent preprint of Mike Shulman's, Set theory for category theory (2008, why is it still unpublished?). In any case, I would advise you (if I ever may) to choose (possibly secretly) one foundational system you're OK with and to fix your terminology in consequence.<|endoftext|> TITLE: An integral cohomology operation related to Steenrod square? QUESTION [6 upvotes]: Let $\beta: H^n(X, \mathbb{Z}_2)\to H^{n+1}(X, \mathbb{Z})$ be the Bockstein homomorphism. Is it possible to define a cohomology operation $f: H^{n+1}(X, \mathbb{Z})\to H^{n+k+1}(X, \mathbb{Z})$ such that \begin{eqnarray*}\beta \text{Sq}^k=f\beta\end{eqnarray*} where $\text{Sq}^k: H^n(X, \mathbb{Z}_2)\to H^{n+k}(X, \mathbb{Z}_2)$ is the $k$-th Steenrod square? Added: It seems that when $k=2$ it is not possible. If we precompose both sides by $\rho_2:H^n(X, \mathbb{Z})\to H^n(X, \mathbb{Z}_2)$, the reduction mod 2 map, then the LHS becomes the third differential of the Atiyah-Hirzebruch spectral sequence which in general does not vanish, but the RHS vanishes because $\beta\rho_2=0$. REPLY [4 votes]: Such an $f$ only exists in two cases: when $k=1$ (when $\beta Sq^1 = 0$) and when $k=0$ (when $\beta Sq^0 = \beta$). Here is a proof, which will take a little work with the Steenrod algebra. Let's suppose that $X$ and $Y$ are either $H\Bbb Z$ or $H\Bbb Z/2$ (not necessarily the same). Then the map $$ [X,Y]_t \to Hom_{A^*}(H^*Y, \Sigma^t H^*X), $$ sending a map to its (graded) effect on cohomology, is an isomorphism for $t \neq 0$. If one of $X$ or $Y$ is $H\Bbb Z/2$, it is an isomorphism for all $t$. Let's prove this. This map is the edge morphism in the Adams spectral sequence $$ Ext_{A^*}^s(H^*Y, \Sigma^t H^*X) \Rightarrow ([X,Y]^\wedge_2)_{t-s} $$ and so it suffices to show that the Adams spectral sequence degenerates. If $Y = H\Bbb Z/2$, then $H^* Y$ is free over $A^*$ and there are no higher Ext-groups. If $Y = H\Bbb Z$, then $H^* Y = A^* / Sq^1$, which has a free resolution $$ \cdots \to A^* \xrightarrow{\cdot Sq^1}A^* \xrightarrow{\cdot Sq^1} A^* \to A^*/Sq^1 \to 0 $$ (using the basis of $A^*$ in terms of admissible monomials). Applying $Hom_{A^*}(-, \Sigma^* H^*(X))$ to this resolution, we get the chain complex $$ 0 \to \Sigma^* H^*(X) \xrightarrow{Sq^1 \cdot} \Sigma^* H^*(X) \xrightarrow{Sq^1 \cdot} \Sigma^* H^*(X) \to \cdots $$ which gives us the Ext-groups (in particular, we could interpret this as the Bockstein spectral sequence from mod-2 cohomology to integral cohomology). If $X = H\Bbb Z/2$, then the basis of admissible monomials shows that this is exact except in degree zero, where it recovers $ker(Sq^1 \cdot -) = Im(Sq^1 \cdot -)$. If $X = H\Bbb Z$, then $H^*(X)$ has a basis consisting of admissible monomials we similarly find that this is exact except in degree zero: $A^* / Sq^1$ is a direct sum of $\Bbb Z/2$ in degree zero with a bunch of free modules over $\Bbb Z/2[Sq^1] / (Sq^1 \cdot Sq^1)$ on the admissible monomials $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$. Therefore, the higher Ext-groups vanish except for those contributing to $[X,Y]_0 = [H\Bbb Z, H\Bbb Z]_0$. The degree-zero part $Hom(A^*/Sq^1, A^*/Sq^1) = ker(Sq^1 \cdot -)$ has a basis consisting of $1$ and admissible monomials of the form $Sq^{2k+1} Sq^{n_1} \cdots Sq^{n_r}$. In particular, we found in the course of the proof that the map $$ [H\Bbb Z, H\Bbb Z]_n \to [H\Bbb Z, H\Bbb Z/2]_n, $$ sending $f$ to $\rho_2 \circ f$, is injective for $n \neq 0$. The image is parametrized by those elements of $A^* / Sq^1$ which are sums of admissible monomials $Sq^{2k+1} Sq^{n_1} \cdots Sq^{n_r}$; this corresponds to the cohomology operation $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2$. Thus: The set $([H\Bbb Z, H\Bbb Z]_n)^\wedge_2$ is $2$-torsion for $n > 0$, and has a basis consisting of the cohomology operations $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2$ over $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$ admissible monomials in the Steenrod algebra with $n_r > 1$. The set $[H\Bbb Z/2, H\Bbb Z]_n$ has a basis of elements of the cohomology operations $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$ over $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$ admissible monomials in the Steenrod algebra. The subset of elements of the form $f \beta$ are thus sums of elements of the form $$ \beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2 \beta= \beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1. $$ The image of right multiplication by $\beta$, therefore, has a basis consisting of sums of elements of the form $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$, where $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$ is an admissible monomial in the Steenrod algebra. In particular, $\beta Sq^n$ is not of this form unless $n=1$ (when it is zero) or $n=0$ (when it is $\beta$).<|endoftext|> TITLE: Monodromy representations and branched covers QUESTION [7 upvotes]: I need to use the following result (that I'm pretty sure is true): Theorem. Let $Y$ be a compact complex manifold and $B \subset Y$ be a connected submanifold of codimension one. Then isomorphism classes of connected analytic covers of degree $n$ $$f \colon X \longrightarrow Y,$$ branched at most over $B$, correspond to group homomorphisms $$\varphi \colon \pi_1(Y - B) \longrightarrow S_n$$ with transitive image, up to conjugacy in $S_n$. When $Y$ is a compact Riemann surface this is very classical and a proof can be found for instance in R. Miranda's book Algebraic curves and Riemann surfaces, Chapter III. In the general case, I was told that this must be contained somewhere in Grauert and Remmert's work. I tried to find it in their celebrated paper Komplexe Räume (Math. Ann. 136, 1958), but it is written in German and it is very long and technical, so it is difficult (at least for me) to extract from there the statement that I want. So my question is Can someone provide a precise reference for the theorem above? REPLY [5 votes]: For a modern treatment of the Grauert Remmert argument see Chapter 4 in the book Several Complex Variables vol 7 by Dethloff and Grauert . For an alternate proof using resolution of singularities see SGA 1 page 255 in the arxiv version . The idea of all these proofs is to restrict to the local case . In the case you are interested in, the local problem is filling a finite covering of a product of a polydisc with a punctured disc .<|endoftext|> TITLE: Why is a Kummer surface simply-connected? QUESTION [10 upvotes]: By a kummer surface, I mean a quotient of two-dimensional complex torus by multiplication by $-1$. It is a K3 surface and known to be simply connected. But it is not clear to me why it is simply-connected from its construction, which is a quotient of a surface with infinite fundamental group. Is there a direct explanation or any references? Thanks. REPLY [4 votes]: I haven't checked details but the following might work. Coverings of the Kummer surface will lift to coverings of the abelian surface (whose fundamental group is clearly $\mathbb{Z}^4$), so the fundamental group of the Kummer surface is abelian. Now it's enough to show that the first homology group is trivial. The latter I expect you will find in textbooks, e.g. as a computation of the Hodge diamond of a K3 surface.<|endoftext|> TITLE: Maximum number of elements in union of subspaces QUESTION [6 upvotes]: Let $V$ be a $m$-dimensional vector space over $\mathbb{F}_q$ and $1<\ell2$. REPLY [5 votes]: Pick naturals $q,m,l,r$. Assume $q$ is a prime power and $2l\leq m$. If $r\leq q^l+1$ then the maximal size of the union of $r$ many $(m-l)$-dimensional sub-vector spaces of $V=\mathbb{F}_q^m$ is $S=rq^{m-l}-(r-1)q^{m-2l}$. If $r\geq q^l+1$ then the size is $q^m$. Note that for $r=q^l+1$ we get $S=q^m$. Since the maximal size of the union is obviously bounded from above by $q^m$ and is non-decreasing as a function of $r$, it is enough then to show that for $r\leq q^l+1$ the maximal size of the union is $S$. Below we do assume $r\leq q^l+1$. Assume having such $r$ subspaces with maximal union size. Since the intersection of each two has dimension bounded from below by $m-2l$, the size of the intersection is bounded from above by $S$ and this is obtained iff they all have a common intersection of dimension $m-2l$. In this case we may pass to the quotient space obtained by moding up this subspace, thus reduce to the special case $m=2l$ which is dealt by the following Lemma: The maximal size of a collection of pairwise trivially intersecting $l$-dimensional subspace of $\mathbb{F}_q^{2l}$ is $q^l+1$. An example of $q^l+1$ many such subspace is obtained by considerng the collection of lines inside a 2-dimensional space over $\mathbb{F}_{q^l}$. On the other hand, if we have $s$ such subspaces and $s>q^l+1$ then the size of their union is $sq^l-(s-1)>q^{2l}$, which is an absurd. Bonus exercise: based on the proof of the lemma above, prove that for general $l$ and $m$, the maximal number of pairwise trivially intersecting $l$-dimensional subspaces of $\mathbb{F}_q^m$ is $\frac{q^{\lfloor \frac{m}{l}\rfloor\cdot l}-1}{q^l-1}$.<|endoftext|> TITLE: How to solve optimization problems on manifolds? QUESTION [6 upvotes]: I encountered some optimization problem on the special Euclidean group SE(3) at work and wonder how to solve it. The current approach of my colleagues was to use a local parametrization of the manifold and then applying a function from a standard optimization toolbox. However I am also aware of the recent developments on optimization problems on manifolds (see e.g. http://www.eeci-institute.eu/GSC2011/Photos-EECI/EECI-GSC-2011-M5/book_AMS.pdf). This way there is no parametrization needed. Is there any other advantage? Have the two approached been compared somewhere? REPLY [6 votes]: To complement Christian Clason's comment: there is usually no need to compute geodesics to optimize over manifolds directly. The usual replacement used for optimization purposes is called a retraction. A retraction is an approximation of the exponential map (which generates geodesics), accurate up to first order. For various manifolds of practical interest, computationally inexpensive retractions are known. For example, for the sphere, retracting the tangent vector $u$ at $x$ is often done as $R_x(u) = \frac{x+u}{\|x+u\|_2}$. For $u$ small, this agrees with the exponential of $u$ at $x$ up to first order. Edit: in addition to Absil et al.'s excellent book, you may also find my introduction to optimization on manifolds helpful: http://www.nicolasboumal.net/book.<|endoftext|> TITLE: Why is Mumford's GIT-quotient so effective? QUESTION [18 upvotes]: According to remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings. As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid. Update: Yes, as Will Sawin said, the question is approximately why one does not iterate it (why there is no need/why nothing better happens)? Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is $$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$ Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$. Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference? REPLY [16 votes]: By my understanding, your question is not "Why is Mumford's construction better than the affine quotient". As you note, Proj is better than Spec of invariants for taking quotients by $\mathbb{G}_m$. Instead, your question is "Why can't we get an even better quotient by going further, involving more characters somehow?" I would say an answer is that any construction involving more characters would not be much different from Mumford's construction, as long as $\chi$ lies in the interior of the cone of all one-dimensional characters of $G$ appearing in $k[X]$. Take a character $\chi$ and consider another character $\chi'$. Suppose there is a function $f$ which transforms according to $\chi'$. Should we incorporate this function into our quotient? No, as it is not an invariant function. On the other hand if we have two functions $f_1,f_2$ that transform according to $\chi'$, we might hope that $f_1/f_2$ descends to a rational function on our quotient. However, as long as there is a natural number $n$ and function $g$ such that $g$ transforms under the character $\chi^n/\chi'$, then the rational function $f_1/f_2$ actually lives on the GIT quotient with character $\chi$, because it can be represented as $(gf_1)/(gf_2)$. So at least birationally Mumford's GIT gives you all the information you could hope to get from the extra characters. For more precise information, one normally studies the spaces you get from varying the character $\chi$ ("variation of GIT") rather than trying to construct a single space incorporating all the information. But this is not very surprising as, gesturing very vaguely, a family of spaces can carry more information than a single space.<|endoftext|> TITLE: 'Strong' density of differential operators in linear operators QUESTION [5 upvotes]: Let $A$ be a commutative, regular, finitely generated $k$-algebra, where $k$ is a field of characteristic 0. Then Grothendieck proved that in this case the algebra of differential operators on $A$ is simply the subalgebra of $End_k(A)$ generated by multiplication by $A$ and $Der_k(A)$, $k$-linear derivations. Denote this algebra by $D(A)$. I have read that $D(A)\subseteq End_k(A)$ is 'strongly dense', i.e. given any finite-dimensional subspace $M\subseteq A$ and any linear operator $L\in End_k(A)$, there exists a differential operator $D\in D(A)$ such that $L|_M=D|_M$. Does anyone know that proof of this fact? My thought process has been to try and work locally first: we know that $Spec(A)$ has a cover by affines $Spec(A_f)$ such that $D(A_f)\cong A_f[\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}]$, where $x_1,\dots,x_n\in A_f$ are a system of parameters at all closed points in $Spec(A_f)$. Here the derivations $\frac{\partial}{\partial x_i}$ are only required to satisfy $x_j\mapsto \delta_{ij}$ (if I understand things correctly). So it's easy to find a differential operator which does anything you like to a subspace generated by monomials in the $x_i$, but I don't see what to do about other elements. REPLY [3 votes]: I think it may be easier if you do not choose the open cover in advance, so that the cover may depend on $M$. Equivalently, let us work with local rings first. Here is a sketch of the argument. I am keeping your notation: $A$ is smooth over a characteristic zero field $k$, $D(A)$ is the algebra of differential operators, and $M\subset A$ is a finite-dimensional subspace. Consider the evaluation map $$ev:D(A)\to Hom_k(M,A).$$ We want to show that $ev$ is surjective. Note that $ev$ is a morphism of $A$-modules, so it is enough to check its surjectivity locally. (Explicitly, this step involves looking at partitions of unity.) That is to say, for every point $x\in Spec(A)$, we consider the local ring $A_x$, an we need to prove surjectivity of the map $$ev_{A_x}:D(A_x)=A_x\otimes_A D(A)\to Hom_k(M,A_x).$$ Moreover, $Hom_k(M,A)$ is a finitely generated $A$-module (in fact, it is free of finite rank), so it suffices to consider closed points $x\in Spec(A)$. Let now $x\in Spec(A)$ be a closed point, and we want to prove surjectivity of $ev_{A_x}$. By Nakayama's Lemma, it suffices to check that the composition $$D(A_x)\to Hom_k(M,A_x)\to Hom_k(M,k_x)$$ is surjective, here $k_x$ is the residue field of $x$. However, for this statement you can replace $A_x$ by its completion, which turns it into a claim about Taylor series. P.S. It may be psychologically easier to extend scalars from $k$ to its algebraic closure at the very beginning... But if you resist the temptation, you get to think about things like $k$-linear differential operators on the ring of Taylor series with coefficients in a finite extension $k_x\supset k$...<|endoftext|> TITLE: rational parameterizations of Lie groups QUESTION [8 upvotes]: What is a good reference for an overview of rational parameterizations of Lie groups (like the Cayley transform for the groups of invariance of non-degenerate bilinear forms)? Is the classification of rational parameterizations known? REPLY [3 votes]: Concerning rational parametrizations of algebraic groups similar to the Cayley transform see this answer, which deals with Cayley maps, i.e., equivariant birational isomorphisms between an algebraic group and its Lie algebra. If you are interested in birational parametrizations which are not necessarily equivariant, look in papers by Chernousov and Platonov, for example: Chernousov, Vladimir I.; Platonov, Vladimir P. The rationality problem for semisimple group varieties. J. Reine Angew. Math. 504 (1998), 1–28.<|endoftext|> TITLE: Gluing Alexandrov spaces along parts of boundary QUESTION [5 upvotes]: I'm familiar with the Petrunin gluing theorem that states that gluing two Alexandrov spaces $M_1,M_2\in Alex(k)$ along their boundaries via an isometry $:\partial M_1\rightarrow \partial M_2$ results in an Alexandrov space of the same lower curvature bound. I'm wondering for what "parts" of the boundary we may restrict such a gluing and still obtain an Alexandrov space. For example, two eighth-spheres of $S^2$ (geodesic triangles with three right angles) can be glued along a single edge, resulting in a quarter-sphere, and all three are $Alex(1)$. I think in this example, the two eighth-spheres could be open and the gluing could be done along the interior of a boundary edge, but I'm interested in what other type of restrictions are possible. That is, under what conditions is it true that if $E_1\subset \partial M_1$ and $E_2\subset \partial M_2$ are isometric, gluing along this isometry produces an $Alex(k)$ space? Of particular interest to me is if $E_1, E_2$ are themselves isometric $Alex(k)$ spaces with non-empty boundary. Does gluing along their interiors produce an $Alex(k)$ space? This question is migrated from SE. REPLY [3 votes]: In your specific question, the answer is no. Take two triangles, as two 2-dimensional Alexandrov surfaces. Suppose they each have a side of a given length. Then these sides can be your $E_1$ and $E_2$ and the glued space is in general a quadrilateral. It is only an Alexandrov space if and only if it is convex. In general, as mentioned by Zimbrón, if you glue along a boundary face, in the sense that it is a codimension one extremal subset, this does work. This covers your spherical example. However, it is not a necessary condition. This is clear from the triangle gluing problem. Nan Li has though a lot about these questions: see http://arxiv.org/pdf/1110.5498v7.pdf You could also check out Woerner's paper in G&T, at http://msp.org/gt/2012/16-4/gt-v16-n4-p12-p.pdf<|endoftext|> TITLE: Box dimension of the set of Pisot numbers? QUESTION [9 upvotes]: A Pisot number is an algebraic integer bigger than $1$ with all of its Galois conjugates having modulus less than $1$. The set of Pisot numbers is known to be countably infinite and is not dense in $(1,\infty)$. What is the box dimension of the set of Pisot numbers? I have been unable to find anything in the literature that would answer this, but I wouldn't be surprised if it follows directly from something that's already known. REPLY [10 votes]: The set $\mathcal P$ of all Pisot numbers is known to be closed (Salem). Its limit points $\mathcal P'$ are also known in $(1,2)$ (Talmoudi). The smallest element of $\mathcal P'$ is the golden ratio $G=\frac{1+\sqrt5}2$, and $\mathcal P\cap (1,G)$ is also known (Dufresnoy-Pisot). In particular, the roots $\eta_n\in(1,2)$ of the polynomials $x^n(x^2-x-1)+1$ are all Pisot numbers, and $\eta_n\uparrow G$ as $n\to\infty$. The question is how fast this convergence is. We have $$ \eta_n^n(\eta_n^2-\eta_n-1)+1=0, $$ and $$ \eta_n^n(G^2-G-1)=0. $$ Subtracting, we get that $\eta_n^{-n}\asymp G-\eta_n$, whence $$ G-\eta_n\asymp G^{-n}. $$ Thus, the convergence is exponential. Now, if we recall the toy example $\Omega=\{2^{-n}\mid n\ge1\}\cup \{0\}$, we have $\dim_B(\Omega)=0$, so my conjecture is $\dim_B(\mathcal P)=0$ as well. To prove this rigorously for $\mathcal P\cap(1,2)$, one probably needs to look at all limit points and all sequences of Pisot numbers converging to them (F. L. Talmoudi, ‘Sur les nombres de S ∩ [1, 2[’, C. R. Acad. Sci. Paris S´er. A–B 287 (1978) A739–A741.) For $\mathcal P'\cap (2,\infty)$, to my best knowledge, not much is known. But I still expect exponential convergence and, consequently, zero box dimension. Update. As I expected, outside $(1,2)$ this is a very tough problem. The main issue is to deal with $\mathcal P''$, i.e., the set of limit points for the limit points. Even at $\theta=2$ (which is its minimum), it is unclear how to show that $\mathcal P\cap (2, 2+\delta)$ has zero box dimension -- simply because not much is known about $\mathcal P'\cap (2,2+\delta)$.<|endoftext|> TITLE: Algebraic groups without torsors QUESTION [12 upvotes]: If $G$ is an algebraic group such that $H^1(S, G) = 0$ for all schemes $S$, must $G$ be the trivial group? My original motivation for the question is the rationale I always give students for studying stacks: when moduli problems involve nontrivial automorphisms, a fine moduli space cannot exist. The reasoning is that if $X$ is an object of a moduli problem $M$ admitting an automorphism group $G$ then one gets an embedding of the classifying stack $\mathrm{B}G$ of $G$-torsors inside of $M$. Two distinct $G$-torsors over a base $S$ give distinct maps to $\mathrm{B}G$ and hence to $M$ that are pointwise identical and constant, hence must be the same in any purported fine moduli space for $M$. This argument requires a nontrivial $G$-torsor to work. It is not difficult to show that a nontrivial $G$-torsor exists for any disconnected group, and there are many familiar examples for connected groups. On the other hand, the analogous question for torsors on $C^\infty$ manifolds under Lie groups has a negative answer, since maps from $X$ to the Lie group $\mathbf R$ correspond to sections over $X$ of $C^\infty(X)$, and $H^1(X, C^\infty(X)) = 0$ because of partitions of unity. REPLY [10 votes]: I am just posting my comment as an answer. If you want to produce a nontrivial torsor, it suffices to produce a nontrivial torsor after base change to the algebraic closure of a residue field of the base scheme. So now you can use the structure theory of algebraic groups. For multiplicative groups, $\mathbb{A}^{n+1}\setminus\{0\} \to \mathbb{P}^n$ is a nontrivial torsor. For additive groups, there is a nontrivial torsor over $\mathbb{A}^2\setminus\{0\}$. For semisimple groups, the $G$-torsor induced over $\mathbb{P}^1$ by a nontrivial torsor for the maximal torus is nontrivial, cf. Grothendieck, et al. Edit. At the request of Jonathan Wise, here are some references. MR0087176 (19,315b) Grothendieck, A. Sur la classification des fibrés holomorphes sur la sphère de Riemann. (French) Amer. J. Math. 79 (1957), 121–138. 53.3X MR0263826 (41 #8425) Harder, Günter Halbeinfache Gruppenschemata über vollständigen Kurven. (German) Invent. Math. 6 (1968), 107–149. 14.50 Jonathan Wise also asks how to deduce nontriviality of torsors from nontriviality for factor torsors. Let $\phi:G\to L$ be a homomorphism of algebraic groups. For every $G$-torsor $E_G$, the quotient of $E_G \times L$ by the "diagonal" action of $G$ is an $L$-torsor $E_L$. If $E_G$ has a section, then so does $E_L$. First, consider the case that $\phi$ is a surjective (thus flat) homomorphism from a solvable group to a multiplicative group over an algebraically closed field. By the structure theory of algebraic groups, there exists a closed subgroup $L'\subset G$ such that $\phi:L'\to L$ is an isogeny. The group of $L$-torsors over $\mathbb{P}^1$ is a free Abelian group. Up to multiplying the torsor by a suitably positive and divisible integer, it is induced from an $L'$-torsor. For the inclusion homomorphism $L'\subset G$, there is an associated $G$-torsor $E_G$ whose $L$-torsor $E_L$ is a multiple of the original $L$-torsor over $\mathbb{P}^1$. In particular, if the original $L$-torsor is nontrivial, then so is the $G$-torsor. Next, consider the case that $\phi:G\to L$ is a surjective (thus flat) homomorphism from $G$ to a reductive group whose kernel is solvable, over an algebraically closed field. Let $T\subset L$ be a maximal torus. Let $\phi^{-1}(T)\subset G$ be the inverse image. By the previous case, after multiplying by a sufficiently positive and divisible integer, every nontrivial $T$-torsor over $\mathbb{P}^1$ lifts to a $\phi^{-1}(T)$-torsor, which then induces a $G$-torsor over $\mathbb{P}^1$. By Grothendieck, if the original $T$-torsor is nontrivial, then so is the $L$-torsor associated to this $G$-torsor. Thus the $G$-torsor is also nontrivial. The only remaining case is when $G$ itself is a unipotent group over an algebraically closed field. A bit more generally, assume that $G$ is any non-reductive group. Let $G\hookrightarrow L$ be any closed embedding of $G$ into a reductive group. Then $L/G$ is not affine by Matsushima-Richardson. If the $G$-torsor $L$ over $L/G$ were split, then the affine variety $L$ would be isomorphic to $G\times (L/G)$, and this is not affine. Edit. I should also say, most of what I know about this I learned from the excellent articles of Johan Martens and Michael Thaddeus. Their proof of the Grothendieck-Harder theorem is very clear and geometric.<|endoftext|> TITLE: The Finslerian version of the Nash embedding theorem QUESTION [10 upvotes]: Is it true to say that every Finslerian manifold can be isometrically embedded in some $M_{n}(\mathbb{R})$ with operator norm? Note that every Riemannian manifold can be embedded in some matrix space isometrically, since the matrix space contains a copy of the standard $\mathbb{R}^{n}$: Hilbert-irreducible Banach space REPLY [12 votes]: Burago and Ivanov have shown that any compact Finsler manifold can be isometrically embedded into a finite-dimensional normed space. They also furnish examples of non-compact Finsler manifolds that can not be isometrically embedded in any finite-dimensional normed space.<|endoftext|> TITLE: Can the integration of integrable sections of a measurable function of two variables ever result in a non-measurable function? QUESTION [10 upvotes]: I spent some time searching MathOverflow for a problem that would resemble the one given below, but it turned out to be a rather futile endeavor. I was led to this problem in my attempts to construct a counterexample refuting a result that had already been published in a peer-reviewed article. Problem. Find measure spaces $ (X,\mathcal{S},\mu) $ and $ (Y,\mathcal{T},\nu) $, at least one of which is not $ \sigma $-finite, and an $ (\mathcal{S} \otimes \mathcal{T}) $-measurable function $ f: X \times Y \to \mathbb{R}_{\geq 0} $ with the following properties: The function $ f(x,\bullet): Y \to \mathbb{R}_{\geq 0} $ belongs to $ {L^{1}}(Y,\mathcal{T},\nu) $ for every $ x \in X $. The function $ \left\{ \begin{matrix} X & \to & \mathbb{R}_{\geq 0} \\ x & \mapsto & \displaystyle \int_{Y} f(x,\bullet) ~ \mathrm{d}{\nu} \end{matrix} \right\} $ is not $ \mathcal{S} $-measurable. Does anyone know if this problem can even be solved? Thank you very much for your time! REPLY [4 votes]: Let $\mathcal{B}$ denote the class of Borel subsets of $[0,1]$ and let $A \subseteq [0, 1]$ be a non Borel set. Let $f$ be the characteristic function of the graph of a bijection from $A$ to $[0, 1]$. Then $f$ is $\mathcal{B} \otimes \mathcal{P}([0, 1])$-measurable (check) and the map $x \mapsto \int f(x, y) d\mu(y)$ is non-zero precisely on $A$ where $\mu$ is counting measure. Edit: I was asked to provide more details so here they are. Suppose $W \subseteq \mathbb{R}^2$ is such that every horizontal section $W^y = \{x : (x, y) \in W\}$ is closed. Then $W \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R})$. To see this, define, for each interval $J$ with rational endpoints, $Y_J = \{y : W^y \cap J = \phi\}$. As each $W^y$ is closed, we have $$ W = \mathbb{R}^2 \Big\backslash \bigcup \{J \times Y_J : J \text{ is an interval with rational endpoints}\} \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R}). $$ It follows that the graph of every partial bijection on $\mathbb{R}$ is in $\mathcal{B} \otimes \mathcal{P}(\mathbb{R})$. Now choose any non Borel set $A \subseteq \mathbb{R}$ and an injection $i: A \to \mathbb{R}$. Define $f: \mathbb{R}^2 \to \mathbb{R}$ to be the characteristic function of the graph of $i$. Let $\mu$ be the counting measure on $\mathbb{R}$. The map $x \mapsto \int f(x, y) d\mu(y)$ is precisely the characteristic function of $A$ and hence is non-Borel.<|endoftext|> TITLE: Finding all automorphisms of $\mathbb{C}(x,y)$ QUESTION [6 upvotes]: The group of $\mathbb{C}$-algebra automorphisms of $\mathbb{C}[x,y]$ is well-known, see, for example, the proof of Dicks or the proof of Mckay and Wang. What can be said about the group of $\mathbb{C}$-algebra automorphisms of $\mathbb{C}(x,y)$? Of course, every $\mathbb{C}$-algebra automorphism of $\mathbb{C}[x,y]$ yields a $\mathbb{C}$-algebra automorphism of $\mathbb{C}(x,y)$, but there are more $\mathbb{C}$-algebra automorphisms of $\mathbb{C}(x,y)$, for example, $x \mapsto x^{-1}, y \mapsto y$. Can one find all of them and characterize that group? EDIT: After letting me know that I am looking for the Cremona group, I wish to quote from wikipedia: "In two dimensions, Max Noether and Castelnuovo showed that the complex Cremona group is generated by the standard quadratic transformation, along with PGL(3, k), though there was some controversy about whether their proofs were correct, and Gizatullin (1983) gave a complete set of relations for these generators". What was the problem in their proof (if at all), and is there another published proof? Thank you very much! REPLY [12 votes]: The proof of Noether that the automorphisms of $\mathbb{C}(x,y)$ are generated by $\mathrm{PGL}(3,\mathbb{C})$ and the standard quadratic transformation had troubles because of infinitely near points. The proof of Castelnuovo (1907 is completely valid and is the first one which is accurate. This is why the theorem is Called "Noether-Castelnuovo theorem". You can find a lot of proofs of this result. See for instance the book "Geometry of the Plane Cremona Maps" of Maria Alberich Carramiñana, which gives a detailed proof and explains the history of thre problem. As YCor said, knowing the generators does not directly give a good understanding of the group. For instance, it is useless here to find conjugacy classes of elements of finite order (but the classification is known using birational geometry). There are many other results on the group that you can find in the literature.<|endoftext|> TITLE: Homologous curves and maps to surfaces QUESTION [5 upvotes]: Say I have two smooth one-cycles $Z, Z'$ on a 3-manifold $M$, i.e. just formal linear combinations of (let us say disjoint) smooth circles. Assume they're homologous. Is there a map to a surface $\pi: M \to S$ and homologous zero-cycles $P, P'$ on the surface such that $\pi$ is a submersion over $P, P'$ and one has $\pi^{-1}(P) = Z$ and $\pi^{-1}(P') = Z'$? REPLY [5 votes]: I suspect the answer to this question is no in general. Suppose $Z$ has two components $Z_1 \cup Z_2$ with different multiplicities (as a formal linear combination of knots) and $Z'$ has one component. Then $P$ must have two points $P=\{p_1,p_2\}$ (ignoring multiplicity), and $P'$ one $P'=\{p'\}$. Then $\pi_1(M-(Z\cup Z'))$ surjects a free group of rank at least two (the meridians of $Z_i$ go to loops about $p_i$, which must generate a subgroup of $\pi_1(S-(P\cup P'))$ of rank at least two). The corank of a group is the largest rank of a free group that it surjects. It is known that there are 3-manifold groups with arbitrarily large $b_1$ and corank 1. What is needed to get a counterexample to your question is a manifold with 3 torus boundary components which has fundamental group of corank 1, and satisfying an appropriate homological condition. I feel like a construction of the sort that Shelly Harvey carries out ought to have the right properties. Addendum: I looked at a paper of Leininger-Reid, which gives an example of a compact 3-manifold $M$ with hyperbolic interior, 4 boundary components, and cut number 1. From this, one can get an example which answers your question in the negative. By half-lives, half-dies, there is an oriented surface $\Sigma\subset M$ which is homologically non-trivial in $H_2(M, \partial M)$, and intersects each boundary component in consistently oriented parallel non-trivial curves (in particular, non-empty). Now, perform Dehn fillings on each boundary component so that the intersection number with $\partial \Sigma$ are different to get a manifold $N$. Partition the cores of the Dehn fillings into $Z$ and $Z'$, with multiplicity determined by the intersection number of $\partial \Sigma$ with the meridian of each Dehn filling, and orienting the components of $Z'$ opposite to the induced orientation of $\partial \Sigma$. This gives the desired pair of cycles which are homologous and which cannot be induced by pullback from 0-cycles on a surface because the link complement has cut number 1.<|endoftext|> TITLE: why this procedure grows asymptotically to $n^2/\pi$ QUESTION [5 upvotes]: Consider a natural number, say $n$. Find the first number which is greater than or equal to $n$ and is a multiple of $n-1$. Again find a number which is greater than or equal to this number and is a multiple of $n-2$. Do this iteratively until a multiple of $2$ occurs and call the last number $x$. (More formally: define numbers $a_1, a_2, \ldots, a_n$ where $a_1 = n$ and $a_{k+1}$ is the least number greater than or equal to $a_k$ that is a multiple of $n-k$. Put $x = a_n$.) .... I want to prove that $x$ is asymptotic to $n^2/\pi$. ..... It's easy to see that $x$ is greater than $n^2/4$ because the first half of procedure can be determined and it achieves $n^2/4$ but the proof must be much harder... Can somebody help me? Thanks... REPLY [10 votes]: This is proved by Kevin Brown. The numbers are tabulated, and many links and references given, at the Online Encyclopedia of Integer Sequences.<|endoftext|> TITLE: Vanishing of characteristic numbers vs vanishing of characteristic classes QUESTION [23 upvotes]: A famous result by Thom states that Oriented Bordism classes are determined by characteristic numbers; specifically, two closed manifolds are orientedly bordant if and only if they have the same Stiefel-Whitney and Pontryagin numbers (I'll just talk about Stiefel-Whitney for brevity). An immediate consequence is that if $M$ is a closed manifold which has a non-vanishing Stiefel-Whitney number involving $w_k$ for some $k$, then $w_k(N)\neq 0$ for any $N$ which is bordant to $M$; in other words this non-vanishing characteristic number provides a "bordism reason" for why a characteristic class should be non-zero. My question concerns the converse. Given an $M$, suppose that for some $k$ every Steifel-Whitney number involving $w_k$ vanishes, so that there is "no bordism reason" for the class to be non-vanishing: is it then possible to find an $N$ which is bordant to $M$ and has $w_k(N)=0$? If so, is it possible to simultaneously eliminate all classes which have no bordism reason to be non-zero? This seems like something which maybe shouldn't be expected since characteristic classes often provide obstructions to doing surgery, and two manifolds are orientedly bordant exactly when they differ by a finite sequence of surgeries. On the other hand the manifold $\mathbb{CP}^n\#\overline{\mathbb{CP}^n}$ is null-bordant even though it has many non-vanishing characteristic classes for $n>1$ (half the time it isn't even spin). REPLY [8 votes]: This is not a complete answer, but rather two specific instances where the answer is yes. If every Stiefel-Whitney number of $M$ involving $w_1$ vanishes, then $M$ is cobordant to an orientable manifold $N$. See Proposition 4 of Wall, C. T. C., Determination of the cobordism ring, Ann. Math. (2) 72, 292-311 (1960). ZBL0097.38801. If every Stiefel-Whitney number of $M$ involving $w_1$ and $w_2$ vanishes, then $M$ is cobordant to a spin manifold $N$. See Corollary 2.4 of Anderson, D. W.; Brown, Edgar H. jun.; Peterson, Franklin P., The structure of the Spin cobordism ring, Ann. Math. (2) 86, 271-298 (1967). ZBL0156.21605.<|endoftext|> TITLE: Reference request: tangent space to moduli space of coherent sheaves is $\operatorname{Ext}^1(E, E)$ QUESTION [7 upvotes]: Is there a standard reference for the fact that, in an appropriate algebraic-geometrical context, the tangent space at the point $[E]$ to the moduli space $\mathcal M$ is something like $\operatorname{Ext}^1(E, E)$? I am primary interested in two situations: $\mathcal M$ is the moduli space of semi-stable coherent torsion-free sheaves on $\mathbb CP^2$ with given Hilbert polynomial, or $\mathcal M$ is the moduli space of semi-stable finite-dimensional complex representations of finite acyclic quiver with given dimension vector, and in both situation, if the coefficients of the Hilbert polynomial/dimension vector are coprime, there is a smooth projective variety that is a fine moduli space $\mathcal M$. Unfortunately, I do not know any deformation theory, so a study book on it that also proves something like $T_{[E]} \mathcal M=\operatorname{Ext}^1(E, E)$ would be welcome. REPLY [5 votes]: Theorem 2.6 (page 9) from Hartshorne Lectures on Deformation Theory (it seems that Hartshorne uses the same notation both for an affine scheme $D$ and its function algebra): Let $X$ be a scheme over $k$, and let $\mathcal F$ be a coherent sheaf on $X$. We define a deformation of $\mathcal F$ over $D=\operatorname{Spec} (k[t]/t^2)$ to be a coherent sheaf $\mathcal F_0$ on $X_0 = X \times D$, flat over $D$, together with a homomorphism $\mathcal F_0 \to \mathcal F$ such that the induced map $\mathcal F_0 \otimes_{k[t]/t^2} k \to \mathcal F$ is an isomorphism. Then the deformations of $\mathcal F$ over $D$ are in natural one-to-one correspondence with the elements of $Ext^1 (\mathcal F, \mathcal F)$, with the zero-element corresponding to the trivial deformation. Now by the universal property of the moduli space $\mathcal M$ of coherent torsion-free sheaves on $\mathbb{CP}^2$ its tangent space at $[\mathcal F]$ consists of the deformations of $\mathcal F$ over $D$. Moreover, I believe that the proof should work as well over "a non-commutative affine scheme" whose function algebra is the path algebra of a finite acyclic quiver.<|endoftext|> TITLE: Can one axiomatize projective lines using the cross-ratio? QUESTION [7 upvotes]: I known axiomatizations of projective spaces of dimension > 2 and also of projective planes (either those obeying the axiom of Pappus, which come from fields, or those obeying the axiom of Desargues, which come from division rings, or the fully general ones, which come from planar ternary rings). How about projective lines? One might try to axiomatize at least those coming from fields using the cross-ratio. If $k$ is a field the cross-ratio gives at least a partially-defined 4-ary function $$ kP^1 \times kP^1 \times kP^1 \times kP^1 \to kP^1 $$ where $kP^1$ is the corresponding projective line. Is the cross-ratio well-defined even when several of the four arguments coincide? Does it give a regular map of projective varieties? Does it obey enough equational laws that we can use these as axioms for an interesting concept of projective line? Some related ideas are discussed here: Mike Shulman, Synthetic projective lines, Math Overflow. Jacob Lurie, Action of PGL(2) on projective space, Math Overflow. The latter turned up this nice result: as is well known, $\mathrm{PGL}(2,k)$ acts in a sharply 3-transitive way on $kP^1$, but more generally, any sharply 3-transitive group arises from a 'KT-field', which is a kind of generalized field. I don't know how KT-fields are related to planar ternary rings. However, all this is---at least superficially---a bit different than trying to axiomatize the cross-ratio. Perhaps it's just another viewpoint on the same thing. REPLY [3 votes]: Is the cross-ratio well-defined even when several of the four arguments coincide? No, it is not. Modeled as homogeneous coordinates over $k^2$ you get the null vector in cases where three or more points coincide, but that's not an element of your line any more. If I understand you correctly, it would be sufficient to have a set of axioms for the projective line from which you can derive all the required axioms of the underlying field, thus demonstrating that there is indeed a field underlying the line. Which means you need to be able to perform arithmetic. At least if the characteristic of the underlying field is different from two, I'd probably approach this using harmonic throws: four points $a,b,c,d$ form a harmonic throw iff $(a,b;c,d)=-1$, so if you have cross ratio you also have harmonic throws. So how can harmonic throws encode arithmetic operations? $(\infty,0;a,-a)=-1$ so you have the additive inverse (except if $a=0$ or $a=\infty$, in which case the cross ratio is undefined as three points coincide). $(\infty,a;0,2a)=-1$ so you can multiply or divide by two (again excepting $a=0$ and $a=\infty$). $(a,b;\infty,\frac{a+b}2)=-1$ so you can compute the average, and by doubling that the sum of two values. You now have addition. The special cases where $a$ and/or $b$ is zero or infinite can be handled explicitely and included here as well. $(-1,1;a,\frac1a)=-1$ so you have the multiplicative inverse (including division by zero resulting in infinity, but excluding $a=\pm 1$). $(-a,a;1,a^2)=-1$ so you can square values (unless $a=\pm1$ or $a=0$). Using $ab = \frac{(a+b)^2-a^2-b^2}2$ you can combine the above to obtain multiplication. The cases where $a$ or $b$ is zero, $\pm1$ or infinity can again be handled explicitely. One could start an axiomatization with “there exists at least three distinct points on the line. Picking three and labeling them as $0,1,\infty$ …” and then use the above to define addition and multiplication, then include the axioms of a field expressed this way. The large number of special cases mentioned above may make this a bit awkward, as does the large number of intermediate steps required to e.g. represent multiplication. So I guess the resulting set of axioms could be trimmed down a bit, but it would be a starting point. I'm currently unsure what to do about characteristic two. You don't have a $-1$ distinct from $1$ there, and you'd have some more undefined cross ratios. I guess there should be a set of axioms which covers characteristic two as well, but I don't know how to get there just now.<|endoftext|> TITLE: Resolutions by free Differential Graded Algebras QUESTION [5 upvotes]: I am struggling to find references for explicit computations of things like symmetric algebras and resolutions in the dg context. (any pointers in this direction would be highly appreciated!) I have the following question. Let k be a field of char zero and let A = k[x,y], the free commutative algebra on two (degree zero) generators. What is an explicit resolution/cofibrant replacement of A in the category of differential graded algebras over k? (I really mean the category of dgas not cdgas) If the standard resolution is huge, is there a way to cut it down to get something smaller and more tractable? REPLY [6 votes]: Let $A=k[x_1,\dots,x_m]$ be the algebra of commutative polynomials in $m$ variables over a field (or commutative ring) $k$, of arbitrary characteristic. Denote by $B=\bigwedge(x_1^*,\dots,x_m^*)$ the exterior algebra in $m$ variables $x^*_1,\dots,x^*_m$ dual to $x_1,\dots,x_m$. Let $C=B^*$ be the dual vector space to $B$; it is naturally a coassociative coalgebra over $k$. In other words, $C$ is the exterior coalgebra cogenerated by the variables/elements $x_1,\dots,x_m$. The algebra $B$ is naturally augmented with the augmentation morphism $b:B\to k$ taking the generators $x_1^*,\dots,x_m^*$ to zero. Let $B_+\subset B$ denote the augmentation ideal, i.e., the kernel of $b$. Passing to the dual coalgebras, we have the coaugmentation morphism $c:k\to C$ with the cokernel $C_+=B_+^*$. The vector space $B_+$ is naturally an associative algebra without unit, and $C_+$ is a coassociative coalgebra without counit. Assuming that $A$ is ungraded, or in other words, when viewed as a graded algebra it sits entirely in degree $0$, the algebra $B$ should be graded so that its generators $x_j^*$ sit in the cohomological degree $1$. Then the cogenerators $x_j$ of the coalgebra $C$ sit in the cohomological degree $-1$. Now we are doing the cobar construction, so the cohomological grading on $C_+$ will be shifted forward by $1$, making the elements $x_j\in C_+$ sit in the cohomological degree $0$. The cobar construction $Cob(C)$ is the free associative algebra generated by the cohomologically graded vector space $C_+[-1]$. The differential $d$ on $Cob(C)$ is induced by the comultiplication in $C$ (with some sign rule). This should be the smallest possible choice of a cofibrant replacement of the algebra $A$ in the model category of associative DG-algebras over $k$. In the case of $m=2$, the DG-algebra $Cob(C)$ is the free associative algebra with two generators $x_1$ and $x_2$ in cohomological degree $0$ and one generator $x_{12}$ in cohomological degree $-1$. The differential is provided by the rule $d(x_{12})=x_1x_2-x_2x_1$. Generally for an integer $m\ge1$, the DG-algebra $Cob(C)$ is a free associative algebra with $2^m-1$ generators.<|endoftext|> TITLE: How to compute the formal group law of a Shimura variety (using its invariant differentials)? QUESTION [10 upvotes]: I have a 3 dimensional abelian variety whose formal group law breaks into a formal summand where one of the pieces is one-dimensional. I am desperately wondering how to compute the $p$-series of this one-dimensional formal group law without finding an explicit model of the abelian variety. Here is the set up of the variety of interest: Let $K= \mathbb{Q}(\zeta_7)$. Then $[K:\mathbb{Q}]= 6$. Let $A$ be the three dimensional abelian variety $\mathbb{C}^3/O_K$ over $\mathbb{C}$ where $O_K$ is embedded in $\mathbb{C}^3$ by the following three homomorphisms: $h_1, h_2, h_3: K \to \mathbb{C}$. $h_a$ is the homomorphism $K \to \mathbb{C}$ which sends $\zeta_7$ to $\zeta_7^a$. Then $A$ has complex multiplication by $O_K=\mathbb{Z}[\zeta_7]$. Let $p$ be a prime number such that $p \mod 7 = 2 \mod 7$. Then by the theory of complex multiplication, for some number field $F$, $A$ is defined over $F$ and of good reduction at a place $v$ of $F$ lying over $p$. At $v$, the formal group associated to $A$ at $v$ is the direct sum of a 1-d formal group of height $3$ and a 2-d formal group of height $2$. I think this is due to the Shimura-Taniyama formula ("Introduction to Shimura varieties" by J.S. Milne), which tells us that the slope of the Frobenius is 1/3 or 2/3. Now we know that this 1-d formal group law of height 3 exists, but what is its $p$-series? Here is my question: How do I compute the $p$-series of the 1-d formal group law associated to this higher dimensional abelian variety $\mathbb{C} \times \mathbb{C} \times \mathbb{C} / \mathbb{Z}[\zeta_7]$? I'm curious if this can be done without finding an explicit model of the variety. I have a vague idea that this can be done using its invariant differentials, but I don't know how to find the invariant differentials in the higher dimensional case. REPLY [10 votes]: I don't know how to do this purely with invariant differentials. One of the main reasons is that, in order to solve this problem, you need to construct an "algebraic" lift of this variety from the complex numbers to a much smaller number field. This lift isn't necessarily unique and different choices will give you different formal group laws (though they'll become isomorphic over extension fields). What I'll describe, then, is a method for determining -- as best I know how -- an algebraic lift of $A$ and the $p$-series in terms of a coordinary on the formal group law. Let's pick a finite field $\Bbb F_q$ where $q = p^k$ and discuss things in terms of the Tate-Honda classification of abelian varieties over $\Bbb F_q$. Your description in terms of the Shimura-Taniyama formula means that our lift of $A$ is going to give us an abelian variety over $\Bbb F_q$ which is $3$-dimensional and whose Newton polygon breaks up into a component of slope $1/3$ and a component of slope $2/3$. This forces $A$ to be simple because the $p$-divisible group can't break up into simpler pieces that still satisfy self-duality. The Frobenius endomorphism $\pi$ of $A$ is an element of the ring of integers ${\cal O}_K$, and it's a Weil $q$-integer: it has absolute value $q^{1/2}$ in any complex embedding. Moreover, the Shimura-Taniyama formula tells you exactly which primes of ${\cal O}_K$ divide $\pi$ and how much. The only indeterminacy left is possibly multiplying $\pi$ by a unit $u$ which has absolute value $1$ in any complex embedding, which forces $u$ to be a root of unity (which we've got a goodly stock of here). Replacing $\pi$ with $u \pi$ replaces your abelian variety by a Galois twist: extending fields from $\Bbb F_q$ to $\Bbb F_{q^k}$ replaces $\pi$ by $\pi^k$, and so this root-of-unity indeterminacy goes away after a sufficient extension of finite fields. We should also note for a moment that while, in theory, the (Galois conjugacy class of) Frobenius $\pi$ only determines $A$ up to isogeny, the constraint that $A$ should have an action of ${\cal O}_K$ actually means that $\pi$ determines $A$ up to isomorphism. This is because any isogenous abelian variety that still has ${\cal O}_K$-multiplication is determined by an ${\cal O}_K$-stable lattice in $\Bbb Q \otimes {\cal O}_K$ -- i.e., a fractional ideal -- and because $K$ has class number one, any such fractional ideal is principal. Multiplication by a generator of the ideal gives an isomorphism between the isogenous variety and $A$. This means that, up to a Galois twist, we can determine $A$ completely over a finite field just from this slope data. We can't get any farther without making some choices, so let's make some. I'll say our prime is $p=2$ for the purposes of illustration. Fix $\zeta \in K$ to be our chosen seventh root of unity, and let $w = \zeta + \zeta^2 + \zeta^4$. Then $w^2 + w + 2 = 0$ so $w = \tfrac{-1\pm \sqrt{-7}}{2}$. The element $w$ has absolute value $2^{1/2}$ in any complex embedding, and $w \bar w = 2$; in $K$, this reflects the splitting of prime ideals $(2) = (w) (\bar w)$. Let $\pi = 2 \bar w$. Then $\pi$ has minimal polynomial $t^2 + 2t + 8$ and has absolute value $8^{1/2}$ in any complex embedding. It's also divisible by $w$ once and by $\bar w$ twice. This means that $\pi$ is a Weil $8$-integer and an abelian variety associated to $\pi$ over $\Bbb F_8$ has an endomorphism ring $End(B) \otimes \Bbb Q$ a central division algebra over $\Bbb Q(\sqrt{-7})$ with invariant $1/3$ at $w$ and $2/3$ at $\bar w$, with trivial invariant at all other places. The field $K$ embeds into this one, so $B$ has complex multiplication by $K$. By replacing $B$ with an isogenous abelian variety $A$, we can get an abelian variety $A$ over $\Bbb F_8$ with an action of ${\cal O}_K$ whose Frobenius (the $8$'th power Frobenius, since we're working over $\Bbb F_8$) satisfies the minimal polynomial $F^2 + 2F + 8$. This abelian variety, having the correct slope data, will be an algebraic lift of the abelian variety you constructed over the complex numbers because of the argument we sketched above. Now let's look at this 1-dimensional summand of the formal group law, which corresponds to a $1$-dimensional summand of the $2$-divisible group. If we look at the $2$-divisible group, the Frobenius acts on it too, and it still satisfies the same minimal polynomial -- but now over $\Bbb Z_2$ instead of integrally. The polynomial $t^2 + 2t + 8$ has two solutions over $\Bbb Z_2$ (which we might call $-1 \pm \sqrt{-7}$): one divisible by $2$ and the other divisible by $4$. Let's call the first root $\alpha$ and the other $8/\alpha$. Our minimal polynomial factors now as $F^2 + 2F + 8 = (F - \alpha)(F - 8/\alpha)$, and the splitting of the $2$-divisible group into a piece of slope $1/3$ and one of slope $2/3$ corresponds to the factors where $F$ acts by $\alpha$ and by $8/\alpha$ respectively. This was a bit of a slog to get to this point, but let's reflect on what we've found: we've found our $1$-dimensional piece of the $2$-divisible group, and we've found that the Frobenius acts on it as multiplication by this element $\alpha = -1 - \sqrt{-7}$. Or, equivalently, we get an identity $$ \tfrac{2}{-1-\sqrt{-7}} F = 2 $$ as endomorphisms of this $1$-dimensional formal group law $G$. But if we write that in a power series expansion, we get $$ \left[\tfrac{2}{-1-\sqrt{-7}}\right]_G(x^8) = [2]_G(x) $$ If we want to push this a little farther, we have to start doing a little more gruntwork. Namely, we've got to write this element $\tfrac{2}{-1-\sqrt{-7}}$ in binary expansion as a $2$-adic integer: $$ \tfrac{2}{-1-\sqrt{-7}} = a_0 + 2 a_1 + 4 a_2 + \dots $$ where $a_i \in \{0,1\}$. Then our identity for the Frobenius becomes $$ \sum_G a_i [2]_G^{(i)}(x^8) = [2]_G(x). $$ And then we can start solving for the 2-series iteratively. It starts with $x^8$. Let's write $f(x)$ for $[2]_G(x)$. I believe the $a_i$ starts with $a_0 = 1, a_1 = 0, a_2 = 1, a_3 = 1, a_4 = 0$, etc. So we find saying $$ f(x) = x^8 +_G f(f(x^8)) +_G f(f(f(x^8))) + \dots $$ and so we can start recursively solving to find $$ [2]_G(x) = x^8 + x^{8^3} + \dots $$ I would imagine that there are ways to be a little more clever about this by finding a series expansion for $2$ in terms of $F$ instead of $F$ in terms of $2$. If we can figure out how to write $$ 2 = \sum b_i \alpha^i \equiv \sum b_i F^{(i)} $$ for $b_i \in \{0,1\}$, this becomes $$ [2]_G(x) = \sum_G b_i x^{8^i}, $$ which tells you directly that the Araki generators satisfy $b_i = v_{3i}$ (and the $v_j$ are zero when $j$ is not a multiple of $3$). While there are a lot of steps in here, the endgame takeaway is not that bad: find an element in ${\cal O}_K$ with appropriate prime divisibility, split it $p$-adically, and then use a $p$-adic power expansion to determine how Frobenius is acting on the appropriate $1$-dimensional summand. Choosing $\pi$ in $\Bbb Q(\sqrt{-7})$ will have a tendency to make your life a little easier. Hope this helps.<|endoftext|> TITLE: Composition of Derangements QUESTION [15 upvotes]: What is the probability that composing two random derangements yields another derangement? A slightly more specific question: suppose we start with a given derangement and compose it with all derangements. What is the distribution of cycle structures? Addendum: perhaps it would be worth explaining why this question popped up in the first place. I am interested (purely out of curiosity) in the expected number of "steps" it takes to reach the identity permutation where each step is composition with a random derangement. See here for a related question (the only different is starting with a random permutation). If you like to have "context": this is like asking how many steps it takes for $n$ penguins/koalas/beings with terrible memory to find their room, which they will recognize upon being inside, but which is otherwise indistinguishable. That's why I originally began with an arbitrary permutation, in any case. REPLY [21 votes]: An array of $k$ permutations of $n$ letters such that each pair are derangements of each other is a $k\times n$ Latin rectangle. If $L(k,n)$ is the number of $k\times n$ Latin rectangles, then the probability you ask for is $n!\, L(3,n)/L(2,n)^2$. The value of $L(2,n)$ is $n!$ times the number of derangements. The value of $L(3,n)$ doesn't have closed form but there are recurrences and summations; see here for a survey. The asymptotic value of $L(k,n)$ is $e^{-\binom k2} (n!)^k$ for fixed $k$, from which you can see that the asymptotic value of the probability is $e^{-1}$. I don't know the answer to your second question, though I rather suspect that the average cycle structure of the product of two random derangements is quite similar to the cycle structure of an unrestricted random permutation.<|endoftext|> TITLE: A Jacobian with a good reduction, which is simple : how is the reduction of the curve? QUESTION [6 upvotes]: Let $C$ be a smooth genus $g>1$ curve defined over a number field (say over $\mathbb Q$). Let $J(C)$ be its jacobian. Suppose that $J(C)$ has good reduction $J_p$ at a prime $p$ and moreover $J_p$ is absolutely simple (ie simple over any field extensions). Is it true that $C$ has good reduction $C_p$ at $p$ (maybe after some finite extension) ? So far in the (few) examples I found in the literature of a curve that has bad reduction and for which the jacobian has good reduction, that reduction is isogeneous to a (non-trivial) product. A question related to that problem is : what is the link between $C_p$ and $J_p$ ? Is there still a non-constant map $C_p \to J_p$ ? Does the image of that map generates $J_p$ ? There is a criteria of Oda for good reduction of curves mentioned here: A curve with bad reduction for which the jacobian has good reduction but I do not see how to apply it to the hypothesis I make on the curve $C$... REPLY [9 votes]: Yes, this is true. One can argue as follows: Since $J(C)$ has good reduction, your other hypothsism implies that the special fibre of its Neron model at $p$ is an absolutely simple abelian variety. By Grothendieck's theorem, $C$ has semi-stable reduction at $p$, so it has a regular model whose fibre at $p$ is a semi-stable curve. By the last line of Theorem 4, p. 267, of the book "Neron Models" by Bosch, Raynaud and Luetkebohmert, the identity component of the Neron model of $J(C)$ is $\mathrm{Pic}^0$ of the regular semi-stable model of $C$. (By a base change one can assume that all components of the special fibre are geometrically irreducible.) By Example 8 on p. 246 of the same book, which gives the structure of $\mathrm{Pic}^0$ of a semi-stable curve over a field, and your assumptions on the reduction of $J(C)$, it follows that $C_p$, the special fibre of a regular semi-stable model of $C$, must have a unique smooth component of genus $g(C)$ and possibly some other smooth rational components. Since the arithemetic genus is constant in flat families it follows that any extra rational component must be a "tail" and can be contracted. We conclude that the special fibre of the stable model of $C$ is a smooth curve of the same genus, i.e., $C$ has good reduction at $p$. (Note that no base change is required to get good reduction.)<|endoftext|> TITLE: Is the set of prime pairs such that $gcd(p−1,q−1)=2$ of positive density? QUESTION [9 upvotes]: Is the set of prime pairs such that $gcd(p−1,q−1)=2$ of positive density? For example, for $p,q≤10^4$ the answer is approximately $1/2$. I was wondering if it were possible to use sieve methods and results such as the Siegel-Walfisz Theorem to give a good approximation of prime pairs of this form. The motivation for the question is for understanding the order of elements in the group $(\mathbb{Z}/pq\mathbb{Z})^∗≃(\mathbb{Z}/p\mathbb{Z})^∗×(\mathbb{Z}/q\mathbb{Z})^*$. REPLY [22 votes]: The answer is $$ \frac{3}{4} \prod_{p>2} \Big(1 -\frac{1}{(p-1)^2} \Big) = 0.4951\ldots . $$ The product above is also known as the twin prime constant. This follows easily from the prime number theorem in arithmetic progressions. Restricting to odd primes $p$ and $q$ below $N$ we want to count $(p-1,q-1)=2$ which by Mobius inversion can be expressed as $$ \sum_{\substack{d| (p-1)/2 \\ d| (q-1)/2 } } \mu(d). $$ Thus we want $$ \sum_{3 \le p, q\le N } \sum_{\substack{d| (p-1)/2 \\ d| (q-1)/2 } } \mu(d) = \sum_{d \le N} \mu(d) \sum_{\substack{p, q\le N \\ p\equiv q\equiv 1 \mod{2d}}} 1 . $$ For $d\le (\log N)^3$, use the prime number theorem in APs (Siegel-Walfisz) to see that these terms are $$ \sim \sum_{d\le (\log N)^3} \frac{\mu(d)}{\phi(2d)^2} \frac{N^2}{(\log N)^2}. $$ For $d> (\log N)^3$ estimate the sums over $p$ and $q$ trivially by $(N/d)^2$. So these terms contribute $$ \ll \sum_{d> (\log N)^3} \frac{N^2}{d^2} \ll \frac{N^2}{(\log N)^3}, $$ which is negligible. So the required density is $$ \sum_{d\le (\log N)^3} \frac{\mu(d)}{\phi(2d)^2}, $$ which tends to the constant given above as $N\to \infty$.<|endoftext|> TITLE: Existence of incomplete Riemannian metrics QUESTION [6 upvotes]: During this question a manifold $M$ is meant to be a smooth connected (second countable, finite dimensional) manifold. A Riemannian metric on a manifold $M$ is called complete if every geodesic is defined for all times. It is well-known that this is equivalent to completeness of the metric space $(M,d)$ where $d$ is the geodesic distance. I want to show the following equivalence: A manifold $M$ is compact if and only if every Riemannian metric on $M$ is complete. (keep in mind that $M$ is connected) It is easy to show the implication from left to right, but I have no clue for the reverse implication. I tried the following argument: Every smooth manifold $M$ can be embedded as a bounded submanifold of $\mathbf R^n$. Assume that $M$ is not compact, then $M$ is not closed in $\mathbf R^n$. So, as a metric space $M$ is not complete when given the metric of the surrounding space $\mathbf R^n$. Then I realized that this does not imply that $M$ with the induced Riemannian metric is not complete since the geodesic distance on $M$ may be totally different from the induced distance function. So my question is: If $M$ is non-compact, does there exist an incomplete Riemannian metric on $M$? REPLY [10 votes]: If $M$ is non-compact, then there exists a sequence of points with no convergent subsequence. Fix a curve passing through all of the points. It suffices to construct a Riemannian metric such that this curve has finite length.<|endoftext|> TITLE: Twisted modular forms of half-integral weight QUESTION [6 upvotes]: I am looking for references (or explainations) about the twist of modular forms of half-integral weight. I try to mimic the proof of the "integral weight case" to prove that the twist of $$ \theta(\tau)=\sum_{n\in \textbf{Z}}{q^{n^2}} \quad (\tau \in \mathcal{H}) $$ by a Dirichlet character $\chi$ of odd modulus, namely $$ \theta_{\chi}(\tau)=\sum_{n\in \textbf{Z}}{\chi(n^2)q^{n^2}} \quad (\tau \in \mathcal{H}), $$ is still a modular form. But this method highly depends on the form of the automorphic factor of the $\theta$ function. 1) Do we know if the twist of a modular form of half-integral weight (defined by the use of a four-sheeted cover of $\operatorname{Gl}_2^{+}(\textbf{Q})$) is still a modular form ? 2) If $\eta$ denotes the usual Dedekind function, is $\eta_{\chi}$ a modular form ? Many thanks ! REPLY [7 votes]: In general, the statement is something like the following. (The following is Proposition 3.12 from Ken Ono's book "The Web of Modularity") Suppose that $g(z) = \sum c(n) q^{n}$ is a half-integer weight modular form for $\Gamma_{0}(4N)$ with character $\chi$. If $\psi$ is a Dirichlet character modulo $m$, then $$ \sum \psi(n) c(n) q^{n}$$ is a half-integer weight modular form for $\Gamma_{0}(4Nm^{2})$ with character $\chi \psi^{2}$. Ono does not provide a proof of this. However, there is a proof given in Shimura's 1973 Annals paper "On modular forms of half-integral weight". The fact you are seeking is Lemma 3.6 on page 466.<|endoftext|> TITLE: Unique predual of a Banach space QUESTION [23 upvotes]: Suppose $E$ is a dual Banach space whose predual is unique, and $E_0$ is a codimension 1 weak* closed subspace of $E$. Is the predual of $E_0$ necessarily unique? Okay, I will reveal the motivation. It is a long-standing problem whether the predual of ${\rm Lip}_0(X)$, for $X$ a pointed metric space, is unique. I can prove this for ${\rm Lip}(Y)$ spaces, an important special case. If the diameter of $X$ is finite then ${\rm Lip}_0(X)$ isometrically embeds as a weak* closed codimension 1 subspace of some ${\rm Lip}(Y)$, so a positive answer to my question would resolve the finite diameter case. I just noticed something else. In my case $E_0$ is also isometrically a quotient of $E$ by a complementary 1-dimensional subspace. Maybe that helps? Editing to note that I've posted a paper using this result on the arXiv. Those 200 points were well spent! REPLY [11 votes]: The following post is an edit of a previous one which was not accurate. It provides a partial answer to the question. As a compromise I decided to include also a bit of a generalization. Thanks to Cameron Zwarich for spotting a gap in my previous post. Thanks to Nik Weaver for suggesting a way around it using a stronger assumption and to Cameron Zwarich for identifying this assumption with the established property of having a "strongly unique predual". That was very helpful. Recall that the Banach space $X$ is called a unique predual if for every isometric isomorphism $\phi:X^*\simeq Y^*$ there exists an isometric isomorphism $\phi_*:Y\simeq X$. The Banach space $X$ is called a strongly unique predual if moreover there exists such $\phi_*$ satisfying $\phi=(\phi_*)^*$. The latter notion seems more natural. Note that the existence of a Banach space which is unique predual but not strongly unique predual seems to be an open problem - at least it was in 1989 - see X.2 in "Existence and uniqueness of isometric preduals: a survey" by Godefroy (Contemp. Math, 131-193, 85, 1989). In particular, it seems that the spaces $\text{Lip}(Y)$ which motivated Nik's question are known (by him) to be strongly unique preduals, thus my answer below, though partial, does help in proving that for a bounded metric space $X$ the space $\text{Lip}_0(X)$ has a strongly unique predual. This is mentioned as an open problem in Nik's book "Lipschitz Algebras" (p. 40). Theorem: Assume $E$ has a strongly unique predual and $E_0\phi_*>> X\\ @V V V @VV V\\ Z^{**} @>\phi^*>> X^{**} \end{CD} Proof of the theorem: Let $E_*$ be the unique predual of $E$. Let $E_0 TITLE: Galois descent for dimension of vector spaces QUESTION [5 upvotes]: Let $L/K$ be a Galois extension (I am interested in $\overline{\mathbb{Q}}/\mathbb{Q}$ so I do not assume it to be finite). Let $V\subset L^n$ be a $L$-subvector space, of dimension $d$, such that $g(V)=V$ for each $g\in \mathrm{Gal}(L/K)$. Do you have a reference for the fact that the dimension of the $K$-vector space $V\cap K^n$ is equal to $d$? It seems to me that it corresponds to a Galois descent and I've looked into books for this notion but I did not find something with this precise statement. REPLY [5 votes]: What you need is the corollary on the bottom of page 60 of Bourbaki, Algèbre, chapitre V : Corps commutatifs ; section 10 (Extensions galoisiennes), subsection 4 (Descente galoisienne). The statement given there is more precise: given an $L$-subspace $W$ of $L^n$ which is stable under Galois, the $K$-subspace $V=W\cap K^n$ of $K^n$ satisfies $W=V\otimes_K L$ and is the unique one with this property. Since dimension is preserved by base change, one has $\dim_L(W)=\dim_K(V)$, as you requested.<|endoftext|> TITLE: The topology of pointwise convergence with the adjoint operator on a von Neumann algebra QUESTION [6 upvotes]: Let $H$ be a Hilbert space and ${\mathcal A}$ a von Neumann algebra in $B(H)$. Let us endow ${\mathcal A}$ with the topology of pointwise convergence with the Hermitian adjoint operator, i.e. a net $A_i\in {\mathcal A}$ tends to 0 in ${\mathcal A}$ if $$ \forall x\in H\qquad A_ix\overset{H}{\underset{i\to\infty}{\longrightarrow}}0 \quad\&\quad A_i^*x\overset{H}{\underset{i\to\infty}{\longrightarrow}}0. $$ The algebra ${\mathcal A}$ with this topology has totally bounded sets $S\subseteq {\mathcal A}$, which I believe, are described by the property $$ \forall x\in H\qquad \text{the sets $\{Ax;\ A\in S\}$ and $\{A^*x;\ A\in S\}$ are totally bounded in $H$} $$ Let $u:{\mathcal A}\to{\mathbb C}$ be a linear functional, which is continuous on every totally bounded set (equivalently, on each compact set) $S\subseteq {\mathcal A}$ in this sense (with respect to this topology induced from ${\mathcal A}$ to $S$). Question: does $u$ belong to ${\mathcal A}_*$ (the predual of ${\mathcal A}$)? EDIT. I must say, even the case of ${\mathcal A}=B(H)$ is not clear for me. And even if we replace the topology on ${\mathcal A}$ with the usual strong operator topology (i.e. if we remove the requirement $A^*_ix\to 0$). We also can change premises in other different ways, for example, we can think that ${\mathcal A}$ is commutative (for me that would be sufficient now). If anybody could cast some light on any weakened conditions, I would appreciate this very much. I need this for my current work, this is related to the study of envelopes of topological algebras. REPLY [2 votes]: This is a comment on the above answer but it will be too long and in any case I am not entitled. The questions you pose are answered in much more detail in a Comptes Rendus article "Topologies sur les espaces des opérateurs dans l'espace hilbertien" (A276 (1973) 1509). This discusses three locally convex topologies on a von Neumann algebra---the finest locally convex ones which coincide with the weak operator, the strong operator and the symmetric strong operator topology on the unit ball. They are all complete, have the same convergent sequences as the above-mentioned ones, have the same dual space (the normal functionals) and have the property that a linear functional thereon is continuous if and only it is continuous on the unit ball for each or any of the above operator topologies. This can be found in the secondary literature (in english) as chapter IV of "Saks spaces and applications to functional analysis" which is readily available online. Particularly relevant is Prop. 2.3 on p. 204. The commutative case is treated in Ch. 3.<|endoftext|> TITLE: Haken's Lemma for Essential Tori QUESTION [7 upvotes]: Let $M$ be a closed, connected, orientable 3-manifold with a Heegaard splitting $(F,H_1,H_2)$. Haken's Lemma states that if $M$ is reducible, i.e. if there is an embedded essential sphere in $M$, then there is an embedded essential sphere $S$ in $M$ such that $S\cap F$ is a simple closed curve, and hence $S\cap H_i$ are disks. Is there a version of this lemma for essential tori embedded in $M$? In other words, can one say that if there is an embedded essential torus in $M$, then there is one which cuts each handlebody $H_i$ in an annulus? REPLY [4 votes]: Yes. If the splitting surface $F$ is strongly irreducible then, after an isotopy you can assume that $T$ meets each of $H_1$ and $H_2$ in collections of disjoint incompressible annuli. This has been independently proved by Kobayashi, Thompson, and Hempel. For references, please see the bibliography of Hempel's paper "3-manifolds as viewed from the curve complex".<|endoftext|> TITLE: Are $G$-limits of a slender group $G$ in the space of marked groups also slender? QUESTION [6 upvotes]: A group $G$ is slender if every subgroup $H \leq G$ is finitely generated. This includes polycyclic-by-finite groups. Such groups are also called noetherian. Suppose that $L$ is a $G$-limit group in the sense that $L$ has a finite generating set $X$ such that for every $n$, the ball $B_n$ of radius $n$ in the Cayley graph is isomorphic to some ball of radius $n$ in a Cayley graph of some subgroup of $G$. Then is it true that $L$ is itself slender? Does it change anything if $G$ is equationally noetherian? What about smallness? If $G$ is small, then must a $G$-limit group also be small? I suspect this may be a tough question, but maybe I'm missing something obvious. Any response would be much appreciated. REPLY [7 votes]: 1) Consider the semidirect product $G=\mathbf{Z}^2\rtimes\mathbf{Z}$ with action by the matrix $A=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$. It has the presentation $$\langle t,x,y\mid txt^{-1}=x^2y,\; tyt^{-1}=xy,\; [x,y]=1\rangle.$$ (It is actually generated by $(t,x)$). Claim: some limit of markings on $G$ fails to be noetherian (although $G$ is) [Added: this result was proved by Luc Guyot (Limits of metabelian groups. Internat. J. Algebra Comput. 22 (2012), no. 4, 1250031; ArXiv link, MR link), showing (2nd item of Theorem A therein) that suitable markings accumulate to the free metabelian group on 2 generators.] Proof of the claim: [Edit: It's enough for purposes here to prove something weaker than Guyot's theorem, namely showing that some limit admits $\mathbf{Z}\wr\mathbf{Z}$ as a subgroup.] Consider the generating subset $(t^{n^2+1},t^n,x)$. There is a unique homomorphism $j_n$ from the wreath product $\langle s,x\mid [s^mxs^{-m}, x]=1,\forall m \in \mathbb{Z}\rangle$ to $G$ mapping $s$ to $s:=t^n$ and $x$ to $x$. We claim that it is injective in a ball of radius $p_n$ going to infinity with $n$. Once this is proved, this shows that some limit of copies of $G$ (any accumulation point for this choice of marking) contains a copy of $\mathbf{Z}\wr\mathbf{Z}$ and hence is not noetherian. To prove the claim: assume the contrary. This means that some nontrivial element $w$ of the wreath product such that $j_n(w)=1$ for infinitely many $n$ (say $n\in I$, infinite set of integers). Clearly $w$ should be in the kernel of the canonical map to $\mathbf{Z}$, and therefore can be written $w=P(x)$ with $P\in\mathbf{Z}[u^{\pm 1}]$ (Laurent polynomial ring). Here for $P=\sum a_nu^n$, $P(x)$ means $\prod s^nx^{a_n}s^{-n}$. (This is a standard identification, since $\mathbf{Z}\wr\mathbf{Z}\simeq\mathbf{Z}[u^{\pm 1}]\rtimes\mathbf{Z}$). That $j_n(w)=1$ means that, in $\mathbf{Z}^2$, $P(A^n)x=0$. Since $A$ is irreducible, that $P(A^n)$ has nontrivial kernel implies that it's zero. So for all $n\in I$, we have $P(A^n)=0$. But the $A^n$ have distinct eigenvalues (except for $n$ and $-n$), so infinitely of them cannot be killed by a single polynomial, contradiction. Note that $G$ is linear over $\mathbf{Z}$, hence equationally Noetherian. 2) Small groups. Consider, for instance, a f.g. group $H$ containing all finite groups as subgroups (e.g., Neumann's group $\mathrm{Sym}_0(\mathbf{Z})\rtimes\mathbf{Z}$, or its finitely presented analogue due to Houghton), with no free subgroup. Claim: some limit of markings on $H$ contains a free subgroup (although $H$ doesn't) Fix a sequence of finite marked groups $(V_n,x_n,y_n)$ tending to the free groups on 2 generators. Then consider a sequence of generating subset of $H$, as a concatenation of a fixed generating subset of $H$, and a pair $(x'_n,y'_n)$ in $H$ generating a subgroup isomorphic to $(V_n,x_n,y_n)$ as marked group. Then any accumulation point of this sequence of marked groups has the property that the last two generators freely generated a free group.<|endoftext|> TITLE: Concept of "Rigidity" in mathematics QUESTION [12 upvotes]: I am reading from the book "Topics in Galois Theory" by Serre. I came across the word "Rigidity". I am not able to understand this concept. If I am not wrong, This term was first used by Thompson, He used this concept("Rigidity Method") to show that monster group can be realized as the Galois group over $\mathbb{Q}$. I don't have any intuition for this concept. I tried reading from the Wikipedia page on "Rigidity", but I got more confused. It says The above statement does not define a mathematical property. Instead, it describes in what sense the adjective rigid is typically used in mathematics, by mathematicians. What is the idea ? One more question, What does the adjective "rigid" means in context of Galois theory. REPLY [6 votes]: Consider a (ramified) Galois covering of the Riemann sphere $f\colon X\to B=\mathbf P^1(\mathbf C)$, of group $G$. Let $b_1,\dots,b_r\in B $ be the (distinct) ramification points. Fix a base point $o\in B\setminus\{b_1,\dots,b_r\}$ and for all $i$, fix an injective $C^1$-path $\gamma_i$ from $o$ to $b_i$ that they do not intersect outside of $o$ (in particular, they avoid the other ramification points). Define an element $\sigma_i$ in $\pi_1(B,o)$ by the following recipe: start from $o$, go along the chosen path until very close to $b_i$, make one turn around $b_i$ in the positive direction, and go back to $o$ along the chosen path. If the cyclic ordering of the points $b_1,\dots,b_r$ is chosen so as to coincide with the angles the paths leaving $o$, then the elements $\sigma_1,\dots,\sigma_r$ generate the group $\pi_1(B,o)$, and there is the single relation $\sigma_1\cdots \sigma_r=e$. (This reflects in particular the fact that the Riemann sphere, and even the complex plane, is simply connected.) The group $\pi_1(B,o)$ acts naturally on $X$ (on the right) via an anti-homomorphism $\pi_1(B,o)\to G$. Fix a point $x\in X$ above $o$. The path $\gamma_i$ lifts to a path from $x$ to some point $a_i$ in $X$. The subgroup $G_i$ of $G$ that fixes $a_i$ is cyclic of order $e_i$ (the ramification index at $i$), generated by the action $[\sigma_i]$ of the loop $\sigma_i$. In particular, $[\sigma_1],\dots,[\sigma_r]$ generate $G$. The elements $[\sigma_1],...,[\sigma_r]$ depend on the choice of the initial paths $\gamma_i$. Changing those paths turns them into conjugate elements $h_i [\sigma_i] h_i^{-1}$. In particular, their conjugacy classes $C_i$ are well-defined, and $C_1,\dots,C_r$ generate $G$. In general, if $G$ is a finite group, one says that a sequence $(C_1,\dots,C_r)$ of conjugacy classes is rigid if: (i). For each $i$, there exists $g_i\in C_i$ such that $g_1\dots g_r=e$ and $(g_1,\dots,g_r)$ generate $G$; (ii). For any two families $(g_1,\dots,g_r)$ and $(g'_1,\dots,g'_r)$ as in (i) are conjugate by an element $h\in G$: $g'_i=h g_i h^{-1}$ for all $i$. The interest in the concept — take the time to study Serre's book if you need so — is that the Inverse Galois Problem is “easy” to solve in these cases. The arguments is basically as follows. First of all, one constructs a covering of the Riemann sphere satisfying these geometric properties, where the points $b_i$ are integers, say. By Riemann's existence theorem, this corresponds to a finite field extension $\mathbf C(T)\subset F$, Galois of group $G$, with inertia subgroups corresponding to the given conjugacy classes. Now, consider an arbitrary automorphism of $\mathbf C$; thanks to the rigidity property, it naturally extends to an automorphism of $F$. All these automorphisms allow to construct a canonical subfield $E$ of $F$ such that there is a finite field extension $\mathbf Q(T)\subset E$, that gives $F$ after extending the scalars from $\mathbf Q$ to $\mathbf C$. (I am slightly cheating here, because some cyclotomic extension may enter the picture. Let's pretend there is none, which happens when the classes $C_i$ are “$\mathbf Q$-rational”.) Let $e\in E$ be a primitive element which is, say, integral over $\mathbf Q[T]$. Its minimal polynomial $P$ belongs to $\mathbf Q[T,X]$, and is irreducible viewed as a polynomial in $\mathbf C[T,X]$. Now apply Hilbert's irreducibility theorem: for many elements $u\in\mathbf Q$ (“most of them”, in some sense), the polynomial $P(u,X)$ is irreducible in $\mathbf Q[X]$ and generates a Galois extension of $\mathbf Q$ of group $G$. The Inverse Galois problem is solved in this case.<|endoftext|> TITLE: Is there a Borel subset of $ \mathbb{R}^{2} $, with finite vertical cross-sections, whose projection onto the first component is non-Borel? QUESTION [14 upvotes]: This question is related to another one that I asked two days ago. Question. Does there exist a Borel subset $ M $ of $ \mathbb{R}^{2} $ with the following two properties? The projection of $ M $ onto the first component is a non-Borel analytic subset of $ \mathbb{R} $. Every vertical cross-section of $ M $ is finite, i.e., the set $ \{ y \in \mathbb{R} \mid (x,y) \in M \} $ is finite for every $ x \in \mathbb{R} $. An affirmative answer to this question will provide a counterexample in the topic of measure theory, as explained in the linked post. It therefore holds some importance. Thank you very much for your help! REPLY [4 votes]: There is a pure recursion theoretical proof of the result. The idea is as follows: By Spector-Gandy theorem, a lightface Borel set $(x,y)$ is an r.e set over $L_{\omega_1^{CK}}[x,y]$. If there are at most countably many $y$'s corresponded to the $x$ for every $x$, then it can be reduced to $L_{\omega_1^{CK}}[x]$. Then the projection to the first section becomes a $\Pi^1_1$ statement.<|endoftext|> TITLE: integral of a "sin-omial" coefficients=binomial QUESTION [104 upvotes]: I find the following averaged-integral amusing and intriguing, to say the least. Is there any proof? For any pair of integers $n\geq k\geq0$, we have $$\frac1{\pi}\int_0^{\pi}\frac{\sin^n(x)}{\sin^k(\frac{kx}n)\sin^{n-k}\left(\frac{(n-k)x}n\right)}dx=\binom{n}k. \tag1$$ I also wonder if there's any reason to relate these with an MO question that I just noticed. Perhaps by inverting? AN UPDATE. I'm extending the above to a stronger conjecture shown below. For non-negative reals with $r\geq s$, a generalization is given by $$\frac1{\pi}\int_0^{\pi}\frac{\sin^r(x)}{\sin^s(\frac{sx}r)\sin^{r-s}\left(\frac{(r-s)x}r\right)}\,dx =\binom{r}{s}. \tag2$$ REPLY [10 votes]: Here is another way to prove it. Surprisingly, for $n$ integral and $k$ real, the integral in question can be written down as an indefinite integral. This gives a direct proof for non-integer $k$, though obviously less clear than the contour method. (In fact, it is convenient to avoid integer $k$ in this method, and extend to integer $k$ by continuity.) Writing $y=x/n$ and $l=n-k$, we have for example for $n=2$: $$\int\frac{\sin^2(2y)}{\sin^k(ky)\sin^l(ly)}dy= \frac{\frac{2}{l-k}\sin((l-k)y)+\sin(2y)} {kl\sin^{k-1}(ky)\sin^{l-1}(ly)}$$ In general for $n$ even ($n$ odd is similar with cosines): $$I_{n,k}(y)=\int\frac{\sin^n(ny)}{\sin^k(ky)\sin^l(ly)}dy= \frac{\sum_{r=0}^{n-1}\sum_{s=0}^{n-1}\lambda_{r,s}\sin(((n-1-2r)k+(n-1-2s)l)y)} {kl\sin^{k-1}(ky)\sin^{l-1}(ly)}$$ where $$\lambda_{r,s}=\begin{cases} \displaystyle \frac{(-1)^r(n-1)^{\underline{r}}}{(r-l)^{\underline{\smash{r-s}}}s!}\lambda_{0,0},\;\;r\ge s\\ \displaystyle \frac{(-1)^s(n-1)^{\underline{s}}}{(s-k)^{\underline{\smash{s-r}}}r!}\lambda_{0,0},\;\;s\ge r \end{cases} $$ $$\lambda_{0,0}=(-1)^{n/2+1}2^{1-n},$$ and $x^{\underline{r}}$ denotes the falling power $x(x-1)\ldots(x-r+1)$. It is easy to check the derivative, $I_{n,k}'(y)$ is correct by considering the coefficient of $\cos((ak+bl)y)/(\sin^k(ky)\sin^l(ly))$ for each $a$, $b$. If $a\neq b$ then you get zero, otherwise for $a=b=n-2r$, $(ak+bl)y=(n-2r)ny$ and you get $\frac{1}{2}(-1)^r\binom{n}{r}\lambda_{0,0}\cos((n-2r)ny)$. Then $\sin^n(ny)$ arises from the binomial expansion: $$(-1)^{n/2}2^{-n}\sum_{r=0}^n (-1)^r\binom{n}{r}\cos((n-2r)ny)=\sin^n(ny).$$ Note that $I_{n,k}(0)=0$ because, being the integral of something well-behaved at $0$, $I_{n,k}(y)$ must be continuous at $0$, so its numerator must vanish to order $n-2$ like its denominator. Using L'H\^{o}pital, taking $n-2$ derivatives of the numerator gives only sines, which themselves vanish at 0. To evaluate $I_{n,k}(\pi/n)$, note that $\sin(k\pi/n)=\sin(l\pi/n)$ and $\sin(((n-1-2r)k+(n-1-2s)l)y)=\sin(2(r-s)k\pi/n)$. Conditioning on $r-s=d>0$, you get (e.g., by considering partial fractions in $k$) $$\sum_{s=0}^{n-1-d}\lambda_{s+d,s}= \frac{(-1)^d(n-1)!\binom{n-2}{d-1}}{(k-1)^{\underline{\smash{n-1}}}}\lambda_{0,0},$$ and similarly for $d<0$ with the opposite sign, and using $-d$ in place of $d$. Using the binomial expansion of $(1-e^{2\pi ik/n})^{n-2}$, you get $$\sum_{d=1}^{n-1}(-1)^d\binom{n-2}{d-1}\sin\left(\frac{2dk\pi}{n}\right)= (-1)^{n/2}2^{n-2}\sin^{n-2}\left(\frac{k\pi}{n}\right)\sin(k\pi)$$ So finally, putting the pieces together, $$\frac{n}{\pi}I_{n,k}\left(\frac{\pi}{n}\right)=\frac{n\sum_{r=0}^{n-1}\sum_{s=0}^{n-1}\lambda_{r,s}\sin\left(\frac{2(r-s)k\pi}{n}\right)} {\pi kl\sin^{k-1}\left(\frac{k\pi}{n}\right)\sin^{l-1}\left(\frac{l\pi}{n}\right)}=\frac{n!\sin(k\pi)}{\pi k^{\underline{\smash{n+1}}}}$$ which (for even $n$) we recognise as $\binom{n}{k}$ by the reflection formula for factorials.<|endoftext|> TITLE: How small can the smallest of the three "weak Goldbach" primes always be? QUESTION [6 upvotes]: I've checked here for discussions of Helfgott's proof of the weak GC and found nothing that helps me with the following; apologies if I missed something. I'm probably being naive here (please enlighten me), but I’m thinking that there is a natural conjecture that is intermediate between the strong and weak Goldbach conjectures. The strong GC can be stated as the weak GC (in its "three odd primes" formulation) plus the additional constraint that the smallest of the three primes can always be 3. My question is: is there a larger number $N$ for which it can already be stated that the smallest of the three primes referred to in the weak GC can always be at most $N$? If not, what can be said about this question? REPLY [3 votes]: Your question can be reformulated as: if $n$ is an odd number then how small the odd prime $p$ should be such that $n-p$ is a Goldbach number. Harman, in Chapter 6 of his book Prime-detecting sieves (Princeton University Press, 2007) proved that in the interval $[n-n^{11/180},n]$ all but $O_A(n^{11/180}\log^{-A}n)$ even numbers are Goldbach numbers. It follows (upon taking any $A>1$) that there exists an odd prime $p\ll n^{11/180}$ with the required property. I think this is the state-of-the-art for your question. Added. Under GRH, much more is known. Kaczorowski-Perelli-Pintz (1993) proved that in the interval $[n-\log^c n,n]$ all but $O(\log^{c/2+3}n)$ even numbers are Goldbach numbers. On the other hand, there are $\gg_c\log^c n/\log\log n$ prime numbers up to $\log^c n$, i.e. more than Goldbach exceptions if $c>6$. It follows that there exists an odd prime $p\ll_\epsilon \log^{6+\epsilon}n$ with the required property.<|endoftext|> TITLE: Euler characteristic of local system depends only on rank? QUESTION [5 upvotes]: Let $X$ be a proper variety over a finite field $k$ of characteristic $p>0$, and let $\mathcal F$ be a finite rank $\mathbb F_\ell$ local system on (the etale site of) $X$. Is it true (and, if so, how does one prove) that $$\chi(X,\mathcal F)=\operatorname{rk}\mathcal F\cdot\chi(X,\underline{\mathbb F_\ell}_X)$$ where $\underline{\mathbb F_\ell}_X$ denotes the constant local system on $X$? Some remarks: If $k=\mathbb C$, then this is true (without assuming properness), but the only proof I know is via analytification and topological arguments. This implies the result for $(X,\mathcal F)$ which lift to characteristic zero, but this leaves out lots of varieties $X$ and local systems $\mathcal F$. Properness is necessary when $\operatorname{char}k>0$. Namely, let $f:\mathbb A^1\to\mathbb A^1$ denote the Artin--Schrier map over $\mathbb F_p$, so $f_\ast\underline{\mathbb F_\ell}_{\mathbb A^1}$ is a local system since $f$ is etale. Now $\operatorname{rk}f_\ast\underline{\mathbb F_\ell}_{\mathbb A^1}=p$, but $\chi(\mathbb A^1,f_\ast\underline{\mathbb F_\ell}_{\mathbb A^1})=\chi(\mathbb A^1,\underline{\mathbb F_\ell}_{\mathbb A^1})$ by the Leray spectral sequence. REPLY [5 votes]: As $X$ is proper, the Swan conductor of $\mathcal{F}$ vanishes. Hence the identity $\chi(X,\mathcal F)=\operatorname{rk}\mathcal F\cdot\chi(X,\underline{\mathbb F_\ell}_X)$ follows from Theorem $4.2.9$ in Kato & Saito's "Ramification theory for varieties over a perfect field ".<|endoftext|> TITLE: Inverse Limits in the category of Perfectoid Spaces QUESTION [11 upvotes]: First I apologize as the following is likely littered with misunderstanding. My understanding at least from the discussion in Scholze-Weinstein is that inverse limits in the category of adic spaces rarely exist because there isn't a canonical way to define a topology on the direct limit of affinoid rings. Because of this, Scholze defines a weak notion of inverse limit which he denotes by $\sim$. If such a weak inverse limit $X$ is perfectoid, the situation is better as $X$ satisfies a universal property and thus is unique in the category of perfectoid spaces. My question is do inverse limits always exist in the category of perfectoid spaces? My guess would be no because of the above. However, I am confused by language immediately before II.3.11 and before III.2.34 in Scholze's torsion paper seeming to suggest otherwise. REPLY [2 votes]: As mentioned by Scholze in this lecture (around 40:00; uniform Tate rings are called "spectral rings" in the lecture), given a directed system of maps of uniform Tate-Huber pairs $(R_i,R_i^+) \to (S,S^+)$, the map $\varinjlim_i R_i \to S$ is continuous for the $\varpi$-adic topology on $\varinjlim_i R_i^+$, where $\varpi$ is the image of a pseudouniformizer of some $R_i$. Here we need to use the fact that $S$ is uniform, so $S^+$ is a ring of definition and it has the $\varpi$-adic topology. So filtered inverse limits should exist in the category of (sous)perfectoid spaces since (sous)perfectoid rings are stably uniform. Unfortunately I do not know of a reference that explicitly proves this, but it does seem to be used implicitly in Etale cohomolgy of diamonds.<|endoftext|> TITLE: Reference request: an example of Bott residue formula's usage QUESTION [6 upvotes]: Could you give me an example of a clear and beautiful application of Bott residue formula in torus-equivariant cohomology (see below)? I found an example calculating a product of Chern classes on singular quadric in section 5.3 of Edidin, Graham Localization in equivariant intersection theory and the Bott residue formula, but in singular case one need equivariant Chow groups instead of equivariant cohomology, and anyway the example is too technical. Let $T=(\mathbb C^*)^n$ be an algebraic torus acting on a smooth complex projective variety $X$ such that there is a finite number of $T$-fixed points of $X$. A technically simple case of Bott residue formula says that for $T$-equivariant vector bundle $E$ with $\operatorname{rk} E=\dim X$ the top Chern class is a number given by $$\int_X c_{top}(X, E)=\sum_{x \in X^T} \frac{c_{top}^T (x, E_x)}{c_{top}^T (x, TX_x)},$$ where the right hand side should be calculated in the quotient field of $T$-equivariant cohomology ring of a point, that is simply $$\operatorname{Quot} H_T^*(x)=\operatorname{Quot} \mathbb Q[t_1, \ldots, t_n]=\mathbb Q(t_1, \ldots, t_n);$$ the $T$-equivariant top Chern class $c_{top}^T(x, V)$ of a $T$-representation $V$ (above it is the $x$-fiber $E_x$ or $TX_x$) is the product of $T$-characters of $V$ as elements $$a_1t_1+\ldots+a_nt_n \in \mathbb Q[t_1, \ldots, t_n];$$ the equality is taken in $\mathbb Q(t_1,\ldots,t_n)$ as the summands lie there, but the sum lies in $\mathbb Q$. So one can derive the formula $\chi(X)=\chi(X^T)$ by choosing $E=TX$, but it is too degenerate. Could you give me an example which is more meaningful but still clear enough? Any reference or idea is also welcome! REPLY [4 votes]: There is the famous paper by Ellingsrud and Strømme, who apply Bott's residue formula to enumerative geometry: Geir Ellingsrud and Stein Arild Strømme. Bott's formula and enumerative geometry. J. Amer. Math. Soc. 9 (1996), 175-193<|endoftext|> TITLE: Cohomological interpretation of G-equivariant line bundles QUESTION [5 upvotes]: In the theory of reductive algebraic groups, there is the following map (notation: $G$ reductive over an algebraically closed field, $T$ a maximal torus, $B$ a Borel, $X(T)$ the characters of $T$, $Pic^G$ denotes the group of isomorphism classes of $G$-equivariant line bundles): $$X(T) \to Pic^G(G/B), \lambda \mapsto \mathcal O(-\lambda) := G \times^B {k_\lambda}$$ My questions: Is (analogously to $Pic(S) = H^1(S, \mathcal O^\times)$) $Pic^G$ expressible as some cohomology group? Is the above map some kind of connection homomorphism or does it have another cohomological interpretation? Thank you! REPLY [6 votes]: See Theorem 4.2.2 in https://www-fourier.ujf-grenoble.fr/~mbrion/lin.pdf In particular, in your example properness of $X=G/B$ simplifies the left part of the sequence, turning it into $$0\to \hat{G}\xrightarrow{\gamma} Pic_G(X)\to Pic(X)\to Pic(G\times X)$$ where $\hat{G}$ is the group of characters of $G$ and $\gamma$ maps $\chi$ to $X\times_G \chi$. The map $Pic_G(X)\to Pic(X)$ is just the forgetting map, for the definition of the last one, see the text. Also, if $G$ is affine, as in your case, $Pic(G\times X)=Pic(X)\times Pic(G)$(I write $=$ instead of $\cong$ to emphasise that the isomorphism is given by projection maps) so, applying Proposition 4.2.3, we get $$0\to \hat{G}\to Pic_G(X)\to Pic(X)\to Pic(G)$$ Edit: As Marc Hoyois mentions, if $X$ is a $G$-bundle over $Y$, then $Pic_G(X)=Pic(Y)$, see lemma 3.3.1<|endoftext|> TITLE: Finite Homomorphic images of infinite products of finite solvable groups QUESTION [6 upvotes]: I conjecture that: Every Finite Homomorphic image of an infinite (with arbitrary cardinality) product of finite solvable groups is solvable -- or at least Not a simple (non-abelian) group. I can see this conjecture in some cases. but the general case seems very complicated. Question: Has this problem been investigated ? Thank you. Edit-- Moreover, I also conjecture that Every Finite Homomorphic image of an INFINITE product G = \prod G_i of finite simple (non-abelian) groups {G_i}, is a product of finite simple groups each of which is isomorphic to some G_i. Y Cor has kindly outlined the proof of my 2 conjectures (in comments below) via the Nikolov-Segal Theorem (Annals of Math., 2007). Moreover, Yilong Yang has just found another proof (of my 2nd conjecture) by using his covering properties of finite groups (introduced in his publication in 2016). REPLY [7 votes]: That's correct: every finite quotient is solvable. Indeed let $G$ be your product of finite solvable groups. Let $p:G\to F$ be a (possibly non-continuous) surjective homomorphism to a finite group $F$. Lift $F$ to a finite subset $\tilde{F}$, and let $H$ be the closed subgroup generated by $\tilde{F}$. Then by the Nikolov-Segal theorem, the restriction of $p$ to $H$ is continuous. But as a closed subgroup of $G$, $H$ is a projective limit of finite solvable groups. So all its quotients by open subgroups are finite solvable groups. More generally, this shows that any abstract finite quotient of a profinite group $G$ is isomorphic to some $H/K$ where $H$ is a closed subgroup of $G$ and $K$ an open normal subgroup of $H$. (In the particular case when $G$ is pro-$p$ for some prime $p$, this appeals to a much older theorem of Serre instead of Nikolov-Segal.) (Edit, to answer a question in a comment.) The previous actually shows a slightly stronger statement: (1) if $G$ is a profinite group and $F$ is a finite quotient of $G$, then there exists a t.f.g. closed subgroup $M$ of $G$ such that for every t.f.g. closed subgroup $H$ of $G$ containing $M$, $F$ is isomorphic to the quotient of $H$ by some closed normal subgroup. [t.f.g. means topologically finitely generated, meaning having a dense finitely generated subgroup] Now I claim that (2) if $G$ is a product $\prod S_i$ of finite simple groups (possibly abelian), then every t.f.g. closed subgroup of $G$ is contained in a t.f.g. closed subgroup that is a product of finite simple groups. If we combine (1) and (2), we deduce that in an arbitrary product of finite simple groups $\prod_{i\in I}S_i$, every abstract finite quotient is isomorphic to some finite product $\prod_{j\in J}S_j$ (although it's not always given by the obvious projection!). Actually while (1) follows from the Nikolov-Segal theorem (every finite index subgroup is open, in a t.f.g. profinite group), the particular case of the Nikolov-Segal theorem for products of finite simple group was proved by Saxl and Wilson, namely: if $(S_n)$ is a sequence of finite simple groups and $\lim_{n\to\infty}|S_n|=\infty$, then every finite index subgroup in $\prod S_n$ is open. (J. Saxl and J. S. Wilson, A note on powers in simple groups, Math. Proc. Cambridge Phil. Soc. 122 (1997), 91-94.) Checking (2) is quite elementary, modulo the fact that finite simple groups have a bounded generating rank, namely every finite simple group is generated by 2 elements, which relies on classification. Gather together all isomorphic simple groups in the product, to write $G=\prod_S G_S$, where each $G_S\simeq S^{I_S}$, and $S$ ranges over non-isomorphic finite simple groups. Let $L$ be a t.f.g. closed subgroup of $G$; so $L$ has a dense subgroup generated by $d$ elements for some $d$. Let $L_S$ be its projection in $S^{I_S}$; it is generated by $d$ elements. Let $M_S$ be the subgroup of $S^{I_S}$ generated by $L_S$ and the diagonal; it is generated by $\le d+2$ elements (because any finite simple group is generated by 2 elements). Also, every finitely generated subgroup of $S^{I_S}$ containing the diagonal is isomorphic to $S^k$ for some $k$. So if we choose a family of $d+2$ elements in $\prod M_S$ whose $S$-component generates $M_S$, we obtain a generating subset for a dense subgroup of $\prod M_S$, and $M_S$ is a product of finite simple groups. This proves (2).<|endoftext|> TITLE: Spheres in simplicial complexes with homotopy type of a wedge of spheres QUESTION [13 upvotes]: Let $K$ be a finite $d$-dimensional simplicial complex which is homotopy equivalent to a wedge of $k$ $d$-spheres. Does $K$ necessarily contain a $d$-sphere as a subcomplex? REPLY [8 votes]: It is false for $d\geq 2$. For an example with $d=2$, start with a circle, and attach two 2-discs to it via (for example) a map of degree 4 and a map of degree 5. If you like to think in terms of group presentations, this is the presentation 2-complex for the presentation $\langle a: a^4=a^5=1 \rangle$ of the trivial group. This 2-complex is homotopy equivalent to a single copy of the 2-sphere, but does not contain any subcomplex homeomorphic to the 2-sphere. Describing an explicit triangulation of a space homeomorphic to this is a bit more complicated, but it can be done. Triangulate the circle using four edges, and triangulate the two discs so that their boundary circles have 16 and 20 edges respectively, and so that there are no `short routes' in the interior of the disc between distinct vertices on the boundaries of the disc. Given any such triangulations, you can glue the discs on to the circle and get a simplicial complex homeomorphic to the CW-complex mentioned above. If you want to find a subcomplex that is homeomorphic to the 2-sphere, it must contain one of the 2-simplices. But once it contains one of the 2-simplices, it has to contain every 2-simplex in the triangulation of one of the two 2-discs, in which case it contains an edge of valency 4 or 5, giving a contradiction. Suspensions of this example (suitably triangulated) will work in higher dimensions by a similar argument.<|endoftext|> TITLE: Finite groups in which all proper subgroups are cyclic QUESTION [8 upvotes]: Is there any classification of finite group in which all proper subgroups are cyclic? Would you please tell me a reference? REPLY [16 votes]: All subgroups of a cyclic group (including the proper ones) are cyclic. And there is the following classification of non-cyclic finite groups, such that all their proper subgroups are cyclic: A finite group $G$ is a minimal noncyclic group if and only if $G$ is one of the following groups: 1) $C_p × C_p$, where $p$ is a prime 2) $Q_8$ 3) $\langle a,b | a^p = b^{q^m} = 1, b^{−1}ab = a^{r}\rangle$, where $p$ and $q$ are distinct primes and $r ≡ 1 \pmod q$, $r^q ≡1 \pmod p$. This theorem first appeared in "Non-abelian groups in which every subgroup is abelian" by G.A.Miller and H.G.Moreno (1903)<|endoftext|> TITLE: Isomorphism classes of sheaves which arise as extensions QUESTION [5 upvotes]: Let $X$ be a proper(say, smooth) variety and $E,F$ are coherent sheaves on it. Extensions of $E$ by $F$ are parametrised by a finite-dimensional vector space $\mathrm{Ext}^1(E,F)$. I am intersted in the isomorphism classes of sheaves which can be realized as such extensions. Q1: Is it true, that the set $S=\{G\in Coh(X)|$there exists a short exact sequence$0\to F\to G\to E\to 0\}$ is finite? Q2: Assuming the positive answer to Q1, is it true that subsets $S_G=\{e\in\mathrm{Ext}^1(E,F)|$middle sheaf of $e$ is $G$ $\}\subset \mathrm{Ext}^1(E,F)$ give a stratification on this affine space? In other words, can $S$ be ordered so that function, assigning to an $\mathrm{Ext}$ class the isomorphism class of the middle sheaf is semi-continous in Zariski topology. I have checked this explicitly for $E=\mathcal{O}(2),F=\mathcal{O}(-2)$ on $\mathbb{P}^1$. There we get $\dim\mathrm{Ext}^1=3$ and $$S_{\mathcal{O}(2)\oplus\mathcal{O}(-2)}=0,S_{\mathcal{O}(1)\oplus\mathcal{O}(-1)} \text{ is a quadratic cone without the origin}, S_{\mathcal{O}\oplus\mathcal{O}}\text{ is the complement to this cone}$$ It seems that for any two line bundles on projective line the answer to $Q2$ is also positive since the splitting type of the extension is controlled by the rank of the connecting homomorphism in the $\mathrm{Ext}$'s exact sequence which is semi-continous. Q3: Even if, in general, my assertions fail, what can be said about the line bundles on $\mathbb{P}^1$ case? I feel that this stratification might be connected to representation therory, as in the above example we get nilpotent cone of $\mathfrak{sl}_2$(though, this may be farfetched). REPLY [3 votes]: As abx explained, the answer to Q1 is negative (I had the same example in mind). Also this shows that the answer to Q2 is negative as well. In linw bundles on $P^1$ case, of course, the stratification is $SL_2$-invariant (and is easy to describe), but it is more coarse than the orbit stratification. Indeed, if the line bundles are $O$ and $O(-d)$, so that $Ext^1 = S^{d-2}(k^2)$, there are only $\lfloor d/2 \rfloor + 1$ isomorphism classes of extension sheaves, while the number of types of orbits is the namber of partitions of $d-2$, and moreover, when $d -2 \ge 4$ there are infinite families of orbits.<|endoftext|> TITLE: Intuition behind the non-Borel Lusin example QUESTION [5 upvotes]: Among the concrete examples of a non-borel subset of $\mathbb{R}$, I know only the Lusin example. This is the set $L$ of all irrational numbers whose continued fraction representation $(a_0,a_1,\cdots)$ is such that there exist an infinite subsequence $(a_{k_0},a_{k_1},\cdots)$ such that each for all $j$, $a_{k_j}$ divides $a_{k_{j+1}}$. Knowing this, I am totally unable to see why this set $L$ would not be Borel, is there some intuition behind this ? What is the idea behind the proof of this fact ? REPLY [2 votes]: I definitely don't think this is a good "first example" of an analytic non-Borel set. It does, however, make more sense when put in comparison with the standard example of a co-analytic, non-Borel set: Given a binary relation $R$ on $\mathbb{N}$, we can code $R$ by a set $X_R\subseteq\mathbb{N}$: let $\langle \cdot,\cdot\rangle$ be a bijection from $\mathbb{N}^2$ to $\mathbb{N}$ (e.g. the Cantor pairing function), and let $$X_R=\{\langle m, n\rangle: mRn\}.$$ Now for a property $\mathfrak{P}$ of relations, let $S_\mathfrak{P}$ be the family of sets coding relations with that property. E.g. $S_{\mbox{linear}}$ is the family of sets which code linear orders of $\mathbb{N}$, and so forth. It's not hard to see that $S_{\mbox{linear}}$ is Borel, indeed closed. It turns out that a good way to get Borel sets of high complexity, and non-Borel sets, is to look at sets of the form $S_\mathfrak{P}$ for more complicated properties $\mathfrak{P}$. One important property of relations is well-foundedness: call a relation $R$ well-founded if there is no infinite sequence $a_0R a_1R a_2R...$. And the associated family of sets is $S_{WF}$. It's not hard to show that $S_{WF}$ is co-analytic. (Strictly speaking, I've described $S_{WF}$ as a subset of $2^\omega$, not $\mathbb{R}$; but it's easy to switch it over, e.g. via the standard bijection between $2^\omega$ and the middle-thirds Cantor set.) But it turns out that $S_{WF}$ is not Borel! Why? Well, remember that the Borel sets are arranged in a hierarchy, indexed by countable ordinals (the $\Sigma^0_\alpha$ and $\Pi^0_\alpha$ sets, as $\alpha\in\omega_1$). Now, a well-founded relation has a rank - a countable ordinal that measures how complicated it is. Bigger ranks mean, intuitively, that the relation is "closer to being ill-founded". The point is that Borel sets have a kind of "overspill" property: sets of fixed Borel rank that contain reals coding well-founded relations of too high rank, also contain reals coding ill-founded relations. And this means that the set $S_{WF}$ is not, in fact, Borel. This is of course not a proof - the details are much more complicated. But this is the idea: there's a connection between the complexity of a Borel set, and its ability to distinguish well-founded relations from ill-founded ones. This motivates the following general intuition: If I have a set defined in terms of quantification over infinite sequences, rather than only over natural numbers, then that set probably isn't Borel. There are of course counterexamples, but this is a good rule of thumb. And the example you mention, of course, is of this kind. It's kind of the opposite of $S_{WF}$: $S_{WF}$ was defined by saying that there is no sequence of a certain form, whereas your set is defined by saying that there is a sequence of a certain form. So a natural thing to do is try to find a Borel isomorphism between your set and the complement of $S_{WF}$.<|endoftext|> TITLE: Principal series representations of $SL(2,\mathbb{R})$: introductory textbooks QUESTION [5 upvotes]: I am interested in introductory books/papers/reports about the (unitary) representation theory of $SL(2,\mathbb{R})$, with particular emphasis on the principal series representations. My background: I have a basic knowledge of Lie groups, I'm not a master in this subject, while I work with unitary representations (of discrete groups), therefore I am familiar with this last topic. I would be grateful for any advice. Thanks. REPLY [4 votes]: As an introduction I can recommand the following two books: Taylor, Michael Eugene. Noncommutative harmonic analysis. No. 22. American Mathematical Soc., 1986. In this book he discusses the unitary irreducible representations for several explicit examples. Among others there is a chapter on $SL(2,\mathbb{R})$ (Chapter 8) and a subsection on the principle series. Taylor does only assume very little prerequisits such as elementary facts on Lie groups and Lie algebras and he classifies all irreducible unitary representations of $SL(2,\mathbb R)$ and provides concrete realizations of them. Knapp, Anthony W. Representation Theory of Semisimple Groups: An Overview Based on Examples (PMS-36). Princeton university press, 2016. Chapter II--Seciton 5 contains a treatment of the unitary irreducible representations of $SL(2\mathbb R)$. If I remember correctly Knap only gives a list of the unitary irreducible representations and describes their realization and does not discuss the question that those are all possible unitary irreducible representations. However the advantage of this book is that it goes much deeper into the represetation theory of semisimple Lie groups in the subsequent chapters. So if you want to understand $SL(2,\mathbb R)$ as a starting point to understand the representation theory of general semisimple Lie groups, then this book could be very usefull.<|endoftext|> TITLE: Is the fixed point set of a self-diffeomorphism of odd order orientable? QUESTION [8 upvotes]: I've asked this on SE, but got no answer. So I try it here: Let $M$ be a smooth oriented manifold and let $f \colon M \to M$ be a self-diffeomorphism with $f^p = \text{id}_M$ for some odd $p > 0$. Then $f$ is orientation preserving and it follows from the slice theorem that the fixed point set $$ F = \{ x \in M : f(x) = x\} $$ is a smooth submanifold of $M$. Is $F$ orientable? REPLY [10 votes]: I believe this is true. Lemma For $\mathbb{Z}_p$, $p$ odd, all nontrivial irreducible real representations are 2-dimensional. (They really come from a complex representation, forgetting the complex structure.) Lemma For two such real representations, there is a non trivial intertwiner if and only if the weights of the corresponding complex representations agree up to sign. (The corresponding intertwiners come from complex conjugation or the identity.) The normal bundle of a fixpoint set as in your question is a real $\mathbb Z_p$-vector bundle $E$ over each connected component of $M^{\mathbb Z_p}$. It has the following properties: $\mathbb{Z}_p$ acts trivially on the base. $E^{\mathbb{Z}_p}_b=\{0\}$ for all $b$ in the base. The point is now that you can put a complex structure on your bundle. Locally, your bundle splits into a direct sum of non-trivial irreducible $\mathbb{Z}_p$-representations $V_i$. They are all 2-dimensional by the first lemma. On each one of them, we can choose a complex structure. Now, because of the second lemma, we can choose in a way consistent with the transition functions. Using the complex structure, we see that the normal bundle is orientable, and since $M$ was orientable, we get an orientation for the submanifold $M^{\mathbb Z_p}$.<|endoftext|> TITLE: Good references on the theory of harmonic mappings between Riemannian manifolds QUESTION [5 upvotes]: I am currently reading the book by Professor Jost, "Riemannian geometry and geometric analysis", the last chapter on harmonic maps. It talks mainly about existence and regularity of the theory of harmonic mappings between Riemannian manifolds (with certain constraint on the target manifolds, e.g., non-positive curvature). I wonder whether there is a book with a comprehensive treatment on the theory of harmonic mappings between Riemannian manifolds and with applications in other problems. Another related question that I would like to know is how many different methods that we have to prove the existence of harmonic mappings from a Riemannian manifolds to a Riemannian manifolds with non-positive curvature. How about the regularity part? The motivation behind the question is that I would like to study the theory of harmonic mappings between singular metric spaces (e.g. the survey book of Professor Lin Fanghua). I know the earlier works of Korevaar-Schoen and Jost around the 1990s, considering such problems in general metric space setting. Any comments and suggestions are warmly welcome! REPLY [5 votes]: Maybe the following book is useful for you: The analysis of harmonic maps and their heat flows (Fanghua Lin & Changyou Wang)<|endoftext|> TITLE: When the sum of positive definite matrices converges, does the sum of the norm of the associate matrices converges? QUESTION [10 upvotes]: Suppose $A_k>0$ (which means they are positive definitive square $n\times n$-matrices with $n>1$). If $\sum_{k=1}^\infty A_k$ exists, then $\sum_{k=1}^\infty \|A_k\| < +\infty$, Where $\|A\|=\sup_{\|x\|\leq 1}\langle Ax,x\rangle$. Is this true? (I am not able to give any counterexample.) Thank you! REPLY [18 votes]: Yes. The norm of a positive definite matrix does not exceed its trace, and the sum of traces is finite, since the sum of diagonal elements is finite for each of $n$ places. REPLY [11 votes]: You can bound $\|A_k\| \leq C(n)\max_{i,j} |(A_k)_{ij}|$ for some function of the dimension only $C(n)$, because all norms are equivalent in finite dimension. If I am not mistaken $C(n)=\sqrt{n}$, but it doesn't really matter here. This maximum is attained on a (positive) diagonal entry, because of positive definiteness. Then you have $$\sum_k \|A_k\| \leq C(n) \sum_k \max_i (A_k)_{ii} \leq C(n)\sum_k \sum_i (A_k)_{ii} = C(n) \sum_i (\sum_k (A_k)_{ii}),$$ which is finite because $\sum_k (A_k)_{ii}$ is finite for each $i$.<|endoftext|> TITLE: Fourth moments of Gaussian processes QUESTION [9 upvotes]: I am working on a topic outside my main research area, so I am afraid I am reproving obvious results, so I would like to ask for a reference. Google didn't help, mostly because I am looking for formulas. Let $v$ be an $n$-variate Gaussian random variable, with $E[v]=0$ and $E[vv^\top]=I$. If I am not mistaken, it follows (with a bit of work) from Isserlis' theorem that for any $a,b\in\mathbb{R}^n$ $$ E[(a^\top v)^2(b^\top v)^2] = (a^\top a)(b^\top b) + 2 (a^\top b)^2. $$ Q1. Do you have a reference for this identity? Now let us consider a generalization: $v_t$, for $t\in \mathbb{Z}$, is a $n$-variate random process, i.e., each $v_t$ is a vector-valued random variable, and $E[v_t]=0$, $E[v_tv_t^\top]=I$, $E[v_tv_s^\top]=0$ for $s\neq t$. Let $L$ be a shift operator on the sequence, i.e., $Lv_t = v_{t-1}$, and $a(L)$ be a polynomial in it with coefficients $a_i\in\mathbb{R}^n$, i.e., $a(L)v_t = a_0 v_t + a_1v_{t-1} + \dots + a_d v_{t-d}$. Then, I imagine that there should be a result of the kind $$E[(a(L)^\top v)^2(b(L)^\top v)^2] = \text{the coefficient of $L^0$ in } (a(L)^\top a(L^{-1}))(b(L)^\top b(L^{-1})) + 2 (a(L)^\top b(L^{-1}))^2.$$ Q1. Is a formula of this kind true / known? Where can I find references on this topic and this kind of algebra on processes? REPLY [8 votes]: The first equality you mention is a special case of Wick's formula or diagram formula. Suppose that you have a Gaussian random vector $X=(X_1,\dotsc, X_n)$ that is centered, i.e., $\newcommand{\bE}{\mathbb{E}}$ $\newcommand{\bR}{\mathbb{R}}$ $$ \bE[X_i]=0,\;\;\forall i=1,\dotsc, n. $$ A special case of Wick's formula computes $\bE[X_1\dotsc X_n]$. Note that this expectation is $0$ if $n$ is odd so we only need to consider the case $n=2k$. A Feynman graph associated to $\{X_1,\dotsc, X_{2k}\}$ is then a graph with vertex set $\{X_1,\dotsc, X_{2k}\}$ and exactly $k$-edges so no two edges have a vertex in common. If $\newcommand{\be}{\boldsymbol{e}}$ $\be$ is an edge of a Feynman graph with endpoints $X_i, X_j$ then we define its weight to be the covariance $\bE[X_iX_j]$. The weight of a Feynman graph $\Gamma$ is then $$w(\Gamma):=\prod_{\be \in E(\Gamma)}w(\be), $$ where $E(\Gamma)$ denotes the set of edges of $\Gamma$. Wick's formula states that $$ \bE[X_1X_2\dotsc X_{2k}]=\sum_{\Gamma} w(\Gamma) $$ where the summation is carried over the set of Feynman graphs with vertex set $\{X_1,\dotsc, X_{2k}\}$. Here is how it applies to your case. Suppose that $Y=(Y_1,Y_2)$ is a centered Gaussian vector. Then the random vector $$ X=(X_1,X_2,X_3,X_4)=(Y_1,Y_1,Y_2,Y_2) $$ is also a centered random Gaussian vector and $$ \bE[X_1X_2X_3X_4]=\bE[Y_1^2Y_2^2]. $$ In this case there are $3$ Feyman graphs: $$ \Gamma_1,\;\;E(\Gamma_1)= \big\{ (1,2), (3,4)\big\},\;\; w(\Gamma_1)= \bE[Y_1^2]\bE[Y_2^2], $$ $$\Gamma_2,\;\;E(\Gamma_2)=\big\{ (1,3), (2,4)\big\} ,\;\; w(\Gamma_2)=\bE[Y_1Y_2]^2, $$ $$\Gamma_2,\;\;E(\Gamma_2)=\big\{ (1,3), (2,3)\big\} ,\;\; w(\Gamma_3)=w(\Gamma_2)=\bE[Y_1Y_2]^2. $$ Wick's formula then implies $$ \bE[Y_1^2Y_2^2]= \bE[Y_1^2]\bE[Y_2^2]+2\bE[Y_1Y_2]^2. $$ The result you mention in $Q_1$ corresponds to $Y_1= a^\top v$, $Y_2=b^\top v$. The fact that $(Y_1,Y_2)$ is a Gaussian vector follows from the following property of Gaussian vectors: if $V=(V_1,\dotsc, V_n)$ is a Gaussian vector and $A: \bR^n\to\bR^m$ is a linear map, then $AV$ is an $m$-dimensional Gaussian vector. If we choose $A: \bR^n\to \bR^2$, $V\mapsto (Y_1,Y_2)=(a^\top V, b^\top V)$ then you see that $(Y_1,Y_2)$ is Gaussian. For the generalization you mention, there are plenty to say if the process is Gaussian. For details see Sections 1 and 2 of these notes. For many more details see Janson's book Gaussian Hilbert Spaces<|endoftext|> TITLE: Long-time decay of heat kernel on compact manifolds QUESTION [8 upvotes]: Let $M$ be a compact Riemannian manifold and let $V \in C^\infty(M)$. Consider the operator $\Delta + V$ and let $p_t(x, y)$ be the corresponding heat kernel. If $\Delta + V$ is a positive operator such that the first eigenvalue is larger than $\varepsilon >0$, we know that the heat kernel satisfies the estimate $$ p_t(x, y) \leq C e^{-t\varepsilon}$$ for some constant $C>0$ and all $t>1$. Question: Can I make the constant $C$ depend only on the volume and the dimension of $M$? That is, does there exist a constant $C = C(n, v)$ such that for all $n$-dimensional Riemannian manifolds $M$ with fixed volume $v$ and all functions $V \in C^\infty(M)$ such that $\Delta+V$ is bounded below by $\varepsilon>0$, the corresponding heat kernel satisfies the above bound? Note that without restriction on the volume, this cannot work, since the heat kernel for $V=0$ converges to the function $1/\sqrt{v}$, where $v$ is the volume of $M$. \Edit: if the answer to the question is no, what are "good" (necessary or sufficient) geometric assumptions that assure such a bound? In other words, what other explicit geometric data should $C$ depend on? REPLY [6 votes]: The paper of Carlen , Kusuoka and Stroock relates Nash inequality to diagonal bounds for the Heat Kernel for both short times and all time.They also explain a method of E B Davies to convert diagonal bounds of the heat kernel to off diagonal bounds.The paper is in Annales Institut Henri Poincare Probabilites et Statistiques 23 no S2 1987 pages 245-287<|endoftext|> TITLE: a sum with binomial coefficients QUESTION [19 upvotes]: Let integers $n,k$ satisfy $0 \le k \le n$. We desire proof that $$ {n\choose k} = \sum {n\choose a}(-1)^a\;{-k\choose b}(-1)^b\;{-(n-k)\choose c}(-1)^c \tag{$*$}$$ where the (finite) sum is over all ordered triples $(a,b,c)$ of nonnegative integers satisfying $$ a\cdot n + b\cdot k + c\cdot(n-k) = k(n-k) \\ \text{(or equivalently}\quad \frac{a+b}{n-k} + \frac{a+c}{k} = 1 \quad). $$ Of course the binomial coefficient for a binomial to a negative power is $$ {-k\choose b} = \frac{(-k)(-k-1)(-k-2)\cdots(-k-b+1)}{b!} . $$ Now two of the terms in ($*$) are $(a,b,c)=(0,n-k,0)$ and $(a,b,c)=(0,0,k)$. Those two terms do, indeed, add to ${n\choose k}$. But then the problem becomes: show that the rest of the terms sum to zero. Note This arose from my attempt to solve integral of a "sin-omial" coefficients=binomial . It is the calculation of the residue at $w=0$ of rational function $$ \frac{(w^n-w^{-n})^n}{(w^k-w^{-k})^k(w^{n-k}-w^{-(n-k)})^{n-k}w} $$ obtained by changing variables in the integral of that problem. REPLY [2 votes]: This is an elementary/self-contained proof based on manipulation of binomials. Let $l=n-k$ and $g=\gcd(n,k)=\gcd(k,l)$. The condition $an+bk+cl=kl$ (where $a$, $b$, and $c$ are as in the original problem description) is equivalent to $(a+c)n=k(l+c-b)$, which means $k/g\mid a+c$, so $a+c=tk/g$, $a+b=l(1-t/g)$, and the only possible values of $t$ are $0,1,\ldots,g$. $t=0$ and $t=g$ correspond to $(a,b,c)=(0,l,0)$ and $(a,b,c)=(0,0,k)$ respectively, which provide ${n\choose k}$ as mentioned in the problem description (by an easy binomial sum). The problem is to show that $t=1,\ldots,g-1$ contribute zero in total to the sum (obvious if $n$ and $k$ are coprime). Helpfully it turns out that the contributions from each value of $t$ are separately zero, not just in aggregate, and the WZ method can be applied to each $t$ as follows. To make the notation slightly prettier, define $\tau=t/g$ and $\tau'=1-t/g$. (There is then a symmetry $k\leftrightarrow l$, $\tau\leftrightarrow\tau'$.) $\tau$ and $\tau'$ will only ever occur multiplied by $k$, $l$ or $n$, so everything is an integer. What needs to be proved is, for fixed $t$ between $1$ and $g-1$, $$\sum_{a=0}^{\min(k\tau,l\tau')} (-1)^a {n\choose a}{-k\choose l\tau'-a}{-l\choose k\tau-a}=0.$$ This is established by the explicit sum, valid for $0<\tau<1$: $$\sum_{a=0}^{r-1} (-1)^a {n\choose a}{-k\choose l\tau'-a}{-l\choose k\tau-a}= (-1)^{r+1} {n\choose r}{-k\choose l\tau'-r}{-l\choose k\tau-r} \frac{r(l+k\tau-r)(k+l\tau'-r)}{kln\tau\tau'},$$ which vanishes for $r>\min(k\tau,l\tau')$. It is routine to check that the delta of the RHS is the LHS.<|endoftext|> TITLE: Analogies between classical geometry on complex surfaces and Arakelov geometry QUESTION [27 upvotes]: This is my first question on this wonderful site. The following question about Arakelov geometry is gonna be quite long and wide; to be clear one of that kind of questions that are usually ignored. The point is that I haven't found any help in the literature and I feel lost. Let's fix an arithmetic surface $p:X\to \operatorname{Spec }O_K$ ($K$ is a number field). The usual concept of invertible sheaf/line bundle on a complex surface, is substituted by the notion of metrized line bundle. In particular a metrized line bundle $\overline{\mathcal L}$ is a couple $(\mathcal L, \{h_\sigma\})$ where $\{h_\sigma\}$ is a finite collection of "admissible" metrics on the pullback holomorphic line bundles $\mathcal L_\sigma$ on the Riemann surfaces at infinity $X_\sigma$. This is very cool since we have a reasonable intersection theory of metrized line bundles which employs the Deligne pairing, in paricular to each couple of metrized line bundles $\overline{\mathcal L}$, $\overline{\mathcal M}$ we associate a metrized line bundle $\overline E=\left<\overline{\mathcal L},\overline{\mathcal M}\right>$ on $\operatorname{Spec } O_K$. Again $\overline E$ is a couple made of a projective $O_K$-module of rank $1$ $E$ and a collection of hermitian inner products on $\mathbb C$, one for each embedding of $O_K$ in $\mathbb C$ (up to conjugation). The intersection pairing can be clearly interpreted as pairing on the group of Arakelov divisors which like in the geometric case correspond exactly to metrized line bundles. Now let's go to the questions: In the usual theory of complex surfaces we have crucial "objects" like the complex vector space $H^0(\mathcal L)$ and the Euler-Poincare chacteristic $\chi$. Let $\overline {\mathcal L}$ be a metrized line bundle on $X$. What is a reasonable notion of $H^0(\overline{\mathcal L})$? It seems that a good choice should be the subspace of $H^0(\mathcal L)$ made of the global sections $s$ such that $||s_\sigma||\le 1$ for any $\sigma$. These are also called small sections. This definition seems very odd to me, indeed if we interpret this in term of Arakelov divisors and try to create $H^0(\overline D)$ for an Arakelov divisor $\overline D$ it seems that we don't care about what happens at the Archimedean places of $\overline D$ because we only consider effective coefficients at infinity. To be precise: if I modify $\overline D$ only in the archimedian part without changing the sign of the coefficients, I'll get the same set $H^0(\overline D)$. This quite useless! Where is my mistake? What is a better definition of $H^0(\overline{\mathcal L})$ (if there is one)? An arithmetic version of $\chi$ can be introduced by means of the determinant of the cohomology. The formula says: $$\chi(\overline{\mathcal L})=\operatorname{deg }\left(\operatorname{det}Rp_\ast{\mathcal L}\right)\,.$$ $ \operatorname{det}R{p_\ast}{\mathcal L}$ is a honest a metrized line bundle on $\operatorname{Spec }O_K$ and a theorem due to Arakelov says that it can be endowed with a canonical metric. At this point we take the (arithmetic) degree and we get our arithmetic $\chi$. Very well, but why do we have such a complicate definition? The usual $\chi$ is a cohomological object but in the arithmetic case we don't have any suitable cohomology of metrized line bundles. Where is the analogy with the geometry? It seems that the only reason is to get a form of the Riemann-Roch theorem which at least formally resembles the classical one. Thank you for you attention. REPLY [12 votes]: These are indeed good questions, and while there is a very good corpus of answers to them, the analogy is not perfect. 0. The non-archimedean analogy First of all, I would like to go back to the relative situation of a surface $\mathcal X\to B$ fibered over a germ of curve $(B,b)$. Then any local function $f$, resp. local section $s$ of a line bundle $\mathcal L$, may vanish along the special fiber with some multiplicity $m$, and this multiplicity is a valuation: the function $f\mapsto m=v(f)$ satisfies $v(f+g)\geq \min(v(f),v(g))$ and $v(fg)=v(f)+v(g)$. The function $f\mapsto \exp(-v(f))$ behaves as a metric, except that it is non-archimedean. Now forget the surface $\mathcal X$ and just remember about the generic fiber, viewed as a curve over the complete valued field $F_b$, completion of the field of functions of $B$ with respect to the $b$-adic absolute value. This furnishes a functor from pairs consisting of a surface $\mathcal X$ over $(B,b)$, line bundles on $\mathcal X$, to pairs consisting of a curve $X$ over $F_b$ and a metrized line bundle on $X$. This extends to vector bundles, in fact. Under this functor, morphisms of vector bundles go to norm non-increasing morphisms of vector bundles. This functor is “essentially” fully faithful (one needs some assumptions on $\mathcal X$, say it is normal). Up to blowing-up $\mathcal X$ along closed subschemes of the special fiber, it is “essentially” essentially surjective. 1. What is $H^0(\overline{\mathcal L})$ ? The answer, which mimicks the above analogy, says that it is the subset of $H^0(\mathcal L)$ of sections of norm $\leq 1$ everywhere. It is a finite set with essentially no algebraic structure. There are two propositions for its “dimension”. The earliest one sets $h^0(\mathcal L)=\log \# (H^0(\overline{\mathcal L}))$. More recently, van der Geer, Schoof, Rössler, Bost, etc. have suggested to consider rather the theta-invariant: $$ h^0_\theta(\mathcal L)= \log \left(\sum_{v\in H^0(\mathcal L)} \exp(-\| v\|^2) \right). $$ To answer one concern expressed in your statement of question 1. If you add $m_\sigma$ to the real component at $\sigma$, this multiplies the metric at $\sigma$ by $\exp(-m_\sigma)$. If $m_\sigma>0$, there will be more global sections, in good analogy with the fact that if you add an effective divisor to a divisor, the space of global sections increases. 2. Arithmetic degrees In any case, rather than this specific $H^0(\overline{\mathcal L})$, it is rather convenient to remember the pair consisting of the free $\mathbf Z$-module $H^0(\mathcal L)$ and of its norm induced by the supremum norm of sections. (One can also introduce useful Euclidean norms by integrating local norms squared agains a fixed volume form; there are comparison results.) This object is the analogue of a vector bundle over a curve, and can be given an arithmetic degree, satisfying good algebraic properties. Minkowski's theorem is the analogue of the Riemann-(Roch) theorem giving non-zero sections in $H^0(\overline{\mathcal L})$ provided its arithmetic degree is large enough. There is an analogue of Serre's duality theorem. Actually, there are two analogues, according to your definition of $H^0$. In the naïve one, it is given by an inequality (Gillet-Soulé, Israel J. Math.). In the theta-version, it comes from the Poisson summation formula and is an exact equality. 3. Arithmetic intersection theory There I touch your second question. When $\mathcal X$ is a projective surface, there is a nice intersection pairing of line bundles. It can be defined geometrically, intersecting divisors. It can also be defined cohomologically, using Euler-Poincaré functions. (See, for example, the first chapter of Debarre's book on Higher dimensional geometry, for a rapid and clear treatment.) The same can be done in Arakelov geometry. For arithmetic surfaces (the scope of your question), it is due to Arakelov, Faltings and Deligne; in general, this is due to Gillet-Soulé and Bismut-Gillet-Soulé. These authors define an arithmetic intersection of metrized line bundles, more generally of “arithmetic cycles”, and prove an arithmetic analogue of Grothendieck's Riemann-Roch theorem. The difficulty with the mere statement of a Riemann-Roch theorem comes from the fact that the naive $H^0$, as we saw, are not algebraic objects, and that there is even no a priori definition of the higher cohomology groups. The idea is then to define the determinant of the cohomology, as a metrized line bundle on the base, whose arithmetic degree would be the analogue of the Euler-Poincaré characteristic. 4. Applications I cannot resisting mentioning a bunch of applications. a) This theory fits extremly well, and makes precise, Weil's theory of heights. In Weil's theory, height functions are defined up to bounded ambiguities; here, one has actual functions under the hand. b) Gillet-Soulé proved an analogue of the Hilbert-Samuel theorem, giving an estimation for the cardinality of $H^0(\overline{\mathcal L}^n)$, where $n\to\infty$. This theorem has been extremly useful, for example in the proof of Bogomolov's conjecture. c) Zhang proved an analogue of the Nakai-Moishezon theorems, that is, criteria on $\overline{\mathcal L}$ implying that large powers will be generated by sections of norm $<1$. (Basically, the height of every closed subvariety has to be $>0$.) d) The precise formulas coming out of the arithmetic Riemann-Roch theorem furnish beautiful arithmetic formulas relating heights and special values of $L$-functions. (See the work of Maillot-Rössler, for example.) e) Bost proved analogues of Lefschetz's hyperplane section theorem on Riemann surfaces, implying for example the triviality of the fundamental groups of some arithmetic surfaces. f) Bost and myself proved generalizations of the Borel-Dwork criterion, viewed as analogues of the Hironaka-Matsumura theory (algebraization of formal schemes). g) The formalism is also very useful to formulate in a geometric way the proofs of transcendental number theory (Bost's slope method). It has been used, for example, by Bost-David (explicit version of Masser-Wüstholz isogeny estimates), Bost (generalization of Chudnovsky/André's results about Grothendieck's conjecture to foliations), Gasbarri and Herblot (Schneider-Lang type theorems), etc.<|endoftext|> TITLE: Conditions for a functor to induce a logical functor between presheaf toposes? QUESTION [15 upvotes]: Let $F\colon\mathcal{C}\to\mathcal{D}$ be a functor between small categories. Question: Under what conditions is the induced functor $$F^*\colon\mathsf{Set}^\mathcal{D}\to\mathsf{Set}^\mathcal{C}$$ a logical functor between presheaf toposes? (I prefer to avoid contravariance if possible, so the "presheaf toposes" I'm referring to here are $\mathrm{Psh}(\mathcal{C}^{op})=\mathsf{Set}^\mathcal{C}$ and $\mathrm{Psh}(\mathcal{D}^{op})=\mathsf{Set}^\mathcal{D}$.) REPLY [4 votes]: As Mike Shulman pointed out, this is the same as requiring the geometric morphism induced by $f$ to be atomic. In Atomic toposes 7.2 you'll find that in the case of $! = F \colon C \to 1$, $F^*$ is logical iff $C$ is a groupoid. What about the general case? Not an answer in any way, but two (UPDATE 2. went nowhere, added 3.) possible paths: Atomic morphisms are locally connected (Elephant C3.5). What do we know about functors for which the induced geometric morphism is locally connected? looking at the Elephant again, Lemma C3.3.5 which gives a sufficient condition, and On functors which are lax epimorphisms, which characterizes connected ones, one is tempted to conjecture that the induced geometric morphism is locally connected iff it is absolutely dense into its Cauchy-image, or something like that. In the spirit of Simon Henry comment above, and given the characterization of atomic sites in Atomic toposes 7.3, or C3.5.8 in the Elephant, maybe it is enough for $F$ to be a fibration in categories where (i) every morphism is an effective epi and (ii) every pair of morphisms with common codomain can be completed to a commutative square (right Ore condition); note that a groupoid obviously satisfies these two conditions. UPDATE not the case, see comment below Look at its hyperconnected-localic factorization, which in this case coincides with the connected-light factorization (or "comprehensive") of a locally-connected morphism. For essential geometric morphisms between presheaf categories, this yields the comprehensive factorization of a functor into a final functor followed by a discrete fibration; Factorization theorems for geometric morphisms II is a reference (at the end of the paper). Now, a geometric morphism is atomic iff both halves of this factorization are atomic (Elephant C3.5.4), so we have reduced it to when both the final and the discrete fibration components induce an atomic morphism.<|endoftext|> TITLE: $n$ such that decimal digits of $2^n$ begins with $n$ QUESTION [9 upvotes]: Are there infinitely many $n$ such that the decimal expansion $2^n$ begins with $n$? For example, $2^6=64$ and $2^{10} =1024$. It can easily be shown that this problem is equivalent to the following. Are there infinitely many $n$ such that $\{n\log_{10}2-\log_{10}n\}<\log_{10}{(1+\frac{1}{n})}$? Here, $\{x\}=x-[x]$, where [x] denotes the largest integral number that is smaller than $x$. More generally, for $\alpha,\beta,\gamma > 0$, are there infinitely many $n$ such that $\{\alpha n - \beta \log n\}<\frac{\gamma}{n}$? I think it would help if I know something about the distribution of $\{\log n\}$. REPLY [5 votes]: Some general information on the problem, which is probably open. As pointed out by Ivan Neretin at math.stackexchange, this is OEIS A100129. There you can find the first 16 numbers with this property. According to this paper, Jan van de Lune, "A note on a problem of Erdős" (1978) the problem goes back to Erdős: "Recently my attention was drawn to the following observation made by P. Erdős: $2^6=64$ and $2^{10}=1024$. Here we have two examples of the phenomenon that the number $2^n$ starts with the same ordered sequence of digits as the natural number $n$ itself." The characterization used in van de Lune's calculations is $$0\leq \{n\log 2\}-\{\log n\}<\log(n+1)-\log n$$<|endoftext|> TITLE: Does isometric immersion map boundary to boundary? QUESTION [26 upvotes]: Let $M$ be a compact, connected, oriented, smooth Riemannian manifold with non-empty boundary. Let $f:M \to M$ be a smooth orientation preserving isometric immersion. Is it true that $f(\partial M) \subseteq \partial M$? Edit: In this paper, it is proved that any "local isometry" from a compact, connected metric space into itself is a homeomoprhism, where by a "local isometry", they mean "locally preserving distance". Thus, if every isometric immersion is locally preserving distance, then our $f$ will be a homeomorphism (hence will map boundary to boundary). Is it true that isometric immersion $\Rightarrow$ locally preserving distance? Motivation: A positive answer to this question would imply that every smooth orientation preserving isometric imersion $M \to M$ is a Riemannian isometry (see details below**), Moreover, if two manifolds can be isometrically immersed in each other, then they are isometric (see remark 1 in the "updates and remarks" of this question). In particular, to refute the conjecture "$f(\partial M) \subseteq \partial M$" it is enough to find such an immersion which is not surjective, or at least not an isometry. Remarks: (1) Compactness is essential. Look at $M=[0,\infty) \,,\,f(x)=x+1$. (2) The claim clearly holds in dimension $1$ ($M$ must be a closed interval). ** Indeed, since $f(\partial M) \subseteq \partial M$ and it's easy to see that $f(M^o) \subseteq M^o$ we get that $f(M^o)$ is clopen in $M^o$, thus $f(M^o)=M^o$. Since $f(M)$ is closed in $M$, and contains $M^o$, we conclude $f(M)=M$, and so $f(\partial M)= \partial M,f(M^o)=M^o$. So, $f$ is a surjective $1$-Lipschitz map from a compact space to itself. (The $1$-Lipshictzity follows since isometric immersions preserve lengths of paths). Thus, it is a metric isometry. Hence, by the positive answer to this question, $f$ is smooth, and in fact a Riemannian isometry. REPLY [8 votes]: The following relies on the answer of Anton Petrunin and his explanations. Set $X:=\cap_{n\geq 1} f^n(M)$. Claim 0: Each $x\in X$ is a limit point of a sequence $f^n(y)$ for some $y\in M$. Moreover, if $x\in X\cap M^\circ$, then $y$ can be chosen in $M^\circ$. Proof of Claim 0: Assume $x\in X$. Then $x=f^n(y_n)$ for some $y_n\in M$ and each $n\geq 1$. After passing to a subsequence, we may assume that $y_{n_l}\rightarrow y$. Fix $\varepsilon>0$ and $K\geq 1$; we show that for some $k\geq K$, $f^k(y)\in B_\varepsilon(x)$. Choose $l\geq 1$, such that $k:=n_l\geq K$ and $y_{k}\in B_\varepsilon(y)$. Then, since $f$ is distance non-increasing, $dist(f^k(y),f^k(y_k))\leq dist(y,y_k)<\varepsilon$. The 'moreover' statement follows, since we can replace $y$ by $f^{k_0}(y)$, and choose $k_0$ such that $f^{k_0}(y)\in M^\circ$; the latter is possible, since a subsequence of $f^n(y)$ converges to $x\in M^\circ$. Claim 1: For any $x\in M^\circ$ and $r>0$ such that $B_r(x)\subset M^\circ$ and $n\geq 1$, we have $f^n(B_r(x))=B_r(f^n(x))$. We defer a proof of Claim 1 to the end. Claim 2: $X^\circ$ is nonempty, indeed it coincides with the set of all limit points of sequences $f^n(y)$ for $y\in M^\circ$. Proof Claim 2: By Claim 0, each $x\in X\cap M^\circ \supset X^\circ$ is such a limit point. Conversely, if $x$ is such a limit point, then $x\in B_r(f^k(y))=f^k(B_r(y))\subset f^k(M)$ for (certain) arbitrary large $k$; here we choose $r$ such that $B_r(y)\in M^\circ$ and use Claim 1. Therefore $x\in X=\cap_{n\geq 1} f^n(M)$, since the intersection is decreasing. Actually the same argument shows, that $B_s(x)\subset X$ for each $s TITLE: Discontinuity of solutions to approximation schemes in the Barles-Souganidis framework QUESTION [7 upvotes]: I have attached pg. 275 and pg. 276 of [BS91]. My concern is with the claim (2.7) on pg. 276. To prove this claim, I require the following additional assumption, which is not made by the authors: Assumption: $u^{\rho}$ is upper semicontinuous for each $\rho$. Remark: In the supersolution case, the analogue of the above assumption is that $u^\rho$ is lower semicontinuous for each $\rho$. My proof of (2.7) under the additional assumption above: Let $(\rho_n,x_n)_n$ be a sequence with $\rho_n\rightarrow 0$, $x_n\rightarrow x_0$, and $u^{\rho_n}(x_n)\rightarrow \overline{u}(x_0)$. Then, by the assumption above, for each $n$, there exists $y_n$ such that $$(u^{\rho_n}-\varphi)(y_n) = \sup_{ \overline{B(x_0;r)} } \{ u^{\rho_n} - \varphi \}.$$ Extract a subsequence of $(\rho_n,x_n,y_n)_n$ converging to some $(0,x_0,\hat{y})$ and relabel it, with a slight abuse of notation, $(\rho_n,x_n,y_n)_n$. It follows that \begin{multline*} 0=(\bar{u}-\varphi)(x_{0})=\lim_{n\rightarrow\infty}(u^{\rho_{n}}-\varphi)(x_{n})\leq\limsup_{n\rightarrow\infty} (u^{\rho_{n}}-\varphi)(y_{n}) \\ \leq\limsup_{\substack{\rho\rightarrow0\\ y\rightarrow \hat{y} } }(u^{\rho}-\varphi)(y)=(\overline{u}-\varphi)(\hat{y}). \end{multline*} Because it was assumed that $x_0$ is a strict local maximum of $\overline{u}-\varphi$, we get $\hat{y}=x_0$. This proves (2.7), with $(\rho_n,y_n)$ being the desired sequence. My problem: I cannot see how to remove the assumption. While the assumption can be relaxed, as far as I can tell, we need to impose something to ensure that the maximum is attained in the proof below. Am I mistaken? [BS91] Barles, Guy, and Panagiotis E. Souganidis. "Convergence of approximation schemes for fully nonlinear second order equations." Asymptotic analysis 4.3 (1991): 271-283. REPLY [2 votes]: After checking the arguments with @Jeff, I am willing to conclude that the issue in the question above is an extremely minor error in the original paper [BS91]. Both Jeff and I came up with "fixes" for the case of $u^{\rho}$ not continuous. Both are detailed below. Update: I had the chance to speak to Professor Barles about this, who was kind enough to suggest a way to improve my fix. The result is below. Parsiad and Professor Barles' fix Redefine the notion of an approximation scheme as a pair of functions $$\underline{S},\overline{S}:(0,\infty)\times\overline{\Omega}\times\mathbb{R}\times B(\overline{\Omega})\rightarrow\mathbb{R}.$$ For fixed $\rho$, a solution of this approximation scheme is a bounded function $u^{\rho}:\overline{\Omega}\rightarrow\mathbb{R}$ such that $$ \underline{S}(\rho,x,(u^{\rho})^{*}(x),(u^{\rho})^{*})\leq0\leq \overline{S}(\rho,x,(u^{\rho})_{*}(x),(u^{\rho})_{*})\text{ for all }x\in\overline{\Omega}. $$ Remark: Intuitively, $(u^{\rho})^*$ and $(u^{\rho})_*$ are simply upper and lower semicontinuous envelopes of the solution $u^{\rho}$ of an approximation scheme in the sense of [BS91]. Monotonicity is extended in the obvious way: an approximation scheme $(\underline{S},\overline{S})$ is monotone if for each $S\in\{\underline{S},\overline{S}\}$, \begin{gathered} S(\rho,x,t,u)\leq S(\rho,x,t,v)\text{ if }u\geq v\\ \text{ for all }\rho>0\text{, }x\in\overline{\Omega}\text{, }t\in\mathbb{R}\text{, and }u,v\in B(\overline{\Omega}). \end{gathered} Stability is also extended in the obvious way: an approximation scheme $(\underline{S},\overline{S})$ is stable if \begin{gathered} \text{for all }\rho>0\text{, there exists a solution }u^{\rho}\in B(\overline{\Omega})\text{,}\\ \text{with }u^{\rho}\text{ bounded independent of }\rho. \end{gathered} The consistency requirement is modified slightly. In particular, we require that $\overline{S}$ (resp. $\underline{S}$) only satisfy (2.4a) (resp. (2.4b)) of [BS91]. That is, an approximation scheme $(\underline{S},\overline{S})$ is consistent if for all $x\in\overline{\Omega}$ and $\varphi\in C_{b}^{\infty}(\overline{\Omega})$, \begin{align*} \limsup_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{\overline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \leq F^{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x)\\ \text{and }\liminf_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{\underline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \geq F_{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x). \end{align*} Proposition: Let $(\underline{S},\overline{S})$ be an approximation scheme that is monotone, stable, and consistent (in the sense above). Assume $F=0$ satisfies a strong comparison principle (i.e., (2.5) of [BS91]). Then, as $\rho\rightarrow0$, the solution $u^{\rho}$ of the scheme converges locally uniformly to the unique viscosity solution of $F=0$ (in the sense of [BS91]). Proof: With the setup above, the proof is essentially identical to that of [BS91] (and avoids the issue of being unable to take extrema). Jeff's fix Proposition: Theorem 2.1 of [BS91] holds if we replace (2.4) of [BS91] by the following modified consistency requirement \begin{align*} \limsup_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{S(\rho,y,\varphi(y)+\xi+e^{-1/\rho},\varphi+\xi)}{\rho} & \leq F^{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4a}\text{)}\phantom{.}\\ \text{and }\liminf_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{S(\rho,y,\varphi(y)+\xi-e^{-1/\rho},\varphi+\xi)}{\rho} & \geq F_{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4b}\text{)}. \end{align*} Proof: While the claim (2.7) cannot be proved, we can instead establish the existence of a sequence $(\rho_{n},y_{n})$ such that $$ \rho_{n}\rightarrow0\text{, }y_{n}\rightarrow x_{0}\text{, and }(u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\geq(u^{\rho_{n}}-\varphi)(\cdot). $$ To see this, let $(\rho_{n},x_{n})_{n}$ be a sequence with $\rho_{n}\rightarrow0$, $x_{n}\rightarrow x_{0}$, and $u^{\rho_{n}}(x_{n})\rightarrow\overline{u}(x_{0})$. Then, for each $n$, there exists $y_{n}$ such that $$ (u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\geq\sup_{\overline{B(x_{0};r)}}\left\{ u^{\rho_{n}}-\varphi\right\} . $$ Extract a subsequence of $(\rho_{n},x_{n},y_{n})_{n}$ converging to some $(0,x_{0},\hat{y})$ and relabel it, with a slight abuse of notation, $(\rho_{n},x_{n},y_{n})_{n}$. It follows that \begin{multline*} 0=(\overline{u}-\varphi)(x_{0})=\lim_{n\rightarrow\infty}(u^{\rho_{n}}-\varphi)(x_{n})\leq\limsup_{n\rightarrow\infty}\left\{ (u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\right\} \\ \leq\limsup_{\substack{\rho\rightarrow0\\ y\rightarrow\hat{y} } }\left\{ (u^{\rho}-\varphi)(y)+e^{-1/\rho}\right\} =\limsup_{\substack{\rho\rightarrow0\\ y\rightarrow\hat{y} } }(u^{\rho}-\varphi)(y)=(\overline{u}-\varphi)(\hat{y}). \end{multline*} Because it was assumed that $x_{0}$ is a strict local maximum of $\overline{u}-\varphi$, we get $\hat{y}=x_{0}$, as desired. Now, the remainder of the proof in [BS91] is modified as follows. Let $\xi_{n}=(u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}$. Then, $u^{\rho_{n}}\leq\varphi+\xi_{n}$ and hence $$ 0=S(\rho_{n},y_{n},u^{\rho_{n}}(y_{n}),u^{\rho_{n}})\geq S(\rho_{n},y_{n},\varphi(y_{n})+\xi_{n}-e^{-1/\rho_{n}},\varphi+\xi_{n}). $$ The desired result now follows by dividing by $\rho$, taking limit inferiors, and applying the consistency condition. $\square$ Remark: the exponential decay is chosen as it does not cause any problems in creating approximation schemes, practically speaking. For example, consider the approximation $$\frac{u^\rho(x+\rho) - u^\rho(x)}{\rho}.$$ If we substitute the corresponding entities in the revised consistency requirement, we get $$\frac{\varphi(x+\rho) - \varphi(x) \pm e^{-1/\rho}}{\rho} = \varphi^\prime(x) + \mathcal{O}(\rho)$$ by a Taylor approximation.<|endoftext|> TITLE: How similar are large cardinals, over $L$? QUESTION [7 upvotes]: EDIT: Joel's answer shows that no $\Sigma_2$ large cardinal property will do the job - however, $\Pi_2$ properties (such as unfoldability and its relatives) may still be useful. Throughout this question, I'm working in "$V=L$". So when I say "large cardinal," I mean "small large cardinal". Informally, I want to ask: How much of the "$L$-theory" of a cardinal does a large cardinal property imply? That is, if I know that $\alpha$ and $\beta$ both have some large cardinal property $(*)$, what sorts of (not necessarily first-order) sentences do I know that $L_\alpha$ and $L_\beta$ agree about? Formally, fix an ordinal $\gamma$. Say that a large cardinal property $(*)$ is "$\gamma$-decisive" if whenever $\alpha, \beta$ have property $(*)$, we have $$L_{\alpha+\gamma}\equiv L_{\beta+\gamma}.$$ My question is: What are some examples of $\gamma$-decisive large cardinal properties? (CAVEAT: Obviously any large cardinal property can be arbitrarily decisive: suppose it has no instances, or exactly one! So really I want to know about large cardinal properties whose decisiveness can be proved in ZFC+V=L, or some strengthening thereof that does not limit the number of relevant large cardinals.) Note that decisiveness looks a lot (at least to me) like indescribability; however, indescribability doesn't quite seem to do the job, unless I'm missing something. I'm also interested in decisiveness with parameters - e.g. for a finite tuple of sets $\overline{c}\in L$, say that a large cardinal property $(*)$ is $\gamma$-decisive over $\overline{c}$ if whenever $\alpha,\beta$ have property $(*)$ with $\overline{c}\in L_\alpha\cap L_\beta$, we have $$(L_\alpha, \overline{c})\equiv(L_\beta,\overline{c}).$$ But understanding the parameter-free version should come first, and besides, I imagine the situations aren't too different. By contrast, something potentially interesting happens if we allow, not tuples of elements of $L$, but tuples $\overline{C}$ of subsets of $L$ which are in $V$ but not necessarily $L$ - coded as unary predicates, of course. A really overkill example of this would be the $0^\#$ as a unary predicate, but already tamer examples could be interesting. For instance, suppose $\overline{C}$ is a finite sequence of mutually Cohen reals over $L$ - then what are some $\gamma$-decisive over $\overline{C}$ large cardinal properties? But this question seems much broader, and much harder, so I'll merely mention it as a curiosity. (Note that in this context, detrivializing things becomes a bit trickier, so this is a more informal question.) REPLY [6 votes]: Updated answer. I claim that none of the familiar large cardinal notions consistent with $V=L$ are provably $\gamma$-decisive for any $\gamma$. This includes the cases of wordly cardinals, inaccessible cardinals, uplifting cardinals, Mahlo cardinals weakly compact cardinals, $\Pi^n_m$-indescribable cardinals, totally indescribable cardinals, unfoldable cardinals and all the others. First, as a warm-up, let's handle the case of $\Sigma_2$-definable large cardinal notions. Theorem. If a large cardinal notion LC is $\Sigma_2$-definable and there are a proper class of them in $L$, then there is no $\gamma$ for which they are $\gamma$-decisive. Proof. To see this, fix any $\gamma$, and let $\kappa$ be the least LC cardinal above $\gamma$. Note that both $\kappa$ and $\gamma$ are definable in $L_{\kappa+\gamma}$, since $\kappa$ is the largest cardinal of this structure, and $\gamma$ is least such that $\kappa+\gamma$ does not exist. Further, since $\kappa$ was the least LC cardinal above $\gamma$, it follows that $L_{\kappa+\gamma}$ will agree that there are no such large cardinals of that type in the interval $(\gamma,\kappa)$, since $\Sigma_2$ definitions express locally verifiable properties (see my blog post about Local properties in set theory). But the corresponding fact will not be true in $L_{\delta+\gamma}$ for some other much larger LC cardinal $\delta$, since once $\delta$ is large enough, then $L_{\delta+\gamma}$ will think that $\kappa$ is an LC cardinal. So $L_{\kappa+\gamma}$ is not elementarily equivalent to all $L_{\delta+\gamma}$, and so the LC large cardinal notion is not $\gamma$-decisive in this situation. QED Next, with a slightly stronger assumption on consistency, we can handle the $\Pi_2$-definable notions. Theorem. Assume that a large cardinal notion LC is $\Pi_2$-definable and there are a sufficiently stationary proper class of them in $L$, then there is no $\gamma$ for which they are $\gamma$-decisive. Proof. Fix any $\gamma$, and assume that the class of LC cardinals in $L$ meets every $\Pi_2$-definable class club. Let $C$ be the club of $\Sigma_2$-correct cardinals $\delta$, those for which $V_\delta\prec_{\Sigma_2} V$. This is $\Pi_2$-definable. Let $\kappa$ be the least LC cardinal in $C$ above $\gamma$. Note that both $\kappa$ and $\gamma$ are definable in $L_{\kappa+\gamma}$. Further, since $\kappa$ is $\Sigma_2$-correct, it must be a beth-fixed-point, and so $L_\kappa=(V_\kappa)^L$. Since $L_\kappa$ is $\Sigma_2$-correct and hence also $\Pi_2$-correct, it is correct about the LC cardinals below $\kappa$. And it is also correct about the $\Sigma_2$-cardinals below $\kappa$. So by the minimality of $\kappa$, the structure $L_\kappa$ can see that there is no $\Sigma_2$-correct LC cardinal below $\kappa$ and above $\gamma$, and this is part of the theory of $L_{\kappa+\gamma}$. But now, if $\delta$ is a larger LC cardinal in $C$, then $L_{\delta+\gamma}$ will be able to see that there is a $\Sigma_2$-correct LC cardinal below $\delta$. And so $L_{\kappa+\gamma}$ and $L_{\delta+\gamma}$ do not have the same theory. So the LC cardinals are not $\gamma$-decisive. QED Finally, we can use this idea to push through an argument for any first-order definable large cardinal notion. Theorem. No first-order definable large cardinal notion LC is provably $\gamma$-decisive for any $\gamma$, if it is consistent that there is a stationary proper class of LC cardinals in $L$. Proof. Suppose that the LC large cardinal notion is $\Sigma_n$-definable, and the class of LC cardinals in $L$ is stationary with respect to all $\Pi_n$-definable class clubs. Fix any $\gamma$, and let $\kappa$ be the least $\Sigma_n$-correct LC cardinal above $\gamma$. There is such a $\kappa$ because the class of $\Sigma_n$-correct ordinals is a $\Pi_n$-definable class club. Both $\kappa$ and $\gamma$ are definable in $L_{\kappa+\gamma}$, and $L_\kappa$ is correct about LC cardinals and about $\Sigma_n$-correctness. So $L_\kappa$ will see no $\Sigma_n$-correct LC cardinals above $\gamma$. But if we use a much larger $\Sigma_n$-correct LC cardinal $\delta$, then $L_\delta$ will see that $\kappa$ is a $\Sigma_n$-correct LC cardinal above $\gamma$. So $L_{\kappa+\gamma}$ and $L_{\delta+\gamma}$ will have different theories. QED<|endoftext|> TITLE: Singular cohomology and birational equivalence QUESTION [8 upvotes]: Let us remenber that we have the following proposition of Artin and Mumford (in "Some elementary examples unirational varieties which are not rational" proposition 1.): "The torsion subgroup $T_2\subset H^3(V,\mathbb{Z})$ is a birational invariant of a complete non-singular complex variety $V$ of any dimension $n$." Question : Are there other any other type of singular cohomological information (even homotopical) which is known to be a birational invariant of complete non-singular complex varieties? REPLY [10 votes]: Basically, any invariant $T$ which satisfies the following purity property will turn into a birational invariant for proper smooth varieties (aka complex compact manifolds): let $X$ be a smooth variety and $U$ be an open subspace of $X$ such that $\mathop{\rm codim}_X(X\setminus U)\geq 2$; then the injection $U\hookrightarrow$ induces an isomorphism $T(X)\simeq T(U)$. The proof is as follows. Let $X,Y$ be proper smooth varieties and let $f\colon X\dashrightarrow Y$ be a birational isomorphism. Because $X$ is smooth and $Y$ is proper, the valuative criterion applies and shows that $f$ is defined on an open subset $U\subset X$ such that $\mathop{\rm codim}_X(X\setminus U)\geq 2$. Similarly, the inverse isomorphism $f^{-1}$ is defined on an open subset $V\subset Y$ such that $\mathop{\rm codim}_Y(Y\setminus V)\geq 2$. Say $T$ is covariant, for definiteness. One gets maps $f_*\colon T(X)\simeq T(U)\to T(Y)$ and $g_*\colon T(Y)\simeq T(V)\to T(X)$. One checks that their composite is the identity. A loot of interesting examples are given in the answer of R. van Dobben de Bruyn. Another class of examples is given by the unramified cohomology. They are important in the study of rationality of varieties.<|endoftext|> TITLE: How can one construct a sparse null space basis using recursive LU decomposition? QUESTION [8 upvotes]: Given an $m$ by $n$ matrix $A$ I'm familiar with the standard method to compute a basis for the null space of $A$ by computing a QR factorization of $A^T$. If $A$ is large and sparse, we can use sparse-QR techniques with pivoting, but the resulting $Q$ matrix (and portion used for the null space basis) may be quite dense. For simple running example, consider a $n \times 1$ row of ones $A = [1 ... 1]$. This matrix has all zero-average vectors in its null space. If I use (MATLAB's) QR factorization ([Q,R,E] = qr(A');) the resulting $Q$ matrix is a dense $n \times n$ matrix. In this case, we know a sparse basis for the null space exists: $N = \left[\begin{array}{c} \begin{array} --1 & -1& \dots & -1\\ \end{array} \\ I \end{array}\right]$, where $I$ is the $n-1 \times n-1$ (sparse) identity matrix. A preliminary question is, 1) how can one construct a(/the most) sparse null space basis given $A$? But, actually I have found an algorithm's implementation online by Pawel Kowal* that computes sparse null spaces very well. Trying to trace through the code it seems to work by recursively applying an LU-decomposition with pivoting. However, I can't completely understand what it's doing and certainly don't understand why it's working. The comments and function names say it is computing an "LUQ" decomposition: function [L,U,Q] = luq(A,do_pivot,tol) % PURPOSE: calculates the following decomposition % % A = L |Ubar 0 | Q % |0 0 | % % where Ubar is a square invertible matrix % and matrices L, Q are invertible. % Is this decomposition well known? Does it go by another name? This algorithm seems to work very well. In the example above, this decomposition produces the "ideal" sparse basis for the null space $N$. So my current question, 2) is there a corresponding academic paper describing this method for computing a sparse basis for the null space of a matrix via recursive LU decomposition? *I've had no luck trying to contact Pawel Kowal for more information. REPLY [4 votes]: On the LUQ decomposition The algorithm implemented in luq (see reference given below) computes bases for the left/right null spaces of a sparse matrix $A$. Unfortunately, as far as I can tell, there seems to be no thorough discussion of this particular algorithm in the literature. In place of a reference, let us clarify how/why it works and test it a bit. The luq routine inputs an $m$-by-$n$ matrix $A$ and outputs an $m$-by-$m$ invertible matrix $L$, an $n$-by-$n$ invertible matrix $Q$ , and an $m$-by-$n$ upper trapezoidal matrix $U$ such that: (i) $A=LUQ$ and (ii) the pivot-less columns/rows of $U$ are zero vectors. For example, $$ \underbrace{\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}}_A = \underbrace{\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}}_L \underbrace{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}}_U \underbrace{\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}}_Q $$ Point (ii) allows one to construct bases for the left/right null spaces of $A$. Bases for Left/Right Null Spaces of $A$ Let $r = \operatorname{Rank}(A)$. Suppose we can compute the exact $LUQ$ decomposition of $A$ as described above. Then, The $n-r$ columns of $Q^{-1}$ corresponding to the pivotless columns of $U$ are a basis for the null space of $A$. This follows from the fact that $\operatorname{null}(A) = \operatorname{null}(A Q^{-1}) = \operatorname{null}(L U)$ and that the pivotless columns of $U$ are zero vectors by construction. The $m-r$ rows of $L^{-1}$ corresponding to the pivotless rows of $U$ are a basis for the left null space of $A$. This follows from the fact that $\operatorname{null}(A^T) = \operatorname{null}((L^{-1} A)^T) = \operatorname{null}( (U Q)^T)$ and that the pivotless rows of $U$ are zero vectors by construction. LUQ Algorithm Assume that $m \ge n$. (If $m < n$, then the lu command mentioned below outputs a slightly different $PA=LU$ factorization. Otherwise the LUQ decomposition is almost the same, and so, we omit this case.) Given an $m$-by-$n$ matrix $A$, the LUQ decomposition calls MATLAB command lu with partial (i.e., just row) pivoting. lu implements a variant of the LU decomposition that inputs $A$ and outputs: $m$-by-$m$ permutation matrix $P$; $m$-by-$n$ lower trapezoidal matrix $\tilde L$ with ones on the diagonal; and, $n$-by-$n$ upper triangular matrix $\tilde U$ such that $PA = \tilde L \tilde U$. Write: $$ \tilde U = \begin{bmatrix} \tilde U_{11} & \tilde U_{12} \\ 0 & \tilde U_{22} \end{bmatrix} $$ where $\tilde U_{11}$ has nonzero diagonal entries, and hence, is invertible. Also, let $e_i$ denote unit $m$-vectors equal to $1$ in the $i$th component and zero otherwise. The algorithm then builds: $$ L = P^T \begin{bmatrix} \tilde L & e_{n+1} & \cdots & e_m \end{bmatrix} $$ which is an $m \times m$ invertible matrix, and $$ U = \begin{bmatrix} \tilde U_{11} & 0 \\ 0 & \tilde U_{22} \\ 0 & 0 \end{bmatrix} $$ which is upper trapezoidal, and $$ Q = \begin{bmatrix} I & \tilde U_{11}^{-1} \tilde U_{12} \\ 0 & I \end{bmatrix} $$ which is an $n$-by-$n$ invertible matrix. To summarize, we obtain: $$ A = L \begin{bmatrix} \tilde U_{11} & 0 \\ 0 & \tilde U_{22} \\ 0 & 0 \end{bmatrix} Q $$ For the most part, that is all the algorithm does. However, if there are any nonzero entries in $\tilde U_{22}$, then the algorithm will call luq again with input matrix containing all of the nonzero entries of $\tilde U_{22}$. This last step introduces more zeros into $U$ and modifies the invertible matrices $L$ and $Q$. To understand this last step, it helps to consider a simple input to luq like $$ A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} $$ The first call to luq with this input trivially gives $U=A$ with $L$ and $Q$ being the $3$-by-$3$ identity matrices. Since $U$ has nonzero entries, a second call is made to luq with input $1$, which outputs $L=U=Q=1$. This second decomposition is incorporated into the first one by making the second column of $L$ the first one and moving all the other columns to the right of it, and similarly, moving the third row of $Q$ to the first row and moving all the other rows below it. This yields, $$ A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} $$ To be sure, consider another simple example $$ A = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & a & 0 & b \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & c \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$ where $a,b,c$ are nonzero reals. In the first pass through luq the algorithm again sets $U=A$ and $L$, $Q$ equal to the $5$-by-$5$ identity matrices. Since $U=\tilde U_{22}$ has nonzero elements, luq is called again with input matrix $$ B = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} $$ This is incorporated into the first decomposition by permuting $L$ and $Q$ as shown: $$ A = \begin{pmatrix} 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} a & b & 0 & 0 & 0 \\ 0 & c & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix} $$ In general, the columns of $L$ and the rows of $Q$ are permuted so that the the zero columns/rows of $\tilde U_{22}$ are moved to the end of the matrix. An LUQ decomposition is then performed on this nonzero sub-block. A full explanation would be notation heavy (requiring index sets for the zero/nonzero elements) and not much easier to understand than the code itself. Simple Test In reality, the algorithm computes an approximate LUQ decomposition and approximate bases, i.e., with rounding errors. These rounding errors might be significant if some of the nonzero singular values of $A$ are too small for the algorithm to detect. Here is a MATLAB script file that tests the luq code. The script is a slight modification of the demo file that the software comes with. I modified the original file so that it inputs a sparse, random, rectangular, rank deficient matrix and outputs bases for the left/right null spaces of this input matrix. Here is a sample output from this demo file. elapsed time = 0.011993 seconds Input matrix: size = 10000x500 true right null space dimension = 23 true left null space dimension = 9523 Output: estimated right null space dimension = 23 estimated left null space dimension = 9523 error in basis for right null space = 0 error in basis for left null space = 2.2737e-13 "Extreme" Test This example is adapted from Gotsman and Toledo [2008]. Consider the $(n+1)$-by-$n$ matrix: $$ A_1 = \begin{pmatrix} 1 & & & & \\ -1 & 1 & & & \\ \vdots & -1 & \ddots & & \\ \vdots & & \ddots & 1 & \\ -1 & -1 & \cdots & -1 & 1 \\ 0.5 & 0.5 & \cdots & 0.5 & 0.5 \end{pmatrix} $$ and in terms of this matrix, define the block diagonal matrix: $$ A = \begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix} $$ where $A_2$ is an $n$-by-$n$ random symmetric positive definite matrix whose eigenvalues are all equal to one except $3$ are zero and one is $10^{-8}$. With this input matrix and $n=1000$, we obtain the following sample output. elapsed time = 1.1092 the matrix: size of A = 2001x2000 true rank of A = 1997 true right null space dimension = 3 true left null space dimension = 4 results: estimated right null space dimension = 3 estimated left null space dimension = 4 error in basis for right null space = 9.2526e-13 error in basis for left null space = 5.9577e-14 Remark There is an option in the luq code to use LU factorization with complete (i.e., row and column) pivoting $PAQ=LU$. The resulting $U$ matrix in the $LUQ$ factorization may better reflect the rank of $A$ in more ill-conditioned problems, but there is an added cost to doing column pivoting. Reference Kowal, P. [2006]. "Null space of a sparse matrix." https://www.mathworks.com/matlabcentral/fileexchange/11120-null-space-of-a-sparse-matrix Gotsman, C., and S. Toledo [2008]. "On the computation of null spaces of sparse rectangular matrices." SIAM Journal on Matrix Analysis and Applications, (30)2, 445-463.<|endoftext|> TITLE: Cover a complete graph by smaller complete graphs QUESTION [13 upvotes]: Suppose we have a complete graph $G$ of size $n$. What is the minimum number of complete graph of size $k, k TITLE: Existence of a uniformly continuous function $g$ on $\mathbb{R}$ where $f = g$ a.e.? QUESTION [8 upvotes]: Suppose $f \in L^\infty(\mathbb{R})$, $f_h(x) = f(x + h)$, and$$\lim_{h \to 0} \|f_h - f\|_\infty = 0.$$Does there exist a uniformly continuous function $g$ on $\mathbb{R}$ such that $f = g$ almost everywhere? REPLY [8 votes]: Yes. Let $\{\phi_n\}$ be a sequence of standard mollifiers, so $\phi_n$ is continuous, $\phi_n \ge 0$, and $\int \phi_n = 1$. I claim the sequence $\{f \ast \phi_n\}$ is uniformly equicontinuous. For let $\epsilon > 0$ and choose $\delta$ so small that $|h| < \delta$ implies $\|f - f_h\|_{L^\infty} < \epsilon$. Then for any such $h$ and any $x \in \mathbb{R}$ we have $$\begin{align*}|(f \ast \phi_n)(x) - (f \ast \phi_n)(x+h)| &= |((f - f_h) \ast \phi_n)(x)| \\ &\le \int |f(y) - f_h(y)| \phi_n(x-y)\,dy \\ &\le \|f - f_h\|_{L^\infty} \cdot \int \phi_n(x-y)\,dy \\ &< \epsilon \cdot 1.\end{align*}$$ Moreover, $\|f \ast \phi_n\|_\infty \le \|f\|_\infty$. Passing to a subsequence if necessary, we have $f \ast \phi_n \to f$ almost everywhere. By Arzela-Ascoli, on any compact interval $[a,b]$ there is a subsequence of $f \ast \phi_n$ converging uniformly to some continuous function $g$, which must equal $f$ almost everywhere on $[a,b]$. By taking larger and larger intervals, and noting that $g$ is well defined, we get $g$ defined on all of $\mathbb{R}$, still having $g$ continuous and $f=g$ almost everywhere. Now since $g$ is continuous, we have $\sup_x |g(x) - g_h(x)| = \|f - f_h\|_{L^\infty} \to 0$ as $h \to 0$. So in fact $g$ is uniformly continuous. REPLY [5 votes]: This is easier when passed to some sort of weak formulation. By Lebesgue differentiation theorem, for almost every x, $\lim_{r\to0} \frac{1}{|B_r|} \int_{x+B_r} f=f(x)$. Replace each f(x) by the left hand side. We have modified f on a set of measure 0, so the left hand side doesn't change further, and we have this equality everywhere. Also, the assumption $\lim_{h\to0} ||f_h-f||_\infty=0$ remains valid. Now we show f is uniformly continuous. For $\epsilon>0$, there is $\delta>0$ such that for all $h<\delta$, $||f_h-f||<\epsilon$. Let $x-x'<\delta/3$. There is $r, r'<\delta/3$ such that $|f(x)-\frac{1}{|B_r|} \int_{x+B_r} f|<\epsilon$ and $|f(x')-\frac{1}{|B_{r'}|} \int_{x+B_{r'}} f|<\epsilon'$. Then $|f(x)-f(x')|<2\epsilon+\frac{1}{|B_r||B_{r'}|} \int_{y\in x+B_r} \int_{z\in x+B_{r'}} |f(y)-f(z)|=2\epsilon+\frac{1}{|B_r||B_{r'}|}\int_{h<\delta} \int_{y\in x+B_r, y+h\in x'+B_{r'}} |f(y)-f(y+h)|<2\epsilon+\frac{1}{|B_r||B_{r'}|}\int_{h<\delta} \int_{y\in x+B_r, y+h\in x'+B_{r'}} \epsilon=2\epsilon+\frac{1}{|B_r||B_{r'}|} \int_{y\in x+B_r} \int_{z\in x'+B_{r'}} \epsilon=3\epsilon$.<|endoftext|> TITLE: Lax's proof of the change of variables theorem QUESTION [7 upvotes]: I am teaching a course on Multivariable Calculus for Graduate students. I came across this nice article by Lax where a special case of the change of variables theorem is proved: Theorem. Let $f:\mathbb{R}^n \to \mathbb{R}$ be a continuous function with compact support and let $\phi:\mathbb{R}^n \to \mathbb{R}^n$ be a continuously differentiable function that is identity outside of some ball. Then $$\int_{\mathbb{R}^n} f = \int_{\mathbb{R}^n} (f \circ \phi)J_\phi,$$ where $J_\phi$ denotes the determinant of the derivative matrix of $\phi$. I presented a differential forms version of Lax's proof given by Ivanov in class. Michael Taylor has also given a differential forms version of Lax's proof and writes that the standard version of the change of variables theorem is easily established using Lax's version. Lax himself has a follow-up article in which a more or less standard version of the change of variables theorem is proved but the proof is quite long. One thing that is not clear to me is how to give a simple proof of the following standard version of change of variables theorem using Lax's theorem: Theorem. Let $\phi:U \to V$ be a $\mathcal{C}^1$-smooth diffeomorphism of open sets in $\mathbb{R}^n$ and let $f:V \to \mathbb{R}$ be a continuous function with compact support. Then $$ \int_V f = \int_U (f \circ \phi)|J_\phi|.$$ I found an article by Kumaresan and Santhanam that claims to deduce the above version from Lax's version but the statement given by them omits the absolute value on $J_\phi$ and their proof is incorrect at the last step. Is it possible to deduce the standard change of variables theorem from Lax's theorem in an easy way? REPLY [2 votes]: We may assume in Th.2 that $U$ is connected, so the (never vanishing) $J_\phi$ has constant sign, that we can assume positive w.l.o.g. We first establish the formula locally, for special $f$ (with "small enough supports"): precisely, assume that $\phi^{-1}(\mathrm{supp}\, f)$ is included in a ball $B\Subset U$ such that (B) there exists a diffeomorphism $\psi:\mathbb{R}^n\to\mathbb{R}^n$ such that $\psi =\phi$ on $B$, and $\psi$ is the identity outside some large ball. In this case, $ \phi( x)= \psi(x)$ for all $x\in B$, and $f(\phi( x))=f(\psi(x))=0$ for all $x\notin B$ (because $\psi$ is injectve) so that by Th1 $$\int_U(f\circ\phi)|J_\phi|=\int_{\mathbb{R}^n}(f\circ\psi)J_\psi=\int_V f.$$ To extend the formula to any $f$, first observe that, given the diffeo $\phi:U\to V$ with $J_\phi>0$, and $x\in U$, there is a ball $B\Subset U$ satisfying (B) (this point is elementary though maybe technical: one can e.g. use the radius as a deformation parameter, and homotop $\phi$ on its differential, and then to the identity map within the linear group). Therefore, for any $y\in V$ there is a nbd $W$ such that the formula of thm2 holds for functions with support in $W$. Since all integrals depend linearly on $f$, the formula holds then for any function with support in $V$, by partition of unity.<|endoftext|> TITLE: Lower bound for total mean curvature among mean convex set in $3D$ with fixed volume QUESTION [5 upvotes]: I learned that the sphere has the smallest total mean curvature among all convex solids with a given surface area. This actually implies the sphere also has the smallest total mean curvature among all convex solids with a given volume. The first result can be proved by using Steiner symmetrization, and the latter result is a consequence of of the first one and the isoperimetric inequality. Now my question is, does sphere have the smallest total mean curvature among all mean convex solids with a given volume? By mean convex set, I mean bounded smooth set with nonnegative mean curvature on the boundary. I searched but didn't find any satisfying result. It seems that people haven't considered minimizing total mean curvature under the mean convex set setting. Also, the condition of fixing the volume should be added, since I believe if only fixing the surface area then the infimum of total mean curvature may not be attained. Any ideas on how to stab on this question? Is it a trivial question? Any comment and ideas would be really appreciated. REPLY [3 votes]: Even with an area constraint, the minimization of total mean curvature for mean-convex surfaces seems to be open. You should look at this recent article http://link.springer.com/article/10.1007/s12220-015-9646-y<|endoftext|> TITLE: $\mathbb{G}_m$-torsors and line bundles QUESTION [9 upvotes]: I recently came across this elegant translation of etale $\mathbb{G}_m$-torsors into line bundles: Let $\text{Spec }R$ be a $\mathbb{G}_m$-torsor over $\text{Spec }A$ for the etale topology, where $\mathbb{G}_m = \text{Spec }A[x,x^{-1}]$, then the action of $\mathbb{G}_m$ on $\text{Spec }R$ is given by a homomorphism of rings (the coaction): $$\mu : R\rightarrow A[x,x^{-1}]\otimes_A R$$ Somehow, this induces a decomposition $R = \bigoplus_{n\in\mathbb{Z}} R_1^{\otimes n}$ making $R$ into a graded $A$-algebra, where $R_1$ is a projective $A$-module. Unfortunately, I don't understand how to prove this decomposition, and why $R_1$ is projective. I can imagine that the etale local triviality of $R$ would somehow result in $R$ being etale-locally free, but I don't see why this would imply Zariski-freeness (ie, projectivity). REPLY [10 votes]: If $G$ is a commutative monoid (for your question we will want $G = \mathbb{Z}$), then $A[G]$-comodules identify with $G$-graded $A$-modules. A reference is Demazure, Gabriel, Introduction to Algebraic Geometry and Algebraic Groups, II, §2, no 2, Example 1. Indeed, if $M$ is an $A$-module with an $A[G]$-module structure given by the coaction $\mu: M \to M\otimes_A A[G]$, the corresponding grading $M = \bigoplus_{g \in G} M_g$ is given by $M_g = \{m \in M : \mu(m)=m\otimes g\}$. Conversely, given a $G$-grading $M = \bigoplus_{g\in G}M_g$, for $m\in M$, let $m_g$ be the projection of $m$ onto $M_g$, then the map $$\mu : M\longrightarrow M\otimes_A A[G],\quad m\mapsto m_g\otimes g$$ is an $A[G]$-coaction on $M$. These functors are mutually inverse, and define an equivalence of symmetric monoidal categories. Hence, $A[G]$-comodule commutative algebras identify with $G$-graded commutative $A$-algebras. Now assume that $R \to \underbrace{R\otimes_A A[\mathbb{Z}]}_{R[x,x^{-1}]} = R\otimes_A \underbrace{A[x,x^{-1}]}_{(\mathbb{G}_m)_{/A}}$ is a comodule commutative algebra. The corresponding $\mathbb{G}_m$-scheme $\mathrm{Spec}(R)$ is a torsor if and only if $R$ is faithfully flat over $A$, The natural morphism $\mathrm{Spec}(R)/\mathbb{G}_m \to \mathrm{Spec}(A)$ is an isomorphism, i.e. the natural morphism $A \to R_0$ is an isomorphism, The natural morphism $\mathbb{G}_m \times_{\mathrm{Spec}(A)} \mathrm{Spec}(R) \to \mathrm{Spec}(R) \times_{\mathrm{Spec}(A)} \mathrm{Spec}(R)$ is an isomorphism, i.e. the natural morphism $R \otimes_A R \to R[x,x^{-1}]$ is an isomorphism. For $n,m \in \mathbb{Z}$ the natural morphism $R_n \otimes_A R_m \to R_{n+m}$ is an isomorphism since this is so when we tensor with $R$ over $A$ (by faithful flatness). It follows that $R_1$ is invertible and that $R_n \cong R_1^{\otimes n}$ for $n \in \mathbb{Z}$.<|endoftext|> TITLE: When do real analytic functions form a coherent sheaf? QUESTION [18 upvotes]: It is known that, in general, the sheaf of real analytic functions on a real analytic manifold is not coherent. However, there are some examples, where we have coherence: for example, if $X$ is a complex analytic manifold, then the sheaf of real analytic functions $A^{\omega}_X$ is the restriction to the diagonal of the sheaf $\mathcal O_{X \times \bar X}$ of complex analytic functions on the product of $X$ and its complex conjugate $\bar X$, so $A^{\omega}_X$ is coherent. Are there other examples of real analytic manifolds for which we can prove the coherence of the structure sheaf? I am especially interested in the case where $X$ has the metric with special holonomy, for example, when it is a $G_2$ manifold. REPLY [15 votes]: For real-analytic manifolds, coherence of the structure sheaf always holds. The 1-sentence reason is that one can pass to real and imaginary parts on Oka's coherence theorem in several complex variables. Before discussing a proper proof of that 1-sentence executive summary, I should address that in the real-analytic setting however an analogue of one of Oka's "coherence" results over $\mathbf{C}$ can fail (leading one to see phrases in the literature such as "non-coherent real-analytic spaces"): there is no real counterpart to the "analytic Nullstellensatz" of Oka. To explain this, recall Oka's result that for a complex-analytic set $X$ in a complex manifold $V$ (i.e., a closed subset that is locally on $V$ given by the vanishing of finitely many holomorphic functions) the ideal sheaf of sections of $O_V$ vanishing on $X$ is always locally finitely generated (or equivalently a coherent $O_V$-module, since $O_V$ is coherent by Oka's big theorem). But this fails in the real-analytic case. In contrast, there exist real-analytic sets $Z$ in real-analytic manifolds $U$ (i.e., a closed subset that is locally on $U$ given by the vanishing of finitely many real-analytic functions) such that the ideal sheaf $I_Z$ of sections of $O_U$ vanishing on $Z$ is not locally finitely generated; such a $Z$ is called "non-coherent" in $U$ (because in such cases the subsheaf $A = O_U/I_Z$ of the sheaf of $\mathbf{R}$-valued continuous functions on $Z$ can fail to be a coherent sheaf of rings, though I have never cared enough to check up on a proof of this consequence in some cases). An explicit example of $Z$ and $U$ with $I_Z$ not coherent inside $O_U$ is given near the start of https://arxiv.org/pdf/math/0612829.pdf. To be more detailed about the proof of coherence of the structure sheaf of a real-analytic manifold, the assertion is of local nature and so only depends on the (local) dimension; i.e., for manifolds of (pure) dimension $n$ it is equivalent to showing that the sheaf $O_U$ of real-analytic functions on every small open ball $U$ around the origin in $\mathbf{R}^n$ is coherent. By definition, this coherence amounts to local finite generation for the kernel of any $O_U$-linear map $\varphi:O_U^{\oplus N} \rightarrow O_U$ for any small $U$. This map locally "extends" over an open in $\mathbf{C}^n$; i.e., by working locally on $U$ we can arrange that that exists an open $V \subset \mathbf{C}^n$ satisfying $V \cap \mathbf{R}^n = U$ and holomorphic $F_1, \dots, F_N$ on $V$ whose restriction to $U$ recovers the $N$ components $f_1, \dots, f_N$ of $\varphi$. By Oka's big coherence theorem (not the analytic Nullstellensatz above), the resulting map $\Phi: O_V^{\oplus N} \rightarrow O_V$ extending $\varphi$ has kernel that is locally finitely generated. Working locally on $V$ around points of $U$, we can thereby arrange that there exist $s_1, \dots, s_r \in O(V)^{\oplus N}$ generating $\ker \Phi$. We claim that the real and imaginary parts of the restrictions to $U$ of these $N$-tuples belong to $\ker \varphi$ and generate it. Our problem is local on $V$ near $U$, so for generation we can focus on $(g_1,\dots,g_N) \in (\ker \varphi)(U)$. By working locally on $V$ around a point in $U$ we can arrange that $V$ is connected and that each $g_j$ extends (necessarily uniquely) to a holomorphic $G_j$ on $V$. Then the global section $\sum G_j F_j \in O(V)$ has restriction to $U$ equal to $\sum g_j f_j = \varphi(g_1,\dots,g_N)=0$, so $\Phi(G_1,\dots,G_N)=\sum G_j F_j = 0$ by connectedness of $V$ and analyticity considerations. Thus, working locally on $V$ some more we can arrange that $(G_1,\dots,G_N) = \sum_{k=1}^r a_k s_k$ for some $a_{1}, \dots, a_{r} \in O(V)$, so restricting to $U$ gives $$(g_1,\dots,g_N) = \sum_{k=1}^r a_{k}|_U \cdot s_k|_U.$$ Each function $g_j$ is $\mathbf{R}$-valued, whereas $a_{k}|_U$ and the $N$ components of $s_k|_U$ are $\mathbf{C}$-valued, and the real and imaginary parts of these various $\mathbf{C}$-valued functions are real-analytic on $U$. Thus, $$(g_1,\dots,g_N) = \sum_{k=1}^r ({\rm{Re}}(a_{k}|_U){\rm{Re}}(s_k|_U) - {\rm{Im}}(a_{jk}|_U) {\rm{Im}}(s_k|_U)).$$ Finally, the condition $s_k \in (\ker \Phi)(V) \subset O(V)^{\oplus N}$ says that the $N$ components $s_{k1},\dots,s_{kN} \in O(V)$ satisfy $\sum F_j s_{kj} = 0$, so $\sum f_j \cdot s_{kj}|_U = 0$ as $\mathbf{C}$-valued functions on $U$. But each $f_j$ is $\mathbf{R}$-valued, so $\sum f_j {\rm{Re}}(s_{kj}|_U)$ and $\sum f_j {\rm{Im}}(s_{kj}|_U)$ both vanish on $U$; i.e., ${\rm{Re}}(s_k|_U)$ and ${\rm{Im}}(s_k|_U)$ belong to $(\ker \varphi)(U)$. Thus, ${\rm{Re}}(s_1|_U), \dots, {\rm{Re}}(s_r|_U), {\rm{Im}}(s_1|_U), \dots, {\rm{Im}}(s_r|_U)$ generate $\ker \varphi$ over the (now shrunken) $U$. This shows that in the initial setup (before we began shrinking $U$) the kernel of $\varphi$ is locally finitely generated.<|endoftext|> TITLE: Colombeau generalized functions QUESTION [5 upvotes]: I'm currently reading some aspects of Colombeau generalized functions, and in almost all of his examples he discuss aspects of Quantum Field Theory, but then I go to some "standard" texts on QFT and I can not find any information (Ziedler only point out some similarities with Hörmander wave front sets; I can't find anything on Weinberg's book, or Folland). Why is that? what is the major drawback about Colombeau "new" generalized functions to make it hard to find on physics books? or are there any standard physics books discussing it? I know about the books "Nonlinear Theory of Generalized Functions" and "The linear theory of Colombeau generalized functions". Thanks. REPLY [2 votes]: This recent textbook might be helpful: Geometric Theory of Generalized Functions with Applications to General Relativity, M. Grosser, M. Kunzinger, M. Oberguggenberger, and R. Steinbauer (2013). Over the past few years a certain shift of focus within the theory of algebras of generalized functions (in the sense of J. F. Colombeau) has taken place. Originating in infinite dimensional analysis and initially applied mainly to problems in nonlinear partial differential equations involving singularities, the theory has undergone a change both in internal structure and scope of applicability, due to a growing number of applications to questions of a more geometric nature. The present book is intended to provide an in-depth presentation of these developments comprising its structural aspects within the theory of generalized functions as well as a (selective but, as we hope, representative) set of applications, in particular in the context of general relativity.<|endoftext|> TITLE: Why there is a Quot-scheme, not a Sub-scheme? QUESTION [26 upvotes]: Let $X$ be a projective variety, and $E$ be a coherent sheaf on $X$. Grothendieck has proven that there is a scheme $\mathrm{Quot}_X(E)$ parametrizing arbitrary quotient sheaves of $E$. It is probably a well-known question, but I found nothing in Google: $$\text{Why not a scheme $\mathrm{Sub}_X(E)$ parametrizing arbitrary subsheaves of $E$?}$$ Does such a scheme exist? If yes, is it purely a question of style, like Grothendieck's definition of the projectivization $\mathbb P(V)$ as the set of one-dimensional quotients of a vector space $V$? If not, is it because cokernels of sheaves require sheafification, while kernels coincide with kernels for pre-sheaves? Moreover, if one considers a reflexive sheaf $E$ (that is the canonical morphism $E \to E^{**}$ is an isomorphism) and its reflexive subsheaves, then there is a Sub-scheme via dualizing and taking a Quot-scheme. Is there a less restrictive constraint? REPLY [19 votes]: For standard universal properties, you need the scheme to behave well under base change, which in these cases would mean tensor products. Tensor product is right exact, so a quotient remain a quotient, not left exact, so a sub may not remain a sub.<|endoftext|> TITLE: How to show that something is not completely metrizable QUESTION [11 upvotes]: I have a Polish space $X$ and a subset $A \subset X$. I know that $A$ is completely metrizable (in its induced topology) if and only if $A$ is a $G_\delta$-set in $X$. This means: If I want to show that $A$ is completly metrizable then it suffices to find a sequence $U_1,U_2,\ldots$ of open sets such that $A=\bigcap_j U_j$. But what if I want to show that $A$ is not completely metrizable? Are there any criteria I can use? The only criterium I know is that if $A$ is not Baire then it is not completely metrizable (This shows that $\mathbb Q$ is not $G_\delta$ in $\mathbb R$) but this does not help if my set $A$ is a Baire space. Also there may be cases where it is not easy to decide wether $A$ is Baire. I hope that something can be said in this generality... Thanks for your help in advance! REPLY [9 votes]: There is an equivalent condition for complete metrizability in terms of a game known as the Choquet game. The game is described, for example, in the book Classical Descriptive Set Theory by Kechris, who calls it the strong Choquet game. The game goes like this. The rounds are labeled with natural numbers. On round $i$, the first player chooses an open set $U_i$ and a point $x_i \in U_i$. The second player then chooses an open set $V_i$ with $x_i \in V_i \subseteq U_i$. At the second and subsequent rounds, the first player must additionally ensure that $U_{i+1} \subseteq V_i$, so the open sets that are chosen are nested $U_1 \supseteq V_1 \supseteq U_2 \supseteq V_2 \supseteq \cdots$. The second player wins if $\bigcap_{i\in\mathbb{N}} U_{i} \not = \emptyset$, which is equivalent to $\bigcap_{i\in\mathbb{N}} V_{i} \not = \emptyset$. It is an easy exercise to prove that if $(A,d)$ is completely metrizable then the second player has a winning strategy. Choquet proved the converse: a space is completely metrizable if and only if the second player has a winning strategy in the game for that space. Only the easy half of the equivalence is needed for our current purpose. So, if you can prove the second player does not have a winning strategy for a space, that is sufficient to disprove complete metrizability. In particular, one possible method in the current situation is to attempt to prove that the first player has a winning strategy in the game.<|endoftext|> TITLE: Polynomials with all but one root inside the unit disc QUESTION [8 upvotes]: Call a real number $\lambda>1$ special if it is a root of a polynomial $f(x)$ such that $f(x)$ is monic with integer coefficients, all roots of $f(x)$ are distinct, for all $z\in\mathbb{C}$, if $f(z)=0$ and $z\neq \lambda$ then $|z|<1$. For example, the Golden Ratio $\frac{1+\sqrt{5}}{2}$ is special. Question: Are there special numbers arbitrarily close to 1? Some Remarks: After messing around on Wolfram Alpha, I was able to beat the Golden Ratio using the unique real root of $x^3-x-1$, which is approximately 1.3247. I have found lots of work on polynomials such that all roots are on the unit disc, and some things where all but two roots are on the unit disc. But nothing close to what I want: exactly one real root not on the unit disc, and the rest strictly inside the unit disc. REPLY [13 votes]: This is a well-known topic. Such numbers are called Pisot numbers. Salem (1945) proved that they form a closed subset $S$ of $(1,+\infty)$, and Siegel (1944) proved that its lowest element is the positive root of $X^3-X-1$. Therefore the answer to your question is No. The Golden ratio is however the smallest among the cluster points of $S$. The derived set $S'$ was described by Dufresnoy & Pisot (1953). A Pisot number $\theta$ (root of $f$ as above) belongs to $S'$ iff there exist another polynomial $g\in{\mathbb Z}[X]$ such that $|g|\le|f|$ over the unit circle, and $\pm g$ is neither $f$, nor its reciprocal polynomial $f^*(X)=X^nf(1/X)$.<|endoftext|> TITLE: Class forcing as set forcing followed by truncation QUESTION [7 upvotes]: My question arises from thinking about how we can obtain class forcing extensions from truncations of set forcing extensions in the presence of a (strongly) inaccessible cardinal in the following sense. Let $\kappa$ be an inaccessible cardinal and let $\mathbb P$ be a forcing such that $\mathbb P \subseteq V_{\kappa}$. Now we can consider $\mathbb P$ simultaneously as a set forcing (in $V$) and as a class forcing over $V_{\kappa}$. Now let $G$ be a $(V;\mathbb P)$-generic filter. Then we can ask how $V[G]$ and $V_{\kappa}[G]$ relate to each other, where in this context $V_{\kappa}[G] := \{ \sigma_G : \sigma\in V^{\mathbb P}\cap V_{\kappa} \}$. A naive approach would be to conjecture that $V_{\kappa}[G] = (V_{\kappa})^{V[G]}$. This is of course not true in general: if $\mathbb P$ is the Levy collapse $\mathrm{Col}(\omega, <\kappa)$, then, e.g., we have $\wp^{V[G]}(\omega) \in (V_{\kappa})^{V[G]}\setminus V_\kappa[G]$, the underlying problem being that $(V_{\kappa})^{V[G]}$ is a model of the power set axiom, whereas $V_{\kappa}[G]$ is not. We can remedy this last problem by considering $H_{\kappa}$ instead of $V_\kappa$ in the above argument (which is the same in the ground model); and indeed we get the following Fact: Let $\kappa$ be inaccessible and $\mathbb P\subseteq H_{\kappa}^V$ a forcing preserving the regularity of $\kappa$. Let $G$ be a $(V;\mathbb P)$-generic filter. Then $H_\kappa^V[G] = (H_\kappa)^{V[G]}$. Proof. The difficult direction is to show that $(H_\kappa)^{V[G]} \subseteq H_\kappa^V[G]$. For this let $x\in (H_\kappa)^{V[G]}$ and assume (inductively) that $x\subseteq H_\kappa^V[G]$. Then let $\sigma\in V^{\mathbb P}$ be such that $\sigma_G = x$ and $\mathrm{dom}(\sigma) \subseteq V^{\mathbb P}\cap H_{\kappa}^V$. In $V[G]$ we define a function $f: x\to \kappa$ by setting $$ f(y) := \min\{ \alpha < \kappa : \exists\langle \tau, p\rangle\in \sigma\cap V_{\alpha+1} (p\in G \wedge \tau_G = y) \} $$ for any $y\in x$. Now note that $f``x$ is bounded below $\kappa$, since $\kappa$ is regular in $V[G]$ and by assumption there can be no surjection from $x$ onto $\kappa$. Thus take $\alpha < \kappa$ such that $f``x\subseteq \alpha$ and let $\sigma' := \sigma\cap V_{\alpha}$. Then $\sigma'_G = \sigma_G = x$ and $\sigma' \in V^{\mathbb P}\cap H^V_{\kappa}$ by inaccessibility of $\kappa$ in $V$. Therefore $x\in H_\kappa^V[G]$. $\Box$ This proof depends on the fact that $\mathbb P$ preserves the regularity of $\kappa$, but on the other hand I could not construct a counterexample for a forcing singularizing $\kappa$. This leads me to ask the following Question: Let $\kappa$ be inaccessible. Can there be a forcing $\mathbb P\subseteq H_{\kappa}^V$ that does not collapse $\kappa$ and such that there is some $(V;\mathbb P)$-generic filter $G$ with $H_{\kappa}^V[G] \neq (H_{\kappa})^{V[G]}$ ? REPLY [6 votes]: No, there can be no such forcing. The point is that such a forcing should have size $\kappa$ and by my proof given in Singularizing forcing of "small" cardinality?, we have $\Vdash_{\mathbb{P}} |\kappa|=cf(\kappa)$, so $\mathbb{P}$ can not change the regularity of $\kappa$ as else it collapses $\kappa.$ Now the result follows from your argument.<|endoftext|> TITLE: Strong ergodicity and spectral gap QUESTION [5 upvotes]: Suppose that $\Gamma \curvearrowright (X,\mu)$ is a measure-preserving action of a discrete, finitely generated group $\Gamma = \left\langle S \right\rangle$ on a probability space $(X,\mu)$. Let $\pi: \Gamma \to L^{2}(X,\mu)$ be the associated Koopman representation $(\pi(\gamma)f)(x) := f(\gamma^{-1}x)$. I've come across two definitions of a spectral gap for such an action and it seems to me that they are not equivalent and this is what I would like to ask about. There exists a constant $\kappa>0$ such that for every measurable set $A \subset X$ we have $\max_{s \in S} \mu(A \Delta sA) \geqslant \kappa \mu(A)(1-\mu(A))$. There exists a constant $\kappa'>0$ such that for every $v\in L^{2}_{0}(X,\mu)$, the subspace of mean-zero functions, we have $\max_{s\in S}\|v - \pi(s)v\|_2 \geqslant \kappa' \|v\|_2$. The first definition implies that the action is strongly ergodic, i.e. if $(A_n)_{n \in \mathbb{N}}$ is a sequence of measurable sets such that $\lim_{n\to \infty} \max_{s\in S}\mu(A_n \Delta sA_n)=0$ then $\lim_{n\to \infty} \mu(A_n)(1 - \mu(A_n))=0$. Here comes the first question: Is the first definition equivalent to strong ergodicity? I know that there do exist strongly ergodic actions that do not admit spectral gap, so an affirmative answer would show that the definitions are not equivalent. If the answer is negative, one has to ask the following question. Is the first definition equivalent to the second one? These things appear naturally in Popa's deformation/rigidity theory, hence the choice of tags. EDIT: I realised that in the context of expanders the way to obtain a spectral gap from a property of indicator functions only is to use Cheeger's inequality. Maybe the right question to ask is: is there a Cheeger-type inequality in this case that shows that the two definitions introduced above are equivalent? REPLY [5 votes]: Yes, conditions 1 and 2 are equivalent (as you already noticed). Let me start by replacing both definitions with their negations, and replace definition 2 with its $L^1$-version. A and B below are the negations of 1 and 2 correspondingly, slightly rewritten: A. There exists a sequence of measurable subsets $\emptyset,X\neq A_n\subset X$ such that for every $s\in S$, $$(*)\quad \frac{\|(\chi_{A_n}-\mu(A_n)\cdot 1)-s(\chi_{A_n}-\mu(A_n)\cdot 1)\|_1}{\|(\chi_{A_n}-\mu(A_n)\cdot 1)\|_1} = \frac{\mu(A_n \Delta sA_n)}{\mu(A_n)(1-\mu(A_n))}\to 0.$$ B. There exists a sequence of vectors $0\neq v_n$ in $L^1_0(X)$ such that for every $s\in S$, $$(**) \quad \frac{\|v_n-sv_n\|_1}{\|v_n\|_1} \to 0.$$ To see that A is the negation of 1 is straightforward. To see that B is the negation of 2, recall that the map $L^2\ni u\mapsto \text{sgn}(u)\cdot |u|^2\in L^1$ is uniformly continuous equivariant map on the unit sphere. Use it and a projection mod constants. I am not elaborating, as this is well known. From now on we work in $L^1$ exclusively, thus $\|\cdot\|=\|\cdot\|_1$. Clearly A $\Rightarrow$ B. The rest of this post is devoted to proving the application B $\Rightarrow$ A. Thus, we assume that $v_n$ is a sequence of unit vectors satisfying $(**)$ and argue to provide a sequence $A_n$ satisfying $(*)$. Let me denote by $B$ the unit ball in $L^1_0(X)$. The idea, as usual, is to map $B$ continuously into a compact space and take a limit point of the image of the sequence $v_n$. A first candidate for such a map that comes to mind is the embedding into the unit ball of $L^1(X)^{**}\simeq L^\infty(X)^*$. This idea will lead us eventually to prove the existence of an invariant mean, as in the answer by Mateusz Wasilewski, but justifying everything down this road seems longer than the path we take below in which we use another map. Consider the space of probability measures on $\mathbb{R}$, $P$. Endow $P$ with the weakest topology for which integration against every function in $C_c(\mathbb{R})$ is continuous. Observe that the map $B\to P$, $f\mapsto f_*\mu$ is continuous and that the image is precompact (use Prokhorov's theorem). Let me denote $\nu_n=(v_n)_*\mu\in P$ and assume as we may that $\nu_n\to \nu$. We consider two different cases: either $\nu$ is a point mass or not. In case $\nu$ is NOT a point mass we can find and fix $t\in \mathbb{R}$ such that $\nu(-\infty,t),\nu(t,\infty)>0$ and then, defining $A_n=v_n^{-1}(t,\infty)$, it easy to check that $(*)$ is satisfied (use the continuity of $B\to P$). From now on we assume $\nu$ is a point mass, that is $\nu=\delta_t$ for some $t\in \mathbb{R}$. By the fact that $\int v_n=0$ and $\int|v_n|=1$ it is easy to conclude that $|t|\leq 1/2$ and it follows that $1/2\leq \|v_n-t\|<3/2$. It is also easy to see that the sequence $|v_n-t|$ is almost invariant in $L^1(X)$ and that $|v_n-t|_*\mu \to \delta_0$. We normalize this sequence, setting $u_n=|v_n-t|/\|v_n-t\|$. To summarize: we found a new almost invariant sequence of positive unit vectors $u_n\in L^1(X)$ such that $(u_n)_*\mu\to \delta_0$. We assume as we may that $\sum_{s\in S}\|su_n-u_n\|< \|u_n\|/n$. The construction of the sequence $A_n$ will follow by the "layer cake decomposition" method: for every positive function $f\in L^1(X)$, $\int_X f=\int_0^\infty f_*\mu(t,\infty)dt$. Applying this decomposition to the functions $|su_n-u_n|$ and $u_n$ we obtain $$ \sum_{s\in S}\int_0^\infty |su_n-u_n|_*\mu(t,\infty)dt=\sum_{s\in S}\int_X |su_n-u_n|=\sum_{s\in S}\|su_n-u_n\| $$ $$ < \|u_n\|/n = 1/n \int_X u_n =1/n \int_0^\infty (u_n)_*\mu(t,\infty)dt $$ and deduce the existence of $t>0$ for which $$ \sum_{s\in S} |su_n-u_n|_*\mu(t,\infty) < 1/n (u_n)_*\mu(t,\infty). $$ We set $A_n=u_n^{-1}(t,\infty)$ and conclude $\sum_{s\in S} \mu(A_n \Delta sA_n) < 1/n \mu(A_n)$. By the strict inequality we get $A_n\neq \emptyset$. Since $(u_n)_*\mu\to \delta_0$ we have $1-\mu(A_n)\to 1$, thus we may assume $A_n\neq X$, and indeed we get $$ \frac{\mu(A_n \Delta sA_n)}{\mu(A_n)(1-\mu(A_n))} \to 0. $$<|endoftext|> TITLE: "Classical" description of automorphic forms on unitary groups QUESTION [5 upvotes]: What I mean by classical: For the case of $GL_2$, the answer to my question would be that the automorphic forms are either Maas forms or modular forms. For $GSp(2n)$ these are the Siegel modular forms. Are there any "classical" objects which appeared in mathematics before this automorphic perspective which align with automorphic forms on $U(n)$ or other unitary groups? Maybe its the case that, historically, automorphic forms on unitary groups came first: then the question is: are there any other "models" of these kinds of functions? REPLY [10 votes]: These go by the name of "Hermitian modular forms". They occur very frequently in papers of Shimura (e.g. his monograph Arithmeticity on the theory of automorphic forms) and in other more recent works. For instance, this paper by Bouganis describes in detail at the beginning of section 2 the classical description of Hermitian modular forms for $U(n, n)$, the adelic description, and the passage between the two. There are similar descriptions available for $U(m, n)$ with $m \ne n$ as well.<|endoftext|> TITLE: Irrelevant parabolics and inner forms of GSp(4) QUESTION [5 upvotes]: In Ralf Schmidt's appendix to "Jacquet-Langlands-Shimizu correspondence for theta lifts to $\mathrm{GSp}(2)$ and its inner forms" by Narita and Okazaki , he computes the representations of $\mathrm{GSp}(1,1)$ by examining the tables in his book and discarding the representations whose Weil-Deligne representations fall in the Siegel parabolic. He obtains the answer of IIa, IVa, IVc, Va, Vb, Vc, and VIc. When I try to replicate the calculation I run into the following problem: representations of type VIa and VIb look almost identical to those of Va. But Va is included and VIa and VIb are not. Furthermore, as written almost all the nilpotent matrices seem to fall neatly into the Siegel parabolic, and the representations are diagonal. My guess is that we have to compute the image of the Galois representation, and that this cannot be easily read off from the Weil-Deligne representation. Unfortunately I've not succeeded in seeing how to do this either: the discussion in chapter $2$ of Local Newforms for $\mathrm{GSp}(4)$ is not helping. REPLY [4 votes]: I've never really got my head around ``relevant $L$-parameters'', but the condition on the $L$-parameter (i.e. the associated WD-rep) which ensures transfer to a given inner form should look something like the condition for being in the discrete series. Roughly speaking, representations whose $L$-parameters have fairly large image should show up on inner forms. I think here you want to discard the Weil--Deligne representations which land in the Levi of the Siegel parabolic. This should explain the third sentence in your second paragraph. For the VIa/b versus Va issue, we want to show that the Weil--Deligne representation for type VIa/b lands inside the Levi of a Siegel parabolic so we can discard it. It doesn't look like it does as written, but we are allowed to conjugate the image by a symplectic matrix, so it might be contained in a conjugate of the Levi of the standard Siegel parabolic. To show that the image is contained in a Siegel parabolic it suffices to show that it leaves stable a two-dimensional isotropic subspace. You can try to cook up one as follows: take an isotropic vector $v$ such that $N(v)$ is also isotropic, and moreover $\langle v, N(v) \rangle = 0$, then take the span of $\{v, N(v)\}$ as your isotropic subspace. Here we can take $v = \sqrt{-1}e_3 + e_4$. The reason that this works in the case in VIa/b but not in case Va is that in VIb/c the Frobenius acts on $e_1$ and $e_2$ (resp. $e_3$ and $e_4$) by the same way so it preserves the isotropic subspace, but in Va the character $\xi$ is necessarily non-trivial, so Frobenius does not preserve the isotropic subspace.<|endoftext|> TITLE: SDE for conditioning on subfiltration QUESTION [6 upvotes]: Setup: Suppose $X_t$ solves the SDE $$ dX_t = \mu(t,X_t)dt +\sigma(t,X_t)dZ_t, $$ where $Z_t$ is a Lévy process on $\mathbb{R}^d$, $g(t,s,x):[0,T]\times[0,1]\times \mathbb{R}^d \rightarrow \mathbb{R}^d$ is $C^{1,2,2}$ and $S_t$ is an SDE solving $$ dS_t = a(t,S_t,X_t)dt + b(t,S_t,X_t)dY_t, $$ where $Y_t$ is a Lévy process as well. Question: Is it possible to obtain an SDE for: $$ \mathbb{E}[g(t,S_t,X_t)|\sigma(X_s)_{0\leq s\leq t}]? $$ That is the projection of $g(t,S_t,X_t)$ onto the information only known throught $X_t$ up to time $t$? Idea: I was considering some type of filtering approach, but I'm not certain how to go about this. REPLY [4 votes]: You can obtain an SPDE (or, rather, an SIDE - stochastic integro-differential equation) for the conditional density of $S$ given the history of $X$. From here you can calculate the quantity you're looking for easily. In the continuous case (i.e. $Z$ and $Y$ are Wiener processes), there are plenty of references, look for Zakai's equation. When you have jumps, the main reference I know of is B. Grigelionis and R. Mikulevicius: Nonlinear filtering equations for stochastic processes with jumps.<|endoftext|> TITLE: Products of double-negation sublocales (and probability distributions on them) QUESTION [5 upvotes]: In locale theory, one can produce from any locale $A$ its double negation sublocale $A_{\neg\neg}$ via the nucleus which maps an open $U$ of $A$ to $\neg \neg U$, which is the interior of the closure of $U$. This quotient frame can be viewed as a sublocale of $A$. For many spaces, such as $\mathbb{R}$, the double-negation sublocale $\mathbb{R}_{\neg\neg}$ has no points. For any locales $A$ and $B$, there is a continuous map, which I believe is mono, $$f : (A \times B)_{\neg\neg} \to A_{\neg\neg} \times B_{\neg\neg}.$$ Does this map have a continuous inverse? That is, are these spaces homeomorphic, and does the double-negation operation commute with products? My hunch is that the answer is no. Taking $A$ and $B$ to both be $\mathbb{R}$, the "diagonal" relation on $\mathbb{R}$ should be clopen in $(\mathbb{R} \times \mathbb{R})_{\neg\neg}$ (in fact, it is equivalent to $\bot$), but I have a feeling that the diagonal of $\mathbb{R}_{\neg\neg} \times \mathbb{R}_{\neg\neg}$ shouldn't be open. If we have probability distributions $\mu_A : \mathcal{R}(A_{\neg\neg})$ and $\mu_B : \mathcal{R}(B_{\neg\neg})$, then the independent product is $$\mu_A \otimes \mu_B : \mathcal{R}(A_{\neg\neg} \times B_{\neg\neg}).$$ Under what conditions is this independent product the image of some probability distribution over $\mathcal{R}((A \times B)_{\neg\neg})$ of the $f$ defined above? That is, when does this product distribution satisfy the (potentially stronger) regularity properties? I think that this is the case when $A$ and $B$ are both $\mathbb{R}$, because I think that for $\mathbb{R}^n$, a probability distribution $\mu$ on $\mathbb{R}^n$ is absolutely continuous with respect to Lebesgue measure if and only if it can be expressed as a distribution over ${\mathbb{R}^n}_{\neg\neg}$, and since product measures preserve absolute continuity. I imagine there might be some analogies to be drawn with Dmitri Pavlov's discussion of measurable spaces in this MathOverflow answer. REPLY [3 votes]: I just want to expand on Simon's answers. Simon was correct regarding my mistaken interest in probability distributions over double-negation locales. Measures, Randomness, and Sublocales by Alex Simpson explains the "right" nucleus to use. For any given measure $\mu$, one can create the "smallest sublocale of measure 1", $\text{Ran}(\mu)$, with the nucleus $$U \mapsto \bigvee \{ V \ |\ U \le V, \mu(U) = \mu(V) \}.$$ If the original locale admits a presentation, then this smallest sublocale is generated by the presentation which adds to the original presentation the relation $\top \le U$ whenever $\mu(U) = 1$. I believe that these locales are not necessarily Boolean algebras. and I don't know how Simon's $\mathbb{R}^\text{Lebesgue}$ relates to $\text{Ran}(\lambda)$, where $\lambda$ is the Lebesgue measure on $\mathbb{R}$. In particular, $\text{Ran}(\lambda)$ is a sublocale of $\mathbb{R}$, whereas it sounds like $\mathbb{R}^\text{Lebesgue}$ isn't. Then, the appropriate question about products is whether, in general, $$ \text{Ran}(\mu \otimes \nu) \cong \text{Ran}(\mu) \times \text{Ran}(\nu).$$ The former is a sublocale of the latter, but we do not expect the latter to be a sublocale of the former. In particular, if $\mu$ and $\nu$ are both the standard normal distribution on $\mathbb{R}$, then the diagonal of the former is $\bot$ but on the latter is positive.<|endoftext|> TITLE: Manifold embedded in $R^{n+1}$ with a submanifold that doesn't embed in $R^n$ QUESTION [32 upvotes]: Suppose you have a closed $m$-dimensional manifold $M$, which embeds in $\mathbb{R}^{n+1}$ for some $n$. Can it have a closed submanifold $N$ (of dimension strictly smaller than $m$) which does not embed in $\mathbb{R}^n$? (By which I mean there is no embedding, not just that the restriction/projection of the first one doesn't work.) I'm not sure how important the dimension of $N$ is; it seems like codimension 1 would be the easiest place to find an example, but I'm also interested in higher codimension examples. In fact, is there an upper limit to the codimension in which such examples can exist? I would particularly like a codimension 1 example where both $M$ and $N$ are orientable. Or an example (orientable or not) for $n=2$ or $3$, but maybe there are obstructions in low dimensions... I am primarily thinking about smooth manifolds, but examples in the topological category would also be interesting. REPLY [17 votes]: Here is a simple example, using $2$-manifolds and $3$-manifolds. A Klein bottle does not embed in $\mathbb R^3$. But it is a submanifold of a 3-manifold that embeds in $\mathbb R^4$. For the $3$-manifold take the unit normal bundle of the standard embedding of the Klein bottle in $\mathbb R^4$. This bundle has a section, so it contains the Klein bottle. This bundle's Euler class is zero.<|endoftext|> TITLE: Distance between distributions and distance of moments QUESTION [9 upvotes]: Let's say I have a sequence of random variables $X_n$ such that $$\mathbf E X_n^k = \mathbf E X^k+O(a_k/\sqrt{n})\quad\text{for all }k\in\mathbb N,\tag{$\ast$}$$ where $X$ is a random variable of standard (zero mean, unit variance) Gaussian distribution and $a_k$ are some constants which typically grow with $k$ (for my purposes they would be $a_k=(k/2)!$, for example). Since the normal distribution is uniquely determined by its moments $(\ast)$ implies that $X_n\Rightarrow X$ weakly, as $n\to\infty$. Is there an appropriate way to formalize the distance $d$ between the distributions of $X_n$ and $X$, such that $(\ast)$ implies $$d(X_n,X)=O(f(n))$$ for some function $f$? I think $(\ast)$ is not strong enough to control distance measures like the Kullback-Leibler divergence or the Hellinger distance, but there might be some appropriate weaker notion? I think that $(\ast)$ implies $$\mathbf E f(X_n)=\mathbf E f(X)+O(1/\sqrt{n})$$ for a small class of test functions $f$, but depending on the growth of $a_k$, this class might be vary small. REPLY [4 votes]: The classical approach to this is called "smoothing" and is described in Chapter 16 of Feller's Introduction to Probability Theory, Vol 2. Basically you use the bounds on the moment differences to bound the difference between the characteristic functions near the origin, then adjust a parameter (called $T$ by Feller) to get the best result.<|endoftext|> TITLE: Do we have "cancellation law" for products of varieties QUESTION [10 upvotes]: Sorry for the naive question. Let $X_1$, $X_2$ and $Y$ be three projective varieties over an algebraically closed field of characteristic zero. If we have $X_1\times Y\cong X_2\times Y$, do we automatically get $X_1\cong X_2$? If not, do we have counter examples? If necessary, we could put stronger conditions (for example, smoothness) on the varieties. REPLY [14 votes]: You may look at T. Fujita's paper "Cancellation Problem of Complete Varieties". He proved that if $X_1$ and $Y$ are "Picard independent" (look at Proposition 3 therein for a definition) then cancellation holds for any $X_2$. In the same paper (Remark 8), the author cites T. Shioda "Some remarks on abelian varieties" in order to stress that cancellation does not hold for abelian varieties, as F. Polizzi said.<|endoftext|> TITLE: What was commutative algebra before (modern) algebraic geometry? QUESTION [27 upvotes]: Reading "H. Matsumura - Commutative Ring theory" I had the impression that the definitions were all made to mean something in algebraic geometry afterwards. I wonder what was commutative algebra before the modern algebraic geometry (which ideas were developed, how sophisticated they were, etc). Moreover, whether the success of algebraic geometry had inspired the expansion of commutative algebra, and how much. Any older reference would also be interesting (not for studying, just to figure out). Thanks REPLY [26 votes]: here is an excerpt from Zariski's talk at the Icm several decades ago: "The arithmetic trend in algebraic geometry is not in itself a radical departure from the past. This trend goes back to Dedekind and Weber who have developed, in their classical memoir, an arithmetic theory of fields of algebraic functions of one variable. Abstract algebraic geometry is a direct continuation of the work of Dedekind and Weber, except that our chief object is the study of fields of algebraic functions of more than one variable. The work of Dedekind and Weber has been greatly facilitated by the previous development of classical ideal theory. Similarly, modern algebraic geometry has become a reality partly because of the previous development of the general theory of ideals. But here the similarity ends. Classical ideal theory strikes at the very core of the theory of algebraic functions of one variable, and there is in fact a striking parallelism between this theory and the theory of algebraic numbers. On the other hand, the general theory of ideals strikes at most of the foundations of algebraic geometry and falls short of the deeper problems which we face in the post- foundational stage. Furthermore, there is nothing in modern commutative algebra that can be regarded even remotely as a development parallel to the theory of algebraic function fields of more than one variable. This theory is after all itself a chapter of algebra, but it is a chapter about which modern algebraists knew very little. All our knowledge here comes from geometry. For all these reasons, it is undeniably true that the arithmetization of algebraic geometry represents a substantial advance of algebra itself. In helping geometry, modern algebra is helping itself above all. We maintain that abstract algebraic geometry is one of the best things that happened to commutative algebra in a long time."<|endoftext|> TITLE: Why are Lagrangian submanifolds called Lagrangian? QUESTION [24 upvotes]: Much of the terminology in symplectic geometry comes from classical mechanics: the symplectic manifold is modeled on a cotangent bundle $T^*N$ of some configuration space $N$ with local position coordinates $q_i$ and momentum coordinates $p_i$ such that $\omega = \sum_i dq_i \wedge dp_i$. Then, a Hamiltonian - a.k.a. a smooth "energy" function - on this phase space induces a flow satisfying Hamilton's equations from classical mechanics: $$\frac{dp_i}{dt} = -\frac{\partial H}{\partial q_i} \hskip{1 in} \frac{dq_i}{dt} = \frac{\partial H}{\partial p_i}$$ In this physical interpretation, what are Lagrangian submanifolds? In particular, why are they named Lagrangian? Is there a relationship between this notion and the Lagrangian formulation of mechanics? (Note: this question has a lot of great answers providing some physical or geoemtric intuition for Lagrangian submanifolds - in a cotangent bundle, both fibers and the images of closed sections (closed $1$-forms on $N$) are motivating examples of Lagrangian submanifolds - but it does not address the source of the terminology.) REPLY [11 votes]: This echos the 2017 comments, but since the question has now been bumped to the front page it might be helpful to give the actual source in Maslov's book [1]. [1] V.P. Maslov, Perturbation Theory and Asymptotic Methods (in Russian, Moscow, 1965); Théorie des pertubations et méthodes asymptotique (French translation, Paris, 1972). The Lagrangian manifold (variété lagrangienne) is introduced on page 114-115:<|endoftext|> TITLE: Over the complex numbers, is there an example of an algebraic but non-representable matroid? QUESTION [5 upvotes]: According to Wikipedia https://en.wikipedia.org/wiki/Algebraic_matroid "For fields of characteristic zero (such as the real numbers) linear and algebraic matroids coincide" I cannot find the two references it provides. There is a theorem that if you are algebraic over F then you are linear over an extension of F. In the case of characteristic zero, or more precisely, the complex numbers, is it true without going to an extension? I checked the examples at the end of Oxley but none seem to have this property. EDIT: According to the review in Mathscinet, this is proved in Ingleton, A. W. Representation of matroids. 1971 Combinatorial Mathematics and its Applications (Proc. Conf., Oxford, 1969) pp. 149–167 Academic Press, London but I cannot access that reference, I just want to double check, or maybe see if there is another reference. REPLY [3 votes]: Algebraic over $\mathbb{C}$ implies realizable over $\mathbb{C}$. If your matroid is algebraic over $\mathbb{C}$, then it is realizable over some $\mathbb{C} \subset K$, so it is realizable over the algebraic closure $\overline{K}$, which explicitly means that a list of equalities and inequalities over $\overline{K}$ has a solution. But then these equations and inequalities have a solution over $\mathbb{C}$ as well.<|endoftext|> TITLE: What are the top journals in applied mathematics and what are the differences between them? QUESTION [7 upvotes]: This question is essentially an applied mathematics version of Which are the best mathematics journals, and what are the differences between them? Unfortunately, unlike the above question I was not able to find out so far what is the community consensus on top journals in applied mathematics (with statistics excluded -- it is a world of its own), and the question on top specialized journals also leaves the applied mathematics as a whole untouched. The problem is further exacerbated by the fact that while in principle many top generalist math journals declare willingness to publish applied mathematics papers, the fact is that such papers are published there only exceptionally, hence my question. To make things clear let me reiterate that I would like to understand what is the community consensus (if any) on what are the top applied mathematics journals rather than being shown various numerical rankings (like, say, those based on the impact factor) of which I am quite aware. REPLY [3 votes]: I don't think you can have a broad answer to this question. The reason is that there is no generalist applied mathematics journal, that everybody is aiming at. So you only have more focused top journals whithin each subject. Maybe you can aim for a list of those. I can give my very limited contribution with respect to fields adjacent to mine: For many areas: All SIAM journals. Optimization: Mathematical Programming Mathematics of Operations Research Numerical Analysis: IMA Journal of Numerical Analysis Journal of Computational Physics Inverse Problems Mathematics of Computation<|endoftext|> TITLE: Integral lattices in Lie group representations QUESTION [7 upvotes]: Let $G$ be a split semisimple algebraic group scheme over $\mathbf{Z}$ (I'm mostly interested in the case $G = Sp_4$). Let $V$ be an irreducible representation of the generic fibre $G_{\mathbf{Q}}$, of some highest weight $\lambda$, and $v$ a choice of highest-weight vector. Then there is a notion of an "admissible $\mathbf{Z}$-lattice" in $V$: it's a $\mathbf{Z}$-lattice $V_{\mathbf{Z}}$ in $V$ for which the structure map $G_{\mathbf{Q}} \to GL(V)$ extends to a map of $\mathbf{Z}$-group schemes $G_\mathbf{Z} \to GL(V_\mathbf{Z})$, and such that the intersection of $V_{\mathbf{Z}}$ with the highest-weight space $\mathbf{Q} \cdot v$ is $\mathbf{Z} \cdot v$. There are finitely many of these, and in particular there's a maximal and a minimal one. My question is this: let $P$ be a parabolic subgroup (containing our fixed choice of Borel) and $M$ its Levi factor. It's easy to see that $V$ breaks up as a direct sum of eigenspaces uder $Z(M)$, and the highest-weight eigenspace is the weight $\lambda$ representation $W$ of $M$, with $v$ as its highest weight vector. Moreover, the image in $W$ of an admissible lattice in $V$ is an admissible lattice in $W$. Is the projection to $W$ of the maximal (or minimal) admissible lattice in $V$ equal to the maximal (resp. minimal) admissible lattice in $W$? (If this works for the minimal lattice it works for the maximal one, and vice versa, because the minimal and maximal lattices are dual to each other.) REPLY [6 votes]: Let $\mathfrak{n}_-$ be the Lie algebra of the derived subgroup $N_-$ of $B_{-}$ where $B_{-}=TN_-$ is a Borel subgroup of $G$ opposite to a Borel subgroup $B_+\subset P$ (here $T$ is a split maximal torus of $G$ defined over $\mathbb{Z}$). If $V_\mathbb{Z}$ is a minimal admissible $\mathbb{Z}$-lattice then there is a highest weight vector $v\in V$ (for $B_+$) such that $V_{\mathbb{Z}}=U^{-}_{\mathbb Z}\cdot v$ where $U^{-}_{\mathbb Z}$ is the Kostant $\mathbb{Z}$-form of the universal enveloping algebra $U(\mathfrak{n}_-)$. It is spanned over $\mathbb{Z}$ by the PBW-monomials in $e_\alpha^n/n!$ where $\alpha$ is a negative root of $G$ with respect to $T$ and $n\in \mathbb{Z}_{\ge 0}$ (the root vectors $e_\alpha$ should come from a Chevalley basis of $\mathfrak{g}={\rm Lie}(G)$). It follows that $V_\mathbb{Z}\cap W$ is spanned over $\mathbb{Z}$ by the elements of the form $X\cdot v$ where $X$ is a PBW monomial in $e_\alpha^n/n!$ and $\alpha$ is a negative root of $P$ with respect to $T$. This shows that $V_\mathbb{Z}\cap W$ is a minimal admissible lattice for a Levi subgroup of $P$.<|endoftext|> TITLE: An Interesting Two Players' Game Involving Cumulative Sum of Uniform Distribution QUESTION [6 upvotes]: $A$ and $B$ are two players, each have exactly one turn. $A$ goes first. $A$ keeps on choosing a random number uniformly distributed over $(0,1)$ and add the values. If at one point it exceeds $1$, $A$ loses. If $A$ thinks his cumulative sum is very close to $1$, hence there is a risk of losing, he stops. Then $B$ starts the same process and add the values separately. If at one point $B$ exceeds $A$'s sum and still below $1$, he wins. What is the optimal strategy for $A$ to stop adding and what is the probability of winning in that case ($B$ knows the value $A$ stopped at)? From simulation It appears that the optimal threshold of $A$'s cumulative sum is approximately $0.5772$, which is very close to the Euler-Mascheroni constant $\gamma$. REPLY [11 votes]: The optimal strategy for A is to play until she reaches 0.570557. This is a root of $3e^x-x-2xe^x=e$. To see this, one can compute that A's probability of winning if she stops at $\alpha$ is $f(\alpha)$, where $f(x)=1-(1-x)e^x$. Her probability of winning if she carries on for one more turn and then stops is $g(\alpha)=\int_\alpha^1f(x)\,dx=1-e+2e^\alpha-\alpha-\alpha e^\alpha$. These two functions are equal when $\alpha=0.570557...$. You might ask why should you only play 1 step more? The answer is that at the critical point, playing one step more will neither increase nor decrease the chances of winning; but playing two additional steps will definitely decrease your chances of winning. EDIT: Here is a derivation of $f$. Consider the game from B's point of view once A has scored $t$. B just keeps generating random numbers until the sum is greater than $t$. If the sum is then less than 1, he wins; otherwise he loses. Let $g_\delta(x)$ be the probability that the cumulative sum lands in the interval $[x,x+\delta]$ (so that B's probability of winning is $1-f(t)=g_{1-t}(t)$). Now for any $x,\delta$ summing to at most 1, $g_\delta(x)=\delta+\int_0^x g_\delta(t)\,dt$: the probability of winning in 1 step plus the probability of winning in multiple steps. You can solve this to get $g_\delta(x)=\delta e^x$. Hence $f(t)=1-(1-t)e^t$.<|endoftext|> TITLE: Is there a categorical notion of reduced commutative algebras? QUESTION [12 upvotes]: A commutative ring $R$ is reduced if $r^2=0 \Rightarrow r=0$ holds for all $r \in R$. Commutative rings are precisely the commutative algebra objects in the symmetric monoidal category $(\mathsf{Ab},\otimes)$. This leads to the following question: Has any notion of "reduced commutative algebra objects" been studied for general linear symmetric monoidal categories (perhaps with some additional properties)? I would like to require that this notion should be naturally definable in the language of linear symmetric monoidal categories. (In particular, saying that an induced ringed space or stack is reduced would not qualify as an answer.) Ideally it should also be first-order and constructive; this rules out the condition that the intersection of prime ideals is zero. If $X$ is a well-behaved ringed space, then $\mathcal{O}_X$ should turn out to be reduced according to the categorical notion if and only if it is in the usual sense, i.e. for all opens $U \subseteq X$ the ring $\Gamma(U,\mathcal{O}_X)$ is reduced. This rules out the naive definition that $R$ is reduced if $r^2=0 \Rightarrow r=0$ holds for morphisms $r : 1 \to R$. Also notice that for arbitrary morphisms $r : T \to R$ (replacing $r^2$ by $r \otimes r$) the property would be far too strong. If there is no such definition, I wonder which additional structure do we need in order to give such a definition. Presumably, $\mathsf{Ab}$ has this extra structure. An idea would be to define $R$ reduced if for all invertible objects $\mathcal{L}$ and all morphisms $r : \mathcal{L} \to R$ with $r \otimes r = 0$ we have $r=0$. But this probably only gives the correct answer for ringed spaces where there are enough invertible objects. REPLY [3 votes]: One can use idals. The unit $1_{\mathcal{C}} \in \mathcal{C}$ is called reduced if for all idals $e : I \to 1_{\mathcal{C}}$ (i.e. morphisms $e$ satisfying $e \otimes I = I \otimes e$) with $e \otimes e = 0$ one has $e=0$. If $\mathcal{C}=\mathsf{Mod}(X)$ for some ringed space $X$, this gives the usual notion. It also gives the usual notion if $\mathcal{C}=\mathsf{Qcoh}(X)$ for a quasi-separated scheme $X$.<|endoftext|> TITLE: Block version of Maurey Pisier theorem QUESTION [6 upvotes]: The Maurey-Pisier theorem states that if $p_X$ is the supremum of those $p$ such that the Banach space $X$ has Rademacher type $p$, then $\ell_{p_X}$ is finitely representable in $X$. For $1\leq p<\infty$, let us say the Schauder basis $(e_i)_{i=1}^\infty$ has \emph{block type} $p$ if there exists a constant $C$ such that for any natural number $n$, any $0=k_0<\ldots TITLE: matrix inequality with orthogonal matrices QUESTION [15 upvotes]: I would like to know if for $A,B\in SO(3)$ the inequality $$ \|AB-BA\|_F\leq \|A-I\|_F\|B-I\|_F $$ holds, where $\|\cdot\|_F$ denotes the Frobenius norm and $I$ the identity matrix. Using the identity $$ AB-BA=(A-I)(B-I)-(B-I)(A-I) $$ one can show the inequality with a factor $2$. However the inequality seems to be true also without this factor. I encountered this problem when I tried to estimate the error of permuting two small rotations. REPLY [9 votes]: I confirm Peter's suggestion that the best constant is $\frac1{\sqrt2}$. For let $\omega$ be this best constant. Then the inequality amounts to writing $$3-{\rm Tr}(A^tB^tAB)\le2\omega^2(3-{\rm Tr}\,A)(3-{\rm Tr}\,B).$$ Let me parametrize $A$ by the angle of rotation $\theta$ and the axis of rotation $u$, a unit vector. Likewise $B$ has an angle of rotation $\alpha$ and an axis $v$. The left-hand side above is $8\omega^2(1-c)(1-c')$, where $c=\cos\theta$ and $c'=\cos\alpha$. Now $B^tAB$ is the rotation of same angle $\theta$ about $w=B^{-1}u$, and $A^t$ the rotation of angle $-\theta$ about $u$. It is an instructive calculus that $${\rm Tr}(A^tB^tAB)=2c+c^2+(1-c)^2(w\cdot u)^2+2s^2w\cdot u.$$ Because of $w\cdot u\ge c'$, we find that $\omega$ is the best constant in $$3-2c-c^2+(1-c)^2c'^2-2s^2c'\le8\omega^2(1-c)(1-c').$$ The left-hand side factorizes as $(1-c)(1-c')(3+c+c'-cc')$, and there remains the inequality $$3+c+c'-cc'\le8\omega^2.$$ Therefore $\omega^2$ is the supremum of $\frac18(3+c+c'-cc')$ over $[-1,1]^2$. Because this quantity if affine in both $c$ and $c'$, the supremum is achieved at a vertex, and its value is $4$. Whence $\omega^2=\frac48=\frac12$.<|endoftext|> TITLE: matroids representable over root of unity partial fields that are not the 6th root of unity partial field QUESTION [5 upvotes]: On reading about matroids representable over partial fields, one learns about the 6th root of unity partial field, but other even-th root of unity partial fields seem to be absent from the standard repertoire of examples. Why is this? Is the matroid theory boring or not useful for those partial fields? I'd be particularly interested in knowing what is known about matroids over $\{1, -1, i, -i\}$ because it might be useful to me in another problem. Thanks. REPLY [2 votes]: Consider the partial field $P:=\{p\in \mathbb{C}: |p|=1\}$. It is a theorem of Whittle that if a matroid $M$ is representable over $P$, then $M$ is sixth-root of unity. $P$ contains each $P_k:=\{k^\text{th}\text{ roots}\}$, so all the $k$-root of unity matroids are also sixth-root-of-unity. To represent $U_{2,4}$ over $P_k$ you need a $p\in P_k$ so that $(1-p)\in P_k$ - which happens only if $k$ is a multiple of $6$. Otherwise, the matroids over $P_k$ are both binary and sixth-root, and hence regular. If you're looking for a really interesting class extending the sixth-root-of-unity matroids, I can recommend the 'quaternionic unimodular' matroids defined here: https://arxiv.org/abs/1106.3088 They are pretty wild, but they should not be completely unmanageable: they do not contain $U_{2,7}$ if I remember right. It is a conjecture in that paper that quaternionic unimodular matroids have the half plane property.<|endoftext|> TITLE: Is the Čech cohomology of an orbifold isomorphic to its singular cohomology? QUESTION [7 upvotes]: Let $\mathcal{O}$ be a finite-dimensional, paracompact, Hausdorff, smooth (and compact, if it helps) orbifold. Is there an isomorphism between the real Čech cohomology and singular cohomology of the underlying space $|\mathcal{O}|$? (NB. The answers of Simon Rose and David C refer to an earlier version of the question and relate the Čech cohomology of the orbifold $\mathcal{O}$ to the singular cohomology of it's underlying space $|\mathcal{O}|$) REPLY [3 votes]: It's a classical result that if $X$ is paracompact and locally contractible, then singular cohomology and Cech cohomology of $X$ coincide, with coefficients in any abelian group. A reference is Spanier's textbook. This applies in particular to the underlying space of an orbifold, in which case a small neighborhood of any point is the cone on the link of the point.<|endoftext|> TITLE: Derived version of equivalence between motives and representations of Motivic galois groups? QUESTION [10 upvotes]: A slight variant $\tilde Mot_{num}(k,\mathbb{Q})$ of the category of pure motives $Mot_{num}(k,\mathbb{Q})$ is a Tannakian category equivalent to a category of representations of some algebraic group $GMot_k$ (the Motivic galois group): $$ Mot_{num}(k,\mathbb{Q}) \simeq Rep(GMot_k) \,. $$ nLab What it is known about the Derived category version (The derived category of motives) of this equivalence? Equivalence between the sub-category of compact objects of (say for example) étale Motivic sheaves and the derived category of representations of finite dimension of (Ayoub) Motivic galois groups REPLY [11 votes]: Let $k$ be a field and $\operatorname{DM}_{gm}(k)_{\mathbb Q}$ the ∞-category of rational geometric motives over $k$. A mixed Weil cohomology theory induces a symmetric monoidal exact functor $$ R: \operatorname{DM}_{gm}(k)_{\mathbb Q} \to D_c(\mathbb Q), $$ where $D_c(\mathbb Q)$ are the compact=dualizable objects in the derived ∞-category of $\mathbb Q$. For abstract reasons, there exists a best possible commutative Hopf algebra $H$ in $D(\mathbb Q)$ such that, if $G=\operatorname{Spec}(H)$, then $R$ lifts to $$ \tilde R: \operatorname{DM}_{gm}(k)_{\mathbb Q} \to \operatorname{Rep}_{D_c(\mathbb Q)}(G). $$ This general construction can be found in this paper by Iwanari. As you know, Ayoub has also constructed $H$, but only in the homotopy category of $D(\mathbb Q)$, which is not enough to define $\tilde R$. Let me now assume that $k\subset \mathbb C$ and that $R$ is the Betti realization. Ayoub then proves that $H$ is connective, so that $\tau_{\leq 0}H$ is a Hopf algebra in rational vector spaces. If $G^{cl}=\operatorname{Spec}(\tau_{\leq 0}H)$, we then have a closed immersion $i\colon G^{cl}\hookrightarrow G$, whence $$ i^*\circ \tilde R: \operatorname{DM}_{gm}(k)_{\mathbb Q} \to \operatorname{Rep}_{D_c(\mathbb Q)}(G^{cl}). $$ Choudhury and Gallauer Alves de Souza proved that the group scheme $G^{cl}$ is isomorphic to Nori's motivic Galois group. The strongest conjecture one can make here is that $i^*\circ\tilde R$ is an equivalence of ∞-categories. This conjecture is equivalent to the conjunction of two conjectures: $\tilde R$ is an equivalence, and $G^{cl}=G$. It also clearly implies the conservativity conjecture (that $R$ is conservative) and the existence of the motivic $t$-structure. Conversely, the existence of a $t$-structure such that $R$ is $t$-exact implies that $G^{cl}=G$. The conservativity of $R$ together with the existence of a suitable $t$-structure should imply that $\tilde R$ is an equivalence, but I cannot find an applicable reference at the moment.<|endoftext|> TITLE: The Hölder inequality for fractional order Sobolev seminorm? QUESTION [11 upvotes]: This question is post on MSE a week ago. I move it here to draw more attention. Let $u\in C^\infty(\bar I)$ be given where $I=(0,1)$. Define $$ t(\alpha):=\left(\int_I\int_I \frac{|u(x)-u(y)|^\alpha}{|x-y|^{1+s\alpha}}\right)^{\frac1\alpha} $$ where $1<\alpha<2$, $0 TITLE: Maximum of a B-spline QUESTION [6 upvotes]: Given $p+2$ nondecreasing (and not all identical) knots $t_0, \ldots, t_{p+1}$ on the real line, consider the normalized B-spline of degree $p$ defined over these knots. We know that the B-spline is nonnegative. Where is its maximum, and what is its value? If the knots are equispaced, the maximum is assumed at $(t_0 + t_{p+1})/2$. What is known about its value? REPLY [4 votes]: Using the recursive derivative formula (see for example here): $$N'_{i,p}(t) = \frac{p}{t_{i+p}-t_i} N_{i,p-1}(t) - \frac{p}{t_{i+p+1}-t_{i+1}} N_{i+1,p-1}(t)$$ We get that the maximum is achieved when: $$f(t) = \frac{N_{0,p-1}(t)}{t_{p}-t_0} - \frac{N_{1,p-1}(t)}{t_{p+1}-t_{1}} = 0$$ This corresponds to the intersection point of two basis functions of degree $p-1$, which lies in the interval $[t_1, t_p]$. Since the ratio between the denominators ${t_{p}-t_0}$ and ${t_{p+1}-t_1}$ can be arbitrary, I don't see a way to narrow it further than that.. So the problem now becomes finding the (single) root of this equation. One way to proceed is to use a bisection method over the $\{t_i\}$ knot values, and once the root knot interval is isolated you're in polynomial territory and can use any univariate polynomial root method you like, or just continue with the bisection inside the interval. Regarding the value of the maximum, for the general case, I don't see any simpler way than evaluating the basis function at the root point. However, for the special case of equispaced knots, the root is at $(t_0 + t_{p+1})/2$ as you correctly noted, and there is some symmetry that might be used to simplify some of the expressions. The basic recursive function for B-Spline basis functions is: $$N_{i,p}(t) = \frac{t-t_i}{t_{i+p}-t_i} N_{i,p-1}(t) + \frac{t_{i+p+1}-t}{t_{i+p+1}-t_{i+1}} N_{i+1,p-1}(t)$$ For the rest of the answer I assume for simplicity that $t_i = i$, this enables to derive expressions that involve only integer numerators and denominators. The basis functions become simpler: $$N_{i,p}(t) = \frac{t-i}{p} N_{i,p-1}(t) + \frac{i+p+1-t}{p} N_{i+1,p-1}(t)$$ Note that this recursion implies that the denominator of the value is always the factorial $p!$ (since you can factor $\frac{1}{p}$ out in every level of the recursion). The case where $p$ is odd is simpler since the root falls on a knot and specifically on the middle knot $t_{\frac{p+1}{2}} = (p+1)/2$. Assigning the above recursive function to the knots (omitting the denominator) results in a nice "Pascal-triangle-like" pattern: $$ \begin{array}{{13}{c}} p=1 &&&&&0&&1&&0&&&&\\ p=2 &&&&0&&1&&1&&0&&&\\ p=3 &&&0&&1&&4&&1&&0&&\\ p=4 &&0&&1&&11&&11&&1&&0&\\ p=5 &0&&1&&26&&66&&26&&1&&0&\\ \end{array} $$ Each row in the triangle above represents the (numerator of the) values of the basis function at the knots. The values for row $p$ in the triangle are computed with the recursive formula (the start and end values are zero by definition): $$V_p[k] = k V_{p-1}[k] + (p+1-k) V_{p-1}[k-1] , k=1..p$$ This recursive relation is derived directly from the basis recursive formula above. The maximum value (finally) for the odd cases is then $\frac{V[(p+1)/2]}{ p!}$. For example, for $p=3$ the maximal value is $\frac{4}{3!} = \frac{2}{3}$ and for $p=5$ it is $\frac{66}{5!} = \frac{11}{20}$. EDIT: Turns out this triangle has a respectable name - "Euler's Triangle". See Wikipedia and this MO answer for its nice properties. For the even case, the derivation is a little more complicated, since the maximal value is achieved at the middle of the middle knot interval, namely at $(p+1)/2$, which is now no longer integer. To compute this value we evaluate the values at half-points (namely at the $p+3$ points $-0.5, 0.5, 1.5, ..., p+0.5, p+1.5$). Similar to the integer solution the recursive formula results in an integer numerator and denominator, however the denominator here is $2^p p!$ since we deal with half-points. The resulting triangle is: $$ \begin{array}{{13}{c}} p=1 &&&&0&&1&&1&&0&&&\\ p=2 &&&0&&1&&6&&1&&0&&\\ p=3 &&0&&1&&23&&23&&1&&0&\\ p=4 &0&&1&&76&&230&&76&&1&&0&\\ \end{array} $$ Each row in the triangle above represents the (numerator of the) values of the basis function at the $p+3$ half points above. The values for line $p$ are computed with the recursive formula (the start and end values are zero by definition): $$V_p[k] = (2k-1) V_{p-1}[k] + (2p+3-2k) V_{p-1}[k-1] , k=1..p+1$$ The maximum value (finally finally) for the even cases is then $\frac{V[(p+2)/2]}{2^p p!}$. For example, for $p=2$ the maximal value is $\frac{6}{2^2 2!} = \frac{3}{4}$ and for $p=4$ it is $\frac{230}{2^4 4!} = \frac{230}{384}$.<|endoftext|> TITLE: Fundamental groups of complements of divisors in $\mathbb P^2$ QUESTION [6 upvotes]: Let $D$ be a divisor in $\mathbb P^2_{\mathbb C}$ and let $X= \mathbb P^2_{\mathbb C} - D$. Under what condition on $D$ is the fundamental group of $X$ infinite? REPLY [11 votes]: I'd leave this as a comment, but I don't have enough reputation. Consider the long exact sequence in homology of the pair $(\mathbb{P}^2, \mathbb{P}^2-D)$. Since $H_1(\mathbb{P}^2,\mathbb{Z}) = 0$ and $H_2(\mathbb{P}^2, \mathbb{Z}) = \mathbb{Z}$, this has a segment $$\mathbb{Z}\rightarrow H_2(\mathbb{P}^2, \mathbb{P}^2-D)\rightarrow H_1(\mathbb{P}^2-D)\rightarrow 0.$$ The middle term is freely generated by the components of $D$. So if $D$ has 2 or more irreducible components, $H_1(\mathbb{P}^2-D, \mathbb{Z})$ has positive rank and $\pi_1(\mathbb{P}^2-D,*)$ is infinite.<|endoftext|> TITLE: MathSciNet review alert? QUESTION [26 upvotes]: Other than reading MathSciNet regularly, is there a way to get notified when one of my papers gets a review on MathSciNet? REPLY [18 votes]: Though only semi-related to the question (couldn't submit it as a comment), I may take the liberty to add that for zbMATH subscribers there is an Atom Feed providing such regular updates for, e.g., subjects or authors. In your case, it would be sufficient to check your author ID at your profile https://zbmath.org/authors/?q=P%C3%A1lv%C3%B6lgyi+D%C3%B6m%C3%B6t%C3%B6r (i.e., Author ID palvolgyi.domotor) - and to use it to subscribe for https://zbmath.org/atom/ai/palvolgyi.domotor with you favourite news reader. Olaf Teschke, Managing Editor, zbMATH<|endoftext|> TITLE: Arrangement of the introduction of a paper QUESTION [14 upvotes]: In my understanding the introduction of a paper is a very important section because it not only gives a concise summary of the results, but it is also the only section most people will read at all. Therefore I often ask myself how to arrange it in such a way that, for example, the results are easy to grasp, the results are clearly motivated, the reader gets convinced of the significance of the results, the reader gets interested to read the rest of the paper. Now there are surely several ways to do this, but in order to keep it focused, for this question I would only like to know the following: Do you prefer to give a summary of the results, including short remarks about how the objects you have studied are defined, and then later give some context and motivation for these results, or do you prefer it vice versa? Or do you prefer to give statement, context and motivation of a result and then move on to the next result? And what are your main reasons for doing so? In particular, how are your goals of an introduction, for example those mentioned above, achieved by these decisions? Is there an arrangement which is more common? REPLY [6 votes]: To some extent this depends on the nature of your result, but the principle I generally follow is this: State your result as soon as possible, subject to the constraint that when readers arrive at the statement of your result, they will have some idea of what question you are answering and why. For example, if you're proving a conjecture that is easy to state, or is likely to be known already to anyone who is interested in your paper, then you can pretty much launch right in and say that your paper proves so-and-so's conjecture, and state the conjecture. If you're proving a conjecture that isn't so well known or easy to state, then I'd start by saying that you're proving so-and-so's conjecture. Then give the minimum necessary background to understand the statement of the conjecture and why it was made. After that you can give further context and motivation if it seems appropriate. If you're not proving someone's conjecture but are proving something that bears a simple relationship to an existing result (e.g., it's the converse of some known theorem, or a generalization), then again I would start by saying that you're (for example) generalizing so-and-so's result. Then state the other result, and state your result. Further context and motivation can come afterwards if appropriate. If your result does not fall neatly into one of these categories but is more of a general theoretical advance, then it may be trickier to decide how much background to give before stating your result. I usually lean towards giving as little background as possible, because it's easy to make the mistake of giving too much background, and losing the interest of the reader. As a first draft, try stating your result without any context, and then ask yourself if your typical reader would react by thinking, "Why on earth would anyone prove such a result?" If so, then back up and give just enough context to pre-empt such a question. One general comment is that if your result is difficult to state precisely because the precise statement is very complicated, then see if you can come up with a simplified version, even if you have to be slightly vague or have to weaken your result to do so. This will allow you to state your result sooner, in accordance with the highlighted principle above. The more precise version of your theorem can come later, after the reader has decided that it's of sufficient interest.<|endoftext|> TITLE: Symmetric group action on Young Tableaux QUESTION [8 upvotes]: Let $S_n$ be the symmetric group and let $\lambda$ be a partition of $n$ and let $S$ be the set of all standard young tableaux of shape $\lambda$. Is there a well defined action of $S_n$ on the set $S$ ? Permuting entries of a standard tableau may not give a standard tableau. May be some kind of rearrangement needed after the permutation action. REPLY [9 votes]: Such an action was defined by Lascoux and Schützenberger in their paper Le monoïde plaxique, http://igm.univ-mlv.fr/~berstel/Mps/Travaux/A/1981-1PlaxiqueNaples.pdf. Another reference is Theorem 1.5 of https://arxiv.org/pdf/q-alg/9709010.pdf. REPLY [7 votes]: To give an alternative reference, see the intro in Crystals For Dummies by M. Shimozono, where you want to be looking at the action $s_i$. This is strongly related to the action as R. Stanley refer to, and also extends naturally to semi-standard tableaux. EDIT: On further reflection, the crystal reflection operators FIX all standard Young tableaux (but act nicely on SSYT). So this action is quite boring considered only on SYT.<|endoftext|> TITLE: Understanding model independently the equivalence of two ways of obtaining homotopy types from categories QUESTION [6 upvotes]: It is well known that any homotopy type can be obtained as the classifying space of a ($1$-)category. The classifying space of a category $\mathcal{C}$ can be interpreted in at least two ways: We can view $\mathcal{C}$ as an object in the $(\infty,1)$-category of $(\infty,1)$-categories, and localise $\mathcal{C}$ along all its morphisms. We can consider the constant functor $\mathcal{C} \to \mathbf{Type}, \; X \mapsto *$ (where $\mathbf{Type}$ denotes the $(\infty,1)$-category of homotopy types), and take its colimit. (Or thus equivalently the free homotopy colimit of the unique functor $\mathcal{C} \to 1$, where $1$ denotes the category with only one morphism; see Dugger). Using various models these two views as well as their equivalence can be made precise as follows: View $\mathcal{C}$ as an object in $\mathbf{SSet}$ equipped with the Joyal model structure. Inverting all morphisms of $\mathcal{C}$ just corresponds to taking its fibrant replacement in the Kan model structure. Taking the colimit in 2. can be formalised by taking the geometric realisation of (the nerve of) $\mathcal{C}$: This is exactly the formula you get for computing the homotopy colimit of the constant functor $\mathcal{C} \to \mathbf{Top}, \; X \to *$ using the simplicial replacement of this functor (described e.g. here). My question then is: Is there a model independent way of seeing that these two constructions are equivalent? So I'm looking for an argument which could be formalised in any model of $(\infty,1)$-categories. REPLY [7 votes]: Here is an argument, which is basically Denis Nardin's comment. To have a model independent proof you need model independent definitions of the hocolim and of the localization. You can define them via adjunctions, but from a pragmatic point of view I am not sure this is so helpful. Ultimately to do any kind of calculation you will have to pick a model and a construction of the localization/hocolim and if your definition is the model independent one, then you have the additional task of proving that they satisfy these definitions. Anyway... There is an ($\infty$-)adjuction of $(\infty,1)$-categories: $$hocolim_C : Cat_{(\infty,0)}^C \leftrightarrows Cat_{(\infty,0)}: const$$ and this is a definition $hocolim_C$. Here $Cat_{(\infty,0)} \simeq Top$ is the usual $(\infty,1)$-category of spaces which the OP called Type. There is another one: $$ ||-||: Cat_{(\infty,1)} \leftrightarrows Cat_{(\infty,0)}: i $$ which defines the localization at all morphisms $||C||$. Here $i$ is the inclusion of $(\infty,0)$-cats (= spaces) into all $(\infty,1)$-categories. Then this means that for all spaces $X$ we have $$ \begin{align} Cat_{(\infty,0)}( hocolim_C const(*) , X) &\simeq Cat_{(\infty,0)}^C( const(*), const(X)) \\ &\simeq Fun(C, Cat_{(\infty,0)}(*, X)) \\ &\simeq Fun( C, iX) \\ & \simeq Cat_{(\infty,0)}( ||C||, X) \end{align} $$ And so by Yoneda you conclude that $hocolim_C const(*) \simeq ||C||$.<|endoftext|> TITLE: Does there exists a complete Kahler metric with positive scalar curvature on Poincare disk? QUESTION [6 upvotes]: Let (M,g) be a Poincare disk. Does there exists a complete Kahler metric h, such that the scalar curvature of h is positive? REPLY [3 votes]: The answer to your question is no.In case of surfaces the scalar curvature and Ricci curvature coincide with Gauss curvature. If an n dimensional M had a complete Riemannian metric h with nonnegative Ricci curvature then by the Bochner Weitzenbock formula all L^2 harmonic one forms would be parallel.By an observation of Calabi M has infinite volume. Therefore all L^2 harmonic forms are zero. Now L^2 condition on one forms is a conformally invariant condition on a Riemann surface. However the disc has many nonzero L^2 harmonic one forms .This leads to a contradiction.<|endoftext|> TITLE: How is compactness related to countable saturation? QUESTION [6 upvotes]: By Cantor's intersection theorem every decreasing nested sequence of nonempty compact sets has a common point. A superficially similar result holds that every decreasing nested sequence of nonempty internal sets in an ultrapower model of a hyperreal field ${}^\ast\mathbb{R}$ has a common point, a property known as countable saturation. Is such a resemblance more than superficial? REPLY [2 votes]: It turns out that there is a direct relationship between Cantor's lemma for nested compact sets $K_n$, on the one hand, and nested sequences of internal sets, on the other. Namely, the latter can be viewed as a special case of the former as follows. The decreasing nested sequence of internal sets, $\langle{}^\ast\!K_n\colon n\in\mathbb N\rangle$, has a common point $x$ by saturation. But for a compact set $K_n$, every point of ${}^\ast\!K_n$ is nearstandard, i.e., infinitely close to a point of $K_n$ (in other words, every point is in the halo of a standard point of $K_n$). In particular, $st(x)\in K_n$ for all $n$, as required.<|endoftext|> TITLE: Integral polynomials dividing N! QUESTION [25 upvotes]: Consider a polynomial $P(X)\in\mathbb Z[X]$. Is it true that $P(N)$ divides $N!$ for infinitely many integer $N$? This question is motivated by the special case where $P(X) = X^2 + 1$ that appeared in a math olympiad. I was wondering if anyone can point me to references of this question. REPLY [16 votes]: In general this is an open problem. A closely related problem (essentially equivalent) is to ask for the values $P(n)$ for $n$ of size $X$ to be $X$ smooth (i.e. composed only of primes below $X$). This is known for quadratic polynomials, but already open for general cubic irreducible polynomials. For some special polynomials (e.g. those of the form $ax^d+b$) such results are known, thanks to Schinzel, Balog and Wooley etc. For a description of related results, and more references, see this paper of Dartyge, Martin and Tenenbaum.<|endoftext|> TITLE: How to check whether a module is an n-th syzygy QUESTION [6 upvotes]: Given a finite dimensional algebra $A$, define $\Omega^{n}(mod-A)$ (modules here are always finite dimensional) to be the full subcategory of projective modules or modules $M$ such that $M \cong \Omega^{n}(N)$ for some other module $N$. How can one check in an easy way to see whether a given module $M$ (we can assume it is indecomposable) is in $\Omega^{n}(mod-A)$? For special subcategories there is sometimes an easy answer. Example: Let $A$ have dominant dimension $d \geq 1$, then $\Omega^{n}(mod-A)$ is equal to the full subcategory of modules having dominant dimension at least $n$. I'm interested in calculating the intersection $\Omega^{\infty}$ of all $\Omega^{n}$ for $n \geq 1$ for special algebras (which can be assume to be representation-finite if that helps) with the GAP package QPA. REPLY [4 votes]: A method for this is now implemented in QPA as IsNthSyzygy. For a given indecomposable module $M$, one checks whether $M$ is a direct summand of $\Omega^n(\Omega^{-n}(M)) \oplus P(M)$ when $P(M)$ is the projective cover of $M$.<|endoftext|> TITLE: Books for (Complex) Algebraic Curves QUESTION [7 upvotes]: I would like suggestions for some good books on (Complex) Algebraic Curves. I am searching a book for Undergraduate-Beginner Level in this part of mathematics: algebraic curves. I found some books like "Plane Algebraic Curves" from Gerd Fischer, "Complex Algebraic Curves" from Frances Kirwan, "Elementary Geometry of Algebraic Curves: An Undergraduate Introduction" from Gibson but these were too difficult for my level. Also, please suggest my a book (or combine chapters of books or notes), which covers the following topics: Plane Curves; The set of points $V(f)$ of a plane curve, Transformations in $\mathbb{C}^2$, Conics, Intersection number, Isolated (abnormal) points, tangent lines, rational curves Intersection Of Curves; The projective space, Lines in projective space, Projective transformations, Εquivalence of projective curves, Intersection number, Irregular poits, Tangent Lines, Flex points, Rational Projective Curves Linear Systems of Curves; Intersection Points, Bézout's theorem Thanks in advance. REPLY [4 votes]: Keith Kendig's A guide to plane algebraic curves is quite elementary and readable. REPLY [3 votes]: Another beginner-friendly book is P.A. Griffiths, Introduction to algebraic curves.<|endoftext|> TITLE: Infinitely many primes $p \in \mathbb{Z}$ where reduced curve $E/\mathbb{F}_p$ has Hasse invariant $1$? QUESTION [6 upvotes]: Let $E$ be an elliptic curve defined over $\mathbb{Q}$, and fix a Weierstrass equation for $E$ having coefficients in $\mathbb{Z}$. Are there infinitely many primes $p \in \mathbb{Z}$ such that the reduced curve $E/\mathbb{F}_p$ has Hasse invariant $1$? REPLY [9 votes]: Yes. In fact, Elkies and Serre [1] have independent proofs that for every $\epsilon>0$ there is a constant $C_\epsilon>0$ such that $$ {\#\{p\le X : E/\mathbb{F}_p \text{ is ordinary}\}} \ge C_\epsilon X^{3/4-\epsilon}. $$ But if you just want to know that there are infinitely many ordinary primes, there is an elementary proof (also suggested by Serre) sketched in [2, Exercise 5.11]. [1] Serre, Quelques applications du theoreme de densite de Chebotarev, IHES Publ. Math. 54 (1981), 323-401. [2] Silverman, The Arithmetic of Elliptic Curves, GTM 106, Springer.<|endoftext|> TITLE: What exactly is a judgement? QUESTION [8 upvotes]: Before formulating my question, let me briefly sum up what I know about the topic (feel free to correct me if something I claimed is false!). This is for you good to see what my state of knowledge is, and maybe some people could learn something. This is what I know An interesting problem in logic is: how can one precisely define a relation $\Gamma\vdash\phi$ between a set $\Gamma$ of formulae and a single formula $\phi$, that formalizes the intuitive notion of "there is a proof of $\phi$ with the axioms $\Gamma$". There are different calculi that solve this problem, for example Hilbert systems and calculi of natural deduction. Without question the more natural solution is a natural deduction calculus which tries to model logical reasoning as it's done by mathematicians. This calculus operates with sequents which are ordered pairs $(\Gamma, \phi)$ and tries to define what a derivable sequent is. This is done through inference rules, which tell us how to get from given sequents to other sequents. For example the rule $$\frac{(\Gamma\cup\{\psi\}, \phi)}{(\Gamma, \psi\rightarrow\phi)}$$ tells us that if $(\Gamma\cup\{\psi\}, \phi)$ is a derivable sequent, then so is $(\Gamma, \psi\rightarrow\phi)$. One can state a bunch of these rules and then define that a sequent is derivable iff it can be shown to be so by a finite number of applications of the given rules. Now we can define our syntactic consequence relation $\vdash$ as follows: $$\Gamma\vdash\phi\quad:\Leftrightarrow\quad \text{the sequent $(\Gamma, \phi)$ is derivable.}$$ When I learned this it was quite confusing to me that $\Gamma\vdash\phi$ can denote two different things: The statement that says that the sequent $(\Gamma, \phi)$ is derivable. The sequent $(\Gamma, \phi)$ itself is sometimes denoted $\Gamma\vdash\phi$. Thus the inference rule above can also be written like that: $$\frac{\Gamma\cup\{\psi\}\vdash \phi}{\Gamma\vdash \psi\rightarrow\phi}.$$ My question What exactly is a judgement? Looking into wikipedia In mathematical logic, a judgment or assertion may be thought of as a statement or enunciation in the meta-language. For example, typical judgments in first-order logic would be that a string is a well-formed formula, or that a proposition is true. Similarly, a judgment may assert the occurrence of a free variable in an expression of the object language, or the provability of a proposition. In general, a judgment may be any inductively definable assertion in the metatheory. If I understand it correctly then judgement is not a technically defined term but is used to describe meta-statements about a formally defined logic. But it seems to me that this interpretation isn't compatible with many things I read. For example on the wikipedia page about natural deduction linked to above it says: The general form of an inference rule is: $$\frac{J_1\quad J_2\quad\cdots\quad J_n}{J}$$ where each $J_i$ is a judgment. But in my way of thinking sequents are not judgement. In particular, I think that the rule $$\frac{(\Gamma\cup\{\psi\}, \phi)}{(\Gamma, \psi\rightarrow\phi)}$$ does not state that if $(\Gamma\cup\{\psi\}, \phi)$ then $(\Gamma, \psi\rightarrow\phi)$ (because this wouldn't make sense, the sequent $(\Gamma\cup\{\psi\}, \phi)$ is a mathematical object and can't imply something; to say "if $3$ then $5$" is bullshit too) but instead I think that this rules says that if the sequent $(\Gamma\cup\{\psi\}, \phi)$ happens to be derivable then $(\Gamma, \psi\rightarrow\phi)$ is derivable. Also, wikipedia says: The most important judgments in logic are of the form "$A$ is true". Shouldn't it be "valid" instead of "true"? The formulae $A$ could be true in some models but false in others, right? So what does it me for a formula $A$ to be "true"? The nlab and wikipedia tell me that "$P$ prop" is a judgement too which says that $P$ is a well-formed formula. Wikipedia for example gives this inference rule $${\frac {A{\hbox{ prop}}\qquad B{\hbox{ prop}}}{(A\wedge B){\hbox{ prop}}}}$$ I wonder why they used the vertical line here. I thought above and under the vertical line there should be mathematical objects, and the rule tells us that if the objects above are derivable in a certain calculus then the mathematical object(s) under the vertical line are derivable objects in the particular calculus. But "$P$ prop" is in my way of thinking not a mathematical object, but a statement in the meta-language. Thus if one was to give a calculus that defines which words over the alphabet are well-formed then the rule should be $${\frac {A\qquad B}{(A\wedge B)}}$$ which says that $A, B\text{ well-formed}\implies (A\wedge B)\text{ well-formed}$ The nlab page also talks about judgements $(x:X)⊢(ϕ\text{ prop})$ (here $\vdash$ obviously means $\vdash$ in the second meaning explained above). This is a judgement in a judgement. Is this allowed? It says that the statement that $\phi$ is well-formed is a consequence of the typing judgement $x:X$, right? Could you please clarify my confusions and explain to me what exactly a judgement is? REPLY [10 votes]: I highly recommend reading Martin-Löf's paper referenced by Ulrik Buchholtz in the comments to your question. Apart from that, here are a couple of point that might help, some of which were already made by Andreas Blass in his answer. A judgement is an act of knowing, or asserting a piece of knowledge about a mathematical object. For instance, a non-traditional single-sorted first-order logic could have the following kinds of judgments: $t\ \mathsf{term}$, intended meaning "$t$ is a well-formed term" $P\ \mathsf{prop}$, intended meaning "$P$ is a well-formed formula" $\Gamma\ \mathsf{hyps}$, indented meaning "$\Gamma$ is a well-formed list of hypotheses" $\Gamma \vdash P\ \mathsf{holds}$, intended meaning "formula $P$ is derivable under hypotheses $\Gamma$". You do not normally see all of these because for first-order logic the first three are very simple and they get relegated to an informal explanation of syntax. But notice that there is a difference between the syntactic object "$P \land Q$" and the assertion "$P \land Q$ is a well-formed formula" (a judgment). If you doubt that there is a difference, consider the syntactic object "$P \land (Q \lor R$". We can speak meaningfully about it ("It's missing a closing parenthesis", "It contains two connectives."), but we do not assert "$P \land (Q \lor R$ is a well-formed formula". Let me emphasize the difference between the sequent $$\Gamma \vdash P$$ and the judgement $$\Gamma \vdash P\ \mathsf{holds}.$$ The sequent is a syntactic object. The judgement is an assertion about the syntactic object. A lot of texts use the same notation for both, and that's where some of the confusion may be coming from. (Also note that many texsts write "$\mathsf{true}$" instead of "$\mathsf{holds}$", but I want to emphasize here that I am not talking about semantic truth, but ratehr about derivability.) Rules of inference explain how we can build evidenece of assertions. The "syntactic" assertions, such as as "$P$ is a well-formed formula" or "variable $x$ does not occur in $P$" can be explained in terms of inference rules, but you won't see them in logic textbook, because they prefer to explain syntax in an informal way, e.g., "If $P$ is a wff and $Q$ is a wff then $P \land Q$ is a wff." If we want to, we can write down the corresponding formal rule: $$\frac{P\ \mathsf{prop} \qquad Q\ \mathsf{prop}}{(P \land Q)\ \mathsf{prop}}.$$ This says: "If $P$ is a well-formed formula and $Q$ is a well-formed formula, then $P \land Q$ is a well-formed formula." It is different from the rule $$\frac{\Gamma \vdash P\ \mathsf{holds} \qquad \Gamma \vdash Q\ \mathsf{holds}}{\Gamma \vdash P \land Q\ \mathsf{holds}}$$ which says: "If $P$ is derived from hypotheses $\Gamma$ and $Q$ is derived from hypotheses $\Gamma$, then $P \land Q$ can also be derived from hypotheses $\Gamma$." To see why it might make sense to have formal rules for establishing that something is a wff, consider the rule: $$\frac{ }{\Gamma \vdash t = t\ \mathsf{holds}}$$ It does not say that $t$ must be a well-formed term. It would be more precise to say: $$\frac{\Gamma\ \mathsf{hyps} \qquad t\ \mathsf{term}}{\Gamma \vdash t = t\ \mathsf{holds}}$$ Now the rule is explicit: reflexivity holds only under well-formed hypotheses, and only for well-formed terms. Of course, we need to give rules for $t\ \mathsf{term}$ and $\Gamma\ \mathsf{hyps}$. They would be something like $$ \frac{ }{()\ \mathsf{hyps}} \qquad\qquad \frac{P\ \mathsf{prop} \qquad \Gamma\ \mathsf{hyps}}{(P, \Gamma)\ \mathsf{hyps}} $$ and (assuming that the language has variables $x_k$, a constant $1$ and a binary operation $+$): $$ \frac{k \in \mathbb{N}}{x_k\ \mathsf{term}} \qquad\qquad \frac{ }{1 \ \mathsf{term}} \qquad\qquad \frac{t\ \mathsf{term}\qquad u\ \mathsf{term}}{(t + u)\ \mathsf{term}} $$ I am fudging details about meta-level natural numbers that are used for indexing variable names. You can devise your own solution (for instance use unary representation of numbers, and give inference rules for such a representation). It is perhaps silly to follow such rigorous formal discipline for first-order logic, but there are formal systems where the discipline is necessary. Dependent type theory is an example, as well as formal systems describing programming languages. These can grow quite complex, and since they also get implemented in practice, it is quite helpful to have all details written down formally.<|endoftext|> TITLE: Producing non-trivial homotopy classes in $\infty$-dimensional vector spaces by removing points QUESTION [7 upvotes]: It is renowned that removing a point from an $\infty-$dimensional vector space $H$ preserves its contractibility (by Kuiper's theorem). Is there any hope that $H\setminus A$ is not contractible, where $A$ is a disjoint union of points (possibly infinitely many, but still countable)? REPLY [9 votes]: In comments it is stated that the space of most interest to the OP is the space of $C^k$ Riemannian metrics on a compact manifold. If you look at the space of $C^k$ Riemannian metrics with $C^k$ topology, where $k$ is finite or infinite, then the space is Fréchet and separable. In any separable Fréchet space $X$ the complement of any countable union of compact sets is homeomorphic to $X$ (this is proved in Bessaga-Pełczyński book, see definition IV.5.1 and section V.6). Perhaps, the OP is interested in the space of $C^\infty$ Riemannian metrics with $C^k$ topology, where $k$ is finite. Theorem 1.1 in https://arxiv.org/abs/1510.07269 determines the homeomorphism type of the space. Namely, if $\Sigma$ denotes the linear span of the Hilbert cube in $l^2$, then the space of the $C^\infty$ Riemannian metrics with $C^k$ compact-open topology is homeomorphic to $\Sigma^\omega$, the product of countably many copies of $\Sigma$. I do not know whether any countable subset of $\Sigma^\omega$ has a contractible complement.<|endoftext|> TITLE: Reduction mod $p$ map is injective? QUESTION [14 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Q}$ with good or multiplicative reduction at $p$. How do I see that the reduction mod $p$ map from the endomorphisms of $E$ over $\overline{\mathbb{Q}}$ to the endomorphisms of the reduction of $E$ mod $p$ over $\overline{\mathbb{F}}$ is injective? REPLY [11 votes]: [The following is a vast simplification of an answer I gave a bit earlier, so I deleted most of the earlier answer. For anyone who cares about the sanctity of upvoting points it is all OK since I am certain anyone who upvoted the earlier answer would upvote this one too.] Any specific endomorphism is defined over an actual number field $K$ inside $\overline{\mathbf{Q}}$, and in case of semistable reduction the formation of the identity component of the special fiber of the Neron model commutes with any local extension of the discrete valuation ring. Thus, we can work relative to the discrete valuation ring $O_{K, \mathfrak{p}}$ (localization at a prime over $p$). It is sufficient to prove more generally that if $A$ and $B$ are abelian varieties (or just elliptic curves) over the fraction field $F$ of a discrete valuation ring $R$ with residue field $k$ and their Neron models $\mathcal{A}$ and $\mathcal{B}$ over $R$ have $k$-fibers with semistable reduction then the natural map $${\rm{Hom}}_F(A,B) = {\rm{Hom}}_R(\mathcal{A}, \mathcal{B}) \rightarrow {\rm{Hom}}_k(\mathcal{A}^0_k, \mathcal{B}^0_k)$$ is injective. In what follows one can work throughout with elliptic curves if that is preferred, and the main idea can be carried out in more elementary terms for elliptic curves, but I give the general case because it is more attractive. The first thing to observe is that if $n$ is a nonzero integer then its effect on the semi-abelian identity components of the special fibers of the Neron models is an isogeny (by viewing such identity components as extensions of abelian varieties by tori, and noting that a nonzero integer acts as an isogeny on any torus and on any abelian variety). The next thing to observe is since an isogeny of abelian varieties admits a map in the opposite direction such that the composition in both directions is multiplication by a common nonzero integer, the preceding observation about the effect of nonzero integers on the identity components of the special fibers implies that an isogeny between abelian varieties over $F$ with semistable reduction induces an isogeny between identity components of special fibers of the Neron model. (This already settles the case of elliptic curves in the question posed above.) Consequently, for our purpose we can replace either of $A$ or $B$ with an $F$-isogenous abelian variety (which always has semistable reduction too: this is well-known for elliptic curves, and follows from Grothendieck's inertial semistable reduction criterion in general). Thus, in case either of $A$ or $B$ are not $F$-simple (though they are $F$-simple for the case of elliptic curves) we can apply the Poincare reducibility theorem to reduce to the case when both are $F$-simple. But then any nonzero $F$-homomorphism between them is an isogeny and hence as we have seen above induces an isogeny between identity components of special fibers of the Neron models. In particular, the induced map between such special fibers is nonzero.<|endoftext|> TITLE: What is the maximal order of the automorphism group of a given Shimura variety? QUESTION [5 upvotes]: Background: Given an elliptic curve $E$, it seems that $max(ord(Aut(E)))$ over the prime 2 is 24, and $(max(ord(Aut(E)))$ over the prime 3 is 12. The endomorphism algebra of an elliptic curve over $\overline{\mathbb{F}}_p$ is, after tensoring with $\mathbb{Q}$, either an imaginary quadratic field or the unique division algebra $D_p$ which is non-split at $p$ and infinity. So, in order to compute the maximal order of the automorphism algebra, we just compute the units of $D_p$, and then count them. Here's my question: Given a Shimura variety, how does one compute the maximal order of its automorphism group? For example, the variety $\mathbb{C}^{\times 3}/ \mathbb{Z}[\zeta_7]^{\times 3}$ where the embedding is given by $\sigma_1 \times \sigma_2 \times \sigma_3: \mathbb{Z}[\zeta_7]^{\times 3} \to \mathbb{C}^{\times 3}$. Here, we look at its automorphisms over $\mathbb{F}_p$ where $p$ is $2$ or $4$ mod $7$. I have heard that Shimura varieties have finite dimensional automorphism groups (analogous to the fact that $Aut(A, p)$ is finite, where $p$ is a polarization of the abelian variety $A$). What is the automorphism group of this Shimura variety? REPLY [3 votes]: Summary: The automorphism group of this variety at least contains $Z/2 \oplus Z/6 \simeq \mathbb{Z}[\zeta_7]^\times$,and is isomorphic to it if the variety is nondegenerate. More generally, I learned today that given $A$ of the form $\mathbb{C}^{\phi(n)/2}/\mathbb{Z}[\zeta_n]$ (with the appropriate choice of embedding); then $Aut(A)$ is indeed simply $\mathbb{Z}[\zeta_n]^\times$ (and $End(A)$ is $\mathbb{Z}[\zeta_n]$. To convince oneself that this is at the very least a subgroup of the endomorphism group, recall that endomorphisms of $\mathbb{C}^g/\Lambda$ are maps which preserve the lattice $\Lambda$ (or at least map the lattice to a scaled version of itself, such as $\Lambda \mapsto 2\Lambda$. For example, for any $x \in \mathbb{Z}[\zeta_n]$, we can multiply component-wise every point, taking the lattice to an $x$ scaled version of itself (where we consider $\zeta_n \in \mathbb{C}$ as $e^{2 \pi i/n}$): $$\mathbb{C}^g \to \mathbb{C}^g$$ $$(p_1, ..., p_g) \mapsto (\sigma_1(x)p_1, ..., \sigma_g(x)p_g)$$<|endoftext|> TITLE: On Tate's "Endomorphisms of Abelian Varieties over Finite Fields", sketch of proof of main result? QUESTION [17 upvotes]: Let $k$ be a field, $\overline{k}$ its algebraic closure, and $A$ an abelian variety defined over $k$, of dimension $g$. For each integer $m \ge 1$, let $A_m$ denote the group of elements $a \in A(\overline{k})$ such that $ma = 0$. let $l$ be a prime number different from the characteristic of $k$, and let $T_l(A)$ denote the projective limit of the groups $A_{l^n}$ with respect to the maps $A_{l^{n + 1}} \to A_{l^n}$ which are induced by multiplication by $l$. It is well known that $T_l(A)$ is a free module of rank $2g$ over the ring $\mathbb{Z}_l$ of $l$-adic integers. The group $G = \text{Gal}(\overline{k}/k)$ operators on $T_l(A)$. Let $A'$ and $A^{\prime\prime}$ be abelian varieties defined over $k$. The group $\text{Hom}_k(A', A^{\prime\prime})$ of homomorphisms of $A'$ into $A^{\prime\prime}$ defined over $k$ is $\mathbb{Z}$-free, and the canonical map$$\mathbb{Z}_\ell \otimes \text{Hom}_k(A', A^{\prime\prime}) \to \text{Hom}_G(T_l(A'), T_l(A^{\prime\prime}))\tag*{$(1)$}$$is injective. Theorem (Tate, 1966). If $k$ is finite, the map $(1)$ is bijective. Question. Could anybody supply an outline of Tate's proof of this theorem and contribute their intuition as to why such a result is true? REPLY [22 votes]: I don't have any contribution for the intuition beyond the fact that, I can't construct something outside the image of (1) so I hope it's surjective. Here is a sketch of the central idea of Tate's proof. Consider $A = A' \times A''$ and try to find an endomorphism of $A$ from an endomorphism of its Tate module. The endomorphism of the Tate module gives you subgroups $G_n$ of $A_{\ell^n}$ for every $n$ and you can consider $A/G_n$. Tate uses that there exists a moduli space of abelian varieties which is a variety over $k$ and thus has finitely many points over $k$ to show that the $A/G_n$ have to repeat, i.e., for some $n, A/G_n$ is isomorphic to some $A/G_m, m \ne n$. This isomorphism produces an element of the endomorphism ring of $A$, tensored with $\mathbb{Q}$. This gives the required element on the right hand side of (1). Incidentally, Faltings's proof that (1) is surjective for number fields follows the same strategy, except he shows that the height of $A/G_n$ is bounded to get that they repeat.<|endoftext|> TITLE: Bochner's formula on surfaces using moving coframes QUESTION [10 upvotes]: I've read many times that moving coframes where a convenient tool for computations in Riemannian Geometry, especially on surfaces, but never really used it. Lately to get a better feel of the method, I've tried to reprove classical results on surfaces using moving coframes and the Cartan structure equations : Cartan's equations for surfaces : if $(\eta^1,\eta^2)$ is a orthonormal coframe on $(\Sigma^2,g)$, then $d\eta^1=\omega\wedge\eta^2$ and $d\eta^2=-\omega\wedge\eta^1$ for a unique $1$-form $\omega$, which moreover satisfies $d\omega=-K_g\eta^1\wedge\eta^2$. What is really nice is that in these three equations we define at the same time the Levi-Civita connection (through $\omega$ which is the connection $1$-form) and the curvature, without ever actually computing covariant derivative. I have been amazed at how some result get much easier in this framework, like the fact that $K_g=0$ implies local isometry to the euclidean metric, and also at how curvature computation become more straight forward. Moreover a lot of the objects of interest in geometric analysis can be defined solely in terms of $1$-forms, the Hodge star $\ast$, exterior differential of forms and wedge product : $g(\nabla u,\nabla v)=\ast(du\wedge \ast dv)$. $\Delta u=\ast\ d\ast d u$. As a geometric analyst, my favorite equation is probably Bochner's formula, which in the case of surfaces reads as : $$\tfrac{1}{2}\Delta |\nabla u|^2=\langle\nabla u,\nabla\Delta u\rangle+|\mathrm{Hess}\, u|^2+K_g|\nabla u|^2.$$ The usual proof of the Bochner formula involves commutations of covariant derivatives. My question is : Question : Is it possible to prove the Bochner formula for surfaces relying on Cartan's structure equations instead of commutation of covariant derivative ? It basically boils down to two subquestions : How does one compute $d\ast d\left(\ast(du\wedge \ast du)\right)$ ? Can one express $|\mathrm{Hess}\, u|^2$ in terms of the Hodge star, the connection form and exterior derivatives ? REPLY [13 votes]: The straightforward way to do this is to note that we have, for some $u_1$ and $u_2$, $$ \mathrm{d}u = u_1\,\eta^1 + u_2\,\eta^2 $$ Taking the exterior derivative of this and using Cartan's Lemma then shows that, for some $u_{11}$, $u_{12}$, and $u_{22}$, $$ \begin{aligned} \mathrm{d}u_1 &= \phantom{-}u_2\,\omega + u_{11}\,\eta^1 + u_{12}\,\eta^2\\ \mathrm{d}u_2 &= -u_1\,\omega + u_{12}\,\eta^1 + u_{22}\,\eta^2 \end{aligned} $$ Then taking exterior derivatives of these equations and using Cartan's Lemma yields $$ \begin{aligned} \mathrm{d}u_{11} &= & 2u_{12}\,&\omega + u_{111}\,\eta^1 + u_{112}\,\eta^2\\ \mathrm{d}u_{12} &= &(u_{22}-u_{11})\,&\omega + u_{121}\,\eta^1 + u_{122}\,\eta^2\\ \mathrm{d}u_{22} &= &-2u_{12}\,&\omega + u_{221}\,\eta^1 + u_{222}\,\eta^2\\ \end{aligned} $$ for some $u_{ijk}$ satisfying $u_{112}-u_{121} = -K\,u_2$ and $u_{122}-u_{221} = K\,u_1$. Now just plug these formulae into the left hand side of Bochner's formula and compute. Comment: I understand that the OP would like to have some kind of exterior algebra 'magic' that just makes the calculation a consequence of $\mathrm{d}^2=0$, but that's an unreasonable request, it seems to me. The strength of the method of calculating by moving coframes is that it does show that all the local identities that you can prove are consequences of $\mathrm{d}^2=0$ and exterior algebra, though you may have to apply this principle multiple times if the identity you want involves higher derivatives. For example, in this particular case, the general structure equations for an orthonormal coframe (i.e., $g = {\eta_1}^2 + \cdots + {\eta_n}^2$) are that there are unique $1$-forms $\eta_{ij}=-\eta_{ji}$ satisfying $$ \begin{aligned} \mathrm{d}\eta_i &= -\eta_{ij}\wedge\eta_j\\ \mathrm{d}\eta_{ij} &= -\eta_{ik}\wedge\eta_{kj} + \tfrac12 R_{ijkl}\,\eta_k\wedge\eta_l \end{aligned}\tag1 $$ where $R_{ijkl}=-R_{ijlk}$ and all terms are summed on repeated indices. Now, if $u$ is a function in the domain of $\eta$, we have $$ \mathrm{d}u = u_i\,\eta_i\tag2 $$ for some functions $u_i$. Applying the exterior derivative and then Cartan's Lemma to (2) yields $$ \mathrm{d}u_i = -u_j\,\eta_{ij} + u_{ij}\,\eta_j\tag3 $$ for some functions $u_{ij}=u_{ji}$. Applying the exterior derivative and then Cartan's Lemma to (3) yields $$ \mathrm{d}u_{ij} = -u_{ik}\,\eta_{kj} -u_{jk}\,\eta_{ki}+ u_{ijk}\,\eta_k\tag4 $$ where $u_{ilk}-u_{ikl} = R_{ijkl}\,u_j$ (and, of course, $u_{ijk}=u_{jik}$). Now note that $|\nabla u|^2 = u_iu_i$ and then plug in the above formulae for derivatives to get $$ \tfrac12\Delta|\nabla u|^2 = u_{ij}u_{ij} + u_iu_{ijj}. $$ Since, by the above formulae, $u_{ijj}=u_{jij} = u_{jji} + R_{jlji}\,u_l$, this yields $$ \tfrac12\Delta|\nabla u|^2 = u_iu_{jji} + u_{ij}u_{ij} + R_{jlji}\,u_iu_l\,, $$ and the rest is just interpreting the three terms, where, in general dimension, the final term involving curvature becomes $$ \mathrm{Ric}_g(\nabla u,\nabla u). $$ Whether you choose to use moving coframes or not is a matter of taste, of course, since you can equally well work with the definition of curvature in terms of the Lie brackets of the dual vector fields. (Of course, the $u_i$, $u_{ij}$ and $u_{ijk}$ are just the result of 'covariantly differentiating' $u$ with respect to these vector fields.)<|endoftext|> TITLE: Kontsevich weights in the complex algebraic setting QUESTION [8 upvotes]: In Kontsevich's Deformation quantization of Poisson manifolds, he gives an explicit formula for the star product: $$ f \star g = fg + \sum_{n=1}^\infty \hbar^n \sum_{\Gamma \in G_n} w_\Gamma B_{\Gamma} (f, g) \tag{$\ast$} $$ where $B_\Gamma (f, g)$ is the bilinear operator associated to a graph $\Gamma$ and $w_\Gamma$ is a "weight", constructed as the integral of some configuration space. I understand that the generalization of Kontsevich formality to the algebraic setting (smooth algebraic variety, possibly non-affine) is more involved and I have the following questions: Is the formula for the weights in the (complex) algebraic context exactly the same as in the real and smooth case? Are the weights computed "globally" or chart-by-chart, so that one has to check compatibility/gluing in an additional step? REPLY [6 votes]: yes, the formula for the weights is the same in whatever setting (differentiable, holomorphic and algebro-geometric). weights are involved in a local formula for a start-product. Indeed, $B_\Gamma$ doesn't even make sense globally on a manifold (being differentiable, holomorphic, etc...). Hence, even in the differentiable setting, Kontsevich's local formula must be globalized. There are standard techniques to do this, but the resulting star-product is not quite explicit. Let me add that in the holomorphic (or algebro-geometric) setting there is an additional issue: a global star-product may not exist. Indeed, it is not always possible to glue the local formula in order to get a sheaf of algebras deforming the structure sheaf of the complex manifold/algebraic variety. As it is explained in this other paper of Kontsevich, one can only hope to get an algebroid stack (this is a linearized analog of a gerbe). There has been a lot of work on these gadgets (by Yekutieli, Kashiwara-Schapira, and many others).<|endoftext|> TITLE: Physical interpretations/meanings of the notion of a sheaf? QUESTION [16 upvotes]: I fairly understand the fiber bundles, both the mathematical concept of fiber bundles and the physics use of fiber bundles. Because the fiber bundles are tightly connected to the gauge field theory in (quantum) field theory, and the general relativity in gravity theory. Here the gauge field theory are very powerful framework to descibe three of fundamental forces in the Standard Model, including (i) the U(1) gauge theory for Maxwell electro-magnetism. (ii) SU(2)$_{weak}$ Yang-Mills gauge theory for the nuclear weak interactions with a parity-violating chiral coupling to the SU(2) doublet of left-moving leptons/quarks particles in 3-generations ($\nu_e, e$), ($u,d$),($\nu_\mu, \mu$), ($c,s$), ($\nu_\tau, \tau$), ($t,b$). (iii) SU(3)$_{color}$ Yang-Mills gauge theory for the nuclear strong interactions, where the SU(3) gauge fields only coupling to quarks which carrying the SU(3) fundamental-representation charges in 3 colors, e.g. commonly called the 3 colors as $r,g,b$ (red, green, blue) --- say the color-triplets for all quarks; $(u_r,u_g,u_b)$, $(d_r,d_g,d_b)$, $(c_r,c_g,c_b)$,$(s_r,s_g,s_b)$,$(t_r,t_g,t_b)$,$(b_r,b_g,b_b)$. And the general relativity using Riemannian geometry on the curved spacetime with the Minskowski signature describes the fourth fundamental forces, the gravity. A remarkable summary between the physics (gauge fields) and the math (fiber bundles) is made by Concept of nonintegrable phase factors and global formulation of gauge fields by Tai Tsun Wu and Chen Ning Yang in Phys. Rev. D 12, 3845, and there is a schematic table in the paper: However, even if I encounter the sheaf several times in different math/physics context, I don't have a clear understanding how this can be tightly connected to certain theory of physics. So my question is: What are the physical interpretations / meanings of the sheaf? Perhaps there can be more than one branch of physics that really requires the use of the concept of sheaf in fundamental and deep ways? Could there be a correspondence table for the sheaf and something-else in physics or anything in our natural real-word, just analogous to the table above? REPLY [7 votes]: I expected someone to give an answer of the following form; since that still has not happened let me give it a shot. I should mention that I am by no means an expert on this topic, and many others here could probably write up a much nicer version of this answer; I welcome any comments and corrections. To start with, the dictionary in the paper by Yang and Wu quoted in the OP is a summary of the way to rephrase gauge theory in terms of differential geometry. Here the latter provides a coordinate-independent (global), and therefore natural, way of thinking about the former. Fiber bundles are the proper way of attaching geometric structures to spaces in order to account for concepts from QFT like gauge groups, spin, and so on. Reversely, gauge theory seems to be "the way" to interpret or understand (or: "the example" of) the geometric notions in the context of theoretical physics. (Edit: One could either add general relativity as a second important example, or argue that it is a gauge theory with gauge group the Poincaré group, albeit the action for the gauge fields has a different form than for Yang-Mills theory.) To me the broadness of the OP's question, asking for such a dictionary for the case of sheaves in general, sounds rather like the question: what are sheaves, really? Let's first try to get this question out of the way. In short I would say that a sheaf enables one to attach more general structures (i.e. than for fiber bundles) to a topological space, again in a nice and global manner. Indeed, it allows one to think of the structure locally (via restriction: this is the presheaf condition) if one wants to understand the global situation (equality can be tested locally, and compatible local data gives rise to global data: these are the sheaf conditions). I'll leave detailed answers to others: see for example Wikipedia on sheaves; nLab on sheaves; the gentle Chapter 1 of Ramanan's book Global Calculus; the more category-theoretic and (but) very nice Section II.1 of Mac Lane and Moerdijk's book Sheaves in Geometry and Logic. Next one might ask for examples of sheaves in theoretical physics to understand the definition better. Different examples are related to different structures that sheaves may account for, and give rise to different dictionaries summarizing it. The crux here is that the notion of a sheaf really is quite general. Even if gauge theory might be the example giving fiber bundles and so on a meaning in theoretical physics, I don't think that we can expect there to be a canonical dictionary for sheaves in theoretical physics as well. This is reflected in the comments and other answers to the question, each pointing to different examples. Let me add two more examples: First of all, back to the table in the OP: many standard notions from differential geometry (and thus gauge theory) can be rephrased in terms of sheaves. For examples of this type I like the last two references above (see also Section II.3 of Mac Lane and Moerdijk's book mentioned above). An important example in theoretical physics where sheaves (as far as I know) have to be used are provided by supermanifolds. You can try Wikipedia and the references there; another classic is the notes written up by Deligne and Morgan of Bernstein's lectures on supersymmetry, of which a slightly more complete version can be found in the first volume of Quantum Fields and Strings: A Course for Mathematicians. Edit: For me personally the latter is the context in physics where I first encountered sheaves. At the time my MSc thesis advisor, André Henriques, told me to learn what a sheaf is by talking with mathematicians. Since the abstract definition of a sheaf is perhaps a bit intransparent when one first sees it, at least as a physicist, this might really be the way to go!<|endoftext|> TITLE: Fields for which there exist multivariable polynomials vanishing at single specified point QUESTION [9 upvotes]: For which fields $k$ does the following hold for all $n \geq 1$? Let $(a_1,\ldots,a_n) \in k^n$. Then there exists a polynomial $f(x_1,\ldots,x_n) \in k[x_1,\ldots,x_n]$ such that $f(a_1,\ldots,a_n) = 0$ but $f(b_1,\ldots,b_n) \neq 0$ for all $(b_1,\ldots,b_n) \neq (a_1,\ldots,a_n)$? This is clearly impossible for $k$ algebraically closed. It does hold for $k$ a subfield of $\mathbb{R}$; indeed, in that case we can take $$f(x_1,\ldots,x_n) = (x_1-a_1)^2 + (x_2-a_2)^2 + \cdots + (x_n-a_n)^2.$$ REPLY [20 votes]: If $k$ is not algebraically closed, such a polynomial always exists (the opposite is also true and is mentioned in the post). We may assume that $a_i=0$ for all $i$. Take an irreducible polynomial $g(x)$ of degree $d>1$, then for the homogeneous form $G(x,y)=y^dg(x/y)$ we have $G(x,y)=0$ if only if $x=y=0$. This solves the case $n=2$, for $n=3$ consider the polynomial $G(G(x,y),z)$, it takes zero value only when $x=y=z=0$, and so on. REPLY [2 votes]: I guess finite fields $k = \mathbb{F}_{q}$ satisfy this property, namely we can take $$ f = ((x_{1}-a_{1})^{q-1}-1) \dotsb ((x_{n}-a_{n})^{q-1}-1) - (-1)^{n} $$ for any $n$. In fact we have a Lagrange interpolation formula for finite fields, see this answer.<|endoftext|> TITLE: Explicit formula of base change for GL(n) QUESTION [5 upvotes]: Let $E/F$ be a quadratic extension of number fields and $v$ is a place of $F$. Let $\chi_1,\chi_2$ be the unramified characters of $F_v^{\times}$. If $B(\chi_1,\chi_2)$ is the unramified principal series representation of $GL_2(F_v)$, what is the $BC(\pi)$, the base change of $\pi$ to $GL_2(E_v)$? I suppose that $BC(\pi)=B(\chi_1 \circ \text{Norm}_{E_v/F_v},\chi_2 \circ \text{Norm}_{E_v/F_v})$. Is this right? REPLY [5 votes]: The answer to your final question should be yes. Let me assume that $v$ is inert in $E$, so that $E_v / F_v$ is a honest quadratic extension. Assuming that $B(\chi_1, \chi_2)$ is irreducible, it corresponds via local Langlands (a theorem for $\mathrm{GL}_2(K)$ and any $p$-adic field $K$) to the 2-dimensional representation of the Weil group of $F_v$ given by the sum of the characters $\chi_1$ and $\chi_2$. The operation of base change on the $p$-adic side corresponds to restriction of Galois representations, thus the base change of $B(\chi_1, \chi_2)$ corresponds via local Langlands to the restriction of $\chi_1 \oplus \chi_2$ to the Weil group of $E_v$. When we restrict $\chi_1 \oplus \chi_2$ to the Weil group of $E_v$ and then make this homomorphism factor through the abelianization $E_v^*$, local class field theory tells us that this corresponds exactly to pre-composition of $\chi_1$ and $\chi_2$ with the Norm map. Now go back to the $p$-adic side and you have your statement.<|endoftext|> TITLE: Kazhdan Lusztig map and Richardson orbits QUESTION [5 upvotes]: Let $g$ be a simple Lie algebra with Weyl group $W$. Kazhdan and Lusztig defined a map $\Phi$: nilpotent orbits in $g$ $\rightarrow$ conjugacy classes in $W$. Let $\eta_p$ be a Richardson orbit associated to a parabolic $p$. Question: What is $\Phi(\eta_p)$? Here is a natural guess: Let $W_p$ denote the (parabolic) subgroup of $W$ corresponding to $p$. Note that we have a canonical map $i_{p,g}$ from conjugacy classes in $W_p$ to conjugacy classes $W$. Let $c_p$ denote the Coxeter conjugacy class of $W_p$. Question: Is it true that $\Phi(\eta_p)=i_{p,g}(c_p)$? Remark: Spaltenstein has studied the KL map extensively. However, I could not find the answer to this question in his papers. REPLY [2 votes]: This is a very nice line of thinking! But I think the question, as stated, is imprecise. As is correctly pointed out in the question, the KL map takes you from nilpotent orbits in $\mathfrak{g}$ to conjugacy classes in the Weyl group $W(\mathfrak{g})$. However, a given Richardson orbit is not uniquely associated to a single parabolic. In fact, it is not even associated to a unique Levi subalgebra. To understand this, it is best to view Richardson orbits as those that are "induced" (in the sense of Lusztig-Spaltenstein) from zero orbits in proper Levi subalgebras. But, a given Richardson orbit can be induced from more than one (non-conjugate) Levi! You can however associate a unique "Dixmier sheet" to a pair $(r,l)$, where $r$ is a Richardson Orbit and $l$ is one of the Levis from which it is induced. Now, I think it is a nice question to ask if one can naturally think of an analog of the KL map that takes you from a Dixmier sheet to a fixed conjugacy class of the Weyl Group and if this conjugacy class can be thought of coming from the Coxeter class of the corresponding parabolic sub-Weyl group. But, I don't know of any work where something like this is pursued. One more potential source of confusion : In the world of Weyl Groups, what is called a parabolic sub-Weyl group really corresponds to what is called a Levi subalgebra in the world of Lie Algebras and not to a "Parabolic" subalgebra. There are far too many Parabolic subalgebras. Even after you fix a Levi, a given Richardson orbit could have multiple "Polarizations" that correspond to choosing a Parabolic with that specific Levi factor. As a reference for Richardson Orbits/Sheets/Polarizations, you can consult a paper of de Graaf and Elashvili, "Induced Nilpotent Orbits...".<|endoftext|> TITLE: Spectral decomposition of a C$^*$algebra with respect to an action of a compact abelian group QUESTION [5 upvotes]: Let $G$ be a compact abelian group (finite dimensional, but not finite) and $A$ be a $C^*$-algebra. Consider an action $\alpha: G\to Aut(A)$. In analogy with the case of finite abelian group, I believe that it is true that $A=\oplus_{\chi\in \hat{G}} A_\chi$, with $A_\chi$ being the subspace made of elements $x\in A$ such that $\alpha_g(x)=\chi(g)x$. Moreover, I think that there exist a family of linear maps (indexed by the characters of the group) $E_\chi: A\to A_\chi$ $$ E_\chi(x)=\int_G \overline{\chi(g)}\alpha_g(x) $$ Can someone tell me if it is true and give me reference to a book or paper? Thanks for the help. REPLY [3 votes]: The facts that $A$ is a $C^*$-algebra and that $G$ is abelian are irrelevant: such a decomposition holds more generally for any Banach space $A$ with a continuous linear action of a compact group. The reference I know for this is Representations of a compact group on a Banach space by Shiga, available here. There are probably earlier references for abelian groups, but I do not know them.<|endoftext|> TITLE: a question about complexity of Boolean problem QUESTION [5 upvotes]: I was thinking in solving the following problem for the general case : **) Given a list of pairs $((n_i, A_i))_{i=1}^k$, where for each $i$ we have that $n_i $ is a non-negative integer, and $A_i$ is a set, does there exist a set $A$ such that $|A\cap A_i|=n_i$? Example 1 : Given as an input the list : $(3, \{a,b,c\}), (4, \{a,b,c,d,e,f\}), (2, \{a,b,c,d\})$ the answer will be negative, because from the first set we are forced to take all the 3 elements $a$,$b$ and $c$ but from the last set we must take just 2 elements but it's already containing 3 elements that we forced to pick. Example 2 : Given as an input the list : $(2, \{a,b,c\}), (3, \{a,b,c,d\}), (4, \{a,b,c,d,e,f\})$ the answer is positive, for instance we can pick $\{a,b,d,f\}$. My question is : is the General case of this problem solvable in Exp,SubExp,Quasi-Poly,Poly time ? space ? REPLY [2 votes]: The problem is visibly in NP, as we can check in polynomial time whether a given set $A$ is a correct witness. In fact, the problem is NP-complete, and one can show this by reduction from the NP-complete problem 1-in-3-SAT: given a 3-CNF in variables $x_1,\dots,x_n$ with clauses $C_1,\dots,C_m$, we reduce it to the list $$(1,\{x_1,\overline{x_1}\}),\dots,(1,\{x_n,\overline{x_n}\}),(1,C_1),\dots,(1,C_m).$$ (The clauses $C_i$ are here treated as subsets of the $2n$-element set of literals $\{x_1,\dots,x_n,\overline{x_1},\dots,\overline{x_n}\}$.) The problem remains NP-complete when restricted to the case where all $n_i=1$, and all the sets $A_i$ have three elements. The reduction above almost achieves this, except that some of the sets have only two elements. This is easy to fix: for example, we may replace a two-element set $(1,\{a,b\})$ with the input $(1,\{a,b,u\}),(1,\{a,b,v\}),(1,\{u,v,w\})$ where $u,v,w$ are three new elements.<|endoftext|> TITLE: Interesting examples of functions that are not orthogonal to the Mobius function? QUESTION [10 upvotes]: There are several recent results on non-trivial bounded functions $f$ that are orthogonal, i.e. $$\sum_{n\leq x} \mu(n)f(n)=o(x),$$ to the Mobius function $\mu(n)$. These include the results of Green and Tao on Nilsequences (which includes information on the history of the problem) and Green on bounded depth circuits. The lecture notes by Peter Sarnak on Mobius randomness is an interesting overview of the subject. Sarnak mentions there that it is easy to construct counter examples, or examples which decay very slowly. Of course, its easy to do so if $f$ is sparsely supported or has partial sums that are the inverse Mellin transform of a function having positive powers of $\zeta(s)$ as factors. What are some examples of bounded, non-vanishing (or at least positive density) functions that are not orthogonal to $\mu(n)$, and which don't come from $\zeta(s)$ in an obvious way? REPLY [4 votes]: This question has been considered by Lemanczyk and others, and Lemanczyk developed a quite general way to produce dynamical systems which are not disjoint from Mobius (unfortunately or fortunately, depending on ones preferences, they are of positive entropy). The basic idea is to use the square free flow constructed by Sarnak (which Terry have mentioned) and to find some symbolic system which "approximates" this flow for large times. See for example the detailed survey - https://arxiv.org/abs/1410.1673 In the end, the fact that there are systems which are not Mobius disjoint is not that surprising - Sarnak has computed the entropy of the Mobius flow and show it positive, general ergodic theory (i.e. the construction of the Bernoulli factor by Sinai/Ornstein) would tell you it is "easy" to correlate with Mobius. Just another related interesting comment - Bourgain has constructed a system with positive entropy which is Mobius disjoint.<|endoftext|> TITLE: The topological duals of spaces of finite measures QUESTION [5 upvotes]: In volume 1 of "Linear Operators", Dunford and Schwartz say that (footnote F1, page 374) "No completely satisfactory representation for the conjugate space of $ba(S, \Sigma)$, $ca(S, \Sigma)$ or $rca(S, \Sigma)$ [...] seems to be known. In the above, $ba(S, \Sigma)$ is the space of finite, finitely additive measures on an algebra $\Sigma$, $ca(S, \Sigma)$ is the space of finite, countably additive measures on a $\sigma$-algebra $\Sigma$, and $rca(S, \Sigma)$ is the space of finite, regular, countably additive measures on a $\sigma$-algebra $\Sigma$ (all of them understood as complex measures). One may also consider the space $rba(S, \Sigma)$ with the obvious meaning. This was in 1957. Today, 60 years later, has our understanding improved? (For my work, I am searching for a space $X$ with a known dual $X^*$, such that $X^*$ should contain $C(S)$, the space of bounded continuous functions on $S$. For the moment, I am working with $X = L^1 (S)$, but the elements of $X^* = L^\infty (S)$ are not regular enough for my needs. I was hoping that the dual of a space of measures might be closer to $C(S)$.) REPLY [8 votes]: The bidual of $C(S)$ can be described as the direct product of von Neumann algebras generated by the cyclic representations associated with positive elements of $C(S)^*$. For a positive finite measure $\mu$ the von Neumann algebra generated by the representation associated with $\mu$ is just $L^\infty(S, \mu)$, so if the elements of that space aren't regular enough for you, then an indescribable direct product of them probably won't be any better. Because $C(S)^{**}$ is a commutative C$^*$-algebra, it is $*$-isomorphic to $C(K)$ for some compact $K$, but as far as I know, no description of the space $K$ has ever occurred. There is a new book coming out on the subject of spaces of continuous functions as dual spaces, which will probably be a lot easier to read than the other literature on the subject.<|endoftext|> TITLE: When is a proper map between smooth varieties a submersion? QUESTION [6 upvotes]: Let $f: X\to Y$ be a submersion between projective manifolds. Then $Rf^i_*(\mathbb{C}_X)$ are local systems on $X$, for all $i$. My question is whether the converse is true. More precisely, let $f: X\to Y$ be a holomorphic map between projective manifolds. Suppose that for all $i$, $Rf^i_*(\mathbb{C}_X)$ are global local systems on $X$. Is $f$ necessarily a submersion? If this is not true, what is a counter-example? REPLY [6 votes]: I am just posting my comment above as an answer. There are counterexamples that depend on the fact that the divisible coefficient sheaf $\mathbb{C}_X$ (for the analytic topology, presumably) does not detect certain torsion phenomena. Begin with a smooth, projective, geometrically integral curve $E$ of genus $1$ and a nontrivial $2$-torsion translation $\tau:E\to E$. Form $Z=E\times \mathbb{A}^1$. Let a cyclic group $G$ of order $2$ act diagonally on $Z$ via $\tau$ on $E$ and via $t\mapsto -t$ on $\mathbb{A}^1$. Let $X$ be $Z/G$, and let $Y$ be $\mathbb{A}^1/G$. Let $f:X\to Y$ be the unique morphism induced by the projection $\text{pr}_2:E\times \mathbb{A}^1 \to \mathbb{A}^1$. Then $f_*\mathbb{C}_X$ is $\mathbb{C}_Y$, $R^2f_*\mathbb{C}_X$ is isomorphic to $\mathbb{C}_Y$, and $R^1f_*\mathbb{C}_X$ is also a localy system isomorphic to $\mathbb{C}_Y^{\oplus 2}$. It is the last one that is a bit more delicate, since for every $y \in Y$ other than the origin, for the fiber $X_y=f^{-1}(\{y\})$, the map on homology, $$H_1(X_y;\mathbb{Z})\to H_1(X;\mathbb{Z}),$$ is not an isomorphism. It is injective and the image is an index $2$ subgroup. However, once we tensor with the divisible group $\mathbb{C}$, the induced map is an isomorphism. To say it differently, $R^1f_*\mu_2$ is not a local system.<|endoftext|> TITLE: Degree $d$ function with boolean inputs with small range is a junta? QUESTION [5 upvotes]: Let $f : \{-1,1\}^n \rightarrow \{-1,1\}$ be a boolean function which is of degree at most $d$ when expressed as a multilinear polynomial ($f(x) = \sum_S \hat{f}(S) \prod_{i \in S} x_i$). It is known that such a function is a $d2^{d-1}$-junta - i.e. $f$ depends on at most $d2^{d-1}$ variables. In fact, this should hold for any range of size $2$. Suppose the range of $f$ consisted of $m$ real numbers and $f$ were still of degree $d$. Is it known whether $f$ is a $g_m(d)$-junta for some function $g_m(d) << n$? REPLY [4 votes]: This is a nice question. I have to think that the answer must appear somewhere, but I'm not sure where. Here is, I think, a solution for $m = 3$. I guess it could be generalized to any $m$, but possibly with a bad dependence on $m$. By scaling and translating, we can assume $f$ takes on the three values $-1$, $+1$, and $c \in (-1, 1)$. Case 1: $c \in [-\frac12, \frac12]$. In this case, $f$'s range is sufficiently "discrete" that the "usual" proof for Boolean-valued $f$ should work. I.e., things should be okay because $f$'s derivatives $D_i f$ take on largish values when they're nonzero. I can elaborate on this case if you want. Case 2: $\frac12 < |c| < 1$. In this case, let $g$ be the Boolean-valued function $\mathrm{sgn}(f)$. Now $f$ is a degree-$d$ "approximating polynomial" for $g$ in the sense of Nisan--Szegedy, so $g$ itself must have degree at most $\mathrm{poly}(d)$; I forget what's best known these days, maybe $d^6$. Now $fg$ is two-valued and of degree $O(d^6)$. By a simple translation/scaling we can get an $h$ of degree $O(d^6)$ which is the Boolean indicator function that $f$ takes on the value $c$. Thus by Nisan--Szegedy $h$ (and also $g$) depend on at most $\widetilde{O}(2^{d^6})$ coordinates. I think by some simple playing around now you can conclude $f$ also depends on at most $\widetilde{O}(2^{d^6})$ coordinates. I wonder if it's possible to get it down to $2^{O(d)}$ coordinates.<|endoftext|> TITLE: Darboux-like theorems QUESTION [15 upvotes]: Related to Kahler version of Darboux's Theorem I know a few theorems that feel like Darboux's theorem. By that, I mean some kind of geometry based around a "pointwise" condition and the existence of a tensor or something generalizing a tensor which vanishes exactly when there is some kind of "flatness", or local coordinates that express the geometry in a canonical form. Examples include: Darboux's theorem: $d\omega$ vanishes exactly when it's possible to choose coordinates that express the symplectic form as $\omega = \sum dx_i \wedge dy_i$ Riemannian geometry: Riemannian curvature $R$ vanishes exactly when it's possible to choose coordinates that express the metric as $g = \sum dx_i \otimes dx_i$ Kahler geometry: Discussed in the link above. Complex geometry: The Nijenhuis tensor vanishes exactly when there are coordinates $x_i, y_i$ such that $J(dx_i) = dy_i$. Are there any other similar theorems? REPLY [20 votes]: While Francois' answer is fine, as far as it goes, I think that it is important to bear in mind the history of this problem. The original problem of 'flatness' or integrability (and more generally, equivalence) of what are now called $G$-structures was formulated by Élie Cartan in his fundamental paper Les sous-groupes des groupes continus de transformations, Annales scientifiques de l'École Normale Supérieure, Sér. 3, 25 (1908), p. 57-194. Moreover, the flatness problem was essentially solved by him in this paper, in the sense that he provided an explicit algorithm for deciding when a given $G$-structure was 'flat' (i.e., locally equivalent to the natural 'constant' $G$-structure on $\mathbb{R}^n$) by describing how to compute a sequence of obstructions to flatness for any given subgroup $G\subset\mathrm{GL}(n,\mathbb{R})$. (These obstructions would nowadays be formulated as 'torsion' and/or 'curvature' tensors of various orders of the $G$-structure. Cartan simply called them 'invariants' of the given $G$-structure.) The 'essentially' refers to two weaknesses: First, Cartan did not provide (nor did he have) a proof that the sequence terminated in a finite number of steps, although it did in all the cases that he (and others) computed (which included symplectic, (pseudo-)Riemannian, conformal, and projective geometries, for example). Second, his conditions (in case the sequence terminated), which were clearly necessary, were mostly known to be sufficient only in the case that the original $G$-structure was real-analytic or 'finite type'. (This was because his main PDE existence theorem was Cartan-Kähler, which only applies in the real-analytic category, or the (ODE) Frobenius theorem, which applies, even for smooth structures, in the 'finite type' case, which includes pseudo-Riemannian geometry, conformal geometry in dimension $n>2$, and projective geometry, as well as many other examples nowadays called 'parabolic geometries'. Symplectic geometry, i.e., Darboux' Theorem is not finite type, but it turns out that one can prove the existence theorem using only ODE techniques anyway, so real-analyticity is not needed in this case. As another example, if $G\subset\mathrm{GL}(n,\mathbb{R})$ is compact, Cartan's algorithm yields that a $G$-structure is flat if and only if its canonical connection is torsion-free and has vanishing curvature. In particular, this covers the Kähler case, in which $G=\mathrm{U}(\tfrac12n)$.) The first weakness was remedied in M. Kuranishi, On É. Cartan's prolongation theorem of exterior differential systems, Amer. J. Math., vol. 79, 1957, p. 1–47, in which Kuranishi re-formulated Cartan's constructions in terms of cohomology groups and proved a finiteness theorem based on Hilbert's theorem about finite generation of modules over polynomial rings. This established that, at least for the 'flatness' or 'integrability' problem, Cartan's algorithm always terminates in a finite number of steps (and hence furnishes sufficient conditions in the real-analytic case and the finite type cases). The second weakness took longer to resolve. For example, for almost-complex structures, Cartan's algorithm states that, for real-analytic almost-complex structures, the vanishing of the Nijnhuis tensor is the necessary and sufficient condition for flatness, but, since this is not a finite type case, for smooth (or even weaker regularity) almost-complex structures, this necessary condition was not known to be sufficient until the work of Newlander and Nirenberg in 1957. The 'flatness' problem for smooth $G$-structures not of finite type was only finally solved in complete generality considerably later, by combined work of Guillemin, Spencer, Goldschimdt, Malgrange and a number of other people. An account of this story can be found in the final chapters of our book "Exterior Differential Systems" (Bryant-Chern-Gardner-Goldschmidt-Griffiths, 1991). Remark 1: 'Flatness' and 'integrability' have not always been synonymous. For example, see Shiing-Shen Chern, Pseudo-groupes continus infinis, Géométrie différentielle., Colloques Internationaux du Centre National de la Recherche Scientifique, Strasbourg, vol. 1953, Centre National de la Recherche Scientifique, Paris, 1953, pp. 119–136. In this paper, a notion of 'integrability' is defined that is more general than 'flatness'. It includes, for example, the Pfaff-Darboux Theorem for contact structures, which are not 'flat' as $G$-structures (i.e., they cannot be expressed as constant coefficient tensors in any coordinate system) but are still 'integrable' in the (original) wider sense. Remark 2: Since Cartan's time, there have been notable improvements in the implementation of Cartan's algorithm. Cartan himself, in his later work on differential geometry in the 1920s and 1930s, noted that, in certain cases of $G$-structures of finite type, the algorithm could be structured in the form of what is now called the construction of a canonical 'Cartan connection' associated to a given $G$-structure, and the necessary and sufficient condition for flatness could be expressed as the vanishing of curvature of that connection. In the 1950s and 1960s, his work was reformulated in a more modern 'global' language by several mathematicians, among them Guillemin, Kobayashi, Spencer, Singer, and Sternberg. In the 1970s, Tanaka found a general construction of a canonical Cartan connection for a wide range of geometries of finite type and greatly clarified the process and the interpretation of the curvature, and this work has had numerous applications, particularly in the study of so-called 'parabolic geometries'. (Still, it should be remarked that it is not always possible to introduce a canonical connection in the finite-type case. It is possible that Cartan himself was aware of this problem, as the first such 'wild' examples already occur in dimension $n=3$, but there is no evidence of this in his published works.)<|endoftext|> TITLE: When are two binary matrices simultaneously equivalent to their transpose? QUESTION [8 upvotes]: For any real square matrix $A$ there is an invertible matrix $P$ such that $A^t = P^{-1}AP$. I have two binary ($0,1$) matrices $A$ and $B$. When does there exist a $P$ such that $A^t = P^{-1}AP$ and $B^t = P^{-1}BP$ hold simultaneously? I am particularly looking for some easy conditions on these matrices $A$ and $B$. A matrix is conjugate to its transpose, see Matrix is conjugate to its own transpose. Using Jordan canonical form to evaluate $P$ is a computationally difficult task. REPLY [6 votes]: Clearly, a necessary condition is that for every word $w$ in two letters, one has $${\rm Tr}\,w(A^t,B^t)={\rm Tr}\,w(A,B).$$ Equivalently, $${\rm Tr}\,\hat w(A,B)={\rm Tr}\,w(A,B),$$ where $\hat w$ is the reverse word. Namely, if $w=x^\ell y^mx^n\cdots$, then $\hat w=\cdots x^ny^mx^\ell$. Unless the word $\hat w$ be conjugated (in the free group ${\mathbb F}_2$) to $w$, which means $w=w_1w_2$ and $\hat w=w_2w_1$, this is a non-trivial condition. I suspect that this necessary condition is also sufficient. I doubt that the assumption that the entries are $0,1$ simplifies anything. By the way, the Jordan form is not a good way to prove that $A^t$ is conjugated to $A$, because it does not show that the change-of-basis matrix $P$ has entries in the smallest field containing the entries of $A$.<|endoftext|> TITLE: Rational cohomology of $LBSO(n)$ QUESTION [8 upvotes]: For which $n$ has the $S^1$-equivariant rational cohomology of $LBSO(n)$ been computed? Here, $SO(n)$ means the isometry group of the round sphere (preserving the orientation), $B$ stands for classifying space and $L$ for free loop space. The $S^1$-action on $LBSO(n)$ is, of course, via rotating the loops. There are two fibrations $$SO(n) \to LBSO(n) \to BSO(n)$$ and $$LBSO(n) \to LBSO(n) \times_{S^1} ES^1 \to BS^1$$ (both admitting a split on the right), and I am interested in knowing $$H^{\ast}_{S^1}(LBSO(n);\mathbb Q) = H^{\ast}(LBSO(n) \times_{S^1} ES^1;\mathbb Q),$$ including the ring structure. So one way to calculate this is to understand what happens in the Serre spectral sequences in the two fibrations above. I could do this (with some effort) for $n = 2,3$ but I do not have much hope figuring out all the differentials for bigger $n$. Is this something that can be found in the literature? REPLY [6 votes]: 1st method) If $X$ is a 1-connected then you can use the fact that you have an isomorphism of graded algebras (J.D.S. Jones "Cyclic homology and equivariant homology" Inventiones Math. 1987): $$H^n_{S^1}(LX;\mathbb{Q})\cong HC_{-n}^{-}(S^*(X;\mathbb{Q}))$$ where the right term is the negative cyclic homology of the cochain algebra of $X$. Moreover if $X$ is formal you have an isomorphism of algebras: $$H^n_{S^1}(LX;\mathbb{Q})\cong HC_{-n}^-(H^*(X;\mathbb{Q})).$$ This is indeed the case for $X=BSO(n)$ which is formal. Now you use the fact $H^*(BSO(n;\mathbb{Q}))$ is a polynomial algebra. 2nd method) Rationnaly $BSO(n)$ is formal and homotopy equivalent to a product of Eilenberg-MacLane spaces of type $K(\mathbb{Q},4k)$ and of a copy of $K(\mathbb{Q},2m)$ if $n=2m$, and now use the fact the for an $H$-space $Y$ we have $LY\cong Y\times \Omega Y$. Thus rationnaly you get that the cohomology of $LBSO(n)$ is isomorphic to $$H^*(BSO(n);\mathbb{Q})\otimes H^*(SO(n);\mathbb{Q}).$$ Edit, Reference: A good reference that will give you a cochain complex that computes $H^n_{S^1}(LX;\mathbb{Q})$ is the paper "A model for cyclic homology and algebraic K -theory of 1-connected topological spaces" by Micheline Vigué-Poirrier and Dan Burghelea (J. Differential Geom. Volume 22, Number 2 (1985), 243-253). In particular the theorem $A$ that gives an algebraic model from the minimal model of $X$.<|endoftext|> TITLE: A criterion for $F,U$ to be adjoint QUESTION [5 upvotes]: I'm trying to prove the following statement: Let $F\dashv U\colon {\cal C}\leftrightarrows {\cal D}$ be an adjunction, and $G \colon {\cal C}^\text{op}\times{\cal D}\to \cal E$ a functor; then there is an isomorphism $$\tag{$\star$} \int_{C\in\cal C} G(C,FC) \cong \int_{D\in\cal D} G(UD,D) $$ This is somewhat surprising, but in retrospective kind of obvious (it follows from the synergy between dinaturality and the zig-zag identities of the adjunction $F\dashv U$). What I want to prove is a converse of this statement: In the notation above, if for every functor $G : {\cal C}^\text{op}\times {\cal D}\to \cal E$ there is an isomorphism like $(\star)$ then $F\dashv U$ form an adjoint pair. Any clue? The strategy is to build two candidates for unit and counit using $(\star)$ and then prove that they satisfy the zig-zag identities; is it possible to do it in a simple (and possibly nifty) way? REPLY [7 votes]: Yes, this is true if we assume that the isomorphisms are natural in $G$. We may even restrict to $\mathcal{E}=\mathsf{Set}$. Here is a sketch of the proof. Consider $G=\mathrm{Hom}_{\mathcal{D}}(F(-),-) : \mathcal{C}^{op} \times \mathcal{D} \to \mathsf{Set}$. Then $$\int_x G(x,Fx) = \int_x \mathrm{Hom}_{\mathcal{D}}(Fx,Fx) = \mathrm{Hom}(F,F)$$ contains the identity $\mathrm{id}_F$, which gets mapped to an element of $$\int_y G(Uy,y) = \int_y \mathrm{Hom}_{\mathcal{D}}(FUy,y) = \mathrm{Hom}(FU,\mathrm{id}_{\mathcal{D}}).$$ This is our counit. In the same way, using $G' =\mathrm{Hom}_{\mathcal{C}}(-,U(-)) : \mathcal{C}^{op} \times \mathcal{D} \to \mathsf{Set}$, we construct our unit. Then the assumed naturality should imply the zig-zag-identities, namely with respect to the morphism $G \to G'$ induced by the the unit and the morphism $G' \to G$ induced by the counit (I haven't checked the details). Details: Let $$\alpha_G : \int_x G(x,Fx) \to \int_y G(Uy,y)$$ be an isomorphism which is natural in $G : \mathcal{C}^{op} \times \mathcal{D} \to \mathsf{Set}$. Then $\varepsilon_y : F(U(y)) \to y$ is by definition $\alpha_G(\mathrm{id}_F)_y$ for $G=\mathrm{Hom}_{\mathcal{D}}(F(-),-)$, and $\eta_x : x \to U(F(x))$ is by definition $\alpha^{-1}_{G'}(\mathrm{id}_U)_x$ for $G'=\mathrm{Hom}_{\mathcal{C}}(-,U(-))$. We have a morphism $h : G \to G'$ defined by mapping $f:F(x) \to y$ to $U(f)\circ \eta_x : x \to U(F(y))$. By naturality we have $\int_y h_{Uy,y} \circ \alpha_G = \alpha_{G'} \circ \int_x h_{x,Fx}$ as morphisms $\int_x G(x,Fx) \to \int_y G'(Uy,y)$, resp. $\mathrm{Hom}(F,F) \to \mathrm{Hom}(U,U)$. Evaluation at $\mathrm{id}_F$ and some $y \in \mathcal{D}$ yields $U(\varepsilon_y) \circ \eta_{U(y)} = \alpha_{G'}(\eta)_y=\mathrm{id}_{U(y)}$, which is the first zig-zag-identity. The other is proven similarly.<|endoftext|> TITLE: A cup product in Galois cohomology of Elliptic curve QUESTION [5 upvotes]: Let $E$ be an elliptic curve over a field $K$. I am mostly interested in the case $K$ is a number field or a local field but the following question is valid for all $K$. Let $p$ be any prime power which is comprime to the characteristics of $K$. Galois cohomology gives a cup product map $$H^{1}(K,E[p]) \times H^{1}(K,E[p]) \to H^{2}(K,E[p] \otimes E[p])$$ Composing this map with the map induced by the Weil pairing gives a map $$ \phi: H^{1}(K,E[p]) \times H^{1}(K,E[p]) \to H^2(K,\mu_{p}). $$ Are there any known examples where the above map $\phi$ is non-trivial? In theory, it is possible to compute both sides explicitly when p is small. However, computing the cup product is quite hard (for me.) REPLY [10 votes]: One can use the exact sequence $$ 0 \to E(K)/mE(K) \to H^1(K,E[m]) \to H^1(K,E)[m] \to 0 $$ to define a pairing $$ E(K)/mE(K) \times H^1(K,E)[m] \to H^2(K,\mu_m) $$ by taking $(Q,\xi)$ to $\phi(Q'\cup \xi')$, where $Q'$ is the image of $Q$ in $H^1(K,E[m])$ and $\xi'$ is any pullback of $\xi$ to $H^1(K,E[m])$. For $K/\mathbb Q_p$, Tate proved that this is a perfect pairing (Tate local duality). Since it's easy to find examples with $E(K)/mE(K)\ne0$, this gives examples where your map $\phi$ is non-trivial. And indeed, if you take $m=p^k$, then you can force $E(K)/p^kE(K)$ to be quite large. (In this case, we have $H^2(K,\mu_m)\cong \frac1m\mathbb Z/\mathbb Z$, so this piece of the Brauer group makes a natural target for a pairing.) For global fields, one does something similar to construct the Cassels-Tate pairing on the Tate-Shafarevich group, but the target space is instead $H^2(K,\mathcal{C}[m])$, where $\mathcal{C}$ is the idele class group. This is preferable to $H^2(K,\mu_m)$, since the Brauer group $H^2(K,\mathbb G_m)$ is huge.<|endoftext|> TITLE: Is Квант the actual source of this problem? QUESTION [7 upvotes]: If I recall correctly, Andreescu & Andrica attributed the olympiad problem which prompted this question by S. Pek to the Russian magazine Kvant. Does anybody here know if the problem actually appeared in the pages of Kvant once? Hope this question of mine doesn't get closed... I dared to ask it here because I believe that several MO users were avid readers of Kvant back in the day. Thanks in advance for your attention! REPLY [12 votes]: Yes, it is problem M618a), published in No. 4, 1980. Part b) claimed that for every $\alpha>0$ there exist infinitely many $n$ with $n^2+1\mid [\alpha n]!$. The problem is attributed to A. Sivatsky (who was a 10th grade student at that time). The solution is in No. 2, 1981. Also, this problem is discussed in an article by V. Senderov and A. Spivak in No. 4, 2002. If you need, I may find a cite to the electronic versions of all these.<|endoftext|> TITLE: A Schur positivity conjecture related to row and column permutations QUESTION [10 upvotes]: The problem Counting cycles after permuting within rows and columns reminds me of the following unpublished conjecture of mine. Let $D$ be any finite planar diagram (in the sense of Young diagram, which is a special case), say with $n$ squares. Put the numbers $1,2,\dots,n$ into the squares of the diagram. Let $R_D$ be the subgroup of the symmetric group $S_n$ permuting elements within each row, and similarly $C_D$ for the columns. Let $\chi$ and $\psi$ be any characters of $S_n$. Define $$ u_D=\sum_{\substack{u\in R_D\\ v\in C_D}} \chi(u)\psi(v)p_{\rho(uv)}, $$ where $p_{\rho(uv)}$ is the power sum symmetric function indexed by the cycle type of $uv$. Then (conjecturally) $u_D$ is Schur-positive. This conjecture is open even for diagrams of partitions when $\chi$ and $\psi$ are the trivial character. (In this case, one can show for hook shapes that $u_D$ is even $h$-positive, but $h$-positivity fails in most other cases.) When $D$ is the diagram of a partition $\lambda$, and where $\chi$ is the trivial character and $\psi$ the sign character, we have $u_D= H_\lambda s_\lambda$, where $H_\lambda$ is the product of the hook lengths of $\lambda$. See the slides of Valentin Féray at http://fpsac.combinatorics.kr/program. REPLY [9 votes]: I heard about this conjecture from Sara Billey at FPSAC, and I think I've got an argument. Let $F : \mathbb{C}[S_n] \to \mathbb{Z}[x_1, \ldots, x_N]^{S_N}$ be the linear map sending $w \mapsto p_{\rho(w)}(x_1, \ldots, x_N)$, and $V$ a complex vector space with $\dim V = N \geq n$. Lemma: If $\alpha \in \mathbb{C}[S_n]$ acts on $V^{\otimes n}$ (on the right) with nonnegative eigenvalues, then $F(\alpha)$ is Schur-positive. Proof: By a density argument we can assume $\alpha$ acts diagonalizably: say $V^{\otimes n} = \bigoplus_{\omega} U_{\omega}$ where $U_{\omega}$ is the $\omega$-eigenspace of $\alpha$. Each $U_{\omega}$ is a left $\operatorname{GL}(V)$-module since the left $\operatorname{GL}(V)$-action commutes with $\alpha$. Let $X \in \operatorname{GL}(V)$ have eigenvalues $x_1, \ldots, x_n$. The trace of $X \times \alpha$ on $V^{\otimes n}$ (i.e. of the map $z \mapsto Xz\alpha$) is, on the one hand, $\sum_{\omega} \omega \operatorname{tr}(X|_{V_\omega})$. Since $V_\omega$ is a $\operatorname{GL}(V)$-module, $\operatorname{tr}(X|_{V_\omega})$ is a Schur-positive polynomial in $x_1, \ldots, x_N$. On the other hand, $\operatorname{tr}(X \times \alpha) = F(\alpha)$. Since $\mathbb{C}[S_n]$ acts faithfully on $V^{\otimes d}$, the eigenvalues of $\alpha$ acting on $\mathbb{C}[S_n]$ or on $V^{\otimes d}$ are the same, ignoring multiplicity (maybe the lemma can be modified to work directly on $\mathbb{C}[S_n]$?). Up to constant factors, $\sum_{u \in R_D} \chi(u)u \sum_{v \in C_D} \psi(v)v$ is the product of two idempotents, which are both Hermitian with respect to the inner product on $\mathbb{C}[S_n]$ for which permutations form an orthonormal basis. The product of two positive semidefinite matrices has nonnegative eigenvalues, so $F(\sum_{u \in R_D} \chi(u)u \sum_{v \in C_D} \psi(v)v)$ is Schur-positive by the lemma.<|endoftext|> TITLE: When do automorphisms of subshifts extend to automorphisms of the full shift? QUESTION [7 upvotes]: Let $A$ be a finite alphabet, $X$ = $(A^\mathbb{Z}, \sigma)$ the full shift, and $Y \subset X$ a subshift. Question: Are there any general results characterizing whether automorphisms of $(Y, \sigma)$ are likely to extend to automorphisms of $X$ (that is, whether automorphisms of $(Y, \sigma)$ are likely to be restrictions of automorphisms of $X$). I suspect that the answer is that automorphisms generally do not extend. For example, it may be the case that for most subshifts (for some appropriate meaning of "most"), the only automorphisms that extend are powers of $\sigma$. Any thoughts or references would be greatly appreciated! REPLY [7 votes]: If $\phi$ is an automorphism of $X$ and $Y$ is the set of points in $X$ of exact period $n$, then $\phi|_Y$ is an automorphism of $Y$. There is a subtle relationship between the sign of the permutation that $\phi$ induces on orbits of lengths $n$ for various $n$, and the value of a transfer function called the nth gyration number of $\phi$, called the sign-gyration compatibility condition, introduced by Boyle and Krieger [Periodic points and automorphisms of the shift, Trans. AMS 302 (1987), 125-149]. Using this, Kim, Roush, and Wagoner [Automorphisms of the dimension group and gyration numbers, Journ. AMS 5 (1992), 191-212] showed that there is a shift of finite type and a permutation of its fixed points that is not the restriction of an automorphism of the shift. These ideas played a key role in the discovery by Kim, Roush, and Wagoner of counterexamples to what was then the main open problem in symbolic dynamics, namely the Shift Equivalence Conjecture. This conjecture was that a relatively easily computed invariant of shifts of finite type, the dimension triple, was a complete invariant for topological conjugacy.<|endoftext|> TITLE: Can you tie up these Laurent sequences? QUESTION [9 upvotes]: Fix an integer $k\geq3$. Define the two families of sequences $\{x_n\}$ and $\{y_n\}$ according to the rules: $$x_n=\frac{x_{n-1}^2+x_{n-2}^2+\cdots+x_{n-k+1}^2}{x_{n-k}} \qquad n\geq k$$ and $$y_n=\frac{(y_{n-1}+y_{n-2}+\cdots+y_{n-k+1})^2}{y_{n-k}} \qquad n\geq k$$ with initial conditions $x_j=y_j=1$ for $j=0,1,\dots,k-1$. QUESTIONS. (1) It seems that both $x_n$ and $y_n$ are always positive integers. (2) It also appears true that $y_n=x_n^2$ for all $n$. Any ideas of a proof? Of course, if (2) holds and $x_n\in\mathbb{N}$ then $y_n\in\mathbb{N}$. REMARK. Special cases are also appreciated, say for $k=3$ etc. REPLY [10 votes]: Suppose we know that $y_j=x_j^2$ for $j=n-1, \ldots, n-k$. Then $$x_n^2=\left(\frac{x_{n-1}^2+x_{n-2}^2+\cdots+x_{n-k+1}^2}{x_{n-k}} \right)^2=\frac{(y_{n-1}+y_{n-2}+\cdots+y_{n-k+1})^2}{y_{n-k}} =y_n.$$ If all $y_n$ are integers then from requrrent relation follows that they are squares. It means that all $x_n$ are also integers. So $\{y_n\}$ are integers iff $\{x_n\}$ are integers. Integrality of $\{x_n\}$ is a special case of more general result (see case (1) in Theorem 3.9 from the article Laurent Phenomenon Sequences by Joshua Alman, Cesar Cuenca and Jiaoyang Huang). REPLY [7 votes]: As to Question 2, the sequence $(x_n^2)$ solves the same recursive relation as $(y_n)$, with the same initial values, therefore they coincide.